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Calculus Advanced /
Fourier Series.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=0.5cm]{geometry} \begin{document} {\large Fourier Series} \begin{align*} \text{\bf Periodic}&\text{ \bf Functions:}\quad f(x+p)=f(x)\quad\text{for all $x$.}\\ &\text{If $f$ is not constant, the smallest positive $p$ for which the above holds is called the \bf primitive period \rm of $f$.}\\ \text{\bf Fourier}&\text{ \bf Series:}\quad \boxed{f(x)=a_0+\sum_{n=1}^\infty(a_n\cos nx+b_n\sin nx)},\quad \text{a periodic function with primitive period $2\pi$.}\\ &\text{To find the coefficients $a_n$ and $b_n$:}\\ &\int_{-\pi}^{\pi}f(x)~dx =\int_{-\pi}^{\pi}a_0+\sum_{n=1}^\infty(a_n\cos nx+b_n\sin nx)~dx =\int_{-\pi}^{\pi}a_0~dx =2\pi a_0.\\ &\therefore\quad\boxed{a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)~dx.}\\ &\because\quad\int_{-\pi}^{\pi}f(x)\cos mx~dx=\pi a_m,\quad\therefore\text{we have}\quad \boxed{a_m=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos mx~dx,\quad m\in\mathbb{N}.}\\ &\quad\text{Proof:}\quad I=\int_{-\pi}^{\pi}f(x)\cos mx~dx =\int_{-\pi}^{\pi}a_0\cos mx~dx+\sum_{n=1}^\infty\left[a_n\int_{-\pi}^{\pi}\cos nx\cos mx~dx+b_n\int_{-\pi}^{\pi}\sin nx\cos mx~dx\right]\\ &\qquad\quad =\int_{-\pi}^{\pi}a_0\cos mx~dx +\sum_{n=1}^\infty\Big[ a_n~\frac{1}{2}\int_{-\pi}^{\pi}\cos(n+m)x~dx +a_n~\frac{1}{2}\int_{-\pi}^{\pi}\cos(n-m)x~dx\\ &\qquad\qquad\qquad+b_n~\frac{1}{2}\int_{-\pi}^{\pi}\sin(n+m)x~dx +b_n~\frac{1}{2}\int_{-\pi}^{\pi}\sin(n-m)x~dx \Big].\\ &\qquad\text{All terms are zero except $\frac{1}{2}\int_{-\pi}^{\pi}\cos(n-m)~dx=\pi$ when $n=m$.} \qquad\therefore I=\pi a_m.\\ &\text{Likewise,}\quad\boxed{b_m=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin mx~dx,\quad m\in\mathbb{N}.}\\ \\ &\text{For periodic function with period $2L$,}\quad \boxed{f(x)=a_0+\sum_{n=1}^\infty\left(a_n\cos\frac{n\pi x}{L}+b_n\sin\frac{n\pi x}{L}\right).}\\ &\qquad\boxed{a_0=\frac{1}{2L}\int_{-L}^{L}f(x)~dx.}\\ &\qquad\boxed{a_m=\frac{1}{L}\int_{-L}^{L}f(x)\cos\frac{m\pi x}{L}~dx,\quad m\in\mathbb{N}.}\\ &\qquad\boxed{b_m=\frac{1}{L}\int_{-L}^{L}f(x)\sin\frac{m\pi x}{L}~dx,\quad m\in\mathbb{N}.}\\ \\ \text{\bf Complex}&\text{ \bf Form:}\quad \boxed{f(x)=\sum_{n=-\infty}^\infty c_n~e^{inx}}~,\qquad \text{where}\quad\boxed{c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)~e^{-inx}~dx.}\\ &\qquad\qquad\text{Note: The resulting $f$ is still a real function.}\\ \\ \text{\bf Linear}&\text{\bf lity:}\quad \text{If $f_1$ and $f_2$ are two Fourier Series, $af_1+bf_2$ is a Fourier Series with coefficients}\\ &\qquad\quad\text{equal to the sums of the corresponding coefficients of $f1$ and $f2$ times $a$ and $b$ respectively.}\\ &\qquad\quad\text{i.e. Let $f_1=\sum_{n=-\infty}^\infty c_{1n}~e^{inx}$\quad and\quad$f_2=\sum_{n=-\infty}^\infty c_{2n}~e^{inx}$,} \quad af_1+bf_2=\sum_{n=-\infty}^\infty(a~c_{1n}+b~c_{2n})~e^{inx}.\\ \\ &\qquad\quad\text{If $f$ is a Fourier Series, $kf$ is a Fourier Series with coefficients}\\ &\qquad\quad\text{equal to $k$ times the corresponding coefficients of $f$. i.e. }\quad kf=\sum_{n=-\infty}^\infty(k~c_n~e^{inx}). \end{align*} \end{document}