Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $?

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.