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\begin{document}


{\bf Question}

This question will demonstrate that a convergent Taylor series can
also be an asymptotic series. Consider the Taylor series expansion
about the origin of a sufficiently differentiable function $f(x)$:

$$f(x)=f(0)+xf'(0)+\ds\frac{1}{2}x^2f''(0)+\cdots+\ds\frac{1}{n!}x^nf^{(n)}(0)+R_n(x)$$

where $R_b(x)$ has the exact representation

$$R_n(x)=\ds\frac{1}{n!}\ds\int_0^x (x-t)^nf^{(n+1)}(t)\, dt.$$

\begin{description}
\item[(i)]
Write down the Poincar\'e definition of an expansion with a gauge
$\{x^n\}$ which is asymptotic to a function $f(x)$ at $x=0$

\item[(ii)]
Using the fact that $\left|\ds\int_0^z g(\zeta)\, d\zeta\right|
\leq \ds\int_0^z |g\zeta|d\zeta$ for any suitable integrable
integrand, show that

$$|R_n(x)| \leq \ds\frac{|x^n|}{n!} \ds\int_0^x
\left(1-\ds\frac{t}{x}\right)^n\left|f^{(n+1)}(t)\right|\,dt.$$

\item[(iii)]
Assuming that $f^{(n+1)}(t)$ is continuous on $[0,x]$ then there
exists a number $M$, such that $|f^{(n+1)}(x)| \leq M$. Given
this, show that,

$$|R_n(x)| \leq \ds\frac{M|x|^{n+1}}{n!}\ds\int_0^1(1-\xi)^n\,
d\xi$$

\item[(iv)]
Hence show that

$$R_n(x)=O(x^{n+1}),$$

and establish that the Taylor expansion is indeed Poincar\'e
asymptotic to $f(x)$ at the origin.
\end{description}



\vspace{.5in}

{\bf Answer}

 \begin{description}
\item[(i)]
$\ds f(x) \sim \sum_{s=0}^\infty a_s x^{+s}$ as $x\to 0^+$ if $\ds
\left[f(x)-\sum_{s=0}^n a_s x^s\right]=0(x^{n+1})$ as $x\to 0^+\ \
\ (A) $

for every \emph{fixed} $n \geq 0$.

With gauge $\{x^n\}$ as $x\to 0$. This is the usual Taylor series
expansion which will have a finite radius of convergence.

\item[(ii)]
Given

$$R_n(x) = \frac{1}{n!}\ds\int_0^x(x-t)^nf^{(n+1)})(t)\, dt$$
$$|R_n(x)| =\frac{1}{n!}\left| \ds\int_0^x (x-t)^nf^{(n+1)})(t)\,
dt \right|$$

we have

\begin{eqnarray*} |R_n(x)| & \leq & \ds\frac{1}{n!} \ds\int_0^x |(x-t)^n
f^{(n+1)}(t)|\, dt\\ & = & \ds\frac{|x^n|}{n!}\ds\int_0^x
\left|\left(1-\ds\frac{t}{x}\right)f^{(n+1)}(t)\right|\, dt
\end{eqnarray*}

But consider the range of the integrand: $t$ runs from $0 \to x$

i.e., $0 \leq t \leq x$.  Therefore
$\left|\left(1-\ds\frac{}{x}\right)^n\right|=\left(1-\ds\frac{t}{x}\right)^n$
for every positive integer $n$.

Therefore $|R_n(x)| \leq \ds\frac{|x^n|}{n!}\ds\int_0^x
\left(1-\ds\frac{t}{x}\right)^n\left|f^{(n+1)}(t)\right|\, dt$ as
required.

\item[(iii)]
Given $f^{(n+1)}(t) \in C[0,x] \Rightarrow$ there exists $M (\geq
0)$ such that $$\left|f^{(n+1)}(t)\right| \leq M.$$

Thus in (ii),

$$|R_n(x)| \leq M\ds\frac{|x^n|}{n!} \ds\int_0^x
\left(1-\ds\frac{t}{x}\right)^n\, dt$$ $$\Rightarrow |R_n(x)| \leq
\ds\frac{M|x^{n+1}|}{n!}\ds\int_0^1 (1-\xi)^n\, d\xi$$

(setting $\frac{t}{x}=\xi$)

\item[(iv)]
$\ds\int_0^1 (1-\xi)^n\, d\xi \stackrel{(\xi=\cos^2 u)}{=}
2\ds\int_0^{\frac{\pi}{2}} \sin^{n+1} u \cos u\,
du=2\left[\frac{\sin^{n+2} u}{n+2}\right]_0^{\frac{\pi}{2}}=2$

Therefore $\left|R_n(x)\right| \leq
\ds\frac{2M}{(n+2)n!}\left|x^{n+1}\right|$

So, by definition of order symbols,

$R_n(x)=O(x^{n+1})\ \left[{\rm implied\ constant\
}=\ds\frac{2M}{(n+2)n!}\right]$

Clearly this satisfies $(A)$ as $x\to 0^+$ so Taylor series is
\emph{also} asymptotic.
\end{description}



\end{document}
