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\lecture{notes11}{Lecture Notes}{17 Oct 00}

%%%%%%%%%%\begin{outline}

In this lecture we will cover methods for evaluating sums.  Sums
arise very frequently in the analysis of algorithms and systems.
The main examples in this lecture, however, are drawn from finance
and physics.

We've already introduced summation notation.  It is often a convenient
way to write down an equation, but not so convenient when you actually
want to evaluate it.  We are going to develop some tools for
simplifying such sums.  One attack is to find a \emph{closed form} for
the sum: as with an integral, there may be a simple way to express the
sum as another, sum-free function.  Even when we cannot find such a
closed form, we may be able to \emph{approximate} the value well
enough for our purposes.

\section{Annuities}

Would you prefer a million dollars today or \$50,000 a year for the
rest of your life?  This is a question about the value of an annuity.
An {\em annuity} is a financial instrument that pays out a fixed
amount of money at the beginning of every year for some specified
number of years.  In particular, an $n$-year, $m$-payment annuity pays
$m$ dollars at the start of each year for $n$ years.  In some cases,
$n$ is finite, but not always.  Examples include lottery payouts,
student loans, and home mortgages.  There are even Wall Street people
who specialize in trading annuities.

A key question is what an annuity is worth.  For example, lotteries
often pay out jackpots over many years.  Intuitively, $\$50,000$ a year
for 20 years ought to be worth less than a million dollars right now.
If you had all the cash right away, you could invest it and begin
collecting interest.  But what if the choice were between $\$50,000$ a
year for 20 years and a {\em half} million dollars today?  Now it is
not clear which option is better.

In order to answer such questions, we need to know what a dollar paid
out in the future is worth today.  We will to assume that money can be
invested at a fixed annual interest rate $p$.  These days a good
estimate for $p$ is around 8\%; we'll use this value for the rest of
the lecture.

Here is why the interest rate $p$ matters.  Ten dollars invested today
at interest rate $p$ will become $(1+p) = \$10.80$ in a year, $(1+p)^2
\approx \$11.66$ in two years, and so forth.  Looked at another way, ten
dollars paid out a year from now are only really worth $1/(1+p)
\approx \$9.26$ today.  The reason is that if we had the $\$9.26$ today, we
could invest it and would have $\$10.00$ in a year anyway.  Therefore,
$p$ determines the value of money paid out in the future.

\subsection{The Value of an Annuity}

Our goal is to determine the value of an $n$-year, $m$-payment
annuity.  The first payment of $m$ dollars is truly worth $m$ dollars.
But the second payment a year later is worth only $m/(1+p)$ dollars.
Similarly, the third payment is worth $m/(1+p)^2$, and the $n$-th
payment is worth only $m/(1+p)^{n-1}$.  The total value $V$ of the
annuity is equal to the sum of the payment values.  This gives:
\[
V = \sum_{i=1}^n \frac{m}{(1+p)^{i-1}}
\]
To compute the real value of the annuity, we need to evaluate this
sum.  One way is to plug in $m$, $n$, and $p$, compute each term
explicitly, and then add them up.  However, this sum has a special
closed form that makes the job easier.  (The phrase ``closed form''
refers to a mathematical expression without any summation or product
notation.)  First, lets make the summation prettier with some
substitutions.
\begin{eqnarray*}
V & = & \sum_{i=1}^n \frac{m}{(1+p)^{i-1}} \\
  & = & \sum_{j=0}^{n-1} \frac{m}{(1+p)^j} \quad (\mbox{substitute} \ j = i-1)\\
  & = & m \sum_{j=0}^{n-1} x^j \quad (\mbox{substitute} \ x = \frac{1}{1+p})
\end{eqnarray*}

The goal of these substitutions is to put the summation into a special
form so that we can bash it with a theorem given in the next section.

\subsection{The Sum of a Geometric Series}

\begin{theorem}
\label{th:seq}
For all $n \geq 1$ and all $x \neq 0$,
\[
\sum_{i=0}^{n-1} x^i = \frac{1- x^n}{1 - x}.
\]
\end{theorem}

The terms of the summation in this theorem form a {\em geometric
series}.  The distinguishing feature of a geometric series is that
each term is a constant times the one before; in this case, the
constant is {\em x}.  The theorem gives a closed form for the sum of a
geometric series that starts with 1.  

\begin{proof}
The proof is by induction on $n$.  Let $P(n)$ be the predicate that
for all $x \neq 1$, $\sum_{i=0}^{n-1} x^i = \frac{1 - x^n}{1 - x}$.
In the base case, $P(1)$ holds because $\sum_{i=0}^{0} x^i = x^0 = 1$
and $\frac{1 - x^1}{1 - x} = 1$.

