Question: Find a power series representation for the function. (Give your power series representation…

Find a power series representation for the function. (Give your power series representation centered at \( x = 0 \).)

\[ f(x) = \frac{x^2}{x^4 + 16} \]

\[ f(x) = \sum_{n=0}^{\infty} \left( \square \right) \]

Determine the interval of convergence. (Enter your answer using interval notation.)

\[ \square \]

Solution

To find a power series representation for the function \( f(x) = \frac{x^2}{x^4 + 16} \) centered at \( x = 0 \), we can follow these steps: \[ f(x) = \frac{x^2}{x^4 + 16} \] First, factor the denominator: \[ x^4 + 16 = 16\left(1 + \frac{x^4}{16}\right) \] Rewrite the function using this factorization: \[ f(x) = \frac{x^2}{16\left(1 + \frac{x^4}{16}\right)} = \frac{x^2}{16} \cdot \frac{1}{1 + \frac{x^4}{16}} \] Recognize the geometric series form \( \frac{1}{1 - r} = \sum_{n=0}^{\infty} r^n \) where \( |r| < 1 \). Here, \( r = -\frac{x^4}{16} \): \[ \frac{1}{1 + \frac{x^4}{16}} = \sum_{n=0}^{\infty} \left(-\frac{x^4}{16}\right)^n \] Multiply by \( \frac{x^2}{16} \) to obtain the power series representation of \( f(x) \): \[ f(x) = \frac{x^2}{16} \sum_{n=0}^{\infty} \left(-\frac{x^4}{16}\right)^n = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n + 2}}{16^{n + 1}} \] Therefore, the power series representation is: \[ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n + 2}}{16^{n + 1}} \] Next, determine the interval of convergence by ensuring the geometric series converges: \[ \left| -\frac{x^4}{16} \right| < 1 \implies \frac{x^4}{16} < 1 \implies |x|^4 < 16 \implies |x| < 2 \] Thus, the interval of convergence is: \[ (-2, 2) \]