%\setcounter{page}{1100}
\chapter{Series of Constants\label{FirstSeriesChapter}}
In this chapter we consider ``infinite sums,'' such as
\begin{equation}
\sum_{n=1}^\infty a_n=\underbrace{a_1}_{n=1}+\underbrace{a_2}_{n=1}
  +\underbrace{a_3}_{n=3}+\cdots.
\label{SeriesFirstIntroducedEquation}\end{equation}
The ``sum'' above begins with $a_1$, but we will often begin
with a term $a_0$, or $a_2$, etc.  It is not the beginning
terms which determine if we can in fact compute such a
sum, but rather it is the infinite ``tail'' of the series.
This is reasonable because we can always, in principle,
add as many terms together as we like, so long as there are
finitely many of them.  As with other calculus concepts,
the tool which breaks the finite/infinite barrier is limit.
Indeed, to make sense of a sum such as (\ref{SeriesFirstIntroducedEquation}),
we consider the $N$th {\it partial sum},
\begin{equation}
S_N=\sum_{n=1}^na_n=a_1+a_2+\cdots+a_{N-1}+a_N,
\label{PartialSumFirstIntroducedEquation}\end{equation}
and then look at the {\it sequence} $S_1,S_2,S_3,\cdots$,
i.e.,
\begin{align*}
S_1&=a_1,\\
S_2&=a_1+a_2,\\
S_3&=a_1+a_2+a_3,\\
S_4&=a_1+a_2+a_3+a_4,\end{align*}
and so on.  To determine if (\ref{SeriesFirstIntroducedEquation})
makes sense is then considered (by definition) equivalent
to determining the behavior of the sequence $\left\{S_N\right\}_{N=1}^\infty$.
We say the series (\ref{SeriesFirstIntroducedEquation})
{\it converges} to $L\in\Re$ if and only if $S_N\longrightarrow L$
as $N\to\infty$.

In a few cases we will actually be able to compute a simple
formula for $S_N$, and thus be able to compute the
series by taking $N\to\infty$.  However, in many cases
we cannot find a compact formula for $S_N$.  In those cases
we have to develop other methods for determining if
the series converges at all, and if so, how to approximate
the value of the series with as much precision as we require,
by determining how large we require $N$ to be so that
we can approximate the full series by $S_N$.

In this text we will take more steps than most other texts in developing
the theory of series, since this topic is the source of much
confusion for students.  Indeed we devote this entire chapter
to the topic of series of constant terms, leaving nonconstant
terms for their own chapter.  The concepts are
intuitive---perhaps deceptively so---but require practice so that,
for example, one can recognize when and where to apply a particular
test of convergence or divergence.

In the next chapter
we will look at functions defined by series
$$f(x)=\sum_{n=0}^\infty a_n(x-a)^n
  =a_0+a_1(x-a)+a_2(x-a)^2+\cdots.$$
In fact most functions we have dealt  with in this text can be written
in the form above, at least on some open intervals, so such functions are
very important theoretically. In fact
these functions also have use in applications, when we do not know
any other method of representing a particular function arising
from a particular application except by such a series.

\newpage
\section{Series and Partial Sums}
As mentioned in the introduction to this chapter, the convergence
of a series is defined  as equivalent to the convergence of its
partial sums.
For convenience, we will define the $N$th partial sum
to be the sum of all terms of the underlying sequence up to
the term whose subscript is $N$.  Thus if the sequence is
series is $\sum_{n=k}^\infty a_n$, with
underlying sequence $\left\{a_n\right\}_{n=k}^\infty$, then
\begin{equation}
S_N=\sum_{n=k}^N a_n=a_k+a_{k+1}+\cdots+a_N.\end{equation}
So for a series $a_0+a_1+a_2+\cdots$, the partial
sum $S_N=a_0+a_1+\cdots+a_N$ would actually have $N+1$ terms,
though we will still call it the $N$th partial sum. (Of course
if $N<k$ we do not define an $N$th partial sum.)

\bex Consider the series
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2+1}.$$
Find the first five partial sums.

\underline{Solution}: We do this directly:
\begin{alignat*}{3}
S_1&=\sum_{n=1}^1\frac{(-1)^n}{n^2+1}
  &&=\frac{(-1)^1}{1^2+1}&&=\frac{-1}2\\
S_2&=\sum_{n=1}^2\frac{(-1)^n}{n^2+1}
  &&=\frac{(-1)^1}{1^2+1}+\frac{(-1)^2}{2^2+1}
  &&=\frac{-1}2+\frac1{5}=-\frac{3}{10}=0.3\\
S_3&=\sum_{n=1}^3\frac{(-1)^n}{n^2+1}
  &&=\frac{(-1)^1}{1^2+1}+\frac{(-1)^2}{2^2+1}+\frac{(-1)^3}{3^2+1}
    &&=-\frac{3}{10}+\frac{-1}{10}=\frac2{10}=0.2\\
S_4&=\sum_{n=1}^4\frac{(-1)^n}{n^2+1}
  &&=\frac{(-1)^1}{1^2+1}+\frac{(-1)^2}{2^2+1}+\frac{(-1)^3}{3^2+1}
         +\frac{(-1)^4}{4^2+1}
  &&=S_3+\frac{1}{17}=\frac{44}{170}\approx0.2588235294\\
S_5&=\sum_{n=1}^5\frac{(-1)^n}{n^2+1}&&
    =S_4+\frac{(-1)^5}{5^2+1}
   &&=\frac{44}{170}+\frac{-1}{26}\approx0.220361991.
\end{alignat*}
\eex
Note we used the simple recursion relationship for partial sums
of a series: given a series
$\ds{\sum_{n=k}^\infty a_n}$, and $N\ge k$ we have
\begin{equation}
S_{N+1}=S_N+a_{N+1},\label{RecursionSn+1=Sn+An+1}
\end{equation}
that is,
\begin{align*}
S_{N+1}&=\sum_{n=k}^{N+1}a_n=\underbrace{a_k+a_{k+1}+\cdots+a_N}_{S_N}
                +a_{N+1}=\sum_{n=k}^Na_n+a_{N+1}\\
       &=S_N+a_{N+1},\text{ q.e.d.}\end{align*}
In a later section
we will see that the series in the above example does in fact
converge, though we can only approximate its exact value
by computing $S_N$ for large values of $N$.

\subsection{Telescoping Series}
Telescoping series do occur on occasion, but the main reason
they are included in most calculus textbooks is that their
partial sums simplify in nice ways, leaving us able to compute
their limits and thus the whole series.  Indeed, the behavior
of telescoping series is unusually ``nice''---rivaled only by
that of the much more important
geometric series we will see later in this section---and therefore
well-suited for early examples of the general notion of series.

The simplest type of telescoping series is one in which
the terms added are themselves sums of two terms, constructed
in such a way that there is cancellation such as the following:
\begin{align}
\sum_{n=1}^\infty a_n&=\sum_{n=1}^\infty
\left[b_n-b_{n-1}\right]\label{SimpleAbstractTelescopingSeries}\\
&=\left(b_1-b_0\right)+\left(b_2-b_1\right)+\left(b_3-b_2\right)
+\left(b_4-b_3\right)+\cdots.\notag\end{align}
After a careful examination of the terms which appear in
(\ref{SimpleAbstractTelescopingSeries}), it seems that
all cancel except for $-b_0$. However we must be even more
careful since there are infinitely many terms we are claiming
we can cancel.  The correct approach is to carefully examine the partial
sums:
\begin{align*}
S_1&=b_1-b_0,\\
S_2&=\not{b_1}-b_0+b_2-\not{b_1}=b_2-b_0,\\
S_3&=\not{b_1}-b_0+\not{b_2}-\not{b_1}+b_3-\not{b_2}=b_3-b_0,\\
S_4&=\not{b_1}-b_0+\not{b_2}-\not{b_1}+\not{b_3}-\not{b_2}
   +b_4-\not{b_3}=b_4-b_0,\end{align*}
and so on, whereby we can conclude that, for this simplest type
of example (\ref{SimpleAbstractTelescopingSeries}), we have
\begin{equation}
S_n=b_n-b_0.\label{SnForSimpleAbstractTelescopingSeries}
\end{equation}
Now such a series will therefore converge if and
only if $\left\{b_n\right\}_{n=1}^\infty$ converges.
If $b_n\longrightarrow B\in\Re$ as $n\to\infty$, then
by (\ref{SnForSimpleAbstractTelescopingSeries}) we have
$S_n\longrightarrow B-b_0$, whence
$\sum_{n=1}^\infty\left[b_n-b_{n-1}\right]=B-b_0$.

More complicated telescoping series also occur, though the
basic idea is that the partial sums can be written
in such a way that all but a few terms found in the partial sums
eventually cancel,
and where we can compute the limits of those terms which do not.\footnote{%%%
%%% FOOTNOTE
It is interesting to visualize why the term {\it telescoping}
is used to describe such a series.
One of the {\it Webster's} dictionaries defines
the intransitive verb form of {\it telescope} as follows:
\begin{quote}
{\it to slide together, or into something else, in the manner of the
tubes of a jointed telescope.}\end{quote}
For another example, a ``telescoping antenna'' comes to mind.
The reader should keep such images in mind as we consider
so-called telescoping series.
%%% END FOOTNOTE
} Rather than memorizing the sample telescoping
forms (\ref{SimpleAbstractTelescopingSeries})
and (\ref{SnForSimpleAbstractTelescopingSeries}), it is better to
consider each example separately, writing out the terms of $S_N$
for enough values of $N$ that the pattern emerges.
\bex Consider the series $\ds{\sum_{n=1}^\infty\left[\frac1{n+1}-\frac1n\right]
}$.  Compute the form of each partial sum $S_N$ (as a function of $N$), and
the value of the series if it converges.

\underline{Solution}: We will write out a few partial sums longhand,
from which the pattern will emerge.
Indeed, all but two terms will cancel in each of the following.
\begin{alignat*}{2}
S_1&=\left[\frac12-1\right]&&=\frac12-1,\\
S_2&=\left[\frac12-1\right]+\left[\frac13-\frac12\right]
     &&=\frac13-1,\\
S_3&=\left[\frac12-1\right]+\left[\frac13-\frac12\right]
   +\left[\frac14-\frac13\right]&&=\frac14-1,\\
S_4&=\left[\frac12-1\right]+\left[\frac13-\frac12\right]
   +\left[\frac14-\frac13\right]+\left[\frac15-\frac14\right]
   &&=\frac15-1.
\end{alignat*}
From this we do indeed see a pattern in which
$$S_N=\frac1{N+1}-1.$$
Taking $N\to\infty$, we see $S_N=\frac1{N+1}-1\longrightarrow0-1=-1$,
and so we conclude that the series converges to $-1$, i.e.,
$$\sum_{n=1}^\infty\left[\frac1{n+1}-\frac1n\right]=-1.$$
\eex

Sometimes we need to do a little more work to detect a
telescoping series, and its formula for $S_N$.
Note that the general term of the added sequence terms,
namely $\left[\frac1{n+1}-\frac1n\right]$,
in our series above looks like a partial fraction decomposition
if the variable is $n$.  For that reason, when
the general term can be written in a PFD, the series may
in fact be telescoping.  This is the case with the following
example.

\bex Consider the series $\ds{\sum_{n=2}^\infty\frac1{n^2-1}}$.
Compute a general formula for the $N$th partial sum $S_N$,
and compute its limit, if $S_N$ converges, thereby computing the series.

\underline{Solution}: Note first that there is no $S_1$ here.
That said, the technique which we will use for this
is to first look at the partial fraction decomposition (PFD)
for $\frac1{n^2-1}$.  Of course we need the denominator factored,
giving us the form
$$\frac1{n^2-1}=\frac1{(n+1)(n-1)}=\frac{A}{n+1}+\frac{B}{n-1}.$$
Multiplying by $(n+1)(n-1)$ in the second equation then gives us
$$1=A(n-1)+B(n+1).$$
Now we use the usual methods for computing the coefficients $A$ and $B$:
\begin{alignat*}{2}
\underline{n=1}:&\qquad &1&=B(2)\implies\boxed{B=\frac12}\\
\underline{n=-1}:&\qquad&1&=A(-2)\implies\boxed{A=-\frac12}.
\end{alignat*}
From this we can rewrite our series
$$\sum_{n=2}^\infty
 \left[\frac{-1/2}{n+1}+\frac{1/2}{n-1}\right]
 =\sum_{n=2}^\infty\left[\frac12\left(\frac{-1}{n+1}+\frac1{n-1}\right)\right].
$$
There is no $S_1$, so we begin with $S_2$.
\begin{alignat*}{2}
S_2&=\frac12\left(\frac{-1}{3}+1\right)
       &&=\frac12\left(\frac{-1}{3}+1\right),\\
S_3&=\frac12\left(\frac{-1}{3}+1\right)
       +\frac12\left(\frac{-1}4+\frac12\right)
       &&=\frac12\left(\frac{-1}{3}+1+\frac{-1}4+\frac12\right),\\
S_4&=\frac12\left(\frac{-1}{3}+1\right)
       +\frac12\left(\frac{-1}4+\frac12\right)
       +\frac12\left(\frac{-1}5+\frac13\right)
       &&=\frac12\left(1+\frac12-\frac14-\frac15\right),\\
S_5&=S_4+\frac12\left(\frac{-1}6+\frac14\right)
       &&=\frac12\left(1+\frac12-\frac15-\frac16\right),\\
S_6&=S_5+\frac12\left(\frac{-1}7+\frac15\right)
       &&=\frac12\left(1+\frac12-\frac16-\frac17\right),\\
S_7&=S_6+\frac12\left(\frac{-1}8+\frac16\right)
       &&=\frac12\left(1+\frac12-\frac17-\frac18\right)
\end{alignat*}
By this point a pattern has clearly  emerged:
$$S_N=\frac12\left(1+\frac12-\frac1N-\frac1{N+1}\right),$$
and so
$S_N\longrightarrow \frac12\left[1+\frac12-0-0\right]=\frac12\cdot\frac32
         =\frac34$ as $N\to\infty$.
We can thus conclude that
$$\sum_{n=2}^\infty\frac1{n^2-1}
  =\sum_{n=2}^\infty
         \left[\frac12\left(\frac{-1}{n+1}+\frac1{n-1}\right)\right]
  =\frac34.$$
\eex
\bex Find $S_N$ and discuss the convergence (or divergence)
     of the series
$$\sum_{n=0}^\infty\left[\sqrt{n+1}-\sqrt{n}\right].$$

\underline{Solution}:
\begin{alignat*}{2}
S_0&=\left[\sqrt1-\sqrt0\right]&&=\sqrt1-\sqrt0,\\
S_1&=\left[\sqrt1-\sqrt0\right]+\left[\sqrt2-\sqrt1\right]&&=\sqrt2-\sqrt0,\\
S_2&=\left[\sqrt1-\sqrt0\right]+\left[\sqrt2-\sqrt1\right]
   +\left[\sqrt3-\sqrt2\right]&&=\sqrt3-\sqrt0,\\
S_3&=\left[\sqrt1-\sqrt0\right]+\left[\sqrt2-\sqrt1\right]
   +\left[\sqrt3-\sqrt2\right]+\left[\sqrt4-\sqrt3\right]
   &&=\sqrt4-\sqrt0,
\end{alignat*}
and so on, so that
$$S_N=\sqrt{N+1}-\sqrt0=\sqrt{N+1}\longrightarrow\infty
\text{ as }N\to\infty.$$
Thus the series diverges (to infinity, to be more descriptive).
\eex
Note that we could simplify our earlier expressions for $S_N$,
since for instance $\sqrt0=0$, $\sqrt1=1$ and $\sqrt4=2$, but
to do so would more likely obscure the pattern of cancellation.

\subsection{Geometric Series}
The class of series considered  here is arguably the most important
we will encounter.  Many important series analyses
depend upon how a particular series compares to, or mimics the
behavior of, an appropriately chosen geometric series.
As with the telescoping series, the geometric series
is one for which we can actually compute a general formula
for $S_N$, from which we can tell if the series converges,
and if so compute its sum.

What makes a series $\sum a_n$ geometric is that there exists a constant
$r\in\Re-\{0\}$ such that
\begin{equation}
(\forall n)\left[\frac{a_{n+1}}{a_n}=r\right].\label{RForGeometricSeries}
\end{equation}
In other words, such a series can be defined recursively
by $a_{n+1}=r\cdot a_n$.  (Note that this is equvialent to
$a_n=r\cdot a_{n-1}$, so long as $a_{n-1}$ is defined.)
Put more colloquially, a geometric series is one in which
the next term added is a constant multiple of the
term immediately prior to it.  Examples of geometric series follow:
\begin{itemize}
\item $\ds{\sum_{n=0}^\infty\left(\frac12\right)^n
   =1+\frac12+\frac14+\frac18+\cdots}$ \qquad($r=1/2$),
\item $\ds{\sum_{n=2}^\infty \frac6{5^n}=\frac6{25}
          +\frac6{125}+\frac6{625}+\frac6{3125}+
       \cdots}$ \qquad($r=1/5$),
\item $\ds{\sum_{n=1}^\infty\frac{2(-1)^n}{3^n}
   =-\frac23+\frac29-\frac2{27}+\frac2{81}-\cdots}$\qquad
     ($r=-1/3$),
\item $\ds{\sum_{n=1}^\infty\frac1{3^{2n}}
   =\frac19+\frac1{81}+\frac1{729}+\frac1{6561}+\cdots}$
   \qquad($r=1/9$).
\end{itemize}
Note that this last series can be
rewritten $\sum_{n=1}^\infty\frac1{9^n}$,
or even $\sum_{n=0}^\infty\left[\frac19\cdot\left(\frac19\right)^n\right]$.
In fact, unlike the
telescoping series, every geometric series can be written in the
same form, namely\footnote{%%
%%% FOOTNOTE
With geometric series, it is understood that ``$r^0$'' represents
$1$, even though technically this is only correct if $r>0$.
In each general setting in which we follow the convention
that $r^0$ is defined to be $1$ (regardless of the sign of $r$),
we will remark on this point.
%%% END FOOTNOTE
}
\begin{equation}
\sum_{n=0}^\infty \alpha r^n
  =\alpha+\alpha r+\alpha r^2+\alpha r^3+\cdots,
\label{GeneralGeometricSeries}
\end{equation}
where
\begin{align}
&\alpha\text{ is the {\it first term} of the series, and}\\
&r\text{ is the {\it constant ratio}, } a_{n+1}/a_n.
\end{align}
In the examples above, the first terms
are $\alpha=1,6/25,-2/3,1/9$ respectively.
Each of the series above can be rewritten
in $\Sigma$-notation in the form (\ref{GeneralGeometricSeries}),
starting with $n=0$.
For instance, the third series above can be rewritten,
using $\alpha=-2/3$ and $r=-1/3$, as
$$
\sum_{n=1}^\infty\frac{2(-1)^n}{3^n}
   =-\frac23+\frac29-\frac2{27}+\frac2{81}-\cdots
   =\sum_{n=0}^\infty \frac{-2}3\left(\frac{-1}3\right)^n.
$$
In fact, once we know a series is geometric (that is,
that $a_{n+1}=r\cdot a_n$ for each $n$), all we need
to do is to identify $\alpha$ and $r$, and we can write
the series in the exact $\Sigma$-notation form
(\ref{GeneralGeometricSeries}).

\bex Write the series $4+\frac23+\frac19+\frac1{54}+\frac1{324}+\cdots$
in the form (\ref{GeneralGeometricSeries}).

\underline{Solution}: Though perhaps not immediately obvious,
in fact each successive term is $\frac16$ times its immediate
predecessor.  The first term is $4$.  We translate these two facts as
$\alpha=4$ and $r=\frac16$, and so
this series is the same as the series
$$\sum_{n=0}^\infty 4\cdot\left(\frac16\right)^n.$$
\eex

As with telescoping series, a geometric series
allows for a simple formula for $S_N$.
To use the formula, however, we need to make two assumptions:
\begin{enumerate}
\item that the series is already written in the form
      $\ds{\sum_{n=0}^\infty\alpha r^n=\alpha+\alpha r+\alpha r^2+
                 \alpha r^3+\cdots}$, and
\item that $r\ne1$.
\end{enumerate}
As we have seen, the first requirement is easy enough to accomplish: we
need only identify $\alpha$ (the first term in the geometric
series) and $r$.  The second requirement is for technical reasons
we will encounter momentarily.  We do not loose much
in assuming $r\ne1$, since in the case $r=1$
the series is simply $\alpha+\alpha+\alpha+\cdots$,
which is clearly a divergent series if $\alpha\ne0$, and trivial if
$\alpha=0$.\footnote{%%%
%%% FOOTNOTE
We will not generally consider the case $\alpha=0$ because
it is trivial, and because we cannot identify a unique $r$.
Indeed, if $\alpha=0$, then any geometric recursion is valid,
but our original method of defining $r$, namely (\ref{RForGeometricSeries})
on page~\pageref{RForGeometricSeries}, is undefined if
$\alpha=0$.
%%% END FOOTNOTE
}
Now we state our theorem.
\begin{theorem} For a geometric series $\ds{\sum_{n=0}^\infty\alpha r^n
=\alpha+\alpha r+\alpha r^2+\alpha r^3+\cdots}$,
assuming $r\ne1$, we have
\begin{equation}
S_N=\frac{\alpha\left(1-r^{N+1}\right)}{1-r}.\label{SnForGeometric}
\end{equation}
\end{theorem}
\begin{proof}
The usual method of proof of (\ref{SnForGeometric}) is to exploit
the geometric nature of the series in the following way:
$$\begin{array}{lccccccccccc}
&S_N&=&\alpha&+&\alpha r&+&\alpha r^2&+&\cdots&+&\alpha r^N\\
\implies
&r\cdot S_N&=&\alpha r&+&\alpha r^2&+&\alpha r^3&+&\cdots&+&\alpha r^{N+1}\\
\hline
\implies&(1-r)S_N&=&\alpha&+&0&+&0&+&0&-&\alpha r^{N+1}\end{array}$$
In the first line we wrote the definition of $S_N$.  In the next
line we multiplied that equation by $r$.  In the third line,
the second line is subtracted from the first. In doing so,
the terms $\alpha r,\alpha r^2,\cdots,\alpha r^{N}$ cancel,
leaving only $\alpha-\alpha r^{N+1}$ on the right-hand side.
This gives us
$$(1-r)S_N=\alpha\left(1-r^{N+1}\right).$$
Since we are assuming $r\ne1$, we can divide by $1-r$ and get
(\ref{SnForGeometric}), as desired.
\end{proof}

To utilize (\ref{SnForGeometric}), one needs to know $\alpha$, $r$ and $N$.
Note that $N$ is not the number of terms, but the highest power
of $r$ which occurs.  In fact there are $N+1$ terms added
to arrive at $S_N$, since the first is $\alpha r^0$.

\bex Consider the series $1+\frac12+\frac14+\frac18+\cdots$.
     Find the sum of the first 9 terms.

\underline{Solution}: What we are seeking here is
$S_8=\frac{\alpha\left(1-r^{8+1}\right)}{1-r}$,
where $\alpha=1$ and $r=\frac12$.  Thus
$$S_8=\frac{1\left[1-\left(\frac12\right)^9\right]}{1-\frac12}
     =\frac{1-\frac1{512}}{\frac12}
     =\frac{1-\frac1{512}}{\frac12}\cdot\frac{512}{512}
     =\frac{512-1}{256}=\frac{511}{256}=1.99609375$$
\eex

The formula (\ref{SnForGeometric}) also works when $r<0$.
\bex Consider the series $1-\frac12+\frac14-\frac18+\cdots$.
Find the sum of the first 9 terms.

\underline{Solution}: Again we want $S_8$, but while $\alpha=1$ as before,
here we have $r=-1/2$.

$$S_8=\frac{1\left[1-\left(\frac{-1}2\right)^9\right]}{1-\left(-\frac12\right)}
     =\frac{1-\frac{-1}{512}}{\frac32}
     =\frac{1+\frac1{512}}{\frac32}\cdot\frac{512}{512}
     =\frac{512+1}{256\cdot3}
     =\frac{513}{256\cdot3}
     =\frac{171}{256}=0.66796875.$$

\eex

\bex Suppose one deposits into an account (without interest)
     one penny (\$0.01) on the first
     day of a month, then deposits two pennies (\$0.02)
     the next day, four pennies the next, and so on,
     each day depositing twice what was deposited the
     day before.  How much money is in the account
     after the first week (7 payments),
     second week, third week, and thirty-first
     day?

\underline{Solution}: This is the same as asking for
     partial sums of the series $0.01+0.02+0.04+0.08+\cdots$.
     This is a geoemtric series (\ref{GeneralGeometricSeries}) with
     $\alpha=0.01$ and $r=2$.  Here we have to be careful
     about $N$, since after the first day $N=0$, after the
     second $N=1$, etc.  Now we compute the total deposit after
     \begin{itemize}
     \item 1 week, i.e., 7 days, we have $N=6$ and
        $$S_6=\frac{0.01\left[1-2^7\right]}{1-2}
             =\frac{0.01\left[1-2^7\right]}{-1}
             =0.01\left(2^7-1\right)=0.01(127)=1.27.$$
     \item 2 weeks, i.e., 14 days, we have $N=13$ and (continuing the
           pattern above)
         $$S_{13}=\frac{0.01\left[1-2^{14}\right]}{1-2}
                 =0.01\left(2^{14}-1\right)=0.01(16383)=163.83.$$
     \item 3 weeks, i.e., 21 days, we have $N=20$ and
         $$S_{20}=\cdots=0.01\left(2^{21}-1\right)=0.01(2,097,151)
                 =20,971.51$$
     \item 31 days, so we have $N=30$, and
         $$S_{30}=\cdots=0.01\left(2^{31}-1\right)
                 =0.01(2,147,483,647)=21,474,836.47.$$
      \end{itemize}
\eex
This latest example illustrates that, when $r>1$, the
function $N\mapsto S_N$ is essentially exponential.
Indeed, {\it as a function of} $N$,
$$S_N=\frac{\alpha}{1-r}\left[1-r^{N+1}\right]
=\frac{\alpha}{1-r}+\frac{-\alpha\cdot r^{N+1}}{1-r}
=\frac{\alpha}{1-r}+\left[\frac{\alpha}{r-1}\cdot r\right]
\cdot r^{N}=A+Br^N,$$
where $A=\frac{\alpha}{r-1}$ and $B=\frac{\alpha r}{r-1}$.
Thus as a function of $N$, $S_N$ is basically
a vertical translation of an
exponential growth $Br^N$, assuming again that $r>1$.\footnote{%%%
%%% FOOTNOTE
If $r\in(0,1)$ we get a translation
of exponential decay; if $r\in(-1,0)$ we get a kind of
``damped  oscillation''; if $r=-1$ we get steady oscillation;
and if $r<-1$ we get a growing oscillation.  Details are left to
the reader.
%%% END FOOTNOTE
}
This partially explains why some use the term ``geometric growth''
when referring to exponential growth.

\subsection{Convergence/Divergence in Geometric Series}
Now we look at necessary and sufficient conditions for a
geometric series to converge.  If a given geometric
series does converge, we compute its sum.  Our result
is the following:

\begin{theorem}
For a geometric series $\ds{\sum_{n=0}^\infty \alpha r^n
   =\alpha+\alpha r+\alpha r^2+\alpha r^3+\cdots}$, where $\alpha\ne0$,
\begin{enumerate}
\item the series {\bf converges} if and only if $|r|<1$, i.e., $r\in(-1,1)$;\\
      equivalently, the series {\bf diverges} if and only if $|r|\ge 1$, i.e.,
      $r\in(-\infty,-1]\cup[1,\infty)$.
\item if $|r|<1$, then the series converges to $\ds{\frac{\alpha}{1-r}}$.
\end{enumerate}
\end{theorem}
Restated, the geometric series converges to $\frac{\alpha}{1-r}$ if $|r|<1$,
and diverges otherwise.

\begin{proof} The proof requires some care, as the various cases
contain their own technicalities.
\begin{itemize}
\item Case $r=1$. In such a case, it is not diffiult to see (we just
      count the terms!) that
$$S_N=\sum_{n=0}^N\alpha=(N+1)\alpha\longrightarrow\infty
       \qquad\text{as }N\to\infty.$$
Thus $r=1$ gives a divergent series.
\item Case $r=-1$. In such a case, we have
$$\sum_{n=0}^\infty\alpha(-1)^n=\alpha-\alpha+\alpha-\alpha+\cdots,$$
and so
$$S_N=\left\{\begin{array}{rl}
\alpha,&\qquad\text{if }n\text{ is even,}\\
0,&\qquad\text{if }n\text{ is odd.}\\
\end{array}\right.$$
In other words, $\left\{S_N\right\}_{N=0}^\infty=\alpha,0,\alpha,0,\alpha,0,
   \cdots$, which is clearly a divergent sequence, i.e.,
   the series itself is divergent (by definition).\footnote{%%%
%%% FOOTNOTE
Recall that the convergence of the series is defined by the convergence
of the (sequence of) partial sums.
%%% END FOOTNOTE
}
\item Case $|r|>1$. Here we can use the formula for the partial sums:
$$S_N=\frac{\alpha\left(1-r^{N+1}\right)}{1-r}.$$
Now there is only one term which is not a fixed constant, and so
the convergence of this expression depends upon only the
convergence that,  $r^{N+1}$-term.  Clearly if $r>1$,
this is an exponential growth, and diverges.  For the general case
$|r|>1$, we get that\footnote{%%%
%%% FOOTNOTE
This follows from
continuity of the function $x\mapsto|x|$ giving us the ``$\implies$.''
See Theorem~\ref{ContFunctCarriesSequenceToF(Limit)},
page~\pageref{ContFunctCarriesSequenceToF(Limit)}.
%%% END FOOTNOTE
}
\begin{equation}
r^{N+1}\text{ converges }\implies \left|r^{N+1}\right|\text{ converges }
\iff|r|^{N+1}\text{ converges.}\label{AConvergenceArgument1}\end{equation}
But for $|r|>1$, we have $|r|^{N+1}$ diverges, so with
the contrapositive of (\ref{AConvergenceArgument1}) we have
\begin{align*}
|r|>1&\implies |r|^{N+1}\text{ diverges }\implies
r^{N+1}\text{ diverges } \\
&\implies S_N=\frac{\alpha\left(1-r^{N+1}\right)}{1-r} \text{ diverges }
\iff \sum_{n=0}^\infty\alpha r^n\text{ diverges.}\end{align*}
\item Case $|r|<1$.
Again we look at the variable part of the formula for $S_N$.
It is enough to show that $|r|<1\implies r^{N+1}$ converges.
One method is to use the sandwich theorem. In the
argument below, note that $|r|<1\implies|r|\in(0,1)
\implies |r|^{N+1}\longrightarrow0$.  The relevant
sandwich theorem application is then (as $N\to\infty$):\footnote{%
%%% FOOTNOTE
Recall that for any $x\in\Re$, we have $-|x|\le x\le |x|$.
%%% END FOONOTE
}
\begin{center}
\begin{pspicture}(0,2)(8.7,4.5)
\rput[l](0,4){$\underbrace{-|r|^{N+1}}=-\left|r^{N+1}\right|$}
  \rput[l](3.8,4){$\le$}
\rput[l](4.6,4){$r^{N+1}$}
  \rput[l](6.2,4){$\le$}
\rput[l](7,4){$\left|r^{N+1}\right|=\underbrace{|r|^{N+1}}$}
\psline{->}(.63,3.5)(.63,2.3)
\psline{->}(8.99,3.5)(8.99,2.3)
 \rput(.63,2){0}
 \rput(9,2){0}
\end{pspicture}
\end{center}
Thus $|r|<1\implies r^{N+1}\longrightarrow 0$ as $N\to\infty$.
We can conclude that
$$|r|<1\implies
S_N=\frac{\alpha\left(1-r^{N+1}\right)}{1-r}
   \longrightarrow \frac{\alpha(1-0)}{1-r}=\frac{\alpha}{1-r}
\text{ (as }N\to\infty).$$
\end{itemize}
This completes the proof.
\end{proof}
The implication above is worth repeating in a summarized form:
\begin{equation}
|r|<1\implies\sum_{n=0}^\infty\alpha r^n=\frac{\alpha}{1-r}.
\label{|r|<1ImpliesSumOfGeometricSeries}\end{equation}
\bex Here are some series computations using
   the theorem and (\ref{|r|<1ImpliesSumOfGeometricSeries}).
\begin{itemize}
\item $\ds{\sum_{n=0}^\infty 2\left(\frac13\right)^n
   =\frac2{1-\frac13}=\frac2{\frac23}=2\cdot\frac32=3}$.
   \qquad($\alpha=2$, $r=\frac13$.)
\item $\ds{\sum_{n=0}^\infty0.99^n=\frac1{1-0.99}=\frac1{.01}=100}$.
   \qquad($\alpha=1$, $r=0.99$.)
\item $\ds{\sum_{n=0}^\infty 1.01^n}$ diverges.\qquad
   ($\alpha=1$, $r=1.01$ so $|r|>1$, and the series diverges.)
\item $\ds{\sum_{n=2}^\infty\frac1{3^n}=\frac{\frac19}{1-\frac13}
    =\frac{\frac19}{\frac23}=\frac19\cdot\frac32=\frac16}$.
    \qquad(First term is $\alpha=\frac19$, $r=\frac13$.)
\item $\ds{1-\frac12+\frac14-\frac18+\frac{1}{16}-\cdots
       =\frac1{1-\left(\frac{-1}2\right)}=\frac1{\frac32}=\frac23}$.
     \qquad($\alpha=1$, $r=-\frac12$.)
\item $\ds{\sum_{n=1}^\infty e^{-n}=\frac{e}{1-\frac1e}
           =\frac{e}{1-\frac1e}\cdot\frac{e}e=\frac{e^2}{e-1}}$.
     \qquad($\alpha=e$, $r=\frac1e$.)
\item $\ds{\sum_{n=1}^\infty\frac5{3^{2n}}
       =\sum_{n=1}^\infty\frac5{9^n}=\frac{5/9}{1-\frac19}
       =\frac{5/9}{8/9}=\frac58.}$
       \qquad($\alpha=\frac59$, $r=\frac19.$)
\end{itemize}

\eex

\newpage
\begin{center}
\underline{\Large{\bf Exercises}}\end{center}

\begin{multicols}{2}

\begin{enumerate}
\item Show that the following series can be
      written as a telescoping series, and discuss
      its convergence: $$\sum_{n=1}^{\infty}\ln\left(\frac{n}{n+1}\right).$$
\item Give an alternative proof of the formula (\ref{SnForGeometric})
      for the partial sums of geoemtric series.
      For this new proof, begin with the formula for $S_N$ as
      in the original proof (page~\pageref{SnForGeometric}),
      and then multiply by $(1-r)$, noting how the right-hand side
      simplifies.  (See also page~\pageref{aN-bN})

\end{enumerate}
\vfill
\end{multicols}
\eject

\section{Elementary Test for Divergence}
Because it is the exceptional case (e.g., geometric, telescoping)
that we can actually find a compact
formula for $S_N$, we have to develop other tests for the convergence
or divergence of series.  There will be several such tests, and
which particular test or tests are expeditious and conclusive
will vary from series to series.  We explore the first of those
tests in this section.  We start with a theorem.

\begin{theorem} Given a series $\ds{\sum_{n=k}^\infty a_n}$.
If the series converges, then $a_n\longrightarrow0$ as $n\to\infty$:
\begin{equation}
\left[\sum_{n=k}^\infty a_n\text{ converges }\right]\implies
\left[\lim_{n\to\infty}a_n=0\right].
\label{SeriesConvergesImpliesTermsShrinkToZero}
\end{equation}
\label{SeriesConvergesImpliesTermsShrinkToZeroTheorem}\end{theorem}
It should be intuitively clear that, if we are going to ``add up''
infinitely many terms, and have the sums approach a finite number,
then the
terms we are adding are going to have to shrink to zero.
The proof uses the fact
that $S_n\to L\implies S_{n-1}\to L$, the latter limit occurring
``one step behind'' the former, but occurring nonetheless
since $n\to\infty\implies n-1\to\infty\implies S_{n-1}\to L$.
Now to the proof.

\begin{proof}
Suppose $\ds{\sum_{n=k}^\infty a_n}$ converges, i.e.,
$\ds{\sum_{n=k}^\infty a_n=L}$ for some $L\in\Re$
(and in particular $L$ is finite).
Then by definition
$$S_N\longrightarrow L\text{ as }N\to\infty.$$
Now recall $S_{n}=a_n+S_{n-1}$, so that $a_n=S_n-S_{n-1}$.
Taking $n\to\infty$ we get
$$a_n=S_n-S_{n-1}\longrightarrow L-L=0,\qquad\text{q.e.d.}$$
\end{proof}
Note that it was important that $L$ be finite in the limit
computation above.  For instance, if $S_n\to\infty$, we
would have $a_n=S_n-S_{n-1}$ giving $\infty-\infty$-form
(which is indeterminant) as
$n\to\infty$.

Again, the intuition behind the theorem is that, in order to
be able to add infinitely many terms---one at a time in the
sense that we compute $S_k$, $S_{k+1}$, $S_{k+1}$, etc.,
and look for a trend towards $L$---the terms that we add,
i.e., $a_k$, $a_{k+1}$, $a_{k+2}$, etc., have to shrink
if the partial sums are to approach a finite number $L$.
In fact, the form of the theorem which we use is the
contrapositive.  Recall
the logical equivalence $P\longrightarrow Q
\iff(\sim Q)\longrightarrow(\sim P)$.\footnote{%%%
%%% FOOTNOTE
To be sure, here $P\longrightarrow Q$ is read,
``$P$ implies $Q$.''  The symbol ``$\sim$'' is still
the ``not,'' or logical negation, operator.
The symbol ``$\iff$'' stands in for logical equivalence.
%%% END FOOTNOTE
}
In this case, $P$ is the statement that the series converges (to
a finite number $L$), while $Q$ is the statement that $a_n\to0$.
The contrapositive for of
Theorem~\ref{SeriesConvergesImpliesTermsShrinkToZeroTheorem}
is our main result in this section, and we dub that result
{\it Elementary Test for Divergence}, or ETFD:
\begin{theorem} {\bf (ETFD)}
If it is {\bf not} the case that $a_n\to0$, then $\ds{\sum_{n=k}^\infty a_n}$
diverges.
Put symbolically,
\begin{equation}
a_n\not{\!\!\to0}\implies\sum_{n=k}^\infty a_n\text{ diverges.}
\label{ETFDequation}
\end{equation}
\label{ETFD}\end{theorem}

\begin{proof} It is enough to say that this is the contrapositive of
Theorem~\ref{SeriesConvergesImpliesTermsShrinkToZeroTheorem},
and therefore also true.
One way to write this symbolically is the following.
$$\underbrace{\sum_{n=k}^\infty a_n\text{ converges }
\longrightarrow \left(a_n\to 0\right)}_{\text{always true by
Theorem~\ref{SeriesConvergesImpliesTermsShrinkToZeroTheorem}}}
\iff
\underbrace{
\left[\sim\left(a_n\to 0\right)\right]
\longrightarrow
\sum_{n=k}^\infty a_n\text{ diverges}}_{\text{statement of
Theorem~\ref{ETFD}}}.$$
The statement on the right must be always true (a tautology),
since it is equivalent to the statement on the left,
which---being the statement of
Theorem~\ref{SeriesConvergesImpliesTermsShrinkToZeroTheorem}---is
itself a tautology.  The statement on the right being a
tautology means that it stands alone, with ``$[\ ]\implies\sum$''
replacing ``$[\ ]\longrightarrow\sum$,'' q.e.d.
\end{proof}
This theorem is undoubtedly one of
the most misunderstood and misapplied results
in all of Calculus I and II.  It is as important to understand
what it does not say, as it is to understand what it says.
The theorem says that if the terms of a series do not
shrink to zero, then the series must diverge.  This is
``elementary,'' hence the name ``Elementary Test for Divergence.\footnotemark''
\footnotetext{%%
%%% FOOTNOTE
Other texts call this the {\it $n$th term test}, or
the {\it $n$th term test for divergence}, for reasons which should be
clear.  The author is not aware if any other text calls
it the elementary test for divergence.  However, both names
are descriptive enough that one is not likely to be misunderstood
in conversation or written reports if using either name for
this result.}
%%% END FOOTNOTE

But it is not as comprehensive as one might think.  Afterall,
if the terms of a series {\it do} shrink to zero, the
theorem is silent!  (Therein lies the unfortunately
very common mistake made by Calculus Students.)
To emphasize this we look at the following examples.

\bex Discuss what Theorem~\ref{ETFD} has to say about
the series \newline
(a) $\ds{\sum_{n=1}^\infty\frac{n}{n+1}}$;\hfill
(b) $\ds{\sum_{n=1}^\infty\frac1{n+1}}$\hfill
(c) $\ds{\sum_{n=1}^\infty\cos\frac1{n}}$;\hfill
(d) $\ds{\sum_{n=1}^\infty\sin\frac1{n}}$;\hfill
(e) $\ds{\sum_{n=1}^\infty\sin\frac1{n^2}}$.

\underline{Solution}:
\begin{enumerate}[(a)]
\item $\ds{\frac{n}{n+1}\longrightarrow1\ne0
        \overset{\text{\rm ETFD}}{\implies}
           \sum_{n=1}^\infty \frac{n}{n+1}}$ diverges.
\item $\ds{\frac1{n+1}\longrightarrow}$.  The
      ETFD is inconclusive.
\item $\ds{\cos\frac1n\longrightarrow\cos0=1\ne0
        \overset{\text{\rm ETFD}}{\implies}
        \sum_{n=1}^\infty\cos\frac1{n}}$ diverges.
\item $\ds{\sin\frac1n\longrightarrow\sin0=0}$.
       The ETFD is inconclusive.
\item $\ds{\sin\frac1{n^2}\longrightarrow\sin0=0}$.
       The ETFD is inconclusive.
\end{enumerate}
\eex

Looking closely at the symbolic statement of ETFD given
in (\ref{ETFDequation}), we see that there is never an
implication of convergence.  Indeed, the test either concludes
divergence, or is inconclusive.  This is a very quick but incomplete test,
which can only detect divergence in certain (still common) circumstances,
namely that $a_n\not{\!\!\to}0$.

