Python中的递归关系

2019年8月3日 134次阅读

我正在处理重复“xn 1 =(13/3)xn – (4/3)xn-1”.我正在尝试编写一个
Python脚本来打印xn的前50个值,给定的值为x0 = 1和x1 = 1/3.这是我的代码目前的样子：

import math
def printRecurrence():
    x = [0]*51 #initialize list of x values
    x[0] = 1
    x[1] = 1/3
    for i in range(1, 51):
        x[i+1] = (13/3)*x[i] - (4/3)*x[i-1]
        print(x[i])

我收到的输出是：

0.3333333333333333
0.11111111111111094
0.03703703703703626
0.012345679012342514
0.004115226337435884
0.0013717421124321456
0.00045724737062478524
0.00015241578946454185
5.0805260179967644e-05
1.6935074827137338e-05
5.644977344304949e-06
1.8814687224716613e-06
6.263946716372672e-07
2.0575194713260943e-07
5.63988753916179e-08
-2.994080281313502e-08
-2.049419793790756e-07
-8.481608402251475e-07
-3.402107668470205e-06
-1.361158544307069e-05
-5.444739336201271e-05
-0.00021778992397796082
-0.0008711598127551465
-0.0034846392899683535
-0.013938557172856001
-0.05575422869575153
-0.22301691478444863
-0.8920676591382753
-3.5682706365532617
-14.2730825462131
-57.092330184852415
-228.36932073940963
-913.4772829576384
-3653.909131830553
-14615.63652732221
-58462.546109288836
-233850.18443715532
-935400.7377486213
-3741602.950994485
-14966411.80397794
-59865647.21591176
-239462588.86364704
-957850355.4545882
-3831401421.8183527
-15325605687.27341
-61302422749.09364
-245209690996.37457
-980838763985.4983
-3923355055941.993

这只适用于前13个打印值.我提供的证据是xn = 3-n,这在很大程度上与我的脚本的值不匹配.我在计算中做错了吗？我一直无法看到它.

最佳答案 这种递归关系可以通过使用分数模块或使用十进制模块的可变精度水平来准确回答,这证明了准确地找到计算50次迭代所需的非常高的精度.

Jean-François关于浮点累积误差的观点是正确的.但是,分数模块似乎无法将多个Fraction对象相乘,因此必须在fraction对象中声明所有数值.感谢他为这个问题使用正确的模块.

确切的答案

import math
import fractions

def printRecurrence():
    x = [0]*51 #initialize list of x values
    x[0] = fractions.Fraction(1,1)
    x[1] = fractions.Fraction(1,3)
    for i in range(1, 50):
        x[i+1] = fractions.Fraction(13*x[i]/3) - fractions.Fraction(4*x[i-1]/3)
        print(float(x[i+1]), 3**(-(i+1)), x[i+1]-(3)**(-(i+1)))

printRecurrence()

打印输出显示计算与证明答案完全匹配.

不精确但有启发性的答案

十进制模块允许自定义级别的精度;要达到50次迭代,需要60点或更高的精度.

import math
from decimal import *
getcontext().prec = 60

def printRecurrence():
    x = [0]*51 #initialize list of x values
    x[0] = Decimal(1)
    x[1] = Decimal(1)/Decimal(3)
    for i in range(1, 50):
        x[i+1] = (Decimal(13)/Decimal(3))*x[i] - (Decimal(4)/Decimal(3))*x[i-1]
        print(float(x[i+1]), 3**(-(i+1)), float(x[i+1]-Decimal(3)**(-(i+1))))

printRecurrence()

和Jean-François的答案一样,我打印了结果,计算出的值为3 ** – n和差异.可以使用精度级别getcontext().prec来查看对结果的影响.

0.111111111111 0.111111111111 -1e-60
0.037037037037 0.037037037037 -4e-60
0.0123456790123 0.0123456790123 -1.57e-59
0.00411522633745 0.00411522633745 -6.243e-59
0.00137174211248 0.00137174211248 -2.4961e-58
0.000457247370828 0.000457247370828 -9.984e-58
0.000152415790276 0.000152415790276 -3.993577e-57
5.08052634253e-05 5.08052634253e-05 -1.59742978e-56
1.69350878084e-05 1.69350878084e-05 -6.38971879e-56
5.64502926948e-06 5.64502926948e-06 -2.5558875048e-55
1.88167642316e-06 1.88167642316e-06 -1.02235500146e-54
6.27225474386e-07 6.27225474386e-07 -4.08942000567e-54
2.09075158129e-07 2.09075158129e-07 -1.63576800226e-53
6.96917193763e-08 6.96917193763e-08 -6.54307200905e-53
2.32305731254e-08 2.32305731254e-08 -2.61722880362e-52
7.74352437514e-09 7.74352437514e-09 -1.04689152145e-51
2.58117479171e-09 2.58117479171e-09 -4.18756608579e-51
8.60391597238e-10 8.60391597238e-10 -1.67502643432e-50
2.86797199079e-10 2.86797199079e-10 -6.70010573727e-50
9.55990663597e-11 9.55990663597e-11 -2.68004229491e-49
3.18663554532e-11 3.18663554532e-11 -1.07201691796e-48
1.06221184844e-11 1.06221184844e-11 -4.28806767185e-48
3.54070616147e-12 3.54070616147e-12 -1.71522706874e-47
1.18023538716e-12 1.18023538716e-12 -6.86090827496e-47
3.93411795719e-13 3.93411795719e-13 -2.74436330998e-46
1.3113726524e-13 1.3113726524e-13 -1.09774532399e-45
4.37124217466e-14 4.37124217466e-14 -4.39098129597e-45
1.45708072489e-14 1.45708072489e-14 -1.75639251839e-44
4.85693574962e-15 4.85693574962e-15 -7.02557007356e-44
1.61897858321e-15 1.61897858321e-15 -2.81022802942e-43
5.39659527735e-16 5.39659527735e-16 -1.12409121177e-42
1.79886509245e-16 1.79886509245e-16 -4.49636484708e-42
5.99621697484e-17 5.99621697484e-17 -1.79854593883e-41
1.99873899161e-17 1.99873899161e-17 -7.19418375532e-41
6.66246330538e-18 6.66246330538e-18 -2.87767350213e-40
2.22082110179e-18 2.22082110179e-18 -1.15106940085e-39
7.40273700597e-19 7.40273700597e-19 -4.60427760341e-39
2.46757900199e-19 2.46757900199e-19 -1.84171104136e-38
8.22526333997e-20 8.22526333997e-20 -7.36684416545e-38
2.74175444666e-20 2.74175444666e-20 -2.94673766618e-37
9.13918148886e-21 9.13918148886e-21 -1.17869506647e-36
3.04639382962e-21 3.04639382962e-21 -4.71478026589e-36
1.01546460987e-21 1.01546460987e-21 -1.88591210636e-35
3.38488203291e-22 3.38488203291e-22 -7.54364842542e-35
1.12829401097e-22 1.12829401097e-22 -3.01745937017e-34
3.76098003645e-23 3.76098003657e-23 -1.20698374807e-33
1.25366001171e-23 1.25366001219e-23 -4.82793499227e-33
4.17886668798e-24 4.1788667073e-24 -1.93117399691e-32
1.39295549185e-24 1.3929555691e-24 -7.72469598763e-32
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