# Find the Fourier series of f on the given interval. f(x)=\left\{\begin{matrix} 0,\;\;\; -\pi...

## Question:

Find the Fourier series of {eq}f {/eq} on the given interval.

{eq}f(x)=\left\{\begin{matrix} 0,\;\;\; -\pi < x<0 & \\ sin\;x, 0 \leq x <\pi & \end{matrix}\right. {/eq}

{eq}f(x) = \;\rule{20mm}{1pt} \;\;+ \sum_{N=2}^{\infty} \left [ (\rule{20mm}{1pt} \right ) \cos \;nx + (\rule{20mm}{1pt}) \sin nx] {/eq}

Give the number {eq}F_0 {/eq} to which the Fourier series converges at a point {eq}x_0 {/eq} of discontinuity of {eq}f {/eq}.

{eq}(x_0, F_0) = \;\rule{20mm}{.5pt} {/eq}

## Fourier Series

To compute the Fourier series of function f(x) on the interval {eq}[-L,L] {/eq}, we apply the general definition of Fourier series:

{eq}f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos(\frac{n\pi x}{L})+b_n\sin(\frac{n\pi x}{L})) {/eq}

Where:

{eq}a_0=\frac{1}{L}\int_{-L}^{L}f(x)dx {/eq}

{eq}a_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos(\frac{n\pi x}{L}) {/eq}

{eq}b_n=\frac{1}{L}\int_{-L}^{L}f(x)\sin(\frac{n\pi x}{L})dx {/eq}

a) We have {eq}f(x)=\left\{\begin{matrix} 0,\;\;\; -\pi < x<0 & \\ sin\;x, 0 \leq x <\pi & \end{matrix}\right. {/eq}

Step 1) Now ...

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