任何人都可以告诉我为什么我的代码显示错误的pi值？

Question

Here's a picture of my output:

inptTol = float(input("Enter the tolerance: "))
print()

term = 1
divNum = 3
npower = 1
sumPi = 0.0
count = 0

while abs(term) > inptTol:
    sumPi += term
    term = -term/(divNum * (3**npower))
    divNum += 2
    npower += 1
    count += 1

sumPi = math.sqrt(12) * sumPi  
pythonPi = math.pi  
approxError = abs (sumPi - pythonPi)  

print("The approximate value of pi is %.14e\n" \
        "       Python's value of pi is %.14e\n"
        "The error in the approximation of pi is %.6e\n"
        "The number of terms used to calculate the value of pi is %g " %
        (sumPi, pythonPi, approxError, count))  

These are the values it is is showing:

The approximate value of pi is 3.08770957930231e+00

Python's value of pi is 3.14159265358979e+00

I want it to show me this :

The approximate value of pi is 3.14159265358979

Python's value of pi is 3.14159265358979

Answer 1

I think you're missing signal . Apparently you tried to do this but changing the previous term signal and using it on the next term. See my code, i tried to do like his. What do you think?

import math
inptTol = float(input("The tolerance: "))

signal = 1.0
term = 1.0
divNum = 3.0
npower = 1.0
sumPi = 0.0
count = 0.0

while inptTol < abs(term):
    signal *= -1.0
    sumPi += term
    term = signal / (divNum * (3.0 ** npower))
    divNum += 2.0
    npower += 1.0
    count += 1.0

sumPi *= math.sqrt(12.0)
pythonPi = math.pi  
approxError = abs(sumPi - pythonPi)  

print("The approximate value of pi is %.14f\n" \
        "       Python's value of pi is %.14f\n"
        "The error in the approximation of pi is %.6e\n"
        "The number of terms used to calculate the value of pi is %g " %
        (sumPi, pythonPi, approxError, count))

Answer 2

As for me problem is because you change term value. it has to be 1 or -1 - sign.

My version - I use for loop

import math

terms_number = float(input("Enter terms number: "))

sign = 1
divNum = 1
npower = 0
sumPi = 0.0
count = 0

for x in range(terms_number):

    sumPi += sign/(divNum * (3**npower))

    # values for next term
    sign = -sign
    divNum += 2
    npower += 1
    count += 1


sumPi = math.sqrt(12) * sumPi  
pythonPi = math.pi  
approxError = abs (sumPi - pythonPi)  

print("The approximate value of pi is %.14e\n" \
        "       Python's value of pi is %.14e\n"
        "The error in the approximation of pi is %.6e\n"
        "The number of terms used to calculate the value of pi is %g " %
        (sumPi, pythonPi, approxError, count))

Result for 7 terms

The approximate value of pi is 3.14167431269884e+00
       Python's value of pi is 3.14159265358979e+00
The error in the approximation of pi is 8.165911e-05
The number of terms used to calculate the value of pi is 7 

Result for 15 terms

The approximate value of pi is 3.14159265952171e+00
       Python's value of pi is 3.14159265358979e+00
The error in the approximation of pi is 5.931921e-09
The number of terms used to calculate the value of pi is 15

---

EDIT: version with your while loop

import math

inptTol = float(input("Enter the tolerance: "))
term = 1

sign = 1
divNum = 1
npower = 0
sumPi = 0.0
count = 0

while abs(term) > inptTol:

    term = sign/(divNum * (3**npower))

    sumPi += term

    # values for next term
    sign = -sign
    divNum += 2
    npower += 1
    count += 1


sumPi = math.sqrt(12) * sumPi  
pythonPi = math.pi  
approxError = abs (sumPi - pythonPi)  

print("The approximate value of pi is %.14e\n" \
        "       Python's value of pi is %.14e\n"
        "The error in the approximation of pi is %.6e\n"
        "The number of terms used to calculate the value of pi is %g " %
        (sumPi, pythonPi, approxError, count))

Answer 3

Your term calculation is wrong:

term = -term/(divNum * (3**npower))

Say term is currently -1/(3*3) . This line won't set term to 1/(5 * 3**2) ; it'll set term to 1/(3*3) / (5 * 3**2) . You're reducing term way more than you're supposed to.

Answer 4

It looks like your trying to define term[i+1] in terms of term[i] . If you change that line to:

term = -term*(divNum-2)/(divNum * 3)

then the recursive definition will produce the appropriate values. This definition will flip the sign, remove the old odd number in the denominator, add the new odd number in the denominator, and add a factor of 3 in the denominator.

Answer 5

You can use this kind of pattern to generate the denominator for the terms in your approximation. I'll let you do the division and summation, and the final multiplication by sqrt(12)

print [(-1)**(i%2)*(3**(i)*(1+i*2)) for i in range(0,10)]

Answer 6

You are using e as the number format which means:

Exponent notation. Prints the number in scientific notation using the letter 'e' to indicate the exponent.

If you want fixed point output you can use f :

Fixed point. Displays the number as a fixed-point number.

Other formats/options can be found in the documentation

---

Found the bug in your code. Each term is calculated using the previous term as the numerator, when really you just want alternating -1 and 1s. Changing the formula for calculating term fixes the issue:

term = ((-1)**npower)/(divNum * (3**npower))

Demo