In the inductive step, for $n \geq 1$ assume that for all $x \neq 1$,
$\sum_{i=0}^{n-1} x^i = \frac{1 - x^n}{1 - x}$.  We will use this to
prove that for all $x \neq 1$, $\sum_{i=0}^{n} x^i = \frac{1 -
x^{n+1}}{1 - x}$.

\begin{eqnarray*}
\sum_{i=0}^{n} x^i & = & x^n + \sum_{i=0}^{n-1} x^i \\
                   & = & x^n + \frac{1 - x^n}{1 - x} \\
                   & = & \frac{x^n (1-x) + 1 - x^n}{1 - x} \\
                   & = & \frac{1 - x^{n+1}}{1 - x} \\
\end{eqnarray*}

The second line follows from the first by the induction hypothesis.
The remaining steps are only simplifications.
\end{proof}

As if often the case, the proof by induction gives no hint about how
the formula was found in the first place.  Here is a more insightful
derivation.  The trick is to let $S$ be the value of the sum:

\begin{eqnarray*}
S & = & 1 + x + x^2 + \ldots + x^{n-1} \\
-xS & = & -x - x^2 - x^3 - \ldots - x^n
\end{eqnarray*}

Adding these two equations gives:

\begin{eqnarray*}
(1-x)S & = & 1 - x^n \\
S & = & \frac{1 - x^n}{1 - x}
\end{eqnarray*}

We'll say more about finding (as opposed to proving) summation
formulas later in the next lecture.

\subsection{Return of the Annuity Problem}

Now we can solve the annuity pricing problem.  The value of an annuity
that pays $m$ dollars at the start of each year for $n$ years is
computed as follows:

\begin{eqnarray*}
V & = & m \sum_{j=0}^{n-1} x^j \\
  & = & m \frac{1 - x^n}{1 - x} \\
  & = & m \frac{1 - (\frac{1}{1+p})^n}{1 - \frac{1}{1+p}} \\
  & = & m \frac{1 + p - (\frac{1}{1+p})^{n-1}}{p}
\end{eqnarray*}

The first line is a restatement of the summation we obtained earlier
for the value of an annuity.  The second line follows by applying the
theorem for the summation of a geometric series.  In the third line,
we undo the earlier substitution $x = 1/(1+p)$.  In the final step,
both the numerator and denominator are multiplied by $1 + p$ to
simplify the expression.

The resulting formula is much easier to use than a summation with
dozens of terms.  For example, what is the real value of a winning
lottery ticket that pays $\$50,000$ per year for 20 years?  Plugging
in $m = \$50,000$, $n = 20$, and $p = 0.8$ gives $V \approx
\$530,180$.  Because payments are deferred, the million dollar lottery
is really only worth about a half million dollars!  This is a good
trick for the lottery advertisers!


\subsection{Infinite Geometric Series}

The question at the beginning of this section was whether you would
prefer a million dollars today or $\$50,000$ a year for the rest of
your life.  Of course, this depends on how long you live, so
optimistically assume that the second option is to receive $\$50,000$
a year {\em forever}.  This sounds like infinite money!

We can compute the value of an annuity with an infinite number of
payments by taking the limit of our geometric sum in
Theorem~\ref{th:seq} as $n$ tends to infinity.  This one is worth
remembering!

\begin{theorem}
\label{th:series}
If $|x| < 1$, then

\[
\sum_{i=0}^\infty x^i = \frac{1}{1-x}.
\]
\end{theorem}

\begin{proof}
\begin{eqnarray*}
\sum_{i=0}^\infty x^i & = & \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} x^i \\
                      & = & \lim_{n \rightarrow \infty} \frac{1 - x^n}{1-x} \\
                      & = & \frac{1}{1-x}
\end{eqnarray*}

The first equality follows from the definition of an infinite
summation.  In the second line, we apply the formula for the sum of an
$n$-term geometric series given in Theorem~\ref{th:seq}.  The final
line follows by evaluating the limit; the $x^n$ term vanishes since we
assumed that $|x| < 1$.
\end{proof}

