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\title{Math 23A SOLUTION SET 11}
\author{Matthew G. Lee}

\begin{document}
\maketitle
\begin{enumerate}
\item
We have $f(x)=e^x$.

If $f\in C^\infty (\R),a\in \R$ we define the Taylor series $P
(a,x)$ for $f$ at $a$ by \begin{equation}P(a,x)=\sum _{m\geq
0}\frac{f^{(m)}(a)}{m!}(x-a)^m.\end{equation}  $P(a,x)$ is the
limit as $m$ approaches $\infty$ of the $m$th degree Taylor
polynomial
\begin{equation}P_m(a,x)=\sum _{0 \leq i \leq
m}\frac{f^{(i)}(a)}{i!}(x-a)^i.\end{equation} Define the remainder
by $R_m(a,x)=f(x)-P_m(x)$.

We know from calculus that $f$ is continuous and differentiable
with $f'(x)=f(x)$ for all $x$.  Inductively, we can see that $f$
is infinitely continuously differentiable.  We need to show that
for all $a,x \in \R$, $\lim_{m \to \infty} P_m(a,x)=e^x$. This is
equivalent to showing that for all $a,x \in \R$, $\lim_{m \to
\infty} R_m(a,x)=0$.

Let $a,x \in \R$.  We showed in class that there is a number $c
\in \R$ between $x$ and $a$ such that
$R_m(a,x)=\frac{f^{(m+1)}(c)}{(m+1)!}(x-a)^{m+1} =
\frac{e^c}{(m+1)!}(x-a)^{m+1}$.

So $|R_m(a,x)| \leq \max_{c \in [a,x] \cup [x,a]}
\frac{e^c}{(m+1)!}|(x-a)^{m+1}| =
\frac{e^{\max_{a,x}}}{(m+1)!}|x-a|^{m+1}$.

Let $M$ be the smallest integer greater than or equal to
$2|x-a|-1$.  So $\frac{|x-a|}{M+1} \leq \frac12$.  Then
$|R_{M}(a,x)| \leq \frac{e^{max_{a,x}}}{(M+1)!}|x-a|^{M+1}$ and
for any $m>M$, $$|R_{m}(a,x)| \leq
\frac{e^{max_{a,x}}}{(M+1)!(M+2) \dots m }|x-a|^{M+1}
|x-a|^{m-(M+1)}$$ $$ \leq
\frac{e^{max_{a,x}}}{(M+1)!(M+1)^{m-(M+1)}}|x-a|^{M+1}|x-a|^{m-(M+1)}$$
$$ \leq \frac{e^{max_{a,x}}}{(M+1)!} |x-a|^{M+1}\left(
{\frac{|x-a|}{M+1}} \right) ^{m-(M+1)} $$ $$ \leq
\frac{e^{max_{a,x}}}{(M+1)!}|x-a|^{M+1}\left({\frac12}\right)^{m-(M+1)}$$
$$ \leq \frac{e^{max_{a,x}}}{(M+1)! 2^{(M+1)}
}|x-a|^{M+1}\left({\frac12}\right)^m.$$

Keep in mind that $M$ depends only on $a$ and $m$, which are
fixed.  So $|R_{M}(a,x)| \leq k \left({\frac12}\right)^m$, where
$k$ is a constant. So, clearly, $\lim_{m \to \infty} |R_{M}(a,x)|
= 0$.  So, for all $x \in \R$, $P(a,x) = \lim_{m \to \infty}
P_m(a,x)=e^x$. QED

\item
Let $f$ be a function on $\R$ such that $f(x)=0$ if $x \leq 0$
and $f(x)=e^{-1/x}$ for $x>0$.
\begin{enumerate}
\item
First we will show that $f$ is infinitely continuously
differentiable as a function defined on the open interval $(0,
\infty)$. For $x>0$, $f(x)=e^{-1/x}$. We will show, inductively,
that $f$ is infinitely differentiable and for all integers $m
\geq 0$ and $x>0$, $f^{(m)}(x)=\frac{Q_m(x)}{x^{n_m}}e^{-1/x}$
for some polynomial $Q_m(x)$ and nonnegative integer $n_m$..

Base case: $f(x)=f^{(0)}(x)=\frac{Q_0(x)}{x^0}e^{-1/x}$, where
$Q_0(x)=1$.

