GoldenOj suggested the exponential as a topic. It seemed like a good important topic, but one that was already well-explored by other people. Then I realized I could spend time thinking about something which had bothered me.

In here I write about “the” exponential, which is a bit like writing about “the” multiplication. We can talk about $2^3$ and $10^2$ and many other such exponential functions. One secret of algebra, not appreciated until calculus (or later), is that all these different functions are a single family. Understanding one exponential function lets you understand them all. Mathematicians pick one, the exponential with base e, because we find that convenient. e itself isn’t a convenient number — it’s a bit over 2.718 — but it has some wonderful properties. When I write “the exponential” here, I am looking at this function where we look at $e^{t}$ .

This piece will have a bit more mathematics, as in equations, than usual. If you like me writing about mathematics more than reading equations, you’re hardly alone. I recommend letting your eyes drop to the next sentence, or at least the next sentence that makes sense. You should be fine.

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Exponential.

My professor for real analysis, in grad school, gave us one of those brilliant projects. Starting from the definition of the logarithm, as an integral, prove at least thirty things. They could be as trivial as “the log of 1 is 0”. They could be as subtle as how to calculate the log of one number in a different base. It was a great project for testing what we knew about why calculus works.

And it gives me the structure to write about the exponential function. Anyone reading a pop-mathematics blog about exponentials knows them. They’re these functions that, as the independent variable grows, grow ever-faster. Or that decay asymptotically to zero. Some readers know that, if the independent variable is an imaginary number, the exponential is a complex number too. As the independent variable grows, becoming a bigger imaginary number, the exponential doesn’t grow. It oscillates, a sine wave.

That’s weird. I’d like to see why that makes sense.

To say “why” this makes sense is doomed. It’s like explaining “why” 36 is divisible by three and six and nine but not eight. It follows from what the words we have mean. The “why” I’ll offer is reasons why this strange behavior is plausible. It’ll be a mix of deductive reasoning and heuristics. This is a common blend when trying to understand why a result happens, or why we should accept it.

I’ll start with the definition of the logarithm, as used in real analysis. The natural logarithm, if you’re curious. It has a lot of nice properties. You can use this to prove over thirty things. Here it is:

$log\left(x\right) = \int_{1}^{x} \frac{1}{s} ds$

The “s” is a dummy variable. You’ll never see it in actual use.

So now let me summon into existence a new function. I want to call it g. This is because I’ve worked this out before and I want to label something else as f. There is something coming ahead that’s a bit of a syntactic mess. This is the best way around it that I can find.

$g(x) = \frac{1}{c} \int_{1}^{x} \frac{1}{s} ds$

Here, ‘c’ is a constant. It might be real. It might be imaginary. It might be complex. I’m using ‘c’ rather than ‘a’ or ‘b’ so that I can later on play with possibilities.

So the alert reader noticed that g(x) here means “take the logarithm of x, and divide it by a constant”. So it does. I’ll need two things built off of g(x), though. The first is its derivative. That’s taken with respect to x, the only variable. Finding the derivative of an integral sounds intimidating but, happy to say, we have a theorem to make this easy. It’s the Fundamental Theorem of Calculus, and it tells us:

$g'(x) = \frac{1}{c}\cdot\frac{1}{x}$

We can use the ‘ to denote “first derivative” if a function has only one variable. Saves time to write and is easier to type.

The other thing that I need, and the thing I really want, is the inverse of g. I’m going to call this function f(t). A more common notation would be to write $g^{-1}(t)$ but we already have $g'(x)$ in the works here. There is a limit to how many little one-stroke superscripts we need above g. This is the tradeoff to using ‘ for first derivatives. But here’s the important thing:

$x = f(t) = g^{-1}(t)$

Here, we have some extratextual information. We know the inverse of a logarithm is an exponential. We even have a standard notation for that. We’d write

$x = f(t) = e^{ct}$

in any context besides this essay as I’ve set it up.

What I would like to know next is: what is the derivative of f(t)? This sounds impossible to know, if we’re thinking of “the inverse of this integration”. It’s not. We have the Inverse Function Theorem to come to our aid. We encounter the Inverse Function Theorem briefly, in freshman calculus. There we use it to do as many as two problems and then hide away forever from the Inverse Function Theorem. (This is why it’s not mentioned in my quick little guide to how to take derivatives.) It reappears in real analysis for this sort of contingency. The inverse function theorem tells us, if f the inverse of g, that:

$f'(t) = \frac{1}{g'(f(t))}$

That g'(f(t)) means, use the rule for g'(x), with f(t) substituted in place of ‘x’. And now we see something magic:

$f'(t) = \frac{1}{\frac{1}{c}\cdot\frac{1}{f(t)}}$

$f'(t) = c\cdot f(t)$

And that is the wonderful thing about the exponential. Its derivative is a constant times its original value. That alone would make the exponential one of mathematics’ favorite functions. It allows us, for example, to transform differential equations into polynomials. (If you want everlasting fame, albeit among mathematicians, invent a new way to turn differential equations into polynomials.) Because we could turn, say,

$f'''(t) - 3f''(t) + 3f'(t) - f(t) = 0$

into

$c^3 e^{ct} - 3c^2 e^{ct} + 3c e^{ct} - e^{ct} = 0$

and then

$\left(c^3 - 3c^2 + 3c - 1\right) e^{ct} = 0$

by supposing that f(t) has to be $e^{ct}$ for the correct value of c. Then all you need do is find a value of ‘c’ that makes that last equation true.

