summaryrefslogtreecommitdiff
path: root/info/maad/maad.tex
blob: 2d32f14621918ef8f703f5084087cccef225a6aa (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
1001
1002
1003
1004
1005
1006
1007
1008
1009
1010
1011
1012
1013
1014
1015
1016
1017
1018
1019
1020
1021
1022
1023
1024
1025
1026
1027
1028
1029
1030
1031
1032
1033
1034
1035
1036
1037
1038
1039
1040
1041
1042
1043
1044
1045
1046
1047
1048
1049
1050
1051
1052
1053
1054
1055
1056
1057
1058
1059
1060
1061
1062
1063
1064
1065
1066
1067
1068
1069
1070
1071
1072
1073
1074
1075
1076
1077
1078
1079
1080
1081
1082
1083
1084
1085
1086
1087
1088
1089
1090
1091
1092
1093
1094
1095
1096
1097
1098
1099
1100
1101
1102
1103
1104
1105
1106
1107
1108
1109
1110
1111
1112
1113
1114
1115
1116
1117
1118
1119
1120
1121
1122
1123
1124
1125
1126
1127
1128
1129
1130
1131
1132
1133
1134
1135
1136
1137
1138
1139
1140
1141
1142
1143
1144
1145
1146
1147
1148
1149
1150
1151
1152
1153
1154
1155
1156
1157
1158
1159
1160
1161
1162
1163
1164
1165
1166
1167
1168
1169
1170
1171
1172
1173
1174
1175
1176
1177
1178
1179
1180
1181
1182
1183
1184
1185
1186
1187
1188
1189
1190
1191
1192
1193
1194
1195
1196
1197
1198
1199
1200
1201
1202
1203
1204
1205
1206
1207
1208
1209
1210
1211
1212
1213
1214
1215
1216
1217
1218
1219
1220
1221
1222
1223
1224
1225
1226
1227
1228
1229
1230
1231
1232
1233
1234
1235
1236
1237
1238
1239
1240
1241
1242
1243
1244
1245
1246
1247
1248
1249
1250
1251
1252
1253
1254
1255
1256
1257
1258
1259
1260
1261
1262
1263
1264
1265
1266
1267
1268
1269
1270
1271
1272
1273
1274
1275
1276
1277
1278
1279
1280
1281
1282
1283
1284
1285
1286
1287
1288
1289
1290
1291
1292
1293
1294
1295
1296
1297
1298
1299
1300
1301
1302
1303
1304
1305
1306
1307
1308
1309
1310
1311
1312
1313
1314
1315
1316
1317
1318
1319
1320
1321
1322
1323
1324
1325
1326
1327
1328
1329
1330
1331
1332
1333
1334
1335
1336
1337
1338
1339
1340
1341
1342
1343
1344
1345
1346
1347
1348
1349
1350
1351
1352
1353
1354
1355
1356
1357
1358
1359
1360
1361
1362
1363
1364
1365
1366
1367
1368
1369
1370
1371
1372
1373
1374
1375
1376
1377
1378
1379
1380
1381
1382
1383
1384
1385
1386
1387
1388
1389
1390
1391
1392
1393
1394
1395
1396
1397
1398
1399
1400
1401
1402
1403
1404
1405
1406
1407
1408
1409
1410
1411
1412
1413
1414
1415
1416
1417
1418
1419
1420
1421
1422
1423
1424
1425
1426
1427
1428
1429
1430
1431
1432
1433
1434
1435
1436
1437
1438
1439
1440
1441
1442
1443
1444
1445
1446
1447
1448
1449
1450
1451
1452
1453
1454
1455
1456
1457
1458
1459
1460
1461
1462
1463
1464
1465
1466
1467
1468
1469
1470
1471
1472
1473
1474
1475
1476
1477
1478
1479
1480
1481
1482
1483
1484
1485
1486
1487
1488
1489
1490
1491
1492
1493
1494
1495
1496
1497
1498
1499
1500
1501
1502
1503
1504
1505
1506
1507
1508
1509
1510
1511
1512
1513
1514
1515
1516
1517
1518
1519
1520
1521
1522
1523
1524
1525
1526
1527
1528
1529
1530
1531
1532
1533
1534
1535
1536
1537
1538
1539
1540
1541
1542
1543
1544
1545
1546
1547
1548
1549
1550
1551
1552
1553
1554
1555
1556
1557
1558
1559
1560
1561
1562
1563
1564
1565
1566
1567
1568
1569
1570
1571
1572
1573
1574
1575
1576
1577
1578
1579
1580
1581
1582
1583
1584
1585
1586
1587
1588
1589
1590
1591
1592
1593
1594
1595
1596
1597
1598
1599
1600
1601
1602
1603
1604
1605
1606
1607
1608
1609
1610
1611
1612
1613
1614
1615
1616
1617
1618
1619
1620
1621
1622
1623
1624
1625
1626
1627
1628
1629
1630
1631
1632
1633
1634
1635
1636
1637
1638
1639
1640
1641
1642
1643
1644
1645
1646
1647
1648
1649
1650
1651
1652
1653
1654
1655
1656
1657
1658
1659
1660
1661
1662
1663
1664
1665
1666
1667
1668
1669
1670
1671
1672
1673
1674
1675
1676
1677
1678
1679
1680
1681
1682
1683
1684
1685
1686
1687
1688
1689
1690
1691
1692
1693
1694
1695
1696
1697
1698
1699
1700
1701
1702
1703
1704
1705
1706
1707
1708
1709
1710
1711
1712
1713
1714
1715
1716
1717
1718
1719
1720
1721
1722
1723
1724
1725
1726
1727
1728
1729
1730
1731
1732
1733
1734
1735
1736
1737
1738
1739
1740
1741
1742
1743
1744
1745
1746
1747
1748
1749
1750
1751
1752
1753
1754
1755
1756
1757
1758
1759
1760
1761
1762
1763
1764
1765
1766
1767
1768
1769
1770
1771
1772
1773
1774
1775
1776
1777
1778
1779
1780
1781
1782
1783
1784
1785
1786
1787
1788
1789
1790
1791
1792
1793
1794
1795
1796
1797
1798
1799
1800
1801
1802
1803
1804
1805
1806
1807
1808
1809
1810
1811
1812
1813
1814
1815
1816
1817
1818
1819
1820
1821
1822
1823
1824
1825
1826
1827
1828
1829
1830
1831
1832
1833
1834
1835
1836
1837
1838
1839
1840
1841
1842
1843
1844
1845
1846
1847
1848
1849
1850
1851
1852
1853
1854
1855
1856
1857
1858
1859
1860
1861
1862
1863
1864
1865
1866
1867
1868
1869
1870
1871
1872
1873
1874
1875
1876
1877
1878
1879
1880
1881
1882
1883
1884
1885
1886
1887
1888
1889
1890
1891
1892
1893
1894
1895
1896
1897
1898
1899
1900
1901
1902
1903
1904
1905
1906
1907
1908
1909
1910
1911
1912
1913
1914
1915
1916
1917
1918
1919
1920
1921
1922
1923
1924
1925
1926
1927
1928
1929
1930
1931
1932
1933
1934
1935
1936
1937
1938
1939
1940
1941
1942
1943
1944
1945
1946
1947
1948
1949
1950
1951
1952
1953
1954
1955
1956
1957
1958
1959
1960
1961
1962
1963
1964
1965
1966
1967
1968
1969
1970
1971
1972
1973
1974
1975
1976
1977
1978
1979
1980
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
2010
2011
2012
2013
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
2024
2025
2026
2027
2028
2029
2030
2031
2032
2033
2034
2035
2036
2037
2038
2039
2040
2041
2042
2043
2044
2045
2046
2047
2048
2049
2050
2051
2052
2053
2054
2055
2056
2057
2058
2059
2060
2061
2062
2063
2064
2065
2066
2067
2068
2069
2070
2071
2072
2073
2074
2075
2076
2077
2078
2079
2080
2081
2082
2083
2084
2085
2086
2087
2088
2089
2090
2091
2092
2093
2094
2095
2096
2097
2098
2099
2100
2101
2102
2103
2104
2105
2106
2107
2108
2109
2110
2111
2112
2113
2114
2115
2116
2117
2118
2119
2120
2121
2122
2123
2124
2125
2126
2127
2128
2129
2130
2131
2132
2133
2134
2135
2136
2137
2138
2139
2140
2141
2142
2143
2144
2145
2146
2147
2148
2149
2150
2151
2152
2153
2154
2155
2156
2157
2158
2159
2160
2161
2162
2163
2164
2165
2166
2167
2168
2169
2170
2171
2172
2173
2174
2175
2176
2177
2178
2179
2180
2181
2182
2183
2184
2185
2186
2187
2188
2189
2190
2191
2192
2193
2194
2195
2196
2197
2198
2199
2200
2201
2202
2203
2204
2205
2206
2207
2208
2209
2210
2211
2212
2213
2214
2215
2216
2217
2218
2219
2220
2221
2222
2223
2224
2225
2226
2227
2228
2229
2230
2231
2232
2233
2234
2235
2236
2237
2238
2239
2240
2241
2242
2243
2244
2245
2246
2247
2248
2249
2250
2251
2252
2253
2254
2255
2256
2257
2258
2259
2260
2261
2262
2263
2264
2265
2266
2267
2268
2269
2270
2271
2272
2273
2274
2275
2276
2277
2278
2279
2280
2281
2282
2283
2284
2285
2286
2287
2288
2289
2290
2291
2292
2293
2294
2295
2296
2297
2298
2299
2300
2301
2302
2303
2304
2305
2306
2307
2308
2309
2310
2311
2312
2313
2314
2315
2316
2317
2318
2319
2320
2321
2322
2323
2324
2325
2326
2327
2328
2329
2330
2331
2332
2333
2334
2335
2336
2337
2338
2339
2340
2341
2342
2343
2344
2345
2346
2347
2348
2349
2350
2351
2352
2353
2354
2355
2356
2357
2358
2359
2360
2361
2362
2363
2364
2365
2366
2367
2368
2369
2370
2371
2372
2373
2374
2375
2376
2377
2378
2379
2380
2381
2382
2383
2384
2385
2386
2387
2388
2389
2390
2391
2392
2393
2394
2395
2396
2397
2398
2399
2400
2401
2402
2403
2404
2405
2406
2407
2408
2409
2410
2411
2412
2413
2414
2415
2416
2417
2418
2419
2420
2421
2422
2423
2424
2425
2426
2427
2428
2429
2430
2431
2432
2433
2434
2435
2436
2437
2438
2439
2440
2441
2442
2443
2444
2445
2446
2447
2448
2449
2450
2451
2452
2453
2454
2455
2456
2457
2458
2459
2460
2461
2462
2463
2464
2465
2466
2467
2468
2469
2470
2471
2472
2473
2474
2475
2476
2477
2478
2479
2480
2481
2482
2483
2484
2485
2486
2487
2488
2489
2490
2491
2492
2493
2494
2495
2496
2497
2498
2499
2500
2501
2502
2503
2504
2505
2506
2507
2508
2509
2510
2511
2512
2513
2514
2515
2516
2517
2518
2519
2520
2521
2522
2523
2524
2525
2526
2527
2528
2529
2530
2531
2532
2533
2534
2535
2536
2537
2538
2539
2540
2541
2542
2543
2544
2545
2546
2547
2548
2549
2550
2551
2552
2553
2554
2555
2556
2557
2558
2559
2560
2561
2562
2563
2564
2565
2566
2567
2568
2569
2570
2571
2572
2573
2574
2575
2576
2577
2578
2579
2580
2581
2582
2583
2584
2585
2586
2587
2588
2589
2590
2591
2592
2593
2594
2595
2596
2597
2598
2599
2600
2601
2602
2603
2604
2605
2606
2607
2608
2609
2610
2611
2612
2613
2614
2615
2616
2617
2618
2619
2620
2621
2622
2623
2624
2625
2626
2627
2628
2629
2630
2631
2632
2633
2634
2635
2636
2637
2638
2639
2640
2641
2642
2643
2644
2645
2646
2647
2648
2649
2650
2651
2652
2653
2654
2655
2656
2657
2658
2659
2660
2661
2662
2663
2664
2665
2666
2667
2668
2669
2670
2671
2672
2673
2674
2675
2676
2677
2678
2679
2680
2681
2682
2683
2684
2685
2686
2687
2688
2689
2690
2691
2692
2693
2694
2695
2696
2697
2698
2699
2700
2701
2702
2703
2704
2705
2706
2707
2708
2709
2710
2711
2712
2713
2714
2715
2716
2717
2718
2719
2720
2721
2722
2723
2724
2725
2726
2727
2728
2729
2730
2731
2732
2733
2734
2735
2736
2737
2738
2739
2740
2741
2742
2743
2744
2745
2746
2747
2748
2749
2750
2751
2752
2753
2754
2755
2756
2757
2758
2759
2760
2761
2762
2763
2764
2765
2766
2767
2768
2769
2770
2771
2772
2773
2774
2775
2776
2777
2778
2779
2780
2781
2782
2783
2784
2785
2786
2787
2788
2789
2790
2791
2792
2793
2794
2795
2796
2797
2798
2799
2800
2801
2802
2803
2804
2805
2806
2807
2808
2809
2810
2811
2812
2813
2814
2815
2816
2817
2818
2819
2820
2821
2822
2823
2824
2825
2826
2827
2828
2829
2830
2831
2832
2833
2834
2835
2836
2837
2838
2839
2840
2841
2842
2843
2844
2845
2846
2847
2848
2849
2850
2851
2852
2853
2854
2855
2856
2857
2858
2859
2860
2861
2862
2863
2864
2865
2866
2867
2868
2869
2870
2871
2872
2873
2874
2875
2876
2877
2878
2879
2880
2881
2882
2883
2884
2885
2886
2887
2888
2889
2890
2891
2892
2893
2894
2895
2896
2897
2898
2899
2900
2901
2902
2903
2904
2905
2906
2907
2908
2909
2910
2911
2912
2913
2914
2915
2916
2917
2918
2919
2920
2921
2922
2923
2924
2925
2926
2927
2928
2929
2930
2931
2932
2933
2934
2935
2936
2937
2938
2939
2940
2941
2942
2943
2944
2945
2946
2947
2948
2949
2950
2951
2952
2953
2954
2955
2956
2957
2958
2959
2960
2961
2962
2963
2964
2965
2966
2967
2968
2969
2970
2971
2972
2973
2974
2975
2976
2977
2978
2979
2980
2981
2982
2983
2984
2985
2986
2987
2988
2989
2990
2991
2992
2993
2994
2995
2996
2997
2998
2999
3000
3001
3002
3003
3004
3005
3006
3007
3008
3009
3010
3011
3012
3013
3014
3015
3016
3017
3018
3019
3020
3021
3022
3023
3024
3025
3026
3027
3028
3029
3030
3031
3032
3033
3034
3035
3036
3037
3038
3039
3040
3041
3042
3043
3044
3045
3046
3047
3048
3049
3050
3051
3052
3053
3054
3055
3056
3057
3058
3059
3060
3061
3062
3063
3064
3065
3066
3067
3068
3069
3070
3071
3072
3073
3074
3075
3076
3077
3078
3079
3080
3081
3082
3083
3084
3085
3086
3087
3088
3089
3090
3091
3092
3093
3094
3095
3096
3097
3098
3099
3100
3101
3102
3103
3104
3105
3106
3107
3108
3109
3110
3111
3112
3113
3114
3115
3116
3117
3118
3119
3120
3121
3122
3123
3124
3125
3126
3127
3128
3129
3130
3131
3132
3133
3134
3135
3136
3137
3138
3139
3140
3141
3142
3143
3144
3145
3146
3147
3148
3149
3150
3151
3152
3153
3154
3155
3156
3157
3158
3159
3160
3161
3162
3163
3164
3165
3166
3167
3168
3169
3170
3171
3172
3173
3174
3175
3176
3177
3178
3179
3180
3181
3182
3183
3184
3185
3186
3187
3188
3189
3190
3191
3192
3193
3194
3195
3196
3197
3198
3199
3200
3201
3202
3203
3204
3205
3206
3207
3208
3209
3210
3211
3212
3213
3214
3215
3216
3217
3218
3219
3220
3221
3222
3223
3224
3225
3226
3227
3228
3229
3230
3231
3232
3233
3234
3235
3236
3237
3238
3239
3240
3241
3242
3243
3244
3245
3246
3247
3248
3249
3250
3251
3252
3253
3254
3255
3256
3257
3258
3259
3260
3261
3262
3263
3264
3265
3266
3267
3268
3269
3270
3271
3272
3273
3274
3275
3276
3277
3278
3279
3280
3281
3282
3283
3284
3285
3286
3287
3288
3289
3290
3291
3292
3293
3294
3295
3296
3297
3298
3299
3300
3301
3302
3303
3304
3305
3306
3307
3308
3309
3310
3311
3312
3313
3314
3315
3316
3317
3318
3319
3320
3321
3322
3323
3324
3325
3326
3327
3328
3329
3330
3331
3332
3333
3334
3335
3336
3337
3338
3339
3340
3341
3342
3343
3344
3345
3346
3347
3348
3349
3350
3351
3352
3353
3354
3355
3356
3357
3358
3359
3360
3361
3362
3363
3364
3365
3366
3367
3368
3369
3370
3371
3372
3373
3374
3375
3376
3377
3378
3379
3380
3381
3382
3383
3384
3385
3386
3387
3388
3389
3390
3391
3392
3393
3394
3395
3396
3397
3398
3399
3400
3401
3402
3403
3404
3405
3406
3407
3408
3409
3410
3411
3412
3413
3414
3415
3416
3417
3418
3419
3420
3421
3422
3423
3424
3425
3426
3427
3428
3429
3430
3431
3432
3433
3434
3435
3436
3437
3438
3439
3440
3441
3442
3443
3444
3445
3446
3447
3448
3449
3450
3451
3452
3453
3454
3455
3456
3457
3458
3459
3460
3461
3462
3463
3464
3465
3466
3467
3468
3469
3470
3471
3472
3473
3474
3475
3476
3477
3478
3479
3480
3481
3482
3483
3484
3485
3486
3487
3488
3489
3490
3491
3492
3493
3494
3495
3496
3497
3498
3499
3500
3501
3502
3503
3504
3505
3506
3507
3508
3509
3510
3511
3512
3513
3514
3515
3516
3517
3518
3519
3520
3521
3522
3523
3524
3525
3526
3527
3528
3529
3530
3531
3532
3533
3534
3535
3536
3537
3538
3539
3540
3541
3542
3543
3544
3545
3546
3547
3548
3549
3550
3551
3552
3553
3554
3555
3556
3557
3558
3559
3560
3561
3562
3563
3564
3565
3566
3567
3568
3569
3570
3571
3572
3573
3574
3575
3576
3577
3578
3579
3580
3581
3582
3583
3584
3585
3586
3587
3588
3589
3590
3591
3592
3593
3594
3595
3596
3597
3598
3599
3600
3601
3602
3603
3604
3605
3606
3607
3608
3609
3610
3611
3612
3613
3614
3615
3616
3617
3618
3619
3620
3621
3622
3623
3624
3625
3626
3627
3628
3629
3630
3631
3632
3633
3634
3635
3636
3637
3638
3639
3640
3641
3642
3643
3644
3645
3646
3647
3648
3649
3650
3651
3652
3653
3654
3655
3656
3657
3658
3659
3660
3661
3662
3663
3664
3665
3666
3667
3668
3669
3670
3671
3672
3673
3674
3675
3676
3677
3678
3679
3680
3681
3682
3683
3684
3685
3686
3687
3688
3689
3690
3691
3692
3693
3694
3695
3696
3697
3698
3699
3700
3701
3702
3703
3704
3705
3706
3707
3708
3709
3710
3711
3712
3713
3714
3715
3716
3717
3718
3719
3720
3721
3722
3723
3724
3725
3726
3727
3728
3729
3730
3731
3732
3733
3734
3735
3736
3737
3738
3739
3740
3741
3742
3743
3744
3745
3746
3747
3748
3749
3750
3751
3752
3753
3754
3755
3756
3757
3758
3759
3760
3761
3762
3763
3764
3765
3766
3767
3768
3769
3770
3771
3772
3773
3774
3775
3776
3777
3778
3779
3780
3781
3782
3783
3784
3785
3786
3787
3788
3789
3790
3791
3792
3793
3794
3795
3796
3797
3798
3799
3800
3801
3802
3803
3804
3805
3806
3807
3808
3809
3810
3811
3812
3813
3814
3815
3816
3817
3818
3819
3820
3821
3822
3823
3824
3825
3826
3827
3828
3829
3830
3831
3832
3833
3834
3835
3836
3837
3838
3839
3840
3841
3842
3843
3844
3845
3846
3847
3848
3849
3850
3851
3852
3853
3854
3855
3856
3857
3858
3859
3860
3861
3862
3863
3864
3865
3866
3867
3868
3869
3870
3871
3872
3873
3874
3875
3876
3877
3878
3879
3880
3881
3882
3883
3884
3885
3886
3887
3888
3889
3890
3891
3892
3893
3894
3895
3896
3897
3898
3899
3900
3901
3902
3903
3904
3905
3906
3907
3908
3909
3910
3911
3912
3913
3914
3915
3916
3917
3918
3919
3920
3921
3922
3923
3924
3925
3926
3927
3928
3929
3930
3931
3932
3933
3934
3935
3936
3937
3938
3939
3940
3941
3942
3943
3944
3945
3946
3947
3948
3949
3950
3951
3952
3953
3954
3955
3956
3957
3958
3959
3960
3961
3962
3963
3964
3965
3966
3967
3968
3969
3970
3971
3972
3973
3974
3975
3976
3977
3978
3979
3980
3981
3982
3983
3984
3985
3986
3987
3988
3989
3990
3991
3992
3993
3994
3995
3996
3997
3998
3999
4000
4001
4002
4003
4004
4005
4006
4007
4008
4009
4010
4011
4012
4013
4014
4015
4016
4017
4018
4019
4020
4021
4022
4023
4024
4025
4026
4027
4028
4029
4030
4031
4032
4033
4034
4035
4036
4037
4038
4039
4040
4041
4042
4043
4044
4045
4046
4047
4048
4049
4050
4051
4052
4053
4054
4055
4056
4057
4058
4059
4060
4061
4062
4063
4064
4065
4066
4067
4068
4069
4070
4071
4072
4073
4074
4075
4076
4077
4078
4079
4080
4081
4082
4083
4084
4085
4086
4087
4088
4089
4090
4091
4092
4093
4094
4095
4096
4097
4098
4099
4100
4101
4102
4103
4104
4105
4106
4107
4108
4109
4110
4111
4112
4113
4114
4115
4116
4117
4118
4119
4120
4121
4122
4123
4124
4125
4126
4127
4128
4129
4130
4131
4132
4133
4134
4135
4136
4137
4138
4139
4140
4141
4142
4143
4144
4145
4146
4147
4148
4149
4150
4151
4152
4153
4154
4155
4156
4157
4158
4159
4160
4161
4162
4163
4164
4165
4166
4167
4168
4169
4170
4171
4172
4173
4174
4175
4176
4177
4178
4179
4180
4181
4182
4183
4184
4185
4186
4187
4188
4189
4190
4191
4192
4193
4194
4195
4196
4197
4198
4199
4200
4201
4202
4203
4204
4205
4206
4207
4208
4209
4210
4211
4212
4213
4214
4215
4216
4217
4218
4219
4220
4221
4222
4223
4224
4225
4226
4227
4228
4229
4230
4231
4232
4233
4234
4235
4236
4237
4238
4239
4240
4241
4242
4243
4244
4245
4246
4247
4248
4249
4250
4251
4252
4253
4254
4255
4256
4257
4258
4259
4260
4261
4262
4263
4264
4265
4266
4267
4268
4269
4270
4271
4272
4273
4274
4275
4276
4277
4278
4279
4280
4281
4282
4283
4284
4285
4286
4287
4288
4289
4290
4291
4292
4293
4294
4295
4296
4297
4298
4299
4300
4301
4302
4303
4304
4305
4306
4307
4308
4309
4310
4311
4312
4313
4314
4315
4316
4317
4318
4319
4320
4321
4322
4323
4324
4325
4326
4327
4328
4329
4330
4331
4332
4333
4334
4335
4336
4337
4338
4339
4340
4341
4342
4343
4344
4345
4346
4347
4348
4349
4350
4351
4352
4353
4354
4355
4356
4357
4358
4359
4360
4361
4362
4363
4364
4365
4366
4367
4368
4369
4370
4371
4372
4373
4374
4375
4376
4377
4378
4379
4380
4381
4382
4383
4384
4385
4386
4387
4388
4389
4390
4391
4392
4393
4394
4395
4396
4397
4398
4399
4400
4401
4402
4403
4404
4405
4406
4407
4408
4409
4410
4411
4412
4413
4414
4415
4416
4417
4418
4419
4420
4421
4422
4423
4424
4425
4426
4427
4428
4429
4430
4431
4432
4433
4434
4435
4436
4437
4438
4439
4440
4441
4442
4443
4444
4445
4446
4447
4448
4449
4450
4451
4452
4453
4454
4455
4456
4457
4458
4459
4460
4461
4462
4463
4464
4465
4466
4467
4468
4469
4470
4471
4472
4473
4474
4475
4476
4477
4478
4479
4480
4481
4482
4483
4484
4485
4486
4487
4488
4489
4490
4491
4492
4493
4494
4495
4496
4497
4498
4499
4500
4501
4502
4503
4504
4505
4506
4507
4508
4509
4510
4511
4512
4513
4514
4515
4516
4517
4518
4519
4520
4521
4522
4523
4524
4525
4526
4527
4528
4529
4530
4531
4532
4533
4534
4535
4536
4537
4538
4539
4540
4541
4542
4543
4544
4545
4546
4547
4548
4549
4550
4551
4552
4553
4554
4555
4556
4557
4558
4559
4560
4561
4562
4563
4564
4565
4566
4567
4568
4569
4570
4571
4572
4573
4574
4575
4576
4577
4578
4579
4580
4581
4582
4583
4584
4585
4586
4587
4588
4589
4590
4591
4592
4593
4594
4595
4596
4597
4598
4599
4600
4601
4602
4603
4604
4605
4606
4607
4608
4609
4610
4611
4612
4613
4614
4615
4616
4617
4618
4619
4620
4621
4622
4623
4624
4625
4626
4627
4628
4629
4630
4631
4632
4633
4634
4635
4636
4637
4638
4639
4640
4641
4642
4643
4644
4645
4646
4647
4648
4649
4650
4651
4652
4653
4654
4655
4656
4657
4658
4659
4660
4661
4662
4663
4664
4665
4666
4667
4668
4669
4670
4671
4672
4673
4674
4675
4676
4677
4678
4679
4680
4681
4682
4683
4684
4685
4686
4687
4688
4689
4690
4691
4692
4693
4694
4695
4696
4697
4698
4699
4700
4701
4702
4703
4704
4705
4706
4707
4708
4709
4710
4711
4712
4713
4714
4715
4716
4717
4718
4719
4720
4721
4722
4723
4724
4725
4726
4727
4728
4729
4730
4731
4732
4733
4734
4735
4736
4737
4738
4739
4740
4741
4742
4743
4744
4745
4746
4747
4748
4749
4750
4751
4752
4753
4754
4755
4756
4757
4758
4759
4760
4761
4762
4763
4764
4765
4766
4767
4768
4769
4770
4771
4772
4773
4774
4775
4776
4777
4778
4779
4780
4781
4782
4783
4784
4785
4786
4787
4788
4789
4790
4791
4792
4793
4794
4795
4796
4797
4798
4799
4800
4801
4802
4803
4804
4805
4806
4807
4808
4809
4810
4811
4812
4813
4814
4815
4816
4817
4818
4819
4820
4821
4822
4823
4824
4825
4826
4827
4828
4829
4830
4831
4832
4833
4834
4835
4836
4837
4838
4839
4840
4841
4842
4843
4844
4845
4846
4847
4848
4849
4850
4851
4852
4853
4854
4855
4856
4857
4858
4859
4860
4861
4862
4863
4864
4865
4866
4867
4868
4869
4870
4871
4872
4873
4874
4875
4876
4877
4878
4879
4880
4881
4882
4883
4884
4885
4886
4887
4888
4889
4890
4891
4892
4893
4894
4895
4896
4897
4898
4899
4900
4901
4902
4903
4904
4905
4906
4907
4908
4909
4910
4911
4912
4913
4914
4915
4916
4917
4918
4919
4920
4921
4922
4923
4924
4925
4926
4927
4928
4929
4930
4931
4932
4933
4934
4935
4936
4937
4938
4939
4940
4941
4942
4943
4944
4945
4946
4947
4948
4949
4950
4951
4952
4953
4954
4955
4956
4957
4958
4959
4960
4961
4962
4963
4964
4965
4966
4967
4968
4969
4970
4971
4972
4973
4974
4975
4976
4977
4978
4979
4980
4981
4982
4983
4984
4985
4986
4987
4988
4989
4990
4991
4992
4993
4994
4995
4996
4997
4998
4999
5000
5001
5002
5003
5004
5005
5006
5007
5008
5009
5010
5011
5012
5013
5014
5015
5016
5017
5018
5019
5020
5021
5022
5023
5024
5025
5026
5027
5028
5029
5030
5031
5032
5033
5034
5035
5036
5037
5038
5039
5040
5041
5042
5043
5044
5045
5046
5047
5048
5049
5050
5051
5052
5053
5054
5055
5056
5057
5058
5059
5060
5061
5062
5063
5064
5065
5066
5067
5068
5069
5070
5071
5072
5073
5074
5075
5076
5077
5078
5079
5080
5081
5082
5083
5084
5085
5086
5087
5088
5089
5090
5091
5092
5093
5094
5095
5096
5097
5098
5099
5100
5101
5102
5103
5104
5105
5106
5107
5108
5109
5110
5111
5112
5113
5114
5115
5116
5117
5118
5119
5120
5121
5122
5123
5124
5125
5126
5127
5128
5129
5130
5131
5132
5133
5134
5135
5136
5137
5138
5139
5140
5141
5142
5143
5144
5145
5146
5147
5148
5149
5150
5151
5152
5153
5154
5155
5156
5157
5158
5159
5160
5161
5162
5163
5164
5165
5166
5167
5168
5169
5170
5171
5172
5173
5174
5175
5176
5177
5178
5179
5180
5181
5182
5183
5184
5185
5186
5187
5188
5189
5190
5191
5192
5193
5194
5195
5196
5197
5198
5199
5200
5201
5202
5203
5204
5205
5206
5207
5208
5209
5210
5211
5212
5213
5214
5215
5216
5217
5218
5219
5220
5221
5222
5223
5224
5225
5226
5227
5228
5229
5230
5231
5232
5233
5234
5235
5236
5237
5238
5239
5240
5241
5242
5243
5244
5245
5246
5247
5248
5249
5250
5251
5252
5253
5254
5255
5256
5257
5258
5259
5260
5261
5262
5263
5264
5265
5266
5267
5268
5269
5270
5271
5272
5273
5274
5275
5276
5277
5278
5279
5280
5281
5282
5283
5284
5285
5286
5287
5288
5289
5290
5291
5292
5293
5294
5295
5296
5297
5298
5299
5300
5301
5302
5303
5304
5305
5306
5307
5308
5309
5310
5311
5312
5313
5314
5315
5316
5317
5318
5319
5320
5321
5322
5323
5324
5325
5326
5327
5328
5329
5330
5331
5332
5333
5334
5335
5336
5337
5338
5339
5340
5341
5342
5343
5344
5345
5346
5347
5348
5349
5350
5351
5352
5353
5354
5355
5356
5357
5358
5359
5360
5361
5362
5363
5364
5365
5366
5367
5368
5369
5370
5371
5372
5373
5374
5375
5376
5377
5378
5379
5380
5381
5382
5383
5384
5385
5386
5387
5388
5389
5390
5391
5392
5393
5394
5395
5396
5397
5398
5399
5400
5401
5402
5403
5404
5405
5406
5407
5408
5409
5410
5411
5412
5413
5414
5415
5416
5417
5418
5419
5420
5421
5422
5423
5424
5425
5426
5427
5428
5429
5430
5431
5432
5433
5434
5435
5436
5437
5438
5439
5440
5441
5442
5443
5444
5445
5446
5447
5448
5449
5450
5451
5452
5453
5454
5455
5456
5457
5458
5459
5460
5461
5462
5463
5464
5465
5466
5467
5468
5469
5470
5471
5472
5473
5474
5475
5476
5477
5478
5479
5480
5481
5482
5483
5484
5485
5486
5487
5488
5489
5490
5491
5492
5493
5494
5495
5496
5497
5498
5499
5500
5501
5502
5503
5504
5505
5506
5507
5508
5509
5510
5511
5512
5513
5514
5515
5516
5517
5518
5519
5520
5521
5522
5523
5524
5525
5526
5527
5528
5529
5530
5531
5532
5533
5534
5535
5536
5537
5538
5539
5540
5541
5542
5543
5544
5545
5546
5547
5548
5549
5550
5551
5552
5553
5554
5555
5556
5557
5558
5559
5560
5561
5562
5563
5564
5565
5566
5567
5568
5569
5570
5571
5572
5573
5574
5575
5576
5577
5578
5579
5580
5581
5582
5583
5584
5585
5586
5587
5588
5589
5590
5591
5592
5593
5594
5595
5596
5597
5598
5599
5600
5601
5602
5603
5604
5605
5606
5607
5608
5609
5610
5611
5612
5613
5614
5615
5616
5617
5618
5619
5620
5621
5622
5623
5624
5625
5626
5627
5628
5629
5630
5631
5632
5633
5634
5635
5636
5637
5638
5639
5640
5641
5642
5643
5644
5645
5646
5647
5648
5649
5650
5651
5652
5653
5654
5655
5656
5657
5658
5659
5660
5661
5662
5663
5664
5665
5666
5667
5668
5669
5670
5671
5672
5673
5674
5675
5676
5677
5678
5679
5680
5681
5682
5683
5684
5685
5686
5687
5688
5689
5690
5691
5692
5693
5694
5695
5696
5697
5698
5699
5700
5701
5702
5703
5704
5705
5706
5707
5708
5709
5710
5711
5712
5713
5714
5715
5716
5717
5718
5719
5720
5721
5722
5723
5724
5725
5726
5727
5728
5729
5730
5731
5732
5733
5734
5735
5736
5737
5738
5739
5740
5741
5742
5743
5744
5745
5746
5747
5748
5749
5750
5751
5752
5753
5754
5755
5756
5757
5758
5759
5760
5761
5762
5763
5764
5765
5766
5767
5768
5769
5770
5771
5772
5773
5774
5775
5776
5777
5778
5779
5780
5781
5782
5783
5784
5785
5786
5787
5788
5789
5790
5791
5792
5793
5794
5795
5796
5797
5798
5799
5800
5801
5802
5803
5804
5805
5806
5807
5808
5809
5810
5811
5812
5813
5814
5815
5816
5817
5818
5819
5820
5821
5822
5823
5824
5825
5826
5827
5828
5829
5830
5831
5832
5833
5834
5835
5836
5837
5838
5839
5840
5841
5842
5843
5844
5845
5846
5847
5848
5849
5850
5851
5852
5853
5854
5855
5856
5857
5858
5859
5860
5861
5862
5863
5864
5865
5866
5867
5868
5869
5870
5871
5872
5873
5874
5875
5876
5877
5878
5879
5880
5881
5882
5883
5884
5885
5886
5887
5888
5889
5890
5891
5892
5893
5894
5895
5896
5897
5898
5899
5900
5901
5902
5903
5904
5905
5906
5907
5908
5909
5910
5911
5912
5913
5914
5915
5916
5917
5918
5919
5920
5921
5922
5923
5924
5925
5926
5927
5928
5929
5930
5931
5932
5933
5934
5935
5936
5937
5938
5939
5940
5941
5942
5943
5944
5945
5946
5947
5948
5949
5950
5951
5952
5953
5954
5955
5956
5957
5958
5959
5960
5961
5962
5963
5964
5965
5966
5967
5968
5969
5970
5971
5972
5973
5974
5975
5976
5977
5978
5979
5980
5981
5982
5983
5984
5985
5986
5987
5988
5989
5990
5991
5992
5993
5994
5995
5996
5997
5998
5999
6000
%\tracingstats=2 % see just how much memory is left
\input pictex
\hsize=5.75in % Set the horizontal size to accomodate CDVI, the public domain
%             % viewer.  Use CDVI-2 for VGA monitors with VGA cards.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 
% Set up an "expressmode" for plots to speed up typing and TeXing
% until a final draft is ready.  This is done as per the PiCTeX manual.
%
\newif\ifexpressmode  % default = \expessmodefalse
%
% See explanations on pages 215ff of the TeXBook.
%
\def\yes{y }
\def\Yes{Y }
\message{Type Y for EXPRESSMODE (no plotting) now: }
\read-1 to\answer
\ifx\answer\yes\expressmodetrue
  \else\ifx\answer\Yes\expressmodetrue\fi\fi % The answer is Yes (Y or y).
\ifexpressmode
  \message {EXPRESSMODE --- No plotting}
\fi
%\expressmodetrue
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%
  \newdimen\xposition
  \newdimen\yposition
  \newdimen\dyposition
  \newdimen\crossbarlength
  \def\puterrorbar at #1 #2 with fuzz #3 {%
    \xposition=\Xdistance{#1}
    \yposition=\Ydistance{#2}
    \dyposition=\Ydistance{#3}
  \setdimensionmode
  \put {$\bullet$} at {\xposition} {\yposition}
  \dimen0 = \yposition %                 ** Determine the y-location of
    \advance \dimen0 by -\dyposition %   **   the lower cross bar.
  \dimen2 = \yposition %                 **  Determine the y-location of
    \advance \dimen2 by  \dyposition %   **   the upper cross bar.
  \putrule from {\xposition} {\dimen0} % **  Place vertical rule.
    to {\xposition} {\dimen2}
  \dimen4 = \xposition
    \advance \dimen4 by -.5\crossbarlength
  \dimen6 = \xposition
    \advance \dimen6 by  .5\crossbarlength
  \putrule from {\dimen4} {\dimen0} to {\dimen6} {\dimen0}
  \putrule from {\dimen4} {\dimen2} to {\dimen6} {\dimen2}
  \setcoordinatemode}
%
\def\dev{\mathop{\rm dev}\nolimits} % Define special math. function.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\nopagenumbers
\hsize=5.75in
\centerline{\quad}
\vskip4in
\centerline{\bf Mathematical Approximation and Documentation}
\bigskip
\centerline{Harry A. Watson, Jr.}
\smallskip
\centerline{Quality Assessment Directorate}
\centerline{Naval Warfare Assessment Center}
\centerline{Corona, California 91718-5000}
\centerline{(909) 273-4787}
\vfill\eject
\centerline{\quad}
\vskip 7.5in
{\narrower\noindent
This book and the associated software fit the description in
the U. S. Copyright Act of a ``United States Government Work.''
It was written as a part of the author's official duties as
a government employee.  This means it cannot be copyrighted.  This
document and the software are freely available to the public for use
without a copyright notice, and there are no restrictions
on its use, now or subsequently.}
\vfill\eject
\footline={\hss\tenrm\folio\hss} %
\pageno=-1
\centerline{\bf Preface}
\centerline{to}
\centerline{\bf Mathematical Approximations and Documentation}
\bigskip
There are several elementary mathematical approximations that occur
on every scientific and engineering project at irregular intervals.
Frequently the same approximations appear again and again as {\it ad hoc\/}
report requests (and often with short notice).
While the programming and documentation is straightforward, the more
elaborate and expensive commercial software is often unable to deliver
suitable output.  This text includes the basic theory and elementary
computer programs in BASIC (Beginners All-purpose Symbolic Instruction
Code\footnote{$^1$}{The coding is actually done in MS-DOS QBasic 
(MicroSoft-Disk Operating System).  
There should be little problem in conversion to
GW-BASIC.}) and ``{C}.''\footnote{$^2$}{The ``C'' language is written for the
{\it Microsoft QuickC Compiler}.}  Documentation, including graphs, 
is provided in \TeX\ %
and \PiCTeX.  A later chapter is devoted to more complex graphs done in
the device-independent software {\bf METAFONT}, which can be included
in \TeX\ files.  No deep knowledge of programming is required, nor is
the user expected to be a ``\TeX-nician;'' indeed, the user is expected to
lift the coding ``as is'' and make the obvious substitutions. 
This book and the associated software fit the description in
the U. S. Copyright Act of a ``United States Government Work.''
It was written as a part of the author's official duties as
a government employee.  This means it cannot be copyrighted.  This
document and the software are freely available to the public for use
without a copyright notice, and there are no restrictions
on its use, now or subsequently.
Anyone using this document or these programs
and files does so entirely at her/his own risk.
Anyone who demands payment for the distribution of this
material must make clear that the charge 
is for distribution only and is in
no sense a license fee or purchase fee.  


The three topics covered in this textbook are (1) Simpson's rule for
numeric integration, (2) Newton's method for approximation of roots,
and (3) least-squares curve fit with a test to verify the 
choice of the underlying statistical distribution.
Many examples are provided with hand-calculations and tables.  It is
expected that the user will find portions of the \TeX\ file useful.
With this in mind, the macros ({\tt \char92 def}) are kept to a minimum
and the displayed portions are self-contained.

As often as possible, references are made to textbooks universally
available, such as outline series books found in any college bookstore
or encyclopedia articles.  The author has attempted to locate examples
of particular interest rather than pathological cases.  This is to
ensure a greater likelihood that an example may be copied directly
into a scientific or engineering paper.
\medskip
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                                                  %
% Here is a first graph that you can cut and paste.%
%                                                  %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
$$\beginpicture                                    % Draw the graph of a
  \setcoordinatesystem units <.78539in,.5in>       % Sine curve using the
  \setplotarea x from -2.25 to 4.5, y from -1 to 1 % macros of PiCTeX.
  \plotheading {The graph of $y = \sin x$}
  \axis bottom shiftedto y=0  label {$\scriptstyle y = \sin x$} ticks in
   withvalues {$\scriptstyle -\pi$} {$-{\pi\over2}$}
              {\quad\ O} {$\pi\over2$} {$\scriptstyle\pi$}
              {${3\pi\over2}$} {$\scriptstyle2\pi$} /
              at -2 -1 0 1 2 3 4 / /
  \axis left shiftedto x=0 /
\ifexpressmode
  \put {\bf EXPRESSMODE} at 2.00 0.5
\else
  \setquadratic
  \plot 0 0   0.2 0.30902  0.4 0.58779  0.6 0.80902  0.8 0.95106
              1.0 1.00000  1.2 0.95106  1.4 0.80902  1.6 0.58779
              1.8 0.30902  2.0 0.00000  2.2 -.30902  2.4 -.58779
              2.6 -.80902  2.8 -.95106  3.0 -1.0000  3.2 -.95106
              3.4 -.80902  3.6 -.58779  3.8 -.30902  4.0 0.00000 / %
  \plot 0 0   -0.2 -0.30902  -0.4 -0.58779  -0.6 -0.80902  -0.8 -0.95106
              -1.0 -1.00000  -1.2 -0.95106  -1.4 -0.80902  -1.6 -0.58779
              -1.8 -0.30902  -2.0 -0.00000  / %

\fi
\endpicture$$
\centerline{\rm F{\sevenrm IGURE} 1}  
\medskip
%
%  The sine (not the "sin" or moral failing) is used over and over
%  in scientific and engineering texts.  It is always a good "filler"
%  in any scientific document.
%
A word about fonts is in order.  The only fonts employed are from the
standard distribution (16 fonts) for \TeX.  All computer program statements
are typeset in the so-called {\tt typewriter font\/} to provide contrast
and to allow alignment of the characters for keyboard input.

