\documentclass[12pt]{article}
\usepackage{wsh}
\usepackage[pdftex]{graphicx}
\def \dfdt {{d f(t) \over dt}}
\def \tm { {\tau} }
\def \tmb { \bar{\tm} }
\begin{document}
\bibliographystyle{plain}
\markright{Motivation for the logistic function --- W.S. Harlan}
\title{Bounded geometric growth: \\
motivation for the logistic function}
\author{William S. Harlan}
\date{August 2007; last revised December, 2016}
\maketitle
\section {Introduction}
The logistic function appears often in simple
physical and probabilistic experiments. A
normalized logistic is also known as an S-curve or
sigmoid function. The first derivative of this
function has a familiar bell-like shape,
but it is not a Gaussian distribution.
Many use a Gaussian to describe data when a
logistic would be more appropriate. The tails
of a logistic are exponential, whereas the
tails of a Gaussian die off very quickly. To
decide which distribution makes more sense,
we must must be aware of the conceptual
model for the underlying phenomena.
In biology, the logistic describes population
growth in a bounded environment, such as
bacteria in a petri dish. In business, a
logistic describes the successful growth of
market saturation. In engineering, the
logistic describes the production of a finite
resource such as an oilfield or a collection
of oilfields.
After discussing examples, we will see how
a bound to exponential growth leads to logistic behavior.
There are other forms of the logistic function with
extra variables that allow more arbitrary shifts and scaling.
First, I limit myself to the form derived most naturally
from the Verhulst equation. Normalizations
clarify the behavior without any loss of generality.
Finally, I use a change of variables
for fitting recorded data in physical units.
\section {Examples}
Exponential (geometric) growth is a widely
appreciated phenomenon for which we already
have familiar mental models. Investments and
populations grow exponentially
(geometrically) when their rate of growth is
proportional to their present size. You can
take almost any example of exponential growth
and turn it into logistic growth by putting a
maximum limit on its size. Just make the rate
of growth also proportional to the remaining room
left for growth.
Why is this such a natural assumption?
\subsection{Growth in a petri dish}
Let us consider the bacteria in a petri dish.
This is an easy way to create a logistic
curve in nature, and the mental model is a
simple one.
A petri dish contains a finite amount of food
and space. Into this dish we add a few
microscopic bits of bacteria (or mold, if you
prefer). Each bacterium lives for a certain
amount of time, eats a certain amount of food
during that time, and breeds a certain number
of new bacteria. We can count the total
number of bacteria that have lived and died
so far, as a cumulative sum; or more
easily, we can count the amount of food
consumed so far. The two numbers should be
directly proportional.
At the beginning these bacteria see an
vast expanse of food, essentially
unlimited given their current size. Their
rate of growth is directly proportional to
their current population, so we expect to see
them begin with exponential growth. At some
point, sooner or later, these bacteria will
have grown to such a size that they have
eaten half the food available. At this point
clearly the rate of growth can no longer be
exponential. In fact, the rate of
consumption of food is now at its maximum
possible rate. If
half the food is gone, then the total
cumulative population over time has also
reached its halfway point. As many bacteria
can be expected to live and die after this
point as have gone before. Food is now the
limiting factor, and not the size of the
existing population. The rate of consumption of
food and the population at any moment are in
fact symmetric over time. Both
decline and eventually approach zero
exponentially, at the same rate at which
they originally increased. After most
of the food has disappeared, the population growth
is directly proportional to the amount of
remaining food. As there are fewer places
for bacteria to find food, then fewer
bacteria will survive and consume a lifetime
of food. Although the population size
is no longer a limit, their individual
rates of reproduction still matter.
The logistic function can be used to describe
either the fraction of the food consumed, or
the accumulated population of bacteria that
have lived and died. The first derivative of
the logistic function describes the rate at
which the food is being consumed, and also
the living population of bacteria at any
given moment. (If you have twice as many
bacteria, then they are consuming food at
twice the rate.) This derivative has an
intuitive bell shape, up and down
symmetrically, with exponential tails. The
logistic is the integral of the bell shape:
it rises exponentially from 0 at the
beginning, grows steepest at the half-way
point, then asymptotically approaches 1 (or
100\%) at later times. The time scale is
rather arbitrary. We can adjust the units of
time or the rates of growth and fit different
populations with the same curve.
Let us quickly examine two slightly messier
examples, to see the analogies.
