# Math Help - Radius of convergence

Find the radius of convergence of:

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

2. ## Re: Radius of convergence

We have a sum of two power series. What is the radius of convergence of each one?

3. ## Re: 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 $(-1)^n(x/2)^{2n}$, the root test should work nicely. The nth root of $(x/2)^{2n}$ is $(x/2)^2$. 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: $[x(x/2)^{2(n+1)}]/[x(x/2)^{2n}]= x/2$. 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.

4. ## Re: Radius of convergence

Originally Posted by HallsofIvy
"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 $(-1)^n(x/2)^{2n}$, the root test should work nicely. The nth root of $(x/2)^{2n}$ is $(x/2)^2$. 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: $[x(x/2)^{2(n+1)}]/[x(x/2)^{2n}]= x/2$. 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.
1) The nth term of the first series is $(-1)^n(x^2/2)^n$ and not $(-1)^n(x/2)^{2n}$.
2) $[x(x^2/2)^{n+1}]/[x(x^2/2)^n]=x^2/2$
So, the radius of convergence is $\sqrt{2}$.