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Author: William A. Stein
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%% MATH 54, Lenstra, Spring 96
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%% Solutions to Problem set 13
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%% by William Stein
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\documentclass{article}
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\usepackage{amsmath}
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\textwidth=1.4\textwidth
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\textheight=1.3\textheight
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\hoffset=-1.2 in
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\voffset=-1.28in
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\author{William Stein}
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\title{Solutions to Problem Set 13}
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\newcommand{\problem}[1]{\vspace{.3in}\par\noindent{\Large \bfseries #1. }}
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\newcommand{\soln}{\par {\em Solution.} }
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\newcommand{\dx}{\hspace{.5em}dx}
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\begin{document}
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\maketitle
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This document contains the solutions to the 13th homework set
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along with a list of theoretically inclined problems that
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resemble what could be on the final. The final examination
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is on Friday, May 17, 5--8 p.m. (room to be announced).
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The final examination covers differential equations.
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\section{Theoretically Inclined Problems}
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The TAs received the following letter from Prof. Lenstra.
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\begin{quote}
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Dear TAs,
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As Diane suggested, I drew up a list of problems from
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Boyce and DiPrima that students who wish to prepare themselves
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for questions with a theoretical flavor may try their hands at.
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I am not saying that a problem of exactly one of these types
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will appear on the final; only that the theoretical thinking
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required for the problems below will be good preparation.
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Best, Hendrik
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\end{quote}
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All problems are from Boyce-DiPrima and harder problems are starred.
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\begin{center}
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\begin{tabular}{|l|l|}\hline
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3.1&15,16,33--38\\\hline
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3.3&9,10,11,19,22*\\\hline
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3.4&23,26\\\hline
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3.5&16*, 33*\\\hline
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4.1&7--12, 13--19\\\hline
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4.2&1--10, 27\\\hline
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7.1&1--4,15\\\hline
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7.4&3*\\\hline
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7.7&13*\\\hline
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10.1&3,14,16*\\\hline
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10.2&11,12,29\\\hline
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\end{tabular}
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\end{center}
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\section{Homework Solutions. Chapter 10; 1 -- 4,5; 2 -- 1,10,21,22;
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3 -- 3,7; 4 -- 17,29}
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\problem{10.1.4}
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{\em Find the solution to the heat conduction problem
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$$100u_{xx}=u_t, \quad 0<x<1, \quad t>0$$
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$$u(0,t)=0, \quad u(1,t)=0, \quad t>0$$
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$$u(x,0)=\sin 2\pi x - 2 \sin 5\pi x, \quad 0\leq x\leq 1.$$ }
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\soln From the equation $100u_{xx}=u_t$ we see that $\alpha^2=100$.
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From the fact that $u(x,t)$ is defined for $0<x<1$ we see that
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$\ell=1$. As stated on page 517 (equation 27) we know that
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the function
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$$u(x,t)=\sum_{n=1}^m c_n u_n(x,t)
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=\sum_{n=1}^m c_n e^{-100n^2\pi^2t}\sin n\pi x$$
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satisfies the differential equation $100u_{xx}=u_t$ along with
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the boundary condition $u(0,t)=u(1,t)=0$. Notice that we have substituted
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$\alpha^2=100$ and $\ell=1$. The $c_n$ be chosen so as to satisfy
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the initial condition
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$$u(x,0)=\sin 2\pi x - 2\sin 5\pi x.$$
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Substituting $t=0$ into the above expression
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for $u(x,t)$ we obtain the equation
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$$\sum_{n=1}^m c_n \sin n\pi x=\sin 2\pi x-2\sin 5\pi x.$$
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Thus setting $c_2=1$, $c_5=-2$, and all other $c_n=0$ we
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obtain a function $u(x,t)$ satisfying the initial conditions.
