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Mathematics 7 Online
OpenStudy (anonymous):

Question about Power Series!

OpenStudy (anonymous):

Suppose that the radius of convergence of the power series \[\sum_{}^{} c_n * x^n\] is R. What is the radius of convergence of the power series \[\sum_{}^{} c_n * x^(2n)\] ?

OpenStudy (anonymous):

*It is x^(2n)

OpenStudy (anonymous):

I got sqrt{R} but I'm not so sure that is the final answer?

OpenStudy (rational):

thats right!

OpenStudy (anonymous):

Oh okay. I explained the problem with words, but I want to find a way to show it "mathematically" is there anyway I could maybe use the Ratio Test?

OpenStudy (rational):

radius of convergence of \(\sum c_n * x^n\) is \(R\) means : \[\lim\limits_{n\to\infty}\left|\frac{c_{n+1}x^{n+1}}{c_nx^n}\right| = |x|\lim\limits_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|=|x|\frac{1}{R}\] right ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

and all of that has to be less than 1...

OpenStudy (anonymous):

and then we plug in x^2 for x since the second series is f(x^2)

OpenStudy (rational):

Now apply ratio test for \(\sum c_n * x^{2n}\) : \[\lim\limits_{n\to\infty}\left|\frac{c_{n+1}x^{2(n+1)}}{c_nx^{2n}}\right| = |x^2|\lim\limits_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|=|x^2|\frac{1}{R}\] right ?

OpenStudy (anonymous):

Oh yes, I get it now... Thanks :)

OpenStudy (rational):

np:)

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