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\noindent Solutions for Homework 7
\vskip .2in

\bigskip
\noindent {\bf Page 243/ \#8:}
\bigskip

Given $n\in\mathbf{N}$, set
$$
g_n(x) = \left\{\begin{array}{cll} \sin x & \text{if} & j = 0\mod4 \\
                                \cos x  & \text{if} & j = 1\mod4 \\
                                -\sin x & \text{if} & j = 2\mod4 \\
                                -\cos x & \text{if} & j = 3\mod4 \\
         \end{array}\right.
$$
Note that $g'_n(x) = g_{n+1}(x)$ for every $n$ and that $|g_n(x)|\leq 1$
for every $n$ and $x$.

If $1\leq N<p$, then I claim that for every $n\leq N-1$, the series
$$
\sum_{j=1}^\infty (2\pi j)^n {a_j} g_n(2\pi j x).
$$
converges uniformly on $\mathbf{R}$ and is equal to $f^{(n)}(x)$.
In particular, $f$ is $N-1$ times differentiable.  I will prove this 
claim by induction on $n$:

When $n=0$ this amounts merely to showing that
\begin{equation}
\label{first}
\sum_{j=1}^\infty a_j \sin(2\pi j x)
\end{equation}
converges uniformly.  The sum is equal to $f(x) = f^{(0)}(x)$ by 
definition.  By hypothesis, we have
$$
|a_j \sin(2\pi j x)| \leq |a_j| \leq C/j^p
$$
for all $x\in\mathbf{R}$.  Furthermore, $p$ is at least two, so 
$\sum_{j=1}^\infty \frac{C}{j^p}$ converges.  Hence the Weierstrass 
$M$-test implies that the series in equation \ref{first}
converges uniformly for all $x\in\mathbf{R}$.  This concludes the
case $n=0$.

Now suppose that the claim is true for $n = k$ and consider the
case $n=k+1$.  Then the series
$$
\sum_{j=1}^\infty (2\pi j)^{k}{a_j}g_k(2\pi j x)
$$
converges uniformly on $\mathbf{R}$ and is equal to $f^{(k)}(x)$.
Differentiating this series term by term gives us the new series
\begin{equation}
\label{newseries}
\sum_{j=1}^\infty (2\pi j)^{k+1} a_j g_{k+1}(2\pi j x)
     = \sum_{j=1}^\infty (2\pi j)^n a_j g_n(2\pi j x).
\end{equation}
We have the following upper bounds for the terms of this new series:
$$
|(2\pi j)^n a_j g_n(2\pi j x)| \leq (2\pi j)^n |a_j| 
                               \leq \frac{(2\pi)^nC}{j^{p-n}}.
$$
Now since $n \leq N-1 < p - 1$, we have that $p - n > 1$ and that 
$\sum_{j=1}^\infty 1/j^{p-n}$ converges.  Hence
the Weierstrass $M$-test tells us that the series (\ref{newseries})
converges uniformly for all $x\in\mathbf{R}$.  By Theorem 6.3.3 then, we 
conclude that this series is equal to  
$
\frac{d}{dx} \frac{d^{k-1}f}{dx^{k-1}} = \frac{d^k f}{dx^k}.
$
That is, the claim is true in the case $n=k+1$ and our proof is complete.
\qed


\newpage

\noindent {\bf Page 244/ \#15:}
\bigskip

Note that $e^{-(x-j)^2}/j^2 \leq e^0/j^2 = 1/j^2$ for all $x\in\mathbf{R}$.
Since $\sum_{j=1}^\infty 1/j^2$ converges, the Weierstrass $M$-test tells us
that 
\begin{equation}
\label{theseries}
f(x) = \sum_{j=1}^\infty \frac{e^{-(x-j)^2}}{j^2}
\end{equation}
converges uniformly on $\mathbf{R}$ to a continuous function $f(x)$.

It remains to show that $\lim_{x\to\pm\infty} f(x) = 0$.  We will
do this for $x\to\infty$ only, since the case $x\to-\infty$ is similar.
Let $S_n(x)$ denote the $n$th partial sum for the series in (\ref{theseries}).
Let $\epsilon > 0$ be any fixed number.  Since $S_n(x)$ converges uniformly 
to $f(x)$ (by definition of convergence for a series), there exists an
$N\in\mathbf{N}$ such that 
$$
\sum_{j=n+1}^\infty \frac{e^{-(x-j)^2}}{j^2} = |f(x) - S_n(x)| < \epsilon
$$
whenever $n\geq N$.  In particular, this inequality holds if we choose
$n=N$.  

