Python : arbitrary precision with floats

Question

I tried to compute math.exp(9500) but encountered an OverflowError: math range error (it's roughly 6.3e4125). From this question it seems like it's due to a too large float, the accepted answer says "(...) is slightly outside of the range of a double, so it causes an overflow".

I know that Python can deal with arbitrarily large integers (long type), is there a way to deal with arbitrarily large floats in the same manner ?

Edit : my original question was about using integers for calculating exp(n) but as Eric Duminil said, the simplest way to do that would be 3**n which doesn't provide any useful result. I know realize this question might be similar to this one.

Answer 1

I don't think it's possible to approximate exp() with integers. If you use 3**n instead of 2.71828182845905**n, your calculations will be completely useless.

One possible solution would be to use Sympy. According to the documentation:

There is essentially no upper precision limit

>>> from sympy import *
>>> exp(9500)
exp(9500)
>>> exp(9500).evalf()
6.27448493490172e+4125

You can also specify the desired precision:

>>> exp(9500).evalf(1000)
6.274484934901720177929867046175406311474380389941415760684209191232450360090766458256588885184199320756050569665785657269735313171886975309933254563488343491718198237894473901620914303565550450204805537225888529509352754121292701357622411614860860409639719786022989336837263283678476008817556351031696366815467221836948040042378034720460820127399855873232167818091083005170669482845098735176209372328114732133251096196535355946589133977397512846130629857604295369747597459602137604440011394793443041829253598478244189078131130488653468669559814695095974271938947640276013215753183113041899037415404445478806695965167014404297848725756879184380559837391976534521522360723388582608454995349380217499779247330557664230806254642768796486899322646423713763772064068933790640394967085887914192401473425799354391464743910233873602389444180426155866237536459654917521713769608318128404177877383203786348495822099924812081683286880293701785567962687838594752986160305764297117036426951203418854463404773701882e+4125

With exp(9500).evalf(5000), you even get the integer closest to exp(9500).