Question

To find the interval of convergence of the series \(f(x) = \sum\limits_{n = 0}^\infty {{c_n}} {x^n}\) and formula for \(f(x)\), where \({c_{n + 4}} = {c_n}\).

Short answer

The interval of convergence of the series is \(I = ( - 1,1)\) and formula for

\(f(x) = \frac{{{c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3}}}{{1 - {x^4}}}\)

Step-by-step solution

Prove that the series is sum of the geometric series with initial term

"The sum of the geometric series with initial term \(a\) and common ratio \(r\) is \(\sum\limits_{n = 0}^\infty a {r^n} = \frac{a}{{1 - r}}\) where \(|x| < 1\) (1)

Express \(f(x) = \sum\limits_{n = 0}^\infty {{c_n}} {x^n}\) as follows.\(\begin{aligned}\sum\limits_{n = 0}^\infty {{c_n}} {x^n} &= {c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3} + {c_0}{x^4} + {c_1}{x^5} + {c_2}{x^6} + {c_3}{x^7} + \cdots \\\sum\limits_{n = 0}^\infty {{c_n}} {x^n} &= \left( {{c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3}} \right) + \left( {{c_0}{x^4} + {c_1}{x^5} + {c_2}{x^6} + {c_3}{x^7}} \right) + \cdots \\\sum\limits_{n = 0}^\infty {{c_n}} {x^n} &= \left( {{c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3}} \right) + {x^4}\left( {{c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3}} \right)\quad + {x^8}\left( {{c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3}} \right) + \cdots \end{aligned}\)By the result (1) the series\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)is sum of the geometric series with initial term \(a = \left( {{c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3}} \right)\) and common ratio \(r = {x^4}\) is

\(\sum\limits_{n = 0}^\infty {{c_n}} {x^n} = \frac{{{c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3}}}{{1 - {x^4}}}\)

Calculate the interval of convergence of f(x)

Since the geometric series converges when \(|r| < 1,\left| {{x^4}} \right| < 1\).

\(\left| {{x^4}} \right| < 1\)

\(|x| < 1\)

\( - 1 < x < 1\)

The radius of convergence is R=1 and the convergence interval is (-1,1).

The radius of convergence is \(( - 1,1)\) and formula for \(f(x) = \frac{{{c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3}}}{{1 - {x^4}}}\).