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\newcommand{\defn}{\bigskip\par Define~}
\newcommand{\thm}{\bigskip\par\noindent Theorem.~~}
\newcommand{\lem}{\bigskip\par\noindent Lemma.~~}
\newcommand{\ex}{\bigskip\par\noindent Example.~~}
\newcommand{\doubleint}{\int\!\!\int }
\newcommand{\cor}{\bigskip\par\noindent Corollary.~~}


\begin{document}
\begin{flushright}
  A. Miller\\
  M521\\
  May 11, 1994
\end{flushright}

\defn $P$ is a partition of the rectangle $R=[a,b]\cross [c,d]$.
\defn $L(P,f)$
\defn $U(P,f)$
\lem If $P_1$ is a partition which refines $P_2$, then
$$L(P_2,f)\leq L(P_1,f)\leq U(P_1,f)\leq U(P_2,f)$$
\lem If $P_1$ and $P_2$ are any two partitions, then
$$L(P_1,f)\leq U(P_2,f)$$

\defn $L$ and $U$
\defn $f:R\arrow \reals$ is Riemann integrable (Darboux definition).
\defn $\doubleint_Rf$

\ex Give an example of a bounded function $f$ on $[0,1]\cross[0,1]$ which
is not Riemann integrable.
\thm If $f:R\arrow\reals$ is continuous, then $f$ is Riemann integrable.
\defn $B\subseteq \reals^2$ has zero area.
\thm If $f$ is bounded on $R$ and continuous except for a set of points
with zero area, then $f$ is Riemann integrable.

\defn For $D\subseteq R$ an arbitrary region, define $\doubleint_{D}f$.
\defn Boundary of $D$ and interior of $D$.

\cor If $f:D\arrow\reals$ is continuous and bounded
on the bounded set $D\subseteq\reals$ and
the boundary of $D$ has zero area, then the Riemann integral
$\doubleint_D f$ exists.
\defn For $f:[a,b]\arrow\reals$ define $f$ is Riemann integrable
and $\int^b_a f$.

\thm (Fund Thm I.) Suppose $f:[a,b]\arrow\reals$ is continuous
and define $F(x)=\int_a^x f$ for $x\in [a,b]$.  Then
$F$ is continuous on $[a,b]$, differentiable on
$(a,b)$ and for every $x\in (a,b)\;\;\;F^\prime(x)=f(x)$.

\thm (Fund Thm II.) Suppose $f:[a,b]\arrow\reals$ is continuous
and $F$ is continuous on $[a,b]$, differentiable on
$(a,b)$ and for every $x\in (a,b)\;\;\;F^\prime(x)=f(x)$.  Then
$\int_a^bf=F(b)-F(a)$.

\thm (Substitution) Suppose $f$ is continuous on $\phi([a,b])$ where
$\phi$ is $C^\prime$ on some open set containing $[a,b]$.
Then
$$\int^b_a f(\phi(t))\phi^\prime(t)dt=\int^{\phi(b)}_{\phi(a)} f(x)dx.$$

\thm (Fubini) Suppose $f:R\arrow \reals$ is continuous
on the rectangle $R=[a,b]\cross[c,d]$ and define $F:[c,d]\arrow\reals$
by $F(x)=\int^d_c f(x,y)dy$.  Then $F$ is Riemann integrable
and $$\int^b_aF(x)dx=\doubleint_R f.$$
Similarly if $G(y)=\int^b_a f(x,y)dx$, then $G$ is Riemann
integrable and $$\int^d_cG(y)dy=\doubleint_R f.$$   Hence,
$$\int^b_a(\int^d_c f(x,y)dy)dx=\int^d_c(\int^b_a f(x,y)dx)dy.$$


\cor (to the proof) In Fubini's Theorem instead
of assuming $f$ is continuous it is enough to assume
$\doubleint_R f$ exists and for all $x\in [a,b]$
the integral $\int^d_c f(x,y)dy$ exists.  Similarly, for the
similarly.

