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 \centerline{\large Math.\ 401, Sec.\ 501
\hfill    Spring, 2006}
                                                                           
\bigskip
                                                                               

 
\centerline{\bfseries\large Homework 3, due February 8} 
\bigskip

\begin{enumerate}

 
 \item Find the first three terms (through order $\epsilon^2$) 
of 
the Taylor series (in $\epsilon$ with $t$ fixed) of
 $$ y(t)= {e^{-\epsilon t}\over \sqrt{1-\epsilon^2}}\,\sin 
(\sqrt{1-\epsilon^2}\,
t).$$
 Compare with the perturbative solution found in class (or 
notes) for the 
damped oscillator equation.
{\slshape Suggestion:\/}  The easiest method is
 to find series for each of the three
factors and then  multiply them together like polynomials.

%\item{2.} (dictionary research) \ Where did the phrase 
%``secular terms'' come from?  
%What does it have to do with the word ``secular'' in 
%``secular humanism''?

% \item{2-7.} Logan, p.\ 49, Ex.\ 1.3, 1.15, 1.16, 1.19.

\item 
{\itshape [Logan, p.\ 49,  Ex.\ 1.3, beginning]\/} \quad
Verify the following order relations:

\begin{enumerate}
\item $\ze^2 \tanh {\ze} = O(\ze^2)$ as $\ze\to \zI$.
\item $\exp (-\ze) = o(1)$ as $\ze\to \zI$.
\item $\sqrt{\ze(1-\ze)} = O\bigl(\sqrt{\ze}\bigr)$ as $\ze \to 0^+$.
\item ${\displaystyle{\sqrt{\ze} \over 1-\cos \ze}}
  = O(\ze^{-3/2})$ as $\ze \to 0^+$.
\end{enumerate}

\item 
{\itshape [Logan, p.\ 49,  Ex.\ 1.3, middle]\/} \zj
Verify the following order relations:

\begin{enumerate}
\item $\ze = O(\ze^2)$ as $\ze\to \zI$.

\item $\exp(\ze) -1 = O(\ze)$ as $\ze\to 0$.
\item $\int_0^\ze \exp(-x^2) \, dx = O(\ze)$ as $\ze \to 0^+$.
\item $\exp(\tan \ze) = O(1)$ as $\ze\to 0$.
\end{enumerate}

\item 
{\itshape [Logan, p.\ 49,  Ex.\ 1.3, conclusion]\/} \zj
Verify the following order relations:


\begin{enumerate}
\item $e^{-\ze} = O(\ze^{-p})$ as $\ze\to \zI$, for all $p>0$.
\item $\ln{\ze} = o(\ze^{-p})$ as $\ze\to 0^+$, for all $p>0$.
 \end{enumerate}


 \item 
{\itshape [Logan, p.\ 49,  Ex.\ 1.16]\/}\zj
Show that the three-term expansion
$$\sin t + \ze \cos t - {\ze^2 \over 2}\,\sin t$$
is a uniformly valid approximation of $\,\sin (t+\ze)$
on $-\zI <t<\zI$.
 To justify your answer, look in a calculus book 
under  ``Taylor's theorem with remainder''\negthinspace. 
 
\vfill\eject

 \item 
{\itshape [Logan, p.\ 49,  Ex.\ 1.15]\/}\zj
Consider the boundary value problem (with $0<\ze \ll 1$)
$$Ly \equiv t^2 y'' + \ze t^2 y' + {\textstyle \frac14} y =0 \zj 
\zT{for $1\le t \le e$},$$
$$y(1) = 1, \zJ y(e) = 0.$$

\begin{enumerate}
\item Use regular perturbation theory to find the leading-order
behavior, $y_0(t)$.
{\slshape Hint:\/}
To solve the unperturbed equation,
 look in your differential equations textbook under ``Euler
equation'' ---
 or make a change of variable $t=e^u$.
\item Compute an upper bound for $|Ly_0|$ on $1\le t \le e$
when $\ze=0.01$.
\end{enumerate}

\item
{\itshape [Logan, p.\ 49,  Ex.\ 1.13]\/} \zj
The equation of motion of a pendulum can be scaled to the form
$$\frac {d^2\theta}{dt^2} + {\sin {A\theta}\over A} =0,$$
$$\theta(0)=1, \zJ \frac{d\theta}{dt}(0) = 0$$
(where $t>0$ and $0<A\ll 1$).
Apply the regular perturbation method to find a two-term expansion.
(Define $\epsilon=A^2$.)
Show that the second term (the one of order $O(A^2)$) is secular 
and comment on the  validity of the approximation.

\end{enumerate}
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