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Intuition for the exponential

I take the fact that the derivative of the exponential function is the exponential function itself as its defining feature and try to “rediscover” the power series expansion and Euler’s formula.

Exponential power series

I’ll make the educated guess that there could be some kind of polynomial ∑k=0nakxk\sum_{k=0}^n a_k x^k with this feature (after all, they do give us smooth curves and being able to add arbitrarily many terms makes them pretty flexible).

What would such a polynomial look like? Well, the limit of a sum is the sum of the limits of its terms, so I’ll look at the derivative of an individual term akxka_k x^k. According to the power rule, the derivative of this term is akkxk−1.a_k k x^{k-1}. I’d like to make sure that the derivative is the same as the initial function, so there are two problems here.

  1. A factor kk pops up that was not there before.
  2. The exponent is decremented by one.

To solve the first problem, it would be convenient if the new factor canceled out. This can be achieved if ak=1k!a_k = \frac{1}{k!}, so that 1k!kxk−1=1(k−1)!xk−1.\frac{1}{k!} k x^{k-1} = \frac{1}{(k-1)!} x^{k-1}. With this little trick the kth term becomes the (k-1)th term after differentiation. The first term x00!=1\frac{x^0}{0!} = 1 is just a constant and disappears. To not lose any terms, we need an infinite number of them. So the “polynomial” is actually a power series.

ex=∑k=0∞xkk! e^x = \sum_{k=0}^\infty \frac{x^k}{k!}

Euler’s formula

The trigonometric functions sine and cosine are continuous and periodic and if you cycle through their derivatives, they eventually re-appear. This seems promising. Maybe they could be used to represent the exponential function?

(cosθ)′=−sinθ(sinθ)′=cosθ \begin{align*} (\cos \theta)' &= -\sin \theta \\ (\sin \theta)' &= \cos \theta \end{align*}

Take a look at the unit circle. For any angle θ\theta, cosθ\cos \theta is the x-value and sinθ\sin \theta the y-value of a point on this circle. Now take the derivative of both functions and stay on the same axis such that you get a new point with −sinθ-\sin \theta as x-value and cosθ\cos \theta as y-value. To get to the original point, all you need is a clockwise rotation of π/2\pi / 2 around the origin.

The complex plane

This idea can be represented nicely in the complex plane, where the x-axis represents the real part and the y-axis the imaginary part of a complex number. To get to a point zz on the circle, add the real and imaginary part (z=cosθ+isinθ).(z = \cos \theta + i \sin \theta). Now take the derivative of this sum.

ddθ(cosθ+isinθ)=−sinθ+icosθ \frac{d}{d \theta} \left( \cos \theta + i \sin \theta \right) = -\sin \theta + i \cos \theta

Multiplication with ii must be a counterclockwise rotation of π/2\pi / 2 around the origin since it takes the point (1,0)(1, 0) to (0,i).(0, i). So I multiply both sides by 1/i1 / i to rotate clockwise.

1i⋅ddθ(cosθ+isinθ)=−sinθ+icosθi=−sinθi+cosθ=cosθ−sinθ(−i)i(−i)=cosθ+isinθ \begin{align*} \frac{1}{i} \cdot \frac{d}{d \theta} \left( \cos \theta + i \sin \theta \right) &= \frac{-\sin \theta + i \cos \theta}{i}\\ &= \frac{-\sin \theta}{i} + \cos \theta \\ &= \cos \theta - \frac{\sin \theta (-i)}{i (-i)} \\ &= \cos \theta + i \sin \theta \end{align*}

Finding the exponential via the differential equation

Now I’m looking for another function ff such that the following equation holds.

1i⋅dfdθ=f \frac{1}{i} \cdot \frac{d f}{d \theta} = f

If you know the chain rule, at this point it becomes quite clear that f(θ)=eiθf(\theta) = e^{i \theta} is such a function because ddθeiθ=eiθi.\frac{d}{d \theta} e^{i \theta} = e^{i \theta} i. So we have two functions that satisfy the same differential equation.

f′=ieiθ=ifg′=−sinθ+icosθ=ig \begin{align*} f' &= i e^{i \theta} &= i f\\ g' &= -\sin \theta + i \cos \theta &= i g \end{align*}

Proving equality

But are they actually the same? Using the quotient rule to compute the derivative of ff divided by gg , we find that the rate of change of their ratio is always zero. (Dividing by g=cosθ+isinθg = \cos \theta + i \sin \theta is fine since it describes the points on the unit circle, so it is never zero.)

(fg)′=f′g−fg′g2=ifg−figg2=0 \begin{align*} \left( \frac{f}{g} \right)' &= \frac{f' g - f g'}{g^2}\\ &= \frac{ifg - fig}{g^2}\\ &= 0 \end{align*}

This means that fg\frac{f}{g} must be constant and we can compute this constant by evaluating the ratio at any point. Zero is easy.

(fg)(0)=ei⋅0cos0+isin0=11+0=1 \begin{align*} \left( \frac{f}{g} \right) (0) &= \frac{e^{i \cdot 0}}{\cos 0 + i \sin 0}\\ &= \frac{1}{1 + 0}\\ &= 1 \end{align*}

The ratio is 11 for any θ\theta, so ff and gg must be equal and we end up with Euler’s famous formula.

eiθ=cosθ+isinθ e^{i \theta} = \cos \theta + i \sin \theta