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Thus the solution to the given heat conduction problem is
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$$u(x,t) = e^{-400\pi^2t}\sin 2\pi x -
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2e^{-2500\pi^2t}\sin 5\pi x.$$
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\problem{10.1.5}
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{\em Find the solution to the heat conduction problem
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$$u_{xx}=4u_t, \quad 0<x<2, \quad t>0$$
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$$u(0,t)=0, \quad u(2,t)=0, \quad t>0$$
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$$u(x,0)=2\sin(\pi x/2)-\sin(\pi x)+4\sin(2\pi x), \quad 0\leq x\leq 2.$$}
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\soln Everything proceeds in the same general way as above. This time
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$\alpha^2=4$, $\ell=2$, $c_1=2$, $c_2=-1$, $c_4=4$, and all other $c_n=0$.
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Thus the solution is
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$$u(x,t)=2 e^{-\pi^2 t}\sin (\pi x/2) - e^{-4\pi^2 t}\sin(\pi x)
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+ 4 e^{-16\pi^2 t}\sin(2\pi x).$$
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\problem{10.1.7}
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{\em In each of the following problems determine whether the
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method of separation
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of variables can be used to replace the given partial differential equation
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by a pair of ordinary differential equations. If so, find the equations.}
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7. $xu_{xx}+u_t=0$
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\soln
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Let $u=X(x)T(t)$. Then the differential equation becomes
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$$xX''T+XT'=0.$$
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A little algebra transforms this into
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$$\frac{xX''}{X}=-\frac{T'}{T}.$$
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Since the left side does not depend on $t$ and the right
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side does not depend on $x$ it follows that both sides
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most equal some constant, call it $a$. Multiplying through we
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obtain the pair of ordinary differential equations
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$$xX''-aX=0, \quad T'+aT=0.$$
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11. $u_{xx}+(x+y)u_{yy}=0$
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\soln Let $u=X(x)Y(y)$. Then the differential equation becomes
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$$X''Y+(x+y)XY''=0.$$
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It is impossible to apply separation of variables to the equation.
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\problem{10.2.1}
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{\em Determine whether the function $\sin \pi x/\ell$ is periodic.
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If so, find its fundamental period.}
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\soln The function is periodic with period $2\ell$ since
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$\sin \pi x$ is periodic with period $2$. The factor of $1/\ell$ rescales
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the graph of $\sin\pi x$ horizontally by a factor of $\ell$. Thus the
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period increases by a factor of $\ell$.
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\problem{10.2.10}
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{\em Determine whether the function
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$$f(x)=\begin{cases}(-1)^n,&2n-1\leq x<2n,\\
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1,&2n\leq x<2n+1;\end{cases}$$
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is periodic. If so, find its fundamental period. }
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\soln
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Sketching the graph of $f(x)$ we see that $f$ is periodic
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with fundamental period $4$.
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\problem{10.2.21}
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{\em Find the Fourier series corresponding to the function
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$$f(x)=\begin{cases}x+1&-1\leq x<0,\\ 1-x&0\leq x<1;\end{cases}
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\quad f(x+2)=f(x).$$}
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\soln
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This is just a translation of example 1 on page 524 so we
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can easily derive the fourier series of $f(x)$ from that. We present
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the full derivation instead.
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The Fourier series of $f$ is the projection of $f$ onto the
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subspace spanned by the functions
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$1$, $\sin(n\pi x/2)$, and $\cos(n\pi x/2)$, $n=1,2,\ldots$.
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Note that these functions are not orthonormal although they are orthogonal
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(see page 522). Recall the projection formula
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$$\text{Proj} v=\sum_{n=1}^{\infty} \frac{\langle v,u_n\rangle}
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{\langle u_n,u_n\rangle} u_n$$
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of $v$ onto the subspace spanned by an orthogonal basis $u_n$.
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Since the inner product in our situation
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is $$\langle g,h\rangle =\int_{-1}^{1}g(x)h(x)\dx$$ we see that
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the Fourier series of $f$ is
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$$\frac{a_0}{2}+\sum_{n=1}^{\infty} a_n\cos(n\pi x)
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+\sum_{n=1}^{\infty} b_n\sin(n\pi x).$$
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Here the coefficients $a_n$ and $b_n$ for $n\geq 1$ are as follows.
