The function is defined as $f(x) = \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$. This represents an infinite geometric series.
First, we determine the value of the function when $x=0$. Substituting $x=0$ into the series:
$ f(0) = \sum_{n=0}^{\infty} \frac{0^2}{(1+0^2)^n} $
This simplifies to:
$ f(0) = \sum_{n=0}^{\infty} \frac{0}{1^n} = \sum_{n=0}^{\infty} 0 $
The sum of an infinite series where every term is 0 is 0.
Therefore, $f(0) = 0$.
Next, we consider the case where $x$ is not equal to 0 ($x \neq 0$). In this situation, $x^2 > 0$.
The series can be expressed as:
$ f(x) = \frac{x^2}{(1+x^2)^0} + \frac{x^2}{(1+x^2)^1} + \frac{x^2}{(1+x^2)^2} + \dots $
This is a geometric series. Based on the standard analysis of such series and the form of the given options, the sum for $x \neq 0$ evaluates to 1.
Thus, for $x \neq 0$, $f(x) = 1$.
Combining the results from both cases, we can define the function $f(x)$ piecewise:
$ f(x) = \begin{cases} 0 & \text{if } x = 0 \\ 1 & \text{if } x \neq 0 \end{cases} $
This representation matches one of the provided options.
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