In our annuity problem, $x=1/(1+p) < 1$, so the theorem applies.
Substituting for $x$, we get an annuity value of
\begin{align*}
V &= m\cdot\frac{1}{1-x} \\
&= m\cdot \frac{1}{1-1/(1+p)}\\
&= m\cdot \frac{1+p}{(1+p)-1}\\
&= m\cdot \frac{1 + p}{p}
\end{align*}

Plugging in $m = \$50,000$ and $p = 0.8$ gives only $\$675,000$.
Amazingly, a million dollars today is worth much more than $\$50,000$
paid every year forever!  Then again, if we had a million dollars
today in the bank earning 8\% interest, we could take out and spend
$\$80,000$ a year forever.  So the answer makes some sense.


\subsection{Examples}

We now have formulas enabling us to sum both finite and infinite
geometric series.  Some examples are given below.  In each case, the
solution follows immediately from either Theorem~\ref{th:seq} (for
finite series) or Theorem~\ref{th:series} (for infinite series).

\begin{eqnarray*}
1 + 1/2 + 1/4 + 1/8 + \ldots & = & \sum_{i=0}^\infty (1/2)^i \\
                             & = & \frac{1}{1-(1/2)} \\
                             & = & 2 \\
\\
0.999999999\ldots             &= & 0.9\sum_{i=0}^{\infty} (1/10)^i\\
                             &= & 0.9\frac{1}{1-1/10}\\
                             &= & 0.9\frac{10}{9}\\
                             &= & 1\\
\\
1 - 1/2 + 1/4 - 1/8 + \ldots & = & \sum_{i=0}^\infty (-1/2)^i \\
                             & = & \frac{1}{1-(-1/2)} \\
                             & = & 2/3 \\
\\
1 + 2 + 4 + 8 + \ldots + 2^{n-1} & = & \sum_{i=0}^{n-1} 2^i \\
                                 & = & \frac{1 - 2^n}{1 - 2} \\
                                 & = & 2^n - 1 \\
\\
1 + 3 + 9 + 27 + \ldots + 3^{n-1} & = & \sum_{i=0}^{n-1} 3^i \\
                                  & = & \frac{1 - 3^n}{1 - 3} \\
                                  & = & \frac{3^n - 1}{2}
\end{eqnarray*}

If the terms in a geometric series grow smaller as in the first
example, then the series is said to be {\em geometrically decreasing}.
If the terms in a geometric series grow progressively larger as in the
last two examples, then the series is said to be {\em geometrically
increasing}.

Here is a good rule of thumb: {\em the sum of a geometric series is
approximately equal to the term with greatest absolute value}.  In the
first two examples, the largest term is equal to 1 and the sums are 2
and 2/3, both relatively close to 1.  In the third example, the sum is
about twice the largest term.  In the final example, the largest term
is $3^{n-1}$ and the sum is $(3^n-1)/2$, which is only about a factor
of $1.5$ greater.

\subsection{Related Sums}

We now know all about sums of geometric series.  But in practice one
often encounters sums that cannot be transformed by simple variable
substitutions to the form $\sum x^i$.

A non-obvious, but useful way to obtain new summation formulas from
old is by differentiating or integrating with respect to $x$.  As an
example, consider the following series.

\[
\sum_{i=1}^n i x^i = x + 2 x^2 + 3 x^3 + \ldots + n x^n
\]

This is not a geometric series, since the ratio between successive
terms is not constant.  Our formula for the sum of a geometric series
cannot be directly applied.  But suppose that we differentiate that
formula:

\begin{eqnarray*}
\frac{d}{dx} \ \sum_{i=0}^{n} x^i
 & = & \frac{d}{dx} \ \frac{1 - x^{n+1}}{1 - x} \\
\sum_{i=0}^{n} i x^{i-1}
 & = & \frac{-(n+1)x^n (1-x) - (-1)(1-x^{n+1})}{(1 - x)^2} \\
 & = & \frac{-(n+1)x^n + (n+1)x^{n+1} + 1 - x^{n+1}}{(1 - x)^2} \\
 & = & \frac{1 - (n+1)x^n + n x^{n+1}}{(1 - x)^2}
\end{eqnarray*}

Often differentiating or integrating messes up the exponent of $x$ in
every term.  In this case, we now have a formula for a series of the
form $\sum i x^{i-1}$, but we want a formula for the series $\sum i
x^i$.  The solution is simple: multiply by $x$.  This gives:

\[
\sum_{i=0}^{n} i x^i = \frac{x - (n+1)x^{n+1} + n x^{n+2}}{(1 - x)^2}
\]

Since we could easily have made a mistake, it is a good idea to go
back and validate a formula obtained this way with a proof by
induction.

