求\(\displaystyle\lim_{n\to\infty}\sqrt[n]{n}\) 的子序列方法.

elim

\(b=2021.999505317722323496481550530470314306352616300084174161565120447212249...,\\
\quad\;\; 0.131366826270076605141017937165094719486686762979646549655120418694728358...\\
a=1.000494682277676503518449469529685693647383699915825838434879552787750544...,\\
\quad\;\; 2022.868633173729923394858982062834905280513313237020353450344879581305271...\\
a^b=2.71828182845904523536028747135266249775724709369995957496696762772407663\ldots\\
\;e=2.71828182845904523536028747135266249775724709369995957496696762772407663\ldots\)

王守恩

1.000494682277676503518449469529685693647383699915825838434879552787750544990556..

1. N[FindInstance[{(2023-x)Log[x]==1},{x}],90]

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2022.868633173729923394858982062834905280513313237020353450344879581305271641376..

1. N[FindInstance[{2023-x==Log[x]^-1},{x}],90]

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王守恩

12#是两个不同的解, 这样也可以。贪心一点:  会有类似于根式解吗？
1.000494682277676503518449469529685693647383699915825838434879552787750544990556..

1. FindRoot[{(x)^(2023-x)==E},{x,1},WorkingPrecision->90]

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0.131366826270076605141017937165094719486686762979646549655120418694728358623791..

1. FindRoot[{(2023-x)^x==E},{x,1},WorkingPrecision->90]

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elim

#10 给出了解这个方程的牛顿方法．可以证明，方程有且仅有两组解.
由于e是超越数，a,b 必非有理数．它们是否能表为初等函数在有理点
的值这个(问题\(\star\))很迷人．
这里无论牛顿法或Mathematica都是数值方法而不是解析方法．所以
还没有理由对(\(\star\)问题)乐观．

王守恩

奢望一并解决。

\((1),\ n=1,2,3,4,5,6,...,\ \ \frac{x+n}{x-n}=e\)

\((2),\ n=2,3,4,5,6,7,...,\ \ \frac{n+x}{n-x}=e\)

elim

1. C:\Users\elim\Projects\pari_gp>gp
   
2. Reading GPRC: C:\Academy\Pari\/gprc.txt
   
3. GPRC Done.
   
4. 
   
5.                                       GP/PARI CALCULATOR Version 2.13.3 (released)
   
6.                               amd64 running mingw (x86-64/GMP-6.1.2 kernel) 64-bit version
   
7.                                compiled: Oct 11 2021, gcc version 8.3-posix 20190406 (GCC)
   
8.                                                 threading engine: single
   
9.                                      (readline v8.0 enabled, extended help enabled)
   
10. 
    
11.                                          Copyright (C) 2000-2020 The PARI Group
    
12. 
    
13. PARI/GP is free software, covered by the GNU General Public License, and comes WITHOUT ANY WARRANTY WHATSOEVER.
    
14. 
    
15. Type ? for help, \q to quit.
    
16. Type ?17 for how to get moral (and possibly technical) support.
    
17. 
    
18. parisize = 8000000, primelimit = 500000
    
19. (15:26) gp > \p111
    
20.    realprecision = 115 significant digits (111 digits displayed)
    
21. (15:26) gp > e=exp(1);
    
22. (15:26) gp > g(x)=x-(e^(1/x)+x-2023)/(1-(e^(1/x))/(x^2));
    
23. (15:27) gp > g(13/99)
    
24. %3 =0.1313667208740331735645963166713599554588808622672617066605374065520420191845117743115045420797178387496
    
25. (15:27) gp > g(%)
    
26. %4 =0.1313668262696701974071477086135849067670817966390215428098015332902587542460795875305860158383422708625
    
27. (15:28) gp > g(%)
    
28. %5 =0.1313668262700766051410118943684918891824812731180448358666172319058168809174832604189064718106874954645
    
29. (15:28) gp > g(%)
    
30. %6 =0.1313668262700766051410179371650947194866867616436971739707487069302631373858900618211393721911343661674
    
31. (15:28) gp > g(%)
    
32. %7 =0.1313668262700766051410179371650947194866867629796465496551204186947283586237919365463307746232171298813
    
33. (15:28) gp > g(%)
    
34. %8 =0.1313668262700766051410179371650947194866867629796465496551204186947283586237919365463308399203408280540
    
35. (15:28) gp > b1=%
    
36. %9 =0.1313668262700766051410179371650947194866867629796465496551204186947283586237919365463308399203408280540
    
37. (15:28) gp > g(100)
    
38. %10 =2022.18410007006006148033179171645240086434441631373889378433278384830296970298081325498807223778559761
    
39. (15:28) gp > g(%)
    
40. %11 =2021.99950531772644768514575517420121937792218323920239769960056298812263443659082526405774376462892845
    
41. (15:29) gp > g(%)
    