Indeed, in (a) and (c) above, ETFD gave us divergence.
However, it said nothing in (b), (d) and (e), as the
``if'' part of the theorem was not true.  In fact,
of these three in which ETFD is silent---(b), (d) and (e)---it
turns out that $(b)$ and $(d)$ are divergent, while
$(e)$ is convergent.  The methods to see this are introduced
in later sections.  An example from the exercises
in the last section gives a case where we can in fact
show that it is possible that $a_n\to0$, but the series diverges:
\bex Consider the series
$\ds{\sum_{n=1}^\infty\ln\left(\frac{n}{n+1}\right)}$.
Determine if it converges or diverges.

\underline{Solution}: First we note ${\ln\left(\frac{n}{n+1}\right)
\longrightarrow\ln1=0}$ as $n\to\infty$.  Thus ETFD is inconclusive.\footnote{
%%% FOOTNOTE
Some textbooks would write that the test ``fails.''
That seems a bit strong.  It is merely inconclusive, so we need
to look deeper at the particular series and perhaps employ some other test
which will be conclusive.}
%%% END FOOTNOTE

Looking closer at this series, we should eventually notice that
it is telescoping.  This becomes clear
if we rewrite it:
$$\sum_{n=1}^\infty\left[\ln(n)-\ln(n+1)\right]
  =\left[\ln 1-\ln 2\right]+\left[\ln-\ln3\right]
    +\left[\ln 3-\ln 4\right]+\cdots.$$
It is not difficult to see that $S_N=\ln1-\ln N=-\ln N$,
and thus
$$S_N=-\ln N\longrightarrow-\infty\text{ as }N\to\infty.$$
Since the partial sums diverge, by definition so does the series.
\eex

On the other hand, there are series which converge.  However the ETFD
is not powerful enough to ever prove it.  Rather than risking confusion
by elaborating this point further, we will end the discussion here.

\newpage\section{The Integral Test for Convergence/Divergence}


\documentclass[a4paper,11pt]{amsbook}
\usepackage{amsmath}
\usepackage{../HBSuerDemir}
\usepackage{fullpage}
\pagestyle{headings}
\usepackage{fancyhdr}

\begin{document}
\hPage{b1p2/250}

\vspace*{10mm}
$$
\begin{vmatrix}
11-x & -6&2 \\
-6 & 10-x & -4\\
2& -4 & 6-x
\end{vmatrix}
=0
$$
\\

is 6,and find the other two roots.

is 6,and find the other two roots.\\
\\
17. If n$\varepsilon$N, show that (b-c)(c-a)(a-b) is a factor of\\
$$
\begin{vmatrix}
1&1&1\\
a&b&c\\
a^{n+2}&b^{n+2}&c^{n+2}
\end{vmatrix}
$$
\\
18.If, $\alpha+\beta+\gamma$=2s, prove\\

$
\begin{vmatrix}
1&\cos\gamma&\cos\beta\\
\cos\gamma&1&\cos\alpha\\
\cos\beta&\cos\alpha&1
\end{vmatrix}
=4\sin{s}\sin{(s-\alpha)}\sin{(s-\beta)}\sin{(s-\gamma)}
$
\\
\\
\\
19.Show that
\\

$
\begin{vmatrix}
1&\cos{x}-\sin{x}&\cos{x}+\sin{x}\\
1&\cos{y}-\sin{y}&\cos{y}+\sin{y}\\
1&\cos{z}-\sin{z}&\cos{z}+\sin{z}
\end{vmatrix}
=2
\begin{vmatrix}
1&\cos{x}&\sin{x}\\
1&\cos{y}&\sin{y}\\
1&\cos{z}&\sin{z}
\end{vmatrix}
$
\\
\\
\\
20.Evaluate the following determinants:
\\

a)
$
\begin{vmatrix}
0&0&1&2&3\\
0&0&4&5&6\\
-1&-4&0&7&8\\
-2&-5&-7&0&9\\
-3&-6&-8&-9&0
\end{vmatrix}
$
\hspace{30mm} b)
$
\begin{vmatrix}
0&0&a&b\\
0&0&c&d\\
-a&-c&0&0\\
-b&-d&0&0
\end{vmatrix}
$
\\

(See Example 7 and 8.)\\ \\
\section*{ANSWERS TO EVEN NUMBERED EXERCISES}
\mbox{ }\\
2. a) $adf+2bce-ae^2-dc^2-fb$ \hspace{20mm} b)-2\\ \\
4. a)$(a^3-1)^2$, \hspace{20mm} b)-575\\ \\
8. 0\\ \\
10. ${(a-b)^2}(2x-a-b)(a+b+2x)$\\
\end{document}


\documentclass{article}  %Need this.

\usepackage{amsmath,amsthm,amssymb}

%\usepackage[margin=1in]{geometry}

\newtheorem*{thm}{Theorem}
\newtheorem*{cnj}{Conjecture}
\newtheorem*{lem}{Lemma}
\newtheorem*{cor}{Corollary}
\newtheorem*{prop}{Proposition}

\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}

\title{Reproducing a \LaTeX Document}
\author{Put your name here}
\date{}

\begin{document}
\maketitle

\begin{thm}
If $x$ and $y$ are odd integers then $x+y$ is an even integer.
\end{thm}

\begin{proof}
We assume that $x$ and $y$ are odd integers and will prove that $x+y$ is an even integer. Since $x$ and $y$ are odd, there exist integers $m$ and $n$ such that $x=2m+1$ and $y=2n+1$.  By substitution and algebra we obtain
\begin{align*}
x+y &= 2m+1 + 2n + 1\\
&= 2m+2n+2\\
&=2(m+n+1).
\end{align*}
Define $q=m+n+1$.  Since $m$ and $n$ are integers and the integers are closed under addition, we conclude that $q$ is an integer. Since $x+y=2q$ for the integer $q$ we conclude that $x+y$ is an even integer.
\end{proof}

\vspace{.15in}

\section*{Challenge Typing}

Suppose that $f:(-1,1)\to \R$ and $f$ is differentiable at $0$. Let sequences $(\alpha_n)_{n\geq1}$ and $(\beta_n)_{n\geq1}$ satisify $-1<\alpha_n<\beta_n<1$ for all $n\geq 1$ and $\displaystyle{\lim_{n\to\infty}} \alpha_n = \displaystyle{\lim_{n\to\infty}} \beta_n = 0$. Set

		$$\lambda_n = \frac{f(\beta_n) -f(\alpha_n)}{\beta_n-\alpha_n} . $$

\end{document}


\title{Teaching the Geometry of Schemes}
\titlerunning{Teaching the Geometry of Schemes}
\toctitle{Teaching the Geometry of Schemes}

\author{Gregory G.~Smith \and Bernd Sturmfels}
\authorrunning{G. G. Smith and B. Sturmfels}
% \institute{Department of Mathematics, University of California,
% Berkeley, California 94720, USA}

\maketitle

%%----------------------------------------------------------
\newtheorem*{problem*}{Problem}{\bfseries\upshape}{\itshape}
\newtheorem*{solution*}{Solution}{\itshape}{\rmfamily}

\newcommand{\Spec}{\operatorname{Spec}}
\newcommand{\Proj}{\operatorname{Proj}}
\newcommand{\codim}{\operatorname{codim}}
%%----------------------------------------------------------

\begin{abstract}
This chapter presents a collection of graduate level problems in
algebraic geometry illustrating the power of \Mtwo as an educational
tool.
\end{abstract}

When teaching an advanced subject, like the language of schemes, we
think it is important to provide plenty of concrete instances of the
theory.  Computer algebra systems, such as \Mtwo, provide students
with an invaluable tool for studying complicated examples.
Furthermore, we believe that the explicit nature of a computational
approach leads to a better understanding of the objects being
examined.  This chapter presents some problems which we feel
illustrate this point of view.

Our examples are selected from the homework of an algebraic geometry
class given at the University of California at Berkeley in the fall of 1999.
This graduate course was taught by the second author with assistance from the
first author.  Our choice of problems, as the title suggests, follows the
material in David Eisenbud and Joe Harris' textbook {\em The Geometry of
  Schemes} \cite{SC:EH}.

%%----------------------------------------------------------
\section{Distinguished Open Sets}

We begin with a simple example involving the Zariski topology of an affine
scheme\index{scheme!affine}. This example also indicates some of the
subtleties involved in working with arithmetic
schemes\index{scheme!arithmetic}.

\begin{problem*}
Let $S = \bbbz[x,y,z]$ and $X = \Spec(S)$.  If $f = x$ and $X_{f}$ is
the corresponding basic open subset in $X$, then establish the
following:
\begin{enumerate}
\item[$(1)$] If $e_{1} = x+y+z$, $e_{2} = xy+xz+yz$ and $e_{3} = xyz$
are the elementary symmetric functions then the set $\{X_{e_{i}}\}_{1
\leq i \leq 3}$ is an open cover of $X_{f}$.
\item[$(2)$] If $p_{1} = x+y+z$, $p_{2} = x^{2}+y^{2}+z^{2}$ and $p_{3}
= x^{3}+y^{3}+z^{3}$ are the power sum symmetric functions then
$\{X_{p_{i}}\}_{1 \leq i \leq 3}$ is {\em not} an open cover of
$X_{f}$.
\end{enumerate}
\end{problem*}

\begin{solution*}
$(1)$ To prove that $\{X_{e_{i}}\}_{1 \leq i \leq 3}$ is an open cover
of $X_{f}$, it suffices to show that $e_{1}$, $e_{2}$ and $e_{3}$
generate the unit ideal in $S_{f}$; see Lemma I-16 in Eisenbud and
Harris~\cite{SC:EH}.  This is equivalent to showing that $x^{m}$
belongs to the $S$-ideal $\langle e_{1}, e_{2}, e_{3} \rangle$ for
some $m \in \bbbn$.  In other words, the saturation\index{saturation}
$\big( \langle e_{1}, e_{2}, e_{3} \rangle : x^{\infty} \big)$ is the
unit ideal if and only if $\{X_{e_{i}}\}_{1 \leq i \leq 3}$ is an open
cover of $X_{f}$.  We verify this in \Mtwo as follows:
\beginOutput
i1 : S = ZZ[x, y, z];\\
\endOutput
\beginOutput
i2 : elementaryBasis = ideal(x+y+z, x*y+x*z+y*z, x*y*z);\\
\emptyLine
o2 : Ideal of S\\
\endOutput
\beginOutput
i3 : saturate(elementaryBasis, x)\\
\emptyLine
o3 = ideal 1\\
\emptyLine
o3 : Ideal of S\\
\endOutput
$(2)$ Similarly, to show that $\{X_{p_{i}}\}_{1 \leq i \leq 3}$ is not
an open cover of $X_{f}$, we prove that $\big( \langle p_{1}, p_{2},
p_{3} \rangle : x^{\infty} \big)$ is not the unit ideal.  Calculating
this saturation, we find
\beginOutput
i4 : powerSumBasis = ideal(x+y+z, x^2+y^2+z^2, x^3+y^3+z^3);\\
\emptyLine
o4 : Ideal of S\\
\endOutput
\beginOutput
i5 : saturate(powerSumBasis, x)\\
\emptyLine
\                            2            2\\
o5 = ideal (6, x + y + z, 2y  + 2y*z + 2z , 3y*z)\\
\emptyLine
o5 : Ideal of S\\
\endOutput
\beginOutput
i6 : clearAll\\
\endOutput
which is not the unit ideal.\qed
\end{solution*}

The fact that $6$ is a generator of the ideal $\big( \langle p_{1},
p_{2}, p_{3} \rangle : x^{\infty} \big)$ indicates that
$\{X_{p_{i}}\}_{1 \leq i \leq 3}$ does not contain the points in $X$
lying over the points $\langle 2 \rangle$ and $\langle 3 \rangle$ in
$\Spec(\bbbz)$.  If we work over a base ring in which $6$ is a unit,
then $\{X_{p_{i}}\}_{1 \leq i \leq 3}$ would, in fact, be an open
cover of $X_{f}$.

%%----------------------------------------------------------
\section{Irreducibility}

The study of complex semisimple Lie algebras gives rise to an
important family of algebraic varieties called nilpotent
orbits\index{nilpotent orbits}.  The next problem examines the
irreducibility\index{scheme!irreducible} of a particular nilpotent
orbit.

\begin{problem*}
Let $X$ be the set of nilpotent complex $3 \times 3$ matrices.  Show
that $X$ is an irreducible algebraic variety.
\end{problem*}

\begin{solution*}
A $3 \times 3$ matrix $M$ is nilpotent if and only if its minimal
polynomial $p(\sf T)$ equals ${\sf T}^{k}$, for some $k \in \bbbn$.
Since each irreducible factor of the characteristic polynomial of $M$
is also a factor of $p(\sf T)$, it follows that the characteristic
polynomial of $M$ is ${\sf T}^{3}$.  We conclude that the coefficients
of the characteristic polynomial of a generic $3 \times 3$ matrix
define the algebraic variety $X$.