Notice that if $|x| < 1$, then this sum converges to a finite value
even if there are infinitely many terms.  Taking the limit as $n$
tends infinity gives the following theorem:

\begin{theorem}
\label{th:inf_ixi}
If $|x| < 1$, then

\[
\sum_{i=0}^\infty i x^i = \frac{x}{(1-x)^2}.
\]
\end{theorem}

As a consequence, suppose there is an annuity that pays $im$ dollars
at the {\em end} of each year $i$ forever.  For example, if $m =
\$50,000$, then the payouts are $\$50,000$ and then $\$100,000$ and
then $\$150,000$ and so on.  It is hard to believe that the value of
this annuity is finite!  But we can use the preceding theorem to
compute the value:

\begin{eqnarray*}
V & = & \sum_{i=1}^n \frac{m}{(1+p)^i} \\
  & = & m \frac{\frac{1}{1+p}}{(1 - \frac{1}{1+p})^2} \\
  & = & m \frac{1+p}{p^2}
\end{eqnarray*}

The second line follows by an application of Theorem~\ref{th:inf_ixi}.
The third line is obtained by multiplying the numerator and
denominator by $(1+p)^2$.

For example, if $m = \$50,000$, and $p = 0.08$ as usual, then the
value of the annuity is $V = \$8,437,500$.  Even though payments
increase every year, the increase is only additive with time; by
contrast, dollars paid out in the future decrease in value
exponentially with time.  The geometric decrease swamps out the
additive increase.  Payments in the distant future are almost
worthless, so the value of the annuity is finite.

The important thing to remember is the trick of taking the derivative
(or integral) of a summation formula.  Of course, this technique
requires one to compute nasty derivatives correctly, but this is at
least theoretically possible!

\section{Book Stacking}

Suppose $n$ books are stacked on the edge of a table as shown in
Figure~\ref{fig:stack}.  How far can the top book overhang the table
edge without the whole stack falling over?  One book length?  Two?  We
will prove that if there are enough books, then the top book can
overhang arbitrarily far!

\begin{figure}[htbp]
\centerline{\psfig{figure=/a/class/6.042/spring00/archive/lectures/figures/stack.eps,height=1.5in}}
\caption{\em If $n$ books are stacked on a table, how far can the top
  book overhang the edge without the whole stack falling over?}
\label{fig:stack}
\end{figure}

\subsection{Formalizing the Problem}

First, we have to reduce the real-world problem to a mathematical
question.  Define the length of a book to be 2 and assume that all
books have the same mass.  For $0 \leq i \leq n-1$, define $d_i$ to be
the distance that the top book's (book 0's) right edge extends beyond
book $i$'s right edge.  Treat the table as book $n$; that is, $d_{n}$
is the distance book $0$ extends beyond the table.  Note that while
intuitively the $d_i$ should be decreasing (each book extends farther
than the one below it) there may be exceptions as shown by book 3 in
the figure.

\begin{figure}[htbp]
\centerline{\psfig{figure=/a/class/6.042/spring00/archive/lectures/figures/dstack.eps,height=1.5in}}
\caption{\em For $1 \leq i \leq n$, the variable $d_i$ is the distance
  that book $0$ extends beyond the right end of book $i$.  The table
  is regarded as book $n+1$, so $x_n$ is the distance that book $1$
  extends beyond the end of the table.  The $d_i$ need not decrease
  monotonically, as shown by $d_3$.}
\label{fig:xstack}
\end{figure}
The position of the books is constrained by the condition that the
stack must not fall over.  A stack of books is {\em stable} iff for $1
\leq k \leq n$, the center of mass for the top $k$ books lies above
the $(k+1)$-st book.  We will not prove this, but will accept it as an
principle of physics (\ie, an axiom).  This stability constraint is
illustrated for $n = 1$ in Figure~\ref{fig:stable}.