The copyright for the program \TeX\ belongs to D.~E.~Knuth; the trademark
\TeX\ belongs to the American Mathematical Society; and {\tt MS-DOS} is
a trademark of the Microsoft Corporation.
\vfill\eject
\pageno=-2
\centerline{\bf TABLE OF CONTENTS}
\bigskip
{\narrower
\settabs 6 \columns
\+ Preface &&&&&\ i\cr
\medskip
\+ Table of Contents &&&&&\/ ii\cr
\medskip
\+  && Chapter 1---Simpson's Rule &&&\cr
\+ 1.0 & Introduction &&&&\ 1\cr
\+ 1.1 & Simpson's Rule &&&&\ 2\cr
\+ 1.2 & Integration Computer Programs &&&&\ 4\cr
\+ 1.3 & Vivasection of the Programs   &&&&\ 7\cr
\+ 1.4 & Numerical versus Exact Integration &&&&\ 9\cr
\+ 1.5 & Truncation Error  &&&&11\cr
\+ 1.6 & An Example &&&&12\cr
\+ 1.7 & Verifying Derivatives &&&&14\cr
\+ 1.8 & Generalizations &&&&19\cr
\medskip
\+  && Chapter 2---Newton's Method &&&\cr
\+ 2.0 & Introduction &&&& 21\cr
\+ 2.1 & Theory &&&&22\cr
\+ 2.2 & Applications &&&&23\cr
\+ 2.3 & The Algorithm &&&&26\cr
\+ 2.4 & A Second Opinion &&&&27\cr
\+ 2.5 & The Quasi-Newton Method &&&&28\cr
\+ 2.6 & Pathological Examples &&&&30\cr
\medskip
\+  && Chapter 3---Linear Least-squares Line &&&\cr
\+ 3.0 & Introduction &&&& 31\cr
\+ 3.1 & Dissection of a Graph &&&& 32\cr
\+ 3.2 & Formula Derivation &&&& 37\cr
\+ 3.3 & Goodness of Fit &&&& 39\cr
\+ 3.4 & More Theory &&&& 42\cr
\+ 3.5 & Problems &&&& 44\cr
\medskip
\+ && Chapter 4---An Adaptive Quadrature Routine &&&\cr
\+ 4.0 & Introduction &&&& 47\cr
\+ 4.1 & A Simple Approach &&&& 48\cr
\+ 4.2 & An Adaptive Quadrature Routine &&&& 49\cr
\medskip
\+ && Chapter 5---{\bf METAFONT} &&&\cr
\+ 5.0 & Introduction &&&& 51\cr
\+ 5.1 & A Metafont Program &&&& 52\cr
\medskip
\+ && Chapter 6---An Important Improper Integral &&&\cr
\+ 6.0 & Introduction  &&&& 54\cr
\+ 6.1 & Theoretical Evaluation &&&& 55\cr
\+ 6.2 & Numerical Evaluation   &&&& 57\cr
\bigskip
\+  Appendix A---Mathematical Preliminaries &&&&& 58\cr
\medskip
\+  Appendix B---References                 &&&&& 61\cr
\medskip
\+  Appendix C---A Geometric Persuasion     &&&&& 62\cr
\medskip
\+  Appendix D---A Contour Integral         &&&&& 64\cr 
\medskip
\+  Appendix E---The Malgrange-Ehrenpreis Theorem &&&&& 66\cr
}
\vfill\eject
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pageno=1
%
\headline={\tenrm\hfill Simpson's Rule}
\centerline{\bf Chapter 1}
\bigskip
\noindent{\bf\llap{1.0\quad}Introduction.} 
When one finally finishes algebra and geometry and embarks into the
realm of calculus, all problems seem solvable.  The universe appears
to be reduced to mathematical constructs and it is only a matter of
time until mathematics can define everything.  There is a conceit
among those who first master the art and discipline of {\bf The Calculus}.
The tyro\footnote{$^3$}{A beginner in learning, a novice.  Don't rush
to your dictionary---footnotes will be provided for many words.}
feels a superiority to those who are ignornant of this
powerful tool, replete with its repertoire of skills, special symbols,
concise language, and geometric persuasions.  The euphoria which is
closely associated with the mastery of {\bf The Calculus} leaves its pupil
craving a repeat experience.  The classical development of calculus
has evolved over several centuries; the problems and examples are thus
arranged so as to direct the student along a path devoid of many of the
harsh pitfalls in applied science and engineering.  Embedded in this
development are the two classic encounters with the dirty reality of 
numerical approximation:  Newton's method for approximating roots and
Simpson's rule for approximating integrals.  These two techniques 
are ubiquitous\footnote{$^4$}{Being found everywhere at the same time.}
in numerical analysis, calculating devices, and
engineering in general.  Their study and development is the subject
of the first two chapters.  In this, the first chapter, the topic of
Simpson's rule is considered.  The coding is done in several different 
computer languages and the text edited in different formats to ensure 
complete generality.  Since Simpson's rule depends on properties of
the parabola, a general graph of a parabola is displayed below.
\bigskip
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  The graph of a parabola.
%
\centerline{PARABOLA}
\medskip
$$\beginpicture
  \setcoordinatesystem units <1cm,1cm>
  \setplotarea x from -2.5 to 5.0, y from -1.5 to 5.0 %
  \axis bottom shiftedto y=0.0 ticks length <0pt>
    withvalues {$O$} {$x$} / at -.25 5.0 / / %
  \axis left shiftedto x=0.0 ticks length <0pt>
    withvalues {$y$} / at 5.0 / / %
  \put {$V$} [t] <2pt,-4pt>  at 1 1 %
  \put {$\scriptstyle\bullet$} at 1 1 %
\ifexpressmode
  \put {\bf EXPRESSMODE} at 3 3 %
\else
  \setquadratic
  \plot -2 5  1 1  4 5 / %  
\fi
\endpicture$$
$$y = f(x) = a x^2 + bx + c,\ a>0,$$
$$\hbox{Vertex } = V = \left(-{b\over2a},\ %
{b^2\over2a}\left({1\over2a}-1\right) + c\right).$$
\medskip
$$\hbox{F{\sevenrm IGURE} 2}$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\vfill\eject
\bigskip
\noindent{\bf\llap{1.1\quad}Simpson's rule.}  Most scientists and engineers
have completed elementary calculus and are aware of the technique for
numerical approximation by fitting arcs of parabolas; however, even
someone who has only completed high school mathematics (algebra and
plane geometry) will be able to apply this technique with success
(and produce a professional output that will amaze even the most
erudite\footnote{$^5$}{Possessing or displaying excessive knowledge
acquired mainly from books.} pedant on the staff).  
There are pitfalls; the expert as well as the tyro can be trapped
by some speciously simple-looking functions.\footnote{$^6$}{One pathological
example is the so-called {\it Quadratrix of Hippias}, 
$y=x\tan\big(\pi y/2\big)$.}  Having said all of this, let us begin the
basic theoretical development of Simpson's rule.\footnote{$^7$}{George B.
Thomas, Jr., {\it Calculus and Analytic Geometry}, (Reading, Massachusetts:
Add\-i\-son-Wes\-ley Publishing Company, Inc., 1966), pages 385-388.}

\medskip
\noindent
First of all, let $[\alpha,\beta]$ be a number interval, that is
$$[\alpha,\beta] = \{\,x\,\mid\,\alpha \le x \le \beta\,\}.$$
We'll use the Greek letters $\alpha$ and $\beta$ because later we'll
want to use $a$, $b$, and $c$ as coefficients of the quadratic
equation $y=ax^2+bx+c$.  
% Moreover, as we develop this method, the
% importance of the endpoints will diminish.
We will let $f$ denote a function, $f\colon [\alpha,\beta] \to {\bf R}$, where
${\bf R}=\{\,x\,\mid\,x$ is a real number $\}$.  For convenience, 
we will use let $y_\alpha = y_0 = f(\alpha)$, 
$y_\beta = y_2 = f(\beta)$, and $y_1 = f\big((\beta+\alpha)/2\big)$.
This is all bookingkeeping.  It is one of those mathematical concepts that
causes ``math anxiety;''  however, it doesn't need to.  Just consider $y$
to be the left-hand side of some expression, for instance
$$ y = x^2 - 4x +2.$$
When $x=2$ we might have $y = 2\cdot2 - 4\cdot2 + 2 = -2$.  This could
be written as $y(2)$ or $y(x=2)$ or even $y\big|_{x=2}$, which is
also written $y\big|_{2}$.  We will just
simplify things and write it as $y_2$.  This is a fundamental idea, if
the reader is confused or lost on this matter, it would be a good idea
to review an algebra book (or consult appendix A).

\medskip
\noindent
Simpson's rule follows from the so-called {\bf prismoidal formula} (for areas)
$$A_p\ =\ {h\over3}\big[\,y_0+4y_1+y_2\big]\ =\ {\beta-\alpha\over6}
\left[f(\alpha) + 4\cdot f\left({\beta+\alpha\over2}\right)
    +f(\beta)\right].$$
We will denote the left-hand side of this equation 
as $A_p$, which stands for 
the area under the arc of the parabola.  This simple formula gives the
exact area if the function, $f(x)$, is a polynomial of degree not
higher than 3.\footnote{$^8$}{Glenn James and James, Robert C., {\it
Mathematics Dictionary}, (Princeton, NJ: D. Van Nostrand Company, Inc., 1964),
pages 358-359.}
In particular, $A_p$, gives the area under the arch of the parabola
$$y\ =\ f(x)\ =\ ax^2 + bx + c$$
between $x=-h$ and $x=+h$, as in Fig. 3.\footnote{$^9$}{Without loss
of generality, we choose the interval $[a,b]$ to be $[-h,h]$.
This simplifies the algebra.  Everything is still completely general because
we can always ``shift'' a graph back and forth along the $x$-axis without
changing the area under the curve.}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%   The area under an arch of a parabola.
%
$$\beginpicture
  \setcoordinatesystem units <1cm,1cm>               % Use PiCTeX to
  \setplotarea x from -2.5 to 2.5, y from -.5 to 2.5 % construct this
  \axis bottom shiftedto y=0 label {F{\sevenrm IGURE} 3} %
     ticks length <0pt> withvalues {$\scriptstyle-h$}
     {$O$\quad} {$\scriptstyle h$} / at -2 0 2 / /
  \axis left shiftedto x=0 /
  \putrule from  -2 0  to  -2 1.5
  \putrule from  2 0  to 2 0.75
  \put {$\scriptstyle x$} at 2.75 0
  \put {$\scriptstyle y$} at 0 2.75
  \put {$\scriptstyle y_0$} at -1.75 0.75
  \put {$\scriptstyle y_1$} at  0.2 0.75
  \put {$\scriptstyle y_2$} at  1.75 0.375
\ifexpressmode
  \put {\bf EXPRESSMODE} at 1 1
\else
  \setquadratic
  \plot -2 1.5   0 1.6  2 0.75 /
\fi
\endpicture$$     
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\bigskip
\noindent
This result is readily established as follows:  in the first place, we have
$$A_p = \int_{-h}^h \left(ax^2+bx+c\right)\,dx =
\  {{ax^3\over3}+{bx^2\over2}+cx\,\bigg|}_{x=-h}^{x=h}
\ =\  {2ah^3\over3} + 2ch.$$
Since the curve passes through the three points $(-h,y_0)$, $(0,y_1)$, and
$(h,y_2)$, we also have
$$y_0 = ah^2-bh+c,\quad y_1=c,\quad y_2=ah^2+bh+c,$$
from which it follows that
$$\eqalign{  c\ &=\ y_1,\cr
ah^2 -bh      \ &=\ y_0-y_1,\cr
ah^2+bh       \ &=\ y_2-y_1,\cr
2ah^2         \ &=\ y_0+y_2-2y_1.\cr}$$
Hence, expressing the area $A_p$ in terms of the ordinates $y_0$,
$y_1$, and $y_2$, we have
$$A_p = {h\over3}\left[2ah^2 +6c\right]
  ={h\over3}\left[\big(y_0+y_2-2y_1\big) +6y_1\right]$$
or
$$A_p = {h\over3}\left[y_0 + 4y_1 + y_2\right]\ =
      \  {h\over3}\left[f(-h)+4\cdot f(0) + f(h)\right].\eqno(1)$$
%
\medskip
\noindent
Simpson's rule is derived by applying the prismoidal formula 
to successive subintervals of $[\alpha, \beta]$.  Each separate
subinterval has length $2h$.  So, we may write $[\alpha,\alpha+h]$
as $[x_0, x_1]$, $[\alpha+h, \alpha+2h]$ as $[x_1, x_2]$, and so on.
Now the curve $y=f(x)$ between $x=\alpha$ and $x=\beta$
is really $n/2$ curves.  Each separate piece of the
curve covers an $x$-subinterval of width $2h$; it is approximated by an
arch of a parabola through its ends and its mid-point.  The area under each
parabolic arc is computed as in Eq. (1) and the resulting
areas are summed together to give the rule
$$\frame <5pt> {$\displaystyle A_S = {h\over3}\big[\, y_0 + 4y_1 + 2y_2 +4y_3
   + \cdots + 2y_{n-2} + 4y_{n-1} + y_n \big].$}\eqno(2)$$
This result, $A_S$, is an approximate value of $\int_a^b f(x)\,dx$.
Recall our bookkeeping.  We let $y_0$, 
$y_1$, $y_2$, $\ldots$, $y_n$ stand for $f(x_0)$, $f(x_1)$, $\ldots$,
$f(x_n)$.  In other words,  $y_0$, $y_1$, $\ldots$, $y_n$ are the 
ordinates of the
curve $y=f(x)$ corresponding to the abscissas $x_0=a$, $x_1=a+h$, $x_2=a+2h$,
$\dots$, $x_n=a+nh=b$. 
Each abscissa point then corresponds to an endpoint of a
 subdivision of the interval
$\alpha \le x \le \beta$ into $n$ equal subintervals each of width 
$h=(\beta-\alpha)/n$.
(See Fig. 4.)  The number $n$ of subdivisions must be an {\it even\/}
integer in order for this method to work.  
Not all authors use this same scheme.  Some rely on subintervals of length
$h$ instead of $2h$.  When consulting a table book, be careful to examine
the setup for this rule.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  Simpson's rule over a curve (after Thomas's Calculus).
%
$$\beginpicture
  \setcoordinatesystem units <1cm,1cm>
  \setplotarea x from -0.5 to 10, y from -0.5 to 3.0
  \axis bottom shiftedto y=0 label {F{\sevenrm IGURE} 4} %
     ticks length <0pt>
     withvalues {$O$} {$\alpha$} {$\leftarrow\! h\!\rightarrow$} {$\beta$} /
     at -0.25 1 3.5 7 / / %
  \axis left shiftedto x=0 / %
  \putrule from 1 0 to 1 2.8 
  \putrule from 2 0 to 2 2.9
  \putrule from 3 0 to 3 2.65
  \putrule from 4 0 to 4 2.3
  \putrule from 5 0 to 5 1.85
  \putrule from 6 0 to 6 1.5
  \putrule from 7 0 to 7 1.4
  \putrule from 3 -0.02 to 3 -0.4
  \putrule from 4 -0.02 to 4 -0.4
  \put {$ y=f(x)$} at 5 3
  \put {$ x$} at 10.25 0
  \put {$ y$} at 0 3.25
  \put {$\scriptstyle 1$} <0pt,2pt> at 1.0 2.9
  \put {$\scriptstyle 4$} <0pt,2pt> at 2.0 3.0
  \put {$\scriptstyle 2$} <0pt,2pt> at 3.0 2.75
  \put {$\scriptstyle 4$} <0pt,2pt> at 4.0 2.4
  \put {$\scriptstyle 2$} <0pt,2pt> at 5.0 1.95
  \put {$\scriptstyle 4$} <0pt,2pt> at 6.0 1.6
  \put {$\scriptstyle 1$} <0pt,2pt> at 7.0 1.5
  \put {$\scriptstyle y_0$} at 0.8 1
  \put {$\scriptstyle y_1$} at 1.8 1
  \put {$\scriptstyle y_2$} at 2.8 1
  \put {$\scriptstyle y_{n-1}$} at 5.7 1
  \put {$\scriptstyle y_{n}$} at 6.8 1
\ifexpressmode
  \put {\bf EXPRESSMODE} at 3 3
\else
  \setquadratic
    \plot  0.5 2.6  1.0 2.8  2.0 2.9  3.0 2.65  4.0 2.3
           5.0 1.85  6.0 1.5  7.0 1.4  7.5 1.5  / % 
\fi
\endpicture$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
It looks like the more subintervals we take, that is the smaller we
let $h$ become,
the more accurate the approximation will be to the area under
the curve (the value of the integral).  This is part of error 
estimation and will be done in a later paragraph.  In the next
paragraph we will do numerical examples and see how the method
behaves numerically.
\vfill\eject
\noindent{\bf\llap{1.2\quad}Integration Computer Programs.}  We'll list two
computer programs to do the same job---one in BASIC and one in {``C''}.
One integral that crops up again and again is the so-called elliptic integral.
We will look at a complete elliptic integral of the second kind, usually
denoted
$$E\left(k,{\pi\over2}\right)\ =\ \int\nolimits_0^{\pi/2}
   \sqrt{1-k^2\cdot\sin^2 \theta}\,d\theta.\eqno(3)$$
Finally, we need to choose a value for $k$.  If we let $e$ denote the
base of the natural logarithms, that is $e \approx 2.7182828459045\dots$,
then assign $k$ to be $k = \sin\big(e/2\big)$, we will have a good example
to experiment with.\footnote{$^{10}$}{Harry A. Watson, Jr., ``An
approximation to Planck's constant,'' {\it Abstracts\/}
of the American Mathematical Society (AMS) \# 81T--181, June 1981.}
\bigskip
{\parindent=0pt\tt\ttraggedright\obeylines
100 DEFDBL A-H, O-Z : REM DEF FNA defines the integrand (function).
105 DEF FNA (x AS DOUBLE) = SQR(1!-(SIN(EXP(1)/2)*SIN(x)){\char94}2) 
110 A = 0!: REM                      This is the lower endpoint of [a,b].
120 B = 3.141592653589793\#/2!: REM  This is the upper endpoint of [a,b].
130 SUM = 0!: REM                    This is an accumulator.
135 REM The user is asked for a number of intervals, n, even and > 0.
140 INPUT "Enter number of intervals (must be even) "; N
145 IF N <= 0 OR (2 * FIX(N / 2) - N < 0) THEN PRINT "Input error": GOTO 140
150 H = (B - A) / N: REM      The sub-interval size h = (b-a)/n.
160 FOR I = 1 TO N STEP 2: REM The "FOR" loop is done n/2 times.
170 SUM = SUM + FNA(A + (I - 1) * H)
175 PRINT A + (I - 1) * H, FNA(A + (I - 1) * H)
180 SUM = SUM + 4! * FNA(A + I * H)
185 PRINT A + I * H, FNA(A + I * H)
190 SUM = SUM + FNA(A + (I + 1) * H)
195 PRINT A + (I + 1) * H, FNA(A + (I + 1) * H)
200 NEXT I: REM After loop, the sum is y\_0+4*y\_1+2*y\_2+...+4*y\_{\char123}n-1{\char125}+y\_n.
210 PRINT "Sum = "; SUM: PRINT "  Value of integral = "; SUM * H / 3
220 REM This short program will integrate a function FNA(x) from x = a to b.
} 
\medskip
\noindent The ``C'' program is more complicated because it has the
``{\tt\#include}'' statements.  However, the ``C'' program executes faster
and is portable to other platforms.  In addition, the programs below
both use a math co-processor, which gives floating-point arithmetic.
Minor differences in the last digit may be due to the print (format) algorithm.
The BASIC program would not accept the longer value of $\pi$.
If one examines the \TeX\ file for these listings, it will be noted that
the control characters are input as {\tt\char92char92} for the backspace
({\tt\char92}), {\tt\char92char123} for the left brace ({\tt\char123}),
{\tt\char92char125} for the right brace ({\tt\char125}), and 
{\tt\char92char94} for the hat or circumflex ({\tt\char94}).
\medskip
{\parindent=0pt\tt\ttraggedright\obeylines
\#include <stdio.h>        /* Header for input/output subroutines. */
\#include <math.h>         /* Header for math subroutines. */
\#include <float.h>        /* Header for floating point subroutines. */
\medskip
\#define pi 3.141592653589793238462643383279     /* Accurate value for pi. */
\medskip
/* Simpson's rule for approximating integrals.
\qquad a:              left endpoint
\qquad b:              right endpoint
\qquad fc:             pointer to function to integrate
\qquad n:              number of subintervals
*/
double fc (double x);
\medskip
main()
{\char123}
double a,b,h,sum,x,y;     /* In 'C' all variables must be assigned */
double p1, p2, p3;
int i, n;
printf("{\char92}007");           /* Sound bell in computer. */
a = (double) 0.0;
printf("{\char92}nLeft end point    = \%.16lf",a);
b = (double) pi/2.0;
printf("{\char92}nRight end point   = \%.16lf",b);
i = -1;
while (i < 0){\char123}
printf("{\char92}nEnter number of subintervals (must be even) ");
scanf("{\char92}\%d",\&n);
i = n/2; i = 2*i - n;   /* Don't allow odd values of n.  */
if (n<=0) i = -1;       /* Don't allow zero or negative. */
{\char125}
printf("{\char92}nNumber of subintervals \%d",n);
h = (double) (b-a)/n;
for (i=1, sum=0.0; i<=n; i = i+ 2){\char123}
\quad p1 = fc((double) a+(i-1)*h);
\quad p2 = fc((double) a+i*h);
\quad p3 = fc((double) a+(i+1)*h);
\quad sum += p1 + 4.0 * p2 + p3;
/* printf("{\char92}n x, f(x) \%.16lf \%.16lf",a+(i-1)*h,p1); */
/* printf("{\char92}n x, f(x) \%.16lf \%.16lf",a+i*h,p2);     */
/* printf("{\char92}n x, f(x) \%.16lf \%.16lf",a+(i+1)*h,p3); */
{\char125}
\medskip
printf("{\char92}nValue of sum      = \%.16lf", (double) sum);
y = (double) h*sum/3.0;
printf("{\char92}nValue of integral = \%.16lf", (double) y);
printf("{\char92}n");
return(0);
{\char125}
\medskip
double fc (double x)
{\char123}
\qquad double y;
\qquad y = sqrt(1.0-sqrt(exp(1.)/2.)*sqrt(exp(1.)/2.)*sqrt(x)*sqrt(x));
\qquad return (y);
{\char125}
\medskip
/* End of file */
}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  Parabolic curve fit for unequally spaced abscissas.
%
$$\beginpicture
  \setcoordinatesystem units <.5cm,.3cm>
  \setplotarea x from -0.25 to 20, y from -0.25 to 10.0
  \axis bottom 
     label {Parabolic curve fits unequally spaced abscissas} 
     shiftedto y=0 / %
  \axis left shiftedto x=0  / %
  \putrule from 2 0     to 2 3.8
  \putrule from 4 0     to 4 3.85
  \putrule from 5.3  0  to 5.3  3.6
  \putrule from 9 0     to 9 4.78571
  \putrule from 10. 0   to 10. 4.95
  \putrule from 14 0    to 14 8.35
  \putrule from 16 0    to 16 8.6
  \put {$O$} at 0.5 -0.75
  \put {$x$} at 19 -0.5
  \put {$f(x)$} [r] at -.25 9
\ifexpressmode
  \put {\bf EXPRESSMODE} at 10 5
\else
  \setquadratic
  \plot 1 3   3 4   7 3 / %
  \plot 5 3.5   8 4.5   12 5.5  / %
  \plot 10 5.0  13 8.0  17 8.5 / %
\fi
\endpicture$$
\centerline{F{\sevenrm IGURE} 5}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\vfill\eject
\bigskip
\noindent We'll make a table of values for various integer values
of $n$, that is, various numbers of subdivisions.  Much can be learned
from comparing the outputs of the two programs and examining their
behavior as $n$ becomes large.
\smallskip
$$\vbox{\offinterlineskip
\hrule
\halign{&\vrule#&\strut\quad\hfil#\quad&\vrule#
        &\quad\hfil#\quad&\vrule#&\quad\hfil#\quad&\vrule#
        &\quad\hfil#\quad&\vrule#\cr
height2pt&\omit&&\omit&&\omit&&\omit&\cr
&$n =$\hfil&&BASIC Program&&``C'' Program\quad&&Difference\quad&\cr
height2pt&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}
height2pt&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}
height2pt&\omit&&\omit&&\omit&&\omit&\cr
&  2   &&1.073441760234934 &&1.0734417602349340&&0.00000000000000&\cr
&  4   &&1.057800052054470 &&1.0578000520544700&&0.00000000000000&\cr
&  8   &&1.054898082319603 &&1.0548980823196030&&0.00000000000000&\cr
& 10   &&1.054750510268630 &&1.0547505102686300&&0.00000000000000&\cr
& 20   &&1.054686162116484 &&1.0546861621164840&&0.00000000000000&\cr
& 40   &&1.054685849666779 &&1.0546858496667790&&0.00000000000001&\cr
& 80   &&1.054685849645436 &&1.0546858496454370&&0.00000000000001&\cr
&100   &&1.054685849645437 &&1.0546858496454360&&0.00000000000001&\cr
&200   &&1.054685849645437 &&1.0546858496454370&&0.00000000000000&\cr
&500   &&1.054685849645437 &&1.0546858496454370&&0.00000000000000&\cr
&1,000 &&1.054685849645437 &&1.0546858496454370&&0.00000000000000&\cr
&2,000 &&1.054685849645436 &&1.0546858496454370&&0.00000000000001&\cr
&10,000 &&1.054685849645435 &&1.0546858496454330&&0.00000000000002&\cr
height2pt&\omit&&\omit&&\omit&&\omit&\cr}
\hrule}$$
\smallskip\noindent
Notice, if you will, two things are happening:  (1) after $n=40$ there
is no change in the first five decimal places; and, (2) after $n=80$
there is almost no change at all!  So, after $n=80$, successive terms
become closer and closer together and the calculated value converges
to a given number.  This phenomenon is called, obviously enough,
convergence.  It would be nice if we could tell {\sl beforehand\/}
just what value of $n$ to choose to ensure the accuracy we need.
Clearly, if all we want is five decimal place accuracy, $n=40$ will
suffice.  If we are using a double-precision calculation, then $n=80$
will do.  Of course, you might say why not just take $n=10000$ to start
with?  Just run the program and see how long it takes to get an answer!
It is possible that time is also a factor---even the fastest computers
take time to do the involved mathematical operations required in numerical
integration.  
\medskip\noindent
The next item to be introduced is error estimation.  It will be shown that
for $n$ subintervals, there is a number $\xi\,\in\,(\alpha, \beta)$
such that the truncation error is
$$-{(\beta-\alpha)^5\over 180n^4}\,f^{(4)}(\xi).$$
If the fourth derivative of $f(x)$ is bounded on $[\alpha, \beta]$, that
is, if there exists a number $M$ such that
$$\big| f^{(4)}(x) \big|\,\le\,M\quad\hbox{for all } x\in [\alpha,\beta],$$
then the error is bounded by
$${(\beta-\alpha)^5\,M \over 180 n^4}.$$
At this point in our development, the importance of the error bound is that
we will be assured that the larger the value of $n$, the closer the answer
will be to the exact solution.  This says that for any given degree of
accuracy, there is a positive integer $N$ such that for all $n\ge N$,
each approximation computed by Simpson's rule with $n$ terms will lie
within the given degree of accuracy to the exact solution.
\medskip\noindent
This error estimation sounds like pure theory, and so it could be skipped
without losing anything.  That's really not the case.  What we have is 
exactly equivalent to the mechanical ``tolerance.''  If you want something
within a particular tolerance, say $\epsilon$, then you must calibrate
your tools to within some accuracy, say $\delta$.  It is exactly this
so-called $\epsilon$-$\delta$ philosophy that is the basis for theoretical
analysis.  One of the great failings of pure analysis has been its inability
to relate to the industrial quality efforts and statistical error estimations.
\vfill\eject
\noindent{\bf\llap{1.3\quad}Vivasection\footnote{$^{11}$}{\rm Disection
of a living (as opposed to inoperable) computer program.} of the
Programs.}  The first item to note in the computer programs is the
definition of the function $f(x)$\footnote{$^{12}$}{Mathematicians
prefer to call a function by its single letter abbreviation, e.g.
$f$, $g$, etc.; however, since the independent variable, say $x$, is
required in the computer definitions, it seems better to use the
convention of complex analysis and keep the notation $f(x)$, $g(x)$, etc.,
so that there is no ambiguity in the computer coding.}
In the BASIC program the function is defined by the statement
\smallskip
{\tt\parindent=0pt\ttraggedright\obeylines
105 DEF FNA (x AS DOUBLE) = SQR(1!-(SIN(EXP(1)/2)*SIN(x)){\char94}2) 
}
\smallskip\noindent
while in the ``C'' program the function is defined as a separate subroutine.
``C'' is fashioned after subroutines, so this should be no surprise.  The
function definition subroutine in ``C'' is thus
\smallskip
{\tt\parindent=0pt\ttraggedright\obeylines
double fc (double x)
{\char123}
\qquad double y;
\qquad y = sqrt(1.0-sin(exp(1.)/2.)*sin(exp(1.)/2.)*sin(x)*sin(x));
\qquad return (y);
{\char125}
}
\smallskip\noindent
Notice, further, that BASIC supports exponentiation, that is 
{\tt A\char94B} means $A^B$ whereas ``C'' does not (it requires a
separate subroutine call to {\tt pow()}).  Now we will focus on
the interval endpoints, $\alpha$ and $\beta$,
of the interval $[\alpha, \beta]$.  Since the letters {\tt A, B}
and {\tt a, b} are not used elsewhere inside the programs, they can be
used for endpoint values.  BASIC is not case sensitive; ``C'' is.  We define
the endpoints (or limits of integration) as follows:
\smallskip
{\tt\parindent=0pt\ttraggedright\obeylines 
110 A = 0!: REM                      This is the lower endpoint of [a,b].
120 B = 3.141592653589793\#/2!: REM  This is the upper endpoint of [a,b].
}
\smallskip
{\tt\parindent=0pt\ttraggedright\obeylines
a = (double) 0.0;
printf("{\char92}nLeft end point    = \%.16lf",a);
b = (double) pi/2.0;
printf("{\char92}nRight end point   = \%.16lf",b);
}
\smallskip\noindent
Notice that in the ``C'' program the value of the endpoints is displayed
via a {\tt printf()} function.  It is always a good practice to display
such variables.   The next item of interest is the input of the number
of intervals, $n$.  This also determines the width (length) of the
subintervals, $h = (\beta-\alpha)/n$.  Each program does a check to ensure 
that the integer $n$ is positive and not odd.  There are many ways to make
such a determination, so it is not necessary to dwell on the matter.  The
main portion of the Simpson's rule is the iteration.  Here a variable
named {\tt sum} or {\tt SUM} is initialed to zero and functions as an
accumulator.  The values
$y_{i-1} + 4\cdot y_{i}+y_{i+1}$ are added to the accumulator for
$i$ between 1 and $n-1$ in steps of 2.  The final value of the
accumulator is multiplied by $h/3$ to get the approximation to the 
integral $\int_\alpha^\beta f(x)\,dx$.
\smallskip
{\tt\parindent=0pt\ttraggedright\obeylines
130 SUM = 0!: REM                    This is an accumulator.
150 H = (B - A) / N: REM      The sub-interval size h = (b-a)/n.
160 FOR I = 1 TO N STEP 2: REM The "FOR" loop is done n/2 times.
170 SUM = SUM + FNA(A + (I - 1) * H)
180 SUM = SUM + 4! * FNA(A + I * H)
190 SUM = SUM + FNA(A + (I + 1) * H)
200 NEXT I: REM After loop, the sum is y\_0+4*y\_1+2*y\_2+...+4*y\_{\char123}n-1{\char125}+y\_n.
}
\smallskip\noindent
Notice in the ``C'' coding below the statement {\tt sum += }$\dots$.  This
is identical to the coding {\tt sum = sum + }$\dots$.
\smallskip
{\tt\parindent=0pt\ttraggedright\obeylines
h = (double) (b-a)/n;
for (i=1, sum=0.0; i<=n; i = i+ 2){\char123}
\quad p1 = fc((double) a+(i-1)*h);
\quad p2 = fc((double) a+i*h);
\quad p3 = fc((double) a+(i+1)*h);
\quad sum += p1 + 4.0 * p2 + p3;
{\char125}
} 
\smallskip\noindent
The final calculation and the display is given below.  First for the
BASIC program and then for the ``C'' program.  The BASIC program does
its calculation in the {\tt PRINT} statement.

\smallskip
{\tt\parindent=0pt\ttraggedright\obeylines
210 PRINT "Sum = "; SUM: PRINT "  Value of integral = "; SUM * H / 3
}
\smallskip\noindent
Notice that the ``C''
program uses a subroutine ({\tt printf()}) to display its output
on the video monitor.
Notice further that in the ``C'' program the last statement is a
print with the argument {\tt "{\char92}n"}.  This is a carriage return/line
feed that is needed after the last print statement.  Otherwise, the
program will type ``{\tt Hit any key to continue}'' on the same line as
the output.
\smallskip
{\tt\parindent=0pt\ttraggedright\obeylines
printf("{\char92}nValue of sum      = \%.16lf", (double) sum);
y = (double) h*sum/3.0;
printf("{\char92}nValue of integral = \%.16lf", (double) y);
printf("{\char92}n");
}
\medskip\noindent
We will examine the error estimation for this integral.  Let's write down the
function and its first four derivatives:
$$\eqalign{f(x) &= \sqrt{1-k^2\sin^2 x}; 
    \quad\hbox{where } k = \sin\big(e/2\big);\cr
f'(x) &= -{k^2\over2}{\sin(2x)\over\sqrt{1-k^2\sin^2x} };\cr
f''(x) &= -k^2\cos2x/\sqrt{X} -k^4\sin^2(2x)/\big(4\cdot X^{3/2}\big);
   \quad\hbox{where } X = \left(1-k^2\sin^2(x)\right)\cr
f^{(3)}(x) &= 2k^2\sin(2x)/\sqrt{X} -3k^4\sin(4x)\big(4\cdot X^{3/2}\big)\cr
  &\qquad    - 3k^6\sin^3(2x)/\big(8\cdot X^{5/2}\big);\cr
f^{(4)}(x) &= 4k^2\cos(2x)/\sqrt{X} + k^4\sin^2(2x)/X^{3/2} 
      -3k^4\cos(4x)/X^{3/2}\cr
 &\qquad  -9k^6\sin(4x)\sin(2x)/\big(8\cdot X^{5/2}\big)
   -18k^6\sin^2(2x)\cos(2x)/\big(8\cdot X^{5/2}\big)\cr
 &\qquad  - 15k^8\sin^4(2x)/\big(16\cdot X^{7/2}\big).\cr}$$
Observing that if $0.0 \le x \le 1.572$, then 
$-313.8 \le f^{(4)}(x) \le 85.0$, we get a bound for
the fourth derivative on $[0,\pi/2]$.  If $M=320$, then
$$\left| f^{(4)}(x)\right| \le
M,\quad\forall\,x\in\,\left[0,{\pi\over2}\right].$$
The error is thus bounded by
$${\pi^5\cdot M\over 180\cdot32\cdot n^4}\ \approx\ {17\over n^4},$$
where $n$ is the number of subintervals of $[\alpha,\beta] = [0, \pi/2]$.
Thus, to ensure that our answer is within say .5\% of the true answer, we
would have to pick $n$ to be greater than 7 (and even!).  But this is a 
conservative estimate.  We have already seen that a much smaller value of
$n$ will suffice.  In the examples chosen for textbooks, the fourth
derivative is generally a constant (a constant-values function, that is,
a function $f^{(4)}(x) = $ constant for all $x$ under consideration).
This builds the student's confidence, but does little to reflect the
real-world situation.  The integral we have chosen is an integral that
crops up again and again in engineering problems.  Is there a satisfactory
resolution to this (common) problem?  Indeed, there is.  The idea is to
examine the approximating values for increasing values of the integer $n$
which determines the step size.  This technique is based on the theoretical
consideration that a Cauchy sequence converges.  In computer programming,
it is called an {\it adaptive quadrature routine\/} (AQR).  We will cover
the adaptive quadrature routine in a later paragraph.
\vfill\eject
\noindent{\bf\llap{1.4\quad}Numerical versus Exact Integration.}
The sophomore who has just completed an introduction to calculus
might be surprised to discover that an integral as simple as
$$\int_0^\pi \sqrt{1+\cos^2 x}\,dx\eqno(4)$$
cannot be solved by elementary techniques.  (The value of this integral
is the length of one ``arch'' of the sine curve, $y=\sin x$.  See Fig. 1
in the preface for a graph of the sine curve.)  It can be expressed in terms
of a complete elliptic integral of the second kind.  If we write
$$E\left(k,{\pi\over2}\right)
\ =\ \int_0^{\pi/2} \sqrt{1-k^2\sin^2\theta}\,d\theta,\eqno(5)$$
then we may make the simple trigonometric substitution
$$\cos^2\theta = 1 - \sin^2\theta$$
to obtain
$$\eqalign{\int_0^\pi \sqrt{1-\cos^2\theta}\,d\theta 
\ &=\ 2\cdot\sqrt{2}\cdot E(\pi/4,\pi/2)\cr
\ &=\ \sqrt{2}\,\int_0^{\pi/2}
  \sqrt{1-\sin^2(\pi/4)\cdot\sin^2(\theta)}\,d\theta\cr
&\qquad +\ \sqrt{2}\,\int_{\pi/2}^\pi 
 \sqrt{1-\sin^2(\pi/4)\cdot\sin^2(\theta)}\,d\theta,\cr}$$
where $\sin(\pi/4) = \sqrt{2}/2$.  (Recall that $\pi\over4$ 
radians\footnote{$^{13}$}{A radian is a unit a angular measure.  Basically,
$360^\circ = 2\pi$ radians.  This means that 1 radian is approximately
57.29578 degrees.}
is $45^\circ$.  Lots of tables books give values in degrees and not
in radians.  You do remember radians, don't you?  If not, check out
a trigonometry book or look in Appendix A for a review.)
We look up the value of 
$E(45^\circ, \pi/2)$ in a tables book\footnote{$^{14}$}{Dr. M. Fogiel,
{\it Handbook of
Mathematical, Scientific, and Engineering Formulas, Tables, Functions,
Graphs, Transforms}, (Piscataway, NJ: Research and Education Association,
1984), page 623.} and find that
$$E\left(45^\circ, {\pi\over2}\right) \approx 1.35006.$$
This says that the value of the integral in equation (6) should be
approximately
$$\int_0^\pi \sqrt{1+\cos^2(\theta)}\,d\theta \approx 3.82007.$$
Now we apply our integration method (Simpson's rule) and see what the
results are:
\smallskip
$$\vbox{\offinterlineskip
\hrule
\halign{&\vrule#&\strut\quad\hfil#\quad&\vrule#
        &\quad\hfil#\quad&\vrule#&\quad\hfil#\quad&\vrule#
        &\quad\hfil#\quad&\vrule#\cr
height2pt&\omit&&\omit&&\omit&&\omit&\cr
&$n =$\hfil&&BASIC Program&&``C'' Program\quad&&Difference\quad&\cr
height2pt&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}
height2pt&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}
height2pt&\omit&&\omit&&\omit&&\omit&\cr
&  2    &&3.575356081779317 &&3.5753560817793170&&0.000000000000000&\cr
&  4    &&3.829178925615088 &&3.8291789256150880&&0.000000000000000&\cr
&  8    &&3.820282406120432 &&3.8202824061204320&&0.000000000000000&\cr
& 16    &&3.820197813575163 &&3.8201978135751630&&0.000000000000000&\cr
& 32    &&3.820197789027719 &&3.8201977890277180&&0.000000000000001&\cr
& 64    &&3.820197789027712 &&3.8201977890277120&&0.000000000000000&\cr
& 128   &&3.820197789027713 &&3.8201977890277120&&0.000000000000001&\cr
& 256   &&3.820197789027711 &&3.8201977890277110&&0.000000000000000&\cr
& 512   &&3.820197789027712 &&3.8201977890277110&&0.000000000000001&\cr
height2pt&\omit&&\omit&&\omit&&\omit&\cr}
\hrule}$$
There is a lesson to be learned here.  If we use a more accurate
value for $E(45^\circ, \pi/2)$ than 1.3506, namely 1.3506438, then we
will see that our computed answer agrees more closely with the theoretical
answer.
Now, lets look at the coding that was changed in the BASIC and in the
``C'' programs:
\smallskip
{\tt\parindent=0pt\ttraggedright\obeylines
105 DEF FNA (x AS DOUBLE) = SQR(1!+COS(x)*COS(x)) 
}
\smallskip
{\tt\parindent=0pt\ttraggedright\obeylines
double fc (double x)
{\char123}
\qquad double y;
\qquad y = sqrt(1.0+cos(x)*cos(x));
\qquad return (y);
{\char125}
}
\smallskip\noindent
We added something else new.  In the BASIC program and in the ``C'' program
we added lines giving the (expected) theoretical value.
\smallskip
{\tt\parindent=0pt\ttraggedright\obeylines
230 PRINT "Theoretical approx = "; 2*SQR(2)*1.35064388 
}
\smallskip
{\tt\parindent=0pt\ttraggedright\obeylines
printf("Theoretical approx  = %.16lf",2.0*sqrt(2.0)*1.35064388);
printf("\char92n");
}
\medskip\noindent
We want to delve\footnote{$^{15}$}{To delve---to search carefully and
painstakingly for information.} more deeply into the whole matter of numerical
versus exact integration.  To do this, we will evaluate an integral exactly
and numerically and compare the results.  Lets integrate one arch of
the cosine curve:
$$\int_0^{\pi/2} \cos(x)\,dx = \sin(x)\Big|_0^{\pi/2} = \sin(\pi/2)
   -\sin(0) = 1.\eqno(6)$$
And, numerically, setting $A_p(n)$ to be the value of Simpson's rule
approximation for $n$ subintervals
$$\vbox{\offinterlineskip
\hrule
\halign{&\vrule#&\strut\quad\hfil#\quad&\vrule#
        &\quad#\hfil\quad&\vrule#&\quad#\hfil\quad&\vrule#
        &\quad#\hfil\quad&\vrule#\cr
height2pt&\omit&&\omit&&\omit&&\omit&\cr
&$n =$\hfil&&Simpson's rule&&$\big|1-A_p(n)\big|$\quad&
  &Error bound\quad&\cr
height2pt&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}
height2pt&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}
height2pt&\omit&&\omit&&\omit&&\omit&\cr
&  2    &&1.00228            &&0.00228           &&0.00332          &\cr
&  4    &&1.00013            &&0.00013           &&0.00021          &\cr
&  6    &&1.00003            &&0.00003           &&0.00004          &\cr
&  8    &&1.00001            &&0.00001           &&0.00001          &\cr
& 10    &&1.0000033922209010 &&0.0000033922209010&&0.000005312      &\cr
& 20    &&1.0000002115465910 &&0.0000002115465910&&0.000000332      &\cr
& 50    &&1.0000000054122520 &&0.0000000054122520&&0.000000008      &\cr
& 100   &&1.0000000003382360 &&0.0000000003382360&&0.0000000005     &\cr
height2pt&\omit&&\omit&&\omit&&\omit&\cr}
\hrule}$$
Now we see how the error estimate really works.  If $f(x)=\cos(x)$, then
$f^{(4)}(x) = \cos(x)$, $-1\le \cos(x)\le 1$, for all real $x$. 
$$\eqalign{f'(x)\ &=\ -\sin(x)\cr
           f''(x)\ &=\ -\cos(x)\cr
           f^{(3)}(x)\ &=\ \sin(x)\cr
           f^{(4)}(x)\ &=\ \cos(x)\cr}\eqno(7)$$
$|f^{(4)}(x)|\le 1$.  We have an error bound
$$\left|{(\beta-\alpha)^5\over180\cdot n^4}f^{(4)}(\xi)\right|
 \le {\big(\pi/2\big)^5\over180\cdot n^4}\,1
\ =\  {\pi^5\over32\cdot180\cdot n^4}.$$

