Question

fourier series expansion of f(x)= xsinx from- pi to pi​

Answer 1

Given:

f(x)= x sinx , from -π to π

To find:

Fourier series expansion of f(x) = x sinx from -π to π

Solution:

f(x) = x sinx ; (α , α+2l) = (-π,π)

⇒ α = -π

⇒ 2l = π - (-π) = 2π

⇒ l = π

f(-x) = (-x) sin(-x) = x sinx = f(x)

∴ f(x) = x sinx is even function

Fourier series for (-π,π) is given by

$f(x) = \frac{a_0}{2} + \sum(a_n cosnx + b_nsinnx)\\\\a_0 = \frac{1}{\pi } \int\limits^\pi _{-\pi } {f(x)} \, dx = \frac{2}{\pi } \int\limits^\pi _0{x inx} \, dx$[∵ x sinx is even]

$a_0 = \frac{2}{\pi }[x(-cosx) - 1(-sinx)] \\\\a_0 = \frac{2}{\pi }[(-\pi cos\pi ) - 0]$

a₀ = 2

$a_n = \frac{1}{\pi }\int\limits^\pi _{-\pi } {f(x).cosnx} \, dx \\\\a_n = \frac{1}{\pi } \int\limits^\pi _{-\pi } {x.sinx.cosnx} \, dx \\\\a_n = \frac{1}{\pi } \int\limits^\pi _{0 } {x.sinx.cosnx} \, dx \\\\a_n = \frac{1 }{\pi } \int\limits^\pi _0 {x[\frac{1}{2} (sin(x+nx)+sin(x - nx)]} \, dx \\\\a_n = \frac{1}{\pi } [x.(\frac{-cos(1+n)x}{1+n }+\frac{-cos(1-n)x}{1-n}) - 1(\frac{-sin(1+n)x}{(1+n)^2} - \frac{sin(1-n)x}{(1-n)^2} ) ]$

$a_n = \frac{1}{\pi } [\pi (\frac{-(-1)^{1+n}}{1+n} - \frac{(-1)^{1-n}}{1-n} )- (0)]\\\\a_n = -[\frac{(-1)^{1+n}}{1+n} + \frac{(-1)^{1-n}}{1-n} ]\\\\a_n = -(-1)^{1-n}[\frac{(-1)^2}{1+n} + \frac{1}{1-n} ]\\\\a_n = -(-1)^{1-n}[\frac{(1-n)+(1+n)}{(1+n)(1-n)} ]\\\\a_n = \frac{2(-1)^{1-n}}{n^2 - 1}$

$a_1 = \frac{1}{\pi }\int\limits^\pi _{-\pi } {x.sinx.cosx} \, dx\\\\a_1 = \frac{1}{\pi } \int\limits^\pi _{-\pi } {x.\frac{sin2x}{2} } \, dx \\\\a_1 = \frac{2}{\pi } \int\limits^\pi _0 {\frac{xsin2x}{2} } \, dx\\\\a_1 = \frac{1}{\pi } [x\frac{(-cos2x)}{2} - 1(\frac{-sin2x}{2}) ]\\\\a_1 = \frac{1}{\pi } [\frac{-\pi cos(2\pi )}{2} - 0 ]$

a₁ = -1/2

bₙ = 0

Fourier series is given by

$x sinx = 1 - \frac{cosx}{2} + 2 \sum\frac{(-1)^{1-n}}{(n^2 - 1)}cosnx$

n is from 2 to ∞

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