Inductive step:  Suppose
$f^{(k-1)}(x)=\frac{Q_{k-1}(x)}{x^{n_{k-1}}}e^{-1/x}$, where
$Q_{k-1}(x)$ is a polynomial and $n_{k-1}$ is a nonnegative
integer.  By the rules for differentiation, $$f^{(k)}(x)=
\frac{Q_{k-1}(x)}{x^{n_{k-1}}}\left(\frac1{x^2}e^{-1/x} \right)+
e^{-1/x}\left(\frac{Q_{k-1}'(x)x^{n_{k-1}}-Q_{k-1}(x)n_{k-1}x^{n_{k-1}-1}}{x^{2(n_{k-1})}}\right)$$
$$=\left(\frac{Q_{k-1}(x)}{x^{n_{k-1}+2}}+
\frac{Q_{k-1}'(x)x^{n_{k-1}}-Q_{k-1}(x)n_{k-1}x^{n_{k-1}-1}}{x^{2(n_{k-1})}}
\right)e^{-1/x}$$ $$=\frac{Q_k(x)}{x^{n_k}}e^{-1/x}$$ for some
polynomial $Q_k(x)$ and nonnegative integer $n_k$.

So, for all integers $m \geq 0$ and all $x>0$,
$f^{(m)}(x)=\frac{Q_m(x)}{x^{n_m}}e^{-1/x}$ for some polynomial
$Q_m(x)$ and nonnegative integer $n_m$.  Further, for all $m \geq
0$, since $f^{(m+1)}$ exists, $f^{(m)}$ is differentiable and
therefore continuous.  So $f$ is infinitely continuously
differentiable on $(0,\infty)$.

Now we will show that $f$ is infinitely continuously
differentiable as a function defined on the open interval
$(-\infty,0)$. $f(x)=0$ for all $x \in (-\infty,0)$.  We see that
$f'(x)=f(x)=0$ for all $x \in (-\infty,0)$.  It is easy to show by
induction that $f^{(m)}(x)=0$ for all integer $m \geq 0$ and all
$x \in (-\infty,0)$.  So $f$ is infinitely continuously
differentiable on $(-\infty,0)$.

Now we will show that, for all integer $m \geq 0$,
$f^{(m)}(0)=0$.  We will show this inductively.

Base case: $f(0)=0$ by definition.

Inductive step: Suppose $f^{(m-1)}(0)=0$.  Then
$$f^{(m)}(0)=\lim_{h \to 0}
\frac{f^{(m-1)}(0+h)-f^{(m-1)}(0)}{h}$$ $$=\lim_{h \to
0}\frac{f^{(m-1)}(h)}{h}.$$

We need to show that for every $\epsilon>0$ we can find a
$\delta>0$ such that whenever $|h-0|<\delta$,
$\frac{f^{(m-1)}(h)}{h}<\epsilon$.  For every $\epsilon>0$ and
$\delta>0$, whenever $-\delta<h<0$,
$\frac{f^{(m-1)}(h)}{h}=0<\epsilon$. So we merely need to show
that for every $\epsilon>0$, there exists a $\delta>0$ such that
whenever $0<h<\delta$, $\frac{f^{(m-1)}(h)}{h}<\epsilon$.  So we
need to show that $\lim_{h \to 0+}\frac{f^{(m-1)}(h)}{h}=0$.

From our treatment of the derivatives of $f$ on $(0,\infty)$, we
know that
 $\lim_{h \to 0+}\frac{f^{(m-1)}(h)}{h}=
 \lim_{h \to 0+}\frac{Q_{m-1}(h)}{h^{n_{m-1}}}e^{-1/h}$ for some
polynomial $Q_{m-1}(x)$ and nonnegative integer $n_{m-1}$.  Let
$t=1/h$.  Then this limit is equal to $$\lim_{t \to
\infty}\frac{Q_{m-1}(1/t)}{(1/t)^{n_{m-1}}}e^{-t}$$ $$=\lim_{t \to
\infty}t^{n_{m-1}}Q_{m-1}(1/t)e^{-t}$$ $$=\lim_{t \to
\infty}\frac{t^{n_{m-1}}A(t)}{{t^c}e^t}$$ expressing
$Q_{m-1}(1/t)$ as the quotient of some polynomial $A(t)$ and
$t^c$, where $c$ is some nonnegative integer.  This limit is
clearly zero, as $e^t$ grows faster than any polynomial or
quotient of polynomials.  So $f^{(m)}(0)=0$.

So, for all integer $m \geq 0$, $f^{(m)}(0)=0$.  For any integer
$m \geq 0$, $f^{(m+1)}(0)$ exists, so $f^{(m)}$ is differentiable
at 0 and therefore continuous at 0.