\cor Suppose $\phi_1:[a,b]\arrow \reals$ and
$\phi_2:[a,b]\arrow \reals$ are continuous with $\phi_1(x)\leq\phi_2(x)$
for all $x\in [a,b]$.  Let
$$D=\{(x,y):a\leq b\rmand \phi_1(x)\leq y\leq\phi_2(x)\}$$
and let $f:D\arrow\reals$ be continuous.  Then
$$\doubleint_Df=\int^b_a(\int^{\phi_2(x)}_{\phi_1(x)}f(x,y)\;dy)\;dx.$$

\lem Suppose $g:D\arrow\reals$ is continuous on the
open set $D\subseteq\reals^2$and for every rectangle $R\subseteq D$
$$\doubleint_R g=0.$$
Then $g(p)=0$ for every $p\in D$.

\thm Suppose $f:D\arrow\reals$ is $C^{\prime\prime}$ on the
open set $D\subseteq\reals^2$.  Then
$${{\partial^2 f}\over{\partial x\partial y}}=
{{\partial^2 f}\over{\partial y\partial x}}$$
at every point in $D$.

\thm (Liebnitz)  Suppose that $f$ is $C^\prime$ on an open
set of the plane containing $(a,b)\cross [c,d]$. Then
$${{\partial }\over{\partial x}}\int^d_c f(x,y)dy=
\int^d_c {{\partial }\over{\partial x}}f(x,y)dy$$
for every $x\in (a,b)$.


\defn $\sum a_n$ converges.

\lem $\lim_{n\arrow\infty}x^n=0$ if $0\leq x<1$ and
$\lim_{n\arrow\infty}x^n=\infty$ if $x>1$.

\thm $\sum_{n=0}^\infty x^n$ converges iff $|x|<1$.  It equals
$1\over{1-x}$ when it converges.

\thm $\sum_{n=1}^\infty {1\over n}$ diverges.


\thm If $\sum a_n$ converges and $\sum b_n$ converges, then
  $\sum (a_n+b_n)$ converges and equals $\sum a_n+\sum b_n$.
\thm If $\sum a_n$ converges and $c\in\reals$, then
  $\sum ca_n$ converges and equals $c\sum a_n$.

\thm (Cauchy Criterion) $\sum a_n$ converges iff
$\forall \epsilon>0\exists N \forall n>m>N$
$$ |\sum_{k=m}^n a_k|=|a_m+a_{m+1}+\cdots+a_n|<\epsilon.$$

\thm $\sum a_n$ converges iff for all $N$
the tail $\sum_{n=N}^{\infty}a_n$ converges iff there
exists $N$ such
that $\sum_{n=N}^{\infty}a_n$ converges.

\thm $\sum a_n$ converges implies $\lim_{n\arrow\infty}a_n=0$.

\thm $\sum |a_n|$ converges implies $\sum a_n$ converges.

\defn Absolute convergence.

\defn Conditional convergence.

\thm  (Alternating series).  Suppose $0<a_{n+1}\leq a_n$ for all $n$.
Then $\sum (-1)^na_n$ converges and its value $S$ lies between any two
partial sums, i.e.,
$S_{2n+1}\leq S\leq S_{2n}$.

\ex Give an example of a conditionally convergent series.

\thm $e$ is irrational.

\defn Rearrangement of a series.

\thm If $\sum a_n$ is absolutely convergent, then every
rearrangement of this series converges (and to the same value).

\thm If $\sum a_n$ is conditionally convergent, then for every
$x\in [-\infty,\infty]$ there is a rearrangement of the series
which converges to $x$.

\thm If $a_n\geq 0$ for all $n$, then $\sum a_n $ converges iff
there exists $B$ such that every partial sum $S_n\leq B$.

\thm (Comparison Test) Suppose $0\leq a_n\leq b_n$ for all $n$.
Then if $\sum b_n$ converges, then $\sum a_n$ converges.

\thm (Limit Comparsion Test) Suppose $a_n$ and $b_n$ are positive
and $\lim {a_n\over b_n}=c>0$.  Then
$\sum a_n$ converges iff $\sum b_n$ converges.