\vfill\eject
%
\noindent{\bf\llap{1.5\quad}Truncation Error.}  In this section we will
derive the truncation error for Simpson's rule.  This section has some 
theory and may be omitted by someone only interested in applications.  One
might note that in the previous section that the actual error was much
less than the error bound.  The reason for the ``looseness'' in the
error bound has to do with bounding the fourth derivative over the entire
interval, $[\alpha,\beta]$.
\medskip\noindent
We will assume that the function $f(x)$ has a Taylor series expansion.  This
is a reasonable assumption for theoretical purposes; however, in the real
world it may present a problem.  Derivation with weaker conditions may be
found in standard texts.  Using equation $(1)$ for $A_p$
$$\eqalign{A_p\ &=\ {1\over3}h\big(y_0+4y_1+y_2\big)\cr
&=\ {1\over3}h\Bigg[ y_0 +4\left(y_0 +hy'_0 + {1\over2}h^2y^{(2)}_0
+{1\over6}h^3y_0^{(3)} + {1\over24}h^2y_0^{(4)} + \dots\right)\cr
&\qquad +\left(y_0 + 2hy'_0 + 2h^2y_o^{(2)} +{4\over3}h^3y_0^{(3)}
   + {2\over3}h^4y_0^{(4)}+\dots\right)\,\Bigg] \cr
&=\ {1\over4}h\left(6y_0 + 6hy_0' + 4h^2y_0^{(2)} 
  +2h^3y_0^{(3)} + {5\over6}h^4y_0^{(4)} + \dots\right)\cr}\eqno(8)$$
Now do the same thing to the integral itself.
$$\eqalign{\int_{x_0}^{x_2} y(x)\,dx\ &=\ F(x_2) - F(x_1)\cr
&=\ 2hF'(x_0) + {1\over2}(2h)^2F^{(2)}(x_0) + {1\over6}(2h)^3F^{(3)}(x_0)\cr
&\qquad+{1\over24}(2h)^4F^{(4)}(x_0) + {1\over120}(2h)^5F^{(5)}(x_0) + \dots\cr
&=\ 2hy_0 + 2h^2y_0 + {4\over3}h^3y_0^2 + {2\over3}h^4y_0^{(3)}
   + {4\over15}h^5y_0^{(4)} + \dots.\cr}\eqno(9)$$
Subtract equation (8) from equation (9) to get
$$\int_{x_0}^{x_2} y(x)\,dx - {1\over3}h\big(y_0+4y_1+y_2\big)
 = \left({4\over15}-{5\over18}\right)h^5y-0^{(4)}
\ =\ -{h^5 y_0^{(4)}\over90}\eqno(10)$$
Equation (10) follows from the arithmetic calculation
$${4\over15}-{5\over18}\ =\ {24-25\over90}\ =\ -{1\over90}.$$
If we have an interval $[\alpha,\beta]$, which can be written
as the union of intervals $[x_0,x_2]$, $[x_2,x_4]$, $\ldots$,
$[x_{n-2},x_n]$, then we may write the total truncation error
as
$$-{h^5\over90}\left(y_0^{(4)} + y_2^{(4)} + \ldots + y_{n-2}^{(4)}\right).$$
Now we make the assumption that the fourth derivative is continuous on
the interval $[\alpha,\beta]$.  $\beta-\alpha = nh$, so we have
$$-{h^5\over90}\left(y_0^{(4)} + y_2^{(4)} + \ldots + y_{n-2}^{(4)}\right)
\ =\ -{(\beta-\alpha)^5\over180\cdot n}\,y^{(4)}(\xi),$$
where $\xi\,\in\,(\alpha,\beta)$ and
$$y_0^{(4)} + y_2^{(4)} + \ldots + y_{n-2}^{(4)}
\ =\ {n\over2}\,\cdot\,y^{(4)}(\xi).$$
\vfill\eject
%
\noindent{\bf\llap{1.6\quad}An Example.}  In this section we will
consider an application to the particular function
$$\frame <5pt> {$f(x)\ =
\ (x+4\pi)\cdot(x+4\pi-1/\pi)\cdot(x+4\pi-2/\pi)$}$$
This remarkable function has properties as 
follows\footnote{$^{16}$}{The superscript before the chemical symbol
gives the value of $Z$, the atomic weight; the subscript before the
symbol gives the value of $A$, the atomic number; and, the superscript
after the symbol gives the ionization. Values taken from {\it Quantum
Physics, 2nd Ed.\/} by Eisberg and Resnick, (NY: John Wiley \& Sons,
1985), page 520.}
\medskip
{\settabs5\columns\thinmuskip=2mu
\+Ionized   
  &\underbar{\raise2pt\hbox{\qquad mass\qquad}}   &   &  &relative \cr
\+Isotope   & electron mass &\qquad $x$      &\qquad $f(x)$ &deviation\cr
\baselineskip=14pt
\+\quad${}^1_1\!{\rm H}^{+}$ & {\tt\ \ 1836.152701} &\qquad $0$      
    & {\tt\ \ 1836.15174}
    & 0.00003\%\cr
\+\quad${}^2_1\!{\rm H}^{+}$ & {\tt\ \ 3670.4830}  &\qquad $\pi$    
   & {\tt\ \ 3643.34824}
   & 0.37\%\cr
\+\quad${}^4_2{\rm He}^{++}$& {\tt\ \ 7294.2995}  &\qquad $\pi+4$  
   & {\tt\ \ 7287.74349}
   & 0.01\%\cr
\+\quad${}^9_4{\rm Be}^{+4}$& {\tt\ 16424.2099}  &\qquad $\pi+10$ 
   & {\tt\ 16364.47397}
   &0.18\%\cr
\+\quad${}^{12}_{\ 6}{\rm C}^{+6}$&{\tt\ 21868.6918} &\qquad $5\pi$
   &{\tt\ 21845.9892}
   &0.05\%\cr
\+\quad${}^{63}_{29}{\rm Cu}^{+29}$&{\tt 114684.6335} &\qquad $9\pi+8$
   &{\tt114237.3148}   &0.19\%\cr
\+\quad${}^{120}_{\ 50}{\rm Sn}^{+6}$&{\tt 218518.1598} &\qquad $14\pi+4$
   &{\tt218491.3263}   &0.05\%\cr}
\medskip\noindent
However, for our example, we will only be interested in computing the
integral of the area between the curve and the $x$-axis between the
roots $\alpha_1 = -4\pi$ and $\alpha_2 = -4\pi + 1/\pi$.  This will involve
an application of Simpson's rule.  This example was chosen because it involves
a polynomial (of degree three) which does not have integer coefficients.
Indeed, the coefficients are not even algebraic numbers---they are
transcendental numbers.  We will observe how nicely our integration formula
functions in this case.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%  Cubic graph
%
$$\beginpicture
  \setcoordinatesystem units <10cm,10000pt>
  \setplotarea x from -13.1 to -12.0, y from -0.015 to  0.015
  \plotheading {Graph of $f(x) =
         (x+4\pi)\cdot(x+4\pi-1/\pi)\cdot(x+4\pi-2/\pi)$}
  \axis bottom shiftedto y=0   % label {F{\sevenrm IGURE} 6}
         ticks numbered from -13 to -12 by 1 /
  \axis left ticks in numbered from -0.015 to 0.015 by 0.005 /
%  \savelinesandcurves on "chap1b.t01"
\put {F{\sevenrm IGURE} 6} at -12.5 -0.015
\ifexpressmode
  \put {\bf EXPRESSMODE} at -12.5 0.01
\else
  \setquadratic
  \inboundscheckon 
  \plot
 -12.65310  -0.02541
 -12.60973  -0.01066 
 -12.56637   0.00000 /
  \inboundscheckoff
  \plot
 -12.56637   0.00000
 -12.53454   0.00552
 -12.50271   0.00929
 -12.47088   0.01151
 -12.43905   0.01238
 -12.40722   0.01209
 -12.37538   0.01084
 -12.34355   0.00880
 -12.31172   0.00619
 -12.27989   0.00319
 -12.24806   0.00000 
 -12.22325  -0.00250
 -12.19845  -0.00490
 -12.17364  -0.00713
 -12.14884  -0.00908
 -12.12403  -0.01066
 -12.09922  -0.01178
 -12.07442  -0.01236
 -12.04961  -0.01229
 -12.02481  -0.01149
 -12.00000  -0.00987 /
\fi
  \putrule from -12.53454 0.0 to -12.53454  0.00552  
  \putrule from -12.50271 0.0 to -12.50271  0.00929
  \putrule from -12.47088 0.0 to -12.47088  0.01151
  \putrule from -12.43905 0.0 to -12.43905  0.01238
  \putrule from -12.40722 0.0 to -12.40722  0.01209
  \putrule from -12.37538 0.0 to -12.37538  0.01084
  \putrule from -12.34355 0.0 to -12.34355  0.00880
  \putrule from -12.31172 0.0 to -12.31172  0.00619
  \putrule from -12.27989 0.0 to -12.27989  0.00319
  \putrule from -12.24806 0.0 to -12.24806  0.00000 
%\ifexpressmode
%  \put {\bf EXPRESSMODE} at -12.5 0.01
%\else
%  \replot "chap1b.t01"
\endpicture$$
%\centerline{F{\sevenrm IGURE} 6}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\medskip\noindent
The value of the integral is calculated to be approximately
{\tt 0.0025664955636711}.  This example points out one of the strong
points of using Simpson's rule.  When the answer is computed, it is
in numeric form.  If one obtained a so-called ``closed-form'' 
solution, then it
would be necessary to plug in the values for $\pi$ to get a numeric
result.
\medskip
\noindent  Following this same argument, let's consider integrating
the same function, $f(x) = (x+4\pi)\cdot(x+4\pi-1/\pi)\cdot(x+4\pi-2/\pi)$,
over the (closed) interval $\big[-4\pi+2/\pi,\,0\big]$.  Because of the
variation of the range of $f$, we will use a logarithmic scale on the
$y$-axis.  The value ``under the curve'' is approximately
{\tt 5618.52714450635}.  We do actually have this much accuracy because
of the very precise value in our ``C'' program for the constant $\pi$.
Recall the expression
\medskip
{\tt\obeylines\parindent=0pt\ttraggedright
\#define pi 3.141592653589793238462643383279     /* Accurate value for pi. */
}
\medskip\noindent
The only changes necessary to the ``C'' program were
\medskip
{\tt\obeylines\parindent=0pt\ttraggedright
a = (double) -4.0*pi+2.0/pi;
printf("{\char92}nLeft end point    = \%.16lf",a);
b = (double) 0.0;
printf("{\char92}nRight end point   = \%.16lf",b);
}
\medskip\noindent
And, now the graph.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Figure 7.
%
$$\beginpicture
  \setcoordinatesystem units <0.5cm,10cm>
  \setplotarea x from -13 to 2, y from 0 to .60206 %
  \plotheading {\lines{ The integral of\cr 
    $f(x)=(x+4\pi)\cdot(x+4\pi-1/\pi)\cdot(x+4\pi-2/\pi)$\cr
    over the interval $\big[-4\pi+2/\pi,\,0\big]$\cr} } %
  \axis bottom shiftedto y=0 ticks 
     withvalues {$\scriptstyle -4\pi+8/\pi$}  {$-8$} {$-6$} 
     {$-4$} {$-2$} {$0$} {$2$} / %
     at -10.01989 -8 -6 -4 -2 0 2 / / %
  \axis left label {\stack {P,o,w,e,r,s, ,of, ,$10$} } %
     ticks andacross logged numbered at 1 2 3 4 / %
     unlabeled length <0pt> at 1.25 1.5 1.75 2.25 2.5 2.75 3.5 / %
     from 1.5 to 3.5 by .5 from 1.25 to 2.75 by .50 / %
  \plot
    -10.01989  0.01489  -9.51890   0.11466  -9.01790   0.18282
    -8.51691   0.23359  -8.01591   0.27351  -7.51492   0.30610
    -7.01392   0.33343  -6.51293   0.35685  -6.01193   0.37724
    -5.51094   0.39523  -5.00995   0.41129  -4.50895   0.42575
    -4.00796   0.43888  -3.50696   0.45088  -3.00597   0.46192
    -2.50497   0.47212  -2.00398   0.48159  -1.50298   0.49041
    -1.00199   0.49868  -0.50099   0.50643   0.00000   0.51374 /
  \linethickness=.8pt
  \putrule from -10.01989 0.0 to -10.01989 0.01489
  \putrule from -9.51890  0.0 to -9.51890  0.11466
  \putrule from -9.01790  0.0 to -9.01790  0.18282
  \putrule from -8.51691  0.0 to -8.51691  0.23359
  \putrule from -8.01591  0.0 to -8.01591  0.27351
  \putrule from -7.51492  0.0 to -7.51492  0.30610
  \putrule from -7.01392  0.0 to -7.01392  0.33343
  \putrule from -6.51293  0.0 to -6.51293  0.35685
  \putrule from -6.01193  0.0 to -6.01193  0.37724
  \putrule from -5.51094  0.0 to -5.51094  0.39523
  \putrule from -5.00995  0.0 to -5.00995  0.41129
  \putrule from -4.50895  0.0 to -4.50895  0.42575
  \putrule from -4.00796  0.0 to -4.00796  0.43888
  \putrule from -3.50696  0.0 to -3.50696  0.45088
  \putrule from -3.00597  0.0 to -3.00597  0.46192
  \putrule from -2.50497  0.0 to -2.50497  0.47212
  \putrule from -2.00398  0.0 to -2.00398  0.48159
  \putrule from -1.50298  0.0 to -1.50298  0.49041
  \putrule from -1.00199  0.0 to -1.00199  0.49868
  \putrule from -0.50099  0.0 to -0.50099  0.50643
  \putrule from  0.00000  0.0 to  0.00000  0.51374
  \put {$\scriptstyle\bullet$} at 0 0.51374
  \put {$\longleftarrow (0,\,1836.15)$} [l] <2pt,0pt> at 0 0.51374
  \put {$\mid$} at -11.92975 0.0  %
  \put {$\uparrow\atop -4\pi+2/\pi$} [t] <0pt,-10pt> at -11.92975 0.0 %
\endpicture$$
\centerline{F{\sevenrm IGURE} 7}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\medskip\noindent
We might also consider some other interesting values for the
function $f(x)$, as defined above, for ratios of subatomic particles'
masses to the mass of the electron.
%$$f(x)=(x+4\pi)(x+4\pi-1/\pi)(x+4\pi-2/\pi)\eqno(\dagger)$$
{\settabs5\columns\thinmuskip=2mu
\+Subatomic   
  &\underbar{\raise2pt\hbox{\qquad mass\qquad}}   &   &  &relative \cr
\+Particle   & electron mass &\qquad $x$      &\qquad $f(x)$ &deviation\cr
\baselineskip=14pt
\+\quad$n$ &{\tt\ 1838.683662}  &\qquad $1/64\pi$           
   &{\tt\ 1838.39048447}
   &0.008\%\cr
\+\quad$\tau^{-}$  &{\tt\ 3491.4} &\qquad $3-1/4\pi$         
    &{\tt\ 3488.5}
    & 0.039\%\cr
\+\quad$\eta^{0}$  &{\tt\ 1073.97}  &\qquad $-2-1/64\pi$      
    &{\tt\ 1073.7}
    & 0.014\%\cr
\+\quad$\eta'$ &{\tt\ 1873.78}  &\qquad $1/4\pi$             
    &{\tt\ 1872.2}
    & 0.069\%\cr
\+\quad$\Sigma^{+}$ &{\tt\ 2327.5} &\qquad $1+1/64\pi$      
    &{\tt\ 2326.8}
    &0.0064\%\cr
\+\quad$\Sigma^{0}$ &{\tt\ 2333.76} &\qquad $1+1/16\pi$      
    &{\tt\ 2334.3}
    &0.034\%\cr
\+\quad$\Sigma^{-}$ &{\tt\ 2343.31} &\qquad $1+1/8\pi$       
    &{\tt\ 2344.8}
    &0.049\%\cr
\+\quad$\Lambda$ &{\tt\ 2183.23} &\qquad $1-1/\pi$           
    &{\tt\ 2160.26}
    &0.53\%\cr
}
%***Other truly remarkable values not included in the tables because****
%***they either had more than two terms or did not meet the criteria.***
%***PARTICLES***
%
% f(-1-1/\pi-1/16\pi)=496.79 Mev  K^0 497.67 Mev
% f(-2-1/\pi-1/8\pi)=493.803 Mev  K^{\pm} 493.646 Mev
% f(-6+1/\pi-1/4\pi)=139.1436 Mev \pi^{\pm} 139.5675 Mev
% f(-6+1/\pi-1/64\pi)=134.0077 Mev \pi^0 134.9739 Mev
% f(\pi/4-1/4\pi-1/64\pi)= 1115.21 Mev  \Lambda 1115.63 Mev
% f(\pi+1/64\pi)=1863.5536 Mev  D^{0} 1864.5 \pm 0.5 Mev
% f(\pi+1/16\pi)=1868.9790 Mev  D^{\pm} 1869.3 \pm 0.4 Mev
% f(3-1/\pi-1/4\pi)=1672.6886 Mev  \Omega 1672.43 \pm 0.32 Mev
%
% f(8-2\pi) = 1390.93 Mev  \omega 1391 \pm 18 Mev
% f(-9+\pi)=132.9758 Mev  \pi^{0} 134.9739
%
\vfill\eject
%
\noindent{\bf\llap{1.7\quad}Verifying Derivatives.}  Calculus is traditionally
partitioned into differential and integral calculus.  This partition
supposes that the two concepts are independent and mutually exclusive.
Then, almost as a gift from the gods, we encounter the so-called
{\bf Fundamental Theorem of Calculus},\footnote{$^{17}$}{{\bf Theorem.}
{\sl Let $f$ be continuous on $[\alpha,\beta\,]$ and $x\in(\alpha,\beta\,)$.
If $F(x)=\int_a^x f(t)\,dt$, then $F'(x) = f(x)$.}} 
which unifies the two concepts
(more or less---mostly less).  In this era of digital computers and the
future directed towards artificial intelligence, it might be a good idea to
discard the artificial divisions of calculus in favor of a more 
pragmatic\footnote{$^{18}$}{Pragmatic---practical, especially in contrast
to idealistic.} development.
In this section we will examine a good technique for determining if
a function, which is supposed to be the derivative of a given function,
really is.  The idea is simple.  We have a given function and a candidate
for its derivative.  Looking at some interval, we choose four points and
use Simpson's rule to evaluate the integral of the derivative on
subintervals, ending at the four points, and compare the results with
the original function.  This is awkward to verbalize but easy to
write ``in mathematics.''  Let $f(x)$ be a given function and
let $g(x)$ be a candidate for the derivative $f'(x)$ of $f(x)$.
In the interval $[\alpha,\,\beta\,]$, choose four points,
$x_1$, $x_2$, $x_3$, and $x_4$.  (Let $\alpha < x_1 < x_2 < x_3
< x_4 < \beta$.)  We can eliminate $g(x)$ if any of
the four approximations are incorrect:
$$f\big(x_1\big) - f\big(\alpha\big)
\ \approx \int_\alpha^{x_1} g(t)\,dt$$
$$f\big(x_2\big) - f\big(\alpha\big)
\ \approx \int_\alpha^{x_2} g(t)\,dt$$
$$f\big(x_3\big) - f\big(\alpha\big)
\ \approx \int_\alpha^{x_3} g(t)\,dt$$
$$f\big(x_4\big) - f\big(\alpha\big)
\ \approx \int_\alpha^{x_4} g(t)\,dt$$
We want to examine this idea, using a non-trivial example.
One such example comes from the four de\-riv\-a\-tives
of the func\-tion $f(x)$ $=$ $\sqrt{1-k^2\sin^2(x)}$.  Recall it was
the fourth derivative that was used in the error estimate.  This
function is easy to write; however, its derivatives become more
and more difficult to expand algebraically.  How can we be sure
that we did not make some arithmetic error?  It has been shown
that, on the average, 1 out of every 300 human calculations is in
error.  This will be done via a computer program.  The program
evaluates all the integrals simultaneously and outputs the results
in a rectangular array of real numbers---a {\sl matrix}.  By visually
comparing the rows of the matrix, we can tell at once if our
derivatives are correct.  Without any more delay, let's write down
the program and its results and then do an analysis of the coding.
\medskip
{\tt\parindent=0pt\ttraggedright\obeylines
100 DEFDBL A-H, O-Z
102 C2=SIN(EXP(1.)/2.)*SIN(EXP(1.)/2.)
105 DEF FNA(x AS DOUBLE)= SQR(1.-C2*SIN(x){\char94}2)
106 DEF FNB(x AS DOUBLE)= -.5*C2*SIN(2.*x)/FNA(x)
107 DEF FNC(x AS DOUBLE)= -C2*COS(2.*x)/FNA(x)
\qquad\qquad -.25* C2*C2*SIN(2.*x){\char94}2/FNA(x){\char94}3
108 DEF FND(x AS DOUBLE)= 2.*C2*SIN(2.*x)/FNA(x)
\qquad\qquad -.75*C2*C2*SIN(4.*x)/FNA(x){\char94}3
\qquad\qquad -.375*C2{\char94}3*SIN(2.*x){\char94}3/FNA(x){\char94}5
109 DEF FNE(x AS DOUBLE)= 4.*C2*COS(2.*x)/FNA(x) 
\qquad\qquad + C2{\char94}2*SIN(2.*x){\char94}2/FNA(x){\char94}3
\qquad\qquad -3.*C2{\char94}2*COS(4.*x)/FNA(x){\char94}3 
\qquad\qquad -1.125*C2{\char94}3*SIN(4.*x)*SIN(2.*x)/FNA(x){\char94}5 
\qquad\qquad -(18./8.)*C2{\char94}3*SIN(2.*x){\char94}2*COS(2.*x)/FNA(x){\char94}5 
\qquad\qquad -(15./16.)*C2{\char94}4*SIN(2.*x){\char94}4/FNA(x){\char94}7
110 A = 0.
112 INPUT "Left endpoint = "; B
114 IF (A>B) OR (B>3.141592653589793/2.) 
\qquad\qquad THEN PRINT "Input out of range": GOTO 112
116 OPEN "92\_12\_14.txt" FOR APPEND AS \#1
120 REM Left endpoint <= 3.141592653589793\# / 2!
130 SUM1 = 0.: SUM2 = 0.: SUM3 = 0.: SUM4 = 0.
150 N = 512: H = (B - A) / N
160 FOR I = 1 TO N STEP 2: REM The "FOR" loop is done n/2 times.
170 SUM1 = SUM1 + FNB(A + (I - 1) * H)
172 SUM2 = SUM2 + FNC(A + (I - 1) * H)
174 SUM3 = SUM3 + FND(A + (I - 1) * H)
176 SUM4 = SUM4 + FNE(A + (I - 1) * H)
180 SUM1 = SUM1 + 4.*FNB(A + I * H)
182 SUM2 = SUM2 + 4.*FNC(A + I * H)
184 SUM3 = SUM3 + 4.*FND(A + I * H)
186 SUM4 = SUM4 + 4.*FNE(A + I * H)
190 SUM1 = SUM1 + FNB(A + (I + 1) * H)
192 SUM2 = SUM2 + FNC(A + (I + 1) * H)
194 SUM3 = SUM3 + FND(A + (I + 1) * H)
196 SUM4 = SUM4 + FNE(A + (I + 1) * H)
200 NEXT I
210 L\$ = "\#\#.\#\#\#\#\#  \#\#.\#\#\#\#\#  \#\#.\#\#\#\#\#  \#\#.\#\#\#\#\#  \#\#.\#\#\#\#\#"
300 PRINT USING L\$; FNA(B)-FNA(A); FNB(B)-FNB(A);
\qquad\qquad FNC(B)-FNC(A); FND(B)-FND(A)
310 PRINT USING L\$; SUM1*H/3.; SUM2*H/3.; 
\qquad\qquad SUM3*H/3.; SUM4*H/3.
320 PRINT \#1, USING L\$; FNA(B)-FNA(A); FNB(B)-FNB(A); 
\qquad\qquad FNC(B)-FNC(A); FND(B)-FND(A)
330 PRINT \#1, USING L\$; SUM1*H/3.; SUM2*H/3.; 
\qquad\qquad SUM3*H/3.; SUM4*H/3.
340 CLOSE \#1
350 STOP
360 END
}
\medskip\noindent
Of course, we want to look at the output.  For input values, we used
$.25$, $.50$, $1.00$, and $1.50$.  Rows 1 and 2 of this 4 by 8 matrix
correspond to $x=.25$, rows 3 and 4 correspond to $x=.5$, and so on.
\medskip
{\tt\settabs 4\columns
\+\quad -0.02969  &\quad-0.23615   &\quad0.03387   &\quad\ 0.27142  \cr
\+\quad -0.02969  &\quad-0.23615   &\quad0.03387   &\quad\ 0.27142  \cr
\+\quad -0.11666  &\quad-0.45528   &\quad0.13655   &\quad\ 0.55428  \cr
\+\quad -0.11666  &\quad-0.45528   &\quad0.13655   &\quad\ 0.55428  \cr
\+\quad -0.43151  &\quad-0.76446   &\quad0.62760   &\quad\ 1.73354  \cr
\+\quad -0.43151  &\quad-0.76446   &\quad0.62760   &\quad\ 1.73354  \cr
\+\quad -0.77883  &\quad-0.30495   &\quad4.81402   &\quad17.17882  \cr
\+\quad -0.77883  &\quad-0.30495   &\quad4.81402   &\quad17.17882  \cr}
\medskip\noindent  All this goes to show that the equations we
coded into the computer program are {\sl probably\/} correct.  This is
not rigorous; however, in our days and times anything that fits this
well would be a good approximation---that is, it would serve as
the derivative {\sl for all practical purposes}.  
A word of caution:  there may be
more than one way to express a given equation.  For instance,
$y=x^2+2x+1$ and $y = (x+1)^2$ are {\sl algebraically\/} the same.
Even though two expressions are algebraically they same, they may
not yield the same value from a computer.  This is due to several
factors: (1) computers may evaluate the equations differently; (2) there
may be more truncation (round-off) error with one equation than with
the other (after all, computers don't have infinite accuracy); (3)
a coding error may affect one equation
(especially doing an integer division where
the fractional part is lost); and, (4) the grouping of successive
multiplications and divisions may affect the answer, especially when
two nearly equal numbers are being subtracted.
\medskip\noindent
There are a few items worth mentioning about the computer program itself.
We ``captured'' or ``saved'' the output by {\sl appending\/}
to a file (in this case named {\tt 92\_12\_14.txt}.  Later, we edited the
file and merged it into this document.  This gave two distinct advantages:
(1) it was quicker and (2) there was no possibility of a human blunder in
re-typing the numbers.  This is always a good technique.  Capturing output
data and inputing the file directly into text ensures a correct copy.  It may
not always possible to do this.  A second alternative is to compare the 
final text against the input via some compare routine.  In Appendix B, a
method of comparing text with output will be devised, developed, and
discussed.  In BASIC\footnote{$^{19}$}{BASIC---Beginners' All-purpose Symbolic
Instruction Code.  This elementary computer code is held in disdain by
many programmers who prefer ``C.''} the coding line is 256 characters long,
more or less (usually less, maybe 254 characters).  In a text document, it
is a good idea to have lines of 80 characters or less.  To accommodate the
document, it is necessary to ``break'' long lines into shorter segments.
Whenever this is done, the line is indented.  The sequence numbers are
retained to aid in determining when a line has been broken.  Some versions
of BASIC write floating point integers using an exclamation point, {\it e.g.},
$3.00$ becomes $3!$.  All floating point integers have been written in
the standard mathematical format using a decimal point.
QBasic\footnote{$^{20}$}{QBasic---the version of BASIC copyrighted by the
Microsoft Corporation.} puts in spaces between the mathematical operators.
While the spaces add to the readability of the code and aid the programmer,
they do little for the documentation, which has to fit on an $8{1\over2}
\times 11$ inch page.  The extra spaces were removed.  For the ``\TeX-nicians,''
they may note that some symbols cause the typesetting program to suffer.
Before merging into a \TeX\ file, the special symbols backslash 
({\tt\char92}) and hat (or circumflex) ({\tt\char94}) were replaced by
their character equivalences.  The pound sign, dollar sign, and underscore
(\#, \$, and \_) were dealt with in the usual manner, by prefixing them
with backslashes ({\tt\char92}).
\medskip\noindent
As usual, we will complement the BASIC program with a ``C'' program.  The
``C'' program also saves its output to a file.
{\tt\obeylines\parindent=0pt\ttraggedright
\#include <stdio.h>        /* Header for input/output subroutines. */
\#include <math.h>         /* Header for math subroutines. */
\#include <float.h>        /* Header for floating point subroutines. */

\#define pi 3.141592653589793238462643383279     /* Accurate value for pi. */
\#define k2 0.9558669573934826

/* Simpson's rule for approximating integrals.
\qquad a:              left endpoint
\qquad b:              right endpoint
\qquad fa:             original function
\qquad fb,fc,fd,fe:    pointers to functions to integrate
\qquad n:              number of subintervals
*/
double fa (double x);
double fb (double x);
double fc (double x);
double fd (double x);
double fe (double x);

main()
\char123
double a,b,h,sum,x,y;     /* In 'C' all variables must be assigned */
double p1, p2, p3;
int i, n;
FILE  *fp;
fp = fopen("92\_12\_13.txt", "a");
printf("{\char92}007");           /* Sound bell in computer. */
a = (double) 0.0;
printf("{\char92}nLeft end point    a = \%.16lf",a);
printf("{\char92}nEnter right end point ");
scanf("\%lf",\&b);
printf("{\char92}nRight end point   b = \%.16lf",b);
n = 512;
printf("{\char92}nNumber of subintervals \%d",n);
h = (double) (b-a)/n;
for (i=1, sum=0.0; i<=n; i = i+ 2)\char123
\quad p1 = fb((double) a+(i-1)*h);
\quad p2 = fb((double) a+i*h);
\quad p3 = fb((double) a+(i+1)*h);
\quad sum += p1 + 4.0 * p2 + p3;
\char125

y = (double) h*sum/3.0;
printf("{\char92}nValue of integral   = \%.16lf", (double) y);
printf("{\char92}nValue of f(b)-f(a)  = \%.16lf", (double) fa(b)-fa(a) );
fprintf( fp, "  \%.5lf \%.5lf", y, fa(b)-fa(a) );

for (i=1, sum=0.0; i<=n; i = i+ 2)\char123
\quad p1 = fc((double) a+(i-1)*h);
\quad p2 = fc((double) a+i*h);
\quad p3 = fc((double) a+(i+1)*h);
\quad sum += p1 + 4.0 * p2 + p3;
\char125

y = (double) h*sum/3.0;
printf("{\char92}nValue of integral   = \%.16lf", (double) y);
printf("{\char92}n  f'(b) - f'(a)     = \%.16lf", (double) fb(b)-fb(a) );
fprintf( fp, "  \%.5lf \%.5lf", y, fb(b)-fb(a) );

for (i=1, sum=0.0; i<=n; i = i+ 2)\char123
\quad p1 = fd((double) a+(i-1)*h);
\quad p2 = fd((double) a+i*h);
\quad p3 = fd((double) a+(i+1)*h);
\quad sum += p1 + 4.0 * p2 + p3;
\char125

y = (double) h*sum/3.0;
printf("{\char92}nValue of integral   = \%.16lf", (double) y);
printf("{\char92}n  f''(b) - f''(a)   = \%.16lf", (double) fc(b)-fc(a) );
fprintf( fp, "  \%.5lf \%.5lf", y, fc(b)-fc(a) );

for (i=1, sum=0.0; i<=n; i = i+ 2)\char123
\quad p1 = fe((double) a+(i-1)*h);
\quad p2 = fe((double) a+i*h);
\quad p3 = fe((double) a+(i+1)*h);
\quad sum += p1 + 4.0 * p2 + p3;
\char125

y = (double) h*sum/3.0;
printf("{\char92}nValue of integral   = \%.16lf", (double) y);
printf("{\char92}n f'''(b)-f'''(a)    = \%.16lf", (double) fd(b)-fd(a) );
fprintf( fp, "  \%.5lf \%.5lf {\char92}n", y, fd(b)-fd(a) );

fclose(fp);

return(0);
\char125

double fa (double x)
\char123
\qquad double y;
\qquad y = (double) sqrt(1.0 -k2*sin(x)*sin(x));
\qquad return (y);
\char125

double fb (double x)
\char123
\qquad double y;
\qquad y = (double) -0.5*k2*sin(2.0*x)/fa(x);
\qquad return (y);
\char125

double fc (double x)
\char123
\qquad double y;
\qquad y = (double) -k2*cos(2.0*x)/fa(x)
\qquad\quad -.25*k2*k2*sin(2.0*x)*sin(2.0*x)/pow(fa(x),3.0);
\qquad return (y);
\char125

double fd (double x)
\char123
\qquad double y;
\qquad y = (double) 2.0*k2*sin(2.0*x)/fa(x)
\qquad\quad -0.75*k2*k2*sin(4.0*x)/pow(fa(x),3.0)
\qquad\quad -.375*k2*k2*k2*pow(sin(2.0*x),3.0)/pow(fa(x),5.0);
\qquad return (y);
\char125

double fe (double x)
\char123
\qquad double y;
\qquad y = (double) 4.0*k2*cos(2.0*x)/fa(x)
\qquad\quad +k2*k2*sin(2.0*x)*sin(2.0*x)/pow(fa(x),3.0)
\qquad\quad -3.0*k2*k2*cos(4.0*x)/pow(fa(x),3.0)
\qquad\quad -1.125*k2*k2*k2*sin(4.0*x)*sin(2.0*x)/pow(fa(x),5.0)
\qquad\quad -2.25*k2*k2*k2*sin(2.0*x)*sin(2.0*x)*cos(2.0*x)/pow(fa(x),5.0)
\qquad\quad -0.9375*k2*k2*k2*k2*pow(sin(2.0*x),4.0)/pow(fa(x),7.0);
\qquad return (y);
\char125

/* End of file */
}
\medskip\noindent
Of course, there is an output file to be examined also.  Here we included
values of $1.51$, $1.52$, and $1.55$.
\medskip
{\tt\settabs 8\columns
\+ -0.02969&-0.02969& -0.23615&-0.23615& 0.03387&0.03387&\ 0.27142&\ 0.27142\cr
\+ -0.11666&-0.11666& -0.45528&-0.45528& 0.13655&0.13655&\ 0.55428&\ 0.55428\cr 
\+ -0.43151&-0.43151& -0.76446&-0.76446& 0.62760&0.62760&\ 1.73354&\ 1.73354\cr 
\+ -0.77883&-0.77883& -0.30495&-0.30495& 4.81402&4.81402& 17.17882& 17.17882\cr
\+ -0.78168&-0.78168& -0.26553&-0.26553& 4.97895&4.97895& 15.74172& 15.74172\cr 
\+ -0.78414&-0.78414& -0.22454&-0.22454& 5.12756&5.12756& 13.91650& 13.91650\cr 
\+ -0.78894&-0.78894& -0.09416&-0.09416& 5.43883&5.43883&\ 6.37637&\ 6.37637\cr
} 
\vfill\eject
\noindent{\bf\llap{1.8\quad}Generalizations.}  Mathematicians love to
take a popular, useful concept and generalize it.  Sometimes much can
be learned by generalizing and by abstracting; more often than not,
a generalization results in a more complicated, theoretical, and generally
useless body of knowledge that exists solely as a requirement for a degree.
This is especially true with Simpson's rule.  Simpson's rule is popular and
efficient.  Its error calculation is straightforward and its convergence
is assured.  There are some examples where Simpson's rule does poorly---but
these are mostly ``pathological'' in nature.  (They are artificially
constructed simply to demonstrate the fallability of the rule.)  There are
also so-called improper integrals.  The improper integrals are either
defined over an interval such as $[\alpha,\,+\infty]$, $[-\infty,\,\beta\,]$,
or $[-\infty,\,+\infty]$, or the function is unbounded in the interval
of integration (or both!)  The improper integrals have to be approached with
common sense.  Take for example the integral
$$\int_0^1 {dx\over \sqrt{x}}\ =\ 2\sqrt{x}\big|_{x=1}
   \ -\ \lim_{x\to0} 2\sqrt{x}\ =\ 2.$$
How could Simpson's rule be applied to such an example?  The answer is
simple---make the computer do the work!  Choose a small number
$\epsilon$, $0 < \epsilon \ll 1$.  By a proper determination of the
number of subintervals and a sufficiently small value of $\epsilon$, the
desired answer can be obtained with the required accuracy.  
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
$$\beginpicture
  \setcoordinatesystem units <1in,.125in>
  \setplotarea x from -2 to 2, y from -0.25 to 18
  \axis bottom shiftedto y=0 label {F{\sevenrm IGURE} 8} / %
  \axis left shiftedto x=0 / %
  \putrule from 0.0625 0.0 to 0.0625 16.0
\ifexpressmode
  \put {\bf EXPRESSMODE} at 0 10.0
\else
  \setquadratic
  \plot 0.0625 16.  0.125 8.  0.25 4.  .375 2.66667  .5 2.  .75 1.3333
        1.0 1.0    1.25 0.8    1.5 0.66667   1.75  0.57142    2.0 0.5 / %
\fi
\endpicture$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In the case of integrals of the form
$${1\over\sqrt{2\pi}}\,\int_0^{\infty} e^{-x^2/2}\,dx$$
(the classic normal curve), one needs only observe that the
integral may be replaced by an integral
$${1\over\sqrt{2\pi}}\,\int_0^{N} e^{-x^2/2}\,dx$$
for a suitable chosen integer $N$.  We note here that $N=6$ generally
suffices for most engineering work.  This is called the 
six-sigma ($6$-$\sigma$).  We can compute this estimate by considering
$$\int_1^N e^{-x^2/2}\,dx \le \int_1^N e^{-x/2}\,dx.$$
If $x\ge1$, then $x^2 \ge x$ and $e^{-x^2/2} \le e^{-x/2}$.
$$\int_N^\infty e^{-x/2}\,dx = 2e^{-N/2}.$$
\centerline{%
 \beginpicture %
   \setcoordinatesystem units <.5in,2.5in> %
   \setplotarea x from -3 to 3, y from 0 to .4 %
   \plotheading {\lines {%
     The density $\varphi(\zeta) = e^{-\zeta^2\!/2}/\sqrt{2\pi}$ of the\cr %
     standard normal distribution.\cr}} %
   \axis bottom ticks numbered from -3 to 3 by 1 %
     length <0pt> withvalues $\zeta$ / at 1.5 /  / %
   \linethickness=.25pt %
   \putrule from  1.5 0  to  1.5 .12952 % (.12952 = density at 1.5)
   \setbox0 = \hbox{$swarrow$}%
   \put {$\swarrow$ \raise6pt\hbox{$\varphi(\zeta)$}} %
     [bl] at 1.5 .12952 %
\ifexpressmode
  \put {\bf EXPRESSMODE} at 0 0.2
\else
   \setquadratic \plot
     0.0      .39894
     0.16667  .39344  0.33333 .37738   0.5  .35207   0.66667  .31945 
     0.83333  .28191  1.      .24197   1.25 .18265   1.5      .12952 
     1.75     .08628  2.      .05399   2.25 .03174   2.5      .01753 
     2.75     .00909  3.0     .00443 /                              
   \setquadratic \plot
     0.0      .39894
    -0.16667  .39344 -0.33333 .37738  -0.5  .35207  -0.66667  .31945 
    -0.83333  .28191 -1.      .24197  -1.25 .18265  -1.5      .12952 
    -1.75     .08628 -2.      .05399  -2.25 .03174  -2.5      .01753 
    -2.75     .00909 -3.0     .00443 /                              
% \setshadegrid span <.025in>
% \vshade  0 0 .39894   0.5 0 .35207   1 0 .24197
%          1.5 0 .12952  2 0 .05399  / %
\fi
\endpicture } %
%\smallskip
\centerline{F{\sevenrm IGURE} 9}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\medskip\noindent
Having mentioned how to generalize this rule to improper integrals, it is time
to examine the position of this integration scheme in mathematicians'
grand scheme of things.
Simpson's rule is the most popular of the so-called {\bf Newton-Cotes
integration formulas}.  The first three of which are given below
$$\eqalign{\int_{x_0}^{x_0+h} f(x)\,dx
\ &=\ {h\over2}\big(y_0+y_1\big) - {h^3\over12}f''(\xi)\cr
\int_{x_0}^{x_0+2h} f(x)\,dx
\ &=\ {h\over3}\big(y_0+4y_1+y_2\big) - {h^5\over90}f^{(4)}(\xi)\cr
\int_{x_0}^{x_0+3h} f(x)\,dx
\ &=\ {3h\over8}\big(y_0+3y_1+3y_2+y_3\big)
   -{3h^5\over90}f^{(4)}(\xi),\cr}$$
where $\xi$ is an intermediate value of $x$.  The Newton-Cotes
formulas, such as those above, are called {\sl closed} because
the interval end-points are used.  If the end-points are not used,
but only the interior points in the interval, the formula is called
open.  The Newton-Cotes formulas belong to a class of formulas known as
polynomial approximations; the class of  polynomial approximations is
contained in a class of formulas of approximation of integrals by
families of analytic functions, and so on.
\medskip
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Figure 10 --- Exponential curve
%
$$\beginpicture
 \setcoordinatesystem units <1in,1in>
 \setplotarea x from 0 to 3, y from 0 to 1
 \normalgraphs
 \axis bottom ticks
   numbered from 0 to 3 by 1
   length <0pt> withvalues $x$ / at .5 /  / 
 \linethickness=.25pt  \putrule from  .5 0  to .5 .60653
 \putrule from  1 0  to 1 .36788  \putrule from 2 0  to 2 .13534
 \put {$\scriptstyle\bullet$} at  .5 .60653
 \put {$e^{-x}$} [rt] <-4pt,-4pt> at .5 .60653
 \put {$e^{-x^2}$} [lb] <4pt,4pt> at .5 .77880
\ifexpressmode
 \put {\bf EXPRESSMODE} at 1.5 0.5
\else
 \setquadratic
 \plot  0 1   .25  .77880   .50  .60653   .75  .47237  1.00  .36788
             1.25  .28650  1.50  .22313  1.75  .17377  2.00  .13534
             2.25  .10540  2.50  .08208  2.75  .06393  3.00  .04979 / %
 \setshadegrid span <.025in>
 \vshade  1 0 .36788   1.5 0 .10540   2 0 .01832 / %
 \linethickness=0.75pt
 \setquadratic
 \plot 
    0.00   1.00000     0.25   0.93941      0.50   0.77880
    0.75   0.56978     1.00   0.36788      1.25   0.20961
    1.50   0.10540     1.75   0.04677      2.00   0.01832
    2.25   0.00633     2.50   0.00193      2.75   0.00052
    3.00   0.00012  / %
\fi
\endpicture $$
\centerline {F{\sevenrm IGURE} 10}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vfill\eject
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\headline={\tenrm\hfill Newton's Method}
\centerline{\bf Chapter 2}
\bigskip
\noindent{\bf\llap{2.0\quad}Introduction.} 
One of the most exciting applications of differential calculus is the
use of Newton's method for the determination of roots of functions.
Most students of algebra think that the only roots of interest
are those of polynomials.  An algebra student who has done all the
assigned homework might
think that it's just a matter of factoring; after all, it was so
clever just completing the square in the quadratic equation.  
(Wasn't it fun being so smart and showing that off?)
That's simply not the case.
In fact, there is no purely algebraic method for finding roots of
polynomials of degree five or higher.  The situation with the trigonometric
functions is even more involved.  What technique would one use to find
the root or roots of $x-\cos x = 0$?
$$\beginpicture
  \setcoordinatesystem units <1in,1in>
  \setplotarea x from -.5 to 2.25, y from -.5 to 3.5
  \plotheading {\lines {Tangent line at $(x_0,y_0)$\cr
                        slope $=$ $f'(x_0)$\cr}}
  \axis bottom shiftedto y=0 ticks unlabeled  short quantity 12  / %
  \axis left shiftedto x=0 ticks unlabeled quantity 9  / %
\ifexpressmode
  \put {\bf EXPRESSMODE} at 0 1
\else
  \setquadratic
  \plot
     -0.25 -0.2344  -0.15 -0.2844  -0.05 -0.3094   0.05 -0.3094   
      0.15 -0.2844   0.25 -0.2344   0.35 -0.1594   0.45 -0.0594   
      0.55  0.0656   0.65  0.2156   0.75  0.3906   0.85  0.5906
      0.95  0.8156   1.05  1.0656   1.15  1.3406   1.25  1.6406   
      1.35  1.9656   1.45  2.3156   1.55  2.6906   1.65  3.0906   
      1.75  3.5156    / %   
  \setlinear
  \plot  2.0 3.4375   0.5 -.3125 /  %
\fi  
  \put {$\scriptstyle\bullet$} at 1.0 0.9375
  \put {$\scriptstyle\bullet$} at .625 0.0
  \put {$(x_0,y_0)$} at 1.25 .9375
  \put {$\nwarrow$\lower6pt\hbox{$x$-intercept}} 
        [lt] <0.5pt, -0.5pt> at .62 0 %
  \put {$f(x)$\lower6pt\hbox{$\searrow$}} 
        [rb]  at 1.45  2.3156
 \put {$\nwarrow$\lower6pt\hbox{tangent line}} [lt] at 1.46 2.1
\endpicture$$
\centerline{F{\sevenrm IGURE} 11}
% f(x) = y = (15/12)*(x-1/2)*(x+1/2)
% f'(x) = y' = (15/6)*x
% when x = 1, y(1) = f(1) = 15/16 = .9375
% when x = 1, y'(1) = f'(1) = 15/6
% when x = 2, the point on the tangent line is 3.4375
% when y = 0, the abscissa on the tangent line is 5/8
%
\smallskip\noindent
All of the important details of Newton's method are found in
the above figure.  The equation of the tangent line is
determined by the so-called point-slope method from elementary algebra:
$${y-y_0 \over x-x_0}\ =\ \hbox{slope } 
=\ {\Delta y\over\Delta x}\ =\ {\hbox{rise}\over\hbox{run}}
\ =\ f'(x_0).$$
Set $y=0$ and determine the $x$-intercept of the tangent line,
$\big(x_0-y_0/f'\big(x_0),0\big)$.  From the
above figure, we see that the point at which the tangent line crosses
the $x$-axis is moving quickly towards the root---the point at which
the curve $f(x)$ crosses the $x$-axis.  If we only had a better
first approximation $x_0$, then the $x_0-y_0/f'(x_0)$ would be even
closer to the root.  Why not just try again and define a new approximation?
$$x_1 = x_0 - {f(x_0)\over f'(x_0)}\eqno(11)$$
\vfill\eject
\noindent{\bf\llap{2.1\quad}Theory.}  This paragraph is concerned with the
theory behind Newton's method.  The user who is interested only in
results may skip this entire section.  This paragraph employs the so-called
Taylor series (also known as the Taylor's series)
$$f(x_0) + f'(x_0)(x-x_0) + {f''(x_0)\over2}(x-x_0)^2
  + \ldots + {f^{(n)}\over n!}(x-x_0)^n + \ldots.\eqno(12)$$
An alternate derivation using the Mean-Value Theorem will be presented
in paragraph {\bf 2.2}.  If one looks in a 
dictionary,\footnote{$^{21}$}{{\it Webster's Ninth New Collegiate
Dictionary}, (Springfield, Massachusetts: Merriam-Webster, Inc., 1984),
page 1209.} equation (12) might be written in the form
$$f(x) = f(a) + {f^{[1]}\over 1!}(x-a)
  +{f^{[2]}\over 2!}(x-a)^2 + \ldots + 
  {f^{n]}\over n!}(x-a)^n + \ldots, \eqno(12')$$
where $f^{[n]}(a)$ is the derivative of the 
$n${\raise2pt\hbox{\sevenrm th}} order of $f(x)$ evaluated at $a$.
Before delving into the mathematics, 
a digression\footnote{$^{22}$}{Digression---A swerving away from the
main subject or the {\it Leitmotif\/}; a turning aside from the main
argument.} might be in order.  Whenever a textbook says ``degree,''
it means the highest exponent or power.  For example: $x^2$ is of
second degree; the polynomial $x^3-x^2+1$ is of third degree;
and, $x^2+y^2 = 1$ is a second degree relation.  Whenever a textbook
says ``order,'' it generally refers to a more abstract concept.  In
general, $x^n$ means the $n${\raise2pt\hbox{\sevenrm th}} power
(degree $n$) of $x$, that is
$$x^n \ =\ \underbrace{x\cdot x\cdots x}_{n\ %
\hbox{\sevenrm times }}.$$
(Like every rule, this one has an exception.  In tensor analysis, the
$i$, $j$, and $k$ in expressions like $x^i$, $y^{jk}$, {\it etc.},
refer to contravariant indices, not powers.)  On the other hand,
whenever one sees such notations as $x^{(n)}$, $y^{[n]}$, and
so on, a red flag ought to go up.  Indeed, if a function does have
a derivative, it is usually denoted by $f'(x)$, the second derivative
(the derivative of the derivative) is denoted by $f''(x)$.  But,
the third derivative is not generally denoted by $f'''(x)$.  The
third derivative of $f$ at $x$ is usually written $f^{(3)}(x)$ and
sometimes written as $f^{[3]}(x)$.  What does $f^{(0)}(x)$ mean?
Recall that (for all $x\not= 0$) $x^0\equiv1$.  $f^{(0)}(x)
\equiv f(x)$.  The symbol made up of three horizontal lines 
($\equiv$) means more
that just equals ($=$); this symbol, ($\equiv$),
means equivalence and that means equal for all the values under consideration.
\medskip
\noindent  One of mathematics' favorite devices, or tricks, is to
truncate\footnote{$^{23}$}{Truncate---To cut off; to make short by trimming
or cutting off.}
a Taylor series.  In the Newton method, the series is truncated at
the quadratic term, that is, at 
${1\over2}f''\big(x_0\big)\cdot\big(x-x_0\big)^2$.  So, the truncated
series becomes
$$\eqalign{f(x)\ &=\ f(x_0) + {f^{[1]}\over 1!}(x-x_0)^1\cr
  &=\ f\big(x_0\big) + f'\big(x_0\big)\cdot\big(x-x_0\big).\cr}\eqno(13)$$
This is the derivation.  So, after all that was said and done, there's
not much to do to get the result.  Just pretend that $f(x_1)=0$ and get
$$f\big(x_1\big)\ =\ 0\ =\  f\big(x_0\big)
\ +\ f'\big(x_0\big)\big(x_1-x_0\big),$$
or
$$x_1\ =\ x_0\ -\ {f(x_0)\over f'(x_0)},\eqno(14)$$
provided, of course, that $f'(x_0) \not= 0$.  If $f'(x_0) = 0$, then
we just choose another $x_0$, say $\tilde x_0$ where 
$f'\big(\tilde x_0\big) \not= 0$ and proceed.
\vfill\eject
\noindent{\bf\llap{2.2\quad}Applications.}  One of the most important
applications of Newton's method is in the calculation of square roots.  Many 
small hand calculators have a ``hard-wired'' square root generator.  The
circuitry follows a simple algorithm.  Since long division of decimal fractions
may not be supported, we will derive an algorithm which does not involve
division and finds the square root for any number $a > 0$.
\medskip \noindent
It is clear that if we let $h(x) = x^2 -a$ and solve this equation for
the root $h(x_0)=0$, this will be the same as finding the square root
of $a$.  Newton's formula becomes $x_{n+1} = {1\over2}\big(x_{n}
- a/x_{n}\big)$.\footnote{$^{24}$}{If the following mathematics looks
too deep, try Schaum's Outline {\it Numerical Analysis, 2nd Edition},
page 334.  In any case, the two BASIC computer programs ought to be
useful.}
\medskip\noindent
The next thing that we will do is to plug the function $F(x)= a - 1/x$
into Newton's method.  Then we will look at the fixed point formula.
To determine the starting value, we'll require $\xi$ to satisfy the expression
$|f'(\xi)| < 1$.  After we've done those things, we will generate a short
little BASIC program to check out some sample values.  Then, to satisfy the
academics, we will quote the necessary theorems and satisfy the hypotheses.
\medskip \noindent
$$F(x) = a - {1\over x}\eqno(15)$$
$$F'(x) = {1 \over x^2}\eqno(16)$$
Applying $f(x) = x + g\big(F(x)\big)$, where $g(y) = -y/y'$,
$$\eqalign{f(x)\ &=\ x - F(x)/F'(x)\cr
                 &=\ x - {a - 1/x \over 1/x^2}\cr
                 &=\ x - ax^2 +x\cr
                 &=\ 2x - ax^2\cr}\eqno(17)$$
And we observe, substituting $s = 1/a$ into $f'(s)$ that
$$f'(x)\ =\ 2 - 2ax,\eqno(18)$$
$$f'(1/a)\ =\ 2 -2a(1/a)\ =\ 2-2\ =\ 0.\eqno(19)$$ 
Recall that $a$ $\not=$ $0$.
$$f''(x)\ =\ -2a\eqno(20)$$
tell us that $f''$ does not vanish at $1/a$ so we have quadratic convergence
in some (yet to be determined) neighborhood of $1/a$.
So, the required function is equation $(17)$ and the iteration procedure is
$$x_{n+1}\ =\ 2x_{n} - ax_{n}^2.\eqno(21)$$
All that remains is to find the necessary $x_0 = x_0(a)$ to get things
started.
\medskip \noindent
We would like to determine an interval $[\alpha,\,\beta\,]$ 
on which $f$ is contractive.
It isn't enough just to know that one exists, we really need to compute it.
But we have an explicit function, $f$, and $f \in C^1$.  By the mean
value theorem ({\sl MVT for derivatives\/}), we seek values for $\xi$
such that $|f'(\xi)| < 1$.  This will ensure
$$\big| f(x_1) - f(x_2)\big|\ =\ |f'(\xi)|\,|x_1 - x_2|\eqno(22)$$
$$|f'(x)|\ <\ 1$$
$$|2 -2ax|\ <\ 1$$
$$-1\ <\ 2 -2ax\ <\ 1$$
$$-3\ <\ -2ax\ <\ -1$$
If we multiply an inequality by $-{1\over2}$, it changes the 'sense' of
the inequality:
$${3\over2}\ >\ ax\ >\ {1\over2}\eqno(23)$$
Either $a > 0$ or $a < 0$, so:
$$\hbox{ interval } 
= \cases{{3\over 2a}\,>\,x\,>\,{1\over2a}, & if $a\,>\,0$\cr
         \null\cr
         {1\over 2a}\,>\,x\,>\,{3\over2a}, & if $a\,<\,0$\cr}\eqno(24)$$
\medskip\noindent
Basic program:
\medskip
{\tt\obeylines\ttraggedright
10 REM ************************************************
20 REM *                         Harry A. Watson, Jr. 
30 REM *                                    Math 219b 
40 REM *                                27 April 1991 
50 REM ************************************************
60 INPUT "a, x\_0 = "; A, X0
70 FOR I = 1 TO 100
80 X1 = 2*X0 - A*X0*X0
90 IF ABS(X1-X0) < .0001 THEN PRINT X1,1/A,I:STOP
100 X0 = X1
110 NEXT I
}
\bigskip\noindent
Now we would like to have an iterative method for computing $\root n \of{a}$,
$a > 0$, which converges locally in second order.  We only want to
employ addition, subtraction, multiplication and division (the
four arithmetic operations of the real number field.
\medskip\noindent
Again, we will be successful in using Newton's method.  The fixed point theorem
will guarantee that the iterative method converges locally in second order, 
that is, quadratically.
$$F(x)\ =\ a - x^n\eqno(25)$$
$$F'(x)\ =\ -nx^{n-1}\eqno(26)$$
$${F(x)\over F'(x)}\ =\ {a-x^n \over -nx^{n-1} }\ =\ -{a\over nx^{n-1}}
   + {x\over n}\eqno(27)$$
$$\eqalign{f(x)\ &=\ x - F(x)/F'(x)\cr
      &=\ x + {a\over nx^{n-1}} -{x\over n}\cr
      &=\ \big(1-{\textstyle{1\over n}}\big)x 
     + {\textstyle {1\over n}}\big(a/x^{n-1}\big)\cr}\eqno(28)$$
We will ensure convergence by computing $f'$ and plugging in the value
$s = \root n \of {a}$.
$$f'(x)\ = (1-1/n) - [a(n-1)]/[nx^n]. \eqno(29)$$
$$\eqalign {f'\big(\root n\of {a}\big)\ &= \big(1+{\textstyle{1\over n}}\big)
  -{\textstyle{ n-1\over n}\,{a\over a}}\cr
  &=\ 1 - {\textstyle {1\over n}} - {\textstyle {n\over n}} 
      + {\textstyle{1\over n}}\cr
  &=\ 0\cr}\eqno(30)$$
Theory assured us this would happen; our $f'$ has been correctly computed.
$$f''(x)\ =\ [a(n-1)]/x^{n+1}\eqno(31)$$
For $a > 0$,
$$f''\big(\root n\of{a}\big)\ =\ {a(n-1) \over a \root n\of{a} }\ =
 \ {n-1 \over \root n \of {a}}.\eqno(32)$$    
$n \not= 1$ in any case.  Therefore we have quadratic convergence in some
(yet to be determined) interval.  Again, requiring $|f'(x)| < 1$:
$$-1\ <\ \left(1-{1\over n}\right) -{(n-1)a \over nx^n}\ <\ 1$$
$$-2 + {1\over n}\ <\ -{(n-1)a\over nx^n}\ <\ {1\over n}$$
$${n\over (n-1)a}\left(2-{1\over n}\right)\ >\ {1\over x^n}\ >\ 
{n\over (n-1)a}{1\over n}$$
$${(n-1)a\over 2n-1}\ <\ x^n\ <\ (n-1)a$$
Therefore,
$$\root n \of {(n-1)a\over 2n-1}\ <\ x\ <\ \root n \of {(n-1)a}\eqno(33)$$
is an interval in which the iteration algorithm converges quadratically.
Finally, we note that $x^2$ is just short-hand for $x\cdot x$ or $x\,\times\,x$,
and $x^3$ stands for $x\cdot x\cdot x$ or $x\,\times\,x\,\times\,x$ and so
on.  For a given, arbitrary (but fixed) value of $n$, the above algorithm
can be written without exponentiation, substituting
$$\underbrace{x\,\cdot\,x\,\cdots\,x}_{n\ \hbox{\sevenrm times}}\ \hbox{ for }
 x^n.$$
\medskip\noindent
BASIC computer program (fast):
\medskip
{\tt\obeylines\ttraggedright
10 REM *****************************************************
20 REM *                              Harry A. Watson, Jr. 
30 REM *                                         Math 219b 
40 REM *                                     27 April 1991 
50 REM *****************************************************
60 INPUT "Root n, n = "; N
70 INPUT "a, x\_0 = "; A, X0
80 FOR I = 1 TO 100
90 X1 = (1-1/N)*X0 +A/(N*X0{\char94}(N-1))
100 IF ABS(X1-X0) < .0001 THEN PRINT X1,A{\char94}(1/N),I:STOP
110 X0 = X1
120 NEXT I
130 REM  ***************************************************
140 REM  * Recall that the cube root of 20 is approximately 
150 REM  * $e\ =\ 2.7182818\ldots$                                
160 REM  * Plug in $n\ =\ 3$, $a\ =\ 20$,                  
170 REM  * and $x_0\ =\ 3$                                 
180 REM  * Answer = 2.714418      2.714418      4 (steps)  
190 REM  ***************************************************
}
\vfill\eject
\noindent{\bf\llap{2.3\quad}The Algorithm.}  The algorithm associated with
Newton's method may be seen graphically as follows:
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  Graph showing Newton's method convergence.
%
%
$$\beginpicture
  \setcoordinatesystem units <1cm,1cm>
  \setplotarea x from 0 to 8, y from 0 to 5
  \axis bottom  /  %
  \axis left   / % 
  \put {$\scriptstyle\bullet$} at 1.1 0  %
  \plot 6 3.5  2.75 0 / %
  \putrule from 2.75 0 to 2.75 0.85  %
  \plot 2.75 0.85   1.5 0 / %
  \put {$\scriptstyle\bullet$} at 2.75 0 %
  \put {$\scriptstyle\bullet$} at 1.5 0  %
  \linethickness=.75pt
\ifexpressmode
  \put {\bf EXPRESSMODE} at 4 2.5 %
\else
  \setquadratic
  \plot .25 -.25   3 1   6  3.5  / %
\fi
\endpicture$$
\centerline{F{\sevenrm IGURE} 12}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\medskip
$$\tan\beta = f'(x_0) = {f(x_0) \over x_0 - x_1}$$
$$x_1 = x_0 -{f(x_0) \over f'(x_0)}$$
Solve the equation $f(x)=0$.  We assume that $f$ has a continuous
first derivative, that is $f \in C^1$, where
$$C^1 = \{\,\hbox{functions with continuous } f'\,\}.$$
\medskip\noindent
We will construct the following feedback diagram:
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%  Feedback diagram
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
$$\beginpicture %
  \setcoordinatesystem units <1in,1in> %
  \setplotarea x from 0 to 4, y from 0 to 3.5 %
  \put {\lines{ Last estimate\cr $x_n$\cr} } at 2 3 %
  \put {$\displaystyle x_{n+1} = x_n - {f(x_n)\over f'(x_n)}$} at 2 2 %
  \put {\lines{ Next estimate\cr $x_{n+1}$\cr} } at 2 1 %
\ifexpressmode
  \putrule from 2.0 2.75 to 2.0 2.25 %
  \putrule from 2.0 1.75 to 2.0 1.25 %
  \putrule from 3.5 3.00 to 3.0 3.00 %
\else
  \arrow <10pt> [.2, .4]  from 2.0 2.75 to 2.0 2.25 %
  \arrow <10pt> [.2, .4]  from 2.0 1.75 to 2.0 1.25 %
  \arrow <10pt> [.2, .4]  from 3.5 3.00 to 3.0 3.00 %
\fi
  \putrule from 3.0 1.0 to 3.5 1.0 %
  \putrule from 3.5 1.0 to 3.5 3.0 %
  \linethickness=1pt
  \putrectangle corners at 1 2.75  and  3 3.25 %
  \putrectangle corners at 1 1.75  and  3 2.25 %
  \putrectangle corners at 1 0.75  and  3 1.25 %
  \put {$n := n+1$} [l] <5pt,0pt> at 3.5 2 %
  \put {$ x_0 \longrightarrow $} [r] <-2pt,0pt> at 1.00 3.00 %
  \put{F{\sevenrm IGURE} 13} [B] at 2.00 0.25 %
\endpicture$$ %
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\noindent
We will see that each error is essentially proportional to the square of
the previous error.  This means that the number of correct decimal places
roughly doubles with each successive approximation.  This is called
quadratic convergence.
\vfill\eject
\noindent{\bf\llap{2.4\quad}A Second Opinion.}  We say a derivation of
Newton's method for finding roots in paragraph 2.1.  There was an implicit
assumption made that the function in question, $f(x)$, had a Taylor series
representation.  This is a very strong assumption and we may find examples
easily which do not satisfy this condition.  One very common engineering
example can be constructed as follows:
$$R'(x)\ =\ \cases{ 1 & if $ 0 \le x < 1$;\cr
                    0 & otherwise.\cr}$$
$$R(x)\ =\ \cases{ 0 & if $x < 0$;\cr
                   x & if $ 0 \le x < 1$;\cr
                   1 & if $x \ge 1$;\cr}$$
The function $R(x)$ is the so-called {\it Ramp function\/} so highly
favored by electrical engineers.  Clearly the derivative is discontinuous
at $0$ and $1$.  We can construct $f(x) = R(x) - {1\over2}$ and this function
has a root, $x_0 = {1\over2}$, $f(1/2) = R(1/2)-1/2 = 1/2-1/2 = 0$.
We can weaken the assumption that $f(x)$ has a Taylor series representation
and keep the same basic proof if we employ the so-called {\bf Extended
Law of the Mean (Mean Value Theorem).}\footnote{$^{25}$}{John M. H. Olmsted,
{\it Advanced Calculus}, (NY: Appleton-Century-Crofts, Inc., 1956),
pages 75-85.}  
\proclaim Theorem. {\bf Extended Law of the Mean (Mean Value Theorem).}
If each of $f(x)$, $f'(x)$, $\ldots$, $f^{(n-1)}(x)$ is continuous
on a (closed) number interval $[a,b]$, and if $f^{(n)}(x)$ exists
in the (open) number interval $(a,b)$, then there exists a number
$\xi\,\in\,(a,b)$ such that
$$f(b) = f(a) + f'(a)(b-a) + {f''(x)\over2!}(b-a)^2
+ \cdots + {f^{(n-1)}(a)\over(n-1)!}(b-a)^{n-1}
+ {f^{(n)}(\xi)\over n!}(b-a)^n.\eqno(34)$$ 

