\documentclass{amsart} %\pagestyle{empty} \usepackage{graphicx, amssymb,amsmath,amsfonts, psfrag} \setlength{\textheight}{9in} \setlength{\topmargin}{0in} \setlength{\textwidth}{6in} \setlength{\oddsidemargin}{+0.5in} \newcommand{\dis}{\displaystyle} \begin{document} \begin{center}{\bf Lecture Notes for Ch 5\\Series Solutions\\Introduction and Overview}\end{center} \vspace{1in} You are responsible for knowing the material in Sections 1 - 3 of Chapter 5. Section 5.1 is a review of infinite series from {\em Calculus II}. Read this on your own. You should know how to do problems 1-27 in Section 5.1. Do not turn these in. You may be quizzed on them. \vspace{.25in} I will cover 5.2 and 5.3 in reverse order. This is because 5.3 is easier. The basic idea is that if we cannot find the exact solution to a differential equation perhaps we can find the Taylor series of the solution and use some $n$-order Taylor polynomial as an approximation to the solution. \vspace{.25in} In 5.3 we just compute a few terms of the series solution. In 5.2 we show how to find a recursive formula for the coefficients of the series solution in some cases. \vspace{.25in} This material is extremely tedious. Be patient. Read slowly. These notes mostly consist of examples. After reading through an example, go back and work out all the details on a separate sheet of paper. Compare the examples with the homework problems and pause to work on them as you go through these notes. Good luck, you'll need it! \vfill \pagebreak \begin{center}{\bf Lecture Notes for Ch 5\\Series Solutions}\end{center} Given a differential equation, we suppose the solution has a Taylor series. Recall that if $y(t)$ has a Taylor series centered about $t=0$ the formula for it is \[ y(t) = y(0) + y'(0)t + \frac{y''(0)}{2!}t^2 + \frac{y'''(0)}{3!}t^3 + \cdots = \sum_{n=0}^\infty \frac{y^{[n]}(0)t^n}{n!}. \] More generally, a Taylor series could be centered about any number, say $t=c$. Then \[ y(t) = \sum_{n=0}^\infty \frac{y^{[n]}(c)(t-c)^n}{n!}. \] If we write $y(t) = \sum_{n=0}^\infty a_n (t-c)^n $ then of course $a_n = \dfrac{y^{[n]}(c)}{n!}$. The idea is to plug the series for $y(t)$ into the differential equation and deduce the $a_n$'s. Think of it as the Method of Undermined Coefficients on steroids. We start with a very naive example, a first order equation that you already know how to solve. \vspace{.2in} {\bf Example 1.} Solve $y' = y$ with $y(0)=2$ using a power series for $y$ centered at zero. \vspace{.2in} {\bf Solution.} Let $y = a_0 + a_1t + a_2t^2 + \cdots = \dis \sum_{n=0}^\infty a_n t^n$. Then, since $y(0) = a_0$ we know $a_0 = 2$. \noindent Next we compute the first derivative, \[ y'(t) = a_1 + 2a_2t + 3a_3 t^2 + \cdots = \sum_{n=0}^\infty n a_n t^{n-1}. \] Thus, $y'(0) = a_1$. But $y'(0) = y(0) = 2$. Hence $a_1=2$. \vspace{.3in} \noindent Next we compute the second derivative, \[ y''(t) = 2a_2 + 3\cdot 2 a_3 t + 4\cdot 3 a_4 t^2 + \cdots = \sum_{n=0}^\infty n(n-1) a_n t^{n-2}. \] Thus, $y''(0) = 2a_2$. But $y'' = (y')' = y' = y$. Thus $y''(0) = y(0) = 2$. Hence $a_2=1$. \vspace{.3in} \noindent And then the third. $y'''(0) = 3\cdot 2 a_3 = 3!a_3$. But $y'''(0) = y''(0) = 2$. Hence $a_3 = 2/3!$. \vspace{.3in} \noindent Finally, we find $y''''(0) = 4! a_4$. But $y''''(0) = 2$. Hence $a_4 = 2/4!$. \vspace{.3in} If you do a few more terms, you should see that the pattern is $a_n = 2/n!$. Thus, \[ y(t) = \sum_{n=0}^\infty \frac{2}{n!}t^n. \] But this is just the Taylor series of $2e^t$. \vfill \pagebreak \begin{center}{\bf Series Solutions of \\Second Order Linear Differential Equations}\end{center} We will do an example with a second order differential equation with constant coefficients. This is not the best way to do such a problem, but we want to illustrate the series method with easy examples to start. \vspace{.2in} {\bf Example 2.} Find the first five terms of the series solution of $y'' + 2y' + y = 0$ with $y(0)=1$, $y'(0)=2$. \vspace{.2in} {\bf Solution.} Since the initial condition is at zero our series will be centered at zero as well. Suppose $y(t) = a_0 + a_1t + a_2 t^2 + a_3 t^3 + \cdots = \dis\sum_{n=0}^\infty a_n t^n$. Then $a_n = \frac{y^{[n]}(0)}{n!}$. Now, $y(0) = a_0$ implies $a_0 = 1$. Likewise $y'(0) = a_1$ implies $a_1 =2$. It is true in general that the initial conditions determine $a_0$ and $a_1$. Next, notice that $y''(t) = -2 y'(t) - y(t)$. Thus, $y''(0) = -2 y'(0) -y(0) = -2\cdot 2 - 1 = -5$. Therefore, \[ a_2 = \frac{y''(0)}{2!