数列 a(1)=1, a(n+1)=a(n)[1-a(n)/3], 求证 5/2<100a(100)<3

Ickiverar

你把a[1]=1后面加个.就行了。  a[1]=1. 试试。

haifeng

Calculator 中的 printRecursiveSeries() 函数可以用于计算由递推公式所确定的数列.

>> help(printRecursiveSeries)
in> help(printRecursiveSeries)
out> 打印由递推公式所定义的数列的前N项
例如:\打印 n^2, n 初值为 1.
则可以输入:
in> printRecursiveSeries(n^2,n,1,10,;)
最后一个参数是分隔符, 默认是逗号.

Usage:
printRecursiveSeries(Iteration_expression,x,initialValue,N,delimiter)


------------------------


>> printRecursiveSeries(a_n-a_n^2/3,a_n,1,100,\n)
in> printRecursiveSeries(a_n-a_n^2/3,a_n,1,100,\n)
0.66666667
0.51851852
0.42889803
0.36758019
0.32254179
0.28786405
0.26024215
0.23766682
0.21883831
0.20287491
0.18915550
0.17722890
0.16675887
0.15748936
0.14922173
0.14179936
0.13509701
0.12901328
0.12346514
0.11838393
0.11371235
0.10940218
0.10541257
0.10170863
0.09826041
0.09504204
0.09203104
0.08920780
0.08655512
0.08405786
0.08170262
0.07947751
0.07737195
0.07537648
0.07348261
0.07168271
0.06996991
0.06833798
0.06678129
0.06529471
0.06387358
0.06251364
0.06121099
0.05996206
0.05876358
0.05761253
0.05650613
0.05544182
0.05441722
0.05343014
0.05247855
0.05156055
0.05067439
0.04981843
0.04899114
0.04819110
0.04741697
0.04666751
0.04594156
0.04523802
0.04455586
0.04389412
0.04325189
0.04262831
0.04202259
0.04143396
0.04086170
0.04030514
0.03976364
0.03923659
0.03872342
0.03822359
0.03773658
0.03726190
0.03679908
0.03634769
0.03590731
0.03547753
0.03505798
0.03464829
0.03424812
0.03385714
0.03347504
0.03310151
0.03273627
0.03237905
0.03202958
0.03168762
0.03135292
0.03102525
0.03070439
0.03039014
0.03008229
0.02978064
0.02948501
0.02919522
0.02891110
0.02863248
0.02835921
0.02809113


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By Calculator
Calculator 使用手册

uk702

改成：

Timing[a[1] = 1.0; a[2] = 2/3.0;
a[n_] := a[n] = a[n - 1] - a[n - 1] a[n - 1]/3.0;
N@a[250]
]

Out = {0., 0.0116879}