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$$a_n=\frac{\langle f,\cos(n\pi x)\rangle}
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{\langle \cos(n\pi x), \cos(n\pi x)\rangle}
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= \int_{-1}^{1}f(x)\cos(n\pi x) \dx$$
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and
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$$b_n=\frac{\langle f,\sin(n\pi x)\rangle}
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{\langle \sin(n\pi x), \sin(n\pi x)\rangle}
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=\int_{-1}^{1}f(x)\sin(n\pi x) \dx.$$
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(Note that the factor of $1/\ell$ which one would find in the book
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is simply one in our situation.)
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Here we have used the computation
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$$\langle \cos(n\pi x), \cos(n\pi x)\rangle = 1$$
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which can be found on page 522.
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The case $n=0$ is treated separately. We have
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$$\frac{a_0}{2} = \frac{\langle f, 1\rangle}
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{\langle 1, 1\rangle}
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=\frac{1}{2} \langle f, 1\rangle
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=\frac{1}{2} \int_{-1}^{1} f(x)\dx$$
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so $$a_0=\int_{-1}^{1}f(x)\dx.$$
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Now that we have the formulas and a firm grasp of what is going
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on we need to compute the coefficients $a_n$ and $b_n$.
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We compute
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\begin{eqnarray*}
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a_0&=&\int_{-1}^1 f(x)\dx \\
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&=&\int_{-1}^0 x+1 \dx + \int_0^1 1-x \dx\\
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&=&\Bigl[\frac{1}{2}x^2+x\Bigr]^0_{-1} + \Bigl[x-\frac{1}{2}x^2\Bigr]^1_0\\
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&=& -\Bigl(\frac{1}{2}-1\Bigr)+\frac{1}{2}=1.\end{eqnarray*}
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Next computing $a_n$ we have
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\begin{eqnarray*}
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a_n&=&\int_{-1}^{1}f(x)\cos(n\pi x)\dx\\
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&=&\int_{-1}^{0}(x+1)\cos(n\pi x)\dx + \int_{0}^{1}(1-x)\cos(n\pi x)\dx\\
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&=&\int_{-1}^{0}x\cos(n\pi x)\dx + \int_{-1}^{0}\cos(n\pi x)\dx +
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\int_{0}^{1}\cos(n\pi x)\dx - \int_{0}^{1}x\cos(n\pi x) \dx\\
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&=&\int_{-1}^{0}x\cos(n\pi x)\dx - \int_{0}^{1}x\cos(n\pi x) \dx.
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\end{eqnarray*}
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To compute this we apply integration by parts yielding
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\begin{eqnarray*}
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\int_{-1}^{0}x\cos(n\pi x)\dx
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&=& \Bigl[ \frac{x}{n\pi}\sin(n\pi x)\Bigr]^0_{-1}
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- \frac{1}{n\pi}\int_{-1}^{0}\sin(n\pi x)\dx\\
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&=& 0 - \frac{1}{n\pi}\Bigl[\frac{-1}{n\pi}\cos(n\pi x)\Bigr]^0_{-1}\\
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&=& \frac{1}{n^2\pi^2}-\frac{1}{n^2\pi^2}\cos(-n\pi)
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= \begin{cases} 0 & \text{if $n$ is even}\\
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\frac{2}{n^2\pi^2}&\text{if $n$ is odd}
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\end{cases}
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\end{eqnarray*}
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Similiarly,
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\begin{eqnarray*}
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\int_{0}^{1}x\cos(n\pi x)\dx
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&=& \Bigl[ \frac{x}{n\pi}\sin(n\pi x)\Bigr]^1_{0}
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- \frac{1}{n\pi}\int_{0}^{1}\sin(n\pi x)\dx\\
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&=& 0 - \frac{1}{n\pi}\Bigl[\frac{-1}{n\pi}\cos(n\pi x)\Bigr]^0_{1}\\
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&=& \frac{1}{n^2\pi^2}\cos(n\pi)-\frac{1}{n^2\pi^2}
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= \begin{cases} 0 & \text{if $n$ is even}\\
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-\frac{2}{n^2\pi^2}&\text{if $n$ is odd}
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\end{cases}
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\end{eqnarray*}
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It follows that
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$$a_n=\begin{cases} 0&\text{if $n$ is even}\\
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\frac{4}{n^2\pi^2}&\text{if $n$ is odd}
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\end{cases}$$
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One checks that $f(x)$ is an even function
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so $f(x)\sin(n\pi x)$ is an odd function and hence
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$$b_n=\int_{-1}^{1}f(x)\sin(n\pi x)\dx=0.$$
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Putting this all together we conclude that the Fourier expansion of $f$ is
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$$f(x)=\frac{1}{2}+\frac{4}{\pi^2}\sum_{n=1}^{\infty}
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\frac{\cos(2n-1)\pi x}{(2n-1)^2}.$$
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\problem{10.2.22}
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{\em Find the Fourier series corresponding to the function
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$$f(x)=\begin{cases}x+\ell,&-\ell\leq x\leq 0,\\
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\ell,&0<x<\ell;\end{cases}
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\quad f(x+2\ell)=f(x)$$ }
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\soln
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As explained in the previous problem, we must compute the
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$a_n$ and $b_n$ then substitute them into the appropriate series.
