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Question

The value of the Fourier coefficient, $A_0$, in the series $\{A_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))\}$ of a function $f(x) = x^2$ with a period $2\pi$ defined over an interval $0 \le x \le 2\pi$ is

The correct answer is
$\frac{4\pi^2}{3}$

Fourier Coefficient A0 Calculation

The Fourier series for a function $f(x)$ with period $T=2\pi$ is given by:

$f(x) = A_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))$

The coefficient $A_0$ is calculated using the formula:

$A_0 = \frac{1}{2\pi} \int_{0}^{2\pi} f(x) dx$

Calculating A0 for f(x) = x2

Given the function $f(x) = x^2$ and the interval $0 \le x \le 2\pi$. Substitute $f(x)$ into the formula for $A_0$:

$A_0 = \frac{1}{2\pi} \int_{0}^{2\pi} x^2 dx$

First, find the definite integral of $x^2$ from $0$ to $2\pi$:

$\int_{0}^{2\pi} x^2 dx = \left[ \frac{x^3}{3} \right]_{0}^{2\pi}$

Evaluate the integral at the limits:

= $\frac{(2\pi)^3}{3} - \frac{(0)^3}{3}$

= $\frac{8\pi^3}{3} - 0$

= $\frac{8\pi^3}{3}$

Now, substitute this result back into the formula for $A_0$:

$A_0 = \frac{1}{2\pi} \times \frac{8\pi^3}{3}$

Simplify the expression:

$A_0 = \frac{8\pi^3}{6\pi}$

$A_0 = \frac{4\pi^2}{3}$

Therefore, the value of the Fourier coefficient $A_0$ for the function $f(x) = x^2$ is $\frac{4\pi^2}{3}$.

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Important Questions from Fourier Series

  1. 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?

  2. 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 ______.

  3. 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:

  4. 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 $k$?

  5. The effective value of the current for the Sawtooth waveform as shown is: 

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