Haskell/GHC - Is there a way around Haskell only returning 16 decimal places?

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I am trying to write a program in Haskell that returns 'e' (Euler's number) to a given decimal place. Here is my code so far:

factorial 0 = 1
factorial n = n * factorial (n - 1)

calculateE a 
    | a == 0 = 1 
    | otherwise = nextLevel
    where nextLevel = (1 / (factorial a)) + calculateE (a-1)

Whenever I call calculateE I only get back 16 decimal places. Is this a limitation of Haskell/My computer? Is there a way to get back any number of decimal places?

eulers-number arbitrary-precision precision ghc haskell
1 Answers

This code already works to arbitrary precision. You just need to use an arbitrary precision type and not the standard Float/Double. Haskell's standard library has Rational for this purpose, which represents rational numbers as pairs of integers.

ghci> calculateE 100 :: Rational
4299778907798767752801199122242037634663518280784714275131782813346597523870956720660008227544949996496057758175050906671347686438130409774741771022426508339 % 1581800261761765299689817607733333906622304546853925787603270574495213559207286705236295999595873191292435557980122436580528562896896000000000000000000000000

The issue now is getting a sequence of digits out of it. I'm not aware of anything in the standard library that does it, so here's a stupid simple (might still be buggy!) implementation:

import Data.List(unfoldr)
import Data.List.NonEmpty(NonEmpty((:|)))
import Data.Ratio

-- first element is integral part (+ sign), rest are positive and < 10 and are digits
-- after the decimal point (for negative numbers, these digits should be seen as having negative value)
longDivision :: Integral a => Ratio a -> NonEmpty a
longDivision x = hi :| unfoldr go (abs lo)
  where (hi, lo) = numerator x `quotRem` denominator x
        go 0 = Nothing
        go lo = Just $ (lo * 10) `quotRem` denominator x

printDigits :: Show a => NonEmpty a -> String
printDigits (x :| xs) = show x ++ "." ++ concatMap show xs

So

ghci> take 100 $ printDigits $ longDivision $ calculateE 100
"2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642"

This approximation actually seems to be good to ~160 digits after the decimal point.

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