Fourier Series Calculation

The problem requires finding the sum of the infinite series $S = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots$ using the Fourier series representation of a given periodic function $f(x)$.

Function Definition

The periodic function is defined as:

$ f(x) = \begin{cases} -2, & -\pi < x < 0 \\ 2, & 0 < x < \pi \end{cases} $

The function has a period of $2\pi$, i.e., $f(x + 2\pi) = f(x)$. This function is odd since $f(-x) = -f(x)$.

Coefficients Fourier Series

Because $f(x)$ is an odd function, its Fourier series expansion contains only sine terms. The general form is $f(x) = \sum_{n=1}^{\infty} b_n \sin(nx)$. The coefficients $a_0$ and $a_n$ are zero.

The formula for the sine coefficient $b_n$ is:

$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx $

Since the integrand $f(x)\sin(nx)$ is an even function, we can simplify the integral:

$ b_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin(nx) dx $

Substituting $f(x)=2$ over the interval $(0, \pi)$:

$ b_n = \frac{2}{\pi} \int_{0}^{\pi} 2 \sin(nx) dx = \frac{4}{\pi} \int_{0}^{\pi} \sin(nx) dx $

Evaluating the integral yields:

$ b_n = \frac{4}{\pi} \left[ -\frac{\cos(nx)}{n} \right]_{0}^{\pi} = -\frac{4}{n\pi} [\cos(n\pi) - \cos(0)] $

Using $\cos(n\pi) = (-1)^n$ and $\cos(0) = 1$:

$ b_n = -\frac{4}{n\pi} [(-1)^n - 1] $

Analysis for $n$:

• If $n$ is even, $n=2k$, then $b_n = -\frac{4}{n\pi} [1 - 1] = 0$.
• If $n$ is odd, $n=2k-1$, then $b_n = -\frac{4}{n\pi} [-1 - 1] = \frac{8}{n\pi}$.

Thus, the non-zero coefficients are $b_{2k-1} = \frac{8}{(2k-1)\pi}$ for $k = 1, 2, 3, \dots$.

Expansion Fourier Series

The Fourier series for $f(x)$ is:

$ f(x) = \sum_{k=1}^{\infty} \frac{8}{(2k-1)\pi} \sin((2k-1)x) $

Expanding the series:

$ f(x) = \frac{8}{\pi} \left( \sin(x) + \frac{1}{3}\sin(3x) + \frac{1}{5}\sin(5x) + \dots \right) $

Evaluation Series Convergence

We are looking for the sum $S = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots$. This pattern arises when evaluating the Fourier series at a specific point.

Consider $x = \frac{\pi}{2}$. From the function definition, $f(\frac{\pi}{2}) = 2$.

Substitute $x = \frac{\pi}{2}$ into the Fourier series expansion:

$ f(\frac{\pi}{2}) = \frac{8}{\pi} \left( \sin(\frac{\pi}{2}) + \frac{1}{3}\sin(\frac{3\pi}{2}) + \frac{1}{5}\sin(\frac{5\pi}{2}) + \frac{1}{7}\sin(\frac{7\pi}{2}) + \dots \right) $

Using the sine values: $\sin(\frac{\pi}{2}) = 1$, $\sin(\frac{3\pi}{2}) = -1$, $\sin(\frac{5\pi}{2}) = 1$, $\sin(\frac{7\pi}{2}) = -1$, and so on:

$ 2 = \frac{8}{\pi} \left( 1 + \frac{1}{3}(-1) + \frac{1}{5}(1) + \frac{1}{7}(-1) + \dots \right) $

$ 2 = \frac{8}{\pi} \left( 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots \right) $

Let the sum be $S = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots$. The equation becomes:

$ 2 = \frac{8}{\pi} S $

Solving for $S$:

$ S = \frac{2\pi}{8} = \frac{\pi}{4} $

The sum $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots$ converges to $\frac{\pi}{4}$.