\thm (Ratio Test) Suppose $a_n>0$ for all $n$ and
$$\lim {a_{n+1}\over a_n}=r.$$
Then $\sum a_n$ converges if $r<1$, and
$\sum a_n$ diverges if $r>1$.

\ex Give two examples of series with positive terms, such that
$r=1$ in the ratio test and one of the sequences converges and
the other diverges.

\thm (Root Test) Suppose $a_n\geq0$ for all $n$ and
$$\limsup {^n\sqrt{a_{n}}}=r.$$
Then $\sum a_n$ converges if $r<1$, and
$\sum a_n$ diverges if $r>1$.  In fact, if $r>1$ there are
infinitely many $n$ with $a_n\geq 1$.

\ex Give two examples of series with positive terms, such that
$r=1$ in the root test and one of the sequences converges and
the other diverges.

\thm (Integral Test) Suppose $f:[1,\infty)\arrow \reals$ is a
continuous function such that $\forall x,y\in [1,\infty)$
$x\leq y$ implies $f(x)\geq f(y)\geq 0$, i.e. $f$ is monotone decreasing.
Then $\sum f(n)$ converges iff $\int_1^\infty f(x)dx$ converges.

\cor $\sum{1\over n^r}$ converges iff $r>1$.

\thm (Radius of Convergence) For any power series $\sum a_n x^n$ there
exists $R\in [0,\infty]$ such that $\sum a_n x^n$ converges absolutely
for every $x$ with $|x|<R$ and diverges for every $x$ with $|x|>R$.
Furthermore, the convergence is uniform on every interval of
the form $[-r,r]$ with $r<R$.

\thm If $f_n:R\arrow\reals$ are continuous ($R$ is
a rectangle in $\reals^2$ or an interval in $\reals$) and $f_n$ converges to
$f$ uniformly on $R$, then
$$\doubleint_R f=\lim_{n\arrow\infty} \doubleint_R f_n.$$


\thm Suppose $f_n$ converges uniformly to $f$ on an open set $D\subseteq\reals$ 
and
each $f_n$ is $C^\prime$ and
$f_n^{\prime}$ converges uniformly to $g$ on $D$. Then $f$ is differentiable
and $f^\prime(x)=g(x)$ for every $x\in D$.

\cor If $f(x)=\sum a_n x^n$ has a nonzero radius of convergence,
then the series $g(x)=\sum na_n x^{n-1}$ has the same radius of convergence
and $f^\prime=g$.  Similarly, If $f(x)=\sum a_n x^n$ has a nonzero radius
of convergence,
then the series $g(x)=\sum {a_n\over {n+1}} x^{n+1}$ has the
same radius of convergence
and $g^\prime=f$.

\cor If $f(x)=\sum a_n x^n$ has a nonzero radius of convergence, then
$f$ has infinitely many continuous derivatives ($C^{\infty}$).

\cor If $f(x)=\sum a_n x^n$ has a nonzero radius of convergence, then
$a_n={f^n(0)\over {n!}}$ for every $n$.

\cor $\ln(1+x)=\sum_{n=1}^\infty (-1)^{n+1} {x^n\over n}$ for $|x|<1$.

\cor $\arctan(x)=\sum_{n=0}^\infty (-1)^n {x^{2n+1}\over 2n+1}$  for $|x|<1$.

\thm (Abel) If $\sum a_n x^n$ has a nonzero radius of convergence $R$,
and $\sum a_n x^n$ converges for $x=R$, then $\sum a_n x^n$ converges
uniformly on the interval $[0,R]$.

\cor If $f(x)=\sum a_n x^n$ has a nonzero radius of convergence $R$,
and $\sum a_n x^n$ converges for $x=R$, then $f$ is continuous
on $[0,R]$.

\cor $\ln(2)=\sum_{n=1}^\infty (-1)^{n+1} {1\over n}$.

\cor ${\pi\over 4}=\sum_{n=0}^\infty (-1)^n {{1}\over 2n+1}$.

\end{document}