# Approximating $x^k e^{-x}$

I want to approximate the function $f(x) = x^k e^{-x}$ with some finite series. One approach would be to use the power series expansion for $e^{-x}$. But in that case, the power series would have to be truncated such that the order of the truncated power series is of order greater than $x^k$ to ensure that the approximation $\hat{f}(x)$ does not blow up as $x$ grows large. But what is the optimum way to choose the order of the truncated power series for $e^{-x}~~ in ~~ f(x)$?
I couldn't find any useful reference to this problem. Any suggestions?

Are there alternative approaches ?

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It is usually done with Padé Approximations. You can quickly obtain the numerator and denominator polynomial coefficients in MATLAB with the pade() command.
If I use Pade approximation $R_{\alpha \beta}(x)$ for $e^{-x}$, I still need to figure out how to choose $\alpha ~~and~~ \beta$. Since $\beta$ is the order of denominator polynomial in $R_{\alpha \beta}(x)$, I guess $\beta \geq L + \alpha$ ? – sauravrt Dec 5 '11 at 3:50
The successive terms in the expansion of $f(x)$ are $(-1)^na_n(x)$ with $a_n(x)=x^{n+k}/n!$. Assuming that $x\geqslant0$, one sees that $a_n(x)\geqslant0$ and $(a_n(x))_n$ is decreasing on the range $n\geqslant n(x)$ with $x-1\leqslant n(x)<x$. Hence, for every $n\geqslant n(x)$, the partial sums up to orders $n$ and $n+1$ bound $f(x)$ from both sides more and more precisely. For example, for every $n$ such that $2n\geqslant x$, $$\sum\limits_{i=0}^{2n-1}(-1)^i\frac{x^{k+i}}{i!}\leqslant f(x)\leqslant\varepsilon_n(x)+\sum\limits_{i=0}^{2n-1}(-1)^i\frac{x^{k+i}}{i!},\qquad\varepsilon_n(x)=\frac{x^{k+2n}}{(2n)!}.$$
I'm slightly confused by your notation $n(x)$. Can you please make it clear? – sauravrt Dec 5 '11 at 2:41
Try the following polinomium as numerator$$\frac{(n-1) z^2 \, _1F_1(2-n;2-2 n;-z)}{2 (2 n-1)}$$ – capea Nov 15 '13 at 10:30