Fourier Series of x in (-π, π)

Answer: The Fourier series of x in (-π, π) is given as follows: f(x) = x = $\sum_{n=1}^{\infty} \frac{2}{n} (-1)^{n+1} \sin(nx)$.

Fourier Series of x

Question: Find the Fourier series of x in the interval (-π, π).

Answer:

We know that the Fourier series of a function ( f(x) ) in the interval (-π, π) is given by:

$f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right)$,

where the Fourier coefficients a₀, a_n, b_n are given as follows.

$a_0 = \dfrac{1}{2\pi} \displaystyle \int_{-\pi}^{\pi} x \, dx = 0$.

Also, $a_n = \dfrac{1}{\pi} \displaystyle \int_{-\pi}^{\pi} x \cos(nx) \, dx = 0$. This is because xcos(nx) is an odd function.

And

$b_n = \dfrac{1}{\pi} \displaystyle \int_{-\pi}^{\pi} x \sin(nx) \, dx$ = $\dfrac{2}{\pi} \displaystyle \int_0^{\pi} x \sin(nx) \, dx$, because xsin(nx) is an even function. = $\dfrac{2}{\pi} \left[ x \cdot \dfrac{-\cos(nx)}{n} + \dfrac{1}{n} \displaystyle \int \cos(nx) \, dx \right]_0^{\pi}$ = $\dfrac{2}{\pi} \left[ \dfrac{-x\cos(nx)}{n} + \dfrac{\sin(nx)}{n^2} \right]_0^{\pi}$ = $-\dfrac{2}{n} (-1)^{n}$, using the fact that sin(nπ) = 0, cos(nπ) = (-1)n. = $\dfrac{2}{n} (-1)^{n+1}$

Therefore, the Fourier series expansion of f(x)=x on (−π,π) is given by f(x) = x = $\sum_{n=1}^{\infty} \frac{2}{n} (-1)^{n+1} \sin(nx)$.

Related Topics: Fourier series of mod x

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