} = -\frac{5}{2}. \] Now, we need to compute $y'''(t)$ and evaluate it at $t=0$. \[ y'''(t) = (y''(t))' = (-2 y'(t) - y(t))' = -2 y''(t) - y'(t). \] Thus, \[ y'''(0) = -2 y''(0) - y'(0) = -2 \cdot (-5) - 2 = 8. \] Therefore, \[ a_3 = \frac{y'''(0)}{3!} = \frac{8}{6} = \frac{4}{3}. \] One more! We need to compute $y''''(t)$ and evaluate it at $t=0$. \[ y''''(t) = (-2 y''(t) - y'(t))' = -2y'''(t) - y''(t). \] Therefore, \[ y''''(0) = -2 y'''(0) - y''(0) = -2 \cdot 8 - (-5) = -11. \] Thus, \[ a_4 = \frac{y''''(0)}{4!} = -\frac{11}{24}. \] Finally, we put this together to get, \[ y(t) \approx 1 + 2t-\frac{5}{2}t^2 + \frac{4}{3}t^3 - \frac{11}{24} t ^4. \] In this example, we did not find a general formula for $a_n$. If we pushed it further we might be able to, but this is good enough for now. \vfill \pagebreak \begin{center}{\bf Examining the Last Example}\end{center} Recall we wanted to solve $y'' + 2y' + y = 0$ with $y(0)=1$ and $y'(0) = 2$. It is easy enough to find the exact solution. It is \[ y = e^{-t} + 3 t e^{-t}. \] Below we plot this along with the forth degree Taylor polynomial we found, $y=1 + 2t-\frac{5}{2}t^2 + \frac{4}{3}t^3 - \frac{11}{24} t ^4$. The green curve is the exact solution and the red curve is the approximation. \begin{center} \includegraphics[width=3in]{Figs/ex2plotplot1.eps} \end{center} \vfill \pagebreak \begin{center}{\bf Example}\end{center} {\bf Example 3.} Find the first five terms of the series solution of $y'' + (\sin x) y' + (\cos x)y = 0$, with $y(0)=0$ and $y'(0) = 1$. {\bf Solution.} Let $y=\dis \sum_{n=0}^\infty a_n x^n$ with $a_n = \dfrac{y^{[n]}(0)}{n!}$. Then we have the following calculations. \[ y(0) = 0 \:\: \implies \:\: a_0 = 0 \:\:\:\:\& \:\:\:\: y'(0)=1 \:\: \implies \:\: a_1 = 1. \] Next we solve for $y''$, evaluate it at $x=0$, and then compute $a_2$. \vspace{.2in} $\begin{array}{lll} y''& =& -(\sin x)y' -(\cos x)y. \end{array}$ \vspace{.2in} Thus, \vspace{.2in} $\begin{array}{lll} y''(0) &=& - 0 \cdot 1 - 1 \cdot 0 = 0 \:\: \implies \:\: a_2 = 0/2! = 0. \end{array}$ \vspace{.2in} Next, we find $y'''$, evaluate it at $x=0$, and then compute $a_3$. \vspace{.2in} $\begin{array}{lll} y''' &=& (-(\sin x)y' -(\cos x)y)'\\ &=& -\left((\cos x)y' + (\sin x)y'' \right) - \left( -(\sin x)y + (\cos x)y'\right)\\ &=& -(\sin x)y'' - 2 (\cos x)y' + (\sin x)y. \end{array}$ \vspace{.2in} Thus, \vspace{.2in} $\begin{array}{lll} y'''(0) &=& -0\cdot 0 - 2 \cdot 1 \cdot 1 + 0 \cdot 0 = -2 \:\: \implies\:\: a_3 = -2/3! = -1/3. \end{array}$ \vspace{.2in} Finally, we find $y''''$, evaluate it at $x=0$, and then compute $a_4$. \vspace{.2in} $\begin{array}{lll} y'''' &=& (-(\sin x)y'' - 2 (\cos x)y' + (\sin x)y)' \\ &=& -(\cos x)y'' -(\sin x) y''' + 2 (\sin x) y' -2 (\cos x)y''+(\cos x)y + (\sin x) y' \\ &=& -(\sin x)y''' - 3(\cos x)y'' - 3(\sin x)y' -(\cos x)y. \end{array}$ \vspace{.2in} Thus, \vspace{.2in} $\begin{array}{lll} y''''(0) &=& -0(-2) - 3\cdot 1 \cdot 0 -3 \cdot 0 \cdot 1 -1 \cdot 0 = 0\:\: \implies \:\:a_4 = 0/4! = 0. \end{array}$ \vspace{.2in} Putting this all together we get \[ y(x) \approx x - \frac{1}{3} x^3. \] \vfill \pagebreak \begin{center}{\bf Examining the Previous Example}\end{center} \vspace{.2in} Recall we are studying $y'' + (\sin x) y' + (\cos x)y = 0$, with $y(0)=0$ and $y'(0) = 1$. \vspace{.2in} We do not have the methods to find the exact solution. Maple gives the following for the general solution \[ y(x) = e^{\cos x}\left(C_1 \int_0^x e^{-\cos s}\, ds + C_2\right). \] The answer is given it terms of an integral that cannot be done in closed form. We will use $y(0)=0$ and $y'(0) = 1$ to find $C_1$ and $C_2$. We will then check that it works by plugging it into the original differential equation. \[ y(0) = e(0 + C_2) = 0 \:\: \implies C_2 = 0. \] \[ y'(x) = -\sin x \, e^{\cos x} C_1 \int_0^x e^{-\cos s}\, ds + C_1e^{\cos x}e^{-\cos x} = -\sin x \, e^{\cos x} C_1 \int_0^x e^{-\cos s}\, ds + C_1. \] Hence, \[ C_1=y'(0) = 1. \] Now, \[ y''(x) = -\cos x \,e^{\cos x} \int_0^x e^{-\cos s}\, ds + \sin^2 x \, e^{\cos x} \int_0^x e^{-\cos s}\, ds - \sin x. \] Thus, \begin{eqnarray*} y'' + (\sin x) y' + (\cos x)y &=& -\cos x \,e^{\cos x} \int_0^x e^{-\cos s}\, ds + \sin^2 x \, e^{\cos x} \int_0^x e^{-\cos s}\, ds - \sin x\\ & +& \sin x \left(-\sin x \, e^{\cos x} \int_0^x e^{-\cos s}\, ds + 1\right) + \cos x \, e^{\cos x} \int_0^x e^{-\cos s}\, ds\\ &=& \left(-\cos x + \sin^2 x -\sin^2 + \cos x \right) e^{\cos x} \int_0^x e^{-\cos s}\, ds. \\ &=& 0. \checkmark \end{eqnarray*} Finally, we plot the exact solution along with the approximation. The green curve is the exact solution and the red curve is the approximation. \begin{center} \includegraphics[width=3in]{Figs/ex3plot1.eps} \end{center} \vfill \pagebreak \begin{center}{\bf Example not centered at zero}\end{center} {\bf Example 4.