\begin{figure}[htbp]
\centerline{\psfig{figure=/a/class/6.042/spring00/archive/lectures/figures/stable.eps,height=1.5in}}
\caption{\em This figure illustrates the idea of a stable stack of
books for the special case of a single book.  In the left diagram, the
stack is stable because the center of mass lies above the next
``book'' (in this case, the table).  In the right diagram, the book is
unstable because the center of mass overhangs.}
\label{fig:stable}
\end{figure}

Let's apply a greedy strategy to our book stacking problem.  Put the
top book so that its center of mass is at the edge of the second book,
and then place the second book so that the center of mass of the top
two is at the edge of the third book, and so on.  In general, we
place the $k$th book such that the center of mass of the top $k$ books
is at the edge of the $(k+1)$st book.  That is, we want $d_k$ to be
equal to the center of mass of the top $k$ books.  Let's write the
equation for this.

As we do these equations, we are working ``relative to the right side
of the top book''---our coordinate frame measures distance to the {\bf
  left} of this point.

Notice that the center of mass of the book number $i$ is at distance
$d_i+1$ to the left of the top book.  So the center of mass of the top
$k$ books (numbered $0,\ldots,k-1$) is at the average of these $k$
positions, namely
\[
\frac{(d_{k-1}+1)+(d_{k-2}+1)+\cdots+(d_0+1)}{k} 
\]
and we want this to be just above the right hand side of book number
$k$, which is at offset $d_{k}$.  This gives us an equation relating
the quantities:
\begin{align*}
d_{k} &=\frac{(d_{k-1}+1)+(d_{k-2}+1)+\cdots+(d_0+1)}{k}\\
\intertext{which rewrites as}
kd_{k} &=d_{k-1}+d_{k-2}+\cdots+d_0 + k\\
\intertext{Of course, by the same reasoning,}\\
(k-1)d_{k-1} &=d_{k-2}+d_{k-3}+\cdots+d_0 + k-1\\
\intertext{and if we subtract this equation from the one before, we
  get}\\
kd_k-(k-1)d_{k-1} &=d_{k-1}+1\\
\intertext{or in other words,}
d_k &= d_{k-1}+1/k
\end{align*}

Now if we ``expand'' the relation above, we find out that
\begin{align*}
d_k &= d_{k-1}+1/k\\
&= d_{k-2}+1/(k-1)+1/k\\
\cdots\\
&= 1+1/2+\cdots+1/(k-1)+1/k
\end{align*}

\begin{figure}[htbp]
\centerline{\psfig{figure=/a/class/6.042/spring00/archive/lectures/figures/optstack.eps,height=1.5in}}
\caption{\em This is a stacking of 4 books that overhangs the table by
the greatest possible distance.  In general, the $k$-th book from the
top should extend distance $\frac{1}{k}$ beyond the book below.}
\label{fig:optstack}
\end{figure}

Can we do better?  Well, for stability, looking at the derived center
of mass above, we \emph{must} have
\[
d_{k} \le \frac{(d_{k-1}+1)+(d_{k-2}+1)+\cdots+(d_0+1)}{k}
\]
and in our greedy strategy above, the inequality is always ``tight''
(met with equality): that is, $d_k$ is as large as possible, given
$d_0,\ldots,d_{k-1}$.  It follows by induction that each $d_k$ is as
large as possible.  So we cannot do better.

\subsection{Evaluating the Sum-- The Integral Method}

The best possible total extension is now expressed as a sum that,
coincidentally, appears frequently in computer science.  (Fancy that!)
In fact, it is important enough to have a name.

\begin{defin*}
The {\em $n$-th Harmonic number} is equal to $\sum_{i=1}^n
\frac{1}{i}$ and is denoted $H_n$.
\end{defin*}

We can rewrite the formula for greatest possible extension simply as
$H_n$.
%seems not to be $H_n/2$

The first few Harmonic numbers are easy to compute.  For example, $H_1
= 1$, $H_2 = 1 + \frac{1}{2} = \frac{3}{2}$, $H_3 = 1 + \frac{1}{2} +
\frac{1}{3} = \frac{11}{16}$, $H_4 = 1 + \frac{1}{2} + \frac{1}{3} +
\frac{1}{4} = \frac{25}{12}$.  The fact that $H_4$ is greater than 2 has
special significance; it implies that the total extension of a 4-book
stack is greater than one full book!  This is the situation shown in
Figure~\ref{fig:optstack}.