\medskip\noindent
We let $n=2$ and write
$$f(x) = f(x_0) + f'(x_0)(x-x_0) + {f''(\xi)\over2}(x-x_0)^2.\eqno(35)$$
We are looking for an $x_1$ such that $f(x_1) = 0$.  This being the
case, we have
$$x_1 = x_0 - {f(x_0)\over f'(x_0)} - {f''(\xi)\over2f'(x_0)}(x_1-x_0)^2.
\eqno(36)$$

\medskip\noindent
We will examine other example of convergence, graphing the steps.  In this
case the function is not a polynomial, we will use
$$f(x) = e^{-(x-1/4)} - 1/4.\eqno(37)$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%   Plot showing convergence of Newton's method on transcendental function.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
$$\beginpicture %
  \setcoordinatesystem units <5cm,5cm> %
  \setplotarea x from 0 to 2, y from -.25 to .75 %
  \axis left / %
  \axis bottom shiftedto y=0  / %
\ifexpressmode
  \put {\bf EXPRESSMODE} at 1 .5 %
\else
%  \savelinesandcurves on "chap2b.t01" %
  \plot .375 0   .375 .63249   1.09171 0   1.09171 .18097
        1.51162 0    1.51162 0.03194 / %
  \linethickness=.75pt % 
  \setquadratic
  \plot .25 .75  .375 .63249  .50 .52880  .625 .43728
        .75 .35653  .875 .28526  1.00 .22236  1.25 .11787
        1.50 .03650  1.75 -0.02686  2.00 -0.07622 / %
%  \replot "chap2b.t01" %
\fi
  \put {$x_0$} <0pt,-6pt> at .375 0 %
  \put {$x_1$} <0pt,-6pt> at 1.09171 0 %
  \put {$x_2$} <0pt,-6pt> at 1.51162 0 %
  \put {$\swarrow$\raise6pt\hbox{$\big(x_0,f(x_0)\big)$}}  %
          [lb] <1pt,1pt> at .375 .63249 %
  \put {$\swarrow$\raise6pt\hbox{$\big(x_1,f(x_1)\big)$}} %
          [lb] <1pt,1pt> at 1.09171 .18097 %
  \put {$\swarrow$\raise6pt\hbox{$\big(x_2,f(x_2)\big)$}} %
          [lb] <1pt,1pt> at 1.51162 .03194 %
  \put {$\scriptstyle\bullet$} at .375 .63249 %
  \put {$\scriptstyle\bullet$} at 1.09171 .18097 %
  \put {$\scriptstyle\bullet$} at 1.51162 .03194 %
  \put {$\scriptstyle\bullet$} at .375 0 %
  \put {$\scriptstyle\bullet$} at 1.09171 0 %
  \put {$\scriptstyle\bullet$} at 1.51162 0 %
  \put {$x \rightarrow$} [l] <2pt,0pt> at 2 0 %
  \put {$y=f(x)$} [r] <-2pt,0pt> at 0 0.75 %
\endpicture$$ %
\centerline{F{\sevenrm IGURE} 14}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\vfill\eject
\noindent{\bf\llap{2.5\quad}The Quasi-Newton Method.} The quasi-Newton method,
also known as the method of {\it regula falsi\/}, does not require the
function for the derivative.  It relies on approximation of the derivative
by the ratio
$${f(x_{n-1})-f(x_{n-2}) \over x_{n-1} - x_{n-2} }.$$
Making the substitution
$$f'(x_{n-1}) \approx {f(x_{n-1})-f(x_{n-2})
  \over x_{n-1} - x_{n-2} }, $$
we obtain
$$x_n\ =\ x_{n-1} - {\big(x_{n-1}-x_{n-2}\big)\,f(x_{n-1})
  \over f(x_{n-1}) - f(x_{n-2}) }.\eqno(38)$$
We will examine a problem from a standard 
textbook\footnote{$^{26}$}{Robert Eisberg and Resnick, Robert, {\it Quantum
Physics, 2nd Edition}, (NY: John Wiley \& Sons, 1985), page 24.} 
on quantum physics which will employ this technique.  
In order to derive the Wien displacement law,
$\lambda_{\hbox{max}}T = 0.2014\,hc/k$,\footnote{$^{27}$}{$h=6.626\times
10^{-34}$ joule-sec is the
Planck constant; $c=2.998\times10^{8}$ m/sec is the speed of
light in vacuum; $k=1.381\times10^{-23}$ joule/$^\circ$K is the
Boltzmann's constant.}
it is necessary to solve the
transcendental equation
$$e^{-x} + x/5 = 1.\eqno(39)$$
We know already that the answer is approximately 5, in fact, it is
approximately 4.965.  We will use the following elementary BASIC
computer program to see how well this method works.
\medskip
{\tt\parindent=0pt\obeylines\ttraggedright
5 DEFDBL A-H, O-Z
6 OPEN "92\_12\_16.txt" FOR OUTPUT AS \#1
10 DEF fna (x) = EXP(-x) + .2 * x - 1!
20 INPUT "Initial approximation = ", x0
25 PRINT \#1, "Initial approximation = ", x0
26 PRINT \#1, "x1   =   x0 + .0001   = ", x0 + .0001
30 h = .0001
40 x1 = x0 + h
50 x2 = x1 - (h * fna(x1)) / (fna(x1) - fna(x0))
60 PRINT x2
65 PRINT \#1, "Next approximation = ", x2
70 h = x2 - x1
80 IF INKEY\$ = "" THEN GOTO 80
90 x0 = x1: x1 = x2: GOTO 50
}
\medskip\noindent
Now, we will look at the output (starting with an initial guess of 5):
\smallskip
\centerline{\bf TYPE 92\_12\_16.TXT}
\medskip
{\tt\parindent=0pt\obeylines
Initial approximation =      5 
x1   =   x0 + .0001   =      5.0001 
Next approximation =         4.965135680044722 
Next approximation =         4.965114168528612 
Next approximation =         4.965114155083961 
Next approximation =         4.965114155083955 
Next approximation =         4.965114155083955 
}
\medskip\noindent
So, it didn't take but four steps to converge!  Now, let's draw a graph and
see just how this looks graphically.  Incidently, in the real world this
particular experimental value is rarely calculated with accuracy better than
10\%.  There is a good approximation to 0.2014, namely,
$${1\over2\pi\big(1-\cos(e/2)\big)} = {1\over4.963222169} = .20148\ldots
\eqno(40)$$
(It is accurate to within 0.019\%.)
\vfill\eject
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% Figure 10 --- Exponential curve with intersecting straight line.
%
$$\beginpicture
 \setcoordinatesystem units <.75in,1in>
 \setplotarea x from 0 to 6, y from 0 to 1
 \normalgraphs
 \axis bottom ticks
   numbered from 0 to 6 by 1
   length <0pt> withvalues $x$ / at .75 /  / 
 \linethickness=.25pt  \putrule from  .75 0  to .75 .47237
 \setdots \putrule from .75 .47237 to .75 .85  \setsolid
 \put {$\scriptstyle\bullet$} at  .75 .47237
 \put {$\scriptstyle\bullet$} at  .75 .85
 \put {$y_1=e^{-x}$} [lb] <4pt,0pt> at .75 .47236
 \put {$\swarrow$\raise6pt\hbox{$y_2=1-x/5$}} [lb] <2pt,2pt> at .75 .85
 \put {$\scriptstyle\bullet$} at 4.96511 0
 \put {$\swarrow$\raise6pt\hbox{$y_1 = y_2$}} [lb] <2pt,2pt> at 4.96511 0 %
\ifexpressmode
 \put {\bf EXPRESSMODE} at 1.5 0.5
\else
 \setquadratic
 \plot  0 1   .25  .77880   .50  .60653   .75  .47237  1.00  .36788
             1.25  .28650  1.50  .22313  1.75  .17377  2.00  .13534
             2.25  .10540  2.50  .08208  2.75  .06393  3.00  .04979
             3.25  .03877  3.50  .03019  3.75  .02352  4.00  .01832
             4.25  .01426  4.50  .01109  4.75  .00865  5.00  .00674
             5.25  .00525  5.50  .00409  5.75  .00318  6.00  .00248  / %
 \setlinear
 \plot 0 1  5.5 -0.1 / %
\fi
\endpicture $$
\centerline {F{\sevenrm IGURE} 15}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\medskip\noindent
Let's take a look at another example, this time an iteration for a
transcendental equation.  We want to find the positive solution
for $2\sin(x) = x$.  The first thing we will want to do is to draw a
graph and estimate the root.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  Graph of y_1=2*sin(x) and y_2 = x
%
%  Figure 11.
%
$$\beginpicture
  \setcoordinatesystem units <1in,.5in>
  \setplotarea x from 0 to 4, y from 0 to 4
  \axis left ticks numbered from 0 to 4 by 1 /
  \axis bottom label {F{\sevenrm IGURE} 16} ticks
     numbered from 0 to 3 by 1 /
\ifexpressmode
  \put {\bf EXPRESSMODE } at 2 2 %
\else
  \setquadratic
  \plot
      0.0 0.00000  0.1 0.19967  0.2 0.39734  0.3 0.59104  0.4 0.77884 
      0.5 0.95885  0.6 1.12928  0.7 1.28844  0.8 1.43471  0.9 1.56665
      1.0 1.68294  1.1 1.78241  1.2 1.86408  1.3 1.92712  1.4 1.97090
      1.5 1.99499  1.6 1.99915  1.7 1.98333  1.8 1.94770  1.9 1.89260
      2.0 1.81859  2.1 1.72642  2.2 1.61699  2.3 1.49141  2.4 1.35093
      2.5 1.19694  2.6 1.03100  2.7 0.85476  2.8 0.66998  2.9 0.47850
      3.0 0.28224 / %
  \setlinear
  \plot 0 0  4 4 /
\fi
  \put {$\longleftarrow y_1 = 2\cdot\sin(x)$} [l] <2pt,0pt> at 2.0 1.81859 %
  \put {$\nwarrow$\lower6pt\hbox{$y_2=x$}} [lt] <-1pt,-1pt> at 2.5 2.5 %
  \put { $y_1=y_2$\lower6pt\hbox{$\downarrow$} }  %
        [rb] <5pt,1pt> at 1.8955 1.8955 %
  \setdots
  \putrule from 1.8955 0.0  to 1.8955 1.8955 /
\endpicture$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\medskip\noindent
Of course, we will examine both the quasi-Newton and the Newton method.
The Newton method is listed to the right and we see at once that it
has converged both faster and to greater precision.  The important point
to note here is that with a modern (fast) computer both methods converge
rapidly.  We can use the two in conjunction to check each other's results.
Moreover, with the Simpson's rule procedure to check the derivative
function, we can be assured that the derivative $f'(x)$ of $f(x)$ is
the correct function.  This comparison of two different methods is 
similar to the adaptive quadrature routine (AQR), which will be studied
in a later section. 
\medskip
%
% 
{\tt\parindent=0pt\obeylines
Initial approximation =      2 
x1   =   x0 - .0001   =      1.99990  \quad 1.9009955942039090 
Next approximation =         1.90099  \quad 1.8955116453795950
Next approximation =         1.89580  \quad 1.8954942672087130
Next approximation =         1.89550  \quad 1.8954952670339810
Next approximation =         1.89549  \quad 1.8954952670339810
Next approximation =         1.89549  \quad 1.8954925670339810
Next approximation =         1.89549  \quad 1.8954925670339810
}
\medskip\noindent
\vfill\eject
\noindent{\bf\llap{2.6\quad}Pathological examples.} Newton's method
will fail for certain situations.  There are an excellent developments of
such examples in many elementary textbooks\footnote{$^{28}$}{George B.
Thomas, Jr., {\it Calculus and Analytic Geometry, 3rd Edition},
(Reading Massachusetts: Addison-Wesley Publishing Company, 1966),
page 455.}   We won't dwell on the analytic details of the failure
of the Newton method; on the other hand, we will present graphs to
illustrate the point.  It has already been pointed out that the very
first thing to do is to construct a graph and then estimate the root.
With all the excellent graphics software available, there is no excuse
for not doing so.  Without further ado, let's first look at a graph
where Newton's method fails to converge.
$$f(x)\ =\ \cases{\sqrt{\left| x-r\right|}& if $x > r$\cr\noalign{\vskip2pt}
                  0 & if $x = 0$\cr
                  -\sqrt{\left| x-r\right|}& if $x < r$\cr}\eqno(41)$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Figure 17.
%
$$\beginpicture
  \setcoordinatesystem units <2in,1in>
  \setplotarea x from -.125 to 2, y from -1 to 1 %
  \plotheading {\lines{ Successive approximants\cr
                       oscillate back and forth.\cr} } %
  \axis bottom shiftedto y=0  / %
  \axis left shiftedto x=0 /
\ifexpressmode
  \put {\bf EXPRESSMODE} at 2 1 %
\else
  \linethickness=1pt
  \setquadratic
  \plot
    0.35 -0.80623  0.40 -0.77460  0.45 -0.74162  0.50 -0.70711  0.55 -0.67082
    0.60 -0.63246  0.65 -0.59161  0.70 -0.54772  0.75 -0.50000  0.80 -0.44721
    0.85 -0.38730  0.90 -0.31623  0.95 -0.22361  0.975 -0.15811
    1.00 0.00000  1.025 0.15811 1.05 0.22361
    1.10 0.31623  1.15 0.38730  1.20 0.44721  1.25 0.50000  1.30 0.54772
    1.35 0.59161  1.40 0.63246  1.45 0.67082  1.50 0.70711  1.55 0.74162
    1.60 0.77460  1.65 0.80623  / %
\fi
  \linethickness=0.4pt
  \put {$\scriptstyle\bullet$} at 0.50 -0.70711 %
  \put {$\scriptstyle\bullet$} at 1.50  0.70711 %
  \put {$\scriptstyle\bullet$} at 0.50 0.0 %
  \put {$\scriptstyle\bullet$} at 1.50 0.0 %
  \put {$x_0$} [b] <0pt,4pt> at 0.50 0.0 %
  \put {$x_1$} [t] <0pt,-4pt> at 1.50 0.0 %
  \put {$O$} [rt] <-2pt,-2pt> at 0 0 %
  \put {$r$} [lb] <4pt,2pt> at 1.0 0.00 %
  \put {$x$} at 2.1 0.0 %
  \put {$y$} at 0.0 1.1 %
  \setdashes
  \putrule from 0.5 0.0  to  0.5 -0.70711 %
  \putrule from 1.5 0.0  to  1.5  0.70711 %
  \setlinear
  \setsolid
  \linethickness=.25pt
  \plot 0.5 -0.70711  1.5 0.0 / %
  \plot 0.5 0.0  1.5 0.70711 / %
\endpicture$$
\centerline {F{\sevenrm IGURE} 17} %
\medskip\noindent
Now, let's look at a graph that really gets crazy!  This time the
successive approximations, $x_0$ and $x_1$ don't oscillate, they
get worse!
$$\beginpicture
  \setcoordinatesystem units <2in,1in>
  \setplotarea x from -.125 to 2, y from -1 to 1 %
  \plotheading {\lines{ Successive approximants\cr
                       diverge from root.\cr} } %
  \axis bottom shiftedto y=0  / %
  \axis left shiftedto x=0 /
\ifexpressmode
  \put {\bf EXPRESSMODE} at 2 1 %
\else
  \linethickness=1pt
  \setquadratic
  \plot
    0.35 -0.86624  0.40 -0.84343  0.45 -0.81932  0.50 -0.79370
    0.55 -0.76631  0.60 -0.73681  0.65 -0.70473  0.70 -0.66943
    0.75 -0.62996  0.80 -0.58480  0.85 -0.53133  0.90 -0.46416
    0.95 -0.36840  0.975 -0.29240 1.00  0.00000  1.025 0.29240
    1.05  0.36840  1.10  0.46416
    1.15  0.53133  1.20  0.58480  1.25  0.62996  1.30  0.66943
    1.35  0.70473  1.40  0.73681  1.45  0.76631  1.50  0.79370
    1.55  0.81932  1.60  0.84343  1.65  0.86624  / %
\fi
  \linethickness=0.4pt
  \put {$\scriptstyle\bullet$} at 0.50 -0.79370 %
  \put {$\scriptstyle\bullet$} at 1.50  0.79370 %
  \put {$\scriptstyle\bullet$} at 0.50 0.0 %
  \put {$\scriptstyle\bullet$} at 1.50 0.0 %
  \put {$x_0$} [b] <0pt,4pt> at 0.50 0.0 %
  \put {$x_1$} [t] <0pt,-4pt> at 1.50 0.0 %
  \put {$O$} [rt] <-2pt,-2pt> at 0 0 %
  \put {$r$} [lb] <4pt,2pt> at 1.0 0.00 %
  \put {$x$} at 2.1 0.0 %
  \put {$y$} at 0.0 1.1 %
  \setdashes
  \putrule from 0.5 0.0  to  0.5 -0.79370 %
  \putrule from 1.5 0.0  to  1.5  0.79370 %
%  \setlinear
%  \setsolid
%  \linethickness=.25pt
%  \plot 0.5 -0.79370  1.5 0.0 / %
%  \plot 0.5 0.0  1.5 0.79370 / %
\endpicture$$
\centerline {F{\sevenrm IGURE} 18} %
\vfill\eject
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\headline={\tenrm\hfill Linear Least-squares Line}
\centerline{\bf Chapter 3}
\bigskip
\noindent{\bf\llap{3.0\quad}Introduction.}  One of the best introductions
to the so-called methods of least-squares 
follows from a simple display of 
a least-squares linear fit graph properly done
(with the error bars and all the statistical appendages) and a
dissection the components  of the graph.  The points in the
given point set are graphed as central dots 
({\raise1pt\hbox{$\scriptstyle\bullet$}})
in the error bars 
$\big(\,{{\lower1pt\hbox{$\scriptstyle\top$}}
    \atop{\raise1pt\hbox{$\scriptstyle\bot$}}}\,\big)$.  
The length of the error bar  is determined by
the standard deviation, $\sigma_y$.  From the central dot to the
cross-bar is one unbiased standard deviation.   
%
  \newdimen\xposition
  \newdimen\yposition
  \newdimen\dyposition
  \newdimen\crossbarlength
  \def\puterrorbar at #1 #2 with fuzz #3 {%
    \xposition=\Xdistance{#1}
    \yposition=\Ydistance{#2}
    \dyposition=\Ydistance{#3}
  \setdimensionmode
  \put {$\bullet$} at {\xposition} {\yposition}
  \dimen0 = \yposition %                 ** Determine the y-location of
    \advance \dimen0 by -\dyposition %   **   the lower cross bar.
  \dimen2 = \yposition %                 **  Determine the y-location of
    \advance \dimen2 by  \dyposition %   **   the upper cross bar.
  \putrule from {\xposition} {\dimen0} % **  Place vertical rule.
    to {\xposition} {\dimen2}
  \dimen4 = \xposition
    \advance \dimen4 by -.5\crossbarlength
  \dimen6 = \xposition
    \advance \dimen6 by  .5\crossbarlength
  \putrule from {\dimen4} {\dimen0} to {\dimen6} {\dimen0}
  \putrule from {\dimen4} {\dimen2} to {\dimen6} {\dimen2}
  \setcoordinatemode}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  Graph for least-squares curve fit with shading, areas, etc.
%
$$\beginpicture
  \setcoordinatesystem units <.4in,.45in>
  \crossbarlength=5pt 
  \setplotarea x from -1 to 10, y from 85 to 95 
  \plotheading { \lines{Missile `A'\cr
       Warrantied in shaded area\cr} } %
  \linethickness=.25pt
  \axis bottom  
    label { \lines{ years $\longrightarrow$\cr \noalign{\vskip3pt} %
          F{\sevenrm IGURE} 19\cr } } %
    ticks in short withvalues $1981$ $1982$ $1983$ $1984$ $1985$
    $1986$ $1987$ $1988$ $1989$ $1990$ $1991$ /
    at 0 1 2 3 4 5 6 7 8 9 10 / /
  \axis left 
    label {\lines{$\uparrow$\cr $R$\/\%\cr }}
    ticks in short numbered from 85 to 95 by 1 /
  \puterrorbar at 0 90.08   with fuzz  0.180
  \puterrorbar at 1 90.57   with fuzz  0.180
  \puterrorbar at 2 90.76   with fuzz  0.180
  \puterrorbar at 3 91.30   with fuzz  0.180
  \puterrorbar at 4 91.57   with fuzz  0.180
  \puterrorbar at 5 92.44   with fuzz  0.180
  \puterrorbar at 6 92.87   with fuzz  0.180
  \puterrorbar at 7 93.68   with fuzz  0.180
  \puterrorbar at 8 93.85   with fuzz  0.180
  \puterrorbar at 9 94.49   with fuzz  0.180
  \puterrorbar at 10 94.88   with fuzz  0.180
\ifexpressmode
  \put {\bf EXPRESSMODE} at 5 90
\else
  \linethickness=.4pt
  \plot 0. 89.9077 10. 94.9086 /
  \setlinear
  \setshadegrid span <4pt>
  \vshade 4 85 95  7 85 95 /
  \vshade 7 85 86.5  9.1 85 86.5 / 
  \vshade 9 85 95 10 85 95 /
  \setlinear
  \setshadegrid span <4pt>
  \hshade 87.5 7 9.1 95 7 9.1 /
\fi
  \put {\frame <3pt> {\lines {Unwarrantied\cr Region\cr}}} at 2 93.5
  \put {\frame <4pt> {$\sigma_y = 0.180$}} at 8 87
\endpicture$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
The underlying principle of the method of least squares is
the determination of a curve which
minimizes the squares of the deviations 
of a given point set to that curve.  This is another
way of saying that we will determine a curve such that
the sum of the squares of the distances from the
points in a given point set to the curve is a minimum.
Since the sought-for curve is a straight line, this
case is frequently referred to as a linear least-squares
curve fit.  The determined line is sometimes called a trend line,
sometimes called a regression line, and sometimes called
a line of best fit.
The technique is known as the {\bf maximum likelihood
criterion}.
\footnote{$^{29}$}{Glenn James and James, Robert C.,
{\it Mathematics Dictionary}, (Princeton, NJ: D. Van Nostrand
Company, Inc., 1964), page 253.}
The theory involves an understanding of partial derivatives;  it will
be covered later in an optional paragraph.  The computer programs
are straightforward to easy to apply;  they will be covered first.
\vfill\eject
\noindent{\bf\llap{3.1\quad}Dissection of a graph.}  The first thing to
consider in any least-squares construction is the underlying point set.
This point set is sometimes called the sample space, sometimes called the
sample data, and sometimes called the sample population.  Confusion arises
because of the various names that are attached to the data points.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  Graph giving only the data points
%
$$\beginpicture
  \setcoordinatesystem units <.3in,.2in>
  \setplotarea x from -1 to 10, y from 85 to 95 
  \plotheading { Sample Data } %
  \linethickness=.25pt
  \axis bottom  %
    label { \lines{ abscissa ($x$-axis) $\longrightarrow$\cr %
    \noalign{\vskip3pt} %
    F{\sevenrm IGURE} 20\cr} } %
    ticks in short withvalues {} {} {} {} {} {} {} {} {} {}
    {} /  at 0 1 2 3 4 5 6 7 8 9 10 / / %
  \axis left 
    label {\lines{$\uparrow$\cr $y(x)$\cr }}
    ticks in short withvalues {} {} {} {} {} {} {} {}
    {} {} {} / at 85 86 87 88 89 90 91 92 93 94 95 / / %
  \put {$\scriptstyle\bullet$} at 0 90.08   
  \put {$\scriptstyle\bullet$} at 1 90.57 
  \put {$\scriptstyle\bullet$} at 2 90.76  
  \put {$\scriptstyle\bullet$} at 3 91.30   
  \put {$\scriptstyle\bullet$} at 4 91.57   
  \put {$\scriptstyle\bullet$} at 5 92.44  
  \put {$\scriptstyle\bullet$} at 6 92.87   
  \put {$\scriptstyle\bullet$} at 7 93.68   
  \put {$\scriptstyle\bullet$} at 8 93.85  
  \put {$\scriptstyle\bullet$} at 9 94.49   
  \put {$\scriptstyle\bullet$} at 10 94.88   
\ifexpressmode
  \put {\bf EXPRESSMODE} at 5 90
\else
%  \linethickness=.4pt
%  \plot 0. 89.9077 10. 94.9086 /
\fi
\endpicture$$
%\centerline{F{\sevenrm IGURE} 21}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
The next thing to consider are the arithmetic quantities associated with the
data points (with the so-called point set).  If we simply add up all the
$x$-values (the abscissas) and divide by the count of the number of points,
we will get the $x$-average, also called the $x$-mean, and usually denoted
as
$$\frame <5pt> {$\displaystyle \mu_x\ =\ {\sum_{i=1}^n x_i \over n}.$}
   \eqno(42)$$
In a similar manner, we compute the $y$-average, $\mu_y$,
$$\frame <5pt> {$\displaystyle \mu_y\ =\ {\sum_{i=1}^n y_i \over n}.$}
   \eqno(43)$$
The averages $\mu_x$ and $\mu_y$ are familiar.  Other computations are also
necessary to obtain the ingredients needed to calculate the trend line, 
shown below.  Because this line is the best possible fit to the data, it is
also known as the {\sl line of best fit}.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  Graph showing the trend line and data only
%
$$\beginpicture
  \setcoordinatesystem units <.3in,.2in>
  \setplotarea x from -1 to 10, y from 85 to 95 
  \plotheading { Sample Data } %
  \linethickness=.25pt
  \axis bottom  %
    label {\lines{ abscissa ($x$-axis) $\longrightarrow$\cr %
    \noalign{\vskip3pt} %
    F{\sevenrm IGURE} 21\cr } } %
    ticks in short withvalues {} {} {} {} {} {} {} {} {} {}
    {} /  at 0 1 2 3 4 5 6 7 8 9 10 / / %
  \axis left 
    label {\lines{$\uparrow$\cr $y(x)$\cr }}
    ticks in short withvalues {} {} {} {} {} {} {} {}
    {} {} {} / at 85 86 87 88 89 90 91 92 93 94 95 / / %
  \put {$\scriptstyle\bullet$} at 0 90.08   
  \put {$\scriptstyle\bullet$} at 1 90.57 
  \put {$\scriptstyle\bullet$} at 2 90.76  
  \put {$\scriptstyle\bullet$} at 3 91.30   
  \put {$\scriptstyle\bullet$} at 4 91.57   
  \put {$\scriptstyle\bullet$} at 5 92.44  
  \put {$\scriptstyle\bullet$} at 6 92.87   
  \put {$\scriptstyle\bullet$} at 7 93.68   
  \put {$\scriptstyle\bullet$} at 8 93.85  
  \put {$\scriptstyle\bullet$} at 9 94.49   
  \put {$\scriptstyle\bullet$} at 10 94.88   
\ifexpressmode
  \put {\bf EXPRESSMODE} at 5 90
\else
  \linethickness=.4pt
  \plot 0. 89.9077 10. 94.9086 /
\fi
  \put {$\nwarrow\lower6pt\hbox{Line of Best Fit}$} [lt] <1pt,1pt> at 4.5 92.0
\endpicture$$
%\centerline{F{\sevenrm IGURE} 21}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\vfill\eject
\noindent Now is a good time to introduce the BASIC computer program.  
Since we have seen several BASIC and ``C'' programs already, this
listing will be provided {\it without comments}.  The statistical
parameters $\mu_x$ and $\mu_y$ will be denoted as {\tt XMU} and
{\tt YMU}, respectively.  The statistical parameters for the
standard deviations, $\sigma_x$, $\sigma_y$, and
$\sigma_{xy}$ (biased and unbiased), will be denoted
{\tt SIGMAX}, {\tt SIGMAY}, and {\tt SIGMAXY}, respectively.
\smallskip
{\tt\obeylines\parindent=0pt\ttraggedright
100 CONST N = 11
110 DEFDBL A-H, O-Z: REM Define everything double precision except the indices.
120 DIM X(N), Y(N)
130 DATA 1981.,1982.,1983.,1984.,1985.,1986.,1987.,1988.,1989.,1990.,1991.
140 DATA 90.08,90.57,90.76,91.30,91.57,92.44,92.87,93.68,93.85,94.49,94.88
150 FOR I = 1 TO N: READ X(I): NEXT I
160 FOR I = 1 TO N: READ Y(I): NEXT I
170 SUMX = 0!: SUMY = 0!: SUMX2 = 0!: SUMXY = 0!: SUMX2 = 0!
180 SIGMAX = 0!: SIGMAY = 0!: SIGMAXY = 0!
190 FOR I = 1 TO N
200 SUMX = SUMX + X(I): SUMY = SUMY + Y(I): SUMX2 = SUMX2 + X(I) * X(I)
210 SUMXY = SUMXY + X(I) * Y(I): SUMY2 = SUMY2 + Y(I) * Y(I)
220 NEXT I
230 DELTA = CDBL(N) * SUMX2 - SUMX * SUMX
240 A = (SUMX2 * SUMY - SUMX * SUMXY) / DELTA
250 B = (CDBL(N) * SUMXY - SUMX * SUMY) / DELTA
260 CLS : PRINT "A (x-intercept) = "; A
270 PRINT "B (slope) = "; B
280 XMU = SUMX / CDBL(N): YMU = SUMY / CDBL(N)
290 FOR I = 1 TO N
300 SIGMAX = SIGMAX + (XMU - X(I)) * (XMU - X(I))
310 SIGMAY = SIGMAY + (YMU - Y(I)) * (YMU - Y(I))
320 SIGMAXY = SIGMAXY + (XMU - X(I)) * (YMU - Y(I))
330 NEXT I
340 R = SIGMAXY / SQR(SIGMAX * SIGMAY)
350 PRINT "Correlation Coefficient r =  "; R: SIGMAY = 0!
360 FOR I = 1 TO N
370 SIGMAY = SIGMAY + (Y(I) - A - B * X(I)) {\char94} 2
380 PRINT USING "\#\#\#\#  \#\#.\#\#\#\#\#  \#\#.\#\#\#\#\#"; X(I); Y(I); A + B * X(I)
390 NEXT I
400 SIGMAY = SQR(SIGMAY / (CDBL(N) - 2!))
410 PRINT "sigma\_y (unbiased) = "; SIGMAY
420 PRINT "sigma\_x (unbiased) = "; SIGMAY * SQR(SIGMAX / DELTA)
430 END
}
\smallskip\noindent  After the above BASIC computer program was executed,
the following data were obtained:
\smallskip
{\tt\parindent=0pt\obeylines\ttraggedright
A (x-intercept) = -900.77236
B (slope) = .50009
Correlation Coefficient r = .99475
Sigma\_y (unbiased) = .17980
Sigma\_x (unbiased) = .05421
}
\smallskip\noindent  The next important topic is the construction of
the error bars.  This is often the most forgotten item in the creation
of a data presentation.  It is the only way properly introduct statistical
uncertainty into a display graph.  In the hard science of Physics, such
information is mandatory.  The distance from the data point to the
crossbars (top and bottom of the errorbar) must equal one unbiased
standard deviation, $\sigma_y$.  There may also be uncertainty in the
abscissa.  In this case, it is necessary to construct an error bar
horizontal to the $x$-axis of length $\sigma_x$.  The error bars tell
us more than just some statistical fact about the individual data points.
If we examine closely Fig. 33, we'll note that the line of best fit
passes through most of the error bars.  This tells us that there
is a good probability that there is an underlying functional
relationship between $x$ and $f(x)$, namely $f(x) \approx a\cdot x
+ b$.  If, on the other hand, the best straight line missed a high
proportion of the error bars, or if it missed any by a large
distance (relative to the length of the error bar), then the
relationship between $x$ and $y \approx f(x)$ would probably not be that of
a straight line (if there was any relationship at all).
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  Error bar graph
%
$$\beginpicture % Always put % here because some word processor
  \setcoordinatesystem units <1cm,1cm> % truncate final space(s).
  \setplotarea x from -2 to 8, y from -2 to 2 %
  \crossbarlength = 5pt %
  \plotheading {Error bars with uncertainties} %
  \puterrorbar at 0 0 with fuzz 1.2 %
  \put { \lines{$\uparrow$\cr $\sigma_y$\cr $\downarrow$\cr} } % 
     <0pt,3pt> at .5 .5 %
  \puterrorbar at 5 0 with fuzz 1.2 %
  \put { \lines{$\uparrow$\cr $\sigma_y$\cr $\downarrow$\cr} } % 
     <0pt,3pt> at 5.5 .5 %
  \putrule from 3.6 0.0  to 6.4 0.0 %
  \putrule from 3.6 0.1 to 3.6 -0.1 %
  \putrule from 6.4 0.1 to 6.4 -0.1 %
  \put { $\leftarrow\sigma_x\rightarrow$ } at 5.7 -0.5 %
\endpicture$$ %
\centerline{F{\sevenrm IGURE} 22}
\smallskip\noindent
We'll conclude this long chapter with a another graph
showing the line of best fit and the companion ``C'' program for the
BASIC program above.  The next graph was suggested by the textbook
{\sl An Introduction to Error Analysis\/} by John R. Taylor
(Mill Valley, CA: University Science Books, 1982), page 26.  It
indicates the uncertainties found in any physical measurement.
The underlying physical law is know as {\bf Hooke's Law}, published
by Robert Hooke in 1678.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% A plot of an extension of a spring $x$ versus the load $m$.
%
%
$$\beginpicture
  \setcoordinatesystem units <.002in,.4in>
  \crossbarlength=5pt 
  \setplotarea x from 0 to 1000, y from 0 to 5 
  \linethickness=.25pt
  \axis bottom  %
    label {\lines{ $m$ (gm) $\longrightarrow$ \cr \noalign{\vskip2pt} %
       F{\sevenrm IGURE} 23\cr} } %
    ticks in numbered from 0 to 1000 by 500
    unlabeled short quantity 11  /
  \axis left 
    label {\lines{$\uparrow$\cr $x$\cr (cm)\cr}}
    ticks in withvalues $5$ / at 5 / 
    quantity 6 /
  \plot 0 5  0 5.5 /
  \plot 1000 0  1050 0 /  
  \linethickness=.4pt
  \plot 0 0  1000 5.5 /
  \puterrorbar at 200 1.1   with fuzz  0.3
  \puterrorbar at 300 1.5   with fuzz  0.3
  \puterrorbar at 400 1.9   with fuzz  0.3
  \puterrorbar at 500 2.8   with fuzz  0.3
  \puterrorbar at 600 3.4   with fuzz  0.3
  \puterrorbar at 700 3.5   with fuzz  0.3
  \puterrorbar at 800 4.6   with fuzz  0.3
  \puterrorbar at 900 5.4   with fuzz  0.3
\endpicture$$
\vfill\eject
{\tt\parindent=0pt\obeylines\ttraggedright
/* Program for mean, standard deviation */
/* Program for least-squares fit to a straight line */
/* Program for computing correlation coefficient "r" */
\medskip
\#include <stdio.h>
\#include <math.h>
\#include <float.h>
\medskip
int main()
{\char123}
/* Initialize all the parameters and type variables */
\quad  FILE *fp;   /* file pointer */
\quad  static double x[] = { 1981., 1982., 1983., 1984., 1985.,
\qquad 1986., 1987., 1988., 1989., 1990.,
\qquad 1991. };   /* x values (years) */
\quad  static double y[] = { 90.08, 90.57, 90.76, 91.30, 91.57,
\qquad 92.44, 92.87, 93.68, 93.85, 94.49,
\qquad 94.88 };   /* y values (reliabilities) */
\quad  /* The variables sum\_{something} are used as accumulators. */
\quad  /* a, b are the y-intercept and slope, respectively of the
\qquad least-squares line. */
\quad  /* r is the correlation coefficient (the so-called Pearson's r). */
  double  Delta, a, b, r, sum\_x, sum\_x2, sum\_y, sum\_y2, sum\_xy;
  /* mu\_{something} is a mean, sigma\_{something} is a standard
     deviation (unbiased for least-squares = n-2 degrees of freedom). */
  double  mu\_x, mu\_y, sigma\_x, sigma\_y, sigma\_xy;
  /* The integer i is an index. (unsigned, short integer) */
  int i;
  /* The long integer n is the count of the number of data points. */
  long int n;
/* Begin computations */
  /* Initialize n, x[n], y[n], and give the output file a name.  Then you
     are ready to begin execution. */
  n = 11;
  /* Make sure the accumulators are all initialized to zero. */
  sum\_x = (double) 0.0;
  sum\_y = (double) 0.0;
  sum\_y2 = (double) 0.0;
  sum\_x2 = (double) 0.0;
  sum\_xy = (double) 0.0;
  sigma\_x = (double) 0.0;
  sigma\_y = (double) 0.0;
  sigma\_xy = (double) 0.0;
  /* The so-called "for loop" computes the sum\_{something}s */
  for (i=0; i < n; ++i){\char123}          /* i marches from 0 to n-1.   */
    sum\_y += (double) y[i];                  /* All the arrays in "C"      */
    sum\_y2 += (double) y[i]*y[i];            /* programs begin with index  */
    sum\_x += (double) x[i];                  /* zero.                      */
    sum\_x2 += (double) x[i]*x[i];
    sum\_xy += (double) x[i]*y[i];
  {\char125}
  /* Display the results of the computation of all the sums, Delta,
\quad  the slope, and the y-intercept for
\quad  y = a + b * x, the least-squares straight line */
  printf("{\char92}n sum\_x = \%.16lf", sum\_x);
  printf("{\char92}n sum\_y = \%.16lf", sum\_y);
  printf("{\char92}n sum\_x2 = \%.16lf", sum\_x2);
  printf("{\char92}n sum\_y2 = \%.16lf", sum\_y2);
  printf("{\char92}n sum\_xy = \%.16lf", sum\_xy);
  Delta = (double) n*sum\_x2 - sum\_x*sum\_x;
  printf("{\char92}n Delta = \%.16lf", Delta);
  a = (double) (sum\_x2*sum\_y - sum\_x*sum\_xy)/Delta;
  b = (double) (n*sum\_xy - sum\_x*sum\_y)/Delta;
  printf("{\char92}n A (y-intercept) = \%.16lf", a); /* Use Capital "A" here    */
  printf("{\char92}n B (slope)       = \%.16lf", b);    /* to be consistent with   */
  mu\_y = (double) sum\_y/n;                     /* the text book (Taylor). */
  mu\_x = (double) sum\_x/n;
  /* Display (print to screen) the means only if needed. */
  /* printf("{\char92}n mu\_y = \%.16lf", mu\_y); */
  /* printf("{\char92}n mu\_x = \%.16lf", mu\_x); */
  for (i=0; i < n; ++i){\char123}
\quad    /* Biased estimates for sigma\_x and sigma\_y */
    sigma\_y += (mu\_y-y[i])*(mu\_y-y[i]);       /* We are just using the   */
    sigma\_x += (mu\_x-x[i])*(mu\_x-x[i]);       /* sigma\_{something}s here */
    sigma\_xy += (mu\_x-x[i])*(mu\_y-y[i]);      /* as accumulators.        */
 {\char125}
  /* compute the correlation coefficient from the values in the
     accumulators.  This is not the actual formula.  */
  r = sigma\_xy/sqrt(sigma\_x*sigma\_y);
  printf("{\char92}n correlation coefficient = \%.16lf", r);
  sigma\_y = 0.0;
  for (i=0; i < n; ++i){\char123}
    sigma\_y += (y[i] - a - b*x[i])*(y[i] - a - b*x[i]);
    /* This is the least-squares data to be plotted. */
    printf("{\char92}n \%d \%.4lf \%.4lf \%.4lf", i, x[i], y[i],a + b*x[i]);
 {\char125}
  /* Unbiased estimates for sigma\_x and sigma\_y */
  /* Note that there are n-2 degrees of freedom (not n-1). */
  sigma\_y = sqrt((double) sigma\_y/(n-2.0));
  printf("{\char92}n sigma\_y (unbiased) = \%.16lf", sigma\_y);
  sigma\_x = sigma\_y * sqrt((double) sum\_x2/Delta);
  printf("{\char92}n sigma\_x (unbiased) = \%.16lf", sigma\_x);
/*  Output to file.  Make sure you have a unique filename. */
  fp = fopen("pgm3.txt","w");     /* Open the file */
  fprintf(fp,"{\char92}n A (y-intercept) = \%.16lf", a);
  fprintf(fp,"{\char92}n B (slope)       = \%.16lf", b);
  fprintf(fp,"{\char92}n correlation coefficient = \%.16lf", r);
  fprintf(fp,"{\char92}n sigma\_y (unbiased) = \%.16lf", sigma\_y);
  fprintf(fp,"{\char92}n  x[i]     y[i]     a+b*x[i]");
  for (i=0; i < n; ++i){\char123}
  fprintf(fp,"{\char92}n \%.4lf \%.4lf \%.4lf",x[i],y[i], a + b*x[i]);
  {\char125}
  fclose(fp);                  /* Close the file */
return(0);
{\char125}
\medskip
/* End Of File */
}
\vfill\eject
% End of first portion for chapter 3.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\noindent{\bf\llap{3.2\quad}Formula Derivation.}  Whenever a mathematicians
are faced with the chore of explaining some difficult point, they just make
a definition.  Then, by choosing a definition to fit their needs, they can
create a theorem which employs the definition and declare the problem solved.
That's basically what we'll do here.  There is a formal derivation based on
determining a minimum of the formula
$$P_{a,b}\big[y_i\big]\ \propto\ {1\over\sigma_y}
   e^{-(y_i-ax_i-b)^2/2\sigma_y^2};\eqno(44)$$
however, we'll approach the problem from a simpler tack and arrive at
the same result.  Now, suppose we have a point set consisting of $n$
points.  Let's call this point set $S$ and observe that it can be
written several ways
$$\eqalign{S\ &=\ \big\{\,(x_1,y_1), (x_2,y_2), \ldots, (x_n, y_n)\,\big\}\cr
  &=\ \big\{\,(x_i,y_i)\,\mid\,i=1,\ldots,n\,\big\}.\cr}$$
If we call the difference $\big| y_i - a\cdot x_i - b\big| = \sqrt{
\big(y_i - ax_i -b\big)^2}$ the deviation, or residual, of $y_i$ from
the value $a\cdot x_i + b$, then we may define
$$\dev^2\ =\ \dev^2(a,b)\ =\ \sum_{i=1}^n \left( y_i-a\cdot x_i  -b\right)^2.
   \eqno(45)$$
There are many different ways to approach this subject matter.  
We will choose the most straightforward.  If one looks in a textbook,
the derivation will the much more elaborate.   This is because
mathematicians love
to generalize things.  The line $y = f(x) = ax + b$ is also a point
set (not finite) and sometimes it will be written as
$$L\ =\ \big\{\,(x,y)\,\mid\,y=a\cdot x + b,\ x\in{\bf R}\,\big\}.$$
In any case, we simply want to {\sl minimize\/} the value for
$\dev^2(a,b)$.  In so doing, we will determine $a$ and $b$ in terms of
the values of $x_i$, $y_i$, and the integer $n$.  There's no way out
of taking a partial derivative here.  Just take it on faith that the
minimum occurs exactly when
$${\partial \dev^2(a,b)\over \partial a} = {\partial \dev^2(a,b)\over\partial b}
= 0.\eqno(46)$$
We compute
$${\partial \dev^2(a,b)\over\partial a} 
= \sum_{i=1}^n -2\big(y_i-ax_i-b\big)\cdot x_i\eqno(47)$$
and
$${\partial \dev^2(a,b)\over\partial b} 
= \sum_{i=1}^n -2\big(y_i-ax_i-b\big).\eqno(48)$$ 
Assume that equations (46), (47), and (48) are 
correct.\footnote{$^{30}$}{for the derivation, see John R. Taylor, 
{\it An Introduction to Error Analysis}, 
(Mill Valley, CA: University Science Books, 1982), page 156}
Let's go through the algebra (carefully) and see what happens.
Setting equation (47) equal to zero and expanding gives
$$\sum_{i=1}^n x_iy_i\ - a\,\sum_{i=1}^n x_i^2
    \ -\ b\,\sum_{i=1}^n x_i\ =\ 0.\eqno(49)$$
Setting equation (48) equal to zero and expanding gives
$$\sum_{i=1}^n y_i\ -\ a\,\sum_{i=1}^n x_i
  \ -\ b\,\sum_{i-1}^n 1\ =\ 0.\eqno(50)$$
A very common error is not realizing that 
$$\sum_{i=1}^n 1\ =\ n.$$
We have been writing $\sum_{i=1}^n$ again and again.  Let's simplify
our notation by agreeing that whenever we use $\sum_{i=1}^n$ we can
just write capital sigma, $\sum$.  Equations (49) and (50) can now
we re-written as
$$\sum x_iy_i\ - a\,\sum x_i^2
    \ -\ b\,\sum x_i\ =\ 0\eqno(49')$$
and
$$\sum y_i\ -\ a\,\sum x_i
  \ -\ nb\ =\ 0.\eqno(50')$$
We want to solve for the values of $a$ and $b$.  To solve for $a$, we
eliminate $b$.  That is done by multiplying equation (49') by $n$ and
by multiplying equation (50') by $\sum x_i$ and subtracting one from
the other (clever). Let's do just that
$$\eqalign{n\sum x_iy_i\ - na\,\sum x_i^2
    \ -\ nb\,\sum x_i\ &=\ 0\cr
\sum y_i\sum x_i\ -\ a\,\left(\sum x_i\right)^2
  \ -\ nb\sum x_i\ &=\ 0.\cr
a\,\left(\left(\sum x_i\right)^2-n\,\sum x_i^2\right)
 \  &=\ \sum x_i\sum y_i\ -\ n\,\sum x_iy_i \cr}$$
So,
$$ a\ =\ {  n\sum x_iy_i - \sum x_i\sum y_i
        \over n\sum x_i^2 - \big(\sum x_i\big)^2 }.\eqno(51)$$
Now all we'll have to do is to plug the value for $a$ into either
(49') or (50') to solve for $b$.  (50') is easier.
$$\eqalign{\sum y_i - a\sum x_i - nb \ &=\ 0,\cr
 \sum y_i\ -\ { n\sum x_iy_i- \sum x_i\sum y_i \over \
  n\sum x_i^2 - \left(\sum x_i\right)^2 } \sum x_i\ -\ nb\ &=\ 0,\cr
n \sum y_i \sum x_i^2 - \sum y_i\left(\sum x_i\right)^2 
- n\sum x_i\sum x_iy_i\qquad   \cr
 +\ \sum x_i\sum x_i\sum y_i 
    -nb\left(n\sum x_i^2 -\left(\sum x_i\right)^2\right)
  \ &=\ 0,\cr
n\sum y_i\sum x_i^2 -n \sum x_i\sum x_iy_i - nb \Delta\ &=\ 0,\cr}$$
where
$$\Delta\ =\ n\sum x_i^2 -\left(\sum x_i\right)^2.$$
Thus,
$$ b\ =\ {\sum y_i\sum x_i^2 - \sum x_i\sum x_iy_i\over
   n\sum x_i^2 -\left(\sum x_i\right)^2 }.\eqno(52)$$
$$ a\ =\ {n\sum x_iy_i- \sum x_i\sum x_i \over \Delta}\eqno(51')$$
$$ b\ =\ {\sum y_i\sum x_i^2 - \sum x_i\sum x_iy_i \over \Delta}\eqno(52')$$ 
This is exactly what we saw in the BASIC computer code:
{\tt\obeylines\parindent=0pt\ttraggedright
230 DELTA = CDBL(N) * SUMX2 - SUMX * SUMX
240 A = (SUMX2 * SUMY - SUMX * SUMXY) / DELTA
250 B = (CDBL(N) * SUMXY - SUMX * SUMY) / DELTA
260 CLS : PRINT "A (x-intercept) = "; A
270 PRINT "B (slope) = "; B
}
\smallskip\noindent
(Here {\tt A} and {\tt B} stand for $b$ and $a$, respectively!  {\tt DELTA}
is still $\Delta$.)
\vfill\eject
\noindent{\bf\llap{3.3\quad}Goodness of Fit.}  We saw how to derive the
least squares line of best fit, now the question is ``just how good of
a fit is it?''  Two things are needed: (1) a correlation coefficient;
and (2) a quantitative measure of the degree of correlation for a
given sample size.  The formula for the correlation coefficient is
simply
$$r\ =\ {\sum_{i=1}^n \big(x_i-\mu_x\big)\big(y_i-\mu_y\big)\over
  \left(\sum_{i=1}^n \big(x_i-\mu_x\big)^2
  \sum_{i=1}^n\big(y_i-\mu_y\big)^2 \right)^{1/2} }.\eqno(53)$$
This correlation coefficient is also called the linear correlation
coefficient or {\it Pearson's\/} $r$.  It turns out that $r$ is a
number lying in the interval $[-1, 1]$.  The closer $r$ is to
$\pm 1$, the more closely the data points lie on a straight line.
On the other hand, a value of $r$ near zero implies little or no
linear correlation.  But this is vague.  What does one mean by
saying ``$|r|$ lies close to 1?''  There must be some statistical
test to determine goodness of fit quantitatively.  There is!  We can
calculate the probability that a coefficent of linear correlation
will be greater than some given value, say $r_0$, by the
integral\footnote{$^{31}$}{E. M. Pugh and G. H. Winslow, {\it The
Analysis of Physical Measurement\/} (Addison-Wesley, 1966), 
Section 12--8.}
$$P_N\left(|r|\ge |r_0|\right)
\ =\ {2\Gamma\left[\big(N-1\big)/2\right] \over \sqrt{\pi}
  \Gamma\left[\big(N-2\big)/2\right] }\,
  \int\limits_{|r_0|}^1 \left(1-r^2\right)^{(N-4)/2}\,dr.\eqno(54)$$
But how to evaluate this integral?  We might look for a table of
values.  But relax, if we employ our Simpson's rule we can solve
this integral accurately and quickly.  A BASIC program is listed
below:
\smallskip
{\tt\obeylines\parindent=0pt\ttraggedright
100 DEFDBL A-H, O-Z
110 INPUT "Number of points, Correlation Coefficient "; P, R
130 IF (P < 4) THEN PRINT "Number of points too small": GOTO 110
140 DEF FNA (x AS DOUBLE) = (1 - x * x) {\char94} ((P - 4!) / 2!)
150 A = 0!: B = 1!: N = 256: SUM = 0!: H = (B - A) / N
160 FOR I = 1 TO N STEP 2
170 SUM = SUM + FNA(A + (I - 1) * H)
180 SUM = SUM + 4! * FNA(A + I * H)
190 SUM = SUM + FNA(A + (I + 1) * H)
200 NEXT I: REM N = 256 gives accuracy to 5 decimal places.
220 PSUM = SUM * H / 3!
240 A = R: B = 1!
250 SUM = 0!:  H = (B - A) / N
260 FOR I = 1 TO N STEP 2
270 SUM = SUM + FNA(A + (I - 1) * H)
280 SUM = SUM + 4! * FNA(A + I * H)
290 SUM = SUM + FNA(A + (I + 1) * H)
300 NEXT I
310 RSUM = 100! * (SUM * H / 3!) / PSUM
320 PRINT USING "Probability |r| >= r\_0 is \#\#.\#\%"; RSUM
}
\smallskip\noindent
But where are all the $\Gamma()$'s and the coefficient 
$2\big/\sqrt{\pi}$? What
we are doing is normalizing the value of the integral by observing
that 
$$P_N\big( |r| \ge 0 \big) \ =\ 1.$$
This way the coefficient,
$${2\Gamma\left[\big(N-1\big)/2\right] \over \sqrt{\pi}
  \Gamma\left[\big(N-2\big)/2\right] }$$
does not have to be computed explicitly.\footnote{$^{32}$}{This is nice
because $\Gamma\big(m+{1\over2}\big)= {1\cdot3\cdot5\cdots(2m-1)\over
2^m}\,\sqrt{\pi}$ for $m=1,2,3,\dots$.}  Working through this program
will give us a deeper insight as to why numerical integration is
important and how it ties in to Probability Theory.  In fact, this
whole topic of calculation of probabilities for correlation coefficients
is one frequently misunderstood and confused by working scientists
and engineers.
\medskip
\noindent  The ``C'' program is listed below.  This program has internal
documentation in the form of comments.  A comment in ``C'' begins with
the character combination ({\tt /*}) and ends with the reverse
character combination ({\tt */}).
\medskip
{\tt\parindent=0pt\ttraggedright\obeylines
\#include <stdio.h>        /* Header for input/output subroutines. */
\#include <math.h>         /* Header for math subroutines. */
\#include <float.h>        /* Header for floating point subroutines. */
\medskip
\#define pi 3.141592653589793238462643383279     /* Accurate value for pi. */
\medskip
/* Simpson's rule for approximating integrals.
\quad a:              left endpoint
\quad b:              right endpoint
\quad fc:             pointer to function to integrate
\quad n:              number of subintervals
*/
double fc (double x);
\medskip
main()
{\char123}
double a,b,h,sum,x,y,p,r,psum;   /* In 'C' all variables must be assigned */
double p1, p2, p3;
int i, n, o;
printf("{\char92}007");           /* Sound bell in computer. */
a = (double) 0.0;         /* The normalizing integral is */
b = (double) 1.0;         /* from 0 to 1.  */
n = 256;  /* This has been tested and seen to give 5 decimal accuracy. */
printf("{\char92}nEnter number of points ");
scanf("%d",&o);
if (o < 5){\char123}
\quad printf("Too few points"); return(o);
{\char125}
p = (double) o;
h = (double) (b-a)/n;
for (i=1, sum=0.0; i<=n; i = i+ 2){\char123}
\quad p1 = fc((double) a+(i-1)*h, p);
\quad p2 = fc((double) a+i*h, p);
\quad p3 = fc((double) a+(i+1)*h, p);
\quad sum += p1 + 4.0 * p2 + p3;
{\char125}
\medskip
psum = (double) h*sum/3.0; /* This normalizes the integral */
/* For a given value of N, P\_N[|r|>=0] (must)) = 1.0 */
/* We assure this (and eliminate the need to compute some */
/* Gamma function) in computing the value in "psum." */
\medskip
printf("{\char92}nEnter correlation coefficient r = ");
scanf("%lf", &r); a = fabs((double) r);
if (a > 1) a = (double) 1.0;
\quad /* r\_0 must be between -1 and 1.  Force a to assume a value */
\quad /* between 0 and 1. */
h = (double) (b-a)/n;
for (i=1, sum=0.0; i<=n; i = i+ 2){\char123}
\quad p1 = fc((double) a+(i-1)*h, p);
\quad p2 = fc((double) a+i*h, p);
\quad p3 = fc((double) a+(i+1)*h, p);
\quad sum += p1 + 4.0 * p2 + p3;
{\char125}
\medskip
y = (double) h*sum/3.0;
printf("{\char92}nPercentage probability that N measurements");
printf("{\char92}nof two uncorrelated variables would give an r");
printf("{\char92}nas large as r\_0, %.2lf, that is ", a);
printf("{\char92}nP\_N[|r|>=r\_0] = %.2lf (per cent).", (double) 100.*y/psum);
printf("{\char92}nIf the number %.2lf is less than 5 (per cent),", 100.*y/psum);
printf("{\char92}nthen the correlation is called significant.");
printf("{\char92}nIf the number %.2lf is less than 1 (per cent),", 100.*y/psum);
printf("{\char92}nthen the correlation is call highly significant.");
printf("{\char92}nGiven this simple program and its speedy calculation");
printf("{\char92}nof the significance of a correlation in absolute terms,");
printf("{\char92}nwe only have to plug in the number of data points and,");
printf("{\char92}nthe value gotten from Pearson's correlation coefficient, r.");
printf("{\char92}n");
return(0);
{\char125}
\medskip
double fc (double x, double p)
{\char123}
\quad double y;
\quad y = (double) pow((double)(1.0-x*x),(double) (p-4.0)/2.0);
\quad return (y);
{\char125}
\medskip
/* End of file */
}
\medskip
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  Figure 24---Kolgoroff-Schmirnoff
%
$$\beginpicture
  \setcoordinatesystem units <4in,2in>
  \setplotarea x from 0 to 1, y from 0 to 1 %
  \plotheading {\lines{ Curve $f(x) = \big( 1-x^2\big)^{(N-4)/2}$,\cr
                        where $N=5$ and $|r_0| = {1\over2}$\cr } } %
  \axis left ticks numbered from 0 to 1 by .2 / %
  \axis bottom label {F{\sevenrm IGURE} 24} 
    ticks numbered from 0 to 1 by .2 / %
\ifexpressmode
  \put {\bf EXPRESSMODE} at .5 .5 %
\else  
  \setquadratic
  \linethickness=1pt
  \plot
     0.00  1.00000  0.05  0.99875  0.10  0.99499  0.15  0.98869
     0.20  0.97980  0.25  0.96825  0.30  0.95394  0.35  0.93675
     0.40  0.91652  0.45  0.89303  0.50  0.86603  0.55  0.83516
     0.60  0.80000  0.65  0.75993  0.70  0.71414  0.75  0.66144
     0.80  0.60000  0.85  0.52678  0.90  0.43589  0.925 0.37997
     0.95  0.31225  0.975 0.22220  1.00  0.00000  / %
\fi
  \setlinear
  \linethickness=0.4pt
  \putrule from  0.50 0.0  to  0.50  0.86603  
  \putrule from  0.55 0.0  to  0.55  0.83516
  \putrule from  0.60 0.0  to  0.60  0.80000
  \putrule from  0.65 0.0  to  0.65  0.75993 
  \putrule from  0.70 0.0  to  0.70  0.71414
  \putrule from  0.75 0.0  to  0.75  0.66144
  \putrule from  0.80 0.0  to  0.80  0.60000
  \putrule from  0.85 0.0  to  0.85  0.52678  
  \putrule from  0.90 0.0  to  0.90  0.43589  
  \putrule from  0.95 0.0  to  0.95  0.31225
  \putrule from 0.525 0.0  to 0.525 0.85110
  \putrule from 0.575 0.0  to 0.575 0.81815
  \putrule from 0.625 0.0  to 0.625 0.78062
  \putrule from 0.675 0.0  to 0.675 0.73782
  \putrule from 0.725 0.0  to 0.725 0.68875
  \putrule from 0.775 0.0  to 0.775 0.63196
  \putrule from 0.825 0.0  to 0.825 0.56513
  \putrule from 0.875 0.0  to 0.875 0.48412
  \putrule from 0.925 0.0  to 0.925 0.37997
  \putrule from 0.975 0.0  to 0.975 0.22220
  \put {$\scriptstyle\bullet$} at 0.5 0.0 
  \put {$|r_0|$} [t] <0pt,-3pt> at 0.5 0.0
\endpicture$$ %
%
\vfill\eject
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\noindent{\bf\llap{3.4\quad}More Theory.}  We will examine the theory behind
the ``method of least squares'' as applied to fitting a straight line
$$y = a\cdot x + b \eqno(55)$$
to the data points $(x_1,y_1)$, $(x_2,y_2)$, $\ldots$, $(x_n,y_n)$.  As before,
we might observe that associated with each value of the abscissa, that is,
associated with each $x_i$ ($1 \le i \le n$) there are two value of $y$.
There is the $y_i$ and the value $\tilde y_i = a\cdot x_i + b$.  If we
take the numerical difference of the two, $d_i = y_i - \tilde y_i$,
we get the amount the predicted value of $y$, $\tilde y$, differs from
the observed value of $y$.  Let's write out some of these differences, which
are called {\it deviations}:
$$\eqalign{ d_1\ &=\ y_1 - (a\cdot x_1 + b) \cr
            d_2\ &=\ y_2 - (a\cdot x_2 + b) \cr
            \vdots\ &\ \qquad \vdots \cr
            d_n\ &=\ y_n - (a\cdot x_n + b) \cr}$$
The set of all the $d_i$'s ($1 \le i \le n$), which is also written
$$\big\{\,d_1,\,d_2,\,\ldots,\,d_n\,\big\}$$
or
$$\{\,d_i\,\mid\,1 \le i \le n\,\}.$$
If it turns out that $d_i = 0$ for all the integers $i$,
$i = 1,2,\ldots,n$, then the straight line $y=ax+b$ passes through
each of the $n$ points.  If this is not the case, then we have to
solve the problem of finding a line of best fit.  A line of best fit
is a line which best fits the data in some sense.  One obvious
sense is to minimize the values of the deviations.  Let's look at
the situation graphically:
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
$$\beginpicture
  \setcoordinatesystem units <1in,1in>
  \setplotarea x from -.25 to 3.00, y from -.25 to 1.50
  \plotheading {Line of Best Fit}
  \axis bottom shiftedto y=0 label {F{\sevenrm IGURE} 25} /
  \axis left shiftedto x=0 /
  \put {$O$} [rt] <-2pt,-2pt> at 0 0 %
  \put {$x$} [l] <4pt,0pt> at 3.00 0.00
  \put {$y$} [b] <0pt,4pt> at 0.00 1.50
  \put {$\scriptstyle\times$} at 0.25 1.00
  \put {$(x_1,y_1)$} [lb] <4pt,6pt> at 0.25 1.00
  \put {$\scriptstyle\times$\lower6pt\hbox{$(x_2,y_2)$} } [lt] at 0.75 0.75
  \put {$\scriptstyle\times$} at 1.6 1.2
  \put {$\scriptstyle\times$\raise6pt\hbox{$(x_n,y_n)$} } [rb] at 2.85 1.45
  \put {$\scriptstyle\bullet$} at 0.25 0.83421
\ifexpressmode
\else
  \setlinear
  \plot 0.25 0.83421  0.25 1.00  / %
  \linethickness=1pt
  \plot 0.10 0.8   2.95 1.45 / %
\fi
\endpicture$$
% 10 x0 = .1
% 20 y0 = .8
% 30 x1 = 2.95
% 40 y1 = 1.45
% 50 x = .25
% 60 y = y0 + (y1 - y0) * (x - x0) / (x1 - x0)
% 70 PRINT x, y
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
If a line comes close  to some of the points, the value of the deviation from
those points will be small.  The values of the deviation for points lying
above the line is positive, while the values of the deviation for points
lying below the line is negative.  To get a realistic picture of what is
meant by the {\sl best\/} fit, we could take the absolute value; however,
it is simpler to take the squares of the deviations.  Written as a function,
this becomes
$$\dev^2 \ =\ (y_1-ax_1-b)^2 + (y_2-ax_2-b)^+ \cdots +
    (y_n-ax_n-b)^2.\eqno(56)$$
Recall the ``Big-Sigma'' notation:
$$\sum_{i=1}^n (y_i-ax_i-b)^2\ =\ (y_1-ax_1-b)^2 + (y_2-ax_2-b)^+ \cdots +
    (y_n-ax_n-b)^2.\eqno(57)$$
So, we have a function, $\dev^2$.  But what is it a function of?  Clearly,
the $x_i$'s and $y_i$'s are fixed data points.  There are only two things
to vary, the constants $a$ and $b$.  And this is a general concept for
much advanced mathematics:  ``hold the variables constant and vary the
constants.''  No where is this more often done than in thermodynamics.  So, we
write
$$\eqalign{\dev^2(a,b)\ &=\ (y_1-ax_1-b)^2 + (y_2-ax_2-b)^2+ \cdots +
    (y_n-ax_n-b)^2\cr
  &=\ \sum_{i=1}^n (y_i-ax_i-b)^2.\cr}\eqno(58)$$
Now, by definition, we will say that the line of best fit is that line
$y = ax + b$ for which the sum of the squares of the deviations is a minimum.
This is a philosophically correct definition as well.  The problem is how
to determine which values of $a$ and $b$ to choose.  The minimum value for
equation (56) can be obtained from partial differentiation.  But we have
already done that in a previous paragraph.  We solved the equations
$${\partial \dev^2(a,b) \over \partial a}\ =\ 0\qquad\hbox{and }\quad %
  {\partial \dev^2(a,b) \over \partial b}\ =\ 0.\eqno(59)$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  Least-squares plot after Thomas.
%
$$\beginpicture
  \setcoordinatesystem units <1cm,1cm>
  \setplotarea x from -.25 to 5.00, y from -.25 to 6.00
  \plotheading {Line of Best Fit}
  \axis bottom shiftedto y=0 label {F{\sevenrm IGURE} 26}
    ticks numbered from 1 to 4 by 1  /
  \axis left shiftedto x=0  ticks numbered from 1 to 5 by 1 /
  \put {$O$} [rt] <-2pt,-2pt> at 0 0 %
  \put {$x$} [l] <4pt,0pt> at 5.00 0.00
  \put {$y$} [b] <0pt,4pt> at 0.00 6.00
  \put {$\scriptstyle\times$} at 0.0 1.0
  \put {$\scriptstyle\times$} at 2.0 2.0
  \put {$\scriptstyle\times$} at 1.0 3.0
  \put {$\scriptstyle\times$} at 3.0 4.0
  \put {$\scriptstyle\times$} at 4.0 5.0
  \put {$(x_1,y_1)$} [l] <4pt,0pt> at 0.0 1.0
  \put {$(x_2,y_2)$} [t] <0pt,-4pt> at 2.0 2.0
  \put {$(x_3,y_3)$} [b] <0pt,4pt> at 1.0 3.0
  \put {$(x_4,y_4)$} [rb] <-2pt,2pt> at 3.0 4.0
  \put {$(x_5,y_5)$} [rb] <0pt,4pt> at 4.0 5.0
\ifexpressmode
\else
  \setlinear
  \linethickness=1pt
  \plot 0.0 1.2  5.0 5.7  / %
\fi
\endpicture$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The data points belong to the set $\{\,(0,1),\,(2,2),\,
(1,3),\,(3,4),\,(4,5)\,\}$.  The equation of the line of
best fit is $y = 0.9x + 1.2$.  Since the data values are integers
and the number of points is small, it is possible to evalute the
deviation function directly.\footnote{$^{33}$}{George B. Thomas,
{\it Calculus and Analytic Geometry}, (Reading, Massachusetts:
Addison-Wesley Publishing Company, Inc., 1966), page 699.}
All of this can be generalized to a polynomial $p(x)$ of degree
$n \ge 2$.  We simply write $y=a_0+a_1x+a_2x^2+\cdots+a_nx^n$ to get
$$\dev^2(a_0,\ldots,a_n)
\ =\ \sum_{i=1}^n \big(y_i -a_0 -a_1x -\cdots-a_nx_i^n\big).\eqno(60)$$
Then it is necessary to minimize $\dev^2({\bf a})$, where 
(${\bf a} = (a_0,a_1,\ldots,a_n)$), as follows:
$${\partial \dev^2(a_0,\ldots,a_n) \over \partial a_i}\ =\ 0.\eqno(61)$$
\vfill\eject
\noindent{\bf\llap{3.5\quad}Problems.}  In this paragraph we will work
problems.\footnote{$^{34}$}{Erwin Kreyszig, {\it Advanced Engineering
Mathematics, 5th Edition}, (NY: John Wiley \& Sons, 1983), pages 820-821.}
One very important consideration is the plotting of the data.
With the digital computer, the very first thing a scientist or engineer 
should do is plot the data.  This ensures proper scaling and gives an
intuitive feeling for the behavior of the line of best fit.  Moreover,
unusual characteristics (such as logarithmic dependence, non-linear
behavior, or sectional continuity) can be determined before the machinery
of the least-squares method comes into play.
\medskip\noindent
The first data set: $\{\,(5,10),\,(10,8.9),\,(15,8.2),\,(20,7.0)\,\}$.  Since
the $x$-values are given as integers, we will assume that $\sigma_x = 0$ for
this problem.  The only error-bar needed is parallel to the $y$-axis.  First
plot the data.  Ten points to the inch look like a good scale for the
$x$-axis and one-fourth inch per unit looks good for the $y$-axis.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
$$\beginpicture
  \setcoordinatesystem units <.1in,.25in>
  \setplotarea x from 5 to 20, y from 5 to 10 %
  \axis bottom label {F{\sevenrm IGURE}  27} ticks numbered
     from 5 to 20 by 5 /
  \axis left ticks numbered from 5 to 10 by 1 /
  \put {$\scriptstyle\bullet$} at 5 10
  \put {$\scriptstyle\bullet$} at 10 8.9
  \put {$\scriptstyle\bullet$} at 15 8.2
  \put {$\scriptstyle\bullet$} at 20 7.0
\endpicture$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Now, we run the program and get the following values for the data on the line
of best fit: $\{\,(5,9.98),\,(10,9.01),\,(15,8.04),\,(20,7.07)\,\}$.  The
equation for the line of best fit is
$$y = -0.194 x + 10.949 \eqno(62)$$
and the correlation coefficient is $r = -.99546$.
This looks like it will be a good linear fit.  Let's plot it now.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
$$\beginpicture
  \setcoordinatesystem units <.1in,.25in>
  \setplotarea x from 0 to 25, y from 5 to 12 %
  \axis bottom label {F{\sevenrm IGURE}  28} ticks numbered
     from 0 to 25 by 5 /
  \axis left ticks numbered from 5 to 12 by 1 /
  \put {$\scriptstyle\bullet$} at 5 10
  \put {$\scriptstyle\bullet$} at 10 8.9
  \put {$\scriptstyle\bullet$} at 15 8.2
  \put {$\scriptstyle\bullet$} at 20 7.0
\ifexpressmode
\else
  \setlinear
  \linethickness=1pt
  \plot 0 10.949  25 6.099 /
\fi
\endpicture$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
This graph looks OK without error bars.  They would just clutter up the
picture, so let's not put them in.  The correlation coefficient is nearly
$-1$, there is no need to do a goodness of fit analysis.  In short, this
data set very closely approximates a linear relationship {\sl over the region
in question}.  We cannot, however, extrapolate outside of the extreme
values of the data;  our only conclusions are valid within the specified
range of data values.  This is a very important point.  Notice that the
line of best fit has been drawn across the entire graph.  This is standard
mathematical practice;  otherwise, the equation would not represent a
function---a graph of a function from the $x$-axis.  But this mathematical
practice should not obscure the underlying assumptions.
\vfill\eject
%
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent
The next data set has a different appearance.  The data points are
$\{\,(4,-17)$, $(15,-4)$, $(30,-7)$, $(100,50)$,
$(200,70)\,\}$.  We will choose
one printer's point per data point.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%  Kreyszig problem 3
%
$$\beginpicture
  \setcoordinatesystem units <1pt,1pt>
  \setplotarea x from 0 to 200, y from -20 to 100
  \axis bottom shiftedto y=0 label {F{\sevenrm IGURE} 29} / %
  \axis left /
  \put {$\scriptstyle\circ$} at 4 -17
  \put {$\scriptstyle\circ$} at 15 -4
  \put {$\scriptstyle\circ$} at 30 -7
  \put {$\scriptstyle\circ$} at 100 50
  \put {$\scriptstyle\circ$} at 200 70
\endpicture$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
After running the computer program, the following data along the line of best