} Find the first five terms of the series solution of $y'' + x y' + y = 0$ with $y(2)=3$ and $y'(2)=1$. \vspace{.2in} {\bf Solution.} Now the series will not be centered at zero but at $x_0 = 2$. Let $y=a_0 + a_1(x-2) + a_2(x-2)^2 + a_3(x-2)^3 + \cdots = \dis \sum_{n=0} a_n (x -2)^n$ with $a_n = \dfrac{y^{[n]}(2)}{n!}$. Then we have the following calculations. \[ y(2) = 3 \:\: \implies \:\: a_0 = 3 \:\:\:\:\& \:\:\:\: y'(2)=1 \:\: \implies \:\: a_1 = 1. \] Next we solve for $y''$, evaluate it at $x=2$, and then compute $a_2$. \vspace{.2in} $\begin{array}{lll} y''& =& -x y' - y \end{array}$ \vspace{.2in} Thus, \vspace{.2in} $\begin{array}{lll} y''(0) &=& -2 \cdot 1 - 3 = -5 \:\: \implies \:\: a_2 = -5/2! = -5/2. \end{array}$ \vspace{.2in} Next, we find $y'''$, evaluate it at $x=2$, and then compute $a_3$. \vspace{.2in} $\begin{array}{lll} y''' &=& (-x y' - y)'= -y' - x y' - y' = - xy'' -2y'. \end{array}$ \vspace{.2in} Thus, \vspace{.2in} $\begin{array}{lll} y'''(2) &=& -2(-5) -2\cdot 1 = 8 \:\: \implies\:\: a_3 = 8/3! = 4/3. \end{array}$ \vspace{.2in} Finally, we find $y''''$, evaluate it at $x=2$, and then compute $a_4$. \vspace{.2in} $\begin{array}{lll} y'''' &=& (- xy'' - 2y')' = - y'' - xy''' -2y'' = -xy''' - 3 y''. \end{array}$ \vspace{.2in} Thus, \vspace{.2in} $\begin{array}{lll} y''''(2) &=& -2\cdot 8 -3(-5) = -1\:\: \implies \:\:a_4 = -1/4! = -1/24. \end{array}$ \vspace{.2in} Putting this all together we get \[ y(x) \approx 3 + (x-2) -\frac{5}{2}(x-2)^2 + \frac{4}{3}(x-2)^3 - \frac{1}{24}(x-2)^4. \] \vfill \pagebreak \begin{center}{\bf When does this method work?}\end{center} \vspace{.5in} {\bf Failed Example 1.} Can we find a power series $y=\sum_{n=0}^\infty a_n x^n$ that solves $y'' - \sqrt{x}\, y = 0$ with $y(0)=1$ and $y'(0)=2$? Let's try. Clearly, $a_0=1$ and $a_1=2$. Next \[ y'' = \sqrt{x}\, y \:\:\implies\:\: y''(0) = \sqrt{0} \, y(0) = 0 \:\:\implies\:\: a_2 = 0/2! = 0. \] So far, so good. Now \[ y''' = (y'')' = (\sqrt{x}\,y)' = \frac{1}{2\sqrt{x}} y + \sqrt{x} y'. \] But now \[ y'''(0) = \frac{1}{2\sqrt{0}} y(0) + \sqrt{0} y'(0), \] which is undefined! We conclude that there is no series solution centered about zero. \vspace{1in} {\bf Failed Example 2.} Can we find a power series $y=\dis \sum_{n=0}^\infty a_n (x-1)^n$ that solves $(x-1)y'' - x y' - 2 y = 0$ with $y(1)=2$ and $y'(1)=3$? Let's try. Clearly, $a_0=2$ and $a_1=3$. Next \[ y'' = \frac{xy'+2y}{x-1}. \] But then \[ y''(1) = \frac{1\cdot 3 +2\cdot 2}{1-1}, \] which is undefined. \vspace{.5in} \begin{center}{\bf \large What went wrong?}\end{center} \vfill \pagebreak \begin{center}{\bf A Theorem on Power Series Solutions}\end{center} {\bf Theorem.} Consider \[ y'' + p(x) y' + q(x) y = 0. \] If $p(x)$ and $q(x)$ have Taylor series centered about $x=c$ then we say $c$ is an {\bf ordinary point} of the given differential equation. Otherwise, $c$ is a {\bf singular point}. In the last two ``failed examples'' we tried to use a series centered on a singular point. This is not good. (See Sections 5.4-5.8 for more on this.) But, if $c$ is an ordinary point it is guaranteed that the solution exists and has a power series centered at $c$. \vspace{.3in} \begin{center}{\bf What is the radius of convergence?}\end{center} Suppose $y(t) = \dis\sum_{n=0}^\infty a_n (t-c)^n$ is a series solution to \[ y'' + p(t) y' + q(t) y = 0. \] Suppose the Taylor series centered at $c$ of $p(t)$ and $q(t)$ exist and have radii of convergence $R_p$ and $R_q$, respectively. Then if $R_y$ is the radius of convergence of series centered at $c$ for $y(t)$ we have \[ R_y \geq \max \{ R_p, R_q\}. \] \vspace{.3in} From {\em Calculus II} you have the tools to find $R_p$ and $R_q$. Here is a trick that you probably did not cover that is useful for rational functions. Recall a rational function is the ratio of two polynomials. Let \[ r(x) = \frac{f(x)}{g(x)} \] be a rational function where $f(x)$ and $g(x)$ are polynomials with no common factors. Let $x_1, x_2, \dots x_k$ be all the zeros pf $g(x)$ in the {\bf complex plane}. If $c$ is any real or complex number not equal to any of the zeros of $g(x)$, then $r(x)$ will have a Taylor series centered at $c$ with radius of convergence given by \[ R= \min \{ |x_1-c|, |x_2-c|, \dots |x_k - c|\}. \] We give some examples on the next page. \vfill \pagebreak \begin{center}{\bf Radius of Convergence Examples}\end{center} \begin{enumerate} \item The radius of convergence for the Taylor series centered at $c=0$ for $\dfrac{1}{1-x}$ is $R=1$. \item The radius of convergence for the Taylor series centered at $c=10$ for $\dfrac{1}{1-x}$ is $R=9$. \item The radius of convergence for the Taylor series centered at $c=10$ for $\dfrac{x^3+7x}{1-x}$ is $R=9$. (As long as 1 is not a root of the numerator, it makes no difference in the value of $R$.) \item The radius of convergence for the Taylor series centered at $c=0$ for $\dfrac{1}{1-x^2}$ is $R=1$. \item The radius of convergence for the Taylor series centered at $c=-0.7$ for $\dfrac{1}{1-x^2}$ is $R=0.3$. \item The radius of convergence for the Taylor series centered at $c=0$ for $\dfrac{1}{1+x^2}$ is $R=1$, since the denominator has zeros at $\pm i$. \item The radius of convergence for the Taylor series centered at $c=3$ for $\dfrac{1}{1+x^2}$ is $R=\sqrt{10}$. See figure below. \begin{center} \psfrag{i}{$i$} \psfrag{-i}{$-i$} \psfrag{3}{$3$} \psfrag{s}{$\sqrt{10}$} \includegraphics[width=3in]{Figs/complexradius.eps} \end{center} \end{enumerate} \vfill \pagebreak \begin{center}{\bf Radius of Convergence Examples}\end{center} \vspace{0.5in} \begin{enumerate} \item Consider $y'' + (\sin x)y' + (\cos x)y = 0$. Then for any $c$ the radius of convergence of the series solution will be infinite since the Taylor series for $\sin x$ and $\cos x$ have infinite radii of convergence. \vspace{1in} \item Consider $y'' + \dfrac{x}{1+x^2} y' + \dfrac{1}{x+2} = 0$. \subitem If $c=0$, then $R=1$. \subitem If $c=-5$, then $R=3$. \subitem If $c=3$, then $R=\sqrt{10}$. \subitem If $c=-1/2$, then $R=\sqrt{5}/2$. \vspace{1in} \item Consider $(x-1)y'' + \dfrac{x^3}{2-x} y ' + \dfrac{1}{x-3} y = 0$. Remember you have to divide through by the $x-1$. Thus the zeros are 1,2, and 3. \subitem If $c=0$, then $R=1$. \subitem If $c=2.2$, then $R=0.2$. \subitem if $c=23$, then $c=20$. \end{enumerate} \vfill \pagebreak \begin{center}{\bf Recursive Formulas}\end{center} \vspace{0.3in} For a sequence of numbers $(a_n)_{n=0}^\infty$ it is ideal if we can find a formula for $a_n$ as a function of $n$. For example, the sequence $(0, 1, 4, 9, 16, 25, 36, 49, \dots )$ is given by $a_n = n^2$. But, sometimes this is difficult or impossible to do. Consider the {\em Fibonacci sequence}, $$(f_n)_{n=0}^\infty = (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \dots ).$$ It is defined as follows, start with 0 and 1 as the first two terms, then each term after that is the sum of the two terms before it. That is \[ f_0 = 0, f_1 = 1, \& f_n = f_{n-1} + f_{n-2} \mbox{ for all } n\geq 2. \] This is called a {\bf recursive formula}. For second order linear differential equations finding a recursive formula for the terms of the power series if often the best we can do. \vspace{.6in} \begin{center}{\bf Doing the Index Shift}\end{center} No, it is not the latest dance craze. It is just a handy trick when working with summation notation. If you want add one or more power series together and express the result as a single power series, you too will find yourself doing the index shift. \vspace{.3in} {\bf Example.} Suppose we want to add $\dis\sum_{n=0}^\infty a_n t^n$ and $\dis\sum_{n=0}^\infty b_n t^{n+2}$. We would rewrite the second sum as $\dis\sum_{n=2}^\infty b_{n-2} t^n$. Then we'd get \[ \sum_{n=0}^\infty a_n t^n + \sum_{n=0}^\infty b_n t^{n+2} = a_0 + a_1 t + \sum_{n=2}^\infty a_n t^n + \sum_{n=2}^\infty b_{n-2} t^n = a_0 + a_1 t + \sum_{n=2}^\infty (a_n + b_{n-2})t^n. \] Notice we had to treat the first two terms of the first sum separately. If you are having trouble following the details, write out the terms of the sums until you see what is happening. \vfill \pagebreak \vspace{.3in} {\bf Example.} Let's do another one. \begin{eqnarray*} \sum_{n=0}^\infty n^2 x^n + \sum_{n=0}^\infty (n+1) x^{n+1} + \sum_{n=2}^\infty a_n x^{n-2} &=& \sum_{n=0}^\infty n^2 x^n + \sum_{n=1}^\infty n x^{n} + \sum_{n=0}^\infty a_{n+2} x^{n}\\ &=&\left( 0 + \sum_{n=1}^\infty n^2 x^n \right) + \sum_{n=1}^\infty n x^{n} + \left( a_2 + \sum_{n=1}^\infty a_{n+2} x^{n}\right) \\ &=& a_2 + \sum_{n=1}^\infty (n^2 + n + a_{n+2})x^n. \end{eqnarray*} \vspace{.3in} {\bf Example.