Python 在分数之间交替使用 + 和 -

Question

I am trying to use the Nilakantha Pi Series formula and a for loop to calculate pi depending on how far into the calculation the user chooses the iterations to be. Here is the website that shows this infinite formula: https://www.mathsisfun.com/numbers/pi.html . I want to display the correct answer for iterations greater than 1, but only the first iteration shows the correct answer. Here is what I have so far:

def for_loop(number):
    n = 4
    pi = 3
    for i in range(1, number + 1):
        den = (n-2) * (n-1) * n
        if (number % 2 == 0):
            pi -= (4 / den)
            print(pi)
        else:
            pi += (4 / den)
            print(pi)
        n = n + 2

Answer 1

The immediate problem is that you are checking if number , not i , is even or odd. But you don't need any such checks. You just have to alternate the numerator between 4 and -4.

def for_loop(number):
    n = 4
    pi = 3
    num = 4
    for _ in range(number):
        den = (n-2) * (n-1) * n
        pi += num/den
        print(pi)
        num *= -1
        n += 2

or

from itertools import cycle

def for_loop(number):

    n = 4
    pi = 3
    for num in cycle([4, -4]):
        den = (n-2)*(n-1)*n
        pi += num/den
        print(pi)
        n += 2

or even

from itertools import cycle, count

def for_loop(number):
    pi = 3
    for num, n in zip(cycle([4,-4]), count(4, 2)):
        den = (n-2)*(n-1)*n
        pi += num/den
        print(pi)

Answer 2

You have to rework the condition of %2

def for_loop(number):
    n = 4
    pi = 3
    for i in range(1, number + 1):
        den = (n-2) * (n-1) * n
        if (i % 2 == 0):# replace number by i , its alternating between even/uneven, however number is all the time the same.
            pi -= (4 / den)
        else:
            pi += (4 / den)
        print(pi)
        n = n + 2 

Answer 3

This is another way of doing it by using the reduce() function of the functools module, to solve the operation in the denominator of the fraction. To change the sign (-1 or 1) for each iteration, you can just multiply it by -1.

from functools import reduce


def nilakantha(n):
    i, sign, start, base, stop = 1, 1, 2, 3, 4
    res = base
    while i < n and n > 1:
        res += sign * (4 / reduce(lambda x, y: x * y, range(start, stop + 1)))
        sign *= -1
        start = stop
        stop = start + 2
        i += 1
    return res

Example output

>>> nilakantha(524)
3.141592655327371