我无法让Muller算法的代码适用于只涉及基本python编程的类。我的程序还没有包含像输出那样的虚数的任何内容，而且我的停止标准也不能正常工作，但是现在我主要关心的是让输出中的数字正确打印。我将输出作为代码发布，因为我是这个网站的新手并且它一直给我一个错误，说我的“代码格式不正确”
非常感谢任何帮助！
When f=(x-5)*(x-4)*(x+7)
and initial guesses are 1,3,and -10
and tolerance = 0.000001
The output of this code should look like this:
The initial estimates to the root are:
f( 1 )= 96
f( 3 )= 20
f( -10 )= -630
0 : estimate to the root is f( 1 )= 96
1 : estimate to the root is f( (-4.102012878968893+0j) )= (213.71097514696675+0j)
2 : estimate to the root is f( (3.463202253458595+0j) )= (8.63161417086553+0j)
3 : estimate to the root is f( (3.6797449479714586+0j) )= (4.515592141362759+0j)
4 : estimate to the root is f( (3.9234514667540585+0j) )= (0.9001820953746444+0j)
5 : estimate to the root is f( (3.9988300605239253+0j) )= (0.012883020219233893+0j)
6 : estimate to the root is f( (3.9999973985379675+0j) )= (2.861615003307639e-05+0j)
The approximation to the root is f( (3.999999999978821+0j) ) = (2.329696435810382e-
10+0j)
这是我的实际代码：
import string
from math import *
from cmath import *
def evalFunction(f,x):
x=eval(f)
return x
def main():
f= input("Input the function: ")
p0=eval(input("Input the first estimate to the root of the function: "))
p1=eval(input("Input the second estimate to the root of the function: "))
p2=eval(input("Input the third estimate to the root of the function: "))
t=eval(input("Enter the tolerance "))
fp0=evalFunction(f,p0)
fp1=evalFunction(f,p1)
fp2=evalFunction(f,p2)
print("The initial estimates to the root are:")
print("f(",p0,")=",fp0)
print("f(",p1,")=",fp1)
print("f(",p2,")=",fp2)
count=0
print(count,": estimate to the root is f(",p0,")=",fp0)
fp3=1
while count<30:
while (abs(fp3))>=t:
fp0=evalFunction(f,p0)
fp1=evalFunction(f,p1)
fp2=evalFunction(f,p2)
#Computes a,b,c
o=fp1-fp2
n=fp0-fp2
s=p1-p2
r=p0-p2
denom=r*s*(p0-p1)
a=(s*n-r*o)/denom
b=((r**2)*o-(s**2)*n)/denom
c=fp2
print()
count=count+1
#Computes the roots
x1= (-2*c)/(b+(b**2-4*a*c)**.5)
x2=(-2*c)/(b-(b**2-4*a*c)**.5)
if b>0:
p3= p2+x1
fp3=evalFunction(f,p3)
print(count,": estimate to the root is f(",p3,")=",fp3)
print()
else:
p3= p2+x2
fp3=evalFunction(f,p3)
print(count,": estimate to the root is f(",p3,")=",fp3)
print()
p2=p3
main()
答案 0 :(得分：1)
不清楚程序的哪个方面对你不起作用，除非可能在满足容差后挂起代码，因为count在内循环内增加，不再是在满足容差后执行，导致外循环while count<30:永远运行。
无论如何，将序列while count<30:和while (abs(fp3))>=t:更改为while count<30 and (abs(fp3))>=t:可以解决问题。请参阅回答结束时的注意事项问题p0和p1未在循环内更新，我想象的问题会破坏收敛速度。
关于输出格式，请参阅下面代码中format函数的使用。请注意，在此代码中，我在大多数等号周围添加了空格，以提高可读性，并通过命令行添加参数输入。仍然会提示未在命令行中输入的参数。我不知道在MS-Windows机器上是否以相同的方式拾取命令行参数。
注意，您可以打印具有不同或相同小数位数的实数和复数部分;例如，c=3-5j; d=4.4+5.5j; print ('d: {.real:7.2f} {.imag:+.3f}j c: {:9.4f}'.format(d,d, c))生成d: 4.40 +5.500j c: 3.0000-5.0000j。有关格式设置的详细信息，请参阅Format Specification Mini-Language文档。
使用下面修订的程序，通过./mullermethod.py '(x-5)*(x-4)*(x+7)' 1 3 -10 0.000001在我的Linux系统上运行它（在提示时输入的参数相同）打印
The initial estimates to a root are:
f( 1.0 ) = 96.0
f( 3.0 ) = 20.0
f( -10.0 ) = -630.0
0: estimate to a root is f( 1.00000000000000) = 96.00000000000000
1: estimate to a root is f( -4.10201287896889) = 213.71097514696675