Note also that, since $S_N(x)$ is a {\it finite} sum, we have
\begin{eqnarray*}
0 & \leq & \lim_{x\to\infty} f(x) 
     = \lim_{x\to\infty} \sum_{j=1}^N \frac{e^{-(x-j)^2}}{j^2}(x) 
     + \lim_{x\to\infty} \sum_{j=n+1}^\infty \frac{e^{-(x-j)^2}}{j^2} \\
  & = & \sum_{j=1}^N \lim_{x\to\infty} \frac{e^{-(x-j)^2}}{j^2}(x) 
     + \lim_{x\to\infty} \sum_{j=n+1}^\infty \frac{e^{-(x-j)^2}}{j^2} \\
  & \leq & 0 + \epsilon
\end{eqnarray*}
The only way that this inequality holds for every $\epsilon > 0$
is if we have $\lim_{x\to\infty} f(x) = 0$. 
\qed



\bigskip
\noindent {\bf Page 251/ \#3:}
\bigskip

If $f(x) = \ln x$, then for $n\geq 1$, 
$f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{x^n}$.  Hence the power series for
$\ln(x)$ centered at $x_0 = 1$ is
$$
\sum_{j=0}^\infty \frac{f^{(j)}(1)}{j!}(x-1)^j 
     = \sum_{j=1}^\infty \frac{(-1)^{j-1}}{j}(x-1)^j.
$$
It's not hard to show that $\lim_{n\to\infty} (1/j)^{1/j} = 1$, so the
radius of convergence for this series is 1.  Now if $S_n(x)$ is the 
$n$th partial sum for the series, then Taylor's theorem tells us that
for some $c$ between $x$ and $1$, we have
$$
f(x) - S_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-1)^{n+1} 
      = \frac{(-1)^n (x-1)^{n+1}}{(n+1)c^{n+1}}.
$$ 
Now if $|x-1|<1/2$, we have that $-1/2 < c < 3/2$, so
$$
|f(x) - S_n(x)| \leq \frac{(1/2)^{n+1}}{(n+1)(1/2)^{n+1}} = \frac{1}{n+1},
$$
In other words, for every $n\in\mathbf{N}$,
$$
-\frac{1}{n+1} \leq f(x) - S_n(x) \leq \frac{1}{n+1}
$$
By the Squeeze Theorem, we conclude that 
$
\lim_{n\to\infty} |f(x) - S_n(x)| = 0.
$
\qed





\bigskip
\noindent {\bf Page 251/ \#8:}  Since
$
e^x = \sum_{j=0}^\infty \frac{x^j}{j!}, 
$
and the series on the right side converges uniformly on any
finite interval, we have 
$
e^{-x^2} = \sum_{j=0}^\infty \frac{(-1)^j x^{2j}}{j!}
$
where the series (again) converges uniformly on any finite interval.
Hence we can integrate the latter series term by term (see Theorem 6.3.2)
to obtain
$$
F(x) = \int e^{-x^2}\,dx = \sum_{j=0}^\infty \frac{(-1)^j x^{2j+1}}{j!(2j+1)}.
$$



\bigskip
\noindent {\bf Page 251/ \#11:}
\bigskip

The radius of convergence is
$$
R = (\limsup (j+1)^{1/j}(j+2)^{1/j})^{-1} = 1.
$$
If we call the series $f(x)$, then the antiderivative of the antiderivative
of $f$ can be obtained by integrating the series twice term by term 
(Theorem 6.3.2 again, along with the fact that the series converges uniformly 
on $(-r,r)$ for any $r < R = 1$.)  Hence the antiderivative of the 
antiderivative of $f$ is given by the geometric series
$$
F(x) = \sum_{j=0}^\infty x^{j+2} = x^2 \sum_{j=0}^\infty x^j = \frac{x^2}{1-x}.
$$
Now we differentiate twice to get back to $f$, arriving at the formula
$$
f(x) = F''(x) = \frac{2}{1-x}^3.
$$
\qed


\bigskip
\noindent {\bf Page 251/ \#12:}
\bigskip

Consider the series $\sum_{j=1}^\infty \frac{x^j}{j^p}$.  This series
has radius of convergence equal to one for every $p$ and therefore
converges when $|x|<1$ and diverges when $|x|>1$.  However, whether the 
series converges or diverges at the points $x=1$ and $x=-1$ depends
on $p$.  At $x=1$, we have
$$
\sum_{j=1}^\infty \frac{x^j}{j^p} = \sum_{j=1}^\infty \frac{1}{j^p}
$$
and at $x=-1$, we have 
$$
\sum_{j=1}^\infty \frac{x^j}{j^p} = \sum_{j=1}^\infty \frac{(-1)^j}{j^p}
$$
If $p=0$, then both of these series diverge, because the terms don't
go to zero in either of them.  If $p=1$, then the first series is the
harmonic series and therefore diverges, whereas the second series is
the alternating harmonic series (example from class) and therefore
converges.  If $p=2$, then the first series converges, and this means
that the second series converges absolutely.  Taken together, these
examples show that a power series can converge at both endpoints of
its interval of convergence, diverge at both endpoints, or converge
at one endpoint, but diverge at the other.

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