fit is obtained:
$$\{\,(4,-11.67),(15,-6.65),(30,0.21),(100,32.20),(200,77.91)\,\}.\eqno(63)$$
The equation of the line of best fit and the correlation coefficient $r$
are
$$y = 0.457 x - 13.503 \quad\hbox{and}\quad r = .96074.\eqno(64)$$
Now, we will plot the data against the curve.  It is now easy to
scale the data and put in numbered tick marks.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
$$\beginpicture
  \setcoordinatesystem units <1pt,1pt>
  \setplotarea x from 0 to 250, y from -20 to 100
  \axis bottom shiftedto y=0 label {F{\sevenrm IGURE} 30 }
    ticks numbered from 50 to 250 by 50 /
  \axis left ticks numbered from -20 to 100 by 20 /
  \put {$\scriptstyle\circ$} at 4 -17
  \put {$\scriptstyle\circ$} at 15 -4
  \put {$\scriptstyle\circ$} at 30 -7
  \put {$\scriptstyle\circ$} at 100 50
  \put {$\scriptstyle\circ$} at 200 70
\ifexpressmode
\else
  \setlinear
  \linethickness=1pt
  \plot 0  -13.503    250 100.747  /
\fi
\endpicture$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Now we run the computer program to get the value for $\sigma_y$ and see how
good our fit is.  $\sigma_y = 12.47154$ and the probability that a random
choice of values would give the same correlation coefficient is 0.9\%---we
have a highly significant correlation.  So, let's put in the error bars to
see just how this graph looks in its final form.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$$\beginpicture
  \crossbarlength=5pt
  \setcoordinatesystem units <1pt,1pt>
  \setplotarea x from 0 to 250, y from -20 to 100
  \axis bottom shiftedto y=0  label {F{\sevenrm IGURE} 31}
    ticks numbered from 50 to 250 by 50 /
  \axis left ticks numbered from -20 to 100 by 20 /
\ifexpressmode
  \put {$\scriptstyle\circ$} at 4 -17
  \put {$\scriptstyle\circ$} at 15 -4
  \put {$\scriptstyle\circ$} at 30 -7
  \put {$\scriptstyle\circ$} at 100 50
  \put {$\scriptstyle\circ$} at 200 70
\else
  \setlinear
  \linethickness=1pt
  \plot 0  -13.503    250 100.747  /
  \puterrorbar at 4 -17 with fuzz 12.47154
  \puterrorbar at 15 -4 with fuzz 12.47154
  \puterrorbar at 30 -7 with fuzz 12.47154
  \puterrorbar at 100 50 with fuzz 12.47154
  \puterrorbar at 200 70 with fuzz 12.47154
\fi
\endpicture$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vfill\eject
\noindent
Now things are starting to look routine.  We will repeat this procedure
again and again to ensure that the idea is clear.  Again, let's start with
a data set
$$\{\,(2.8,30), (2.9,26), (3.0,33), (3.1,31), (3.2,33),
   (3.2,35), (3.2,37), (3.3,36), (3.4,33)\,\}.\eqno(65)$$
Notice that one $x$-value has several $y$-values.  This won't cause the
least-square program to have any problems.  First the graph
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%  Kreyszig, page 821, problem 5
%
$$\beginpicture
  \setcoordinatesystem units <10cm,3mm>
  \setplotarea x from 2.8 to 3.4, y from 26 to 38
  \axis bottom label {F{\sevenrm IGURE}  32} ticks
    numbered from 2.8 to 3.4 by 0.2 /
  \axis left ticks numbered from 26 to 38 by 2 /
  \put {$\scriptstyle\times$} at 2.8 30
  \put {$\scriptstyle\times$} at 2.9 26
  \put {$\scriptstyle\times$} at 3.0 33
  \put {$\scriptstyle\times$} at 3.1 31
  \put {$\scriptstyle\times$} at 3.2 33
  \put {$\scriptstyle\times$} at 3.2 35
  \put {$\scriptstyle\times$} at 3.2 37
  \put {$\scriptstyle\times$} at 3.3 36
  \put {$\scriptstyle\times$} at 3.4 33
\endpicture$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
We execute the computer program and get a line of best fit with an
equation
$$y = 12.06767 x - 5.01128 \quad\hbox{and}\quad r = .69155.\eqno(66)$$
With nine points and a value of the correlation coefficient of only
$0.69$ just how significant is our data? ---3.9\%---significant (under 5\%)
but not highly significant.  Now the finished plot, with error bars:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$$\beginpicture
  \setcoordinatesystem units <10cm,3mm>
  \setplotarea x from 2.5 to 3.5, y from 25 to 40
  \crossbarlength=5pt
  \axis bottom label {F{\sevenrm IGURE}  33} ticks
    numbered from 2.5 to 3.5 by 0.5 /
  \axis left ticks numbered from 25 to 40 by 5 /
\ifexpressmode
  \put {$\scriptstyle\times$} at 2.8 30
  \put {$\scriptstyle\times$} at 2.9 26
  \put {$\scriptstyle\times$} at 3.0 33
  \put {$\scriptstyle\times$} at 3.1 31
  \put {$\scriptstyle\times$} at 3.2 33
  \put {$\scriptstyle\times$} at 3.2 35
  \put {$\scriptstyle\times$} at 3.2 37
  \put {$\scriptstyle\times$} at 3.3 36
  \put {$\scriptstyle\times$} at 3.4 33
\else
  \puterrorbar at 2.8 30  with fuzz  2.59
  \puterrorbar at 2.9 26  with fuzz  2.59
  \puterrorbar at 3.0 33  with fuzz  2.59
  \puterrorbar at 3.1 31  with fuzz  2.59
  \puterrorbar at 3.2 33  with fuzz  2.59
  \puterrorbar at 3.2 35  with fuzz  2.59
  \puterrorbar at 3.2 37  with fuzz  2.59
  \puterrorbar at 3.3 36  with fuzz  2.59
  \puterrorbar at 3.4 33  with fuzz  2.59
  \setlinear
  \linethickness=1pt
  \plot 2.5 25.16805  3.5 37.23572 / %
\fi
\endpicture$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
Now try your hand at the data set $\{\,(0,100)$, $(3,130)$, $(5,140)$,
$(8,170)$, $(10,190)\,\}$, which will give $y=8.8216 x + 100.12739$
and have a correlation coefficient of $r= +0.99666$.  What is $\sigma_y$?
This correlation is significant; is it highly significant?
\vfill\eject
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%                                                                     %%%%%
%%%%%                          Chapter 4                                  %%%%% 
%%%%%                                                                     %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\headline={\tenrm\hfill Adaptive Quadrature Routine}
\centerline{\bf Chapter 4}
\bigskip
\noindent{\bf\llap{4.0\quad}Introduction.}  Sometimes it's not easy to tell
just how many subdivisions are needed to do a numerical integration. 
The reasons for this uncertainty are many.
Functions exhibiting bizarre behavior at a point or in a 
subinterval require special attention.  
Some continuous functions fail to have derivatives.
Moreover,  a function can be integrable
even if it is not be continuous at every point. 
For these cases and many others it's nice to
have some kind of automatic system for determining just when an approximation
is ``good enough.''  Fast modern computers may still be too slow.
In particular, if a function is
defined in terms of an integral its numerical 
evaluation may be time consuming; moreover, functions so defined may
be prone to significant error propagation.
In general, a function defined over an interval may be easier to integrate
over some subintervals than over others.  The methods which have been 
devised to deal with this phenomenon are call {\sl adaptive quadrature
routines}, or {\sl AQRs}.  Adaptive methods make use of error estimates to
automatically determine the number of subintervals, $n$, and
the interval size $h$.  These methods may also use smaller values of the
interval size $h$ over subintervals where it is needed.  There are many
systematic ways of doing this.  There is one method which is easily 
programmed (in BASIC), but which is theoretically involved.  This method
entails comparing Simpson's rule with the so-called trapezoid rule.  Another
method, not so easily programmed and requiring recursion, compares two 
different estimates from Newton-C\^otes' formulas.  In the following
paragraphs, we'll develop computer programs for the simplest 
computers---the programmable calculators---and compare results.
\bigskip
\bigskip
\hrule
\bigskip
$$\beginpicture
  \setcoordinatesystem units <1in,.2in>
  \setplotarea x from -.25 to 3.25, y from -5 to 6
  \plotheading {\lines{ An Adaptive Quadratic Routine,
                        yielding unequal intervals, to integrate\cr
                        the function $f(x)=2x\,\sin x^2$
                        over the interval $[0,\pi]$.\cr}}
  \axis bottom shiftedto y=0  / %
  \axis left shiftedto x=0 ticks withvalues
    {$-5$} {$-4$} {$-3$} {$-2$} {$-1$} {$1$} {$2$} {$3$} {$4$} {$5$} / %
    at -5 -4 -3 -2 -1 1 2 3 4 5 / / %
  \put {$x$} at 3.35 0
  \put {$y=2x\,\sin(x^2)$} [l] at 0.1 6.0
  \putrule from 0.398 0.0  to  0.398  0.12556
  \putrule from 0.635 0.0  to  0.635  0.49833
  \putrule from 0.832 0.0  to  0.832  1.06205
  \putrule from 1.012 0.0  to  1.012  1.72904
  \putrule from 1.193 0.0  to  1.193  2.36008
  \putrule from 1.439 0.0  to  1.439  2.52579
  \putrule from 1.586 0.0  to  1.586  1.85901
  \putrule from 1.717 0.0  to  1.717  0.66035
  \putrule from 1.843 0.0  to  1.843 -0.92998
  \putrule from 1.976 0.0  to  1.976 -2.73115
  \putrule from 2.143 0.0  to  2.143 -4.25521
  \putrule from 2.255 0.0  to  2.255 -4.20048
  \putrule from 2.357 0.0  to  2.357 -3.13566
  \putrule from 2.444 0.0  to  2.444 -1.49136
  \putrule from 2.545 0.0  to  2.545  0.98048
  \putrule from 2.699 0.0  to  2.699  4.54638
  \putrule from 2.795 0.0  to  2.795  5.58508
  \putrule from 2.880 0.0  to  2.880  5.21203
  \putrule from 2.962 0.0  to  2.962  3.59141
  \putrule from 3.045 0.0  to  3.045  0.92665
  \putrule from 3.14159 0.0 to 3.14159 -2.70366
\ifexpressmode
  \put {\bf EXPRESSMODE} at 1.5 2.0
\else
  \linethickness=1.5pt
  \setquadratic
  \plot 0 0  .398 .12556  .635 .49883  .832 1.06205  1.012 1.72904
    1.193 2.36008  1.439 2.52579  1.586 1.95901  1.717 .66035
    1.843 -.92998  1.976 -2.73115  2.143 -4.25521  2.255 -4.20048
    2.357 -3.13566  2.444 -1.49136  2.545 .98048  2.699 4.54638
    2.795 5.58508  2.880 5.21034  2.962 3.59141  3.045 .92665
    3.07 0  3.14159 -2.70366 / %
\fi
\endpicture$$
\centerline{\rm F{\sevenrm IGURE} 34}
\vfill\eject
\noindent{\bf\llap{4.1\quad}A Simple Approach.}  The best introduction to this
subject is to work simple examples and then develop the theory as needed. 
Excellent numerical results can be obtained using a very simple approach.
We are given a function, $f(x)$, an interval $[\alpha,\,\beta\,]$, and an
error bound $\epsilon>0$.
First, approximate the integral with two (different) approximations:
the trapezoid rule and Simpson's rule.  Begin with $n=10$ subintervals.
Second, compute the error estimate (relative) as follows
$$E\ =\ \left| {A_T - A_S \over A_T + A_S} \right|, \eqno(67)$$
where $A_T$ is the approximation obtained from the trapezoid rule and
$A_S$ is the value from Simpson's rule.  If $E < \epsilon$, then the
results are OK.  If not, then set $n=20$ ($n=2\cdot n$) subintervals
and repeat the process.  Continue until either $E < \epsilon$ or until $n$
gets too large, say $n > 1000$.  This is easy to program and fast
to run.  The first prototype would look like this:
\medskip
{\tt\parindent=0pt\obeylines
10 \ DEF FNA(X) = SQR(1.0-SIN(EXP(1)/2){\char94}2*SIN(X){\char94}2)
20 \ N=10 : A=0.0 : B=2.0*ATN(1.0)  
30 \ H=(B-A)/N : S = 0.0 : T = 0.0
40 \ FOR I=1 TO N STEP 2
50 \ X=A+(I-1)*H : S = S + FNA(X) : T = T + FNA(X)
60 \ X=A+I*H : T = T + 4.0*FNA(X)
70 \ X=A+(I+1)*H : S = S + FNA(X) : T = T + FNA(X)
80 \ NEXT I : S1 = H*S : S2 = H*S/3.0 : E = ABS((S1-S2)/(S1+S2))
90 \ IF (E>0.000001) THEN N = 2*N : GOTO 30
100 PRINT "Integral = "; T, N
}
\medskip\noindent
The answer is given immediately as
\smallskip
{\tt\parindent=0pt\obeylines
Integral = 1.054686 \qquad 20
}
\smallskip\noindent
From our previous calculations, we know that this is precisely the
correct number of subintervals for this precision.  It is a good idea
to examine an alternative means of coding the same program.  It runs
faster if we use a {\tt GOSUB} construction instead of the defined
function.  The answers will be identical and we can consult the table
prepared in paragraph 1.2 for numerical results.
\medskip
{\tt\parindent=0pt\obeylines
10 \ D=0.0001 : INPUT "A,B = ";A,B : N=10
20 \ N=N+10 : H=(B-A)/N : X=A : S=0.0 : T=0.0  
30 \ FOR I=1 TO N STEP 2
40 \ GOSUB 100 : S = S + Y*H/3 : T = T + Y*H/2 : X = X + H
50 \ GOSUB 100 : S = S + 4*Y*H/3 : T = T + Y*H : X = X + H
60 \ GOSUB 100 : S = S + Y*H/3 : T = T + Y*H/2 : NEXT I
70 \ IF ABS((S-T)/(S+T)) > D THEN 20
80 \ PRINT "Integral = "; T, N
90 \ PAUSE : INPUT "Error \% = ";D : GOTO 20
100 Y = SQR(1.0-SIN(EXP(1)/2){\char94}2*SIN(X){\char94}2) : RETURN
}
\medskip\noindent
This program gives one the ability to increase the error estimate and
thereby determine closely the exact number of steps required.
Using this program with a hand-held calculator will give a very
good qualitative insight into the workings of the algorithm.  The 
interested reader will later be invited to experiment with other
adaptive quadrature routines to determine which is better.
\vfill\eject
\noindent{\bf\llap{4.2\quad}An Adaptive Quadrature Routine.}  Suppose that we
have a function $f(x)$ defined over an interval $[a,b]$.  We would like to
compute the integral close to the actual value.  Let's assume that we know
that the function is bounded, $|f(x)| \le M$ for all $x\in[a,b]$.  Then, for
an answer within say 0.01\% of the actual error, we would choose
$$\epsilon = {0.0001 \over M + 1}.\eqno(68)$$
With an overall error of $\epsilon$, we want to divide up the interval
$[a,b]$ so that the total error of integration is less than $\epsilon$.
There is an error formula which depends on the value of the fourth 
derivative of $f(x)$ on $[a,b]$, but we will depend on another criterion.
If we choose Simpson's rule
$$A_2 = {h\over3}\big( y_0 + 4y_1 + 2y_2 + 4y_3 + y_4\big),\eqno(69)$$
then the doubled interval rule
$$A_1 = {2h\over3}\big( y_0 + 4y_2 + y_4\big),\eqno(70)$$
is a good approximant.  The estimate of the error is\footnote{\dag}{Francis
Scheid, {\it Schaum's Outline Series Theory and Problems of Numerical
Analysis 2nd edition}, (NY: McGraw-Hill Book Company, 1988), page 124.}
$$E_2 \simeq {|A_2-A_1| \over 15 }.\eqno(71)$$
This follows from a comparison of the two error estimates:
$$E_1 = -{(b-a)(2h)^4y^{(4)}(\xi_1)\over180}\qquad
  E_2 = -{(b-a)h^4y^{(4)}(\xi_2)\over180},\eqno(72)$$
where $E_1$ is the error estimate for Simpson's rule with an interval length
of $2h$ and $E_2$ is the error estimate for Simpson'r rule with interval
length of $h$.
We can accept a value of $A_2$ whenever $|A_2-A_1| \le 15\epsilon/2^k$
is reached, for an interval being halved $k$ times.
An alternate derivation\footnote{\ddag}{Mike Stimpson, ``Numerical Integration
Using Adpative Quadrature,'' {\it The C Users Journal}, May 1992, page 35.}
supposes that $A$ is the actual value of the integral and requires
$E_1 = |A-A_1|$, $E_2 = |A-A_2|$, where $A_1$ is Simpson's rule and $A_2$ is
the so-called five-point approximation (Simpson's rule with interval length
of $h$).  From the triangle inequality and the inequality
$$E_2 \le {1\over16}\,E_1,\eqno(73)$$
the result follows.  The programming makes use of a technique known as
{\sl recursion}.  Recursion occurs when a computer subroutine calls
itself.  The situation in analysis occurs when a function, say $f(x)=\sin(x)$,
is used to define $g(x) =\sin\big(\sin(x)\big)=f\big(f(x)\big)$. Now we will
examine a ``C'' program which uses recursion to do adaptive quadrature:
\medskip
{\tt\obeylines\parindent=0pt
\#include <math.h>
\#include <float.h>
\#include <stdio.h>
double adaptive (double a, double b, double (*f)(double x), double *err);
double f(double x);
int main() {\char123}
\ double a,b,*err,int1;
\ a = (double) 0.0;
\ b = (double) 2.0*atan(1.0);
\ *err = (double) 0.00000001;
\ printf("{\char92}nleft endpoint \ = \%.16lf",a);
\ printf("{\char92}nright endpoint = \%.16lf",b);
\ printf("{\char92}nerror estimate = \%.16lf",*err);
\ int1 = adaptive(a, b, f, err);
\ printf("{\char92}nIntegral \ \ \ \ \ \ = \%.16lf",int1);
\ printf("{\char92}nTheoretical \ \ \ = \%.16lf",1.0);
\ return(0);
{\char125}
double f(double x) {\char123}
\  double y;
\  y = sin(x);
\  return (y);
{\char125}
double adaptive (double a, double b, double (*f)(double x), double *err) {\char123}
\  double h, s1, s2, s3, s4, s5, error, err1, err2, t1, t2;
\  h = b - a;
\  s1 = (*f)(a);  s2 = (*f)(a+0.25*h);  s3 = (*f)(a+0.5*h);
\  s4 = (*f)(b-0.25*h);  s5 = (*f)(b);
\  t1 = h*(s1+4.0*s3+s5)/6.0;
\  t2 = h*(s1+4.0*s2+2.0*s3+4.0*s4+s5)/12.0;
\  error = fabs(t1-t2)/15.0;
\  if (error < *err) {\char123}
\quad    *err = error;  return(t2);
\quad {\char125}
\  else {\char123}
\quad   err1 = err2 = *err/2.0;  t2 = adaptive(a, a+0.5*h, f, \&err1);
\quad    t2 += adaptive(a+0.5*h, b, f, \&err2);  return(t2);
\quad  {\char125}
{\char125}
/* End of File */
}
\medskip
\noindent The output of the program consists of the endpoints, the error
estimate (maximum), the computed answer and the theoretical answer.  Here, we
know that $\int_0^{\pi/2} \sin(x)\,dx = 1$.  The quadrature should return an
answer close to 1; we see that it does.
\medskip
{\tt\obeylines
left endpoint \ = 0.0000000000000000
right endpoint  = 1.5707963267948970
error estimate  = 0.0000000100000000
Integral \ \ \ \ \ \ = 1.00000000025973520
Theoretical \ \ \ = 1.0000000000000000
}
%
\medskip
\noindent While the adaptive quadrature routine (AQR) is impressive, it often
fails to evaluate improper integrals.  If the integrand of an improper
integral experiences an anomoly (``blows up'') at a point, it is necessary
to compute successive approximations for values approaching that point.  There
is no ``quick and easy'' solution to improper integrals.  One cardinal rule is
to graph the function first.  It is frequently possible to determine the
correct approach from a graph when so-called adaptive methods fail.
\vfill\eject
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%                                                                     %%%%%
%%%%                   Chapter 5---Metafont                              %%%%%
%%%%                                                                     %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\headline{\tenrm\hfill METAFONT}
\centerline{\bf Chapter 5}
\bigskip
\noindent{\bf\llap{5.0\quad}Introduction.}  The creation of fonts is
usually something not approached lightly.  With the typesetting software of
\TeX\ this is especially true.  The problem of merging text with graphics
has been resolved in the earlier chapters using the macros of \PiCTeX, with
varying degrees of success.  Now it is necessary to investigate the more
sophisticated means of doing graphics.  \PiCTeX\ can be compared to assembly
language; {\bf METAFONT} to machine code.  \PiCTeX\ is straightforward and
easy to use; however, it uses a great deal of randon access memory (RAM) and
it is slow.  {\bf METAFONT}, on the other hand, creates a font which is 
read in directly.  It uses little memory and is fast.  There is a price which
must be paid for using {\bf METAFONT}---it has to be prepared in its own
language and pre-processed through its own processor.
\medskip
\noindent
The public domain version of {\bf METAFONT} does not include a user-friendly
view program.  It is necessary to produce files and then process them through
a \TeX\ program named {\tt testfont.tex}.  The program {\tt testfont.tex} is
well-written and clever.  It produces excellent output; however, one must
have a commercial previewer to use it on-line.  Otherwise, the output must be
printed on a printer.  The public domain previewer, {\tt cdvi12.arc}, only
employs the sixteen basic \TeX\ fonts; {\bf METAFONT} actually produces 
{\sl new\/} fonts---that is, the graphics from {\bf METAFONT} are fonts.  Each
graph is actually a single letter or blocks of letters ``tiled'' together.
Donald E. Knuth is correct in saying, ``Geometric design are rather easy;
$\dots$.''  {\bf METAFONT} does allow the user a great deal of freedom in
constructing graphs.  There are pitfalls, however, and some experimentation
is needed to construct the desired output.
\medskip
\noindent
In {\bf METAFONT}, the subscripted variables $x_1$ and $y_1$ may be
written either as $x[1]$, $y[1]$ or as $x1$, $y1$; likewise, $x_2$, $y_2$ may be
written either as $y[2]$, $y[2]$ or $x2$, $y2$, and so on.  The
letter $z$ stands for the ordered pair $(x,y)$ (just as is the case
in complex analysis).  Thus $z_1$ is equivalent to $(x_1, y_1)$,
$z_2$ is equivalent to $(x_2, y_2)$, etc.  When we write $z3$ we mean
$(x3, y3)$ or $(x[3], y[3])$.  This shorthand aids in construction.
For those who recall FORTRAN (the FORmula TRANSlator of the 1960's),
the symbol {\tt **} meant exponentiation.  $x${\tt **}$y$ meant
$x^y$.  This was the first encounter many of us had with a ``two
character'' operator.  (The binary operators $+$, $-$, $*$ (times),
and $/$ were well understood.)  In BASIC (Beginner's All-purpose
Symbolic Instruction Code), FORTRAN's {\tt **} has been replaced by
the single character operator {\tt \char94}.  {\bf METAFONT} has
re-introduced the two character operator.  This time $z1${\tt..}$z2$
means draw a (curved) line from $z1$ ($z[1]$) to $z2$ ($z[2]$) and
$z1--z2$ means draw a straight line from $z1$ to $z2$.  The symbol
``$:=$'' means assignment while ``$=$'' means simple equality
(replacement).  Variables followed by the hash mark (American pound
sign) ({\tt\#}) are called ``sharped'' and have absolute values
(values that are resolution-independent) 
as opposed to other variables whose values are derived and
determined programmatically.
\vfill\eject
\noindent{\bf\llap{5.1\quad}A Metafont File.}  We will begin by
listing a simple {\bf METAFONT} file that will produce a small
Gaussian distribution. 
\medskip
\noindent
{\tt\obeylines\parindent=0pt\ttraggedright
\% The statement "mode\_setup" adapts METAFONT to the current task.
mode\_setup;
\ em\#:=30pt\#; cap\#:=10pt\#;
\% The statement "define\_pixels( , )" converts the values
\% \quad of its (two) arguments into pixel (picture element) units.
define\_pixels(cap, cap);
\% The "beginchar" operation assigns values to the variables
\% \quad w, h, and d, which represent the width, height,
\% \quad and deptgh of the current character's bounding rectagle.
beginchar("A",em\#,em\#,0); "A for Gaussian Distribution";
\ pickup pencircle scaled 0.4pt; \quad \% Use a pen of diameter 0.4 point.
\ z1=(0,0.39894b); 
\ z2=(0.1a,0.39695b);
\ z3=(0.2a,0.39104b);
\ z4=(0.3a,0.38138b);
\ z5=(0.4a,0.36827b);
\ z6=(0.5a,0.35206b); 
\ z7=(0.6a,0.33322b);
\ z8=(0.7a,0.31225b);
\ z9=(0.8a,0.28969b);
\ z10=(0.9a,0.26608b);
\ z11=(a,0.24197b); 
\ z12=(1.2a,0.19418b);
\ z13=(1.4a,0.14972b);
\ z14=(1.6a,0.11092b); 
\ z15=(1.8a,0.07895b);
\ z16=(2a,0.05399b); 
\ z17=(2.5a,0.01752b);
\ z18=(3a,0.00443b);
\ z19=(3.5a,0.00087b);
\ z20=(4a,0.0001338b);
\ a=w; b=8h;
\ draw z1..z2..z3..z4..z5..z6..z7..z8..z9..z10
\qquad        ..z11..z12..z13..z14..z15..z16..z17..z18..z19..z20;
\ for k=2 step 1 until 20: 
\ z[20+k]=(-x[k],y[k]); endfor
\ draw z1..z22..z23..z24..z25..z26..z27..z28..z28..z30
\qquad        ..z31..z32..z33..z34..z35..z36..z37..z38..z39..z40;
\ draw (0,-0.02b)--(0,0);
\ draw (1a,-0.02b)--(1a,0);
\ draw (2a,-0.02b)--(2a,0);
\ draw (3a,-0.02b)--(3a,0);
\ draw (4a,-0.02b)--(4a,0);
\ draw (-1a,-0.02b)--(-1a,0);
\ draw (-2a,-0.02b)--(-2a,0);
\ draw (-3a,-0.02b)--(-3a,0);
\ draw (-4a,-0.02b)--(-4a,0);
\ draw (-4a,0)--(4a,0);
\ draw (1.5a,0)--(1.5a,0.12952b);
\ endchar; \quad \% Write character to output file.
}
\medskip\noindent
It is not possible to display the above graph using the public domain
view program {\bf CDVI12.arc}.  {\tt CDVI-2.com} and the other previewers
use only the sixteen basic \TeX\ fonts.  Some discipline is needed to use
public domain software; it is not as flexible as the commercially procured
products.  However, there is much to be said for the persons and organizations
that devote their time and energy to the creation of public domain
instruments.  Without such volunteerism, many scholars and scientists would
simply be unable to afford needed tools.  Certain activities are not able
to budget for commercial software; governmental agencies are only able
to procure software within the strict limits of their charters---and then
only with ADPE (Automatic Data Processing Equipment) approval from their MIS
(Management Information System) directors.  Public domain materials are an
invaluable tool for research and development.  
\medskip\noindent
Since a public domain \TeX\ previewer is to be employed and only sixteen
fonts are allowed, one cannot view a newly created font.  The graphs
produced by {\bf METAFONT} are, essentially, new fonts.  In particular,
each graph is one letter in a font.  Some display of the {\bf METAFONT}
graph is needed;
therefore, a \PiCTeX\ picture is provided
to give the reader an idea of the construction.
%\vbox{\narrower\narrower\narrower\noindent{\bf 25.} The main point here
%is that you can't have \PiCTeX\ draw a curve by simply giving it the formula
%for that curve --- you have to supply explicit coordinate points.  Here
%is the code that produced the figure.}
%
\bigskip
\centerline{%
 \beginpicture %
   \setcoordinatesystem units <.5in,2.5in> %
   \setplotarea x from -3 to 3, y from 0 to .4 %
   \plotheading {\lines {%
     The density $\varphi(\zeta) = e^{-\zeta^2\!/2}/\sqrt{2\pi}$ of the\cr %
     standard normal distribution.\cr}} %
   \axis bottom ticks numbered from -3 to 3 by 1 %
     length <0pt> withvalues $\zeta$ / at 1.5 /  / %
   \linethickness=.25pt %
   \putrule from  1.5 0  to  1.5 .12952 % (.12952 = density at 1.5)
   \setbox0 = \hbox{$swarrow$}%
   \put {$\swarrow$ \raise6pt\hbox{$\varphi(\zeta)$}} %
     [bl] at 1.5 .12952 %
\ifexpressmode
  \put {\bf EXPRESSMODE} at 0 .2 %
\else
   \setquadratic \plot
     0.0      .39894
     0.16667  .39344  0.33333 .37738   0.5  .35207   0.66667  .31945 
     0.83333  .28191  1.      .24197   1.25 .18265   1.5      .12952 
     1.75     .08628  2.      .05399   2.25 .03174   2.5      .01753 
     2.75     .00909  3.0     .00443 /                              
   \setquadratic \plot
     0.0      .39894
    -0.16667  .39344 -0.33333 .37738  -0.5  .35207  -0.66667  .31945 
    -0.83333  .28191 -1.      .24197  -1.25 .18265  -1.5      .12952 
    -1.75     .08628 -2.      .05399  -2.25 .03174  -2.5      .01753 
    -2.75     .00909 -3.0     .00443 /                              
\fi
\endpicture } %
\smallskip
\centerline{\rm F{\sevenrm IGURE} 35}
%
%
\vfill\eject
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%                                                                     %%%%%
%%%%             Chapter 6---An Important Improper Integral              %%%%%
%%%%                                                                     %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\def\Res{\mathop{\rm Res}} % Residue defined for use in integral.
\headline{\tenrm\hfill An Important Improper Integral}
\centerline{\bf Chapter 6}
\bigskip
\noindent{\bf\llap{6.0\quad}Introduction.}  An integral can be interpreted
geometrically as the area under a curve.  More specifically, it is the area
bounded by the $x$-axis, the lines $x=\alpha$ and $y=\beta$, and the curve.
Several things might go ``wrong'' with this idea: (1) the function might
not be continuous; (2) either $\alpha=-\infty$ or $\beta=+\infty$, or
both; or (3) the function to be integrated, known as the {\sl integrand},
might ``blow up'' (the function might increase without bound behaving as
$1/x$ near $0$).  In the first case, if the function is composed of
several continuous pieces and there is only a finite collection of
``jump'' discontinuities,
that is, if the function is piece-wise continuous,
then our usual methods apply.  For pathological cases of discontinuous
functions, there is an entire discipline of mathematics, Real Analysis.  The
principal thrust of Real Analysis follows trying to answer the question:
``under what conditions does $\int_\alpha^\beta f(x)\,dx$ make sense?''
Aside from some graphs and geometric persuasions, we won't concern ourselves
with this matter.  This chapter is concerned with the concept of the
improper integral.  One improper integral of particular interest has as
an integrand the function
$$f(x) = {1\over\sqrt{2\pi}}\,e^{-x^2/2}.\eqno(6.1)$$
On first glance, this would seem to be the ideal integral to focus on.
However,  integral has been tabulated and is available in standard tables books.
Another important integral, which will be the motivation for this chaper,
is
$$\int\limits_{\quad0}^{\qquad\infty}\, {\ln x\,dx \over x^2+a^2}.
\eqno(6.2)$$
This integral is important in physics, engineering, and applied
mathematics.  Its evaluation is found in some tables books; however,
it is not treated as a classical example in most texts.  For this reason,
an evaluation and an approximation will be done.  The computed value for
this {\sl definite\/} integral can be found in Schaum's Outline {\it Complex
Variables}\footnote{\dag}{Spiegel, Murray R., {\it Schaum's Outline Series
Theory and Problems of Complex Variables with an application to conformal
mapping and its applications}, (NY: McGraw-Hill, Inc., 1964), page 197.}
\smallskip\noindent
84. Prove that
$$\int_0^\infty {\ln x\,dx\over x^2+a^2}\ =\ {\pi\ln a\over 2a}.$$
\medskip
\hrule
\medskip
$$\beginpicture
  \setcoordinatesystem units <1in,1in>
  \setplotarea x from 0 to 3.0, y from 0 to 1.95
  \plotheading {An Integrable, Discontinuous Function} %
  \axis bottom label {F{\sevenrm IGURE} 36}
    ticks withvalues $\alpha$ $\beta$ / at .25 2.75 / / %
  \axis left ticks withvalues $y_{\rm min}$ $y_{k-1}$
    $y_k$ {\raise4pt\hbox{$y_{k+1}$}} 
    $y_{\rm max}$ / at .5 1.4 1.5 1.6 1.9 / / %
  \setshadegrid span <.02in> %
  \vshade 2.0 0.0 1.5  2.1 0.0 1.5 / %
  \vshade 1.25 0.9 1.0  1.5 0.9 0.9 / %
  \setdashes <2.5pt>
  \putrule from 2.75 0 to 2.75 1.9
  \putrule from 0 1.4  to 2.75 1.4
  \putrule from 0 1.5  to 2.75 1.5
  \putrule from 0 1.6  to 2.75 1.6
  \putrule from 0 1.9  to 2.75 1.9
  \putrule from 0 0.5  to 2.75 0.5
  \putrule from 2.0 0  to 2.0 1.5
  \putrule from 2.1 0  to 2.1 1.5
  \putrule from 0.25 0  to 0.25 1.9
  \put {$x$} at 3.15 0
  \put {$y$} at 0.1 2.0
  \setsolid
  \setquadratic
  \plot 0.75 1.0  1.0 1.9  1.25 1.0 / %
  \plot 0.25 1.0  0.5 0.5  0.75 1.0 / % 
  \setlinear
  \plot 1.5 0.9  2.3 1.7 / %
  \putrule from 2.3 1.7  to  2.6  1.7
  \putrule from 2.6 1.8  to  2.75 1.8 
\endpicture$$
\vfill\eject
\noindent{\bf\llap{6.1\quad}Theoretical Evaluation.}  
$$\beginpicture
  \setcoordinatesystem units <1in,1in>
  \setplotarea x from -2 to 2, y from -.5 to 2
  \axis bottom shiftedto y=0.0 ticks length <0pt> %
        withvalues $-R$ $\rightarrow$ $-\epsilon$ $\epsilon$ %
        $\rightarrow$ $R$ $x$ / at -1.5 -.75 -.25 .25 .75 1.5 2.0 /  / %
  \axis left shiftedto x=0.0 /
  \linethickness=1pt
\ifexpressmode
  \put {\bf EXPRESSMODE} at 1 1 %
\else
  \circulararc 180 degrees from 1.5 0 center at 0 0 %
  \circulararc 180 degrees from .25 0 center at 0 0 %
\fi
  \putrule from -1.5 0 to -.25 0 %
  \putrule from .25 0 to 1.5 0 %
  \put {$y$} [rb] at 0.1 2.0 %
  \put {Contour $\Gamma$} [lB] at 0.25 -0.50
  \put {F{\sevenrm IGURE} 37} [rB] at -0.25 -0.50
  \put {$\nwarrow$} at 1.25 1.25
  \put {$\swarrow$} at -1.25 1.25
\endpicture$$
Evaluate the (improper) integral
$$\int_0^\infty {\ln x\,dx\over x^2+a^2}\eqno(6.3)$$
using the methods of complex analysis.  This is the preferred means of
evaluation of such integrals by physicists and engineers.  We will consider
the above contour, $\Gamma$.  By a contour, we mean a simple closed curve,
in this case $\Gamma$, which is traversed in the positive sense.
If someone walked around this curve, he would not cross his own path
(the curve is simple); he would eventually return to his starting point
(the curve is closed); and, he would notice that the area contained within 
the curve would always be to his left (traversing in the positive sense).
The curve is also made up of pieces, each of which
is well-behaved in some mathematical sense. 
Let $C_1$ denote the outer semi-circle (of radius
$R$) and let $C_2$ denote the inner semi-circle (of radius $\epsilon$).
Then $\Gamma$ is composed of four parts:  (1) the interval along the
$x$-axis from $-R$ to $-\epsilon$, which is denoted
as $[-R,-\epsilon]$; (2) the semi-circle
$C_2$; (3) the interval $[\epsilon,R]$; and, (4) the semi-circle $C_1$.
We will use the so-called calculus of residues to compute
$$\int_{\Gamma}{\ln z\,dz\over z^2+a^2}\ =\ 
   2\pi i \Res_{z=ia}\left[{\ln(z)\over z^2+a^2}\right]
\ =\ {i\pi^2\over2a}\ +\ {\pi\cdot\ln a\over a}.\eqno(6.4)$$
$$\Res_{z=ia}\left[{\ln(z)\over z^2+a^2}\right]\ =\ \lim_{z\to ia}
(z-ia){\ln(z)\over(z-ia)(z+ia)}\ =\ {\ln(ia)\over2ia}
\ =\ {\ln(i)\over2ia}+{\ln(a)\over2ia}\ =\ {\pi\over4a}+{\ln a\over2ia}.$$
$$\int_{\Gamma}{\ln z\,dz\over z^2+a^2}
\ =\ \int_{-R}^{-\epsilon}{\ln z\,dz\over z^2+a^2}
\ +\ \int_{C_1}{\ln z\,dz\over z^2+a^2}
\ +\ \int_{\epsilon}^{R}{\ln x\,dx\over x^2+a^2}
\ +\ \int_{C_1}{\ln z\,dz\over z^2+a^2}.\eqno(6.5)$$
$$\int_{-R}^{-\epsilon}{\ln z\,dz\over z^2+a^2}
\ =\ \int_{-R}^{-\epsilon}{\ln|z|+i\arg(z)\over z^2+a^2}\,dz
\ =\ \int_{\epsilon}^{R}{\ln x\,dx\over x^2+a^2}
\ +\ i\pi\,\int_{\epsilon}^{R}{dx\over x^2+a^2},\eqno(6.6)$$
where $\arg z = \theta$.  We apply the well-known 
formula\footnote{$^*$}{Handbook of Mathematical, Scientific, and
Engineering, {\it etc.}, (Piscataway, NJ: REA, 1980), page 323}
%
$$\eqalignno{\int {dx\over x^2+a^2}
  \ &=\ {1\over a}\,\arctan\left({x\over a}\right)&(6.7)\cr
  \lim_{\epsilon\to0}\,\lim_{R\to\infty}\,\int_\epsilon^R
  {dx\over x^2+a^2}\ &=\ \lim_{R\to\infty}{1\over a}\arctan %
  \left({R\over a}\right)\ -\ \lim_{\epsilon\to0}{1\over a}\arctan %
  \left({\epsilon\over a}\right)\cr
  &=\ {1\over a}\,\arctan\left(\lim_{R\to\infty}{R\over a}\right)
  \ =\ {\pi\over2a}&(6.8)\cr}$$
\bigskip
\noindent
Claim
$$\lim_{\epsilon\to0}\,\int_{C_2} {\ln z\,dz\over z^2+a^2}\ =\ 0
\qquad\hbox{and}\qquad \lim_{R\to\infty}\,\int_{C_1}{\ln z\,dz\over %
z^2+a^2}\ =\ 0.\eqno(6.9)$$
This is an interesting problem because one inequality will suffice for
both integral estimates:
$$\left|\ln a\right|\ =\ \big|\ln\left|z\right| + i\arg z\big|
\ =\ \sqrt{(\ln\rho)^2 + \theta^2},\eqno(6.10)$$
where $z=\rho e^{i\theta}$ and $\theta=\arg z$.  The variable $\theta$
lies in the interval $[0,\pi]$, that is $0\le\theta\le\pi$.  When
$\rho=R$, $z=Re^{i\theta}$, we choose $R>\pi$ so that
$$\big|\ln z\big|\ =\ \sqrt{(\ln R)^2 + \theta^2}
  \le \sqrt{(\ln R)^2+\pi^2} \le \sqrt{2}\,\ln R.\eqno(6.11)$$
We also require that $R>|a|$ so that
$$0 < {1\over R^2+|a|^2} \le {1\over R^2-|a|^2},\eqno(6.12)$$
which follows from the inequality $\big|R^2-|a|^2\big|
\le \big|R^2e^{2i\theta} + a^2\big|$.  The modulus (absolute value) of
the contour integral
$$\int_0^\pi {\ln Re^{i\theta}\,dz\over R^2e^{2i\theta} + a^2}
  \eqno(6.13)$$
can thus be bounded
$$\eqalign{ \left|\int_{C_1} {\ln z\,dz\over z^2+a^2}\right|
\ &=\ \left|\int_0^\pi {\ln(z)Re^{i\theta}\,d\theta \over 
R^2e^{2i\theta}+a^2}\right|\cr 
  &\le\ \int_0^\pi {|\ln(z)|\cdot R\,d\theta\over R^2-|a|^2}\cr
  &\le\ \int_0^\pi {\sqrt{2}\ln(R)\cdot R\,d\theta \over 
  R^2 -|a|^2}\cr
  &=\ {\pi\sqrt{2}\,\ln(R)\cdot R\over R^2-|a|^2}.\cr}\eqno(6.14)$$
And, taking the limit
$$\lim_{R\to\infty} {\pi\sqrt{2}\,\ln(R)\cdot R\over R^2-|a|^2}
\ =\ 0.\eqno(6.15)$$
In the second contour,
$$\eqalign{\left|\int_{C_2} {\ln z\,dz\over z^2+z^2}\,\right|
\ &=\ \left|\int_0^\pi {\ln(Re^{i\theta})\cdot Re^{i\theta}\,d\theta
\over \big(R^2e^{2i\theta}+a^2\big)}\right|
\ \le\ \int_0^\pi {\sqrt{2}\,|\ln(1/\epsilon)|\cdot \epsilon\,d\theta 
  \over \big(|a|^2 - \epsilon^2\big)}\cr
 &=\ {\pi\sqrt{2}\cdot\ln(1/\epsilon)\cdot\epsilon
  \over \big(|a|^2 - \epsilon^2\big)},\cr}\eqno(6.16)$$
by choosing $1 > \epsilon > 0$ so that $|a| > \epsilon$ and
$\ln\epsilon < -\pi$.  This follows from
$$z=\epsilon e^{i\theta},\qquad |z| = \sqrt{|\ln\epsilon|^2+\theta^2},
\quad \hbox{for}\quad 0\le\theta\le\pi$$
and
$$\big|\ln z\,\big| =  \sqrt{|\ln\epsilon|^2+\theta^2}
\le \sqrt{|\ln\epsilon|^2+\pi^2} \le \sqrt{|\ln\epsilon|^2+\epsilon^2}
= \sqrt{2}\,\ln(1/\epsilon).\eqno(6.17)$$
And, in the limit as $\epsilon\to0$,
$$\lim_{\epsilon\to0}{\pi\sqrt{2}\cdot\ln(1/\epsilon)\cdot\epsilon
\over\big(|a|^2-\epsilon^2\big)}\ =\ 0\quad\equiv\quad
\lim_{\delta\to0}{\ln(1/\delta)\over1/\delta}\ =\ 0.\eqno(6.18)$$
\vfill\eject
\noindent{\bf\llap{6.2\quad}Numerical Evaluation.}  
We will apply a program using the adaptive quadrature routine.\footnote{\dag}
{Mike Stimpson, ``Numerical Integration Using Adaptive Quadrature,''
{\it The C Users Journal}, May 1992, pages 31--36.}  For the value of
$a$ in equation (6.3), let $a=2$ so that the
theoretical solution will be $\pi\ln(a)/(2a)$ $=$ 
$\arctan(1)\cdot\ln(2)$.  The output from a computer program is
\medskip
{\tt\obeylines
left endpoint \ = 0.000000500000000
right endpoint = 10000.00000000000000000
error estimate = 0.0000000010000000
Integral \ \ \ \ \ \ = 0.5442733355868506
Theoretical \ \ \ = 0.5443965225759005
Program returned \ (0). \ Press any key
}
\medskip
\noindent We must employ ``C'' because BASIC does not support recursion.  The
program listing is as follows:
\smallskip
{\tt\parindent=0pt\obeylines
\#include <math.h>
\#include <float.h>
\#include <stdio.h>
double adaptive (double a, double b, double (*f)(double x), double *err);
double f(double x);
int main() {\char123} 
\ double a,b,*err,int1;
\ a = (double) 0.0000005;  b = (double) 100000.0;
\ *err = (double) 0.00000001;
\ printf("{\char92}nleft endpoint \ = \%.16lf",a);
\ printf("{\char92}nright endpoint = \%.16lf",b);
\ printf("{\char92}nerror estimate = \%.16lf",*err);
\ int1 = adaptive( a, b, f, err);
\ printf("{\char92}nIntegral \ \qquad \   = \%.16lf",int1);
\ printf("{\char92}nTheoretical \quad\  = \%.16lf",atan(1.0)*log(2.0));
\ return(0); 
{\char125}
double f(double x) {\char123}   
\quad double y;
\quad  y = log(x)/(x*x+4.0);
\quad  return (y); 
{\char125}
double adaptive (double a, double b, double (*f)(double x), double *err) {\char123}
\  double h, s1, s2, s3, s4, s5, error, err1, err2, t1, t2;
\  h = b - a;
\  s1 = (*f)(a);  s2 = (*f)(a+0.25*h);  s3 = (*f)(a+0.5*h);
\  s4 = (*f)(b-0.25*h);  s5 = (*f)(b);
\  t1 = h*(s1+4.0*s3+s5)/6.0;
\  t2 = h*(s1+4.0*s2+2.0*s3+4.0*s4+s5)/12.0;
\  error = fabs(t1-t2)/15.0;
\  if (error < *err)  {\char123}  *err = error;  return(t2); {\char125}
\  else  {\char123}
\quad   err1 = err2 = *err/2.0;  t2 = adaptive(a, a+0.5*h, f, \&err1);
\quad   t2 += adaptive(a+0.5*h, b, f, \&err2);  return(t2); {\char125}
{\char125}
/* End of File */
}
\vfill\eject
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%                                                                     %%%%%
%%%%                   APPENDICES                                        %%%%%
%%%%                                                                     %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\headline{\tenrm\hfill Mathematical Preliminaries}
\centerline{\bf Appendix A}
\bigskip
\noindent
Sometimes the most advanced concepts in mathematics are most quickly grasped
by those with the least formal training.  This may be because the tyro has
less conditioning and has formed fewer bad habits.  This is particularly true
as regards variables with subscripts and superscripts.  The algebra student
who first encounters subscripted and superscripted variables is drilled on
the idea that the superscript denotes successive multiplication, that is
$$x^n = \underbrace {x\cdot x\cdots x}_{n\ \hbox{\sevenrm times}}.
\eqno{\rm(A.1)}$$
This follows from the explanation that $x^2 = x\cdot x$, $x^3 = x\cdot x
\cdot x$, and so on.  The idea that $x^i$, for $i$ an {\it index\/} and not
an integer appears later (in tensor analysis).  Then there's the subscript.
If we have a collection of data, 
where $x$ stands for a single data element, then
it is customary to denote the data by subscripting the symbol $x$.  So, the
set of data is formed by collecting together $x_1$, $x_2$, $\dots$.  There
is a need to group this set, so braces ({\tt \char123 \char125}) are used.
And, for appearance sake, the braces are typeset slightly larger than
the symbols, ($\{ \}$).  The variables and other mathematical entities are
usually set in {\it italics\/} and the vectors in {\bf bold face}.  Functions,
such as $\sin$ and $\cos$, are usually typeset in the standard roman font.
All these things are recognized by the student and accepted, usually without
needing to be verbalized.  As a student progresses from algebra through
calculus, she/he develops more and more of a mind-set towards the symbols
and their appearance.  Then there is the rude awakening of tensor analysis.
Here superscript means coordinate and not power and there is a special system
for arranging the symbols.  The beginning algebra student would have no
problem whatsoever with this arrangement;  however, the math major suffers
no end of frustration.  Those familiar with chemistry and physics are aware
of the conventions associated with isotopes.  There are not only subscripts
and superscripts following the element symbol, but preceeding it as well.
Something like ${}^2_1{\rm H}^{+}$ means the hydrogen (deuterium) ion in 
chemistry.  The idea that elements can be arranged according to their atomic
number via subscripts preceeding the element symbol is easy for the beginner.
Later, learning such conventions becomes harder and harder.
$$ {}_1{\rm H},\,{}_2{\rm He},\,{}_3{\rm Li},\,\hbox{\it etc.}
\qquad {}^1_1{\rm H},\,{}^2_1{\rm H},\,{}^3_1{\rm H},\,
{}^3_2{\rm He},\,{}^4_2{\rm He},\,\dots.\eqno{\rm(A.2)}$$
Mathematicians and theoretical physicists sometimes align the
the subscripts and superscripts in tensor notation to indicate a
priority.  The symbol here is $S$ and it's contravariant parts are
indexed $j,k$ while its covariant parts are indexed $i,,,\ell$.
$$S_i{}^{jk}{}_\ell.$$
This is very different from
$$S_i{}^j{}_{k\ell}.$$
\bigskip
\noindent
Enough on the topic of subscripts and superscripts.  Now let's consider a
more difficult situation.  Small children understand so well that a 
number such as 0.9999999999 is close to 1---but it is not exactly one.
Only with the introduction of the ellipsis ($\dots$) comes true confusion
and misunderstanding.  And just what is this ellipsis?  An ellipsis is
a useful notation for something, clearly understood, that is left out.
If we wish to write the alphabet, for example, we all know the letters
$$\hbox{\tt ABCDEFGHIJKLMNOPQRSTUVWXYZ}.$$
So, we usually just write {\tt ABC$\ldots$XYZ} or even {\tt ABC$\ldots$Z}.
And what permits us to do this?  Well, as long as we know exactly what
the omitted objects are, there is no problem.  Let's dwell on this matter
for a while.  If someone writers {\tt ABC$\ldots$Z}, then it's clear that
the first letter omitted in the ellipsis is {\tt D} and the last letter
omitted is {\tt Y}.  So, we have the situation that {\tt ABC$\dots$Z}
is exactly the same as {\tt ABCD$\ldots$YZ}.  But this is a giant step.
The next big jump in using an ellipsis is leaving out the closing
element.  Consider {\tt ABC$\dots$}.  Now {\tt ABC$\dots$} means
{\tt ABCD$\dots$}, and so on.  But there is no terminal letter present,
no {\tt Z}.  Maybe the last letter in the ellipsis is {\tt Z}, maybe
not.  There is uncertainty.  So we feel uncertain about {\tt ABC$\dots$}
but not about {\tt ABC$\ldots$XYZ}.  And this uncertainty carries over to
such things as $x_1, x_2, \dots$ as opposed to $x_1, x_2, \ldots, x_{12},
x_{13}$.  Now let's apply all of this to numbers.  Consider
$0.4999\dots$.  I claim that this number is exactly the same as
$0.5$ and $0.5000\dots$.  Maybe it's not correct to say {\sl exactly the
same}, maybe it's more correct to say ``equivalent to.''
$$\eqalign{x\ &=\ 0.4999\dots\cr
           10x\ &=\ 4.9999\dots\cr
           10x-x\ &=\ 4.5000\dots\cr
           9x\ &=\ 4.5\dots\cr
            x\ &=\ {4.5\over9}\ =\ 0.5\cr}\eqno{\rm(A.3)}$$
Is there any difference between $0.5000\dots$ and $0.5$?  Clearly, there
is a difference in the way the two are {\sl presented}.  But, aside from
that, is there any real difference?  Anyone would say, ``no.''  In fact,
scientific calculators all substitute trailing blanks for trailing
zeros for ``exact calculations.''  So, it's easy to understand that
$0.5000\dots$ and $0.5$ are exactly the same---it is not at all 
transparent that $0.5$ and $0.4999\dots$ are the same.  There seems to be
something else going on here.  And that's just what we're going to explore.
But first, there is never any problem understanding that $0.5$, $.5$,
$+0.5$, and $+.5$ represent exactly the same number, $1\over2$. The leading
plus sign ($+$) is always understood, in fact it looks funny when it
is present---like a printout from some old manual calculator.
There is a problem with understanding why $0.5$ and $0.4999\dots$
are equivalent.  It is deep, subtle, and elusive.  The mathematician
has a name for this phenomenon---{\it limit}.  This is some finality in
that word also, as in the expression ``that's the limit.''
It's easy to see that $0.5$ can also be written $1\over2$.  What about
$0.4999\dots$?  This will prove to be the key:
$$\eqalign{0.49\ &=\ {1\over2} - {1\over100}\cr
           0.499\ &=\ {1\over2} - {1\over1000}\cr
           0.4999\ &=\ {1\over2} - {1\over10000}\cr
           \vdots\ &=\ \vdots\cr
           0.4999\dots\ &=\ {1\over2} -\lim_{n\to\infty} {1\over10^n}\cr}
\eqno{\rm(A.4)}$$
This all looks very complicated, especially the gaggle of symbols
$$\lim_{n\to\infty} {1\over10^n}.\eqno{\rm(A.5)}$$
Here is that funny symbol ``$\infty$'' which is known as the infinity
sign, or lemniscate, or ``lazy-eight'' in Texas-talk.  This lemniscate is
also seen in the occult hovering over the head of the magician.  And
what is one to make of the arrow in the expression $n\to\infty$?  Well, if
it is true that $0.4999\dots$ $\equiv$ $0.5$, then that cluster of symbols
better be equal to zero.  It turns out that it is:
$$\lim_{n\to\infty} {1\over10^n}\ =\ 0.\eqno{\rm(A.6)}$$
This looks too deep for mathematical preliminaries.  The point is that
the innocuous, harmless equivalence $0.4999\dots$ $\equiv$ $1\over2$
is more than an oddity.  It is a model, or paradigm, of a much deeper
mathematical situation.  And this reasoning is at the very foundation of
Simpson's Rule and Newton's Method.  As soon as the ellipsis is removed,
the answer is no longer exact, it is an approximation.  Only in the world
of theory does the ellipsis make sense.  What has happened is that
mathematicians have taken a literary device and extended it's application
into another realm.  And it all looked so simple with {\tt ABC\dots}!
\medskip
\noindent
It's easy to convince engineers that $1\over3$ is $0.333\dots$ and
$2\over3$ is $0.666\dots$.  However, it's no so easy to convince
them that $1\equiv0.999\dots$.  Even so, consider
$$\eqalign{  {1\over3}\ &=\ 0.333\dots\cr
             {2\over3}\ &=\ 0.666\dots\cr
           1\ =\ {1\over3}+{2\over3}\ &=\ 0.333\dots\ +\ 0.666\dots
                     \ =\ 0.999\dots\cr}\eqno{\rm(A.7)}$$
\vfill\eject
\headline{\tenrm\hfill References}
\centerline{\bf Appendix B}
\medskip
\centerline{\bf References}
\bigskip
{\narrower\noindent
\llap{[1]\quad}Eisberg, Robert and Robert Resnick, {\it Quantum Physics of
Atoms, Molecules, Solids, Nuclei, and Particles, Second Edition}.
NY: John-Wiley \& Sons, 1985.
\medskip\noindent
\llap{[2]\quad}{\it Handbook of Mathematical, Scientific, and Engineering
Formulas, Tables, Functions, Graphs, Transforms}. Piscataway, NJ: Research \&
Education Assoc., 1972.
\medskip\noindent
\llap{[3]\quad}James, Glenn and Robert C. James, 
{\it Mathematics Dictionary---Multilingual Edition}.
Princeton, NJ: D. Van Nostrand Company, Inc., 1959.
\medskip\noindent
\llap{[4]\quad}Knuth, Donald E., {\it The \TeX book}. Reading, Massachusetts:
Addison-Wesley Publishing Company, 1970.
\medskip\noindent
\llap{[5]\quad}Kreyszig, Erwin, {\it Advanced Engineering Mathematics, 5th
Edition}.  NY: John Wiley \& Sons, 1983
\medskip\noindent
\llap{[6]\quad}{\it Numerical Analysis, 2nd Edition}. Schaum's Outline Series,
edited by Francis Scheid.
NY: McGraw-Hill Book Company, 1968.
\medskip\noindent
\llap{[7]\quad}Olmsted, John M. H., {\it Advanced Calculus}.
NY: Appleton-Century-Crofts, Inc., 1961.
\medskip\noindent
\llap{[8]\quad}Stoer, J. and R. Bulirsch, {\it Introduction to Numerical
Analysis}.  NY: Springer-Verlag, 1980.
\medskip\noindent
\llap{[9]\quad}Taylor, John R., {\it An Introduction to Error Analysis---The
Study of Uncertainties in Physical Measurements}.
Mill Valley, CA: University Science Books, 1982.
\medskip\noindent
\llap{[10]\quad}Thomas, George B., {\it Calculus and Analytic Geometry, 2nd
Edition}.  Reading, Massachusetts: Addison-Wesley Publishing Company, Inc.,
1966.
\medskip\noindent
\llap{[11]\quad}{\it Webster's Ninth New Collegiate Dictionary}. Springfield,
Massachusetts: Merriam-Webster, Inc., 1984.
\medskip
}
{\noindent\hskip-6pt
[12]\quad{Wichura, Michael J., {\it The\/} \PiCTeX\ {\it Manual},
\TeX\-niques, Publication for the \TeX\ Com- \hfill}\break
\indent  munity, Number 6.  Providence, RI: \TeX\ User Group, 1987.
}
\vfill\eject
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%                                                                     %%%%%
%%%%                   Appendix C                                        %%%%%
%%%%                                                                     %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\headline{\tenrm\hfill A Geometric Persuasion}
\centerline{\bf Appendix C}
\medskip
\centerline{\bf A Geometric Persuasion}
\bigskip
\rightline{ 6/28/89}
\rightline{ H. Watson}
\rightline{ Corona, CA}
\medskip
\noindent
Assuming that mass is volume times density, consider first a sphere
of radius $r_0$,
$$m = \rho V = \rho\, {4\over3}\, \pi r_0^3. \eqno{\rm(C.1)}$$
$$\beginpicture
  \setcoordinatesystem units <0.5in,0.5in>
  \setplotarea x from -1.0 to 1.0, y from -1.0 to 1.0
\ifexpressmode
  \put {\bf EXPRESSMODE} at 0 0
\else
  \circulararc 360 degrees from 1.0 0.0 center at 0.0 0.0
  \arrow <10pt> [.2,.4] from 0 0 to 0.7071 0.7071
\fi
  \put {$r_0$} [b] at 0.0 0.2
\endpicture $$
For a charged sphere of radius $r_0$, suppose a standing wave exists
on its surface.  Then, due to energy absorption the wave ``splits'' forming
two surfaces:  an {\sl inner surface\/} whose circumference is twice the
radius, $r_0$, and an {\sl outer\/} whose radius is twice the circumference,
$2\pi r_0$.  Thus
$$R = 2\cdot2\pi r_0 = 4\pi r_0 \eqno{\rm(C.2)}$$
and
$$r = {2 r_0 \over 2 \pi} = {r_0 \over \pi}. \eqno{\rm(C.3)}$$
$$\beginpicture
  \setcoordinatesystem units <1in,1in>
  \setplotarea x from -1.0 to 1.0, y from -1.0 to 1.0
\ifexpressmode
  \put {\bf EXPRESSMODE} at 0 0
\else
  \circulararc 360 degrees from 1.0 0.0 center at 0.0 0.0
  \circulararc 360 degrees from 0.2 0.0 center at 0.0 0.0
  \arrow <10pt> [.2,.4] from 0 0 to 0.7071 0.7071
  \arrow <5pt>  [.2,.4] from 0 0 to 0.2 0.0
  \setdashes
  \circulararc 360 degrees from 0.5 0.0 center at 0.0 0.0
\fi
  \put {$R$} [rb] <-2pt,0pt> at 0.5 0.5
  \put {$r$} [t] <0pt,-3pt> at 0.1 0.0
\endpicture $$
The resulting configuration ``collapses'' to a figure whose
radius is $R_0 = 4\pi r_0 - {r_0\over\pi}$ and containing
a ``cone'' whose base is $4\pi r^2 = 4\pi \big(r_0/\pi\big)^2
= 4r_0^2/\pi$ and height $R_0$.  The (area of the) base of the
cone equals the surface (area) of the inner sphere:
$4\pi r^2$ $=$ $4\pi\big(r_0/\pi\big)^2$ $=$
$4r_0^2/\pi$.  The volume of the collapsed figure
is, then,
$$ V_1 = {4\over3}\pi R_0^3 - {1\over3}bh
     = {4\over3}\pi\left[ 4\pi r_0 - {r_0\over\pi}\right]^3
    - {4\over3}\pi\left[{\big(r_0/\pi\big)}^2\right]
       \cdot\left[4\pi r_0 - {r_0\over\pi}\right]. \eqno{\rm(C.4)}$$
$$V_1 = {4\over3}\pi\left[{\left(4\pi - {1\over\pi}\right)}^3
    - \left(4\pi-{1\over\pi}\right){\left({1\over\pi}\right)}^2\right]
     \,r_0^3\eqno{\rm(C.5)}$$
Define $V_0 = {4\over3}\pi r_0^3$ and look at the ratio
$$\eqalign{{V_1\over V_0}\ &=\ \left(4\pi-{1\over\pi}\right)^3
  - \left(4\pi-{1\over\pi}\right)\left({1\over\pi^2}\right)\cr
\ &=\ \left(64\pi^3 -48\pi+{12\over\pi}-{1\over\pi^3}\right)
  - {4\over\pi} + {1\over\pi^3}\cr
  = 64 \pi^3 - 48 \pi + {8\over\pi}.\cr}\eqno{\rm(C.6)}$$
Compute (C.6) numerically:
$${V_1\over V_0} = 1836.151739256348 \eqno{\rm(C.7)}$$
which lies between the (two) most recent values for the 
proton-electron ratio: 
$$1836.15152\ <\ {V_1\over V_0}\ <\ 1836.152701.\eqno{\rm(C.8)}$$
\medskip
\noindent
$$ 64\pi^3 - 48\pi + {8\over\pi} =
   \left(4\pi\right)\cdot\left(4\pi-{1\over\pi}\right)\cdot
   \left(4\pi-{2\over\pi}\right) \eqno{\rm(C.9)}$$
\medskip
\hrule
\medskip
\noindent Looking at $A= 4r_0^2/\pi$ as the surface area of the inner sphere and
$S=4\pi R_0^2$ as the surface area of the larger sphere, we can also compute
$V_1/V_0$ by assuming that that portion of the solid sphere of radius $R_0$
subtended from the center to the area $A$ does not contribute to the total
volume.  This is to say that the volume element whose base 
region {\sl lying on the surface of the sphere\/} has the same volume as a
right circular cone, with equal base area.
$$ {V_1\over V_0}\ =\ \left(1-{A\over S}\right)\cdot R_0^3.\eqno{\rm(C.10)}$$
Direct numerical calculation will reveal that
$$\left(1-{4/\pi\over4\pi(4\pi-1/\pi)^2}\right)\cdot
  \left(4\pi-{1\over\pi}\right)
  \ \approx\ 1836.151738,\eqno{\rm(C.11)}$$
or the same result as equation (C.7).  
\vfill\eject
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%                                                                     %%%%%
%%%%                   Appendix D                                        %%%%%
%%%%                                                                     %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\headline{\tenrm\hfill A Contour Integral}
\centerline{\bf Appendix D}
\medskip
\centerline{\bf A Contour Integral}
\bigskip
\rightline{20 Feb 92}\par
\vskip6pt\noindent
The residue of $f(z)$ at infinity is defined as follows:  If $f(z)$ has an
isolated singularity at infinity, and if $C$ is a large circle which encloses
all the singularities of $f(z)$ except $z=\infty$, then the residue at
$z=\infty$ is defined to be
$${1\over2\pi i}\,\int_C f(z)\,dz\eqno{\rm (D.1)}$$
taken around $C$ in the negative sense (negative with respect to the
origin), provided the integral has a definite value.  If we apply the
transformation $z=1/\zeta$ to the integral it becomes
$${1\over2\pi i}\,\int -f\left({1\over\zeta}\right)\,{d\zeta\over\zeta^2}
\eqno{\rm (D.2)}$$
taken positively around a small circle centered at the origin.
%
$$\beginpicture    
%\ninepoint
    \setcoordinatesystem units <1.00in,1.00in>
    \setplotarea x from -2.25 to 2.25, y from -2.25 to 2.25 %
    \plotheading {\lines{%
      The Beta function \cr
      $\int_0^1 x^\alpha(1-x)^{1-\alpha}\,dx$.\cr}}
    \put {$\scriptscriptstyle\cdot$} at 0 0 %
    \put {$\scriptscriptstyle\cdot$} at 1 0 %
%    \put {$\curvearrowleft$} at -.01 0.02 %
%    \put {$\curvearrowleft$} at 0.99 0.02 %
%    \replot "beta.t01" %
%    \savelinesandcurves on "beta.t01"
%    \writesavefile {Feb 20, 1992 Homework}
\ifexpressmode
  \put {\bf EXPRESSMODE} at -1 -1
\else
    \circulararc 360 degrees from 2 0 center at 0 0 %
    \circulararc 350 degrees from 0.9 -.01 center at 1 0 %
    \circulararc 350 degrees from 0.1 0.01 center at 0 0 %
    \plot  1.1 0.0     2   0.0   / %
    \plot  0.9 0.01    0.1 0.01  / %
    \plot  0.9 -.01    0.1 -.01  / %
\fi
    \put {$\leftharpoonup$} [b] at 1.5 .015 %
    \put {$\rightharpoondown$} [t] at 1.5 -.02 %
    \put {$\leftharpoonup$} [b] at 0.5 .015 %
    \put {$\rightharpoondown$} [t] at 0.5 -.03 %
    \put {$\uparrow$} [l] at -2 0 %
    \put {$\searrow$} at 1.38 1.38 %
    \put {$\swarrow$} at 1.38 -1.38 %
    \put {${\bf C}$ $=$ Complex Plane} at 2 2 %
    \put {$f(z)=z(1/z-1)^{1-\alpha}$} at 0 1 %
    \put {$\partial D(0,R)$\lower6pt\hbox{$\searrow$}} [rb] at -1.42 1.42 %
\endpicture $$
\vskip6pt\noindent
$$\int\limits_{\partial D(0,R)} z^\alpha(1-z)^{1-\alpha}\,dz
  - \int_0^1 x^\alpha(1-x)^{1-\alpha}\,dx + e^{2\pi i\alpha}\,
           \int_0^1 x^\alpha(1-x)^{1-\alpha}\,dx = 0.
\eqno({\rm D.3})$$
Let $I= \int_0^1 x^\alpha(1-x)^{1-\alpha}\,dx$.  
$$\eqalign{I - e^{2\pi i\alpha}\,I 
\ &=\ \int\limits_{\partial D(0,R)} z^\alpha(1-z)^{1-\alpha}\,dz\cr
  &=\ 2\pi i \,\Res_{\infty}(f).\cr}\eqno{\rm (D.4)}$$
Let $h(w)=(w-1)^\beta$, where $w=1/z$ and $\beta=1-\alpha$.
$$\eqalign{h(w)\ &=\ (w-1)^\beta\cr
          h'(w)\ &=\ \beta(w-1)^{\beta-1}\cr
         h''(w)\ &=\ \beta(\beta-1)(w-1)^{\beta-2}\cr
                 &\vdots\cr
     h^{(n)}(w)\ &=\ \beta(\beta-1)\cdots \cr
                 &\qquad(\beta-n+1)(w-1)^{\beta-n}\cr
                 &\vdots\cr}
\qquad
\eqalign{h(0)\ &=\ (-1)^\beta\cr
        h'(0)\ &=\ (-1)^{\beta-1}\beta\cr
       h''(0)\ &=\ (-1)^{\beta-2}\beta(\beta-1)\cr
               &\vdots\cr
   h^{(n)}(0)\ &=\ (-1)^{\beta-n}\beta(\beta-1)\cdots\cr
               &\qquad(\beta-n+1)\cr
               &\vdots\cr}$$
$$h(w)=h(0)+wh'(0) + {w^2\over2}h''(0)+ \ldots+{w^n\over n!}h^{(n)}(0)+\dots$$
$$\eqalign{h(w)
\ &=\ (-1)^\beta + w\beta(-1)^{\beta-1}+{w^2\over2}\beta(\beta-1)+\ldots
    +(-1)^{\beta-n}{w^n\over n!}\beta(\beta-1)\cdots(\beta-n+1)+\dots\cr
\ &=\ (-1)^\beta\left(1 -w\beta+{w^2\over2}\beta(\beta-1)
           +\ldots+(-1)^n{w^n\over n!}
     \beta(\beta-1)\cdots(\beta-n+1)+\dots\right)\cr}$$
$$h(1/z)
=\ (-1)^\beta\left(1 -{\beta\over z}+{\beta(\beta-1)\over 2z^2}
      +\ldots+(-1)^n{\beta(\beta-1)\cdots(\beta-n+1)\over z^n n!}
        +\dots\right)$$

