Question

To find the radius of convergence of the power series \(\sum\limits_{n = 0}^\infty {{c_n}} {x^{2n}}\) if the radius of convergence of the power series \(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\) is \(R\)

Short answer

The radius of convergence of the power series \(\sum\limits_{n = 0}^\infty {{c_n}} {x^{2n}}\) is \(\sqrt R \).

Step-by-step solution

Use the Definition to calculation  

Since the power series converges whenever\(|x| < R,\sum\limits_{n = 0}^\infty {{c_n}} {x^{2n}} = \sum\limits_{n = 0}^\infty {{c_n}} {\left( {{x^2}} \right)^n}\)converges whenever\(\left| {{x^2}} \right| < R\). Thus,\(\left| {{x^2}} \right| < R\).

That is, \(|x| < \sqrt R \).

Hence, the radius of convergence is \(\sqrt R \).

Note: Both power series \(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\) and \(\sum\limits_{n = 0}^\infty {{c_n}} {x^{2n}}\) have a centre of \(a = 0\).