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I have a function $g(x)=f(x)e^{-x}$ and i want to consider the following integral: $\int_{0}^{\infty}g(x)dx$.

Since $f(x)$ is a complicated, but monotonic decreasing, function in the interval $[0,\infty[$ and it is multiplied by $e^{-x}$, the integrand $g(x)$ is a very rapidly decreasing function.

If I consider $T_{n}(x)$ a Taylor expansion of the function $f(x)$ around $x=0$ truncated at a certain order $n$, the integral $\int_{0}^{\infty}T_{n}(x)e^{-x}dx$ converges for every $n$.

What is the error committed in approximating $\int_{0}^{\infty}g(x)dx$ with $\int_{0}^{\infty}T_{n}(x)e^{-x}dx$?

and what would it be if $g(x)=f(x)\cdot x(e^{-x})$?

thanks in advance

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Hint

If you have the Taylor expansion $$f(x)=\sum_{i=0}^n a_i x^i$$, you then have $$\int_0^\infty f(x)e^{-x}dx=\sum_{i=0}^n a_i i!$$ since $\int_0^\infty x^n e^{-x}=n!$.

I am sure that you can take from here.

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  • $\begingroup$ if i consider the error $E$ as: $ E = \int_{0}^{\infty} f(x) e^{-x} dx - \sum_{i=0}^n a_i i!$ rewriting a little the expression and using the following equivalence: $\int_{0}^{\infty} f(x) e^{-x} dx = \lim_{N \rightarrow \infty} \sum_{k=0}^N a_k k!$ I can rewrite the total error as: $E = 1 -\lim_{N \rightarrow \infty} (\frac{\sum_{i=0}^n a_i i!}{\sum_{k=0}^N a_k k!})$ now of course this is a decreasing function of $n$ and this should be compared with the value of the defined integral $\int_{0}^{\infty}g(x)$ is this correct? $\endgroup$
    – SSC Napoli
    Sep 23 '14 at 8:30

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