\subsection{Market share}
The market share of a given product can be
expressed as a fraction, from 0\% to 100\%.
All markets have a maximum size of some kind,
at least the one imposed by a finite number
of people with money. Let us assume someone
begins with a superior product and that the
relative quality of this product to its
rivals does not change over time. The early
days of this product on the market should
experience exponential growth, for several
reasons. The number of new people exposed to
this new product depends on the number who
already have it. The ability of a business
to grow, advertise, and increase production
is proportional to the current cash-flow. A
exponential is an excellent default choice,
in the absence of other special circumstances
(which always exist).
Clearly, when you have a certain fraction of the
market, geometric growth is no longer possible.
Peter Norvig coined this as Norvig's Law:
``Any technology that surpasses 50\% penetration
will never double again (in any number of months).''
But let's also assume we have no regulatory limits
and no one abusing a larger market share (bear with me).
This product should still naturally tend to a
saturated monopoly of the market. Such
market saturation is typically drawn as a
sigmoid much like a logistic. In fact it
is a logistic, given no other mechanisms.
After saturation, the rate of
change of market share is proportional to the
declining number of new customers. In
any given month, a consistent fraction of the
remaining unconverted customers will convert
to the superior product. That is, we have
a geometric or exponential decline in new customers
for each reporting period.
\subsection{Mining and oil}
Finally, let us examine the discovery
and exhaustion of a physical resource, such
as mining a mountain range, or exploration
and production of oil in an field.
The logistic has long been used to predict
the production history, the number of barrels
of oil produced a day, in any oil field.
The curve also accurately handles a collection
of oilfields, including all the oil fields
in a given country.
Such a calculation was first used by King Hubbert
in 1957 to predict correctly the peak of
total US oil production in the early 1970's.
Earliest oil production is easily exponential,
like many business ventures. As long
as there is vastly more oil to be produced
than available, then previously produced oil
can proportionally fund the exploration and
production of new oil wells.
Success also increases our understanding of an area and
improves our ability to recognize and exploit
new prospects, so long as there is no noticeable
limit to those prospects. At some point
though, the amount of oil in a given field
becomes the limiting factor. Like bacteria
in a petri dish, fewer oil wells find a viable spot in the
oil field in order to produce a full lifetime.
The maximum rate of production is achieved,
very observably, when half of the oil has been
produced that will ever been produced.
(That is not to say that oil does not remain
in the ground, but it cannot be produced
economically, using less energy than obtained
from the new oil.)
Oil production from individual oil fields do
often show asymmetry, falling more rapidly
or more gently after a peak than expected from the rise.
Petroleum engineers have learned that deliberately
slowing production increases the ultimate recoverable
oil from a field. Gas production of a single field
tends to maintain a more constant rate of production
until the pressure abruptly fails, dropping production
to nothing. But while individual fields
may have unique production curves, collections of fields in
a region or country tend to follow a more predictable
logistic trend, with the expected symmetry.
\subsection{A simple population game}
We can contrive a simple numerical game that should simulate such growth.
We have a resource that can support
a maximum population of 1000 creatures.
We will begin by dropping 10 creatures into this resource.
All are likely to find an unoccupied location.
With every generation, each existing creature has a 10 percent
chance of spawning a new creature. These new creatures
drop again at random into one of the 1000 possible locations.
If the location is not previously used, then the creature survives.
If the location is already occupied, then the new creature dies.
In early generations, 99 percent of the possible
locations are still free, so each new creature will almost
certainly survive. We expect early generations to show 10 percent
geometric growth. As the population increases, however,
available locations decrease and we see more collisions.
By the time 500 of the available locations are filled, only
half of each new generation will survive, dropping the rate
of growth to about 5 percent. We will stop the game when
990 locations are full, when each new creature has only one
percent chance of survival.
Here is a short Java program to simulate this population growth.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{verbatim}
import java.util.Random;
import java.util.BitSet;
public class LogisticGrowth {
public static void main(String...args) {
Random random = new Random(1);
int capacity = 1000;
BitSet resource = new BitSet(capacity);
for (int i=0; i< 10; ++i) { resource.set(i); }
int population = 0, generation = 0;
while ( (population=resource.cardinality()) < capacity-10) {
System.out.println(generation+" "+population);
for (int spawn=0; spawn 0. \nonumber
\end{eqnarray}
Units of time are fairly arbitrary for such problems.
