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Let $f(x)$ and $g(x)$ be two Taylor series such that: $$ f(x)= \sum_{n=0}^{\infty}(-1)^{n} a(n) x^{n} $$ and $$ g(x)= \sum_{n=0}^{\infty} b(n) x^{n} $$, for $ a(n) >0 $ and $b(n) > 0 $.

My question is, can we extract the asymptotic behavior of these two taylor series for $ x \rightarrow \infty $?

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We have to suppose the radius of convergence to be infinite. What do you exactly mean by asymptotic behavior? What are the properties that interest you? – Davide Giraudo Jun 4 '12 at 9:23

Asymptotic expansions are essentially different from Taylor and Laurent expansions and cannot be easily deduced from one another. To call an expansion of $f(z)$ as $z\to z_0$ asymptotic you need to find an infinite sequence of functions ${w_n(z)}$ such that $\lim_{z\to z_0}|w_{n+1}(z)/w_n(z)|=0$, or $w_{n+1}=o(w_n(z))$ and obtain expansion of the form $$f(z)=\sum_{n=1}^Na_nw_n(z)+o(w_N)$$ where coefficients $a_n$ are determined as follows: $$a_n=\lim_{z\to z_0}\left\{\frac{f(z)-\sum_{n=1}^{N-1}a_n w_n(z)} {w_N(z)}\right\}$$ To get a notion of the difference between the domains where power series converges and an asymptotic expansion is valid, consider the following example. $$I=\int_0^{\infty}\frac{e^{-zt}}{1+t^2}dt$$ where $\Re(z)>0$.Expanding the denominator in the geometric series which converges in the circle $|t|<1$ and integrating term by term we obtain: $$I=\frac{1}{z}-\frac{2!}{z^3}+\frac{4!}{z^5}-...+\frac{(-1)^{n-1}(2n-2)!}{z^{2n-1}}+R_n(z)$$ where $$R_n(z)=(-1)^n\int_0^{\infty}\frac{e^{-zt}t^{2n}}{1+t^2}dt$$ Now $$|R_n(z)|\le\int_0^{\infty} e^{-xt}t^{2n}dt=\frac{(2n)!}{x^{2n+1}}$$ Let $z\ne 0$ and $-(\pi/2)+\alpha\le \arg{z}\le (\pi/2)-\alpha$, $0<\alpha<\pi/2$. Then $x\ge \sin\alpha$, hence $$|R_n(z)|\le\frac{(2n)!}{(\sin\alpha)^{2n+1}}\frac{1}{|z|^{2n+1}}$$ For a fixed $n$, $\lim_{|z|\to\infty}|R_n(z)|=0$. Hence, even that the series does not converge, for large $|z|$ and a finite number of terms it is a good approximation to the function. In contrast working with power series, for a given $z$ approximation is better, the more terms of the series we take. The purpose of the example os also to show that whereas Taylor (Laurent) expansion converges within a circle (annulus), the domain where asymptotic expansion is valid has the shape of an angle in the complex plane.

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no no my comment was a bit different , i mean given a taylor series with infinite radius of convergence so it represents a certain function on the interval $ (0, \infty ) $ obtaine the function $ f(x) $ from its Taylor series – Jose Garcia Jun 6 '12 at 8:50

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