用mma怎么编程

uk702

uk702 发表于 2023-1-14 17:02
3# 的方法就是大招。更进一步，似有
\[
a_n = \frac{2}{n} + \frac{2 Log[n]}{3 n^2} + \frac{2 Log[n]^ ...

I am very sorry! #7 楼的结果没能通过实际数据检验。

1. 
   
2. n = 100000;
   
3. a = 1.0`100; For[i = 0, i < n, a = Log[1 + a]; i++]; a1000000 = a; Print[N[u, 100]];
   
4. f[n_] := 2/n + (2/3 Log[n] - c1)/(n^2); Solve[f[n] == a]
   
5. g[n_] := 2/n + (2/3 Log[n] - c2)/(n^2) + (2 Log[n] Log[n] - 2 Log[n])/(9 n^3); Solve[g[n] == a]
   
6. 解得
   
7. c1 = 4.513881773021752...
   
8. c2 = 4.514150738637968...
   
9. 
   
10. n = 1000000;
    
11. a = 1.0`100; For[i = 0, i < n, a = Log[1 + a]; i++]; a1000000 = a; Print[N[u, 100]];
    
12. f[n_] := 2/n + (2/3 Log[n] - c1)/(n^2); Solve[f[n] == a]
    
13. g[n_] := 2/n + (2/3 Log[n] - c2)/(n^2) + (2 Log[n] Log[n] - 2 Log[n])/(9 n^3); Solve[g[n] == a]
    
14. 解得
    
15. c1 = 4.5139127439173691673462937...
    
16. c2 = 4.513952088988795669136132843492
    

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故(猜想)
\[
a_n = \frac{2}{n} + \frac{2/3 Log[n] - c}{n^2}  + o(n^{-2})
\]

其中 c 是一个与 \( a_1 \) 相关的常数，当  \( a_1 = 1 \)  时，\( c = 4.513... \)

wayne

根据3#的思想，得到$n->\infty,  a(n)->2\sum _{k=1}^{\infty } \frac{1}{2 k-1} (\frac{1}{n})^{2 k-1}= 2 \tanh ^{-1}(\frac{1}{n})$,  也就是原题极限为0.

笨笨

wayne 发表于 2023-1-15 09:18
根据3#的思想，得到$n->\infty,  a(n)->2\sum _{k=1}^{\infty } \frac{1}{2 k-1} (\frac{1}{n})^{2 k-1}= 2 ...

你好，原极限结果等于2/3，2楼的程序可否能改进一下

uk702

\[
a_n = \frac{2}{n} + \frac{2/3 Log[n] - c}{n^2} + \frac{ 2/9 Log[n]^2 - 2(1 + 3 c)/ 9 Log[n] + (2 + 6 c + 9 c^2)/18}{n^3}
\]

1. n = 100000;
   
2. a = 1.0`100; For[i = 0, i < n, a = Log[1 + a]; i++]; Print[N[a, 100]];
   
3. h[n_] = 2/n + (2/3 Log[n] - c) / n^2 + ( 2/9 Log[n]^2 - 2(1 + 3 c)/ 9 Log[n] + (2 + 6 c + 9 c^2)/18 ) / n^3;
   
4. Solve[h[n] == a]
   
5. 解得 c = 4.513922317288945344462772324200553...
   
6. 
   
7. n = 1000000;
   
8. a = 1.0`100; For[i = 0, i < n, a = Log[1 + a]; i++]; Print[N[a, 100]];
   
9. h[n_] = 2/n + (2/3 Log[n] - c) / n^2 + ( 2/9 Log[n]^2 - 2(1 + 3 c)/ 9 Log[n] + (2 + 6 c + 9 c^2)/18 ) / n^3;
   
10. Solve[h[n] == a]
    
11. 解得 c = 4.5139223177270656672830726730444912910404826975...
    
12. 
    
13. n = 10000000;
    
14. a = 1.0`100; For[i = 0, i < n, a = Log[1 + a]; i++]; Print[N[a, 100]];
    
15. h[n_] = 2/n + (2/3 Log[n] - c) / n^2 + ( 2/9 Log[n]^2 - 2(1 + 3 c)/ 9 Log[n] + (2 + 6 c + 9 c^2)/18 ) / n^3;
    
16. Solve[h[n] == a]
    
17. 解得 c = 4.51392231774456470363880866511722354631999512440581836558125598...
    
18. 这时  a = 2.0000006231476527853498662133367261421598872948834435881757796... *10^-7
    
19.   h[n] = 2.00000062314765278534986621333672614215988729488344358817577963496395 *10^-7
    
20. 
    

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注意到 h[n] == a，上面的结果其实没有意义，要考察的是 c 对不同的 n 的稳定性，由于 c 的精度超过了 1/n，介于 1/n 和  1/n^2 之间，故判断上述公式给出的 1/n^3 的系数正确，而 （未给出的） 1/n^4 的系数可能较大。