For the function to approach a value of 1
asymptotically, time must continue to positive infinity.
To avoid a small non-zero value to begin growth,
we can allow the function to begin arbitrarily
early at negative infinity, where it can approach 0.
The scale of time units, whether seconds or days, is also
arbitrary. We will choose a scale that most conveniently
measures a consistent change in the function.
Let us put the halfway point, at zero time so that
\begin{eqnarray}
\label{eq:half}
f(0) = 1/2.
\end{eqnarray}
For earliest values of $t$, we expect $f(t)$
to increase geometrically. That is,
we expect the rate of increase to be
proportional to the current value:
\begin{eqnarray}
f(t) &\rightarrow& 0 , \mbox{ and } \nonumber \\
\label{eq:propp}
\dfdt &\propto& f(t), \\
\mbox{ as } t &\rightarrow& - \infty . \nonumber
\end{eqnarray}
Similarly, as time increases and our
function approaches unity, we expect
the rate of growth to be proportional
to the remaining fractional capacity.
\begin{eqnarray}
f(t) &\rightarrow& 1 , \mbox{ and } \nonumber \\
\label{eq:propn}
\dfdt &\propto& 1 - f(t), \\
\mbox{ as } t &\rightarrow& \infty . \nonumber
\end{eqnarray}
This assumption is worth dwelling upon
in light of our previous examples.
Given an almost complete saturation of our available
capacity, growth cannot be limited any longer by the
existing population. The only remaining limitation
to continued growth is the size of the remaining
opportunities for growth. If the remaining opportunities
shrink by half, then the chance of our getting
one of those opportunities must also decline by half.
Here, I find the dartboard analogy very helpful.
Let us combine these two proportions
(\ref{eq:propp}) and (\ref{eq:propn}) into
a single equation that respects both:
\begin{eqnarray}
\dfdt \propto f(t) [1-f(t)] . \nonumber
\end{eqnarray}
For appropriate time units, we can
avoid any scale factors and write
\begin{eqnarray}
\label{eq:verhulst}
\dfdt = f(t) [1-f(t)] .
\end{eqnarray}
This is slightly simplified version of the
Verhulst equation, which originated in studies
of populations.
The rate of growth at any time is
proportional to the population and to the
remaining available fraction. Both factors
are always in play, though one factor
dominates when the value of the function approaches
either 0 or 1.
By centering this equation at zero time
with (\ref{eq:half}),
we can rearrange the Verhulst equation (\ref{eq:verhulst})
and integrate for $f(x)$ with
\begin{eqnarray}
\left [ {1 \over f(t)} + {1 \over 1-f(t)} \right ] df(t)/dt &=& 1 ,\nonumber \\
{d \over dt} \{ \log f(t) - \log [1-f(t)] \} &=& 1 ,\nonumber \\
\log f(t) - \log [1-f(t)] &=& t , \nonumber \\
\label{eq:logit}
\log \left [ {f(t) \over 1-f(t)} \right ] &=& t , \mbox{ and } \\
\label{eq:success}
{f(t) \over 1-f(t)} &=& \exp(t) .
\end{eqnarray}
Finally, we arrive at the simplest form of
a logistic function:
\begin{eqnarray}
\label{eq:logistic}
f(t) = {\exp(t) \over 1 + \exp(t)} = {1 \over 1 + \exp(-t)} .
\end{eqnarray}
See figure \ref{fig:logistic}.
\begin{figure}[t]
\begin{center}
\includegraphics[width=6in]{fig1}
\end{center}
\caption{The logistic function}
\label{fig:logistic}
\end{figure}
Some versions include include arbitrary scale
factors for time or for the fraction itself.
We have avoided those by normalization to fractions
and convenient time units.
Later we will use a change of variables
useful for fitting physical data.
First notice that this equation is anti-symmetric,
with an additive constant:
\begin{eqnarray}
1- f(t) = 1/[ 1 + \exp(t)] &=& f(-t) ; \nonumber \\
f(t) + f(-t) &=& 1.
\end{eqnarray}
The asymptotic growth at the beginning
mirrors the asymptotic limit at the end.
We can think of the used capacity or
remaining capacity as mirror images of each other.
This is particularly striking because our
rate of uncontrolled growth in the beginning also
determines our rate of diminishing returns
in the end. To lose this symmetry,
we would need to introduce different (fractional)
powers in our original proportions
(\ref{eq:propp}) and (\ref{eq:propn}).
