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November 9th, 2016, 12:29 PM   #1
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Fourier series expansion without shifting to the given interval

Given the function $f(x)=f(x+9)$ with $f(x)=x^2-11x+10$

The Fourier series expansion without shifting $f(x)$ on the working interval

I will solve the Fourier series on the interval [1,10] thus the Period, P=9 and initial value, c=1

Calculating $a_0$

using the equation, $a_0= \frac{2}{P} \int_{c}^{c+P}f(x)dx$

thus, $a_0= \frac{2}{9} \int_{1}^{10}(x^2-11x+10) dx=27$

Calculating $a_n$

using the equation, $a_n= \frac{2}{P}\int_{c}^{c+P}f(x).cos( \frac{n.2 \pi.x}{P})dx$

thus, $a_n= \frac{2}{9}\int_{1}^{10}(x^2-11x+10).cos( \frac{n.2\pi.x}{9})dx= \frac{81.cos( \frac{11.\pi.n}{9})(\pi.n.cos(\pi.n)-sin(\pi.n))}{\pi^3.n^3}$

Calculating $b_n$

using the equation, $a_n=\frac{2}{P}\int_{c}^{c+P}f(x).sin(\frac{n.2 \pi.x}{P})dx$

thus, $a_n=\frac{2}{9}\int_{1}^{10}(x^2-11x+10).sin(\frac{n.2\pi.x}{9})dx=\frac{81.sin( \frac{11.\pi.n}{9})(\pi.n.cos(\pi.n)-sin(\pi.n))}{\pi^3.n^3}$

The Fourier series presentation:

using the equation, $f(x)\tilde{} \frac{1}{2}a_0+\sum_{n=1}^{ \infty} [a_n.cos( \frac{n.2 \pi.x}{P})+b_n.cos(\frac{n.2 \pi.x}{P})]$

we have,
$f(x)\tilde{}\frac{27}{2}+\sum_{n=1}^{\infty} \left[ \frac{81 cos( \frac{11.\pi.n}{9})(\pi.n.cos(\pi.n)-sin(\pi.n))}{\pi^3.n^3} cos(\frac{n.2\pi.x}{9})+\frac{81.sin(\frac{11.\pi. n}{9})(\pi.n.cos(\pi.n)-sin(\pi.n))}{\pi^3.n^3} cos( \frac{n.2\pi.x}{9})\right]$


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Can the Fourier series of f(x) be represented like this, without shifting the original function to the working interval?

Please demonstrate how i can achieve the Fourier series represented by shift the original function

Last edited by msegling; November 9th, 2016 at 12:47 PM.
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expansion, fourier, fourier series, interval, periodic functions, series, shifting



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