The Fourier series of the periodic function $f(x) = |x|$, $-1 < x < 1$, $f(x + 2) = f(x)$, $x \in \mathbb{R}$ is given by $$\frac{1}{2} - \sum_{n=1}^{\infty} \frac{4 \cos((2n-1)\pi x)}{(2n-1)^2 \pi^2}$$ Using the above, the sum of the infinite series $1 + \frac{1}{3^2} + \frac{1}{5^2} + \dots$ is
Fourier Series of $f(x)=|x|$
The Fourier series for the periodic function $f(x) = |x|$, with period 2, is given as:
$f(x) = \frac{1}{2} - \sum_{n=1}^{\infty} \frac{4 \cos((2n-1)\pi x)}{(2n-1)^2 \pi^2}$We need to find the sum $S = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \dots$. This is the sum of reciprocals of squares of odd numbers.
Evaluating Fourier Series at $x=0$
Consider the function value at $x=0$. Since $f(x)=|x|$, we have $f(0)=|0|=0$.
Substitute $x=0$ into the Fourier series:
$f(0) = \frac{1}{2} - \sum_{n=1}^{\infty} \frac{4 \cos((2n-1)\pi \cdot 0)}{(2n-1)^2 \pi^2}$Using $\cos(0) = 1$:
$0 = \frac{1}{2} - \sum_{n=1}^{\infty} \frac{4}{(2n-1)^2 \pi^2}$Calculating the Sum $1 + \frac{1}{3^2} + \frac{1}{5^2} + \dots$
Isolate the summation term:
$\sum_{n=1}^{\infty} \frac{4}{(2n-1)^2 \pi^2} = \frac{1}{2}$Factor out constants:
$\frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{1}{2}$Solve for the summation:
$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{1}{2} \times \frac{\pi^2}{4} = \frac{\pi^2}{8}$This summation represents the series $1 + \frac{1}{3^2} + \frac{1}{5^2} + \dots$.
Thus, the sum of the series is $\frac{\pi^2}{8}$.
Important Questions from Fourier Series
The Fourier series representation of a square wave is shown in the figure below. The fluctuations seen near $x = \pm 1$ are named after which one of the following scientists?
The Fourier series expansion of the function $f(x) = |\cos(x)|$ in the interval $(-\pi, \pi)$ is $\alpha + \beta \left[ \frac{1}{3}\cos(2x) - \frac{1}{15}\cos(4x) + \dots \right]$.
The values of $\alpha$ and $\beta$ respectively are ______.
Periodic function $f(x)$ is given below. $$f(x) = \begin{cases} -1, & \text{when } -\pi < x < 0 \\ 1, & \text{when } 0 < x < \pi \end{cases} ; f(x + 2\pi) = f(x)$$ The CORRECT option representing the Fourier series expansion of $f(x)$ is:
A time-limited waveform $g(x)$ is specified as follows: $$g(x) = \begin{cases} -k, & -\pi < x \le 0 \\ +k, & 0 < x \le \pi \\ 0, & \text{otherwise} \end{cases}$$ A new waveform $f(x)$ is constructed from $g(x)$ as follows: $$f(x) = \sum_{m=-\infty}^{\infty} g(x + 2\pi m), \quad \text{for all } x \in \mathbb{R}$$ The sum of the coefficients of the third harmonics of the sine and cosine terms in the trigonometric Fourier series expansion of $f(x)$ is $\frac{2}{3\pi}$.
What is the value of $k$?
The effective value of the current for the Sawtooth waveform as shown is:
