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.

Besides of computational aspects what's the difference between a Taylor series around a point $\alpha$ and a Taylor series around a point $\beta$?

share|improve this question
Depends on what you mean by "computational aspect." That can cover a lot of ideas. –  Thomas Andrews Jun 6 '11 at 6:06
I meant how hard (by hand) is to get numerical values depending on its Taylor series expansion around distinct points. –  Vicfred Jun 6 '11 at 6:14

2 Answers 2

up vote 4 down vote accepted

The series may have different radii of convergence, for example. I assume you start with a reasonably nice function $f$. Suppose $f$ has a pole, I mean, you are "dividing by 0" somewhere, say, at $x=a$. If nothing else happens, then the radius of convergence at $\alpha$ cannot be larger than the distance from $a$ to $\alpha$, but it could be precisely this distance. Same with $\beta$.

To be specific, take $f(x)=1/(1-x)$. The series at 0 is $\displaystyle \sum_n x^n$ which converges for $|x|<1$. The series at $-1$ is $\displaystyle \sum_n \frac1{2^{n+1}} (x+1)^n$; it converges for $|x+1|<2$.

This explains why even in the case of reasonably well behaved functions we may want to compute the Taylor series at different points depending of the region we are interested in studying.

share|improve this answer
could you explain/expand the "depending on the region" part? I did not quite understand. –  Vicfred Jun 6 '11 at 6:11
Say you want to study the $f$ from the example in a neighborhood of -1.5 using Taylor series. Even though the series at 0 is very well-behaved, you cannot use it for this task. Instead, you need to pick a point for which the series will converge in a neighborhood of -1.5 (for example, -1). In complex analysis this is an important issue. Analytic functions are locally given by power series, but typically there is not a power series that gives us the function over its whole domain. –  Bruce George Jun 6 '11 at 6:41

Suppose $f(z)=\sum_n a_n(z-z_0)^n$ on some open subset $S$ within the disc of convergence $B(z_0,\rho)$ (we could take the expansion to be valid on the whole disc too). Then given another $w \in S \subseteq B(z_0,\rho)$, we can find a $B(w,\delta) \subseteq S \subseteq B(z_0,\rho)$ such that $f$ can be expanded as a power series with $w$ as a center. i.e

$f(z)=\sum_{n=0}^{\infty}\hat{a}_n(z-w)^n$ valid on $B(w,\delta)$ Where $\hat{a}_n=\sum_{i=n}^{\infty}\binom{i}{n} a_i(w-z_0)^{n-i}$

[One use this to show that the derivative of $f$ at a point $w$ inside the disc of convergence is just the first coefficient of the expansion around that point.]

share|improve this answer

Your Answer


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.