Lale A. answered • 07/26/24
Mastering Calculus: Expert Guidance for Your Success
Step 1: Rewrite the Function
First, factor out the constant from the denominator:
\[ f(x) = \frac{x^2}{x^4 + 16} = \frac{x^2}{16 \left(1 + \frac{x^4}{16}\right)} \]
This can be rewritten as:
\[ f(x) = \frac{x^2}{16} \cdot \frac{1}{1 + \frac{x^4}{16}} \]
Step 2: Geometric Series Expansion
Recall the geometric series expansion:
\[ \frac{1}{1 - y} = \sum_{n=0}^{\infty} y^n \]
Here, you need to express our function in a similar form. Note that:
\[ \frac{1}{1 + \frac{x^4}{16}} = \frac{1}{1 - \left(-\frac{x^4}{16}\right)} \]
Let \( y = -\frac{x^4}{16} \). The geometric series expansion gives you:
\[ \frac{1}{1 + \frac{x^4}{16}} = \sum_{n=0}^{\infty} \left(-\frac{x^4}{16}\right)^n \]
Step 3: Substitute Back into the Function
Now, you substitute this series back into our expression for \( f(x) \):
\[ f(x) = \frac{x^2}{16} \cdot \sum_{n=0}^{\infty} \left(-\frac{x^4}{16}\right)^n \]
Step 4: Simplify the Series
Distribute \( \frac{x^2}{16} \) across the series:
\[ f(x) = \sum_{n=0}^{\infty} \frac{x^2}{16} \left(-\frac{x^4}{16}\right)^n \]
\[ f(x) = \sum_{n=0}^{\infty} \frac{x^2}{16} \left(-1\right)^n \frac{x^{4n}}{16^n} \]
Combine the terms inside the series:
\[ f(x) = \sum_{n=0}^{\infty} \left(-1\right)^n \frac{x^{4n + 2}}{16^{n+1}} \]
Step 5: Power Series Representation
Thus, the power series representation of the function \( f(x) \) centered at \( x = 0 \) is:
\[ f(x) = \sum_{n=0}^{\infty} \left(-1\right)^n \frac{x^{4n + 2}}{16^{n+1}} \]