To prove that $X$ is irreducible over $\bbbc$, we construct a rational
parameterization\index{rational parameterization}.  First, observe
that ${\rm GL}_{3}(\bbbc)$ acts on $X$ by conjugation.  Jordan's
canonical form theorem implies that there are exactly three orbits;
one for each of the following matrices:
\[
N_{(1,1,1)} =\left[ \begin{smallmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 &
0 & 0 \end{smallmatrix} \right], \quad
N_{(2,1)} = \left[ \begin{smallmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0
& 0 \end{smallmatrix} \right] \text{ and }
N_{(3)} = \left[ \begin{smallmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 &
0 \end{smallmatrix} \right] \enspace .
\]
Each orbit is defined by a rational parameterization, so it suffices
to show that the closure of the orbit containing $N_{(3)}$ is the
entire variety $X$.  We demonstrate this as follows:
\beginOutput
i7 : S = QQ[t, y_0 .. y_8, a..i, MonomialOrder => Eliminate 10];\\
\endOutput
\beginOutput
i8 : N3 = (matrix \{\{0,1,0\},\{0,0,1\},\{0,0,0\}\}) ** S\\
\emptyLine
o8 = | 0 1 0 |\\
\     | 0 0 1 |\\
\     | 0 0 0 |\\
\emptyLine
\             3       3\\
o8 : Matrix S  <--- S\\
\endOutput
\beginOutput
i9 : G = genericMatrix(S, y_0, 3, 3)\\
\emptyLine
o9 = | y_0 y_3 y_6 |\\
\     | y_1 y_4 y_7 |\\
\     | y_2 y_5 y_8 |\\
\emptyLine
\             3       3\\
o9 : Matrix S  <--- S\\
\endOutput
To determine the entries in $G \cdot N_{(3)} \cdot G^{-1}$, we use the
classical adjoint\index{classical adjoint} to construct the matrix
$\det(G) \cdot G^{-1}$.
\beginOutput
i10 : classicalAdjoint = (G) -> (\\
\           n := degree target G;\\
\           m := degree source G;\\
\           matrix table(n, n, (i, j) -> (-1)^(i+j) * det(\\
\                     submatrix(G, \{0..j-1, j+1..n-1\}, \\
\                          \{0..i-1, i+1..m-1\}))));\\
\endOutput
\beginOutput
i11 : num = G * N3 * classicalAdjoint(G);\\
\emptyLine
\              3       3\\
o11 : Matrix S  <--- S\\
\endOutput
\beginOutput
i12 : D = det(G);\\
\endOutput
\beginOutput
i13 : M = genericMatrix(S, a, 3, 3);\\
\emptyLine
\              3       3\\
o13 : Matrix S  <--- S\\
\endOutput
The entries in $G \cdot N_{(3)} \cdot G^{-1}$ give a rational
parameterization of the orbit generated by $N_{(3)}$.  Using
elimination theory\index{elimination theory} --- see section~3.3 in
Cox, Little and O`Shea~\cite{SC:CLO} --- we give an ``implicit
representation'' of this variety.
\beginOutput
i14 : elimIdeal = minors(1, (D*id_(S^3))*M - num) + ideal(1-D*t);\\
\emptyLine
o14 : Ideal of S\\
\endOutput
\beginOutput
i15 : closureOfOrbit = ideal selectInSubring(1, gens gb elimIdeal);\\
\emptyLine
o15 : Ideal of S\\
\endOutput

Finally, we verify that this orbit closure equals $X$
scheme-theoretically.  Recall that $X$ is defined by the coefficients
of the characteristic polynomial of a generic $3 \times 3$ matrix {\tt
M}.
%% was X = ideal submatrix( (coeff-icients({0}, det(M - t*id_(S^3))))_1,
%% {1,2,3} ), but 'coeff-icients' is to be redesigned, and 'contract' is
%% more self-explanatory, anyway.
\beginOutput
i16 : X = ideal substitute(\\
\              contract(matrix\{\{t^2,t,1\}\}, det(t-M)),\\
\              \{t => 0_S\})\\
\emptyLine
o16 = ideal (- a - e - i, - b*d + a*e - c*g - f*h + a*i + e*i, c*e*g - $\cdot\cdot\cdot$\\
\emptyLine
o16 : Ideal of S\\
\endOutput
\beginOutput
i17 : closureOfOrbit == X\\
\emptyLine
o17 = true\\
\endOutput
\beginOutput
i18 : clearAll\\
\endOutput
This completes our solution.\qed
\end{solution*}

More generally, Kostant shows that the set of all nilpotent elements
in a complex semisimple Lie algebra\index{Lie algebra} form an
irreducible variety.  We refer the reader to Chriss and
Ginzburg~\cite{SC:CV} for a proof of this result (Corollary~3.2.8) and
a discussion of its applications in representation theory.

%%----------------------------------------------------------
\section{Singular Points}

In our third question, we study the singular locus\index{singular
locus} of a family of elliptic curves\index{elliptic curve}.

\begin{problem*}
Consider a general form of degree $3$ in $\bbbq[x,y,z]$:
\[
F = ax^{3} + bx^{2}y + cx^{2}z + dxy^{2} + exyz + fxz^{2} + gy^{3} +
hy^{2}z + iyz^{2} + jz^{3} \enspace .
\]
Give necessary and sufficient conditions in terms of $a, \ldots, j$
for the cubic curve $\Proj\big( \bbbq[x,y,z] / \langle F \rangle
\big)$ to have a singular point.
\end{problem*}

\begin{solution*}
The singular locus of $F$ is defined by a polynomial of degree $12$ in
the $10$ variables $a, \dotsc, j$.  We calculate this polynomial in two
different ways.

Our first method is an elementary but time consuming elimination.
Carrying it out in \Mtwo, we have
\beginOutput
i19 : S = QQ[x, y, z, a..j, MonomialOrder => Eliminate 2];\\
\endOutput
\beginOutput
i20 : F = a*x^3+b*x^2*y+c*x^2*z+d*x*y^2+e*x*y*z+f*x*z^2+g*y^3+h*y^2*z+\\
\                   i*y*z^2+j*z^3;\\
\endOutput
\beginOutput
i21 : partials = submatrix(jacobian matrix\{\{F\}\}, \{0..2\}, \{0\})\\
\emptyLine
o21 = \{1\} | 3x2a+2xyb+y2d+2xzc+yze+z2f |\\
\      \{1\} | x2b+2xyd+3y2g+xze+2yzh+z2i |\\
\      \{1\} | x2c+xye+y2h+2xzf+2yzi+3z2j |\\
\emptyLine
\              3       1\\
o21 : Matrix S  <--- S\\
\endOutput
\beginOutput
i22 : singularities = ideal(partials) + ideal(F);\\
\emptyLine
o22 : Ideal of S\\
\endOutput
\beginOutput
i23 : elimDiscr = time ideal selectInSubring(1,gens gb singularities);\\
\     -- used 64.27 seconds\\
\emptyLine
o23 : Ideal of S\\
\endOutput
\beginOutput
i24 : elimDiscr = substitute(elimDiscr, \{z => 1\});\\
\emptyLine
o24 : Ideal of S\\
\endOutput
On the other hand, there is also an elegant and more useful
determinantal formula for this discriminant\index{discriminant}; it is
a specialization of the formula (2.8) in section~3.2 of Cox, Little
and O`Shea~\cite{SC:CLO2}.  To apply this determinantal formula, we
first create the coefficient matrix {\tt A} of the partial derivatives
of $F$.
%% was A = (coeff-icients({0,1,2}, submatrix(jacobian matrix{{F}}, {0..2}, {0})))_1;
%% but 'coeff-icients' is deprecated.
\beginOutput
i25 : A = contract(matrix\{\{x^2,x*y,y^2,x*z,y*z,z^2\}\},\\
\              diff(transpose matrix\{\{x,y,z\}\},F))\\
\emptyLine
o25 = \{1\} | 3a 2b d  2c e  f  |\\
\      \{1\} | b  2d 3g e  2h i  |\\
\      \{1\} | c  e  h  2f 2i 3j |\\
\emptyLine
\              3       6\\
o25 : Matrix S  <--- S\\
\endOutput
We also construct the coefficient matrix {\tt B} of the partial
derivatives of the Hessian\index{hessian} of $F$.
\beginOutput
i26 : hess = det submatrix(jacobian ideal partials, \{0..2\}, \{0..2\});\\
\endOutput
%% was B = (coeff-icients({0,1,2}, submatrix(jacobian matrix{{hess}}, {0..2}, {0})))_1;
%% but 'coeff-icients' is deprecated.
\beginOutput
i27 : B = contract(matrix\{\{x^2,x*y,y^2,x*z,y*z,z^2\}\},\\
\              diff(transpose matrix\{\{x,y,z\}\},hess))\\
\emptyLine
o27 = \{1\} | -24c2d+24bce-18ae2-24b2f+72adf               4be2-16bdf-48 $\cdot\cdot\cdot$\\
\      \{1\} | 2be2-8bdf-24c2g+72afg+16bch-24aeh-8b2i+24adi 4de2-16d2f-48 $\cdot\cdot\cdot$\\
\      \{1\} | 2ce2-8cdf-8c2h+24afh+16bci-24aei-24b2j+72adj 2e3-8def-24cf $\cdot\cdot\cdot$\\
\emptyLine
\              3       6\\
o27 : Matrix S  <--- S\\
\endOutput
To obtain the discriminant, we combine these two matrices and take the
determinant.
\beginOutput
i28 : detDiscr = ideal det (A || B);\\
\emptyLine
o28 : Ideal of S\\
\endOutput
Finally, we check that our two discriminants are equal
\beginOutput
i29 : detDiscr == elimDiscr\\
\emptyLine
o29 = true\\
\endOutput
and examine the generator.
\beginOutput
i30 : detDiscr_0\\
\emptyLine
\            2   4 3 2             5 3 2           6 3 2          2 2 2 $\cdot\cdot\cdot$\\
o30 = 13824c d*e f g  - 13824b*c*e f g  + 13824a*e f g  - 110592c d e  $\cdot\cdot\cdot$\\
\emptyLine
o30 : S\\
\endOutput
\beginOutput
i31 : numgens detDiscr\\
\emptyLine
o31 = 1\\
\endOutput
\beginOutput
i32 : # terms detDiscr_0\\
\emptyLine
o32 = 2040\\
\endOutput
\beginOutput
i33 : clearAll\\
\endOutput
Hence, the singular locus is given by a single polynomial of degree
$12$ with $2040$ terms.\qed
\end{solution*}

For a further discussion of singularities and discriminants see
Section~V.3 in Eisenbud and Harris~\cite{SC:EH}.  For information on
resultants and discriminants see Chapter~2 in Cox, Little and
O`Shea~\cite{SC:CLO2}.

%%----------------------------------------------------------
\section{Fields of Definition}

Schemes\index{scheme!over a number field} over non-algebraically
closed fields arise in number theory.  Our fourth problem looks at one
technique for working with number fields in \Mtwo.

\begin{problem*}[Exercise~II-6 in  \cite{SC:EH}]
An inclusion of fields $K \hookrightarrow L$ induces a map
$\mathbb{A}_{L}^{n} \to \mathbb{A}_{K}^{n}$.  Find the images in
$\mathbb{A}_{\bbbq}^{2}$ of the following points of
$\mathbb{A}_{\overline{\bbbq}}^{2}$ under this map.
\begin{enumerate}
\item[$(1)$] $\langle x - \sqrt{2}, y - \sqrt{2} \rangle ;$
\item[$(2)$] $\langle x - \sqrt{2}, y - \sqrt{3} \rangle ;$
\item[$(3)$] $\langle x - \zeta, y - \zeta^{-1} \rangle$ where $\zeta$
is a $5$-th root of unity $;$
\item[$(4)$] $\langle \sqrt{2}x- \sqrt{3}y \rangle ;$
\item[$(5)$] $\langle \sqrt{2}x- \sqrt{3}y-1 \rangle$.
\end{enumerate}
\end{problem*}

\begin{solution*}
The images can be determined by using the following three step
algorithm: (1) replace the coefficients not contained in $K$ with
indeterminates, (2) add the minimal polynomials of these coefficients
to the given ideal in $\mathbb{A}_{L}^{2}$, and (3) eliminate the new
indeterminates.  Here are the five examples:
\beginOutput
i34 : S = QQ[a,b,x,y, MonomialOrder => Eliminate 2];\\
\endOutput
\beginOutput
i35 : I1 = ideal(x-a, y-a, a^2-2);\\
\emptyLine
o35 : Ideal of S\\
\endOutput
\beginOutput
i36 : ideal selectInSubring(1, gens gb I1)\\
\emptyLine
\                     2\\
o36 = ideal (x - y, y  - 2)\\
\emptyLine
o36 : Ideal of S\\
\endOutput
\beginOutput
i37 : I2 = ideal(x-a, y-b, a^2-2, b^2-3);\\
\emptyLine
o37 : Ideal of S\\
\endOutput
\beginOutput
i38 : ideal selectInSubring(1, gens gb I2)\\
\emptyLine
\              2       2\\
o38 = ideal (y  - 3, x  - 2)\\
\emptyLine
o38 : Ideal of S\\
\endOutput
\beginOutput
i39 : I3 = ideal(x-a, y-a^4, a^4+a^3+a^2+a+1);\\
\emptyLine
o39 : Ideal of S\\
\endOutput
\beginOutput
i40 : ideal selectInSubring(1, gens gb I3)\\
\emptyLine
\                       2    2               3    2\\
o40 = ideal (x*y - 1, x  + y  + x + y + 1, y  + y  + x + y + 1)\\
\emptyLine
o40 : Ideal of S\\
\endOutput
\beginOutput
i41 : I4 = ideal(a*x+b*y, a^2-2, b^2-3);\\
\emptyLine
o41 : Ideal of S\\
\endOutput
\beginOutput
i42 : ideal selectInSubring(1, gens gb I4)\\
\emptyLine
\             2   3  2\\
o42 = ideal(x  - -*y )\\
\                 2\\
\emptyLine
o42 : Ideal of S\\
\endOutput
\beginOutput
i43 : I5 = ideal(a*x+b*y-1, a^2-2, b^2-3);\\
\emptyLine
o43 : Ideal of S\\
\endOutput
\beginOutput
i44 : ideal selectInSubring(1, gens gb I5)\\
\emptyLine
\             4     2 2   9  4    2   3  2   1\\
o44 = ideal(x  - 3x y  + -*y  - x  - -*y  + -)\\
\                         4           2      4\\
\emptyLine
o44 : Ideal of S\\
\endOutput
\beginOutput
i45 : clearAll\\
\endOutput
\qed
\end{solution*}

It is worth noting that the points in $\mathbb{A}_{\bbbq}^{n}$ correspond
to orbits of the action of ${\rm Gal}(\overline{\bbbq}/\bbbq)$ on the
points of $\mathbb{A}_{\overline{\bbbq}}^{n}$.  For more examples and
information, see section~II.2 in Eisenbud and Harris~\cite{SC:EH}.

