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\begin{center}
UNM Dept. of Mathematics and Statistics \\
Math. 570, Singular Perturbations \\
Problem Set II \\[.2in]
\end{center}
Spring 2011 \\[.2in]
\begin{enumerate}

\item (See: Olver, {\em Asymptotics and Special Functions}) The complementary
incomplete gamma function is defined as
\[ \Gamma(\alpha,x) = \int_x^\infty e^{-t}t^{\alpha-1} dt \ . \]
\begin{enumerate} 
\item Using repeated integration by parts show
\[ \Gamma(\alpha,x) = e^{-x}x^{\alpha -1} 
\left\{ 1 +\frac{\alpha-1}{x} + \frac{(\alpha-1)(\alpha -2)}{x^2} +
\cdots \frac{(\alpha-1)\cdots (\alpha -n+1)}{x^{n-1}}\right\} + \epsilon_n(x) \]
with $\epsilon_n(x)$, the n-th error term.

\item For $n\ge \alpha-1$, show that the error $\epsilon_n$ has the same sign
and is bounded in absolute value by $(\alpha-1)\cdots (\alpha -n)/ x^n$
(the "$(n+1)$th term" of the series). Hence deduce that
\[ \Gamma(\alpha,x) \thicksim\!\!\!\!\!\!^{{}^{\textrm{asy}}} e^{-x}x^{\alpha -1} 
\sum_{s=0}^\infty \frac{(\alpha-1)\cdots (\alpha -s)}{x^s} \]
for fixed $\alpha$ as $x \rightarrow \infty$. Especially, for $\alpha \le 1$,
$x>0$, show that $\Gamma(\alpha,x) \le e^{-x} x^{\alpha - 1}$.
\item If $n<\alpha -1 $ show that $\epsilon_n(x)$ is bounded by the sum of the
first $\lfloor{\alpha}\rfloor - n + 1$ neglected terms in this case.\\
({\em Hint:} $\epsilon_n(x) s = (\alpha-1)\cdots (\alpha -n) e^{-x}
x^{\alpha -n -1} + (\alpha -1)\cdots (\alpha -n -1) \int_x^\infty
e^{-t} t^{\alpha -n -2} dt$, so both RHS terms are $\ge 0$ if $n<\alpha-1$.
This is also true of the first $\lfloor{\alpha}\rfloor - n + 1$).
\end{enumerate}

\item
Using the results of Prob. 1, show that the complementary error function defined
by
\[ erfc(x) = \frac{2}{\pi}\int_x^\infty e^{-t^2} dt \]
has the asymptotic expansion
\[ erfc(x) \thicksim\!\!\!\!\!\!^{{}^{\textrm{asy}}} \frac{e^{-x^2}}{x\sqrt{\pi}} \sum_0^\infty (-1)^n
\frac{ 1\cdot 3\cdots(2n-1)}{ \left(2x^2\right)^n}
\ \ \mbox{as  } x\rightarrow \infty \ .\]

\item (Watson's Lemma): Consider a function $q(t)$, ($t>0$) with
asymptotic expansion
\[ q(t) \thicksim\!\!\!\!\!\!^{{}^{\textrm{asy}}}
\sum_{s=0}^\infty a_s t^{(s+\lambda - \mu)/\mu}\ \ 
\mbox{as  } t\rightarrow 0\ ,\ \ (\lambda, \mu >0) \ \ .\]
Let
\[ \phi_n(t) = q(t) - 
\sum_{s=0}^{n-1} a_s t^{(s+\lambda - \mu)/\mu}\ \ (t>0) \ . \] 
The integral
\[ I(x) = \int_0^\infty e^{-xt} q(t) dt \]
can thus be written
\[ I(x) = \sum_{s=0}^{n-1} \Gamma\left(\frac{s+\lambda}{\mu}\right)
\frac{a^s}{x^{(s+\lambda)/\mu}} + \int_0^\infty e^{-\lambda t} \phi_n(t) dt \ .\]
(Recall: $\Gamma(\alpha) = \int_0^\infty e^{-t} t^{\alpha -1} dt$, $\alpha>0$).\\
Assume that $I(x)$ converges for $x$ suficiently large; then 
the same is true for $I_n(x) = \int_0^\infty e^{-xt} \phi_n(t) dt$.

\begin{enumerate}
\item
Since $\phi_n(t) = {\cal O}\left( t^{(n+\lambda - \mu)/\mu} \right)$ as
$t \rightarrow 0$, deduce that there exist positive constants $k_n, K_n$,
such that
\[ \left| \phi_n(t) \right| \le K_n t^{(n + \lambda - \mu)/\mu }\ \ ;\ \ 
\left(0 < t \le k_n\right) \ .\]
Hence, show that
\[ \left| \int_0^{k_n} e^{-xt}\phi_n(t) dt \right| <
\Gamma\left(\frac{n+\lambda}{\mu} \right) \frac{K_n}{x^{(n+\lambda)/\mu}} \ \].

\item If $I_n(x)$ converges, let
\[ \Phi_n(t) = \int_{k_n}^t e^{-X\nu}\phi_n(\nu) d\nu \ . \]
Integrating by parts show, for $x>X$, that
\[ \left| \int_{k_n}^\infty e^{-(x-X)t} \phi_n(t) dt \right|
\le L_n e^{-(x-X)k_n} \]
where $\left|\Phi_n(t)\right| \le L_n$ for $k_n\le t < \infty$.\\
($L_n$ is guaranteed to exist by convergence of $I_n(x)$).

