数列渐近式问题
本帖最后由 pizza49 于 2023-11-11 17:12 编辑$x_{n+1}=x_{n}-x_{n}^{4}$,当$x_{1}>1$或$x_{1}<0$时,$x_{n}\rightarrow -\infty $,怎么估计$x_{n}$渐近式? 当$x_1 > 1$时,$x_2 < 0$,所以只考虑$x_1 < 0$即可。
又因为$-x_{n+1} = -x_n + (-x_n)^4$,所以设$y_n = -x_n$,问题就等价变成$y_1 > 0$时有$y_{n+1} = y_n^4 + y_n$。
可以参考 https://bbs.emath.ac.cn/forum.php?mod=viewthread&tid=5794&page=1#pid55470 。 282842712474 发表于 2023-11-17 13:44
当$x_1 > 1$时,$x_2 < 0$,所以只考虑$x_1 < 0$即可。
又因为$-x_{n+1} = -x_n + (-x_n)^4$,所以设$y_n...
$y_n=(y_1^4 + y_1)^(4^(n - 2))$ 应该比较接近了 s=Table][]/4^(n-1),{n,10}],{k,10}]
t=Table[{k,2^s[]},{k,10}]
0. 0.25 0.26062 0.260624 0.260624 0.260624 0.260624 0.260624 0.260624 0.260624
1. 1.04248 1.0425 1.0425 1.0425 1.0425 1.0425 1.0425 1.0425 1.0425
1.58496 1.59808 1.59808 1.59808 1.59808 1.59808 1.59808 1.59808 1.59808 1.59808
2. 2.00559 2.00559 2.00559 2.00559 2.00559 2.00559 2.00559 2.00559 2.00559
2.32193 2.3248 2.3248 2.3248 2.3248 2.3248 2.3248 2.3248 2.3248 2.3248
2.58496 2.58663 2.58663 2.58663 2.58663 2.58663 2.58663 2.58663 2.58663 2.58663
2.80735 2.8084 2.8084 2.8084 2.8084 2.8084 2.8084 2.8084 2.8084 2.8084
3. 3.0007 3.0007 3.0007 3.0007 3.0007 3.0007 3.0007 3.0007 3.0007
3.16993 3.17042 3.17042 3.17042 3.17042 3.17042 3.17042 3.17042 3.17042 3.17042
3.32193 3.32229 3.32229 3.32229 3.32229 3.32229 3.32229 3.32229 3.32229 3.32229
1 {1.,1.18921,1.19799,1.198,1.198,1.198,1.198,1.198,1.198,1.198}
2 {2.,2.05977,2.05979,2.05979,2.05979,2.05979,2.05979,2.05979,2.05979,2.05979}
3 {3.,3.0274,3.0274,3.0274,3.0274,3.0274,3.0274,3.0274,3.0274,3.0274}
4 {4.,4.01553,4.01553,4.01553,4.01553,4.01553,4.01553,4.01553,4.01553,4.01553}
5 {5.,5.00997,5.00997,5.00997,5.00997,5.00997,5.00997,5.00997,5.00997,5.00997}
6 {6.,6.00693,6.00693,6.00693,6.00693,6.00693,6.00693,6.00693,6.00693,6.00693}
7 {7.,7.0051,7.0051,7.0051,7.0051,7.0051,7.0051,7.0051,7.0051,7.0051}
8 {8.,8.0039,8.0039,8.0039,8.0039,8.0039,8.0039,8.0039,8.0039,8.0039}
9 {9.,9.00308,9.00308,9.00308,9.00308,9.00308,9.00308,9.00308,9.00308,9.00308}
10 {10.,10.0025,10.0025,10.0025,10.0025,10.0025,10.0025,10.0025,10.0025,10.0025}
以前学过好像有个叫牛顿迭代法的
页:
[1]