Let fx - - if - - x - 0- if 0 - x -be a periodic function of period 2 - The coefficient of sin 5x in the Fourier series expansion of fx in the interval - - is

Fourier series is f(x) = a_02 + ∑ a_n cos nx + ∑ b_n sin nx where b_n = 1π∫_ π^π f(x) sin nxdx Hence to find coefficient of sin 5x, we need to calculate value o ...

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Standard XII

Mathematics

Properties of Modulus

Let fx = -π, ...

Question

Let $f (x) = {\begin{matrix} - π, if & - π < x \leq 0 π, if & 0 < x \leq π \end{matrix}$
be a periodic function of period $2 π$ . The coefficient of $sin 5 x$ in the Fourier series expansion of f(x) in the interval $[- π, π]$ is

$\frac{4}{5}$

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$\frac{5}{4}$

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$\frac{4}{3}$

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$\frac{3}{4}$

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Solution

The correct option is A
$\frac{4}{5}$

Fourier series is f(x) $= \frac{a_{0}}{2} + \sum a_{n} cos n x + \sum b_{n} sin n x w h e r e b_{n} = \frac{1}{π} \int_{- π}^{π} f (x) sin n x d x$
Hence to find coefficient of sin 5x, we need to calculate value of $b_{5} i . e$
$b_{5} = \frac{1}{π} \int_{- π}^{π} f (x) sin 5 x d x = \frac{1}{π} [\int_{- π}^{0} (- π sin 5 x d x)] + [\int_{0}^{π} π sin 5 x d x]$
$= {(\frac{cos 5 x}{5})}_{- π}^{0} - {(\frac{cos 5 x}{5})}_{0}^{π} = \frac{4}{5}$

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Let f(x)={−π,if−π<x≤0π,if0<x≤π be a periodic function of period 2π. The coefficient of sin5x in the Fourier series expansion of f(x) in the interval [−π,π] is

The correct option is A 45 Fourier series is f(x) =a02+∑ancosnx+∑bnsinnxwherebn=1π∫π−πf(x)sinnxdx Hence to find coefficient of sin 5x, we need to calculate value of b5 i.e b5=1π∫π−πf(x)sin5xdx=1π[∫0−π(−πsin5xdx)]+[∫π0πsin5xdx] =(cos5x5)0−π−(cos5x5)π0=45

Let $f (x) = {\begin{matrix} - π, if & - π < x \leq 0 π, if & 0 < x \leq π \end{matrix}$
be a periodic function of period $2 π$ . The coefficient of $sin 5 x$ in the Fourier series expansion of f(x) in the interval $[- π, π]$ is