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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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2 Answers 2

It is usually done with Padé Approximations. You can quickly obtain the numerator and denominator polynomial coefficients in MATLAB with the pade() command.

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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
    
@sauravrt Yes, but that's your choice. If computation is cheap then you can increase the orders, if not, you can keep it simple. Read more on here for example calculation of the first few terms. –  user13838 Dec 5 '11 at 4:13

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 $2n\geqslant x$, $$ \sum\limits_{i=0}^{2n-1}(-1)^i\frac{x^{k+i}}{i!}\leqslant f(x)\leqslant\varepsilon_n(x)+\sum\limits_{i=0}^{2n-1}(-1)^i\frac{x^{k+i}}{i!},\qquad\varepsilon_n(x)=\frac{x^{k+2n}}{(2n)!}. $$

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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

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