Rational function that approximates e^x

Ledsnyder
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Is there a rational function,not series, that approximates e^x
?
for example (x+1)/(x+3)
 
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What do you mean by "approximates"? A given function, rational or not, has a specific difference from e^x. We can always find a function, for example by truncating the Taylor's series to a polynomial, that "approximates" e^x to a desired degree of accuracy. The difference is that the series, \sum_{n=0}^\infty does NOT 'approximate' e^x, it is exactly equal to it.
 
That is not at all the question you asked before. However, the standard Taylor's series for the exponential, \sum_{n=0}^\infty \frac{x^n}{n!}
is a very "rapid" approximation to the exponential.
 
HallsofIvy said:
That is not at all the question you asked before. However, the standard Taylor's series for the exponential, \sum_{n=0}^\infty \frac{x^n}{n!}
is a very "rapid" approximation to the exponential.

Truncating the series after a fixed number of terms is not a good method of evaluating e^x numerically for large |x|, since one loses precision from calculating the large values of x^n and n! and the remainder may be large.

To the OP: Woolfram gives a couple of continued fraction expressions for e^x; truncating these will give you a rational function approximation. But it is better to use a library function for exp if at all possible.
 
1+x.
 
It depends what type of approximation you want. One common one is an approximation such that the function and its derivatives match the approximation up to some degree

$$
e^{-x}\sim P(x)/Q(x)\\
\text{where}\\
P(x)=\sum_{k=0}^m \frac{(m+n-k)!m!}{(m+n)!k!(m-k)!}x^k
\\
Q(x)=\sum_{k=0}^n \frac{(m+n-k)!n!}{(m+n)!k!(n-k)!}x^k
$$

matches up to the n+m+1 derivative

see
http://mathworld.wolfram.com/PadeApproximant.html
http://en.wikipedia.org/wiki/Padé_table#An_example_.E2.80.93_the_exponential_function
http://wwwhome.math.utwente.nl/~vajtam/publications/temp00-pade.pdf
 

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