Question

Find a power series for the function, centered at \(c,\) and determine the interval of convergence. $$ f(x)=\frac{4}{4+x^{2}}, \quad c=0 $$

Short answer

The power series representation for the given function centered at 0 is \( \sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{4^n} \). The interval of convergence is \(-2 < x < 2\).

Step-by-step solution

Rewrite the function

Start by rewriting the given function as: \( f(x)=\frac{4}{4}\times\frac{1}{1+\frac{x^{2}}{4}} \). This can be further simplified to \( f(x)=\frac{1}{1-(\frac{-x^{2}}{4})} \). This is in the form of the geometric series \( \frac{1}{1-u} \) where \( u=-\frac{x^{2}}{4} \).

Generate the Power Series

For a geometric series \( \frac{1}{1-u} \), the power series representation is \( \sum_{n=0}^{\infty}u^n \). We substitute \( u=-\frac{x^{2}}{4} \) to this formula and obtain the power series: \( \sum_{n=0}^{\infty} (-1)^n (\frac{x^{2n}}{4^n}) \). This can be simplified to: \( \sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{4^n} \). This is the power series centered at 0 for the given function.

Find the Interval of Convergence

For a geometric series, the part of the series \(u=-\frac{x^{2}}{4}\) must be less than 1 in magnitude for the series to converge. That is, \( |-x^2/4| < 1 \). Solving this inequality, we get \(-4 < x^2 < 4\), which simplifies to \(-2 < x < 2\). Therefore, the interval of convergence is \(-2 < x < 2\).