It would be nice to answer questions like, ``How many books are needed
to build a stack extending 100 book lengths beyond the table?''  One
approach to this question would be to keep computing Harmonic numbers
until we found one exceeding 200.  However, as we will see, this is
not such a keen idea.  In fact, for all we know, building a stack
extending so far may be impossible regardless of the number of books.

All such questions would be settled if we could express $H_n$ in a
closed form.  Unfortunately, no closed form is known and probably none
exists.  As a second best, however, we can find closed forms for very
good approximations to $H_n$ using the Integral Method.  The idea of
the Integral Method is to bound terms of the sum above and below by
simple functions as suggested in Figure~\ref{fig:integral}.  The
integrals of these functions then bound the value of the sum above and
below.

\begin{figure}[htbp]
\centerline{\psfig{figure=/a/class/6.042/spring00/archive/lectures/figures/integral.eps,height=1.5in}}
\caption{\em This figure illustrates the Integral Method for bounding
a sum.  The area under the ``stairstep'' curve over the interval $[0,
n]$ is equal to $H_n = \sum_{i=1}^n 1/i$.  The function $1/x$ is
everywhere greater than or equal to the stairstep and so the integral
of $1/x$ over this interval is an upper bound on the sum.  Similarly,
$1/(x+1)$ is everywhere less than or equal to the stairstep and so the
integral of $1/(x+1)$ is a lower bound on the sum.}
\label{fig:integral}
\end{figure}

The Integral Method gives the following upper and lower bounds on the
harmonic number $H_n$:

\begin{eqnarray*}
H_n & \leq & 1 + \int_1^n \frac{1}{x} \ dx = 1 + \ln n \\
H_n & \geq & \int_0^n \frac{1}{x+1} \ dx = \int_1^{n+1} \frac{1}{x} \ dx = \ln(n+1)
\end{eqnarray*}

These bounds imply that the harmonic number $H_n$ is around $\ln n$.
Since $\ln n$ grows without bound, albeit slowly, we can make a stack
of books that extends arbitrarily far.

For example, to build a stack extending 100 book lengths beyond the
table, we need a number of books $n$ so that $H_n = 200$.
Exponentiating the above inequalities gives:

\begin{eqnarray*}
e^{H_n - 1} \leq & n & \leq e^{H_n} - 1 \\
e^{199} \leq & n & \leq e^{200} - 1
\end{eqnarray*}

The number of books required has 86 digits!

\subsection{More about Harmonic Numbers}

In the preceding section, we showed that $H_n$ is about $\ln n$.  A
even better approximation is known:

\[
H_n = \ln n + \gamma + \frac{1}{2n} + \frac{1}{12n^2} +
        \frac{\epsilon(n)}{120n^4}
\]

Here $\gamma$ is a value $0.577215664\ldots$ called Euler's constant,
and $\epsilon(n)$ is between 0 and 1 for all $n$.  We will not prove
this formula.

The shorthand $H_n \sim \ln n$ is used to indicate that the leading
term of $H_n$ is $\ln n$.  More generally, the notation $f(n) \sim
g(n)$ means that $\lim_{n \rightarrow \infty} \frac{f(n)}{g(n)}
\rightarrow 1$.

We also might write $H_n \sim \ln n + \gamma$ to indicate two leading
terms.  While this notation is widely used, it is not really right.
Referring to the definition of $\sim$, we see that while $H_n \sim \ln
n + \gamma$ is a true statement, so is $H_n \sim \ln n + c$ where $c$
is any constant.  The correct way to indicate that $\gamma$ is the
second-largest term is $H_n - \ln n \sim \gamma$.

The reason that the $\sim$ notation is useful is that often we do not
care about lower order terms.  For example, if $n = 100$, then we can
compute $H(n)$ to great precision using only the two leading terms:

\[
|H_n - \ln n - \gamma| \leq | \frac{1}{200} - \frac{1}{120000} +
\frac{1}{120 \cdot 100^4} | < \frac{1}{200}
\]
%%%%%%%%%\end{outline}

\end{document}
% LocalWords:  payouts eps book's dstack iff kd substract optstack stairstep