$$\eqalign{f(z)\ &=\ zh(1/z)\cr
\ &=\ (-1)^\beta\left(z -\beta+{\beta(\beta-1)\over2}{1\over z}
       +\ldots+(-1)^n{\beta(\beta-1)\cdots(\beta-n+1)\over z^{n-1}n!}
       +\dots\right)\cr}$$
Therefore,
$$\eqalign{\Res_{z=\infty} f(z)\ =\ \Res_{\infty} (f)
\ &=\ -a_{-1}=(-1)^{\beta-1}{\beta(\beta-1)\over2}\cr
\ &=\ (-1)^{-\alpha}{\alpha(1-\alpha)\over2}\cr
\ &=\ e^{-i\pi\alpha}{\alpha(1-\alpha)\over2}}\eqno{\rm (D.5)}$$
Now we apply the results of our calculation of the residue in
equation (D.5) to equation (D.1):
$$I\cdot(1-e^{i2\pi\alpha}) = 2\pi
i\cdot e^{-i\pi\alpha}\cdot\left({\alpha(1-\alpha)\over2}\right)$$
Now we equate the real parts:
$$I\cdot\big(1-\cos(2\pi\alpha)\big) = \pi\sin(\pi\alpha)\alpha(1-\alpha)$$
$$1-\cos 2\theta = 2\sin^2\theta$$
$$I\cdot\big(2\sin^2(\pi\alpha)\big) 
= \pi\sin(\pi\alpha)\alpha(1-\alpha)$$
$$I = {\pi\alpha(1-\alpha)\over2\sin(\pi\alpha)}.\eqno{\rm (D.6)}$$
Equation (D.6) is the sought-for result.
\vfill\eject
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%                                                                     %%%%%
%%%%                   Appendix E                                        %%%%%
%%%%                                                                     %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\headline{\tenrm\hfill Malgrange-Ehrenpreis Theorem}
\centerline{\bf Appendix E}
\medskip
\centerline{\bf The Malgrange-Ehrenpreis Theorem}
\bigskip
\mathsurround=1pt
\def\bull{\vrule height .9ex width .8ex depth -.1ex} % square bullet
\proclaim The Malgrange-Ehrenpreis Theorem.  If $p(D)$ is a
constant coefficient partial differential operator on ${\bf R}^n$,
then there is a distribution $E\in\cal D'$ such that
$p(D)E =\delta$.  {\rm (See [5], page 48.)}\par

