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Chapter 7: Problem 70

Prove that if the power series \(\sum_{n=0}^{\infty} c_{n} x^{n}\) has a radius of convergence of \(R\), then \(\sum_{n=0}^{\infty} c_{n} x^{2 n}\) has a radius of convergence of \(\sqrt{R}\).

Short Answer

Expert verified
The radius of convergence of the series \(\sum_{n=0}^{\infty} c_{n} x^{2n}\) is \(\sqrt{R}\).

Step by step solution

01

Preliminaries about power series

In general, a power series of the form \(\sum_{n=0}^{\infty} c_{n} x^{n}\) converges for all \(x\) such that \(|x|<R\), where \(R\) is the radius of convergence. In other words, the series converges if \(x\) is in the open interval \((-R, R)\).
02

Considering the new power series

Now consider the power series \(\sum_{n=0}^{\infty} c_{n} x^{2n}\). If we set \(y=x^2\), this series can be rewritten as \(\sum_{n=0}^{\infty} c_{n} y^n\), which is the same form as the original series.
03

Determine the radius of convergence of the new series

Since \(y=x^2\), the original interval of convergence \(-R<x<R\) becomes \(-\sqrt{R}<x<\sqrt{R}\) or \(0<y<R\). Therefore, the new series converges if \(y=x^2\) or \(|x|<\sqrt{R}\). Hence, the radius of convergence of this series is \(\sqrt{R}\).

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