05
What is $e$?
core concept
The number $e$ is defined as the limit of $(1 + \frac{1}{n})^n$ as $n$ grows without bound. The table below shows how this converges:
As $n$ increases, $(1 + \frac{1}{n})^n$ approaches $e$ from below. It never exceeds $e$, this is why continuously compounded interest hits a ceiling. The function $y = e^x$ is special because $\dfrac{d}{dx}(e^x) = e^x$ exactly, no scaling factor, no remainder. That's what makes it the natural base for calculus.
$$e = \lim_{n \to \infty} \left(1 + \dfrac{1}{n}\right)^n \approx 2.71828$$
Do this one with a graphing application. The result below is the reason $e$ is the base calculus uses, and it is worth seeing rather than being told.
Exploration. Plot $y = 2^x$, then $y = 3^x$, on the same axes. Both pass through $(0, 1)$, so compare them at that point only.
Draw the tangent at the $y$-intercept of each and read its gradient. For $y = 2^x$ the gradient at $(0, 1)$ is about $0.69$; for $y = 3^x$ it is about $1.10$. Neither is $1$.
Now raise the base slowly from $2$ towards $3$ and watch that gradient climb through $1$. The base at which the tangent at $(0, 1)$ has gradient exactly $1$ is $e$.
That is the definition worth carrying: $e$ is the base whose graph rises at exactly the rate of its own height. The limit $\left(1 + \frac{1}{n}\right)^n$ and this gradient property describe the same number.
Evaluating with technology. Use the $e^x$, $\ln$ and $\log$ keys to evaluate $e^{1.7}$, $\ln 12$ and $\log_2 40$, the last by change of base. Notice which results are exact and which are not: $\ln 1 = 0$ and $\log_2 8 = 3$ are rational, while $e^{1.7} \approx 5.4739$ and $\ln 12 \approx 2.4849$ are irrational and must be rounded. Rounding is a decision you make and state, not something the calculator decides for you.
The graph of $y = e^x$ is an increasing exponential with asymptote $y = 0$, $y$-intercept $(0,1)$, and it passes through $(1, e) \approx (1, 2.718)$. Its mirror image $y = e^{-x}$ is a decreasing exponential, reflected in the $y$-axis.
Why $e$ matters beyond finance. In physics, radioactive decay follows $N = N_0 e^{-\lambda t}$. In biology, population growth uses $P = P_0 e^{kt}$. In every case, the reason $e$ appears is that it is the only base where the rate of change equals the current quantity, a statement that would be more complicated with any other base.
Beyond the syllabus. The two applications just named, radioactive decay and population growth, are Year 12 modelling. MAV-11-07 asks you to investigate the gradients of $y = a^x$ graphically, to identify $e$ as the base where that gradient equals the function itself, and to establish $\frac{d}{dx}(e^x) = e^x$. That is what is assessed here. The decay and growth models are shown only to say why $e$ turns up outside finance, so read them for the reason and not as formulas to memorise for this focus area.
$e = \lim_{n \to \infty}(1 + \frac{1}{n})^n \approx 2.71828$, irrational like $\pi$; $\frac{d}{dx}(e^x) = e^x$, the self-derivative property
Pause, copy the definition of $e$ ($\approx 2.71828$, irrational) and the self-derivative property $\frac{d}{dx}(e^x) = e^x$, the reason $e$ is the natural base into your book.
