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\title{Some Mathematics for Essay 3}
\date{2010 November 9}
\author{Fred Thulin}
\begin{document}
  \maketitle
Recall you are examining the function
\[
  flat(x) =
  \begin{cases}
  e^{-\frac 1{x^2}} & \text{for } x \ne 0 \\
  0 & \text{for } x = 0
  \end{cases}
\]
%$flat$ is a $\bf{C}^\infty(\mathbb{R})$ function, meaning that
%$flat\colon\mathbb{R}\to\mathbb{R}
%$ and possesses derivatives of all orders.
and want to prove that the Maclaurin series of $flat$
does not converge to $flat$. Do this by showing that $flat$ and all its derivatives have
value 0 at $x=0$ and conclude the Maclaurin series is the identically 0 series, which
converges to the identically 0 function and not to $flat$. Below is an outline of a suggested
proof.

{\bf Lemma 1.} {\em For $x\ne0$, $flat^{(n)}(x)=flat(x)p_n(\frac1x)$ for all $n=0,1,2,\dots$
where $p_n$ is a polynomial (of degree 3 greater than that of $p_{n-1}$ when $n\ge1$).}

{\em Proof outline.} Use mathematical induction on $n$, the degree of the derivative.

 {\em Base case ($n=0$):} $flat^{(0)}(x)=flat(x)=flat(x)\cdot1$ and $1$ is a polynomial.

{\em Inductive step:} Show
$flat^{(k)}(x)=flat(x)p_k(\frac1x)\Rightarrow flat^{(k+1)}(x)=flat(x)p_{k+1}(\frac1x)$,
where $p_{k+1}$ is a polynomial with degree 3 more than the degree of the polynomial $p_k$.
Start by differentiating as follows:
\begin{align*}
flat^{(k+1)}(x)&=\frac{d}{dx}flat^{(k)}(x)\\
	&=\frac{d}{dx}\left(flat(x)p_k(\frac1x)\right)\\
	&=flat(x)\left(\frac{2}{x^3}p_k(\frac1x)-p'_k(\frac1x)\cdot\frac1{x^2}\right)\\
	&=flat(x)p_{k+1}(\frac1x)\text{, where }p_{k+1}(\frac1x)=
		\frac{2}{x^3}p_k(\frac1x)-p'_k(\frac1x)\cdot\frac1{x^2}\text.
\end{align*}
You will need to supply more details and argue that the degree of $p_{k+1}$
is 3 more than that of $p_{k}$.

{\bf Lemma 2.} {\em $flat^{(n)}(0)=0$ for all $n=0,1,2,\dots$.}

{\em Proof outline.} Again use induction on n.

{\em Base case ($n=0$):} $flat(0)=0$ by definition.

{\em Inductive step:} Use the definition of the derivative as the limit of a difference quotient,
along with the inductive assumption that $flat^{(k)}(0)=0$ to show $flat^{(k+1)}(0)=0$.
\begin{align*}
flat^{(k+1)}(0)&=\lim_{h\to0}\frac{flat^{(k)}(0+h)-flat^{(k)}(0)}{h}\\
	&=\lim_{h\to0}\frac{flat^{(k)}(h)}{h}\\
	&=\lim_{h\to0}\frac{flat(h)p_k(\frac1h)}{h}\text{, and now letting $u=\frac1h$,}\\
	&=\lim_{u\to\pm\infty}\frac{up_k(u)}{e^{u^2}}=0\text.
\end{align*}
Again, supply more details. You should now be able to conclude, using the definition of
Maclaurin series (perhaps cite a calculus book), the following:

{\bf Proposition.} {\em The Maclaurin series of flat does not converge to flat except at $x=0$.}

For suggested topic 1, you can use an approach like:
\begin{align*}
  f(x) &:=
  \begin{cases}
  flat(x+a)flat(x+b)& \text{for } -b<x<-a \\
  0 & \text{otherwise,}
  \end{cases}\\
 g(x)&:=\int_{-\infty}^x\! f(t)\,dt\text{,}\\
 h(x)&:=\frac{g(x)}{g(0)}\text{ and}\\
 bump(x)&:=h(x)h(-x)\text{.}
\end{align*}
Though a formal proof in not required, give an argument for why $bump$ has the
desired properties of being a $\bf{C}^\infty$ bump function (at 0).



For suggested topic 2, you can show the ideal $I_n=\langle flat,flat',\dots,flat^{(n)}\rangle$
does not contain $flat^{(n+1)}$ so that $I_n\subsetneq I_{n+1}$.
Use the fact from Lemma 1 that the degree of $p_{n+1}$ is 3 more than that of
$p_n$. This means that the ring
of $\bf{C}^\infty(\mathbb{R})$ functions contains an infinite proper ascending chain of
ideals $I_0\subsetneq I_1 \subsetneq\dots\subsetneq I_n\subsetneq \dots$, thus showing
$\bf{C}^\infty(\mathbb{R})$ fails to satisfy the ascending chain condition. The ascending
chain condition (ACC) states that any such chain must be finite, and Noetherian rings must
satisfy the ACC [2], so $\bf{C}^\infty(\mathbb{R})$ is not Noetherian.

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{\Large References}

1. Ideal (ring theory). (2010, August 24). In Wikipedia, The Free Encyclopedia.
Retrieved 15:47, October 22, 2010, from\\ \url{http://en.wikipedia.org/w/index.php?title=Ideal_(ring_theory)&oldid=380799258}.

2. Noetherian ring. (2010, October 13). In Wikipedia, The Free Encyclopedia.
Retrieved 19:36, October 21, 2010, from\\ \url{http://en.wikipedia.org/w/index.php?title=Noetherian_ring&oldid=390563549}.

  
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