The derivative $df(t)/dt$ is often a
more interesting quantity than $f(t)$ itself.
For example, in oil production, this might be
the number of barrels produced a day (with an
appropriate scale factor). It could be the annual
growth in market share, the rate at
which a population grows, or the rate of
consumption of food.
\begin{eqnarray}
\label{eq:dfdt}
\dfdt &=& { \exp(-t) \over [ 1 + \exp(-t) ]^2 }
= {1 \over [ \exp(t/2) + \exp(-t/2) ]^2} , \\
{d f (0) \over dt} &=& 1/4 , \mbox { and }
{d f ( \pm \infty ) \over dt} = 0 . \nonumber
\end{eqnarray}
The maximum rate of increase, by design,
occurs at time zero. It is also
a perfectly symmetric bell-shape, rising
from zero to a maximum value of 1/4,
then declining again, with exponential tails.
In this form you can see more clearly how
the exponential on one side eventually overwhelms
the one on the other. See figure \ref{fig:derivative}.
\begin{figure}[t]
\begin{center}
\includegraphics[width=6in]{fig2}
\end{center}
\caption{The derivative of the logistic function}
\label{fig:derivative}
\end{figure}
In this form, the derivative (\ref{eq:dfdt}) has
unit area, integrating to 1. The equation
is also useful as the probability distribution
function (pdf) that a given resource (food,
oil, or customer) will be
used at a particular moment in time.
\subsection{Fitting real-world data}
Assume you have some data that you think
might be described by a logistic curve.
You have the data up to a certain point
in time. You might not be halfway yet.
Can you see how well the data are described
by a logistic?
Can you predict the area under the curve,
or the halfway point?
From a partial dataset, we do not
yet know the ultimate true capacity,
and we use real time units.
Let us use another form of the Verhulst
equation more useful for real-world
measurements.
To get a form similar to that used by
Verhulst for his population model,
we replace
\begin{eqnarray}
t \equiv r (\tm - \tmb ) ,
\end{eqnarray}
with $\tm$ for measurable time units,
with $r$ for an unknown time scaling, and
with $\tmb$ for an unknown reference time.
We also substitute
\begin{eqnarray}
f(t) \equiv Q(\tm)/k ,
\end{eqnarray}
where $Q(\tm)$ is a measurable capacity
or population, and $k$ is an unknown upper
limit, called the ``carrying capacity.''
The reference time $\tmb$ is when we expect
to reach half of the maximum capacity:
\begin{eqnarray}
\label{eq:halfp}
Q(\tmb ) \equiv k / 2 .
\end{eqnarray}
With these substitutions, we rewrite the
Verhulst equation (\ref{eq:verhulst}) as
\begin{eqnarray}
{d Q(\tm ) \over d\tm } &=& r [1-Q(\tm )/k] Q(\tm ) ; \nonumber \\
\label{eq:linear}
{d Q(\tm ) \over d\tm } / Q(\tm ) &=& r - (r/k) Q(\tm ) .
\end{eqnarray}
Notice that the measurable quantities
on the left of (\ref{eq:linear}) are a linear function of the
measurable quantities on the right.
The slope of the line is $r/k$, and
the vertical intercept of the line
is $r$.
The quantity on the left of equation ($\ref{eq:linear}$)
could be called the fractional rate of growth.
It is the current rate of growth divided by
the cumulative value so far. We do not need
to know ultimate rates, capacities, or
reference times to calculate this quantity.
At earliest times, when $Q(\tm )$ is small relative to $k$,
the fractional rate of growth (\ref{eq:linear}) achieves a maximum
value of $r$.
We can make a graph with this fractional rate of growth
on the vertical axis, and with the cumulative value
$Q(\tm )$ on the horizontal. For every time at which
we measure these two quantities, we can place a point
on the graph. All values are positive and fall
inside the upper-right quadrant.
If the data fit a logistic curve,
then we should be able to draw a straight line
through them. The slope and vertical intercept
of the line allow us to estimate the unknown constants
$r$ and $k$. The vertical intercept, where $Q(- \infty ) = 0$,
is the rate $r$,
and the horizontal intercept is the maximum carrying
capacity $Q(\infty ) = k$.
So what about the reference time, $\tmb$?
As time increases our data points move along this
line, but not uniformly. Time units do not appear
explicitly, except as a sampling parameter.