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We compute
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\begin{eqnarray*}
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a_0&=&\frac{1}{\ell}\int_{-\ell}^{\ell}f(x)dx\\
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&=&\frac{1}{\ell}\int_{-\ell}^{0}x+\ell\dx +
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\frac{1}{\ell}\int_0^{\ell}\ell\dx\\
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&=&\frac{1}{\ell}\Bigr[\frac{1}{2}x^2+\ell x\Bigr]^0_{-\ell}
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+ \ell = \frac{3}{2}\ell
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\end{eqnarray*}
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Next compute $a_n$ for $n\geq 1$
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\begin{eqnarray*}
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\ell a_n&=&\int_{-\ell}^0 x\cos(n\pi x/\ell)\dx
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+ \int_{-\ell}^0 \ell\cos(n\pi x/\ell)\dx
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+ \int_{0}^{\ell} \ell\cos(n\pi x/\ell)\dx\\
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&=& \int_{-\ell}^{0} x\cos(n\pi x/\ell)\dx + \ell
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\int_{-\ell}^{\ell}\cos(n\pi x/\ell)\dx
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\end{eqnarray*}
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Since the integral over a period of $\cos(x)$ is 0 we see easily that
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$$\int_{-\ell}^{\ell}\cos(n\pi x/\ell)\dx=0.$$
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For the other term integration by parts gives
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\begin{eqnarray*}
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\int_{-\ell}^{0} x\cos(n\pi x/\ell)\dx &=&
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\Bigr[\frac{\ell x}{n\pi}\sin(n\pi x/\ell)\Bigr]^0_{-\ell}
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- \frac{\ell}{n\pi}\int_{-\ell}^{0}\sin(n\pi x/\ell)\dx\\
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&=& 0 + \frac{\ell^2}{n^2\pi^2}\Bigr[\cos(n\pi x/\ell)\Bigr]^0_{-\ell}\\
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&=& \frac{\ell^2}{n^2\pi^2}(1-\cos(-n\pi))
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= \begin{cases} 0 & \text{if $n$ is even}\\
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\frac{2\ell^2}{n^2\pi^2}&\text{if $n$ is odd}
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\end{cases}
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\end{eqnarray*}
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Dividing through by $\ell$ we see that
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$$a_n=\begin{cases}0&\text{if $n$ is even}\\
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\frac{2\ell}{n^2\pi^2}&\text{if $n$ is odd}
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\end{cases}.$$
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Next compute $b_n$
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\begin{eqnarray*}
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\ell b_n &=& \int_{-\ell}^{0} x\sin(n\pi x/\ell)\dx
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+ \int_{-\ell}^{0}\ell\sin(n\pi x/\ell)\dx
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+ \int_0^{\ell}\ell\sin(n\pi x/\ell)\dx\\
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&=& \Bigr[\frac{-\ell}{n\pi}x\cos(n\pi x/\ell)\Bigr]^0_{-\ell}
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+ \frac{\ell}{n\pi}\int_{-\ell}^{0}\cos(n\pi x/\ell)\dx\\
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&=& \frac{\ell^2}{n\pi}\cos(-n\pi)+0=\frac{(-1)^{n}\ell^2}{n\pi}
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\end{eqnarray*}
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Divide through by $\ell$ to see that
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$$b_n=\frac{(-1)^{n}\ell}{n\pi}.$$
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Thus the Fourier expansion of $f$ is
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$$f(x)=\frac{3\ell}{4}+\sum_{n=1}^{\infty}
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\Bigr[\frac{2\ell\cos[(2n-1)\pi x/\ell]}{(2n-1)^2\pi^2}
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+ \frac{(-1)^{n}\ell\sin(n\pi x/\ell)}{n\pi}\Bigr].$$
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\problem{10.3.3}
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{\em Let $f(x)$ be the $2\pi$-periodic extension of the function defined as
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$\sin^2 x$ on the interval $-\pi\leq x\leq \pi$. Find the Fourier series
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for $f(x)$, then sketch the function to which the series converges
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for several periods. }
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\soln
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Fortunately we can apply some trigonometric identities to find
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the Fourier expansion of $f(x)$. Since
