Question

Let \(f(x)=\sum_{n=0}^{\infty} c_{n} x^{n},\) where \(c_{n+3}=c_{n}\) for \(n \geq 0\). (a) Find the interval of convergence of the series. (b) Find an explicit formula for \(f(x)\).

Short answer

The interval of convergence for the given power series is \(|x| < 1\). The explicit formula for \(f(x)\) is \(f(x) = \frac{c_0}{1-x^3} + \frac{c_1x}{1-x^3} + \frac{c_2x^2}{1-x^3}\).

Step-by-step solution

Breaking the infinite series into three separate series

The infinite series \(f(x)=\sum_{n=0}^{\infty} c_{n} x^{n}\) can be divided into three separate series to account for the coefficients \(c_{n+3}=c_{n}\). Hence, we write it as follows: \(f(x) = c_{0}\sum_{n=0}^{\infty} (x^3)^n + c_1\sum_{n=0}^{\infty} x(x^3)^n + c_2\sum_{n=0}^{\infty} x^2(x^3)^n\)

Application of the ratio test to find the interval of convergence

By applying the ratio test to each of these series, it can be determined that the interval of convergence for all three series is \(|x| > 1\). Consequently, the interval of convergence for the original series is the same, \(|x| < 1\).

Finding explicit formula for \(f(x)\)

To find the explicit formula for \(f(x)\), apply formula for sum of an infinite geometric series to each of the three series. This yields \(f(x) = \frac{c_0}{1-x^3} + \frac{c_1x}{1-x^3} + \frac{c_2x^2}{1-x^3}\).