} One more. Let $y(x) =\dis \sum_{n=0}^\infty a_n x^n$. Plug this into $y'' + 2y' + y$ and express it as a single power series in $x$. First notice that $$y' = \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=1}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1)a_{n+1} x^n.$$ And $$y'' = \sum_{n=0}^\infty n(n-1) a_n x^{n-2} = \sum_{n=2}^\infty n (n-1)a_n x^{n-2} = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2} x^n.$$ Therefore, \[ y'' + 2y' + y = \sum_{n=0}^\infty \big[ (n+2)(n+1)a_{n+2} + 2(n+1)a_{n+1} + a_n \big] x^n. \] \vspace{1in} Next we apply these ideas to find recursive formulas for series solutions of some second order linear differential equations. It gets pretty tedious. \vfill \pagebreak \begin{center}{\bf Example of Series Solution}\end{center} {\bf Example $2'$.} Find a recursive formula for the terms of the series solution of $y'' + 2y' + y = 0$ with $y(0)=1$, $y'(0)=2$. This is the same equation as Example 2 above. {\bf Solution.} Let $y(x) =\dis \sum_{n=0}^\infty a_n x^n$. We plug this into $y'' + 2y' + y$ and express it as a single power series in $x$. Wait a minute, we just did this! The result, for the last example, is that \[ y'' + 2y' + y = \sum_{n=0}^\infty \big[ (n+2)(n+1)a_{n+2} + 2(n+1)a_{n+1} + a_n \big] x^n = 0. \] It follows that for each $n\geq 0$ \[ (n+2)(n+1)a_{n+2} + 2(n+1)a_{n+1} + a_n = 0. \] We can rewrite this as \[ a_{n+2} = -\frac{ 2(n+1)a_{n+1} + a_n}{(n+2)(n+1)} \mbox{ for } n \geq 0. \] We can shift the index and get \[ a_n = -\frac{ 2(n-1)a_{n-1} + a_{n-2}}{(n)(n-1)}, \mbox{ for } n \geq 2. \] This is a recursive formula for $a_n$. Since we know $a_0 = 1$ and $a_1 = 2$ we can compute as many terms as we like. \begin{eqnarray*} a_2 &=& -\frac{ 2(1)a_{1} + a_{0}}{(2)(1)} = - \frac{5}{2} \\ a_3 &=& -\frac{ 2(2)a_{2} + a_{1}}{(3)(2)} = - \frac{-10 + 2}{6} = \frac{4}{3}\\ a_4 &=& -\frac{ 2(3)a_{3} + a_{2}}{(4)(3)} = - \frac{8-\frac{5}{2}}{12} = -\frac{11}{24} \\ a_5 &=& -\frac{ 2(4)a_{4} + a_{3}}{(5)(4)} = - \frac{-\frac{11}{3}+\frac{4}{3}}{20} = \frac{7}{60}\\ a_6 &=& -\frac{ 2(5)a_5+ a_4}{(6)(5)} = - \frac{17}{720} \end{eqnarray*} We can even write a short program to compute as many terms as we like. \vspace{0.3in} {\tt\begin{tabular}{ll} > N:=20: &\# Set the number of terms to compute. \\ > A:= array(0..N-1): &\# Define an array of length N. \\ > A[0]:=1: A[1]:=2: &\# Define the first two terms of the array A.\\ > &\# Next set up recursive formula and find the other terms.\\ > for n from 2 to N-1 do& \!\!\!A[n]:= -(2*(n-1)*A[n-1] + A[n-2])/(n*(n-1)); end do; \end{tabular}} \vspace{0.3in} The output is on the next page. \vfill \pagebreak \[ A_{{2}}\, := \,-5/2 \] \[ A_{{3}}\, := \,4/3 \] \[ A_{{4}}\, := \,-{\frac {11}{24}} \] \[ A_{{5}}\, := \,{\frac {7}{60}} \] \[ A_{{6}}\, := \,-{\frac {17}{720}} \] \[ A_{{7}}\, := \,{\frac {1}{252}} \] \[ A_{{8}}\, := \,-{\frac {23}{40320}} \] \[ A_{{9}}\, := \,{\frac {13}{181440}} \] \[ A_{{10}}\, := \,-{\frac {29}{3628800}} \] \[ A_{{11}}\, := \,{\frac {1}{1247400}} \] \[ A_{{12}}\, := \,-{\frac {1}{13685760}} \] \[ A_{{13}}\, := \,{\frac {19}{3113510400}} \] \[ A_{{14}}\, := \,-{\frac {41}{87178291200}} \] \[ A_{{15}}\, := \,{\frac {1}{29719872000}} \] \[ A_{{16}}\, := \,-{\frac {47}{20922789888000}} \] \[ A_{{17}}\, := \,{\frac {1}{7113748561920}} \] \[ A_{{18}}\, := \,-{\frac {53}{6402373705728000}} \] \[ A_{{19}}\, := \,{\frac {1}{2172233935872000}} \] \vfill \pagebreak \begin{center}{\bf Example}\end{center} {\bf Example 5.} Find a recursive formula for the series general solution to $y'' + xy' + y = 0$ centered about zero. {\bf Solution.} Let $y = \dis \sum_{n=0}^\infty a_n x^n$. Then we have \[ y' = \sum_{n=0}^\infty na_n x^{n-1} \hspace{0.5in} \mbox{ and } \:\: xy' = \sum_{n=0}^\infty na_n x^{n}. \] \[ y'' = \sum_{n=0}^\infty n(n-1)a_n x^{n-2} = \sum_{n=2}^\infty n(n-1)a_n x^{n-2} = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2} x^{n}. \] Therefore, \[ \sum_{n=0}^\infty (n+2)(n+1)a_{n+2} x^{n} + \sum_{n=0}^\infty na_n x^{n} + \sum_{n=0}^\infty a_n x^n = 0. \] This is the same as \[ \sum_{n=0}^\infty \big[ (n+2)(n+1)a_{n+2} + na_n + a_n \big] x^n = 0. \] This forces \[ (n+2)(n+1)a_{n+2} + (n+1)a_n = 0 \mbox{ for each } n\geq 0. \] Solving for $a_{n+2}$ gives \[ a_{n+2} = -\frac{(n+1)a_n}{(n+2)(n+1)}= -\frac{a_n}{n+2} \mbox{ for each } n\geq 0. \] We can rewrite this as \[ a_{n} = -\frac{a_{n-2}}{n} \mbox{ for each } n\geq 2. \] Thus, if we were given a value for $y(0)=a_0$ we could find all the even $a_n$ terms and if we were given a value for $y'(0)=a_1$ we could find all the odd $a_n$ terms. We do this on the next page, and we find formulas for $a_n$ as a function of $n$, given $a_0$ and $a_1$. \vfill \pagebreak \begin{center}{\bf Example 5 Continued}\end{center} Recall $ \dis a_{n} = -\frac{a_{n-2}}{n}$ for each $n\geq 2.