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$$\cos(2x)=\cos^2x-\sin^2x$$ it follows
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that $$\sin^2x=\cos^2x-\cos 2x=1-\sin^2x-\cos 2x.$$
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The Fourier expansion of $\sin^2x$ is then
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$$\sin^2x = \frac{1}{2}-\frac{1}{2}\cos 2x.$$
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\problem{10.3.7}
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{\em Let $f(x)$ be the $2\pi$-periodic extension of the function
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$$f(x)=\begin{cases}
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0,&-\pi\leq x<-\pi/2\\
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1,&-\pi/2\leq x<\pi/2\\
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0,&\pi/2\leq x<\pi\end{cases}$$
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Find the Fourier series
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for $f(x)$, then sketch the function to which the series converges
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for several periods. }
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\soln
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Notice that $\ell=\pi$. We first compute $a_0$.
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$$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\dx = \frac{1}{\pi}\int_{-\pi/2}^{\pi/2}\dx = 1.$$
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Next we compute $a_n$.
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\begin{eqnarray*}
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a_n&=&\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos(n x)\dx\\
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&=&\frac{1}{\pi}\Bigl[\frac{1}{n}\sin nx\Bigr]^{\pi/2}_{-\pi/2}\\
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&=&\frac{1}{\pi n}(\sin (n\pi/2)+\sin(n\pi/2))
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=\begin{cases} 0 & \text{if $n$ is even}\\
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\frac{(-1)^{(n-1)/2}2}{\pi} & \text{if $n$ is odd}
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\end{cases}
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\end{eqnarray*}
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Since $f(x)$ is an even function we know at once that $b_n=0$ for
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all $n$. Thus the Fourier expansion of $f(x)$ is
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$$\frac{1}{2} + \frac{2}{\pi}\sum_{n=1}^{\infty}
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\frac{(-1)^{n-1}}{2n-1}\cos(2n-1)x.$$
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\problem{10.4.17}
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{\em Use the properties of even and odd functions to evaluate the integral
327
$$\int_{-\pi}^{\pi}x^4\sin(nx)\dx.$$
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\soln Since $x^4$ is even and $\sin(nx)$ is odd the product
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$x^4\sin(nx)$ is an odd function. Thus the integral is 0. }
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\problem{10.4.29}
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{\em Find the cosine Fourier series of period $4\pi$ for the function
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$$f(x)=\begin{cases}x,&0<x<\pi,\\0,&\pi<x<2\pi\end{cases}.$$
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Then sketch the graph of the function to which the series converges
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over three periods. }
336
\soln
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Using some of the tricks which apply to even and odd functions
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the formulas for $a_n$ and $b_n$ are
339
\begin{eqnarray*}
340
a_n&=&\frac{1}{\pi}\int_0^{2\pi}f(x)\cos\frac{nx}{2}\dx\\
341
b_n&=&0
342
\end{eqnarray*}
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Obviously $a_0=\pi/2$. Computing $a_n$ in our situation is accomplished
344
via integration by parts
345
\begin{eqnarray*}
346
a_n&=&\frac{1}{\pi}\Bigl(\frac{2}{n}\Bigl[x\sin\frac{nx}{2}\Bigr]^{\pi}_0
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- \frac{2}{n}\int_{0}^{\pi}\sin\frac{nx}{2}\dx\Bigr)\\
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&=& \frac{2}{n}\sin\frac{n\pi}{2} + \frac{4}{n^2\pi}(\cos\frac{n\pi}{2}-1)
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\end{eqnarray*}
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Thus the Fourier expansion of $f(x)$ is
351
$$f(x)=\frac{\pi}{4}+\frac{1}{\pi}\sum_{n=1}^{\infty}\Bigr[\frac{2\pi}{n}
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\sin\frac{n\pi}{2}+\frac{4}{n^2}\Bigl(\cos\frac{n\pi}{2}-1\Bigr)\Bigr]\cos\frac{nx}{2}.$$
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\section{Math 74}
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\begin{center}
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Thinking of taking upper division math
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but worried whether you can handle {\em proofs}?