2: estimate to a root is f( 3.46320225345860) = 8.63161417086548
3: estimate to a root is f( 3.90342357843416) = 1.15470992046251
4: estimate to a root is f( 3.98002449809592) = 0.22371275706982
5: estimate to a root is f( 3.99575149263503) = 0.04691400247817
6: estimate to a root is f( 3.99909106270745) = 0.01000657113716
7: estimate to a root is f( 3.99980529453210) = 0.00214213924168
8: estimate to a root is f( 3.99995828046651) = 0.00045893227349
9: estimate to a root is f( 3.99999106024078) = 0.00009833815063
10: estimate to a root is f( 3.99999808434378) = 0.00002107225517
11: estimate to a root is f( 3.99999958950253) = 0.00000451547380
12: estimate to a root is f( 3.99999991203627) = 0.00000096760109
然后程序退出。
#!/usr/bin/env python3.2
from math import *
from cmath import *
def evalFunction(f,x):
return eval(f)
def main():
from sys import argv
f = argv[1] if len(argv)>1 else input("Input the function: ")
p0 = float(argv[2]) if len(argv)>2 else eval(input("Input the first estimate to a root of the function: "))
p1 = float(argv[3]) if len(argv)>3 else eval(input("Input the next estimate to a root of the function: "))
p2 = float(argv[4]) if len(argv)>4 else eval(input("Input the third estimate to a root of the function: "))
toler = float(argv[5]) if len(argv)>5 else eval(input("Enter the tolerance: "))
fp0 = evalFunction(f,p0)
fp1 = evalFunction(f,p1)
fp2 = evalFunction(f,p2)
print("The initial estimates to a root are:")
print("f(",p0,") = ",fp0)
print("f(",p1,") = ",fp1)
print("f(",p2,") = ",fp2)
count = 0
print ('{:2}: estimate to a root is f({:18.14f}) = {:18.14f}'.format(count,p0,fp0))
fp3 = 1e9
while count<30 and abs(fp3) >= toler:
fp0 = evalFunction(f,p0)
fp1 = evalFunction(f,p1)
fp2 = evalFunction(f,p2)
#Computes a,b,c
o = fp1-fp2
n = fp0-fp2
s = p1-p2
r = p0-p2
denom = r*s*(p0-p1)
a = (s*n-r*o)/denom
b = ((r**2)*o-(s**2)*n)/denom
c = fp2
count += 1
#Compute roots
x1 = (-2*c)/(b+(b**2-4*a*c)**.5)
x2 = (-2*c)/(b-(b**2-4*a*c)**.5)
if b>0:
p3 = p2+x1
else:
p3 = p2+x2
fp3=evalFunction(f,p3)
print ('{:2}: estimate to a root is f({:18.14f}) = {:18.14f}'.format(count,p3,fp3))
p2 = p3
main()
注意，我看不到p0和p1在循环内部发生变化（因此fp0和fp1也不会在循环中发生变化）。我对wikipedia的印象是，在您说p0, p1, p2 = p1, p2, p3时，您应该执行p2 = p3之类的操作。此外，为了减少功能评估，您可以删除循环中的fp0，fp1和fp2评估，而不是p0, p1, p2 = p1, p2, p3您说的p0, p1, p2, fp0, fp1, fp2 = p1, p2, p3, fp1, fp2, fp3。< / p>
使用p0, p1, p2 = p1, p2, p3代替p2 = p3，可以在一半的迭代次数中找到更接近的结果（针对不同的根）：
tini ~/sp/math > ./mullermethod.py '(x-5)*(x-4)*(x+7)' 1 3 -10 0.000001
The initial estimates to a root are:
f( 1.0 ) = 96.0
f( 3.0 ) = 20.0
f( -10.0 ) = -630.0
0: estimate to a root is f( 1.00000000000000) = 96.00000000000000
1: estimate to a root is f( -4.10201287896889) = 213.71097514696675
2: estimate to a root is f( -6.34710329529507) = 76.65637351948691
3: estimate to a root is f( -6.95941163108092) = 5.31984100233008
4: estimate to a root is f( -7.00059085626200) = -0.07800105634618
5: estimate to a root is f( -6.99999988135762) = 0.00001566079365
6: estimate to a root is f( -6.99999999999998) = 0.00000000000281