Supposing that the answer has this convenient form may remind you of searching for the lost keys over here where the light is better. But we find so many keys in this good light. If you carry on in mathematics you will never stop seeing this trick, although it may be disguised.

In part because it’s so easy to work with. In part because exponentials like this cover so much of what we might like to do. Let’s go back to looking at the derivative of the exponential function.

$f'(t) = c\cdot f(t)$

There are many ways to understand what a derivative is. One compelling way is to think of it as the rate of change. If you make a tiny change in t, how big is the change in f(t)? So what is the rate of change here?

We can pose this as a pretend-physics problem. This lets us use our physical intuition to understand things. This also is the transition between careful reasoning and ad-hoc arguments. Imagine a particle that, at time ‘t’, is at the position $x = f(t)$ . What is its velocity? That’s the first derivative of its position, so, $x' = f'(t) = c\cdot f(t)$ .

If we are using our physics intuition to understand this it helps to go all the way. Where is the particle? Can we plot that? … Sure. We’re used to matching real numbers with points on a number line. Go ahead and do that. Not to give away spoilers, but we will want to think about complex numbers too. Mathematicians are used to matching complex numbers with points on the Cartesian plane, though. The real part of the complex number matches the horizontal coordinate. The imaginary part matches the vertical coordinate.

So how is this particle moving?

To say for sure we need some value of t. All right. Pick your favorite number. That’s our t. f(t) follows from whatever your t was. What’s interesting is that the change also depends on c. There’s a couple possibilities. Let me go through them.

First, what if c is zero? Well, then the definition of g(t) was gibberish and we can’t have that. All right.

What if c is a positive real number? Well, then, f'(t) is some positive multiple of whatever f(t) was. The change is “away from zero”. The particle will push away from the origin. As t increases, f(t) increases, so it pushes away faster and faster. This is exponential growth.

What if c is a negative real number? Well, then, f'(t) is some negative multiple of whatever f(t) was. The change is “towards zero”. The particle pulls toward the origin. But the closer it gets the more slowly it approaches. If t is large enough, f(t) will be so tiny that $c\cdot f(t)$ is too small to notice. The motion declines into imperceptibility.

What if c is an imaginary number, though?

So let’s suppose that c is equal to some real number b times $\imath$ , where $\imath^2 = -1$ .

I need some way to describe what value f(t) has, for whatever your pick of t was. Let me say it’s equal to $\alpha + \beta\imath$ , where $\alpha$ and $\beta$ are some real numbers whose value I don’t care about. What’s important here is that $f(t) = \alpha + \beta\imath$ .

And, then, what’s the first derivative? The magnitude and direction of motion? That’s easy to calculate; it’ll be $\imath b f(t) = -\beta + \alpha\imath$ . This is an interesting complex number. Do you see what’s interesting about it? I’ll get there next paragraph.

So f(t) matches some point on the Cartesian plane. But f'(t), the direction our particle moves with a small change in t, is another poiat whatever complex number f'(t) is as another point on the plane. The line segment connecting the origin to f(t) is perpendicular to the one connecting the origin to f'(t). The ‘motion’ of this particle is perpendicular to its position. And it always is. There’s several ways to show this. An easy one is to just pick some values for $\alpha$ and $\beta$ and b and try it out. This proof is not rigorous, but it is quick and convincing.

If your direction of motion is always perpendicular to your position, then what you’re doing is moving in a circle around the origin. This we pick up in physics, but it applies to the pretend-particle moving here. The exponentials of $\imath t$ and $2 \imath t$ and $-40 \imath t$ will all be points on a locus that’s a circle centered on the origin. The values will look like the cosine of an angle plus $\imath$ times the sine of an angle.

And there, I think, we finally get some justification for the exponential of an imaginary number being a complex number. And for why exponentials might have anything to do with cosines and sines.

You might ask what if c is a complex number, if it’s equal to $a + b\imath$ for some real numbers a and b. In this case, you get spirals as t changes. If a is positive, you get points spiralling outward as t increases. If a is negative, you get points spiralling inward toward zero as t increases. If b is positive the spirals go counterclockwise. If b is negative the spirals go clockwise. $e^{(a + \imath b) t}$ is the same as $e^{at} \cdot e^{\imath b t}$ .

This does depend on knowing the exponential of a sum of terms, such as of $a + \imath b$ , is equal to the product of the exponential of those terms. This is a good thing to have in your portfolio. If I remember right, it comes in around the 25th thing. It’s an easy result to have if you already showed something about the logarithms of products.