# series. Reply

**S.D.Mechveliani
**
mechvel@math.botik.ru

*Wed, 15 Aug 2001 12:45:30 +0400*

Luis Pablo Michelena <lmichele@multi.com.uy> writes
on the subject `series'
>* where to find a haskell program that calculates the number e, that is
*>* the list of infinite digits?
*>* [..]
*>* what i am looking for is something like the ertostenes sifts, that
*>* prints every prime number until it run out of memory ...
*
In what way the Heratosphenes sieve for prime numbers may relate to
finding approximations of number e ?
As to finding the infinite list of digits for e = lim (1 + 1/n)^n,
n -> infinity
here the program is suggested for finding (eAppr n) :: Rational
such that
|e - (eAppr n)| < 1/2^(n-3) :
import Ratio
eAppr :: Integer -> Rational
eAppr n = appr 0 (1%1) (0%1)
where
appr k member res = --- member = 1/(k!),
if --- res = sum [1/i! | i <- [0..k]]
k==n then res
else appr (k+1) (member/(fromInteger (k+1))) (res+member)
Several decimal digits can be obtained, then, like this:
fromRational (eAppr 6) :: Double --> 2.716666666666667
fromRational (eAppr 20) :: Double --> 2.718281828459045
Explanation. According to Calculus, we have
e = lim (eApp n), where eAppr n = sum [1/k! | k <- [0..n]]
n -> infinity
and |e - (eAppr n)| < 3/n! <= 1/2^(n-3).
Therefore, eAppr(n+3) differs from e in less than 1/2^n.
I believe, this fact will help us to find first true k digits of e
for any given k.
If people would not give other good solution and if you ask me to
complete this task, then I'll try to do this.
Regards,
-----------------
Serge Mechveliani
mechvel@botik.ru