\item Combining (a) and (b) show that
\[ I_n(x) = {\cal O} \left(x^{-(n+\lambda)/\mu}\right)\ \ \mbox{as  }
x\rightarrow \infty \ .\]
\end{enumerate}

\item Show that
\[ \int_0^\infty e^{-x\cosh{t}} dt \thicksim\!\!\!\!\!\!^{{}^{\textrm{asy}}}
\left( \frac{\pi}{2x}\right)^{1/2} e^{-x}
\sum_{s=0}^\infty (-1)^s\frac{1^2\cdot 3^2 \cdots (2s-1)^2}{s!(8x)^s}\ \ 
\mbox{as  } x\rightarrow \infty \ .\]


\item 
Show that the following sequences are asymptotic and combine them to form a 
single sequence:\\
($1$) $1,\epsilon,\epsilon^2,\ldots, \epsilon^n,\ldots $ \\
($2$) $\epsilon |ln \epsilon|, \epsilon^2 |ln \epsilon|,\ldots,\epsilon^n|ln \epsilon|,\ldots, $ \\
$\ldots \ldots \ldots$ \\
($m+1$) $\epsilon |ln \epsilon|^m, \epsilon^2 |ln \epsilon|^m,\ldots,\epsilon^n|ln \epsilon|^m,\ldots, $ \\
$\ldots \ldots \ldots$ \\
($\aleph_0+1$) $|ln \epsilon|^{-1}, |ln \epsilon|^{-2},\ldots,|ln \epsilon|^{-n},\ldots, $ \\
($\aleph_0+2$) $e^{-1/\epsilon}, e^{-2/\epsilon},\ldots,e^{-n/\epsilon},\ldots, $.

\item
Study the equation
\[ \epsilon^p = e^{-1/\epsilon}\ \ ,\ \ \epsilon>0 \ .\]
Determine the values (ranges) of $p$ for which it has $0$, $1$ or $2$ solutions.
Using some numerical scheme (i.e. Newton's method, fixed point iteration etc.)
find the roots for $p=4$.\\
Is the statement $ \epsilon >\!\!\!\!\!\!^{{}^{\textrm{asy}}} e^{-1/\epsilon}
\ (\epsilon \rightarrow 0)$ uniformly valid in $n$ as $n\rightarrow \infty$?
(\begin{small} ANSWER: NO \end{small}).

\item Show that the large positive roots of the equation $\tan x = x$
are given by
\[ x = \mu - \mu^{-1} - \frac{2}{3} \mu ^{-3} + {\cal O}(\mu ^{-5}) \ \ 
(\mu \rightarrow \infty) \ , \]
where $\mu = (n+1/2)\pi$, $n$ being a positive integer.\\
Hint: let $x_n = (n\pi + \pi/2) - \xi_n$ with $0\le \xi <\pi$ for $n$
large. Use the Taylor expansion:
\[ \tan {\xi} = \xi + \frac{1}{3}\xi^3
+\frac{2}{3\cdot 5} \xi^5 + {\cal O}\left( \xi ^7\right) \ \]
and assume (by examining the graph of $\tan x = x$) that $\xi = o(1)$
as $n \rightarrow \infty$.

\item (Discussion of the convergence of the Dirichlet series--from Knopp,
{\em The Theory and Application of Infinite Series}, p.516, Ex.200).\\
\begin{enumerate}
\item Show that for a convergent series $\sum(-1)^n a_n$ we have
({\bf Euler's transformation}):
\[ \sum_{n=0}^\infty (-1)^n a_n = \sum_{n=0}^\infty
\frac{\Delta^n a_0}{2^{n+1}} \ ,\]
where the {\em Finite Difference} symbol $\Delta^n$ is defined as
\[ \Delta^0 a_k := a_k\ \ ;\ \ \Delta^{n+1} a_k :=\Delta^n a_k - \Delta^n a_{k+1}\ \ ;\ \ n=0,1,\ldots \ \ .\]
\item Show that if the series $S = \sum(-1)^n a_n$ has monotone decreasing
terms, and if also the differences $\Delta^n a_k$, $0\le n \le p+1$ with
$p\ge 1$ are monotone decreasing then
\[ 
\frac{a_0}{2} + \frac{\Delta a_0}{2^2} +\cdots
+\frac{\Delta^{p-1}}{2^p} 
< S <
\frac{a_0}{2} + \frac{\Delta a_0}{2^2} +\cdots
+\frac{\Delta^{p-1}a_0}{2^p} 
+\frac{\Delta^{p} a_0}{2^p} \ .\]
\item
Use the previous result to show that the Dirichlet series $\sum (-1)^{n+1}/n^x$
defined for ${\bf R}(x) > 0$ has limit as $x\rightarrow 0^+$:
\[ \lim\!\!\!\!\!\!\!\!\!_{{}_{x\rightarrow 0^+}} 
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^x} = \frac{1}{2} \ .\]
 
\end{enumerate}

\end{enumerate}
\end{document}