\vskip6pt\noindent
{\bf Definitions:}  We are using the standard multi-index notation.  
A multi-index 
$$\alpha = \langle \alpha_1,\ldots,\alpha_n\rangle$$
is an $n$-tuple of nonnegative integers.  The collection of all such
multi-indices will be denoted by $I_{+}^n$.  The symbols $|\alpha|$,
$x^\alpha$, $D^{\alpha}$, and $x^2$ are defined thusly:
$$\eqalign{|\alpha|\ &=\ \sum_{i=1}^n \alpha_i\cr
x^\alpha\ &=\ x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}\cr
D^{\alpha}\ &=\ {\partial^{|\alpha|} \over \partial x_1^{\alpha_1}
\partial x_2^{\alpha_2}\cdots\partial x_n^{\alpha_n}}\cr
x^2\ &=\ \sum_{i=1}^n x_i^2\cr}$$

\vskip6pt\noindent
We define the set ${\cal S}({\bf R}^n)\subset C^{\infty}({\bf R}^n)$ such that
$f\in{\cal S}$ iff
$$\sup_{x\in{\bf R}^n}\left|\, x^\beta D^\alpha f(x)\right|
  < \infty\qquad\left(x^\beta = \prod_{j=1}^n x_j^{\beta_j}\right)$$
for every $\alpha=\langle \alpha_1,\alpha_2,\ldots,\alpha_n\rangle$
and $\beta=\langle\beta_1,\beta_2,\ldots,\beta_n\rangle$ with nonnegative
integers $\alpha_j$ and $\beta_k$.  Such functions are called rapidly
decreasing (at $\infty$) or functions of rapid decrease.  We
call ${\cal S}({\bf R}^n)$ the Schwartz space of $C^{\infty}$ functions of rapid
decrease or simply the Schwartz space.

\vskip6pt\noindent
{\bf Definition:}  Let $f\in{\cal S}({\bf R}^n)$.  The Fourier Transform
of $f$ is the function
$$\hat f(\lambda) = {1\over(2\pi)^{n/2}}\int_{{\bf R}^n} e^{-ix\cdot\lambda}
      f( x)\,dx\eqno{\rm (E.1)}$$
where $x\cdot\lambda = \sum_{i=1}^n x_i\lambda_i$.  On occasion it will
be necessary to write $\hat f= {\cal F} f$.  The inverse Fourier Transform
of $f$ is the function
$$\check f(\lambda) = {1\over(2\pi)^{n/2}}\int_{{\bf R}^n} e^{ix\cdot\lambda}
     f(x)\,dx.\eqno{\rm (E.2)}$$
We will occasionally write $\check f$ $=$ ${\cal F}^{-1} f$.

\vskip6pt\noindent  We define the polynomial $p(x) = p(x_1,\ldots,x_n)$
to be a polynomial of total degree $k$ in $n$ variables.  Recall that the
total degree, $k$, of a polynomial $p$ is the maximum of the degrees, $k_i$,
of the factors $a_ix_1^{k_{i1}}\cdots x_n^{k_{in}}$ such that 
$k_i = k_{i1}+\ldots+k_{in}$ and $a_i\not= 0$.  If $x=\langle
x_1,\ldots,x_n\rangle$, then $p(D)$ denotes the partial differential operator
obtained by substituting $\partial/\partial x_i$ for $x_i$ whenever it
occurs in the polynomial $p(x)$.

\vskip6pt\noindent{\bf Definition:}  Let $b\in{\bf R}$.  The symbol $\delta_b$
is the linear functional $\delta_b(\varphi) = \varphi(b)$.  $\delta_b\in
{\cal S}'({\bf R})$.
$$\delta_b(\varphi) = \int \varphi(x)\delta(x-b)\,dx\eqno{\rm (E.3)}$$
$\delta(x)$ is not a function; it is a symbolic expression known as the
(Dirac) delta function, sometimes written $T_{\delta_0}$.  

\vskip6pt
We will now discuss a {\sl distribution\/} or {\sl generalized function}.
To do this, some background is necessary.  Let $\Omega$ be an open
connected set in ${\bf R}^n$ and $C_{0}^{\infty}(\Omega)$ be the family of
functions with compact support in $\Omega$.  (Recall that if $f(x)$ is a
complex-valued function defined on $\Omega$, we call the support of $f$,
written supp($f$), to be the closure of 
$\big\{\,x\in\Omega\,\big|\,f(x)\not=0\,\big\}$ in the space $\Omega$.
It may be equivalently
defined as the smallest closed set of $\Omega$ outside which $f$ vanishes
identically.)  Let $K_n$ be an increasing family of compact sets with
$\bigcup K^{o}_n = \Omega$, that is
$$K_1 \subseteq K_2 \subseteq K_3 \subseteq \ldots K_n \subseteq \dots$$
where $K^{o}_n$ is the interior of $K_n$ (the largest open subset of $K_n$).
It is not hard to show that each of the $C_0^\infty(K_n)$ is a linear space,
where
$$(f_1+f_2)(x) = f_1(x) + f_2(x), \qquad (\alpha f)(x)=\alpha f(x).
\eqno{\rm(E.4)}$$
For each compact set $K_n$ and each multi-index $\alpha$, 
we can define the semi-norm $\Vert\cdot\Vert$ by
$$\Vert f\Vert_{\alpha,\infty} = \sup \big\{\,|\,D^\beta f(x)|\,
  \big|\, |\beta| \le \alpha, f\in C_0^\infty, x\in K_n\,\big\}.
\eqno{\rm (E.5)}$$
Define
$$\Vert D^\alpha f\Vert_\infty = \Vert f\Vert_{\alpha,\infty}.
\eqno{\rm (E.6)}$$
Put the Fr\'echet topology on $C_0^\infty(K_n)$ generated by the
$\Vert D^\alpha f\Vert_\infty$ norms.  The set $C_0^\infty(\Omega)$
with the inductive limit topology obtained by $C_0^\infty(\Omega)$ $=$
$\bigcup C_0^\infty(K_n)$ is denoted ${\cal D}_\Omega$.  If $\Omega$ is the
space ${\bf R}^n$, then we denote ${\cal D}_{{\bf R}^n}$ by ${\cal D}$.
The space of all continuous linear functionals on ${\cal D}_\Omega$ is
denoted ${\cal D}'_\Omega$, where ${\cal D}'_{{\bf R}^n}$ $=$ ${\cal D}'$.
It can be shown that the topology on ${\cal D}_\Omega$ is independent of
the choice of the $K_n$.  It can also be shown that a sequence 
$f_m\in{\cal D}$ converges to $f\in{\cal D}$ iff all the $f_m$ are in some
$K_n$ and $f_m\to f$ in the topology of $K_n$.  We finally have all the
preliminaries to define a distribution.

\vskip6pt
\proclaim Definition.  A distribution or generalized function is a continuous
linear functional on ${\cal D}_\Omega$ (${\cal D}$ when $\Omega = {\bf R}^n$).
The space of all continuous linear functionals on ${\cal D}_\Omega$ is denoted
by ${\cal D}'_\Omega$ (${\cal D}'$ when $\Omega = {\bf R}^n$).

\vskip6pt
\proclaim Definition.  Suppose $f,g$ $\in$ ${\cal S}\big({\bf R}^n\big)$.  Then
the convolution of $f$ and $g$, denoted $f\ast g$ is the function
$$(f\ast g)(y) = \int_{{\bf R}^n} f(y-x)g(x)\,dx.
\eqno{\rm (E.7)}$$\par


\vskip6pt\noindent
The partial differential operator $p(D) =\sum_{|\alpha|\le k} a_\alpha
x^\alpha$ extends to ${\cal D}'$ by the formula
$$\big(p(D)T\big)(\varphi) = T\left(\sum_{|\alpha|\le k} (-1)^{|\alpha|}
D^\alpha\big(a_\alpha \varphi\big)\right).\eqno{\rm (E.8)}$$

\vskip6pt
\proclaim Definition.  A fundamental solution for the partial differential
operator $p(D)$ is a distribution $E\in{\cal D}'$ 
such that $p(D)E=\delta$.\par

\vskip6pt\noindent
The reason for studying fundamental solutions is that if we define
$u=E\ast f$, where $f\in C_0^\infty({\bf R}^n)$, then
$$\eqalign{p(D)u\ &=\ p(D)(E\ast f)\ =\ p(D)E\ast f\cr
                  &=\ \delta\ast f\ =\ f.\cr}\eqno{\rm (E.9)}$$
Thus, if we can find a fundamental solution, then we have an existence theorem
for all the partial differential equations $p(D)u=f$ where 
$f\in C_0^\infty({\bf R}^n)$.  Furthermore, if we can find an expression for
the distribution $E$, then we have an explicit representation of a solution,
namely
$$u = E\ast f.\eqno{\rm (E.10)}$$
The convolution operation, $E\ast f$, for distributions and functions can
be accomplished on digital computers.  Note that, in general, partial  
differential equations will have many fundamental solutions in ${\cal D}'$,
and sometimes have more than one in ${\cal S}'({\bf R}^n)$.
Convolutions frequently occur when one uses
Fourier transforms because the Fourier transform takes products into
convolutions.  In particular, for distributions and functions, 
if $T$ is a distribution ($T\in{\cal D}'$) and $f\in{\cal D}$, let 
$f^{\dagger}(x)$ $=$ $f(-x)$.
$$\big( T\ast f\big)(\varphi) = T(f^{\dagger}\ast\varphi)\qquad
\forall\varphi\in{\cal D}.\eqno{\rm (E.11)}$$
The Malgrange-Ehrenpreis theorem states that every constant coefficient
partial differential operator $p(D)$ has a fundamental solution.
The proof of this theorem breaks into two parts:  the first part is a 
complex variables argument and the second is an application of the
Hahn-Banach theorem.

\vskip12pt\noindent
{\bf Proof of the Malgrange-Ehrenpreis Theorem:}  Define 
$p^{\dagger}(x) = p(-x)$ and $q(x) = p^{\dagger}(ix)$, with $i=\sqrt{-1}$.
Let $\varphi \in C_0^\infty({\bf R}^n)$, where $C_0^\infty({\bf R}^n)$
is the space of all infinitely differentiable functions with compact
support. The first thing we observe is that
$${\cal F}\big(p^{\dagger}(D)\varphi(x)\big) = \hat\varphi(s)q(s).
\eqno{\rm (E.12)}$$
This beautiful and exciting result follows immediately from the following:
\proclaim Lemma 1.  The maps ${\cal F}$ and ${\cal F}^{-1}$ are continuous
linear transformations of ${\cal S}({\bf R}^n)$ into 
${\cal S}({\bf R}^n)$.  Furthermore, if $\alpha$ and $\beta$ are
multi-indices, then
$$\big((i\lambda)^\alpha D^\beta\hat f\big)(\lambda)
 = {\cal F}\left(\,D^\alpha\big((-ix)^\beta f(x)\big)\right)
\eqno{\rm (E.13)}$$\par

\vskip6pt\noindent
If we assume that $f\in C_0^\infty({\bf R}^n$), then for all
$\zeta = \langle\zeta_1,\ldots,\zeta_n\rangle\in {\bf C}^n$,
the integral
$$\hat f(\zeta) 
= (2\pi)^{n/2}\int\limits_{{\bf R}^n} e^{-i\zeta\cdot x}f(x)\,dx
 \eqno{\rm (E.14)}$$
is well defined.  Furthermore, $\hat f(\zeta)$ is an entire analytic function
of the $n$ complex variables $\zeta_1$, $\zeta_2$, $\ldots$, $\zeta_n$ since
we can differentiate under the integral sign.  The formal differentiation
under the integral sign is permitted since the right hand side is
uniformly convergent in $\zeta$.  This is contained in the following:
\proclaim Paley-Weiner theorem.  An entire analytic function of $n$
complex variables $g(\zeta)$ is the Fourier transform of a 
$C_0^\infty({\bf R}^n)$ function with support in the ball
$\{\,x\,|\,|x|\le R\,\}$ if and only if for each $N$ there is a
$C_N$ such that
$$|g(\zeta)| \le {C_N e^{R|\hbox{\sevenrm Im}(\zeta)|} 
      \over\left(a+|\zeta|\right)^N}
  \eqno{\rm (E.15)}$$
for all $\zeta\in{\bf C}^n$.\ {\rm (See [5], page 16.)}\par

\vskip6pt\noindent  
We apply the Paley-Weiner theorem to the partial differential
operator $p^{\dagger}(D)$ acting on the function 
$\varphi\in C_0^\infty({\bf R}^n)$.
For each $y=\langle y_1,\ldots,y_n\rangle\in {\bf R}^n$, we have
$$\left({\cal F}\big(p^{\dagger}(D)\varphi\big)\right)(y+\zeta)
 = \hat\varphi(y+\zeta)\,q(y+\zeta)\eqno{\rm (E.16)}$$
is an entire (analytic) function of $\zeta\in{\bf C}^n$. Now define $Q(x)$ as
follows:
$$Q(x) = \sum_\alpha |\,D^\alpha q(x)|;\eqno{\rm (E.17)}$$
notice that $Q$ is positive and bounded away from zero.  If $k$ is the total
degree of $p$, then there is some multi-index $\alpha$ such that
$|\alpha|=k$.  $D^\alpha q(x) = a_\alpha$, where $a_\alpha$ is the coefficient
of $x^\alpha$, hence $\alpha_k\not=0$.  Therefore,
$$\sum_\alpha |D^\alpha q(x)| \ge |a_\alpha| > 0.\eqno{\rm (E.18)}$$
Having defined $Q$, we will have to use some results from complex analysis. 
We will state four lemmas:
\proclaim Lemma 2 (Malgrange).  If $f(z)$ is a holomorphic function of a 
complex variable $z$ for $|z|\le 1$ and $p(z)$ is a polynomial
$$p(z)=a_mz^m + a_{m-1}z^{m-1} + \ldots + a_1z + a_0\eqno{\rm (E.19)}$$
of degree $m$, then
$$\left| a_mf(0) \right| 
\le {1\over 2\pi}\,\int_0^{2\pi} 
\left|f\big(e^{i\theta}\big)p\big(e^{i\theta}\big)\right|\,d\theta.
\eqno{\rm (E.20)}$$\par
\proclaim Lemma 3.  If $f(z)$ is a holomorphic function of a 
complex variable $z$ for $|z|\le 1$ and $p(z)$ is a polynomial
$$p(z)=a_mz^m + a_{m-1}z^{m-1} + \ldots + a_1z + a_0\eqno{\rm(E.21)}$$
of degree $m$, then
$$\left| f(0)\, p^{(k)}(0) \right| 
\le {m!\over (m-k)!\,2\pi}\,\int_0^{2\pi} 
\left|f\big(e^{i\theta}\big)p\big(e^{i\theta}\big)\right|\,d\theta.
\eqno{\rm (E.22)}$$\par
\proclaim Lemma 4.  Let $F(\zeta)$ be an entire function of $n$ complex
variables and let $P(\zeta)$ be a polynomial of degree $m$.  Suppose that
$g(\zeta)$ is a nonnegative integrable function with compact support
depending only on $|\zeta_1|$, $|\zeta_2|$, $\ldots$, $|\zeta_n|$.  
$$\left|F(0)D^\alpha P(0)\right|\,\int_{{\bf C}^n} |\zeta|^\alpha
   g(\zeta)\,d\zeta \le C_0\,\int_{{\bf C}^n} \left|F(\zeta)P(\zeta)\right|
    \,g(\zeta)\,d\zeta,\eqno{\rm(E.23)}$$
where $d\zeta$ denotes the Lebesgue measure on ${\bf C}^n$ and $C_0$ is a
constant depending on $\alpha$ and $m$.  $d\zeta$ is the measure denoted
$d\xi_1d\eta_1\cdots d\xi_nd\eta_n$, where $\zeta_k = \xi_k+i\eta_k$; and,
$m$ is regarded as the multi-index $\langle m,m,\ldots,m\rangle$.\par
\proclaim Lemma 5.  Suppose $\tilde q(x)$ $=$ 
$\big(\sum_{|\alpha|\ge k}|q^{(\alpha)}(x)|^2\big)^{1/2}$, where
$q^{(\alpha)}(x)=D_x^{\alpha}q(x)$, $k$ is the total degree of $q$, 
$\varphi\in C_0^{\infty}({\bf R}^n)$, and $\epsilon>0$.  Then
$$\left|\tilde q(x)\hat\varphi(x)\right| \le C_1\,\int_{|\zeta|\le\epsilon}
\left|\hat\varphi(x+\zeta)\,q(x+\zeta)\right|\,d\zeta.
\eqno{\rm (E.24)}$$\par
\vskip6pt\noindent
The first step in the proof uses the Cauchy integral formula to show that
$$\eqalign{\left|Q(x)\hat\varphi(x)\right|
\ &\le\ C_1\,\int_{|\zeta|\le\epsilon}
    \left|\hat\varphi(x+\zeta)\,q(x+\zeta)\right|\,d\zeta\cr
  &=\ C_1\,\int_{|\zeta|\le\epsilon}\left|{\cal
        F}\big(p^{\dagger}(D)\varphi\big)(x+\zeta)\right|\,d\zeta\cr}
\eqno{\rm (E.25)}$$
where $C_1$ depends on $\epsilon$ but is independent of $\varphi$.  As
defined in Lemma 4, 
$d\zeta$ is the Lebesgue measure on ${\bf C}^n$.
To derive equation (E.25), first let $u\in C_0^\infty({\bf R}^n)$ and
$\varphi = p(D)u$.  We will make the following substitutions in equation
(E.23):
$$g(\zeta)\ =\ \cases{1, & if $|\zeta| \le \epsilon$,\cr 
                      0, & otherwise;\cr}\eqno{\rm (E.26)}$$
$$F(\zeta)=\hat\varphi(x+\zeta)\quad\hbox{and }\ P(\zeta)=q(x+\zeta).
\eqno{\rm (E.27)}$$
Equation (E.23) immediately becomes
$$\left|\hat\varphi(x)D^\alpha q(x)\right|\,\int_{|\zeta|\le\epsilon}
   |\zeta|^\alpha \,d\zeta \le C_0\,\int_{|\zeta|\le\epsilon} 
    \left|\hat\varphi(x+\zeta)q(x+\zeta)\right|\,d\zeta.
\eqno{\rm (E.28)}$$
Now we may define the constant (scalar) $C_1$ as
$$C_1\ =\ {C_0\cdot|\alpha|\over\int_{|\zeta|\le\epsilon} 
      |\zeta|^\alpha\,d\zeta},
   \eqno{\rm (E.29)}$$
where $|\alpha|$ $=$ $\sum_\alpha 1$ $=$ $\alpha_1+\alpha_2+\ldots+\alpha_n$.
Recall that $C_0$ is a constant depending only on $\alpha$ and $m$; 
therefore, $C_1$ is a constant depending only on $\alpha$, $m$, and 
$\epsilon$.  In particular, $C_1$ does {\sl not\/} depend on 
the choice of the function $\varphi$.  Now, re-write equation (E.28):
$$\left|\hat\varphi(x)D^\alpha q(x)\right|
   \le {C_1\over|\alpha|}\,\int_{|\zeta|\le\epsilon} 
    \left|\hat\varphi(x+\zeta)q(x+\zeta)\right|\,d\zeta.
  \eqno{\rm (E.30)}$$
$$\eqalign{\left|\hat\varphi(x)Q(x)\right|
\ &\le\ \sum_\alpha \left|\hat\varphi(x)D^\alpha q(x)\right|\cr
  &\le {C_0\over\int_{|\zeta|\le\epsilon} |\zeta|^\alpha\,d\zeta}
   \,\left(\sum_\alpha 1\right)\,\int_{|\zeta|\le\epsilon} 
    \left|\hat\varphi(x+\zeta)q(x+\zeta)\right|\,d\zeta\cr
  &=\ C_1\,\int_{|\zeta|\le\epsilon}
     \left|\hat\varphi(x+\zeta)q(x+\zeta)\right|\,d\zeta\cr}
\eqno{\rm (E.31)}$$
%
We will substitute formula (E.10) into the right-hand side of equation
(E.32) to get
$$\left|\hat\varphi(x)Q(x)\right|
   \le C_1\,\int_{|\zeta|\le\epsilon} 
    \left|{\cal F}\big(p^{\dagger}(D)\varphi\big)(x+\zeta)\right|\,d\zeta.
    \eqno{\rm (E.32)}$$


\vskip12pt\noindent
From Fourier's (inversion) theorem, we have that
$$\varphi(0) = {1\over(2\pi)^{n/2}}\int_{{\bf R}^n} 
\hat \varphi(y)\,dy\eqno{\rm (E.33)}$$ 
$$|\varphi(0)| \le 
{1\over(2\pi)^{n/2}}\int_{{\bf R}^n} |\hat \varphi(y)|\,dy
\eqno{\rm (E.34)}$$ 
$$\eqalign{|\varphi(0)|\ 
&\le\ {1\over(2\pi)^{n/2}}\int_{{\bf R}^n} |\hat \varphi(y)|\,dy\cr
&\le\ C_2\,\int_{{\bf R}^n}\left(\int_{|\zeta|\le\epsilon} 
    \left| {\cal F}\big(p^{\dagger}(D)\varphi\big)(y+\zeta)\right|
    \,\left|Q(y)\right|^{-1}\,d\zeta\right)\,dy\cr
&=\ C_2\,\int_{{\bf R}^n}\int_{|\lambda|^2+|\mu|^2\le\epsilon^2} 
    \left| {\cal F}\big(p^{\dagger}(D)\varphi\big)(y+\lambda+i\mu)\right|
    \,\left|Q(y)\right|^{-1}\,d\lambda d\mu dy\cr}
\eqno{\rm (E.35)}$$
For $|\lambda|\le\epsilon$, we will demonstrate a constant $C_3$ such that
$Q(y+\lambda)\big(Q(y)\big)^{-1}\le C_3$
independently of $y$.
Take the generalized Taylor series expansion for
$D^\alpha q(y+\lambda)$.  $\lambda = \langle \lambda_1,
\lambda_2, \ldots, \lambda_n\rangle$ $\in$ ${\bf R}^n$ 
and $|\lambda| \le \epsilon$. 
$$D^\alpha q(y+\lambda) 
= \sum_\beta {\lambda^\beta\over\beta!}\ D^{\alpha+\beta} q(y),
\eqno{\rm (E.36)}$$
where $q$ is a polynomial of total degree $k$; therefore, $D^{\alpha+\beta} q(y)
= 0$ whenever $|\alpha+\beta| > k$.  Therefore, for all $\alpha$,
$|\alpha| \le k$
$$\left|D^\alpha q(y+\lambda)\right| \le \sum_{|\alpha+\beta|\le k}
      \left|{\lambda^\beta \over \beta !} D^{\alpha+\beta} q(y)\right|;
\eqno{\rm (E.37)}$$
$$\lambda^\beta =
       \lambda_1^{\beta_1}\lambda_2^{\beta_2}\cdots\lambda_n^{\beta_n}.$$
$$|\lambda| \le \epsilon \ \Longrightarrow\ |\lambda|^{|\beta|} \le
       \epsilon^{|\beta|}$$
$$|\beta| \le k \ \Longrightarrow\ |\lambda|^{|\beta|} \le \epsilon^k\ 
\hbox{for all multi-indices } \beta.$$
Therefore,
$$\eqalign{\left|D^\alpha q(y+\lambda)\right| 
\ &\le\ \sum_{|\alpha+\beta|\le k}
      \left|{\lambda^\beta \over \beta !}\right|
      \left| D^{\alpha+\beta} q(y)\right|\cr
\ &\le\ \sum_{|\alpha+\beta|\le k}
      \left|{\epsilon^k \over \beta !}\right|
      \left| D^{\alpha+\beta} q(y)\right|\cr
\ &\le\ \epsilon^k\cdot
      \left(\sum_{|\alpha+\beta|\le k} \left|D^{\alpha+\beta}
             q(y)\right|\right)\cr}\eqno{\rm (E.38)}$$
$$\eqalign{\left|Q(y+\lambda)\right|
\ &=\ \sum_{|\alpha|\le k} \left|D^\alpha q(y+\lambda)\right|\cr
  &\le\ \sum_{|\alpha|\le k}\ \sum_{|\alpha+\beta|\le k}
      \left|{\lambda^\beta\over\beta!}\right|\cdot\left|D^{\alpha+\beta}
               q(y)\right|\cr
  &\le\ \sum_{|\alpha|\le k} \epsilon^k\sum_{|\alpha+\beta|\le k}
         \left|D^{\alpha+\beta} q(y)\right|\cr
  &\le\ \left(\epsilon^k\,\sum_{|\beta|\le  k} 1\right)\cdot
         \sum_{|\alpha|\le k} |D^\alpha q(y)|\cr
  &=\ \left(\epsilon^k\,\sum_{|\beta|\le  k} 1\right)\cdot
         Q(y).\cr}\eqno{\rm (E.39)}$$
Therefore, since $|Q(y+\lambda)| = Q(y+\lambda)$,
$${|Q(y+\lambda)|\over |Q(y)|} \le \epsilon^k\cdot\sum_{|\beta|\le k} 1.$$
Let $C_3 = \epsilon^k\cdot\sum_{|\beta|\le k} 1$.
Setting $C_4=C_2\cdot C_3$ and substituting,
$$\eqalign{|\varphi(0)|
\ &\le\ C_4\,\int_{{\bf R}^n}\int_{|\lambda|^2+|\mu|^2\le\epsilon^2}
  \left|{\cal F}\big(p^{\dagger}(D)\varphi\big)(y+\lambda+i\mu)\right|
   \,\big(Q(y+\lambda)\big)^{-1}\,d\lambda d\mu dy.\cr
  &\le\ C_5\,\int_{{\bf R}^n}\int_{|\mu|^2\le\epsilon^2}
  \left|{\cal F}\big(p^{\dagger}(D)\varphi\big)(y+i\mu)\right|
   \,\big(Q(y)\big)^{-1}\,d\mu dy.\cr}\eqno{\rm (E.40)}$$
We now define the norm $\Vert\cdot\Vert_Q$ as follows
$$\Vert\varphi\Vert_Q = \int_{{\bf R}^n} \int_{|\mu|\le\epsilon}
   \left|\hat\varphi(y+i\mu)\big(Q(y)\big)^{-1}\right|\,d\mu dy.
   \eqno{\rm (E.41)}$$
We will first show that $\Vert\cdot\Vert_Q$ is a continuous norm on
${\cal D}$.   To see that $\Vert\cdot\Vert_Q$ is a norm, observe that
$$\Vert\cdot\Vert_Q\colon\ C_0^\infty({\bf R}^n) \to [0,\infty)
\eqno{\rm(E.42)}$$
and that $\varphi\not=0$ implies that
$$\Vert\varphi\Vert_Q > 0.\eqno{\rm (E.43)}$$
This follows immediately from the fact that $\Vert\hat\varphi\Vert_2$ $=$
$\Vert\varphi\Vert_2$.  This is a corollary to the Fourier
inverson theorem.
If each of $\varphi$ and $\psi$ are in $C_0^\infty({\bf R}^n)$, then we 
know (from the Paley-Wiener theorem) that each of the following integral
exist:
$$\hat\varphi(\lambda) 
= {1\over2\pi}\,\int_{{\bf R}^n} e^{-i\lambda\cdot x} \varphi(x)\,dx,$$
$$\hat\psi(\lambda)
= {1\over2\pi}\,\int_{{\bf R}^n} e^{-i\lambda\cdot x} \varphi(x)\,dx,$$
$$\Vert\varphi\Vert_Q 
= \int_{{\bf R}^n}\int_{|\mu|\le\epsilon} \left|\hat\varphi(y+i\mu)
   \big(Q(y)\big)^{-1}\right|\,d\mu dy,$$
$$\Vert\psi\Vert_Q 
= \int_{{\bf R}^n}\int_{|\mu|\le\epsilon} \left|\hat\psi(y+i\mu)
   \big(Q(y)\big)^{-1}\right|\,d\mu dy,$$
$$\Vert\varphi+\psi\Vert_Q 
= \int_{{\bf R}^n}\int_{|\mu|\le\epsilon} 
   \left|\big(\hat\varphi+\hat\psi\big)(y+i\mu)
   \big(Q(y)\big)^{-1}\right|\,d\mu dy.\eqno{\rm (E.44)}$$
From the so-called triangle inequality, we have that
$|\hat\varphi(\zeta) + \hat\psi(\zeta)|$ $\le$ $|\hat\varphi(\zeta)|$
$+$ $|\hat\psi(\zeta)|$.  This implies immediately that
$$\Vert\varphi+\psi\Vert_Q \le \Vert\varphi\Vert_Q + \Vert\psi\Vert_Q
   \eqno{\rm (E.45)}$$
Next, we have to show that $\Vert\cdot\Vert_Q$ is continuous.  Since
$Q$ is bounded from below, there is a positive number $C_\ell$ such
that $C_\ell < Q(y)$ for all $y\in {\bf R}^n$.  
$$\big(Q(y)\big)^{-1} < {1\over C_\ell} 
   = C_6\qquad\hbox{for all }y\in C_0^\infty({\bf R}^n). 
  \eqno{\rm (E.46)}$$
$$\eqalign{\Vert\varphi\Vert_Q 
\ &\le\  C_6\,\int_{{\bf R}^n}\int_{|\mu|\le\epsilon}
    \left|\hat\varphi(y+i\mu)\right|\,d\mu dy\cr
\ &\le\ C_7\, \sup_{\scriptstyle y\in{\bf R}^n 
                          \atop \scriptstyle|\mu|\le\epsilon}
    \left|(1+y^2)^{n+1}\hat\varphi(y+i\mu)\right|\cr
\ &\le\ C_7\,\sup_{|\mu|\le\epsilon} \Vert (I-\Delta)^{n+1}e^{\mu\cdot x}
   \varphi(x)\Vert_1.\cr}\eqno{\rm (E.47)}$$
If we define $p_1(D)$ $=$ $1-\Delta$ $=$ 
$1-\sum_{j=1}^n \partial^2/\partial x_j^2$, then we will have $p_1(x)$
$=$ $1-x^2$, $q_1(x)=p_1^{\dagger}(ix) = 1+x^2$ and
$${\cal F}\left((1-\Delta)^{n+1}e^{\mu\cdot x}\varphi(x)\right)
 = (1+y^2)^{n+1}\hat\varphi(y+i\mu).\eqno{\rm (E.48)}$$
We implicitly made use of the fact that for $\alpha\in{\bf C}$
if $h(x)=g(x)e^{i\alpha x}$ then $\hat h(s) = \hat g(t-\alpha)$.
In equation (E.47), we have used the fact that if $g\in L^1$, then
$\hat g\in C_0$ and
$$\Vert\hat g\Vert_\infty \le \Vert g\Vert_1.\eqno{\rm (E.49)}$$
The inequality is a direct consequence of the definition of the Fourier
transform.  The function space inclusions are is  consequence of the 
Lebesgue dominated convergence theorem.
The right-hand side is a continuous norm on $C_0^\infty(K)$ for each
compact set $K\subset {\bf R}^n$.  Since ${\cal D}$ has the inductive
limit topology, $\Vert\cdot\Vert_Q$ is a continuous norm on ${\cal D}$.
The basic estimate (equation (E.34)) shows that the map
$$\tilde E\colon p^{\dagger}(D)\varphi \to \varphi(0) \eqno{\rm(E.50)}$$
is well-defined.  That is, if $p^{\dagger}(D)\varphi_1$ 
$=$ $p^{\dagger}(D)\varphi_2$,
then $\varphi_1(0) = \varphi_2(0)$.  $\tilde E$ is continuous since
$\Vert\cdot\Vert_Q$ is a continuous norm.  
We now recall the Hahn-Banach Theorem:
\proclaim The Hahn-Banach Theorem.  If $M$ is a subspace of a normed linear
space $X$ and if $\tilde T$ is a bounded linear functional 
on $M$, then $\tilde T$ can be extended to a bounded linear functional $T$
on $X$ such that $\Vert T\Vert$ $=$ $\Vert\tilde T\Vert$.  
\ {\rm (See [6], page 104.)}\par
\noindent
Thus by the Hahn-Banach theorem
there is an $E$ in ${\cal D}'$ which extends $\tilde E$.  Since
$$\big(p(D)E\big)(\varphi) = E\big(p^{\dagger}(D)\varphi\big) = \varphi(0),
  \eqno{\rm(E.51)}$$
we have found a fundamental solution for $p(D)$.  \bull
\bigskip
\proclaim Lemma 1.  The maps ${\cal F}$ and ${\cal F}^{-1}$ are continuous
linear transformations of ${\cal S}({\bf R}^n)$ into 
${\cal S}({\bf R}^n)$.  Furthermore, if $\alpha$ and $\beta$ are
multi-indices, then
$$\big((i\lambda)^\alpha D^\beta\hat f\big)(\lambda)
 = {\cal F}\left(\,D^\alpha\big((-ix)^\beta f(x)\big)\right)
\eqno{\rm(E.52)}$$\par

