I'm trying to write a thing in python to compute the digits of pi with the chudnovsky formula

however i'm having problems because the numbers are getting too small and giving an overflow error after just 20 or so iterations of the summation thingy.

is there a way to solve this problem?

## numbers too small / chudnovsky formula / python

**Moderators:** phlip, Moderators General, Prelates

### Re: numbers too small / chudnovsky formula / python

Can you post your code? Are you using pure Python?

I'm not entirely sure how Python handles floating point numbers internally, but it shouldn't have problems with anything that is within the range of a 64-bit double precision floating point number.

Ok, maybe you're kind of pushing it

I'm not entirely sure how Python handles floating point numbers internally, but it shouldn't have problems with anything that is within the range of a 64-bit double precision floating point number.

Code: Select all

`n = 20`

>>> math.factorial(6*n) / (math.factorial(3*n)*(math.factorial(n))**3)

55827110111954369003114678433725205384304901681068121718830400L

Ok, maybe you're kind of pushing it

### Re: numbers too small / chudnovsky formula / python

i've split the thing to the right of the summation into three parts here just because it was easier to write it out.

output:

specifically, the 640320^3n part is too long to convert to a float, and even if i had a workaround for that another part would probably break after a few thousand more numbers.

i know the formula is right, because when i add up all the terms, multiply by the thing to the left of the summation symbol, and then find the inverse, i get pi accurate to many decimal places (each iteration adds 14 more decimal places iirc). Except i get negative pi for some reason, but that doesn't seem serious.

Code: Select all

`def sumthingy(n):`

first = (13591409.0/545140134.0+n)

second = math.factorial(6*n)/((math.factorial(3*n)*math.factorial(n)**3)*1.0)

third = ((-1**n)*1.0)/(640320**(3*n))

return first*second*third

for x in xrange(0,100):

print x, sumthingy(x)

os.system('pause')

output:

Code: Select all

`0 -0.0249....`

1 -4.68....e-16

...

...

17 -1.735...e-243

18 (and then it breaks here)

specifically, the 640320^3n part is too long to convert to a float, and even if i had a workaround for that another part would probably break after a few thousand more numbers.

i know the formula is right, because when i add up all the terms, multiply by the thing to the left of the summation symbol, and then find the inverse, i get pi accurate to many decimal places (each iteration adds 14 more decimal places iirc). Except i get negative pi for some reason, but that doesn't seem serious.

### Re: numbers too small / chudnovsky formula / python

Code: Select all

`third = ((-1**n)*1.0)/(640320**(3*n))`

All your terms would be negative because you are doing -1^n instead of (-1)^n.

Code: Select all

`third = (((-1)**n)*1.0)/(640320**(3*n))`

### Re: numbers too small / chudnovsky formula / python

>-) wrote:output:Code: Select all

`0 -0.0249....`

1 -4.68....e-16

...

...

17 -1.735...e-243

18 (and then it breaks here)

You're running into the the limits of what a double precision floating point number can handle, which is about 5e-324 (see here). In this thread it is suggested to use the Decimal module, although I'm not entirely sure if this will resolve your issue. Perhaps mpmath can be of use for you here?

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### Re: numbers too small / chudnovsky formula / python

You can always try the rational numbers library in the "fractions" module... it doesn't have any range limits (as long as the numerator and denominator fit in RAM) though it can get a lot slower as the numerator/denominator get bigger...

Code: Select all

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void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

### Re: numbers too small / chudnovsky formula / python

thanks for the answers i will try to use the mpmath thingy to fix the problem

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