42. %12 =2021.99950531772232349648155053047031636535184241841516813831398479890501795806229813423299166447063818
    
43. (15:29) gp > g(%)
    
44. %13 =2021.99950531772232349648155053047031430635261630008417416156512044721224945500995682043320756459894010
    
45. (15:29) gp > g(%)
    
46. %14 =2021.99950531772232349648155053047031430635261630008417416156512044721224945500944361469720750228058217
    
47. (15:29) gp > g(%)
    
48. %15 =2021.99950531772232349648155053047031430635261630008417416156512044721224945500944361469720750228058217
    
49. (15:29) gp > b2=%
    
50. %16 =2021.99950531772232349648155053047031430635261630008417416156512044721224945500944361469720750228058217
    
51. (15:29) gp > (2023-b1)^b1
    
52. %17 =2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642742
    
53. (15:30) gp > (2023-b2)^b2
    
54. %18 =2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642742
    
55. (15:30) gp > e
    
56. %19 =2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642742
    
57. (15:30) gp >

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\p111 表示显示111位有效数字
e = exp(1) 表示将 2.718281828... 赋值到 e
g(x)=x-(e^(1/x)+x-2023)/(1-(e^(1/x))/(x^2)); 表示牛顿迭代公式的定义
% 代表最新计算结果，于是 g(%) 就是下一步迭代。
a = % 表示将最新计算结果赋值到 a.

elim

题：设方程\(F(x)=0\)有解作序列\(x_{n+1}=x_n-\frac{F(x_n)}{F'(x_n)},\;x_1=a\)
\(\qquad\)分析在什么条件下\(\{x_n\}\)收敛于方程的零点.

王守恩

\((1),\ n=1,2,3,4,5,6,...,\ \ \frac{x+n}{x-n}=e\)
a(1)=0,   n=1,x=0是最接近答案的整数,
a(2)=1,   n=2,x=1是最接近答案的整数,
a(3)=1,   n=3,x=1是最接近答案的整数,
a(4)=2,   n=4,x=2是最接近答案的整数,
a(5)=2,   n=5,x=2是最接近答案的整数,
a(6)=3,   n=6,x=3是最接近答案的整数,

得到这样一串数: (参考OEIS--A075355)
0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 6, 7, 7, 8, 8, 9, 9,10, 10, 11, 11, 12, 12, 12, 13, 13, 14, 14, 15, 15,
16, 16, 17, 17, 18, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 24, 25, 25, 26, 26, 27, 27,
28, 28, 29, 29, 30, 30, 30, 31, 31, 32, 32, 33, 33, 34, 34, 35, 35, 36, 36, 37, 37, 37, 38, 38, 39, 39,
40, 40, 41, 41, 42, 42, 43, 43, 43, 44, 44, 45, 45, 46, 46, 47, 47, 48, 48, 49, 49, 49, 50, 50, 51, 51,
52, 52, 53, 53, 54, 54, 55, 55, 55, 56, 56, 57, 57, 58, 58, 59, 59, 60, 60, 61, 61, 61, 62, 62, 63, ......

\((2),\ n=1,2,3,4,5,6,...,\ \ \frac{n+x}{n-x}=e\)
a(1)=02,   n=1,x=02是最接近答案的整数,
a(2)=04,   n=2,x=04是最接近答案的整数,
a(3)=06,   n=3,x=06是最接近答案的整数,
a(4)=09,   n=4,x=09是最接近答案的整数,
a(5)=11,   n=5,x=11是最接近答案的整数,
a(6)=13,   n=6,x=13是最接近答案的整数,

得到这样一串数: (参考OEIS--A351631)
2, 4, 6, 9, 11, 13, 15, 17, 19, 22, 24, 26, 28, 30, 32, 35, 37, 39, 41, 43, 45, 48, 50, 52, 54, 56, 58,
61,63, 65, 67, 69, 71, 74, 76, 78, 80, 82, 84, 87, 89, 91, 93, 95, 97, 100, 102, 104, 106, 108,110,
113, 115, 117, 119, 121, 123, 126, 128, 130, 132, 134, 136, 138, 141, 143, 145, 147, 149, 151,
154, 156, 158, 160, 162, 164, 167, 169, 171, 173, 175, 177, 180, 182, 184, 186, 188, 190, 193,
195, 197, 199, 201, 203, 206, 208, 210, 212, 214, 216, 219, 221, 223, 225, 227, 229, 232,  ......

elim！你可有兴趣, 把这2串数放到OEIS去申报？

王守恩

凑个“根式解”!!!

\(e=\sqrt[ln(n)]{n}\ \ \ \ n=2,3,4,5,6,7,8,9,...\)

\(证:\ \ ∵\ e=\sqrt[ln(n)]{n},\ \ ∴\ e≡\sqrt[ln(n)]{n}\)

elim

求序列极限的子列方法还有什么有趣的例子？