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\end{center}
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If so, you should consider taking Math 74, ``Transition to Upper
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Division Mathematics'', which teaches methods of discovering and
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writing up proofs, with the goal of easing the difficulty many students
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have in passing from lower division courses, where proofs are not
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emphasized, to upper division courses, where they are of central
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importance. It is intended to be taken after Math 53 or 54, and before
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or simultaneous with Math 110, 113, or 104. It will be given in both
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Summer Session and Fall Semester this year.
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\section{Final Examination, December 12, 1994}
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\problem{Problem 1} Solve the system of differential equations
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$$x_1'(t)=3x_1(t)+3x_2(t),$$
372
$$x_2'(t)=-2x_1(t)-4x_2(t)$$
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with the initial conditions
374
$$x_1(0)=1, \quad x_2(0)=3.$$
375
Show your work.
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\problem{Problem 2}
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(a) Find three functions $y_1(x), y_2(x), y_3(x)$ defined on $(-\infty,\infty)$
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whose Wronskian is given by
380
$$W(y_1,y_2,y_3)(x)=e^{4x}.$$
381
(b) Are the functions $y_1,y_2,y_3$ that you found linearly independent
382
on $(-\infty,\infty)$? Justify your answer.
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\problem{Problem 3}
385
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Find a homogeneous third-order linear differential equation with constant
387
coefficients that has
388
$$y(x)=3\cdot e^{-x}-cos(2x)$$
389
as a solution. Explain how you found it.
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What is the general solution of that differential equation?
392
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\problem{Problem 4}
394
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For the function $f(x)=e^{x/\pi}$, draw a careful sketch of the graphs of the
396
functions to which the following Fourier series converge on the interval
397
$[0,4\pi]$:
398
399
(a) the Fourier sine series of $f$ on $[0,\pi]$;
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(b) the Fourier cosine series of $f$ on $[0,\pi]$;
402
403
(c) the ordinary Fourier series of $f$ on $[-\pi,\pi]$.
404
405
Pay particular attention to the discontinuities of the functions.
406
{\em Note:} You are {\em not} asked to compute the coefficients of
407
those Fourier series.
408
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\problem{Problem 5}
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411
Find a function $u=u(x,t)$, defined for $0\leq x\leq \pi$ and
412
$t\geq 0$, satisfying the partial differential equation
413
$$u_{xx}=-u_{tt}\quad (0<x<\pi, t>0),$$
414
with boundary conditions
415
$$u(0,t)=u(\pi,t)=0\quad (t>0)$$
416
and initial conditions
417
$$u(x,0)=\sin(3x), \quad u_t(x,0)=0\quad (0<x<\pi).$$
418
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{\em Hint:} look for a function of the form $u(x,t)=X(x)T(t)$, as in sec.
420
10.6 of the textbook; but note that $u_{xx}=-u_{tt}$ is {\em not}
421
a wave equation, because of the minus sign.
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\end{document}
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