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\title{Infinitely Differentiable Functions}
\author{Antoine jackson}
\begin{document}
\maketitle
\tableofcontents


\section{Introduction}
\label{sec:Introduction}
I will discuss the section of infinitely differentiable functions, their properties, and some useful examples to show the difference between functions that are infinitely differentiable, and those that aren't.

  
\section{Differentiable functions}
\label{sec: Differentiable functions}
\definition: A real function is said to be $differentiable$ at a point if its derivative exists at that point. The notion of differentiablity can also be extended to complex functions (leading to the Cauchy-Riemann equations and the theory of holomorphic functions)


\section{Infinitely Differentiable Functions}
\label{sec: Infinitely Differentiable Functions}
\definition If $f(x)\in C^n$ on the interval $[a,b]$ for n = 0,1,2..., then $f$ is called Infinitely Differentiable on [a,b]. We shall write \textbf{$C^\infty[a,b]$} for the class of such functions.  

\medskip
Use the definition to find the Taylor series (centered at c) for the function
\begin{ex}$f(x)=e^x, where$ $c=0$
\end{ex}


\begin{ex}$f(x)=e^{-2x}$
\end{ex}


\begin{ex}$f(x)=cos(x), where$ $c=\frac{\pi}{4}$
\end{ex}


\begin{ex}$f(x)=lnx, where$ $c=1$
\end{ex}


\begin{ex}$f(x)=\frac{1}{1 + x^2}$ is $C^\infty$
\end{ex}


\section{Taylor Series}
\label{sec: Taylor Series}
\definition : If a function $f$ has derivatives of all orders at x=c, then the series 

$$
\sum_{n=0}^\infty(x)=f(c)+f'(c)(x-c)+....+\frac{f^n(c)}{n!}(x-c)^n+...
$$

is called the Taylor series. If $c=0$ then the series is a \textbf {Mclaurin Series for $f$}

\begin{displaymath}
\lim_{n \rightarrow \infty} \frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1} = 0
\end{displaymath}

for every $x$ in $I$.


\medskip

\section{Summary of Taylor Series}
\label{sec: summary of Taylor series}

$$
f{x)=\sum_{n=0}^\infty\frac{f^n(c)}{n!}(x-c)^n}
$$
\medskip
the coefficients $x_n$ are given by:
\medskip
$x_n=\frac{f^n(c)}{n!}$
\medskip
where $f^n(c)$ is the $nth$ derivative of $f$ evaluated at a point $c$.
\medskip

The functions of class $C^\infty[a,b]$ form a linear spce.
\medskip
If $f \in C^\infty[a,b]$ and $x_0 \in [a,b]$ we may form the Taylor expansion. The Taylor expansion for a given x, this series may or may not converge. If it converges, it may or may not converge to $f(x)$.

\medskip

\proof


\medskip  



\textbf{Facts to prove:}

\begin{ex} $f$ is $continuous$\end{ex}



\begin{ex} $f$ is $differentiable$ and f'(0)=0\end{ex}

$f^n(0)=0$

\medskip

Let $p(x),q(x)\in \textbf{R}$ be polynomials and define 


if $f \in C^\infty$, then we can certainly write a taylor series for $f$. However, it requires that this Taylor series actually converge (at least across some radius of convergence, such that the series converges absolutely for all real or complex numbers.)

\medskip




$$f(x)=
\left\{
\begin{array}{rl}
e^{\frac{-1}{x^2}}  & \hbox{  if  } x \ne 0 \\
0,                  & \hbox{  if  } x = 0 \\
\end{array}
\right.$$

Then $f \in C^\infty$, and for any n greater than or equal to 0, $f^{(n)}(0)=0$. So the Taylor series around $0$ is $0$; since $f(x) > 0$ for all $x$ not equal to 0, clearly it does not converge to $f$.

\begin{ex} $f$ is $C^\infty (\textbf{R})$  and  $f^n(0)=0$\end{ex}





\proofend


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