我在 Wallis 产品上的 python 代码有什么问题？

Question

So, I was asked to code the Wallis product and it wasn't supposed to be really complicated. So I made a code but apparently, it could only work for Wallis(1) and not the rest. Could anyone help me? Thank you!

def Wallis (n):
    result = 1
    for count in range(2, n+2, 2):
        result = result * (count**2/((count-1)*(count+1)))
    return result

Formula to compute Wallis product

(2*2)/(3*5) * (4*4)/(5*7) * (6*6)/(7*9) and so on until ((n*2) ** 2)/ ((n-1)* (n+1))

Output

Wallis(1) = (2*2)/(3*5) = 0.267
Wallis(2) = Wallis(1) * (4*4)/(5*7) = 0.122

Answer 1

Mainly your range command is wrong:

range(2, 2*n+2, 2)

Alternatively, you can move the complexity to the formula, ie:

for count in range(1, n):
    result = result * (4*count*count/((2*count-1)*(2*count+1)))

Answer 2

The problem with your code is that count is equal to 2*n in your equation so in (count**2/((count-1)*(count+1))) , count**2 is the same as (2*n)**2 however the following count-1 should be the same as n-1 but rather it is actually (2*n)-1 . The same goes for count+1 .

I made my own version which should help you (although I used a different equation which is at https://www.wikiwand.com/en/Wallis_product )

def wallis(limit):
    result = 1
    for x in range(2, limit, 2):
        result *= (x / (x - 1)) * (x / (x + 1))
    return result

As limit gets higher, it converges closer to half pi.

Answer 3

As others said, your range() needs to be changed.

For n=3,

for count in range(2, n+2, 2):
    print(i)

would print

2
4

but you needed 6 as well.

As Python documentation says

class range(start, stop[, step]) 

For a positive step, the contents of a range r are determined by the formula r[i] = start + step*i where i >= 0 and r[i] < stop.

meaning stop is not included.

Here step is 2 , you could use

for count in range(2, 2*(n+1), 2):

And exponentiation is costlier than multiplication. So

count*count

is better than

count**2

So the modified version could be

def Wallis(n):
    result=1
    for count in range(2, 2*(n+1), 2):
        result*=((count*count)/((count-1)*(count+1)))
    return result

Note that the value returned by Wallis is only half the value of Pi.

You could modify the return statement to

return result*2

if you want.