So $f \in C^\infty(\R)$.
\item
As we just showed, $f^{m}(0)=0$ for all $m \geq 0$.  So the
Taylor series for $f$ at $a$ is $P(a,x)=\sum _{m\geq
0}\frac{f^{(m)}(a)}{m!}(x-a)^m=\sum_{m \geq
0}\frac0{m!}(x-a)^m=\sum_{m \geq 0}0$, a sum which obviously
converges to 0, regardless of $x$.
\item
Since all of the derivatives of $f$ at 0 are 0, the Taylor series
$P(0,x)$ for $f$ at 0 converges to the constant function 0. But
for $x>0$, $f(x)>0=P(0,x)$.  The reason for this nonconvergence
is that having all derivatives equal to 0 at a point does not
mean that a function is constant.
\end{enumerate}
\item
\begin{enumerate}
\item
Since $V$ is a finite-dimensional vector space over $\R$, we know
that for some $n$, $V$ is isomorphic to $\R^n$ as an algebraic
structure.  We will assume that limits exist on $V$, so we can do
calculus on $V$.

To avoid specifying $v$ as a variable, we will assume that we are
considering the Taylor polynomial at $v$, which is a function of
$x$.

For this case we don't need the isomorphism. We define the linear
Taylor polynomial $P_1(x)=f(v)+Df(v)(x-v)$. We easily verify that
this is the Taylor polynomial as follows:

$P_1(x)=f(v)+Df(v)(x)-Df(v)(v)$.  Since the first and third terms
are constant and $Df(v)(x)$ is a linear function of $x$,
$DP_1(v)=Df(v)$.

So $P_1(x)$ is the Taylor polynomial of $f$ at $v$.
\item A special case of below.
\item
Let $V$ be a finite-dimensional vector space over $\R$, $f \in
C^{\infty}(V)$, $v \in V$.  A Taylor polynomial of degree $k$ is a
polynomial $P_k(x)$, not necessarily homogeneous, of degree $\leq
k$ such that $\lim_{h \to 0} \frac{f(v+h)-P_k(v+h)}{|h|^{k}}=0$.
We will show that it suffices to find unique $Q_1, \dots, Q_n$,
where each $Q_m$ is either a homogeneous polynomial of degree $m$
or the zero polynomial, such that
\begin{equation}\lim_{h \to 0} \frac{f(v+h) - f(v) - Q_1(h) -
Q_2(h) - \dots - Q_k(h)}{ |h|^{k}}=0. \end{equation}  Let
$Q(h)=f(v)+Q_1(h)+\dots+Q_k(h)$.  Since is $Q_m(h)$ is either a
homogeneous polynomial of degree $m$ or the zero polynomial and
$f(v)$ is a homogeneous polynomial of degree 0 in h, $Q(h)$ is a
polynomial of degree $\leq m$.  Since $v$ is a constant,
$P_k(x)=Q(x-v)$ is a polynomial of degree $\leq m$ which, by
definition, satisfies $\lim_{h \to 0}
\frac{f(v+h)-P_k(v+h)}{|h|^{k}}=0$.

Choose a basis $\{e_1, \dots, e_n \}$ of $V$.  For
$h=h_1e_1+\dots+h_ne_n$, let $Q_m(h)=\sum_{I \in I^m_n}
\frac1{I!}D_If(v)h^I$, where the $I=(i_1,\dots,i_n)$ are
multi-exponents with $n$ entries and total degree $m$,
$D_If=D_1^{i_1} D_2^{i_2} \dots D_n^{i_n}f$, $D_if$ is the
derivative with respect to $e_i$ of $f$, and $h^I=h_1^{i_1}
h_2^{i_2} \dots h_n^{i_n}$.  Since $i_1+ \dots + i_n=m$, $Q_i(h)$
is a homogeneous polynomial of degree $m$ by the 2nd definition
of homogeneous polynomial from class.


Considering $V$ as isomorphic to ${\R}^n$ through the isomorphism
that maps $\{e_1, \dots, e_n \}$ to the elementary basis in
${\R}^n$, $P_k(x)$ is the polynomial in $V$ described by
Definition 3.3.15.  Theorem 3.3.18 tells us that $P_k$ is the
unique polynomial of degree $\leq k$ in the basis $\{e_1, \dots,
e_n \}$ satisfying $\lim_{h \to 0}
\frac{f(v+h)-P_k(v+h)}{|h|^{k}}=0$.

Now comes the part that I forgot in my first solution set:
showing that this Taylor polynomial is basis-independent.  Repeat
the above construction with another basis $\{d_1, \dots ,d_n\}$ to
generate a new Taylor polynomial $S_k(x)$.  Since each $e_i$ can
be expressed as a linear combination of $\{d_1, \dots ,d_n\}$,
$S_k$ is a polynomial of degree $\leq k$ in the basis $\{e_1,
\dots, e_n \}$ satisfying $\lim_{h \to 0}
\frac{f(v+h)-P_k(v+h)}{|h|^{k}}=0$.  By the uniqueness of $P_k$,
$S_k$ is identically equal to $P_k$.  So our definition of Taylor
polynomial is basis-independent.
\end{enumerate}
\end{enumerate}


\end{document}