The time $\tmb$ corresponds to the data
point with half of the ultimate capacity, as in (\ref{eq:halfp}).
We may not have enough data to identify this point
from this graph.
Another drawback to this particular way of graphing data
is that early times will show much greater
scatter than later times. When
$d Q(\tm ) / d\tm$ and $Q(\tm )$ are small,
their ratio will show greater variation
for small variations in either. This particular
linearization is more suitable for an age of graph paper.
I prefer to fit the logistic more directly.
Using the $\tmb$ definition (\ref{eq:halfp}) as a boundary condition,
we can also rewrite the logistic function (\ref{eq:logistic})
in measurable units:
\begin{eqnarray}
\label{eq:logisticp}
Q(\tm ) = {k \over 1 + \exp[-r (\tm -\tmb )]} .
\end{eqnarray}
Here we can see more clearly that $k$ is the ultimate
maximum value of $Q(\tm )$.
If we fit $Q(\tm )$ directly, our fit should improve with
time. The value is a cumulative one, integrating measurements
over longer periods of time. Again, we can expect more
variation at earlier times.
Instead, let us examine an absolute rate of increase $P(\tm )$
that we can also measure:
\begin{eqnarray}
\label{eq:derivativep}
P(\tm ) \equiv
{d Q(\tm ) \over d\tm } =
{k r \over \{\exp[r (\tm -\tmb )/2] + \exp[-r (\tm -\tmb )/2]\}^2} .
\end{eqnarray}
Note the peak value is $P(\tmb ) = k r /4$.
Now we have a function with more consistent variations
over time. The incremental change during a short interval
of time will tend to follow the underlying distribution,
with greater deviations as we shorten the interval.
Actually, it is not difficult simply to scan reasonable
values for all three parameters $k$, $r$, and $\tmb$
and minimize some misfit to $P(\tm )$.
You can also plot the misfit as contours of multiple parameters
and get a better idea of your sensitivity to each.
Choosing a best measure of misfit is still necessary.
Least-squares, the default choice for many, makes sense only
if you think that errors
in your measurements are Gaussian and consistent over time.
This seems unlikely. Lower magnitudes have less potential
for absolute variation than larger ones. We could instead
minimize errors in the ratio of a measured magnitude of $P(\tm )$
to the expected magnitude. Or equivalently,
we can minimize errors in the logarithm of $P(\tm )$.
If we minimize the square of those errors, then we
are assuming that variations in our measurements are multiplicative,
following a log Gaussian distribution. This is much
better, but I think still not optimum.
Another way to think of the problem is that the logistic
derivative $P(\tm )$ in (\ref{eq:derivativep})
describes a probability of a particular quantity being exploited
or consumed at a particular point in time. A given customer,
bacterium, or barrel of oil, is most to appear near the peak
time $\tmb$ rather than near the tails. Given a certain realization
of that probability, our recorded data, what parameters maximize
the probability of that data? It turns out that this likelihood
is maximized by a minimum cross-entropy.
Let our recorded data be pairs of samples $\{P^i , \tm^i \}$ indexed by $i$.
Then the best distribution $P(\tm)$ should minimize
\begin{eqnarray}
\label{eq:derivativep}
\min_{k, r, \tmb} \sum_i \left \{ P^i ~ \log [ P^i ~/ ~ P( \tm^i ) ]\right \} .
\end{eqnarray}
$P(\tm )$ is a function of these three unknown parameters
$(k, r, \tmb)$.
The $P(\tm)$ that minimizes this cross-entropy is the one
that makes the actually recorded data most probable.
Because we have not necessarily sampled the entire function,
we should renormalize both $P(\tm)$ and $P^i$
over the range of available $\tm^i$ before evaluating.
Normalization effectively ignores the unknown capacity $k$
and fits only the local shape of the curve.
The remaining two degrees of freedom $r$ and $\tmb$ can be exhaustively
searched with dense sampling. Once these are known, the best
$k$ can be calculated without normalization.
\section{Comparison to logistic regression}
Neural networks and machine learning algorithms often use the same family of
S curves for ``logistic regression,'' but motivate the equations differently.
Logistic regression attempts to estimate the probability of an event
with a binary outcome, either true or false. The probability is expressed
as a function of some ``explanatory variable.''
For example, what is the probability of a light bulb failing
after a certain number of hours of use?
Maybe more relevant, what is the probability a given drilling program
will be economic, given some measurement of effort?