%%----------------------------------------------------------
\section{Multiplicity}

The multiplicity\index{multiplicity} of a zero-dimensional scheme $X$
at a point $p \in X$ is defined to be the length of the local ring
$\mathcal{O}_{X,p}$.  Unfortunately, we cannot work directly in the
local ring in \Mtwo.  What we can do, however, is to compute the
multiplicity by computing the degree of the component of $X$ supported
at $p$; see page 66 in Eisenbud and Harris~\cite{SC:EH}.

\begin{problem*}
What is the multiplicity of the origin as a zero of the polynomial
equations $x^{5}+y^{3}+z^{3} = x^{3}+y^{5}+z^{3} = x^{3}+y^{3}+z^{5} =
0$?
\end{problem*}

\begin{solution*}
If $I$ is the ideal generated by $x^{5}+y^{3}+z^{3}$,
$x^{3}+y^{5}+z^{3}$ and $x^{3}+y^{3}+z^{5}$ in $\bbbq[x,y,z]$, then
the multiplicity of the origin is
\[
\dim_{\bbbq} \frac{\bbbq[x,y,z]_{\langle x,y,z \rangle}}
{I \bbbq[x,y,z]_{\langle x,y,z \rangle}} \, .
\]
It follows that the multiplicity is the vector space dimension of the
ring $\bbbq[x,y,z] / \varphi^{-1}(I \bbbq[x,y,z]_{\langle x,y,z
\rangle})$ where $\varphi \colon \bbbq[x,y,z] \to
\bbbq[x,y,z]_{\langle x,y,z \rangle}$ is the natural map.  Moreover,
we can express this using ideal quotients:
\[
\varphi^{-1}(I \bbbq[x,y,z]_{\langle x,y,z \rangle}) \,\,= \,\,
\big(I : (I : \langle x,y,z \rangle^{\infty})\big) \, .
\]
Carrying out this calculation in \Mtwo, we obtain:
\beginOutput
i46 : S = QQ[x, y, z];\\
\endOutput
\beginOutput
i47 : I = ideal(x^5+y^3+z^3, x^3+y^5+z^3, x^3+y^3+z^5);\\
\emptyLine
o47 : Ideal of S\\
\endOutput
\beginOutput
i48 : multiplicity = degree(I : saturate(I))\\
\emptyLine
o48 = 27\\
\endOutput
\beginOutput
i49 : clearAll\\
\endOutput
Thus, we conclude that the multiplicity is $27$.\qed
\end{solution*}

There are algorithms (not yet implemented in \Mtwo) for working
directly in the local ring $\bbbq[x,y,z]_{\langle x,y,z \rangle}$.  We
refer the interested reader to Chapter~4 in Cox, Little and
O`Shea~\cite{SC:CLO2}.

%%----------------------------------------------------------
\section{Flat Families}

Non-reduced schemes\index{scheme!non-reduced} arise naturally as flat
limits\index{flat limit} of a family of reduced
schemes\index{scheme!reduced}. Our next problem illustrates how a
family of skew lines in $\bbbp^{3}$ gives rise to a double line with
an embedded point\index{embedded point}.

\begin{problem*}[Exercise~III-68 in \cite{SC:EH}]
Let $L$ and $M $ be the lines in $\bbbp^{3}_{k[t]}$ given by $x=y=0$
and $x-tz = y+t^{2}w =0$ respectively.  Show that the flat limit as $t
\to 0$ of the union $L \cup M$ is the double line $x^{2} = y = 0$ with
an embedded point of degree $1$ located at the point $(0:0:0:1)$.
\end{problem*}

\begin{solution*}
We first find the flat limit by saturating\index{saturation} the
intersection ideal and setting $t = 0$.
\beginOutput
i50 : PP3 = QQ[t, x, y, z, w];\\
\endOutput
\beginOutput
i51 : L = ideal(x, y);\\
\emptyLine
o51 : Ideal of PP3\\
\endOutput
\beginOutput
i52 : M = ideal(x-t*z, y+t^2*w);\\
\emptyLine
o52 : Ideal of PP3\\
\endOutput
\beginOutput
i53 : X = intersect(L, M);\\
\emptyLine
o53 : Ideal of PP3\\
\endOutput
\beginOutput
i54 : Xzero = trim substitute(saturate(X, t), \{t => 0\})\\
\emptyLine
\                   2        2\\
o54 = ideal (y*z, y , x*y, x )\\
\emptyLine
o54 : Ideal of PP3\\
\endOutput
Secondly, we verify that this is the union of a double line and an
embedded point of degree $1$.
\beginOutput
i55 : Xzero == intersect(ideal(x^2, y), ideal(x, y^2, z))\\
\emptyLine
o55 = true\\
\endOutput
\beginOutput
i56 : degree(ideal(x^2, y ) / ideal(x, y^2, z))\\
\emptyLine
o56 = 1\\
\endOutput
\beginOutput
i57 : clearAll\\
\endOutput
\qed
\end{solution*}

Section~III.3.4 in Eisenbud and Harris~\cite{SC:EH} contains several
other interesting limits of various flat families.

%%----------------------------------------------------------
\section{B\'{e}zout's Theorem}

B\'{e}zout's Theorem\index{Bezout's Theorem@B\'ezout's Theorem} --- Theorem~III-78 in
Eisenbud and Harris~\cite{SC:EH} --- may fail without the
Cohen-Macaulay\index{Cohen-Macaulay} hypothesis.  Our seventh problem
is to demonstrate this.

\begin{problem*}[Exercise~III-81 in \cite{SC:EH}]
Find irreducible closed subvarieties $X$ and $Y$ in $\bbbp^{4}$ such
that
\begin{align*}
\codim(X \cap Y) &= \codim(X) + \codim(Y) \\
\deg(X \cap Y) &> \deg(X) \cdot \deg(Y) \, .
\end{align*}
\end{problem*}

\begin{solution*}
We show that the assertion holds when $X$ is the cone over the
nonsingular rational quartic curve\index{rational quartic curve} in
$\bbbp^{3}$ and $Y$ is a two-plane passing through the vertex of the
cone.  First, recall that the rational quartic curve is given by the
$2 \times 2$ minors of the matrix $\left[ \begin{smallmatrix} a &
b^{2} & bd & c \\ b & ac & c^2 & d \end{smallmatrix} \right]$; see
Exercise~18.8 in Eisenbud~\cite{SC:E}.  Thus, we have
\beginOutput
i58 : S = QQ[a, b, c, d, e];\\
\endOutput
\beginOutput
i59 : IX = trim minors(2, matrix\{\{a, b^2, b*d, c\},\{b, a*c, c^2, d\}\})\\
\emptyLine
\                         3      2     2    2    3    2\\
o59 = ideal (b*c - a*d, c  - b*d , a*c  - b d, b  - a c)\\
\emptyLine
o59 : Ideal of S\\
\endOutput
\beginOutput
i60 : IY = ideal(a, d);\\
\emptyLine
o60 : Ideal of S\\
\endOutput
\beginOutput
i61 : codim IX + codim IY == codim (IX + IY)\\
\emptyLine
o61 = true\\
\endOutput
\beginOutput
i62 : (degree IX) * (degree IY)\\
\emptyLine
o62 = 4\\
\endOutput
\beginOutput
i63 : degree (IX + IY)\\
\emptyLine
o63 = 5\\
\endOutput
which establishes the assertion.\qed
\end{solution*}

To understand how this example works, it is enlightening to express
$Y$ as the intersection of two hyperplanes; one given by $a = 0$ and
the other given by $d = 0$.  Intersecting $X$ with the first
hyperplane yields
\beginOutput
i64 : J = ideal mingens (IX + ideal(a))\\
\emptyLine
\                      3      2   2    3\\
o64 = ideal (a, b*c, c  - b*d , b d, b )\\
\emptyLine
o64 : Ideal of S\\
\endOutput
However, this first intersection has an embedded point;
\beginOutput
i65 : J == intersect(ideal(a, b*c, b^2, c^3-b*d^2), \\
\           ideal(a, d, b*c, c^3, b^3)) -- embedded point\\
\emptyLine
o65 = true\\
\endOutput
\beginOutput
i66 : clearAll\\
\endOutput
The second hyperplane passes through this embedded
point\index{embedded point} which explains the extra intersection.

%%----------------------------------------------------------
\section{Constructing Blow-ups}

The blow-up\index{blow-up} of a scheme $X$ along a subscheme $Y$ can
be constructed from the Rees algebra\index{Rees algebra} associated to
the ideal sheaf of $Y$ in $X$; see Theorem~IV-22 in Eisenbud and
Harris~\cite{SC:EH}.  Gr\"{o}bner basis techniques allow one to
express the Rees algebra in terms of generators and relations.  We
illustrate this method in the next solution.

\begin{problem*}[Exercises~IV-43 \& IV-44 in \cite{SC:EH}]
Find the blow-up $X$ of the affine plane\index{scheme!affine}
$\mathbb{A}^{2} = \Spec\big( \bbbq[x, y] \big)$ along the subscheme
defined by $\langle x^{3}, xy, y^{2} \rangle$.  Show that $X$ is
nonsingular and its fiber over the origin is the union of two copies
of $\bbbp^{1}$ meeting at a point.
\end{problem*}

\begin{solution*}
We first provide a general function which returns the ideal of
relations for the Rees algebra.
\beginOutput
i67 : blowUpIdeal = (I) -> (\\
\           r := numgens I;\\
\           S := ring I;\\
\           n := numgens S;\\
\           K := coefficientRing S;\\
\           tR := K[t, gens S, vars(0..r-1), \\
\                     MonomialOrder => Eliminate 1];\\
\           f := map(tR, S, submatrix(vars tR, \{1..n\}));\\
\           F := f(gens I);\\
\           J := ideal apply(1..r, j -> (gens tR)_(n+j)-t*F_(0,(j-1)));\\
\           L := ideal selectInSubring(1, gens gb J);\\
\           R := K[gens S, vars(0..r-1)];\\
\           g := map(R, tR, 0 | vars R);\\
\           trim g(L));\\
\endOutput
Now, applying the function to our specific case yields:
\beginOutput
i68 : S = QQ[x, y];\\
\endOutput
\beginOutput
i69 : I = ideal(x^3, x*y, y^2);\\
\emptyLine
o69 : Ideal of S\\
\endOutput
\beginOutput
i70 : J = blowUpIdeal(I)\\
\emptyLine
\                           2         2          3     2\\
o70 = ideal (y*b - x*c, x*b  - a*c, x b - y*a, x c - y a)\\
\emptyLine
o70 : Ideal of QQ [x, y, a, b, c]\\
\endOutput
Therefore, the blow-up of the affine plane along the given subscheme
is
\[
X = \Proj\left( \frac{(\bbbq[x,y])[a,b,c]}{\langle yb-xc, xb^{2}-ac,
x^{2}b-ya, x^{3}c-y^{2}a \rangle} \right) \, .
\]
Using \Mtwo, we can also verify that the scheme $X$ is
nonsingular\index{singular locus};
\beginOutput
i71 : J + ideal jacobian J == ideal gens ring J\\
\emptyLine
o71 = true\\
\endOutput
\beginOutput
i72 : clearAll\\
\endOutput
Since we have
\[
\frac{(\bbbq[x,y])[a,b,c]}{\langle yb-xc, xb^{2}-ac, x^{2}b-ya,
x^{3}c-y^{2}a \rangle} \otimes \frac{\bbbq[x,y]}{\langle x, y \rangle}
\cong \frac{\bbbq[a,b,c]}{\langle ac \rangle} \, ,
\]
the fiber over the origin $\langle x,y \rangle$ in $\mathbb{A}^{2}$ is
clearly a union of two copies of $\bbbp^{1}$ meeting at one point.  In
particular, the exceptional fiber is not a projective space.\qed
\end{solution*}

Many other interesting blow-ups can be found in section~II.2 in
Eisenbud and Harris~\cite{SC:EH}.

%%----------------------------------------------------------
\section{A Classic Blow-up}

We consider the blow-up\index{blow-up} of the projective plane
$\bbbp^{2}$ at a point.

\vbox{
\begin{problem*}
Show that the following varieties are isomorphic.
\begin{enumerate}
\item[$(a)$] the image of the rational map from $\bbbp^{2}$ to
$\bbbp^{4}$ given by
\[
(r:s:t) \mapsto (r^{2}:s^{2}:rs:rt:st) \, ;
\]
\item[$(b)$] the blow-up of the plane $\bbbp^{2}$ at the point
$(0:0:1)$;
\item[$(c)$] the determinantal variety\index{determinantal variety}
defined by the $2 \times 2$ minors of the matrix $\left[
\begin{smallmatrix} a & c & d \\ b & d & e \end{smallmatrix} \right]$
where $\bbbp^{4} = \Proj\big( k[a,b,c,d,e] \big)$.
\end{enumerate}
This surface is called the {\em cubic scroll}\index{cubic scroll} in
$\bbbp^{4}$.
\end{problem*}
}