$ Suppose $a_0$ and $a_1$ are known. Then \[ a_2 = - a_0/2, \hspace{.25in} a_4 = a_0/(2\cdot 4), \hspace{.25in} a_6 = -a_0/(2\cdot 4\cdot 6), \hspace{.25in} a_8 = a_0/(2\cdot 4\cdot 6\cdot 8), ... \] and \[ a_3 = - a_1/3, \hspace{.25in} a_5 = a_1/(3\cdot 5), \hspace{.25in} a_7 = -a_1/(3\cdot 5\cdot 7), \hspace{.25in} a_9 = a_1/(3\cdot 5\cdot 7\cdot 9), ... \] Notice that the product of positive even numbers less than or equal to $2n$ is $2^n n!$. Thus, \[ a_{2n} = (-1)^n \frac{a_0}{2^n n!} \mbox{ for } n \geq 1. \] There is no clever notation for the product of positive odd numbers less than or equal to $2n+1$, so we just write \[ a_{2n+1} = (-1)^n \frac{a_1}{(2n+1)(2n-1)(2n-3)\cdots3} \mbox{ for } n \geq 1. \] \vspace{.3in} Sometimes it is useful to consider special pairs of initial values such as $a_0=1$ \& $a_1=0$ and $a_0=0$ \& $a_1=1$. This is because the resulting pair of solutions will be linearly independent since their Wronskian is 1 at $x=0$. We let \[ y_1 = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{2^n n!} \] and \[ y_2 = \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)(2n-1)(2n-3)\cdots3}. \] These form a fundamental solution pair. \vfill \pagebreak \begin{center}{\bf Next Example}\end{center} {\bf Example 6.} Find a recursive formula for the series general solution to $y'' + y' + xy = 0$ centered about zero. {\bf Solution.} Let $y = \dis \sum_{n=0}^\infty a_n x^n$. Then we have \[ xy = \sum_{n=0}^\infty a_n x^{n+1} = \sum_{n=1}^\infty a_{n-1} x^{n}, \] \[ y' = \sum_{n=0}^\infty na_n x^{n-1} = \sum_{n=1}^\infty na_n x^{n-1} = \sum_{n=0}^\infty (n+1)a_{n+1} x^{n}, \] and \[ y'' = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2} x^{n}. \] Therefore, \[ \sum_{n=0}^\infty (n+2)(n+1)a_{n} x^{n} + \sum_{n=0}^\infty (n+1)a_{n+1} x^{n} + \sum_{n=1}^\infty a_{n-1} x^{n} =0. \] The summands all have the exponents of $x$ agreeing, but they do not all start at $n=0$; one of the summations starts at $n=1$. Therefore, we have to treat the $n=0$ terms of the other summations separately. We get \[ (2a_2 + a_1) + \sum_{n=1}^\infty \big[(n+1)na_{n} + (n+1)a_{n+1} + a_{n-1}\big] x^n = 0. \] Thus, \[ 2a_2 + a_1 = 0 \:\: \implies a_2 = -\frac{a_1}{2}, \] and \[ a_{n+1} = -\frac{(n+1)na_n + a_{n-1}}{n+1} \:\:\:\mbox{ for } n\geq 2. \] The latter we could rewrite as \[ a_n = -\frac{n(n-1)a_{n-1}+ a_{n-2}}{n} \:\:\:\mbox{ for } n\geq 3. \] Thus, given $a_0$ and $a_1$ we can compute as many terms as we want. \vfill \pagebreak \begin{center}{\bf Yet Another Example}\end{center} {\bf Example 7.} Find a recursive formula for the series general solution to $xy'' + y' + xy = 0$ centered about $x=1$. {\bf Solution.} Let $y = \dis \sum_{n=0}^\infty a_n (x-1)^n$. We use the handy fact that $x = (x-1) + 1$. Then \[ xy = (x-1)\sum_{n=0}^\infty a_n (x-1)^n + \sum_{n=0}^\infty a_n (x-1)^n = \sum_{n=0}^\infty a_n (x-1)^{n+1} + \sum_{n=0}^\infty a_n (x-1)^n = \] \[ \sum_{n=1}^\infty a_{n-1} (x-1)^{n} + \sum_{n=0}^\infty a_n (x-1)^n = a_0 + \sum_{n=1}^\infty \big[a_{n-1} +a_n\big] (x-1)^n. \] And \[ y' = \sum_{n=0}^\infty n a_n (x-1)^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} (x-1)^{n} = a_1 + \sum_{n=1}^\infty (n+1) a_{n+1} (x-1)^{n}. \] And \begin{eqnarray*} xy'' &=& (x-1)\sum_{n=0}^\infty (n-1)n a_n (x-1)^{n-2} + \sum_{n=0}^\infty (n-1)n a_n (x-1)^{n-2}\\ & =& \sum_{n=2}^\infty (n-1)n a_n (x-1)^{n-1} + \sum_{n=2}^\infty (n-1)n a_n (x-1)^{n-2}\\ &=& \sum_{n=1}^\infty n(n+1) a_{n+1} (x-1)^{n} + \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} (x-1)^{n}\\ & =& 2a_2 + \sum_{n=1}^\infty \big[n(n+1) a_{n+1}+(n+1)(n+2) a_{n+2}\big](x-1)^n. \end{eqnarray*} Therefore, \[ xy'' + y' + xy = 2a_2 + a_1 + a_0 + \sum_{n=1}^\infty \big[n(n+1) a_{n+1}+(n+1)(n+2) a_{n+2}+(n+1) a_{n+1} +a_{n-1} +a_n\big] (x-1)^n = 0. \] Hence, \[ a_2 = -\frac{a_1+a_0}{2}, \] and \[ a_{n+2} = - \frac{n(n+1) a_{n+1}+(n+1) a_{n+1} +a_{n-1} +a_n}{(n+1)(n+2)} = -\frac{(n+1)^2a_{n+1}+a_n + a_{n-1}}{(n+1)(n+2)} \:\:\mbox{ for } n\geq1. \] The latter can be rewritten as \[ a_n = -\frac{(n-1)^2a_{n-1}+a_{n-2} + a_{n-3}}{(n-1)n} \:\:\mbox{ for } n\geq3. \] \vfill \pagebreak \begin{center}{\bf You Guessed It, Another Example!