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Question

Let $$f(x) = \begin{cases} (\pi + x), & - \pi \le x < 0, \\ 0 & 0 \le x < \pi, \end{cases}$$ with $f(x + 2\pi) = f(x)$. If $F(x)$ represents the Fourier series of $f(x)$, then the value of $F(-\frac{\pi}{2}) + F(0)$ is

The correct answer is
$\pi$

Fourier Series Value Calculation

We need to compute $F(-\frac{\pi}{2}) + F(0)$, where $F(x)$ is the Fourier series of the given piecewise function $f(x)$.

The function is defined as:

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

with period $2\pi$. The Fourier series is $F(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))$.

Fourier Coefficients Calculation

$a_0$ Coefficient

  • $a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx = \frac{1}{\pi} \int_{-\pi}^{0} (\pi + x) dx$
  • $a_0 = \frac{1}{\pi} \left[ \pi x + \frac{x^2}{2} \right]_{-\pi}^{0} = \frac{1}{\pi} [0 - (-\pi^2 + \frac{\pi^2}{2})] = \frac{\pi}{2}$

$a_n$ Coefficients

  • $a_n = \frac{1}{\pi} \int_{-\pi}^{0} (\pi + x) \cos(nx) dx$
  • Integration by parts yields: $a_n = \frac{1}{\pi} \left[ \frac{1 - (-1)^n}{n^2} \right]$.
  • $a_n = 0$ for even $n$, and $a_n = \frac{2}{\pi n^2}$ for odd $n$.

$b_n$ Coefficients

  • $b_n = \frac{1}{\pi} \int_{-\pi}^{0} (\pi + x) \sin(nx) dx$
  • Integration by parts yields: $b_n = \frac{1}{\pi} \left[ -\frac{\pi}{n} \right] = -\frac{1}{n}$.

Fourier Series Evaluation

The Fourier series is $F(x) = \frac{\pi}{4} + \sum_{k=0}^{\infty} \frac{2}{\pi(2k+1)^2} \cos((2k+1)x) - \sum_{n=1}^{\infty} \frac{1}{n} \sin(nx)$.

Value at $F(-\frac{\pi}{2})$

  • At $x = -\frac{\pi}{2}$, the cosine terms are zero since $\cos((2k+1)(-\frac{\pi}{2})) = 0$.
  • $F(-\frac{\pi}{2}) = \frac{\pi}{4} - \sum_{n=1}^{\infty} \frac{1}{n} \sin(-\frac{n\pi}{2}) = \frac{\pi}{4} + \sum_{n=1}^{\infty} \frac{1}{n} \sin(\frac{n\pi}{2})$
  • The sum $\sum_{n=1}^{\infty} \frac{1}{n} \sin(\frac{n\pi}{2}) = 1 - \frac{1}{3} + \frac{1}{5} - ... = \frac{\pi}{4}$ (Leibniz formula).
  • Therefore, $F(-\frac{\pi}{2}) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$.

Value at $F(0)$

  • The function $f(x)$ has a jump discontinuity at $x=0$.
  • $F(0)$ converges to the average of the left and right limits: $\frac{f(0^-) + f(0^+)}{2}$.
  • $f(0^-) = \lim_{x \to 0^-} (\pi + x) = \pi$.
  • $f(0^+) = \lim_{x \to 0^+} 0 = 0$.
  • $F(0) = \frac{\pi + 0}{2} = \frac{\pi}{2}$.

Final Sum Calculation

Summing the values:

  • $F(-\frac{\pi}{2}) + F(0) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$.
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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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