Find the radius of convergence of:

$\displaystyle c_0\left(1-\frac{x^2}{2}+\frac{x^4}{2^2}-...\right)+c_1\left(x-\frac{x^3}{2}+\frac{x^5}{2^2}-...\right)$

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- Nov 5th 2011, 08:47 PMalexmahoneRadius of convergence
Find the radius of convergence of:

$\displaystyle c_0\left(1-\frac{x^2}{2}+\frac{x^4}{2^2}-...\right)+c_1\left(x-\frac{x^3}{2}+\frac{x^5}{2^2}-...\right)$ - Nov 6th 2011, 04:02 AMgirdavRe: Radius of convergence
We have a sum of two power series. What is the radius of convergence of each one?

- Nov 6th 2011, 04:42 AMHallsofIvyRe: Radius of convergence
"Radius of convergence" only applies to a power series, not a sum of two power series. The function itself will exist on the intersection of the two intervals of convergence of the two power series.

You can find the radius of convergence of these power series by using the "ratio test" or "root test". Here, because the nth term in the first series is $\displaystyle (-1)^n(x/2)^{2n}$, the root test should work nicely. The nth root of $\displaystyle (x/2)^{2n}$ is $\displaystyle (x/2)^2$. For what x is that less than 1? In order to avoid the "nth root of x", I would use the ratio test for the second: $\displaystyle [x(x/2)^{2(n+1)}]/[x(x/2)^{2n}]= x/2$. For what x is that less than 1? The "ratio of converence" for the entire sum (although I don't think that's the correct term- the interval on which the function exists) is the smaller of those two. - Nov 6th 2011, 07:46 AMalexmahoneRe: Radius of convergence