\begin{solution*}
We find the ideal in part~$(a)$ by elimination
theory\index{elimination theory}.
\beginOutput
i73 : PP4 = QQ[a..e];\\
\endOutput
\beginOutput
i74 : S = QQ[r..t, A..E, MonomialOrder => Eliminate 3];\\
\endOutput
\beginOutput
i75 : I = ideal(A - r^2, B - s^2, C - r*s, D - r*t, E - s*t);\\
\emptyLine
o75 : Ideal of S\\
\endOutput
\beginOutput
i76 : phi = map(PP4, S, matrix\{\{0_PP4, 0_PP4, 0_PP4\}\} | vars PP4)\\
\emptyLine
o76 = map(PP4,S,\{0, 0, 0, a, b, c, d, e\})\\
\emptyLine
o76 : RingMap PP4 <--- S\\
\endOutput
\beginOutput
i77 : surfaceA = phi ideal selectInSubring(1, gens gb I)\\
\emptyLine
\                                          2\\
o77 = ideal (c*d - a*e, b*d - c*e, a*b - c )\\
\emptyLine
o77 : Ideal of PP4\\
\endOutput
Next, we determine the surface in part~$(b)$.  We construct the ideal
defining the blow-up of $\bbbp^{2}$
\beginOutput
i78 : R = QQ[t, x, y, z, u, v, MonomialOrder => Eliminate 1];\\
\endOutput
\beginOutput
i79 : blowUpIdeal = ideal selectInSubring(1, gens gb ideal(u-t*x, \\
\           v-t*y))\\
\emptyLine
o79 = ideal(y*u - x*v)\\
\emptyLine
o79 : Ideal of R\\
\endOutput
and embed it in $\bbbp^{2} \times \bbbp^{1}$.
\beginOutput
i80 : PP2xPP1 = QQ[x, y, z, u, v];\\
\endOutput
\beginOutput
i81 : embed = map(PP2xPP1, R, 0 | vars PP2xPP1);\\
\emptyLine
o81 : RingMap PP2xPP1 <--- R\\
\endOutput
\beginOutput
i82 : blowUp = PP2xPP1 / embed(blowUpIdeal);\\
\endOutput
We then map this surface into $\bbbp^{5}$ using the Segre
embedding\index{Segre embedding}.
\beginOutput
i83 : PP5 = QQ[A .. F];\\
\endOutput
\beginOutput
i84 : segre = map(blowUp, PP5, matrix\{\{x*u,y*u,z*u,x*v,y*v,z*v\}\});\\
\emptyLine
o84 : RingMap blowUp <--- PP5\\
\endOutput
\beginOutput
i85 : ker segre\\
\emptyLine
\                                2\\
o85 = ideal (B - D, C*E - D*F, D  - A*E, C*D - A*F)\\
\emptyLine
o85 : Ideal of PP5\\
\endOutput
Note that the image under the Segre map lies on a hyperplane in
$\bbbp^{5}$.  To get the desired surface in $\bbbp^{4}$, we project
\beginOutput
i86 : projection = map(PP4, PP5, matrix\{\{a, c, d, c, b, e\}\})\\
\emptyLine
o86 = map(PP4,PP5,\{a, c, d, c, b, e\})\\
\emptyLine
o86 : RingMap PP4 <--- PP5\\
\endOutput
\beginOutput
i87 : surfaceB = trim projection ker segre\\
\emptyLine
\                                          2\\
o87 = ideal (c*d - a*e, b*d - c*e, a*b - c )\\
\emptyLine
o87 : Ideal of PP4\\
\endOutput
Finally, we compute the surface in part~$(c)$.
\beginOutput
i88 : determinantal = minors(2, matrix\{\{a, c, d\}, \{b, d, e\}\})\\
\emptyLine
\                                          2\\
o88 = ideal (- b*c + a*d, - b*d + a*e, - d  + c*e)\\
\emptyLine
o88 : Ideal of PP4\\
\endOutput
\beginOutput
i89 : sigma = map( PP4, PP4, matrix\{\{d, e, a, c, b\}\});\\
\emptyLine
o89 : RingMap PP4 <--- PP4\\
\endOutput
\beginOutput
i90 : surfaceC = sigma determinantal\\
\emptyLine
\                                          2\\
o90 = ideal (c*d - a*e, b*d - c*e, a*b - c )\\
\emptyLine
o90 : Ideal of PP4\\
\endOutput
By incorporating a permutation of the variables into definition of
{\tt surfaceC}, we obtain the desired isomorphisms
\beginOutput
i91 : surfaceA == surfaceB\\
\emptyLine
o91 = true\\
\endOutput
\beginOutput
i92 : surfaceB == surfaceC\\
\emptyLine
o92 = true\\
\endOutput
\beginOutput
i93 : clearAll\\
\endOutput
which completes the solution.\qed
\end{solution*}

For more information of the geometry of rational normal scrolls, see
Lecture~8 in Harris~\cite{SC:H}.

%%----------------------------------------------------------
\section{Fano Schemes}

Our final example concerns the family of Fano schemes\index{Fano
scheme} associated to a flat family of quadrics.
Recall that the $k$-th Fano scheme $F_{k}(X)$ of a
scheme $X \subseteq \bbbp^{n}$ is the subscheme of
the Grassmannian parametrizing $k$-planes
contained in $X$.

\begin{problem*}[Exercise~IV-69 in \cite{SC:EH}]
Consider the one-parameter family\index{one-parameter family} of
quadrics tending to a double plane with equation
\[
Q = V(tx^{2}+ty^{2}+tz^{2}+w^{2}) \subseteq \bbbp^{3}_{\bbbq[t]} =
\Proj\big(\bbbq[t][x,y,z,w]\big) \enspace .
\]
What is the flat limit\index{flat limit} of the Fano schemes
$F_{1}(Q_{t})$?
\end{problem*}

\begin{solution*}
We first compute the ideal defining $F_{1}(Q_{t})$, the scheme
parametrizing lines in $Q$.
\beginOutput
i94 : PP3 = QQ[t, x, y, z, w];\\
\endOutput
\beginOutput
i95 : Q = ideal( t*x^2+t*y^2+t*z^2+w^2 );\\
\emptyLine
o95 : Ideal of PP3\\
\endOutput
To parametrize a line in our projective space, we introduce
indeterminates $u, v$ and $A, \dotsc, H$.
\beginOutput
i96 : R = QQ[t, u, v, A .. H];\\
\endOutput
We then make a map {\tt phi} from {\tt PP3} to {\tt R} sending the
variables to the coordinates of the general point on a line.
\beginOutput
i97 : phi = map(R, PP3, matrix\{\{t\}\} | \\
\              u*matrix\{\{A, B, C, D\}\} + v*matrix\{\{E, F, G, H\}\});\\
\emptyLine
o97 : RingMap R <--- PP3\\
\endOutput
\beginOutput
i98 : imageFamily = phi Q;\\
\emptyLine
o98 : Ideal of R\\
\endOutput
For a line to belong to $Q$, the {\tt imageFamily} must vanish
identically.  In other words, $F_{1}(Q)$ is defined by the
coefficients of the generators of {\tt imageFamily}.
%% removing a final use of 'coefficients'
%% coeffOfFamily = (coefficients ({1,2}, gens imageFamily))_1;
\beginOutput
i99 : coeffOfFamily = contract(matrix\{\{u^2,u*v,v^2\}\}, gens imageFamily)\\
\emptyLine
o99 = | tA2+tB2+tC2+D2 2tAE+2tBF+2tCG+2DH tE2+tF2+tG2+H2 |\\
\emptyLine
\              1       3\\
o99 : Matrix R  <--- R\\
\endOutput
Since we don't need the variables $u$ and $v$, we get rid of them.
\beginOutput
i100 : S = QQ[t, A..H];\\
\endOutput
\beginOutput
i101 : coeffOfFamily = substitute(coeffOfFamily, S);\\
\emptyLine
\               1       3\\
o101 : Matrix S  <--- S\\
\endOutput
\beginOutput
i102 : Sbar = S / (ideal coeffOfFamily);\\
\endOutput
Next, we move to the Grassmannian\index{Grassmannian} $\mathbb{G}(1,3)
\subset \bbbp^{5}$.  Recall the homogeneous coordinates on
$\bbbp^{5}$ correspond to the $2 \times 2$ minors of a $2 \times 4$
matrix.  We obtain these minors using the {\tt exteriorPower} function
in \Mtwo.
\beginOutput
i103 : psi = matrix\{\{t\}\} | exteriorPower(2, \\
\                   matrix\{\{A, B, C, D\}, \{E, F, G, H\}\})\\
\emptyLine
o103 = | t -BE+AF -CE+AG -CF+BG -DE+AH -DF+BH -DG+CH |\\
\emptyLine
\                  1          7\\
o103 : Matrix Sbar  <--- Sbar\\
\endOutput
\beginOutput
i104 : PP5 = QQ[t, a..f];\\
\endOutput
\beginOutput
i105 : fanoOfFamily = trim ker map(Sbar, PP5, psi);\\
\emptyLine
o105 : Ideal of PP5\\
\endOutput
Now, to answer the question, we determine the limit as $t$ tends to $0$.
\beginOutput
i106 : zeroFibre = trim substitute(saturate(fanoOfFamily, t), \{t=>0\})\\
\emptyLine
\                         2   2                   2                     $\cdot\cdot\cdot$\\
o106 = ideal (e*f, d*f, e , f , d*e, a*e + b*f, d , c*d - b*e + a*f, b $\cdot\cdot\cdot$\\
\emptyLine
o106 : Ideal of PP5\\
\endOutput
Let's transpose the matrix of generators so all of its elements are visible
on the printed page.
\beginOutput
i107 : transpose gens zeroFibre\\
\emptyLine
o107 = \{-2\} | ef       |\\
\       \{-2\} | df       |\\
\       \{-2\} | e2       |\\
\       \{-2\} | f2       |\\
\       \{-2\} | de       |\\
\       \{-2\} | ae+bf    |\\
\       \{-2\} | d2       |\\
\       \{-2\} | cd-be+af |\\
\       \{-2\} | bd+ce    |\\
\       \{-2\} | ad-cf    |\\
\       \{-2\} | a2+b2+c2 |\\
\emptyLine
\                 11         1\\
o107 : Matrix PP5   <--- PP5\\
\endOutput
We see that $F_{1}(Q_{0})$ is supported on the plane conic $\langle d,
e, f, a^{2}+b^{2}+c^{2} \rangle$.  However, $F_{1}(Q_{0})$ is not
reduced\index{scheme!non-reduced}; it has
multiplicity\index{multiplicity} two.  On the other hand, the generic
fiber is
\beginOutput
i108 : oneFibre = trim substitute(saturate(fanoOfFamily, t), \{t => 1\})\\
\emptyLine
\                          2    2    2                                  $\cdot\cdot\cdot$\\
o108 = ideal (a*e + b*f, d  + e  + f , c*d - b*e + a*f, b*d + c*e, a*d $\cdot\cdot\cdot$\\
\emptyLine
o108 : Ideal of PP5\\
\endOutput
\beginOutput
i109 : oneFibre == intersect(ideal(c-d, b+e, a-f, d^2+e^2+f^2), \\
\            ideal(c+d, b-e, a+f, d^2+e^2+f^2))\\
\emptyLine
o109 = true\\
\endOutput
Hence, for $t \neq 0$, $F_{1}(Q_{t})$ is the union of two conics lying
in complementary planes and $F_{1}(Q_{0})$ is the double conic
obtained when the two conics move together.\qed
\end{solution*}

% Local Variables:
% mode: latex
% mode: reftex
% tex-main-file: "chapter-wrapper.tex"
% reftex-keep-temporary-buffers: t
% reftex-use-external-file-finders: t
% reftex-external-file-finders: (("tex" . "make FILE=%f find-tex") ("bib" . "make FILE=%f find-bib"))
% End:


\documentclass[a4paper]{article}

\def\npart{II}

\def\ntitle{Principles of Quantum Mechanics Summary}
\def\nlecturer{D.\ B.\ Skinner}

\def\nterm{Michaelmas}
\def\nyear{2017}

\input{header}

\renewcommand*{\H}{\mathcal{H}}

\newcommand*{\bk}{\braket}
\newcommand*{\bkt}{\braketthree}

%parity operator
\newcommand*{\parity}{\mathcal{P}}
\theoremstyle{definition}
\newtheorem*{postulate}{Postulate}

%stretch table row height
\renewcommand{\arraystretch}{1.5}

\usepackage{pdflscape}

\begin{document}

\input{titlepage}

\section{Introduction}

\begin{definition}[Hilbert space]
  A complete inner product space.
\end{definition}

\begin{definition}[Linear Operator]
  A linear map \(A: \H\to \H\).
\end{definition}

\begin{definition}[(Continuous) dual space]
  \(\H^* = \mathcal{B}(\H,\C)\).
\end{definition}

Note \(\H \cong \H^*\).

\begin{notation}[Dirac notation]
  Write an element of \(\H\) as \(|\psi\rangle\), a \emph{ket} and an element of \(\H^*\) as \(\langle \phi|\), a \emph{bra}.
\end{notation}

Given an orthonormal basis \(\{|e_1\rangle,\dots,|e_n\rangle\}\) of \(\H\), any vector can be expressed as
\[
  |u\rangle = \sum_{i=1}^{n}u_i|e_i\rangle.
\]
The inner product is then
\[
  \langle v| u\rangle = \sum_{i,j=1}^{n}\conj v_j u_i \langle e_j|e_i\rangle = \sum_{i=0}^{n}\conj v_a u_a.
\]
If the basis elements are indexed by a continuous family then we simply change summation above to integration:
\[
  |\psi\rangle = \int_{ }^{ } \psi(a) |a\rangle da
\]
where an basis element \(|a\rangle\) is normalised using Dirac \(\delta\)-function: \(\langle a'|a\rangle = \delta(a'-a)\).

For example, if we use the \emph{position basis} \(\{|x\rangle\}_{x\in\R}\) then a state can be expressed as
\[
  |\psi\rangle = \int_{\R}^{ } \psi(x')|x'\rangle dx'
\]
where
\[
  \psi(x) = \langle x| \psi\rangle
\]
is the \emph{position space wavefunction}.

\begin{definition}[Commutator]
  Given two operators \(A\) and \(B\), the \emph{commutator} \([A,B] = AB- BA\) measures the degree to which they are incompatible.
\end{definition}

\begin{definition}
  \(|\psi\rangle \in \H\) is an \emph{eigenstate} of an operator \(A\) if \(A|\psi\rangle = a_\psi|\psi\rangle\). \(a_\psi\in\C\) is the \emph{eigenvalue}.

  The set of all eigenvalues of \(A\) is called the \emph{spectrum} of \(A\).