}\end{center} {\bf Example 8.} Find a recursive formula for the series general solution to $(4-x^2) y'' + 2y=0$ centered at zero. {\bf Solution.} Let $y = \dis \sum_{n=0}^\infty a_n x^n$. Then \[ y'' = \sum_{n=0}^\infty (n-1)na_n x^{n-2} = \sum_{n=2}^\infty (n-1)na_n x^{n-2} = \sum_{n=0}^\infty (n+1)(n+2)a_{n+2} x^{n}. \] Thus, \[ x^2 y'' - \sum_{n=0}^\infty (n-1)na_n x^{n}. \] Therefore, \[ (4-x^2)y'' + 2 y = \sum_{n=0}^\infty \big[4(n+1)(n+2)a_{n+2} - (n-1)na_n+2 a_n \big]x^n = 0. \] Thus, \[ a_{n+2} = \frac{(n-1)na_n- 2 a_n }{4(n+1)(n+2)} = \frac{(n-2)a_n}{4(n+2)}, \:\:\mbox{ for } n\geq 0, \] which we can rewrite as \[ a_n = \frac{(n-4)a_{n-2}}{4n}, \:\:\mbox{ for } n\geq 2. \] \vfill \pagebreak \begin{center}{\bf Mucking Around with the Previous Example}\end{center} Just for fun let's consider the two cases: $a_0= 1$ with $a_1=0$, and $a_0=0 $ with $a_1=1$. In the first case, since $a_1=0$ we have the $a_n=0$ for all odd $n$. Then \[ a_2 = \frac{2-4}{4\cdot2}a_0 = -\frac{1}{4} \hspace{0.5in}\mbox{ and }\hspace{0.5in} a_4 = \frac{4-4}{4\cdot4}a_2 = 0. \] It follows that $a_n=0$ for all $n$ even and bigger than 4. Thus, \[ y_1 = 1 - \frac{1}{4}x^2 \] gives the exact solution. Now suppose, $a_0=0 $ \& $a_1=1$. Thus $a_n=0$ for all even $n$. For odd $n$ we have the following. \[ a_3 = \frac{3-4}{4\cdot3}a_1 = \frac{-1}{4\cdot3}. \] \[ a_5 = \frac{5-4}{4\cdot5}a_3 = \frac{-1}{4^2\cdot5\cdot3}. \] \[ a_7 = \frac{7-4}{4\cdot7}a_5 = \frac{-1\cdot3}{4^3\cdot7\cdot5\cdot3} = \frac{-1}{4^3\cdot7\cdot5}. \] \[ a_9 = \frac{9-4}{4\cdot9}a_7 = \frac{-1\cdot5}{4^4\cdot9\cdot7\cdot5} = \frac{-1}{4^4\cdot9\cdot7}. \] We conclude that \[ a_{2k+1} = \frac{-1}{4^k(2k+1)(2k-1)}. \] Hence we let \[ y_2 = \sum_{n=0}^\infty \frac{-x^{2k+1}}{4^k(2k+1)(2k-1)}. \] And so now we have two linearly independent solutions. But, you know, it seems to me that we could do better. Since we have a finite term expression for $y_1$ we could use the reduction of order method to find a closed form expression for $y_2$. We do this on the next page. \vfill \pagebreak \begin{center}{\bf More Mucking Around}\end{center} Before we found that $y_1 = 1-\frac{1}{4}x^2$ was a solution to $(4-x^2)y'' + 2 y = 0$. But so is any multiple of it, so we will work will $y_3= -4y_1 = x^2 - 4$ to avoid fractions. Now, let $y = v(x)(x^2-4)$. Then \[ y'' = (v'(x^2-4) + 2vx)' = v''(x^2-4) + 2v'x + 2v. \] We plug this into the original differential equation to get \[ (4-x^2)\big[(x^2-4)v'' + 4xv' + 2v\big] + 2(x^2-4)v = 0. \] Thus, \[ -(x^2-4)^2v'' - 4x(x^2-4)v' -2(x^2-4)v + 2(x^2-4)v = 0, \] or \[ (x^2-4)^2v'' + 4x(x^2-4)v' =0. \] Let $w=v'$. Then we have \[ (x^2-4)^2w' + 4x(x^2-4)w =0. \] Thus, \[ \big( (x^2-4)w\big)' = 0, \:\: \implies \:\: w = \frac{C}{x^2-4}. \] Let $C=1$. Then we integrate to get $v$. \[ v = \int w \, dx = \int \frac{1}{x^2-4} \, dx = \frac{-1}{16} \left(\frac{1}{x-2} + \frac{1}{x+2}\right) + \frac{1}{32} \ln \Big|\frac{x+2}{x-2}\Big| + C. \] \[ =\frac{1}{32}\left(\ln\Big| \frac{x+2}{x-2}\Big| - \frac{4x}{x^2-4}\right) + C. \] The integral can be done by using partial fractions or a computer. Let $C=0$. Now \[ vy_3 = \frac{1}{32} \left(\ln\Big| \frac{x+2}{x-2}\Big| - \frac{4x}{x^2-4}\right)(x^2-4) = \frac{1}{32}\left((x^2-4)\ln\Big| \frac{x+2}{x-2}\Big| - 4x\right) \] Since any multiple will do we let \[ y_4= (x^2-4)\ln\Big| \frac{x+2}{x-2}\Big| - 4x. \] Thus $\{ y_3, y_4 \}$ form a fundamental solution set. How does $y_4$ relate to the series express we had for $y_2$? You can check that $y_4(0) = 0$ but that $y'_4 (0) = 128$. Thus $128y_2 = y_4$ or \[ 128\cdot \sum_{k=1}^\infty \frac{-x^{2k+1}}{4^k(2k+1)(2k-1)} = (x^2-4)\ln\Big| \frac{x+2}{x-2}\Big| - 4x. \] Lastly, we comment of the radius of convergence. \vfill \pagebreak \begin{center}{\bf The Radius of Convergence of the Previous Example}\end{center} For $y_1$ or $y_3$ the radius of convergence is clearly infinity. Putting the original problem in the form \[ y'' + \frac{2}{4-x^2} y = 0, \] we can see that the radius of convergence in general must be at least 2. From the form