James Humes wrote:

>I would use the fact that e^x = SUM(n = 0, inf, [(x^n)/n!]). ie.. e^x =
>1+x+(x^2/2!)+(x^3/3!)+(x^4/4!)+---
> Now just use -at for x and iterate that sum to desired precision. Then with
>a negative exponent, the error would be between 0 and abs(x)^(n+1) /(n+1)!.
>So, the more precise you need it, the larger you make n. Using fixed point
>for the fractions here is no problem.
> That can be implemented in a simple for loop and could be written to avoid
>all the multiplies. Lots of times it only takes a few iterations to be
>somewhat close. If you're not familiar with summations, I could pseudo code
>it.
> James
>
> 
>

What he said.
Seriously, I was just going to write about the same thing.

-- 
Martin Klingensmith
http://wwia.org/
http://nnytech.net/

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