My Math Forum Fourier series expansion without shifting to the given interval

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 November 9th, 2016, 12:29 PM #1 Newbie   Joined: Nov 2016 From: South Africa Posts: 2 Thanks: 0 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, $f(x)\tilde{}\frac{27}{2}+\sum_{n=1}^{\infty} \left[ \frac{81 cos( \frac{11.\pi.n}{9})(\pi.n.cos(\pi.n)-sin(\pi.n))}{\pi^3.n^3} cos(\frac{n.2\pi.x}{9})+\frac{81.sin(\frac{11.\pi. n}{9})(\pi.n.cos(\pi.n)-sin(\pi.n))}{\pi^3.n^3} cos( \frac{n.2\pi.x}{9})\right]$ ---------- 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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