We start with a probability $p$ of one outcome --- say a successful well or a good lightbulb.
That leaves us with a probability of $1-p$ for the alternative --- a bad well, or a bad lightbulb.
Our explanatory variable $x$ could be a unit of time, as before, or some other factor.
We expect the probability $p(x)$ either to increase or to decrease strictly as a function of $x$.
Logistic regression uses the logit function, which is the logarithm
of the ``odds.'' The odds are the ratio of the chance of success to failure.
\begin{eqnarray}
\label{eq:logitp}
\mbox{logit}(p) \equiv \log \left ( {p \over 1-p} \right ) .
\end{eqnarray}
This logit function already appeared in equation~(\ref{eq:logit}), if you interpret $p$ as $f(t)$.
The $\mbox{logit}$ function and the logistic function (\ref{eq:logistic}) are inverses
of each other.
Unlike our earlier derivation, we are not going to assume that our explanatory $t$ has been normalized and
shifted for our convenience, so we will use different symbols. Determining that scaling and shifting is
the work of logistic regression.
Logistic regression assumes that the $\mbox{logit}$ function (\ref{eq:logit}) is a linear
function of the explanatory variable $x$.
\begin{eqnarray}
\label{eq:logitx}
\log \left [ {p(x) \over 1-p(x)} \right ] &=& \beta_0 + \beta_1 x .
\end{eqnarray}
Estimating these two constants $\beta_0$ and $\beta_1$
finds the appropriate horizontal scaling and the midpoint of our curve,
so that we could redefine a normalized $t \equiv \beta_0 + \beta_1 x$
and use our previous equations.
Fitting data with this curve (\ref{eq:logitx}) is still best addressed as a maximum likelihood optimization.
We have a record of successes and failures, each with different values of the
explanatory variable $x$. We adjust the constants until the computed probability (\ref{eq:logitx})
of these events is maximized.
Alternatively, we are fitting a straight line to a graph with a value of $x$ as the horizontal
abscissa and the logit function (\ref{eq:logitx}) as the vertical ordinate.
But a normal least-squares linear regression
will not distribute the errors as correctly as a maximum-likelihood optimization.
The log-odds might seem like an arbitrary quantity to fit, but it has a connection to information theory.
The entropy $H$ of a single binary outcome with probability $p$ is defined as
\begin{eqnarray}
\label{eq:entropy}
H(p) \equiv - p \log p - (1 - p) \log (1 - p) .
\end{eqnarray}
This entropy has a maximum value of $\log(2)$ for the probability $p=\frac{1}{2}$, which is
the most unpredictable distribution.
When the probability is low (near 0) or high (near 1), then the entropy approaches a minimum value of 0.
A lower entropy is a more predictable outcome, with 0 giving us complete certainty.
The derivative of the entropy with respect to $p$ gives us the negative of the logit function:
\begin{eqnarray}
\label{eq:dentropy}
{d H(p) \over dp} = - \mbox{logit}(p).
% dH(p)/dp = -log(p) - 1 + log(1-p) + (1-p)/(1-p) = log[(1-p)/p] = -log[p/(1-p)]
\end{eqnarray}
If we assume the $\mbox{logit}$ is a linear function of the variable $x$ then the
entropy is a second-order polynomial, with just enough degrees of freedom for a
single maximum and an adjustable width.
\end{document}
(Neural networks, and logistic regression prefer a different
motivation for their S-curve. Notice that the logarithm of this ratio
(\ref{eq:logit}) is called the logit function.
This particular logit function is a simple linearly increasing
function of time.
The logit function is also equal to the negative derivative of
the binary entropy function, which measures the uncertainty
of two possible outcomes. Integrating shows that the
binary entropy is changing as a negative square of time, a parabola
convex down, centered at zero time.)
As a curiosity, this derivation also
shows that the ratio (\ref{eq:success})
of used capacity to the remaining capacity increases
exponentially for all times.
This suggests an alternative derivation.
The ability to improve this ratio
is proportional to the ratio itself.
This might explain the unjustified optimism that
sometimes accompanies the exhaustion of a
depleting resource. The ratio of success to failure is
still growing geometrically!
This might also motivate somewhat the use of the S-curve
in neural networks. The ratio (odds) of certainty
to uncertainty is allowed to grow exponentially
with new information. Overall, however, I do not find
this behavior very helpful to intuition.
%\bibliography{wsh}