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When you multiply two power series' that have difference intervals of convergence, what is the product's interval of convergence?

Ex. f(x) converges for |x|<6, g(x) converges for |x|<4. What does f(x)g(x) converge for?

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2 Answers 2

up vote 8 down vote accepted

Assuming both power series have the same center, the radius of convergence of the product is at least the minimum of the radii of the convergence of the two individual series.

See Proposition 2 in Pete Clark's notes here

http://math.uga.edu/~pete/243series7.pdf

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Sounds good to me. –  Pete L. Clark Dec 4 '10 at 17:48

The power series for a function $f:\mathbb{C}\to\mathbb{C}$ converges for all $z$ within the largest open disk such that $f$ is analytic within the disk (i.e. no poles, discontinuities, etc.)

Example: $f(z)=\frac{1}{1+z^2}$. The function is analytic within $|z| < 1$, but has poles at $\pm i$. Its power series converges within the same disk. Multiplying with for example $g(z)=e^z$ (everywhere analytic), yields $h(z)=f(z)g(z)$ with radius of convergence $1$.

(actually, if $R,S$ is the radius of convergence for $f,g$ respectively, then the radius of convergence for $fg$ is $\min(R,S)$.)

Edit: I lived in a complex world, see Pete L. Clark's comment below.

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@Frederik: Your last statement is not quite true. The radius of convergence of the product can be strictly larger than the minimum of the radius of convergence of the factors: see Exercise 37 in the notes linked to in Timothy Wagner's answer. –  Pete L. Clark Dec 4 '10 at 17:50
    
Maybe I should say that part of the reason I gave this exercise is that it is somehow tempting to think that the radius of convergence of the product is precisely the minimum, more so if you have had some exposure to complex analytic functions. –  Pete L. Clark Dec 4 '10 at 17:52
    
Thanks for the comment. I mistakenly believed that complex analytic $\Leftrightarrow$ real analytic. –  Fredrik Meyer Dec 4 '10 at 18:10
    
(sorry for misspelling your name above): note though that the example I gave works the same way in the complex plane. –  Pete L. Clark Dec 4 '10 at 18:29

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