To determine the interval of convergence and the power series representation of the function $f(x) = \frac{x^5}{16 - x^2}$, we can follow these steps:

1. Rewrite the function in a form suitable for a power series expansion.

   We start with: $f(x) = \frac{x^5}{16 - x^2}$

   To express this in a form that allows for a power series, rewrite the denominator as: $f(x) = x^5 \cdot \frac{1}{16 - x^2} = x^5 \cdot \frac{1}{16} \cdot \frac{1}{1 - \frac{x^2}{16}}$ which simplifies to: $f(x) = \frac{x^5}{16} \cdot \frac{1}{1 - \frac{x^2}{16}}$

2. Expand using the geometric series.

   Recall that the function $\frac{1}{1 - u}$ can be expanded as a geometric series: $\frac{1}{1 - u} = \sum_{n=0}^{\infty} u^n, \quad \text{for } |u| < 1$

   Here, let $u = \frac{x^2}{16}$. Then: $\frac{1}{1 - \frac{x^2}{16}} = \sum_{n=0}^{\infty} \left( \frac{x^2}{16} \right)^n$ This gives us: $\frac{1}{1 - \frac{x^2}{16}} = \sum_{n=0}^{\infty} \frac{x^{2n}}{16^n}$

3. Multiply by $x^5 / 16$ to get the power series for $f(x)$:

   Substitute back into the expression for $f(x)$: $f(x) = \frac{x^5}{16} \sum_{n=0}^{\infty} \frac{x^{2n}}{16^n}$ Simplify by combining terms: $f(x) = \sum_{n=0}^{\infty} \frac{x^{2n + 5}}{16^{n+1}}$

   Thus, the power series representation of $f(x)$ is: $f(x) = \sum_{n=0}^{\infty} \frac{x^{2n + 5}}{16^{n+1}}$

4. Determine the interval of convergence.

   Since we used the geometric series with $u = \frac{x^2}{16}$, convergence requires: $\left| \frac{x^2}{16} \right| < 1$ which simplifies to: $|x^2| < 16 \quad \Rightarrow \quad |x| < 4$

   Therefore, the radius of convergence is $R = 4$, and the interval of convergence is: $-4 < x < 4$

Summary

• Power series for $f(x)$: $f(x) = \sum_{n=0}^{\infty} \frac{x^{2n + 5}}{16^{n+1}}$

• Interval of convergence: $(-4, 4)$

Would you like further details on any step?

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Further Questions for Exploration:

1. What is the interval of convergence for a different value in the denominator, such as $25 - x^2$?
2. How would the series change if the function were $f(x) = \frac{x^3}{16 - x^2}$?
3. Can we represent $f(x) = \frac{x^5}{16 - x^2}$ using a Taylor series centered at $x = 0$?
4. How would the interval of convergence be affected if we had $f(x) = \frac{x^5}{1 - x^4}$?
5. How do the coefficients of this power series behave as $n \to \infty$?

Tip: When converting a function to a power series, start by rewriting it in a form that resembles a geometric series if possible.