我正在处理重复＆＃34; x n + 1 =（13/3）x n - （4/3）x n -1 ＆＃34 ;.我正在尝试编写一个Python脚本来打印x n 的前50个值，给定的值为x 0 = 1且x 1 = 1/3。这就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
仅对前13个打印值有效。我提供的证据是x n = 3 -n ，这大部分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。
答案 0 :(得分：4)
这种递归关系可以通过使用fractions模块或使用decimal模块的可变精度水平来准确回答，这证明了查找计算所需的极高精度50迭代准确。
fraction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。
不精确但有指导性的回答
decimal模块允许自定义级别的精度;要达到50次迭代，需要60分或更高的精度。
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()
与Jean-François的答案一样，我打印了结果，计算出的3 ** - n值和差值。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
答案 1 :(得分：3)
您遇到浮点累积错误。
由于当前结果取决于之前的结果，并且浮点类型只是一个近似值，您运行的重复次数越多，累积误差就越大。
由于您正在处理有理数，我建议fraction模块。
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(13/3)*x[i] - fractions.Fraction(4/3)*x[i-1]
print(float(x[i]),3**(-i),x[i]-3**(-i))
printRecurrence（）
我打印了结果，3**-n的计算值和差异：
0.3333333333333333 0.3333333333333333 0.0
0.11111111111111109 0.1111111111111111 -1.3877787807814457e-17
0.037037037037036924 0.037037037037037035 -1.1102230246251565e-16
0.01234567901234521 0.012345679012345678 -4.683753385137379e-16
0.004115226337446681 0.00411522633744856 -1.878705524482882e-15
0.001371742112475337 0.0013717421124828531 -7.516123140538511e-15
0.0004572473707975519 0.0004572473708276177 -3.006584781486965e-14
0.00015241579015560884 0.00015241579027587258 -1.2026374362518466e-13
5.080526294423582e-05 5.080526342529086e-05 -4.810550354871108e-13
1.6935085884210096e-05 1.6935087808430286e-05 -1.9242201893822884e-12
5.645021572595982e-06 5.645029269476762e-06 -7.696880780399043e-12
1.8816456356357935e-06 1.8816764231589208e-06 -3.0787523127313645e-11
6.27102324293796e-07 6.272254743863069e-07 -1.2315009251094865e-10
2.0858255775872449e-07 2.0907515812876897e-07 -4.926003700444828e-10
6.772131789607814e-08 6.969171937625632e-08 -1.9704014801781827e-09
1.534896720470598e-08 2.3230573125418773e-08 -7.881605920712794e-09
-2.378289930771161e-08 7.743524375139592e-09 -3.15264236828512e-08
-1.2352451993969162e-07 2.581174791713197e-09 -1.2610569473140483e-07
-5.035623873283815e-07 8.603915972377324e-10 -5.044227789256192e-07
-2.0174043185033973e-06 2.8679719907924413e-10 -2.0176911157024764e-06
-8.070668863743547e-06 9.559906635974805e-11 -8.070764462809906e-06
-3.228302598488417e-05 3.186635545324935e-11 -3.228305785123962e-05
...
20次复发后，差异会增加，但已经好多了。
编辑：大卫回答让我意识到浮动转换的分数会破坏它。 fraction模块仅执行2个精确计算整数之间的除法，但转换为float会破坏精度。