  The number of linearly independent eigenstates having the same eigenvalue is call the \emph{degeneracy} of the eigenvalue.
\end{definition}

\begin{definition}[Adjoint]
  The adjoint \(A^\dag\) of an operator \(A\) is such that
  \[
    \langle\phi|A^\dag|\psi\rangle = \conj{\langle\psi|A|\phi\rangle}
  \]
  for all \(|\phi\rangle, |\psi\rangle \in \H\).
\end{definition}

\begin{definition}[Self-adjoint operator]
  If \(Q = Q^\dag\) then \(Q\) is \emph{self-adjoint}, or \emph{Hermitian}.
\end{definition}

For a self-adjoint operator \(Q\), the set of eigenstates \(\{|n\rangle\}\) with eigenvales \(\{q_n\}\) form an orthonormal basis so we can write
\[
  Q = \sum_{n}^{ }q_n |n\rangle \langle n|.
\]
From this we can define functions of operators:
\[
  f(Q) := \sum_{n}^{ }f(q_n) |n\rangle \langle n|.
\]

\begin{postulate}[Postulates of Quantum Mechanics]\leavevmode
  \begin{enumerate}
  \item Measurement: if a state is prepared to be in some general state\
    \[
      \ket \psi = \sum_ac_a\ket \phi_a
    \]
    then the \emph{probability} that the measurement will yield an outcome corresponding to some state \(\ket \phi_b\) is
    \[
      \P(\ket \psi\to \ket\phi_b) = \frac{|\bk{\phi_b}{\psi}|^2}{\bk{\psi}{\psi}\bk{\phi_b}{\phi_b}} = |c_b|^2 \frac{\bk{\phi_b}{\phi_b}}{\bk{\psi}{\psi}}.
    \]
    Note that sum of probabilities of all outcomes is \(1\):
    \[
      \sum_b\P(\ket\psi \to \ket\phi_b) = \sum_b\frac{|c_b|^2\bk{\phi_b}{\phi_b}}{\bk{\psi}{\psi}} = \frac{\bk{\psi}{\psi}}{\bk{\psi}{\psi}} = 1.
    \]
    In the case where the states are \emph{normalised}, i.e. \(\bk{\psi}{\psi} = 1\) then this simplifies to
    \[
      \P(\ket\psi \to \ket\phi_b) = |\bk{\phi_b}{\psi}|^2
    \]
    where \(\bk{\phi_b}{\psi} \in\C\) is the \emph{probability amplitude}.

    The states are represented by elements of the projective Hilbert space \(\mathbb{P}\H\), i.e. two vectors in \(\H\) differing by a constant factor (known as phase factor) represent the same state.
  \item Observable: observable quantities are represented by self-adjoint operators. Upon measurement by an operator \(A\), a state is certain to return the definite value \(a\) if and only if it is an eigenstate of \(A\) with eigenvalue \(a\).
  \end{enumerate}
\end{postulate}

\begin{proposition}[Generalised uncertainty relation]
  Given two Hermitian operators \(A\) and \(B\),
  \[
    (\Delta A)_\psi^2(\Delta B)_\psi^2 \geq \frac{1}{4}|\langle[A,B]\rangle_\psi|^2
  \]
\end{proposition}

\begin{proof}
  For any Hermtian operators \(A\) and \(B\), for all \(\lambda \in \R\), by positive-definiteness of inner-product we have
  \[
    0 \leq \|(A+i\lambda B)|\psi\rangle\| = \langle A^2\rangle_\psi + \lambda\langle[A,B]\rangle_\psi + \lambda^2\langle B^2\rangle_\psi.
  \]
  Treat this as a quadratic in \(\lambda\) and notice the discriminant to get
  \[
    \langle A^2\rangle_\psi \langle B^2\rangle_\psi \geq \frac{1}{4}|\langle[A,B]\rangle_\psi|^2.
  \]
  Finally use the fact that \(A'=A-\langle A\rangle_\psi\) is also Hermitian and that \(\langle A'^2\rangle_\psi = (\Delta A)_\psi^2\).
\end{proof}

\section{Transformations and Symmetries}

A transformation is an operator \(U:\H\to \H, |\psi\rangle\mapsto |\psi'\rangle = U|\psi\rangle\).

\begin{proposition}
  \(U\) as above is unitary, i.e. \(U^{-1} = U^\dag\).
\end{proposition}

\begin{proof}
  After the tranformation we are expected to still find the state somewhere. Thus
  \[
    1 = \langle \psi|\psi\rangle = \langle \psi'|\psi' \rangle = \langle \psi |U^\dag U|\psi \rangle \text{ for all } |\psi\rangle \in \H.
  \]

  polarisation identity
\end{proof}

An operator gives a homomorphism from the group of all such transformations to the group of operators of all such transformations, i.e.\(U: g\mapsto U(g)\) and \(U(g_2)\compose U(g_1) = U(g_2\cdot g_2\).

\begin{definition}[Generator]
  Given a \emph{continuous} transformation \(U(\theta)\), write
  \[
    U(\delta\theta) = 1 - i \delta\theta T + \bigO(\delta\theta^2)\footnotemark
  \]
  \footnotetext{The \(-i\) is a historical convention.}
\end{definition}

By unitarity of \(U\), \(T=T^\dag\). A transformation can be seen as the limit of
\[
  U(\theta) = \lim_{N\to \infty}\Big( 1- i \frac{\theta}{N}T \Big)^N = e^{-i\theta T}.
\]
It follows that
\[
  i \frac{\partial |\psi\rangle}{\partial \theta} = T |\psi\rangle.
\]

Instead of view a transformation as ``changing'' the state, we can also see it as ``changing'' the operator by \emph{similarity transform}:
\[
  A \mapsto A' = U^\dag A U = U^{-1} A U.
\]
Note that
\[
  [A',B'] = U^{-1}[A,B]U
\]
and for a continuous transformation,
\[
  U^{-1}(\delta\theta) A U(\delta\theta) = A + i\delta\theta [T,A] + \bigO(\delta\theta^2)
\]
so the rate of change of states are given by the generator while the rate of change of operators is given by the commutator of the generator with the operator.

\subsection{Continuous Transformations}

Translation and rotation are summarised in Table~\ref{tab:continuous}. Here are some additional discussion.

\subsubsection{Translation}

Suppose \(\ket{\V x}\) is a position eigenstate with eigenvalue \(\V x\), i.e.
\[
  \V X\ket{\V x} = \V x\ket{\V x},
\]
representing a particle definitely located at \(\V x\). Then
\begin{align*}
  \V X U(\V a)\ket{\V x} &= ([\V X,U(\V a)]+U(\V a) \V X)\ket{\V x} \\
                         &= (U(\V a)\V a + U(\V a)\V X)\ket{\V x} \\
                         &= (\V x+ \V a)U(\V a)\ket{\V x}
\end{align*}
Thus \(U(\V a)\ket{\V x}\) is an eigenstate of \(\V X\) with eigenvalue \(\V x+ \V a\). Write
\[
  U(\V a)\ket{\V x} = c\ket{\V x+ \V a}
\]
since they represent the same state. Take inner product with \(\bra{\V x'}\),
\begin{align*}
  \bk{\V x'}{\V x} &= c \delta^3(\V x'-\V x-\V a) = c \bk{\V x'}{\V x+ \V a} = \bkt{\V x'}{U(\V a)}{\V x} \\
                  &= \big( U(\V a)^{-1}\ket{\V x'} \big)^\dag \ket{\V x} = \Big( \frac{1}{c}\ket{\V x'-\V a} \Big)^\dag \ket{\V x} = \frac{1}{\conj c}\delta^3(\V x'- \V a - \V x)
\end{align*}
so \(|c|^2=1\).

Similarly for a state in position basis
\[
  \ket{\psi} = \int_{\R^3}^{ } \bk{\V x}{\psi}\ket{\V x} d^3x = \int_{\R^3}^{ } \psi(\V x)\ket{\V x} d^3x,
\]
the wavefunction of the translated space \(\psi'(\V x)\) is

\begin{align*}
  \psi'(\V x) &= \bkt{\V x}{U(\V a)}{\psi} = \bkt*{\V x}{U(\V a) \int_{\R^3}^{} \psi(\V x') d^3x'}{\V x'} \\
              &= \big(U^{-1}(\V a)\ket{\V x} \big)^\dag \int_{\R^3}^{ } \psi(\V x') \ket{\V x} d^3 x' \\
              &= \bkt*{\V x- \V a}{\int_{\R^3}^{ } \psi(\V x') d^3x}{\V x'} \\
              &= \int_{\R^3}^{ } \psi(\V x')\bk{\V x - \V a}{\V x'} dx' \\
              &= \psi(\V x- \V a)
\end{align*}
where we used the unitarity \(U^{-1} = U^\dag\) in \(U^\dag(\V a)\ket{\V x} = \ket{\V x- \V a}\).

In particular, for an infinitesimal translation \(\delta\V a\), we have
\[
  \psi'(\V x) - \psi(\V x) = -\delta\V a\cdot \nabla \psi(\V x)
\]
as well as
\[
  \psi'(\V x) - \psi(\V x) = \bkt*{\V x}{-\frac{i}{\hbar}\delta\V \alpha\cdot \V P}{\psi} = -\frac{i}{\hbar}\delta\V a\cdot \bkt{\V x}{\V P}{\psi}
\]
so the momentum operator \(\V P\) acts in the position representation as
\[
  \bkt{\V x}{\V P}{\psi} = -i\hbar\nabla\psi(\V x).
\]

For example, for a state \(\ket{\V p}\) satisfying \(\V P \ket{\V p} = \V p \ket{\V p}\),
\[
  \psi_{\V p}(\V x- \V a) = \bk{\V x- \V a}{\V p} = \bkt{\V x}{U(\V a)}{\V p} = \bkt{\V x}{e^{-i\V a\cdot\V P/\hbar}}{\V p} =  e^{-i\V a\cdot \V p/\hbar}\bk{\V x}{\V p} = e^{-i\V a\cdot \V p/\hbar}\psi_{\V p}(\V x).
\]
Thus the position wavefunction for momentum eigenstates is
\[
  \psi_{\V p}(\V x) = \frac{1}{(2\pi \hbar)^{3/2}} e^{i\V p\cdot \V x/\hbar}.
\]

\subsubsection{Rotation}

A anticlockwise rotation around the axis \(\hat \V{\alpha}\) by an amount \(|\V \alpha\) is a linear transformation
\begin{align*}
  \V R(\V \alpha): \R^3 &\to \R^3 \\
  \V v &\mapsto \V v' = \V R(\V \alpha)\V v
\end{align*}
that obeys
\[
  \norm{\V v} = \norm{\V v'}, \, \det{\V R(\V \alpha)} = 1.
\]

For an infinitesimal rotation,
\[
  \V v' = \V v + \delta\V \alpha\times \V v + \bigO(\delta\V\alpha^2).
\]

\subsubsection{Time translation}

Since \(U(t)\) represents time translation, if a particle is in some state \(\ket{\psi(0)}\) before translation, translating forward to time \(t\) will put the particle in state
\[
  \ket{\psi(t)} = U(t) \ket{\psi(0)} = e^{-iHt/\hbar}\ket{\psi(0)}.
\]
Differentiating, we get
\[
  i\hbar \frac{\partial}{\partial t} \ket{\psi(t)} = H\ket{\psi(t)},
\]
the \emph{Time Dependent Schrödinger Equation}

\subsection{Discrete Transformation}

\subsubsection{Parity}

Parity transformation is an example of discrete transformation, acting as
\begin{align*}
  \parity : \V x &\mapsto - \V x \\
  t &\mapsto t
\end{align*}

It satisfies the \emph{anti-commutator} relation
\pagestyle{empty}
\begin{landscape}

\begin{table}[htbp]
  \centering
  \begin{tabular}{|c|p{8cm}|p{8cm}|}
    \hline
    & Translation & Rotation \\ \hline
    Governing equation & \(U^{-1}(\V a)\V X U(\V a) = \V X + \V a\) where \(\V X\) is position operator & \(U^{-1}(\V \alpha)\V V U(\V\alpha) = \V R(\V\alpha)\V V\) for any vector \(\V V\) \\ \hline
    Generator & \(U(\delta\V a) = 1- \frac{i}{\hbar}\delta\V a \cdot \V P + \bigO(\delta a^2)\) & \(U(\delta\V\alpha) = 1- \frac{i}{\hbar}\delta\V \alpha \cdot \V J + \bigO(\delta \alpha^2)\) \\ \hline
    Formula & \(U(\V a) = \exp(-\frac{i}{\hbar}\V a\cdot \V P)\) & \(U(\V \alpha) = \exp(-\frac{i}{\hbar}\V \alpha\cdot \V J)\) \\ \hline
    Commutator relation & \(\begin{array}{cc} & \frac{i}{\hbar}[\delta\V a\cdot\V P, \V X] = \delta\V a \\ \Rightarrow & [X_i,P_j] = i\hbar\delta_{i,j} \end{array}\) & \(\begin{array}{cc} & \frac{i}{\hbar}[\delta\V\alpha\cdot \V J, \V V] = \delta\V\alpha\times \V V \\ \Rightarrow & [J_i,V_j] = i\hbar \epsilon_{ijk}V_k \end{array}\) \\ \hline
    Transformation group & \(\R^3\) & \(SO(3)\) \\ \hline
    & abelian & non-abelian \\ \hline
    & \(U(\V a)U(\V b) = U(\V b)U(\V a)\) & \([\V R(\delta\V \alpha), \V R(\delta\V \beta)]\V x = \V R(\delta\V \alpha \times \delta\V \beta)\V x\) \\
    & & applying homomorphism, \([U(\delta\V\alpha), U(\delta\V\beta)] = U(\delta\V\alpha \times \delta\V\beta)\) \\
    Property derived from abelianess & \([P_i,P_j] = 0\) & \([J_i,J_j] = i\hbar \epsilon_{ijk}V_k\) (can also be derived by viewing \(\V J\) as a vector) \\ \hline
  \end{tabular}
  \caption{Translation and rotation}
  \label{tab:continuous}
\end{table}

\begin{table}[htbp]
  \centering
  \begin{tabular}{|c|p{8cm}|p{8cm}|}
    \hline
    & Parity & Time translation \\
    Governing equation & \(\parity^{-1}\V X \parity = -\V X\) & \(\ket{\psi(t)} = U(t) \ket{\psi(0)}\) \\ \hline
    Generator & None (discrete) & \(U(\delta t) = 1- \frac{i}{\hbar}H\delta t + \bigO(\delta t^2\) \\ \hline
    Formula & & \(U(t) = \exp(-\frac{i}{h}Ht)\) \\ \hline
    Commutator relation & \(\{, \parity, \V X \} = 0\)& \\ \hline
    Transformation group & \(\Z/2\Z\) & \\ \hline
    & abelian & abelian \\ \hline
  \end{tabular}
  \caption{Parity and time translation}
  \label{tab:}
\end{table}
\end{landscape}
\pagestyle{plain}

\end{document}