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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 ?

2 Answers 2

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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.

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    @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](http://mathworld.wolfram.com/PadeApproximant.html) for example calculation of the first few terms.2011-12-05
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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). 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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    Try the following polinomium as numerator$\frac{(n-1) z^2 \, _1F_1(2-n;2-2 n;-z)}{2 (2 n-1)}$2013-11-15