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develop this function in an equivalent geometric series. 

The function provided is:

\[
f(x) = \frac{1}{1 + x^2}, \quad |x| < 1
\]

To express this in terms of a geometric series, we recall the standard geometric series expansion for \( \frac{1}{1 - z} \), which is valid for \( |z| < 1 \):

\[
\frac{1}{1 - z} = \sum_{n=0}^{\infty} z^n, \quad \text{for} \ |z| < 1.
\]

To apply this to the given function \( f(x) = \frac{1}{1 + x^2} \), we rewrite it as:

\[
f(x) = \frac{1}{1 - (-x^2)}.
\]

Now we can use the geometric series expansion with \( z = -x^2 \):

\[
f(x) = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n}, \quad \text{for} \ |x| < 1.
\]

Thus, the geometric series representation of the function is:

\[
f(x) = \sum_{n=0}^{\infty} (-1)^n x^{2n}, \quad \text{for} \ |x| < 1.
\]

Would you like more details on how to use this expansion or any further questions?

Here are some related questions you might find useful:
1. What is the radius of convergence of a geometric series?
2. How can you find the sum of a geometric series?
3. How would this series change if the function were \( f(x) = \frac{1}{1 - x^2} \)?
4. Can we use the series expansion to approximate values of \( f(x) \) for small \( x \)?
5. How does this series expansion relate to Taylor or Maclaurin series?

**Tip:** When dealing with series expansions, always ensure the range of convergence (in this case \( |x| < 1 \)) is valid before applying it.

Develop this function in an equivalent geometric series.

Learn how to develop the function f(x) = 1 / (1 + x^2) as a geometric series. This detailed explanation covers the geometric series expansion theorem and demonstrates how to apply it for |x| < 1. Suitable for advanced high school and early college students.

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