\vskip6pt\noindent
{\bf Proof:}  The map ${\cal F}$ is clearly linear.  We directly compute
the Fourier transform of $\big(\lambda^\alpha D^{\beta} \hat f\big)(\lambda)$
from the definition.
$$\eqalign{\big(\lambda^\alpha D^{\beta} \hat f\big)(\lambda)\ 
 &=\ {1\over(2\pi)^{n/2}}\,\int_{{\bf R}^n} 
   \lambda^\alpha_x(-ix)^\beta e^{-i\lambda\cdot x}f(x)\,dx\cr
 &=\ {1\over(2\pi)^{n/2}}\,\int {1\over(-i)^\alpha}
    \big(D_x^\alpha e^{-i\lambda\cdot x}\big)(-ix)^\beta f(x)\,dx\cr
 &=\ {(-i)^\alpha\over(2\pi)^{n/2}}\,\int_{{\bf R}^n}
   e^{-i\lambda\cdot x}D_x^\alpha\big((-ix)^\beta f(x)\big)\,dx\cr}
\eqno{\rm(E.53)}$$

\vskip6pt\noindent
The above formulas require some justification.  First of all, we observe that
for 
$$\hat f(\lambda) = {1\over(2\pi)^{n/2}}\,\int_{{\bf R}^n}
        e^{-ix\cdot\lambda}f(x)\,dx,\eqno{\rm(E.54)}$$
formal differentiation under the integral sign is permitted because the right
hand side is uniformly convergent.
$$D^\beta\hat f(\lambda) = {1\over(2\pi)^{n/2}}\,\int_{{\bf R}^n}
        e^{-ix\cdot\lambda}(-i)^{|\beta|} x^\beta f(x)\,dx.
  \eqno{\rm(E.55)}$$
We now multiply both sides by $(i\lambda)^\alpha$.
$$(i\lambda)^\alpha D^\beta\hat f(\lambda) 
 = {1\over(2\pi)^{n/2}}\,\int_{{\bf R}^n}
     (i\lambda)^\alpha\, e^{-ix\cdot\lambda}(-i)^{|\beta|} x^\beta f(x)\,dx.
 \eqno{\rm(E.56)}$$
The next operation is integration by parts.  We define a function
$\phi^{(\alpha)}$ $=$ $(-i\lambda)^\alpha e^{-i\lambda\cdot x}$
$=$ $D_x^\alpha e^{-i\lambda\cdot x}$.  Again, by
uniform convergence we have
$${1\over(2\pi)^{n/2}}\,\int_{{\bf R}^n}
   (i\lambda)^\alpha\, e^{-ix\cdot\lambda}(-i)^{|\beta|} x^\beta f(x)\,dx$$
$$\qquad\ =\ (-1)^\alpha{1\over(2\pi)^{n/2}}\,\int_{{\bf R}^n}
     (-1)^{-\alpha}\, e^{-i\lambda\cdot x}\,
      D_x^\alpha \big((-ix)^\beta f(x)\big)\,dx.\eqno{\rm(E.57)}$$
The final result is
$$\eqalign{(i\lambda)^\alpha D^\beta\hat f(\lambda) 
\ &=\ {1\over(2\pi)^{n/2}}\,\int_{{\bf R}^n}
      e^{-i\lambda\cdot x}\,
      D_x^\alpha\big( (-ix)^\beta f(x)\big)\,dx\cr
  &=\ {\cal F}D_x^\alpha\big( (-ix)^\beta f(x)\big).\cr}
\eqno{\rm(E.58)}$$
This is the sought-for result.
Now we observe that
$$\eqalign{\Vert\hat f\Vert_{\alpha,\beta}\ 
&=\ \sup_\lambda \left|\lambda^\alpha\big(D^\beta\hat f\big)(\lambda)\right|\cr
&\le\ {1\over(2\pi)^{n/2}}\int_{{\bf R}^n} \big| D_x^\alpha (x^\beta
f)\big|\,dx\ <\ \infty.\cr}\eqno{\rm(E.59)}$$
Thus ${\cal F}$ takes ${\cal S}({\bf R}^n)$ into ${\cal S}({\bf R}^n)$.
There is a positive integer $k$ such that
$$\int_{{\bf R}^n} \big(1+x^2)^{-k}\,dx\ <\ \infty,\eqno{(E.60)}$$
where $dx=dx_1dx_2\cdots dx_n$, $x=\langle x_1,x_2,\ldots,x_n\rangle$.
$$\eqalign{\Vert \hat f\Vert_{\alpha,\beta}
\ &\le\ {1\over(2\pi)^{n/2}}\,\int_{{\bf R}^n}
  {\big(1+x^2\big)^{-k}\over\big(1+x^2\big)^{-k}}\left| D_x^\alpha(-ix)^\beta
              f(x)\right|\,dx\cr
  &\le\ {1\over(2\pi)^{n/2}}\,\left(\int_{{\bf R}^n} \big(1+x^2\big)^{-k}
      \,dx\right)\ \sup_{x}\left\{\big(1+x^2\big)^{+k}\left|D_x^\alpha
          (-ix)^\beta f(x)\right|\,\right\}\cr}\eqno{\rm(E.61)}$$
Using Leibnitz's rule we easily conclude that there exist multi-indices
$\alpha_j$, $\beta_j$ and constants $c_j$ such that
$$\Vert\hat f\Vert_{\alpha,\beta} \ \le\ \sum_{j=1}^M
            c_j\cdot\Vert
            f\Vert_{\alpha_j,\beta_j}\eqno{\rm(E.62)}$$
Therefore, ${\cal F}$ is bounded.  A bounded linear functional is always
continuous.  A symmetric argument holds for the inverse operator 
${\cal F}^{-1}$.\quad \bull
\bigskip
\proclaim Lemma 2 (Malgrange).  If $f(z)$ is a holomorphic function of a 
complex variable $z$ for $|z|\le 1$ and $p(z)$ is a polynomial
$$p(z)=a_mz^m + a_{m-1}z^{m-1} + \ldots + a_1z + a_0\eqno{\rm(E.63)}$$
of degree $m$, then
$$\left| a_mf(0) \right| 
\le {1\over 2\pi}\,\int_0^{2\pi} 
\left|f\big(e^{i\theta}\big)p\big(e^{i\theta}\big)\right|\,d\theta.
\eqno{\rm(E.64)}$$\par

\vskip6pt\noindent
{\bf Proof:}  For $j=1,\ldots,m$, let $z_j$ be a zero of the polynomial 
$p(z)$.  Order the roots such that for $j=1,\ldots,k$, $z_j$ is a zero
of the polynomial interior to the unit circle, that is $|z_j| < 1$
for $j=1,\ldots,k$ and $|z_j|\ge 1$ for $k<j\le m$.  Define $q(z)$ as follows
$$q(z) = p(z)\,\prod_{j=1}^k{\bar z_j z -1 \over z-z_j}.
\eqno{\rm(E.65)}$$
Clearly, $p(z) = p_1(z)\cdot(z-z_1)\cdots(z-z_k)$ so that
$$q(z) = p_1(z)\cdot\prod_{j=1}^k \left(\bar z_j z-1\right)
=\left(a_m\cdot\prod_{j=k+1}^m(z-z_j)\right)\cdot\prod_{j=1}^k(\bar z_jz-1).
\eqno{\rm(E.66)}$$
Since $q(z)$ is a polynomial in ${\bf C}$, $q(z)$ is analytic.  Moreover,
for $0\le \theta<2\pi$, $\big|e^{i\theta}\big|=1$, and
$$\eqalign{\left|q\big(e^{i\theta}\big)\right|\  
  &=\ \left|p_1\big(e^{i\theta}\big)\cdot(\bar z_1e^{i\theta}-1) 
      (\bar z_2e^{i\theta}-1)\cdots(\bar z_ke^{i\theta}-1)\right|\cr 
  &=\ \left|p_1\big(e^{i\theta}\big)\right|
       \cdot\prod_{j=1}^k\left|\bar z_je^{i\theta}-1\right|\cr
  &=\ \left|p_1\big(e^{i\theta}\big)\right|
       \cdot\prod_{j=1}^k\left|\bar z_j -e^{-i\theta}\right|\cr
  &=\ \left|p_1\big(e^{i\theta}\big)\right|
       \cdot\prod_{j=1}^k\left|z_j - e^{i\theta}\right|\cr
  &=\ \left|p\big(e^{i\theta}\big)\right|.}\eqno{\rm(E.67)}$$
Thus $q(z)$ is regular in the unit circle and $|p(z)|$ $=$ $|q(z)|$
for $|z| = 1$.  We apply the Cauchy integral formula, 
$\int_{\partial\Omega} g(\xi)\,d\xi/(\xi-z)$ $=$ $2\pi i\cdot g(z)$, to the
function $g(z)$ $=$ $f(z)q(z)$ to obtain
$$\eqalign{{1\over2\pi}\,\int_0^{2\pi}
  \left|f\big(e^{i\theta}\big)\right|\cdot
  \left|p\big(e^{i\theta}\big)\right|\,d\theta  
  &=\ {1\over2\pi}\,\int_0^{2\pi} \left|f\big(e^{i\theta}\big)
        p\big(e^{i\theta}\big)\right|\,d\theta\cr
  &=\ {1\over2\pi}\,\int_0^{2\pi} \left|f\big(e^{i\theta}\big)
        q\big(e^{i\theta}\big)\right|\,d\theta\cr
  &\ge\ {1\over2\pi}\,\left|\ \int_0^{2\pi} f\big(e^{i\theta}\big)
        q\big(e^{i\theta}\big)\,d\theta\right|\cr
  &=\ \left| f(0)q(0) \right|.\cr}\eqno{\rm(E.68)}$$
Now consider $q(0)$.  
$|\,q(0)|$ $=$ $|p_1(0)|$ $=$ $|a_m|\cdot\prod_{j=k+1}^m |z_j|$  
$\ge$ $|a_m|$.
$${1\over2\pi}\,\int_0^{2\pi}
  \left|f\big(e^{i\theta}\big)p\big(e^{i\theta}\big)\right|\,d\theta  
  \ge\ \left| f(0)q(0) \right| \ge \left| a_mf(0)\right|.
\eqno{\rm(E.69)}$$
\noindent This completes the lemma.\quad\bull
\bigskip
$$\beginpicture
%    \ninepoint
    \setcoordinatesystem units <1.00in,1.00in>
    \setplotarea x from -1.75 to 1.75, y from -1.25 to 1.25 %
    \plotheading {\lines{%
      $p(z) = a_7z^7+\ldots+a_1z+a_0$ \cr
      a polynomial of degree $7$.\cr}}
    \put {$\scriptstyle\bullet$} at 0 0 %
    \put {$0$} [rt] <-2pt,-2pt> at 0 0 %
%    \replot "feb15.t01" %
%    \savelinesandcurves on "feb15.t01"
%    \writesavefile {Feb 15, 92 Homework}
\ifexpressmode
  \put {\bf EXPRESSMODE } at .7071 .7071
\else
    \circulararc 360 degrees from 1 0 center at 0 0 %
    \arrow <10pt> [.1, .4] from 0 0 to .7071 .7071 %
\fi
    \put {$\scriptstyle\circ$} at -.6 0 %
    \put {$\scriptstyle\circ$} at .6 -.2 %
    \put {$\scriptstyle\circ$} at -.1 -.6 %
    \put {$\scriptstyle\circ$} at 0 .5 %
    \put {$\scriptstyle\circ$} at 1.2 1.0 %
    \put {$\scriptstyle\circ$} at -1.2 .8 %
    \put {$\scriptstyle\circ$} at -1.6 -.6 %
    \put {$z_1$\raise3pt\hbox{$\,$}} [rt] at -.6 0 %
    \put {$z_2$\raise3pt\hbox{$\,$}} [rt] at .6 -.2 %
    \put {$z_3$\raise3pt\hbox{$\,$}} [rt] at -.1 -.6 %
    \put {$z_4$\raise3pt\hbox{$\,$}} [rt] at 0 .5 %
    \put {$z_5$\raise3pt\hbox{$\,$}} [rt] at 1.2 1.0 %
    \put {$z_6$\raise3pt\hbox{$\,$}} [rt] at -1.2 .8 %
    \put {$z_7$\raise3pt\hbox{$\,$}} [rt] at -1.6 -.6 %
    \put {$1$} at .4 .2 %
    \put {$\scriptstyle\circ$ zeros } [lb] at 1.5 .4 %
    \put {$m = 7 $} [lb] at 1.5 0 %
    \put {$k= 4 $} [lb] at 1.5 -.4 %
    \put {${\bf D} = $ unit disk} [lb] at 1 -1.2 %
\endpicture $$
\vskip6pt
$$q(z) = p(z)\cdot\prod_{j=1}^k {\bar z_jz-1 \over z-z_j}$$
$$q(z) = p_1(z)\,\prod_{j=1}^k(\bar z_jz-1)$$
$$p_1(z) = a_m\cdot\prod_{j=k+1}^m (z-z_j)\quad\hbox{above } 
    p_1(z)= a_7(z-z_5)(z-z_6)(z-z_7)$$
$$p_1(0) = a_m\cdot\prod_{j=k+1}^m z_j$$
$$|p_1(0)| = |a_m|\cdot\prod_{j=k+1}^m|z_j| \ge |a_m|\cdot 1= |a_m|$$
$$\Longrightarrow\ |q(0)| \ge |a_m|$$
$${1\over2\pi}\int_0^{2\pi}|f(e^{i\theta})p(e^{i\theta})|\,d\theta
  \ge {1\over2\pi}\left|\int_0^{2\pi}
            f(e^{i\theta})q(e^{i\theta})|d\theta\right|
  = |f(0)q(0)| \ge |a_mf(0)|$$
\bigskip
\proclaim Lemma 3.  If $f(z)$ is a holomorphic function of a 
complex variable $z$ for $|z|\le 1$ and $p(z)$ is a polynomial
$$p(z)=a_mz^m + a_{m-1}z^{m-1} + \ldots + a_1z + a_0\eqno{\rm(E.70)}$$
of degree $m$, then
$$\left| f(0)\, p^{(k)}(0) \right| 
\le {m!\over (m-k)!\cdot2\pi}\,\int_0^{2\pi} 
\left|f\big(e^{i\theta}\big)p\big(e^{i\theta}\big)\right|\,d\theta.
\eqno{\rm(E.71)}$$\par

\vskip6pt\noindent
{\bf Proof:}  Let $p$ be a monic polynomial of degree $m$ such that
$$\eqalign{p(z)\ &=\ z^m + a_{m-1}z^{m-1} + \ldots + a_1z +a_0\cr
                 &=\ \prod_{j=1}^m (z-z_j),\cr}\eqno{\rm(E.72)}$$
where $\big\{\,z_j\,\big|\,j=1,2,\ldots,m\,\big\}$ are the roots of $p$.
Apply the previous lemma to the polynomial of degree $k$
$$\prod_{j=1}^k (z-z_j)\eqno{\rm(E.73)}$$
and the holomorphic (analytic) function
$$f(z)\cdot\prod_{j=k+1}^m (z-z_j),\eqno{\rm(E.74)}$$
for some $1\le k\le m$.  This yields the formula
$$\left| f(0)\prod_{j=k+1}^m z_j\right| \le
   {1\over2\pi}\,\int_0^{2\pi} \left|f\big(e^{i\theta}\big)
        p\big(e^{i\theta}\big)\right|\,d\theta.\eqno{\rm(E.75)}$$
We need to consider the derivatives of $p(z)$.
$$p'(z) = p^{(1)}(z) = (z-z_2)\cdots(z-z_m) + (z-z_1)(z-z_3)\cdots(z-z_m)
    +\ldots$$
$$\qquad\ + (z-z_1)(z-z_2)\cdots(z-z_{m-1});\eqno{\rm(E.76)}$$
the first derivative has $m$ terms and each term has $m-1$ factors of the
form $(z-z_j)$.
The second derivative can be written
$$\eqalign{p''(z)\ &=\ p^{(2)}(z)\cr 
  &=\ (z-z_3)\cdots(z-z_m)+(z-z_1)(z-z_4)\cdots(z-z_m)\cr
  &\qquad +\ldots+(z-z_1)\cdots(z-z_{m-2})+(z-z_2)(z-z_4)\cdots(z-z_m)\cr
  &\qquad +(z-z_1)(z-z_3)(z-z_5)\cdots(z-z_m)+\dots.\cr}\eqno{\rm(E.77)}$$
The second derivative has $m(m-1)$ terms each of which is a product
of $(m-2)$ factors of the form $(z-z_j)$.
$$p^{(k)}(z) = (z-z_{k+1})(z-z_{k+2})\cdots(z-z_m) +\ldots
         +(z-z_1)(z-z_2)\cdots(z-z_{m-k+1}).\eqno{\rm(E.78)}$$
The $k$th derivative has $m\cdot(m-1)\cdots(m-k+1)$ terms and each
term has $m-k$ factors.  
Recall that
$$m\cdot(m-1)\cdots(m-k+1) = {m!\over(m-k)!}.\eqno{\rm(E.79)}$$
Each of the $m!/(m-k)!$ terms has the form
$$\prod_{j=k+1}^m \left(z-w_j\right),\eqno{\rm(E.80)}$$
where $\{ w_{k+1},\ldots,w_m\}$ is one of the $m!/(m-k)!$
possible subsets of $m-k$ elements from $\{z_1,z_2,\ldots,z_m\}$.
Thus, we have that $p^{(k)}(0)$ is the sum of $L$ $=$ $m!/(m-1)!$
terms of the form
$$(-1)^{m-k}\,\prod_{j=k+1}^m w_{\ell,j},\eqno{\rm(E.81)}$$
for some $\ell\in L$.  This yields
$$\eqalign{\left|f(0)\cdot p^{(k)}(0)\right|\ 
&=\ \left|\, f(0)\cdot \sum_{\ell\in L} (-1)^{m-k} 
    \prod_{j=k+1}^m w_{\ell,j}\right|\cr
&\le\ \sum_{\ell\in L} \left|\,f(0)\cdot \prod_{j=k+1}^m w_{\ell,j}\right|\cr
&\le\ {m!\over(m-k)!}\,\left|\,f(0)\cdot\prod_{j=k+1}^m w_j\right|\cr
&\le\ {m!\over(m-k)!\cdot2\pi}\,\int_0^{2\pi} \left|f\big(e^{i\theta}\big)
         p\big(e^{i\theta}\big)\right|\,d\theta.\cr}\eqno{\rm(E.82)}$$
For a general polynomial $p$, multiply both sides of the inequality
by $|a_m|>0$.  This proves the lemma.\quad\bull 
\bigskip
\proclaim Lemma 5.  Suppose $\tilde q(x)$ $=$ 
$\big(\sum_{|\alpha|\le k}|q^{(\alpha)}(x)|^2\big)^{1/2}$, where
$q^{(\alpha)}(x)=D_x^{\alpha}q(x)$, $k$ is the total degree of $q$, 
$\varphi\in C_0^{\infty}({\bf R}^n)$, and $\epsilon>0$.  Then
$$\left|\tilde q(x)\hat\varphi(x)\right| \le C_1'\,\int_{|\zeta|\le\epsilon}
\left|\hat\varphi(x+\zeta)\,q(x+\zeta)\right|\,d\zeta.\eqno{\rm(E.83)}$$\par

\vskip6pt\noindent
{\bf Proof:}  Put $p(D) u = v$, where $u\in C_0^{\infty}({\bf R}^n)$.
Then $q(\zeta)\hat u(\zeta) = \hat v(\zeta)$.  Apply the previous lemma
(Lemma 4) with $F(\zeta) = \hat v(\xi+\zeta)$, with 
$p(\zeta)$ replaced by $p(\xi+\zeta)$ and with $|g(\zeta)|=1$ when
$|\zeta|\le\epsilon$ and $=0$ otherwise.  Since $\tilde q(\xi)$ $\le$ 
$\sum_\alpha |D^\alpha q(\xi)|$, we obtain, from the previous lemma,
$$\eqalign{\left|\hat u(\xi)\tilde q(\xi)\right| 
\ &\le\ C_1'\,\int_{|\zeta|\le\epsilon} \left|\hat u(\xi+\zeta)\,
   p(\xi+\zeta)\right|\,d\zeta\cr
  &=\ C_1'\,\int_{|\zeta|\le\epsilon} \left|\hat v(\xi+\xi)\right|\,d\zeta.
  \cr}\eqno{\rm(E.84)}$$
We then apply Fourier's integral theorem,
$$\eqalign{|u(0)|
\ &\le\ (2\pi)^{-n/2}\,\int_{{\bf R}^n} |\hat u(\xi)|\,d\xi\cr
  &\le\ C_1''\,\int_{|\zeta|\le\epsilon}\left(\int_{{\bf R}^n}
        |\hat v(\xi+\zeta)|/\tilde q(\xi)\,d\xi\right)\,d\zeta\cr
  &\le\ C_1''\,\int_{{\bf F}^n}
    \left(\int_{\zeta^{\prime2}+\eta^{\prime2}\le\epsilon^2}
     |\hat v(\xi+\xi'+i\eta')|/\tilde q(\xi)\,d\zeta' d\eta'\right)\,d\zeta.
            \cr}\eqno{\rm(E.85)}$$
On the other hand, we have
$$\tilde q(\xi+\xi')/\tilde q(\xi) \le C_2'\quad\hbox{when}\quad
              |\zeta'|\le\epsilon,\eqno{\rm(E.86)}$$
because
$$D^\alpha q(\xi+\xi') = \sum_\beta {(\xi')^\beta\over\beta !}\,
      D^{\alpha+\beta} q(\xi),\eqno{\rm(E.87)}$$
so that $|D^\alpha q(\xi+\xi')|/\tilde q(\xi)$ is bounded when
$|\xi'|\le\epsilon$.  Therefore,
$$\eqalign{|u(0)|
\ &\le\ C_1''C_2'\,\int_{{\bf R}^n} 
   \left(\int_{\zeta^{\prime2}+\eta^{\prime2}\le\epsilon^2}
      |\hat v(\xi+\xi'+i\eta')|/\tilde q(\xi+\xi'))\,d\xi' d\eta'\right)
         \,d\xi\cr
\ &\le\ C_3'\,\Vert v\Vert^{\prime}\cr}\eqno{\rm(E.88)}$$
where
$$\Vert v\Vert^{\prime} = \int_{|\eta|\le\epsilon}
      \left(\int_{{\bf R}^n} |\hat v(\xi+i\eta)|\big(\tilde q(\xi)\big)\right)
          \,d\eta.\eqno{\rm(E.89)}$$
Recall that $u\in C_0^\infty({\bf R}^n)$.  Now observe that $C_3''$ is a
constant which depends only on $\epsilon$ (and, of course, on $k$ and $m$,
the fixed parameters).\quad\bull
\vfill\eject
\centerline{\bf References for Appendix E}
\vskip24pt\noindent
\llap{[1]\quad }Ehrenpreis, Leon; Solutions of some problems of division.  {\sl Amer. J. 
    Math.} {\bf 76}, 883--903 (1954)\par
\vskip6pt\noindent
\llap{[2]\quad }H\"ormander, Lars; Local and global properties of fundamental solutions.
  {\sl Math. Scand.} {\bf 5}, 27--39 (1957)\par
\vskip6pt\noindent
\llap{[3]\quad }Lewy, Hans; An example of a smooth linear 
partial differential equation
without solutions. {\sl Ann. of Math.} {\bf 66}, 155--158 (1957)\par
\vskip6pt\noindent
\llap{[4]\quad }Malgrange, B.; Existence et approximation 
des solutions des \'equations 
aux d\'eriv\'ees partielles et des \'equations de convolution.
{\sl Ann. Inst. Fourier} {\bf 6}, 271--355 (1955--56).\par
\vskip6pt\noindent
\llap{[5]\quad }Reed, Michael and Barry Simon; {\sl Methods of Modern Mathematical 
Physics II: Fourier Analysis, Self-Adjointness} (San Diego: Academic Press, 
Inc., 1975)\par
\vskip6pt\noindent
\llap{[6]\quad }Rudin, Walter; 
{\sl Real and Complex Analysis--Third Edition\/} (NY: McGraw-Hill Book
Company, 1987)\par
\vskip6pt\noindent
\llap{[7]\quad }Yosida, K\^osaku; {\sl Functional Analysis---Sixth Edition\/} 
(Berlin:  Springer-Verlag, 1980)\par

\vfill\eject\end








%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%                                                                     %%%%%
%%%%%                          Chapter 4                                  %%%%% 
%%%%%                                                                     %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\headline={\tenrm\hfill Adaptive Quadrature Routine}
\centerline{\bf Chapter 4}
\bigskip
\noindent{\bf\llap{4.0\quad}Introduction.}  Sometimes it's not easy to tell
just how many subdivisions are needed to do a numerical integration. 
The reasons for this uncertainty are many.
Function exhibiting bizarre behavior at a point or in a 
subinterval require special attention.  
Some continuous functions fail to have derivatives.
Moreover,  a function can be integrable
even if it is not be continuous at every point. 
For these cases and many others it's nice to
have some kind of automatic system for determining just when an approximation
is ``good enough.''  Fast modern computers may still be too slow.
In particular, if a function is
defined in terms of an integral its numerical 
evaluation may be time consuming; moreover, functions so defined may
be prone to significant error propagation.
In general, a function defined over an interval may be easier to integrate
over some subintervals than over others.  The methods which have been 
devised to deal with this phenomenon are call {\sl adaptive quadrature
routines}, or {\sl AQRs}.  Adaptive methods make use of error estimates to
automatically determine the number of subintervals, $n$, and
the interval size $h$.  These methods may also use smaller values of the
interval size $h$ over subintervals where it is needed.  There are many
systematic ways of doing this.  There is one method which is easily 
programmed (in BASIC), but which is theoretically involved.  This method
entails comparing Simpson's rule with the so-called trapezoid rule.  Another
method, not so easily programmed and requiring recursion, compares two 
different estimates from Newton-C\^otes' formulas.  In the following
paragraphs, we'll develop computer programs for the simplest 
computers---the programmable calculators, and compare results.
\vfill\eject
\noindent{\bf\llap{4.1\quad}A Simple Approach.}  The best introduction to this
subject is to work simple examples and then develop the theory as needed. 
Excellent numerical results can be obtained using a very simple approach.
We are given a function, $f(x)$, an interval $[\alpha,\,\beta\,]$, and an
error bound $\epsilon>0$.
First, approximate the integral with two (different) approximations:
the trapezoid rule and Simpson's rule.  Begin with $n=10$ subintervals.
Second, compute the error estimate (relative) as follows
$$E\ =\ \left| {A_T - A_S \over A_T + A_S} \right|, \eqno(67)$$
where $A_T$ is the approximation obtained from the trapezoid rule and
$A_S$ is the value from Simpson's rule.  If $E < \epsilon$, then the
results are OK.  If not, then set $n=20$ ($n=2\cdot n$) subintervals
and repeat the process.  Continue until either $E < \epsilon$ or $n$
get too large, say $n > 1000$.  This is easy to program and fast
to run.  The first prototype would look like this:
\medskip
{\tt\parindent=0pt\obeylines
10 \ DEF FNA(X) = SQR(1.0-SIN(EXP(1)/2){\char94}2*SIN(X){\char94}2)
20 \ N=10 : A=0.0 : B=2.0*ATN(1.0)  
30 \ H=(B-A)/N : S = 0.0 : T = 0.0
40 \ FOR I=1 TO N STEP 2
50 \ X=A+(I-1)*H : S = S + FNA(X) : T = T + FNA(X)
60 \ X=A+I*H : T = T + 4.0*FNA(X)
70 \ X=A+(I+1)*H : S = S + FNA(X) : T = T + FNA(X)
80 \ NEXT I : S1 = H*S : S2 = H*S/3.0 : E = ABS((S1-S2)/(S1+S2))
90 \ IF (E>0.000001) THEN N = 2*N : GOTO 30
100 PRINT "Integral = "; T, N
}
\medskip\noindent
The answer is given immediately as
\smallskip
{\tt\parindent=0pt\obeylines
Integral = 1.054686 \qquad 20
}
\smallskip\noindent
From our previous calculations, we know that this is precisely the
correct number of subintervals for this precision.  It is a good idea
to examine an alternative means of coding the same program.  It runs
faster if we use a {\tt GOSUB} construction instead of the defined
function.  The answers will be identical and we can consult the table
prepared in paragraph 1.2 for numerical results.
\medskip
{\tt\parindent=0pt\obeylines
10 \ D=0.0001 : INPUT "A,B = ";A,B : N=10
20 \ N=N+10 : H=(B-A)/N : X=A : S=0.0 : T=0.0  
30 \ FOR I=1 TO N STEP 2
40 \ GOSUB 100 : S = S + Y*H/3 : T = T + Y*H/2 : X = X + H
50 \ GOSUB 100 : S = S + 4*Y*H/3 : T = T + Y*H : X = X + H
60 \ GOSUB 100 : S = S + Y*H/3 : T = T + Y*H/2 : NEXT I
70 \ IF ABS((S-T)/(S+T)) > D THEN 20
80 \ PRINT "Integral = "; T, N
90 \ PAUSE : INPUT "Error \% = ";D : GOTO 20
100 Y = SQR(1.0-SIN(EXP(1)/2){\char94}2*SIN(X){\char94}2) : RETURN
}
\medskip\noindent
This program gives one the ability to increase the error estimate and
thereby determine closely the exact number of steps required.
Using this program with a hand-held calculator will give a very
good qualitative insight into the workings of the algorithm.  The 
interested reader will later be invited to experiment with other
adaptive quadrature routines to determine which is better.
\vfill\eject
\noindent{\bf\llap{4.2\quad}An Adaptive Quadrature Routine.}  Suppose that we
have a function $f(x)$ defined over an interval $[a,b]$.  We would like to
compute the integral close to the actual value.  Let's assume that we know
that the function is bounded, $|f(x)| \le M$ for all $x\in[a,b]$.  Then, for
an answer within say 0.01\% of the actual error, we would choose
$$\epsilon = {0.0001 \over M + 1}.\eqno(68)$$
With an overall error of $\epsilon$, we want to divide up the interval
$[a,b]$ so that the total error of integration is less than $\epsilon$.
There is an error formula which depends on the value of the fourth 
derivative of $f(x)$ on $[a,b]$, but we will depend on another criterion.
If we choose Simpson's rule
$$A_2 = {h\over3}\big( y_0 + 4y_1 + 2y_2 + 4y_3 + y_4\big),\eqno(69)$$
then the doubled interval rule
$$A_1 = {2h\over3}\big( y_0 + 4y_2 + y_4\big),\eqno(70)$$
is a good approximant.  The estimate of the error is\footnote{\dag}{Francis
Scheid, {\it Schaum's Outline Series Theory and Problems of Numerical
Analysis 2nd edition}, (NY: McGraw-Hill Book Company, 1988), page 124.}
$$E_2 \simeq {|A_2-A_1| \over 15 }.\eqno(71)$$
This follows from a comparison of the two error estimates:
$$E_1 = -{(b-a)(2h)^4y^{(4)}(\xi_1)\over180}\qquad
  E_2 = -{(b-a)h^4y^{(4)}(\xi_2)\over180},\eqno(72)$$
where $E_1$ is the error estimate for Simpson's rule with an interval length
of $2h$ and $E_2$ is the error estimate for Simpson'r rule with interval
length of $h$.
We can accept a value of $A_2$ whenever $|A_2-A_1| \le 15\epsilon/2^k$
is reached, for an interval being halved $k$ times.
An alternate derivation\footnote{\ddag}{Mike Stimpson, ``Numerical Integration
Using Adpative Quadrature,'' {\it The C Users Journal}, May 1992, page 35.}
supposes that $A$ is the actual value of the integral and requires
$E_1 = |A-A_1|$, $E_2 = |A-A_2|$, where $A_1$ is Simpson's rule and $A_2$ is
the so-called five-point approximation (Simpson's rule with interval length
of $h$).  From the triangle inequality and the inequality
$$E_2 \le {1\over16}\,E_1,\eqno(73)$$
the result follows.  The programming makes use of a technique known as
{\sl recursion}.  Recursion occurs when a computer subroutine calls
itself.  The situation in analysis occurs when a function, say $f(x)=\sin(x)$,
is used to define $g(x) =\sin\big(\sin(x)\big)=f\big(f(x)\big)$. Now we will
examine a ``C'' program which uses recursion to do adaptive quadrature:
\medskip
{\tt\obeylines\parindent=0pt
\#include <math.h>
\#include <float.h>
\#include <stdio.h>
double adaptive (double a, double b, double (*f)(double x), double *err);
double f(double x);
int main() {\char123}
\ double a,b,*err,int1;
\ a = (double) 0.0;
\ b = (double) 2.0*atan(1.0);
\ *err = (double) 0.00000001;
\ printf("{\char92}nleft endpoint \ = \%.16lf",a);
\ printf("{\char92}nright endpoint = \%.16lf",b);
\ printf("{\char92}nerror estimate = \%.16lf",*err);
\ int1 = adaptive(a, b, f, err);
\ printf("{\char92}nIntegral \ \ \ \ \ \ = \%.16lf",int1);
\ printf("{\char92}nTheoretical \ \ \ = \%.16lf",1.0);
\ return(0);
{\char125}
double f(double x) {\char123}
\  double y;
\  y = sin(x);
\  return (y);
{\char125}
double adaptive (double a, double b, double (*f)(double x), double *err) {\char123}
\  double h, s1, s2, s3, s4, s5, error, err1, err2, t1, t2;
\  h = b - a;
\  s1 = (*f)(a);  s2 = (*f)(a+0.25*h);  s3 = (*f)(a+0.5*h);
\  s4 = (*f)(b-0.25*h);  s5 = (*f)(b);
\  t1 = h*(s1+4.0*s3+s5)/6.0;
\  t2 = h*(s1+4.0*s2+2.0*s3+4.0*s4+s5)/12.0;
\  error = fabs(t1-t2)/15.0;
\  if (error < *err) {\char123}
\quad    *err = error;  return(t2);
\quad {\char125}
\  else {\char123}
\quad   err1 = err2 = *err/2.0;  t2 = adaptive(a, a+0.5*h, f, \&err1);
\quad    t2 += adaptive(a+0.5*h, b, f, \&err2);  return(t2);
\quad  {\char125}
{\char125}
/* End of File */
}
\medskip
\noindent The output of the program consists of the endpoints, the error
estimate (maximum), the computed answer and the theoretical answer.  Here, we
know that $\int_0^{\pi/2} \sin(x)\,dx = 1$.  The quadrature should return an
answer close to 1; we see that it does.
\medskip
{\tt\obeylines
left endpoint \ = 0.0000000000000000
right endpoint  = 1.5707963267948970
error estimate  = 0.0000000100000000
Integral \ \ \ \ \ \ = 1.00000000025973520
Theoretical \ \ \ = 1.0000000000000000
}
%
\medskip
\noindent While the adaptive quadrature routine (AQR) is impressive, it often
fails to evaluate improper integrals.  If the integrand of an improper
integral experiences an anomoly (``blows up'') at a point, it is necessary
to compute successive approximations for values approaching that point.  There
is no ``quick and easy'' solution to improper integrals.  One cardinal rule is
to graph the function first.  It is frequently possible to determine the
correct approach from a graph when so-called adaptive methods fail.
\vfill\eject
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%                                                                     %%%%%
%%%%                   Chapter 5---Metafont                              %%%%%
%%%%                                                                     %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\headline{\tenrm\hfill METAFONT}
\centerline{\bf Chapter 5}
\bigskip
\noindent{\bf\llap{5.0\quad}Introduction.}  The creation of fonts is
usually something not approached lightly.  With the typesetting software of
\TeX\ this is especially true.  The problem of merging text with graphics
has been resolved in the earlier chapters using the macros of \PiCTeX, with
varying degrees of success.  Now it is necessary to investigate the more
sophisticated means of doing graphics.  \PiCTeX\ can be compared to assembly
language; {\bf METAFONT} to machine code.  \PiCTeX\ is straightforward and
easy to use; however, it uses a great deal of randon access memory (RAM) and
it is slow.  {\bf METAFONT}, on the other hand, creates a font which is 
read in directly.  It uses little memory and is fast.  There is a price which
must be paid for using {\bf METAFONT}---it has to be prepared in its own
language and pre-processed through its own processor.
\medskip
\noindent
The public domain version of {\bf METAFONT} does not include a user-friendly
view program.  It is necessary to produce files and then process them through
a \TeX\ program named {\tt testfont.tex}.  The program {\tt testfont.tex} is
well-written and clever.  It produces excellent output; however, one must
have a commercial previewer to use it on-line.  Otherwise, the output must be
printed on a printer.  The public domain previewer, {\tt cdvi12.arc}, only
employs the sixteen basic \TeX\ fonts; {\bf METAFONT} actually produces 
{\sl new\/} fonts---that is, the graphics from {\bf METAFONT} are fonts.  Each
graph is actually a single letter or blocks of letters ``tiled'' together.
Donald E. Knuth is correct in saying, ``Geometric design are rather easy;
$\dots$.''  {\bf METAFONT} does allow the user a great deal of freedom in
constructing graphs.  There are pitfalls, however, and some experimentation
is needed to construct the desired output.
\medskip
\noindent
In {\bf METAFONT}, the subscripted variables $x_1$ and $y_1$ may be
written either as $x[1]$, $y[1]$ or as $x1$, $y1$; likewise, $x_2$, $y_2$ may be
written either as $y[2]$, $y[2]$ or $x2$, $y2$, and so on.  The
letter $z$ stands for the ordered pair $(x,y)$ (just as is the case
in complex analysis).  Thus $z_1$ is equivalent to $(x_1, y_1)$,
$z_2$ is equivalent to $(x_2, y_2)$, etc.  When we write $z3$ we mean
$(x3, y3)$ or $(x[3], y[3])$.  This shorthand aids in construction.
For those who recall FORTRAN (the FORmula TRANSlator of the 1960's),
the symbol {\tt **} meant exponentiation.  $x${\tt **}$y$ meant
$x^y$.  This was the first encounter many of us had with a ``two
character'' operator.  (The binary operators $+$, $-$, $*$ (times),
and $/$ were well understood.)  In BASIC (Beginner's All-purpose
Symbolic Instruction Code), FORTRAN's {\tt **} has been replaced by
the single character operator {\tt \char94}.  {\bf METAFONT} has
re-introduced the two character operator.  This time $z1${\tt..}$z2$
means draw a (curved) line from $z1$ ($z[1]$) to $z2$ ($z[2]$) and
$z1--z2$ means draw a straight line from $z1$ to $z2$.  The symbol
``$:=$'' means assignment while ``$=$'' means simple equality
(replacement).  Variables followed by the hash mark (American pound
sign) ({\tt\#}) are called ``sharped'' and have absolute values
(values that are resolution-independent) 
as opposed to other variables whose values are derived and
determined programmatically.
\vfill\eject
\noindent{\bf\llap{5.1\quad}A Metafont File.}  We will begin by
listing a simple {\bf METAFONT} file that will produce a small
Gaussian distribution. 
\medskip
\noindent
{\tt\obeylines\parindent=0pt\ttraggedright
\% The statement "mode\_setup" adapts METAFONT to the current task.
mode\_setup;
\ em\#:=30pt\#; cap\#:=10pt\#;
\% The statement "define\_pixels( , )" converts the values
\% \quad of its (two) arguments into pixel (picture element) units.
define\_pixels(cap, cap);
\% The "beginchar" operation assigns values to the variables
\% \quad w, h, and d, which represent the width, height,
\% \quad and deptgh of the current character's bounding rectagle.
beginchar("A",em\#,em\#,0); "A for Gaussian Distribution";
\ pickup pencircle scaled 0.4pt; \quad \% Use a pen of diameter 0.4 point.
\ z1=(0,0.39894b); 
\ z2=(0.1a,0.39695b);
\ z3=(0.2a,0.39104b);
\ z4=(0.3a,0.38138b);
\ z5=(0.4a,0.36827b);
\ z6=(0.5a,0.35206b); 
\ z7=(0.6a,0.33322b);
\ z8=(0.7a,0.31225b);
\ z9=(0.8a,0.28969b);
\ z10=(0.9a,0.26608b);
\ z11=(a,0.24197b); 
\ z12=(1.2a,0.19418b);
\ z13=(1.4a,0.14972b);
\ z14=(1.6a,0.11092b); 
\ z15=(1.8a,0.07895b);
\ z16=(2a,0.05399b); 
\ z17=(2.5a,0.01752b);
\ z18=(3a,0.00443b);
\ z19=(3.5a,0.00087b);
\ z20=(4a,0.0001338b);
\ a=w; b=8h;
\ draw z1..z2..z3..z4..z5..z6..z7..z8..z9..z10
\qquad        ..z11..z12..z13..z14..z15..z16..z17..z18..z19..z20;
\ for k=2 step 1 until 20: 
\ z[20+k]=(-x[k],y[k]); endfor
\ draw z1..z22..z23..z24..z25..z26..z27..z28..z28..z30
\qquad        ..z31..z32..z33..z34..z35..z36..z37..z38..z39..z40;
\ draw (0,-0.02b)--(0,0);
\ draw (1a,-0.02b)--(1a,0);
\ draw (2a,-0.02b)--(2a,0);
\ draw (3a,-0.02b)--(3a,0);
\ draw (4a,-0.02b)--(4a,0);
\ draw (-1a,-0.02b)--(-1a,0);
\ draw (-2a,-0.02b)--(-2a,0);
\ draw (-3a,-0.02b)--(-3a,0);
\ draw (-4a,-0.02b)--(-4a,0);
\ draw (-4a,0)--(4a,0);
\ draw (1.5a,0)--(1.5a,0.12952b);
\ endchar; \quad \% Write character to output file.
}
\medskip\noindent
It is not possible to display the above graph using the public domain
view program {\bf CDVI12.arc}.  {\tt CDVI-2.com} and the other previewers
use only the sixteen basic \TeX\ fonts.  Some discipline is needed to use
public domain software; it is not as flexible as the commercially procured
products.  However, there is much to be said for the persons and organizations
that devote their time and energy to the creation of public domain
instruments.  Without such volunteerism, many scholars and scientists would
simply be unable to afford needed tools.  Certain activities are not able
to budget for commercial software; governmental agencies are only able
to procure software within the strict limits of their charters---and then
only with ADPE (Automatic Data Processing Equipment) approval from their MIS
(Management Information System) directors.  Public domain materials are an
invaluable tool for research and development.  
\medskip\noindent
Since a public domain \TeX\ previewer is to be employed and only sixteen
fonts are allowed, one cannot view a newly created font.  The graphs
produced by {\bf METAFONT} are, essentially, new fonts.  In particular,
each graph is one letter in a font.  Some display of the {\bf METAFONT}
graph is needed;
therefore, a \PiCTeX\ picture is provided
to give the reader an idea of the construction.
%\vbox{\narrower\narrower\narrower\noindent{\bf 25.} The main point here
%is that you can't have \PiCTeX\ draw a curve by simply giving it the formula
%for that curve --- you have to supply explicit coordinate points.  Here
%is the code that produced the figure.}
%
\bigskip
\centerline{%
 \beginpicture %
   \setcoordinatesystem units <.5in,2.5in> %
   \setplotarea x from -3 to 3, y from 0 to .4 %
   \plotheading {\lines {%
     The density $\varphi(\zeta) = e^{-\zeta^2\!/2}/\sqrt{2\pi}$ of the\cr %
     standard normal distribution.\cr}} %
   \axis bottom ticks numbered from -3 to 3 by 1 %
     length <0pt> withvalues $\zeta$ / at 1.5 /  / %
   \linethickness=.25pt %
   \putrule from  1.5 0  to  1.5 .12952 % (.12952 = density at 1.5)
   \setbox0 = \hbox{$swarrow$}%
   \put {$\swarrow$ \raise6pt\hbox{$\varphi(\zeta)$}} %
     [bl] at 1.5 .12952 %
\ifexpressmode
  \put {\bf EXPRESSMODE} at 0 .2 %
\else
   \setquadratic \plot
     0.0      .39894
     0.16667  .39344  0.33333 .37738   0.5  .35207   0.66667  .31945 
     0.83333  .28191  1.      .24197   1.25 .18265   1.5      .12952 
     1.75     .08628  2.      .05399   2.25 .03174   2.5      .01753 
     2.75     .00909  3.0     .00443 /                              
   \setquadratic \plot
     0.0      .39894