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## anonymous 4 years ago |x-a|<R. I need the detailed explanation of this which is a condition of Taylor Series. Thanks in advance.

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1. anonymous

If you have a power series $\sum_{n = 0}^\infty a_n (x-a)^n$ Via the ratio test, it converges iff $\left| \frac{a_{n+1} (x-a)^{n+1}}{a_n (x-a)^n}\right| < 1$ This puts the condition on x-a that $|x-a| < \left| \frac{a_n}{a_{n+1}} \right| = R$

2. cristiann

|x-a|<R means -R<x-a<R or -R+a<x<R+a In a Taylor series context, a should stand for point around which you discuss the series while R should stand for the radius of the development The theory says that the series (as a function series) uniformly converges on any compact interval [c,d] inside (a-R,a+R) and diverges outside (a-R, a+R) You have to study for each case the behavior on the boundary, i.e. x=a-R and x=a+R The interval (a-R, a+R) with the boundaries modified by the above study gives you the convergence domain

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