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\begin{center}

{\bf Asymptotic Expansions}

\end{center}

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If $\ds f(z)=a_0+\frac{a_1}{z}+\cdots+\frac{a_n}{z^n}+R_n(z)$

where $\ds\lim_{z\to\infty}z^nR_n(z)=0$, (and $\arg z$ is suitably
restricted), then the series is said to be the asymptotic
expansion of $f(z)$ (valid in the appropriate sector of the
$z$-plane).

We write $\ds f(z)\sim a_0+\frac{a_1}{z}+\frac{a_2}{z^2}+\cdots$

$\ds a_0=\lim_{z\to\infty}f(z) \hspace{0.5in}
a_1=\lim_{z\to\infty}(f(z)-a_0)z \hspace{0.5in} \cdots$

$\ds a_n=\lim_{z\to\infty}\left(f(z)-a_0-\frac{a_1}{z}-\cdots
-\frac{a_{n-1}}{z^{n-1}}\right)$

This is not necessarily the simplest way of finding the $a_n$ in
special cases.  In many cases $a_0+\frac{a_1}{z}+\cdots$ taken to
$\infty$ diverges.

${}$

Properties

\begin{itemize}
\item[i)]
If $f(z)$ possesses such an expansion it is unique (since the
$a_n$ are uniquely determined).

\item[ii)]
Non-uniqueness of a function with a given expansion.

Consider $\ds e^{-z} \hspace{0.5in} |\arg
z|\leq\frac{\pi}{2}-\delta$

$\ds\lim_{z\to\infty}z^ne^{-z}=0 \hspace{0.5in}$ for all $n$

Hence $e^{-z}$ has no asymptotic expansion in the sense that the
coefficients are all zero.

Hence if $\ds f(z)\sim a_0+\frac{a_1}{z}+\frac{a_2}{z^2}+\cdots$

Then $\ds f(z)+e^{-z}\sim
a_0+\frac{a_1}{z}+\frac{a_2}{z^2}+\cdots$

\item[iii)]
Multiplication

If $\ds f(z)\sim\sum_{n=0}^\infty\frac{a_n}{z^n} \hspace{0.5in}$
and $g(z)\sim\sum_{n=0}^\infty\frac{b_n}{z^n}$

Then $\ds f(z)g(z)\sim\sum_{n=0}^\infty\frac{c_n}{z^n}
\hspace{0.5in} c_n=a_nb_0+a_{n-1}b_1+\cdots+a_0b_n$

Write $\ds f(z)=a_0+\frac{a_1}{z}+\cdots+\frac{a_n}{z^n}+R_n(z)$

$\ds\hspace{0.5in}
g(z)=b_0+\frac{b_1}{z}+\cdots+\frac{b_n}{z^n}+S_n(z)$

(Multiplying and collecting terms gives the result.)

\item[iv)]
Integration

If $\ds f(z)\sim\sum_{n=0}^\infty\frac{a_n}{z^n}$

we consider $\ds F(z)=\left[f(z)-a_0-\frac{a_1}{z}\right]\sim
\sum_{n=2}^\infty\frac{a_n}{z^n}$

$\ds\int_z^\infty F(w)dw\sim
\sum_{n=2}^\infty\int_z^\infty\frac{a_n}{w^n}dw=
\sum_{n=2}^\infty\frac{a_n}{n-1}\frac{1}{z^{n-1}}$

\item[v)]
Differentiation

It is not true in general that if

$\ds f(z)\sim\sum_{n=0}^\infty\frac{a_n}{z^n}$ then $\ds
f'(z)\sim\sum_{n=0}^\infty\frac{d}{dz}\frac{a_n}{z^n}$

e.g. $f(x)=e^{-x}\sin(e^x) \hspace{0.3in} x\geq0$

$f(x)$ has no asymptotic expansion (all coefficients are zero).

$f'(x)=\sin(e^x)-e^{-x}\sin(e^x).$

The term $\sin(e^x)$ does not tend to a limit as $x\to\infty$ and
the coefficients do not exist.

\end{itemize}


Example

$\ds F(z)=\int_0^\infty\frac{e^{-t}}{t+z}dt$

$\ds\frac{1}{t+z}= \frac{1}{z}+\frac{(-t)}{z^2}+\cdots+
\frac{(-t)^{n-1}}{z^n}+\frac{(-t)^n}{(z+t)z^n}$

$\ds F(z)=
\sum_{r=0}^{n-1}\int_0^\infty\frac{e^{-t}(-1)^rt^r}{z^{r+1}}dt+
R_n(z)$

where $\ds R_n(z)=\int_0^\infty\frac{e^{-t}(-t)^n}{z^n(z+t)}dt$

$\ds|z^nR_nz|\leq\int_0^\infty\frac{e^{-t}t^n}{|z+t|}dt\leq
\int_0^\infty\frac{e^{-t}t^n}{|z|}dt\,\,\to0$ as $|z|\to\infty$ if
$R(z)&gt;0$

If $z=x$, real and positive

$\ds R_n(x)=(-1)^n\int_0^\infty\frac{t^ne^{-t}}{x^n(x+t)}dt$

$\ds |R_n(x)|\leq\int_0^\infty
\frac{t^ne^{-t}}{x^{n+1}}=\frac{n!}{x^{n+1}}$

$\ds |R_n(x)|\leq |$next term in the series$|$

The series $\ds \sum_{r=0}^{n-1}\frac{(-1)^rr!}{z^{r+1}}$ diverges
(by the ratio test).

${}$

{\bf Euler-Maclaurin Formula}

$\ds\int_0^{mh}F(x)dx=
h\left[\frac{1}{2}F(0)+F(1)+\cdots+F(m-1)+\frac{1}{2}F(m)\right]$

$\ds\hspace{0.7in}-\frac{h^2}{2!}B_1[F'(mh)-F'(0)]+
\frac{h^4B_2}{4!}[F'''(mh)-F'(0)]$

$\ds\hspace{0.7in}+(-1)^n\frac{h^{2n}B_n}{(2n)!}
[F^{(2n-1)}(mh)-F^{(2n-1)}(0)]+R_n$

$\ds
R_n=\frac{h^{2n+3}}{(2n+2)!}\int_0^{mh}\Phi_{2n+2}(t)F^{(2n+2)}(ht)dt$

Where $\Phi_n(t)=\phi_n(t)\hspace{0.3in}0\leq t\leq1
\hspace{0.3in}$ (Bernoulli polynomial)

$\hspace{0.25in}\Phi_n(t+1)=\Phi_n(t)$


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