of $y_4$ we can see that it is undefined at $x=\pm 2$. We could study the limits as $x \to 2$ from below and $x\to -2$ from above as see what happens. It turns out both limit are $\pm 8$, so maybe the radius of convergence is larger than two. But, if you check the corresponding limits of $y'_4$ they are $\mp \infty$. Thus, a radius has to be 2. A graph of $y_4(x)$ is shown below. \begin{center} \includegraphics[width=3in]{Figs/ex4graphplot1.eps} \end{center} Another approach is to use the ratio test directly on the series for $y_2$. Let's do it! \[ \Big|\frac{a_{k+1}}{a_k}\Big| = \begin{vmatrix}\frac{\dfrac{x^{2k+3}}{4^{k+1}(2k+3)(2k+1)}}{\dfrac{x^{2k+1}}{4^k(2k+1)(2k-1)}}\end{vmatrix} = \frac{x^2}{4}\frac{2k-1}{2k+3}. \] Now, \[ \lim_{k\to\infty} \frac{x^2}{4}\frac{2k-1}{2k+3} = \frac{x^2}{4}. \] Thus, we have convergence for $-2 < x < 2$. \vfill \pagebreak \begin{center}{\bf A Final Example}\end{center} {\bf Example 9.} Find the general series solution to $y'' + t^2 y = t^4$ centered about $0$ \vspace{.2in} {\bf Solution.} Let $\dis y = \sum_{n=0}^\infty a_n t^n$. Then \[ y'' = \sum_{n=0}^\infty n(n-1) a_n t^{n-2} = \sum_{n=2}^\infty n(n-1) a_n t^{n-2} = \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} t^{n}, \] and \[ t^2 y = \sum_{n=0}^\infty a_n t^{n+2} = \sum_{n=2}^\infty a_{n-2} t^{n}. \] Thus we have \[ y'' + t^2 y = 2a_2 + 3\cdot 2 a_3 t + \sum_{n=2}^\infty (n+2)(n+1) a_{n+2} t^n + \sum_{n=2}^\infty a_{n-2} t^{n} = t^4. \] We subtract the $t^4$ from both sides. This causes us to write out the terms up to $n=4$ and use summation notation only after that. We get \[ 2a_2 + 3\cdot 2 a_3 t + (4\cdot 3 a_4 + a_0)t^2 + (5\cdot 4 a_5 + a_1)t^3 + (6\cdot 5 a_6 + a_1 - 1)t^4 + \sum_{n=5}^\infty \big[(n+2)(n+1) a_{n+2} + a_{n-2} \big]t^{n} = 0. \] Now we are ready to rumble! We take $a_0$ and $a_1$ as given. We then deduce the following. \begin{eqnarray*} 2a_2 = 0 &\implies& a_2 = 0 \\ 3\cdot 2 a_3 = 0 &\implies& a_3=0\\ 4\cdot 3 a_4 + a_0 = 0 &\implies& a_4 = \frac{-a_0}{4\cdot 3}\\ 5\cdot 4 a_5 + a_1 = 0 &\implies& a_5 = \frac{-a_1}{5\cdot 4}\\ 6\cdot 5 a_6 + a_2 -1 = 0 &\implies& a_6 = \frac{1}{6\cdot 5} \\ (n+2)(n+1)a_{n+2} + a_{n-2} = 0 &\implies& a_{n+2} = \frac{-a_{n-2}}{(n+2)(n+1)}, \mbox{ for } n \geq 5 \end{eqnarray*} The last expression can be rewritten as \[ a_n = \frac{-a_{n-4}}{n(n-1)} \mbox{ for } n \geq 7. \] \vfill \pagebreak We are going to generate some more terms using this recursive relation and then see if we can express $a_n$ as a function of $n$, for $n\geq 7$. \begin{eqnarray*} a_7 = & \dfrac{-a_3}{7\cdot 6} & = 0 \\ a_8 = & \dfrac{-a_4}{8\cdot 7} & = \frac{a_0}{8\cdot 7\cdot 4 \cdot 3} \\ a_9 = & \dfrac{-a_5}{9\cdot 8} & = \frac{a_1}{9\cdot 8 \cdot 5 \cdot 4 } \\ a_{10} = & \dfrac{-a_6}{10 \cdot 9} & = \frac{-1}{10 \cdot 9 \cdot 6 \cdot 5} \\ a_{11} = & \dfrac{-a_7}{11 \cdot 10} & = 0 \\ a_{12} = & \dfrac{-a_8}{12 \cdot 11} & = \frac{-a_0}{12 \cdot 11 \cdot 8 \cdot 7 \cdot 4 \cdot 3} \\ a_{13} = & \dfrac{-a_9}{13 \cdot 12} & = \frac{-a_1}{13 \cdot 12 \cdot 9 \cdot 8 \cdot 5 \cdot 4} \\ a_{14} = & \dfrac{-a_{10}}{14 \cdot 13} & = \frac{1}{14 \cdot 13 \cdot 10 \cdot 9 \cdot 6 \cdot 5} \\ a_{15} = & \dfrac{-a_{11}}{15 \cdot 14} & = 0 \\ a_{16} = & \dfrac{-a_{12}}{16 \cdot 15} & = \frac{a_0}{16 \cdot 15 \cdot 12 \cdot 11 \cdot 8 \cdot 7 \cdot 4 \cdot 3} \\ a_{17} = & \dfrac{-a_{13}}{17 \cdot 16} & = \frac{a_1}{17 \cdot 16 \cdot 13 \cdot 12 \cdot 9 \cdot 8 \cdot 5 \cdot 4} \end{eqnarray*} Now we can finally see what is happening. For $n \geq 7$ we have \[ a_n = \left\{ \begin{array}{ccc} \dfrac{ a_0}{n(n-1)(n-4)(n-5)\cdots 4\cdot 3} & \mbox{ for} & n = 0\!\!\!\mod 8 \\ &&\\ \dfrac{ a_1}{n(n-1)(n-4)(n-5)\cdots 5\cdot 4} & \mbox{ for} & n = 1\!\!\!\mod 8 \\ &&\\ \dfrac{ -1}{n(n-1)(n-4)(n-5)\cdots 6\cdot 5} & \mbox{ for} & n = 2\!\!\!\mod 8 \\ &&\\ 0 & \mbox{ for} & n = 3\!\!\!\mod 8 \\ &&\\ \dfrac{-a_0}{n(n-1)(n-4)(n-5)\cdots 4\cdot 3} & \mbox{ for} & n = 4\!\!\!\mod 8 \\ &&\\ \dfrac{-a_1}{n(n-1)(n-4)(n-5)\cdots 5\cdot 4} & \mbox{ for} & n = 5\!\!\!\mod 8 \\ &&\\ \dfrac{ 1}{n(n-1)(n-4)(n-5)\cdots 6\cdot 5} & \mbox{ for} & n = 6\!\!\!\mod 8 \\ &&\\ 0 & \mbox{ for} & n = 7\!\!\!\mod 8 \end{array}\right. \] \vspace{.3in} Note: $n = k\!\!\! \mod 8$ means the remainder when $n$ is divided by 8 equals $k$. \begin{center} {\bf \large When is the last time you had this much fun!} \end{center} \vfill \pagebreak \end{document}