# Rational function that approximates e^x

1. Aug 2, 2014

### Ledsnyder

Is there a rational function,not series, that approximates e^x
?
for example (x+1)/(x+3)

2. Aug 2, 2014

### HallsofIvy

Staff Emeritus
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.

3. Aug 2, 2014

4. Aug 3, 2014

### HallsofIvy

Staff Emeritus
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.

5. Aug 3, 2014

### pasmith

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.

6. Aug 7, 2014

1+x.

7. Aug 9, 2014

### lurflurf

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