# Euler's Formula

Euler's formula, the most remarkable formula in mathematics, according to Richard Feynman and many others, states that for any real number $$x$$,

$e^{ix} = \cos x + i \sin x$

In this section we will derive Euler's formula to get a better understanding of how Euler's formula can be used to describe rotations in space and time, eventually including the Lorentz transformation.

##### Skip if you prefer...
The derivation of Euler's formula may be a little complicated if you've not touched calculus in a while. Feel free to skip this section for now and go straight to the next section, coming back later as you need.

## Finding a function whose derivative is the function itself

Euler's formula comes about by first considering a function of $$x$$, for which the derivative of the function for any value of $$x$$ has the same value as function itself. That is:

$\frac{d}{dx}f(x) = f(x)$

If we start with a guess of $$f(x) = 1 + x$$ then we see it is not quite right as

$\frac{d}{dx}(1 + x) = 1 \ne (1 + x)$

but we can improve it by adding another term

$f(x) = 1 + x + \frac{x^2}{2}$

which is a little closer in that

$\frac{d}{dx}(1 + x + \frac{x^2}{2}) = 1 + x \ne f(x)$

but still not the same. So we improve it again with

$f(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{2\times 3}$

and so on, until we end up with an infinite sum of terms:

$f(x) = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ... = \sum_{k=0}^{\infty} \frac{x^k}{k!}$

We now have a function satisfying $$\frac{d}{dx} f(x) = f(x)$$.

##### The derivative equals the function
Take a couple of minutes to see if you can show that the derivative of our function: $$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$ is itself.

Solution

We can work out the derivative of each term in the infinite sum of terms,

\begin{aligned} \frac{d}{dx} f(x) &= \frac{d}{dx} \sum_{k=0}^{\infty} \frac{x^k}{k!} \\ &= \sum_{k=0}^{\infty} \frac{kx^{k-1}}{k!} \\ \end{aligned}

so we can completely remove the term for $$k=0$$ (as zero times anything is still zero), and continue:

\begin{aligned} \frac{d}{dx} f(x) &= \sum_{k=1}^{\infty} k\frac{x^{k-1}}{k!} \\ &= \sum_{k=1}^{\infty} \frac{x^{k-1}}{(k - 1)!} \\ &= \sum_{k=0}^{\infty} \frac{x^k}{k!} \\ &= f(x) \end{aligned}

Done!

Great, but where does $$e$$ fit in?

## Understanding what an exponent is

We learn in high school that $$3^2 \times 3 = 3^3$$ and $$2^2 \times 2 \times 2 = 2^4$$ and later that $$x^2 \times x^2 = x^4$$ etc., but the general definition of exponentiation is that when multiplying two numbers with the same base ($$3^2 \times 3^1$$), the powers (or exponents) are added ($$3^{2+1} = 3^3$$):

$b^m \times b^n = b^{m + n}$

## Our function is an exponential

If we take our function which satisfies the property $$\frac{d}{dx}f(x) = f(x)$$ we can show that it behaves just like an exponent should, in that:

$$f(m) \times f(n) = f(m+n)$$

To show this requires quite a bit of re-arranging:

\begin{aligned} f(m) \times f(n) &= \sum_{k=0}^{\infty} \frac{m^k}{k!} \times \sum_{k=0}^{\infty} \frac{n^k}{k!} \\ &= (\frac{m^0}{0!} + \frac{m^1}{1!} + \frac{m^2}{2!} + \frac{m^3}{3!} + ...) \times (\frac{n^0}{0!} + \frac{n^1}{1!} + \frac{n^2}{2!} + \frac{n^3}{3!} + ...) \\ &= 1 + (\frac{m^1}{1!} + \frac{n^1}{1!}) + (\frac{m^2}{2!} + \frac{m^1 n^1}{1! 1!} + \frac{n^2}{2!}) + (\frac{m^3}{3!} + \frac{m^2 n^1}{2! 1!} + \frac{m^1 n^2}{1! 2!} + \frac{n^3}{3!}) ... \\ &= 1 + (\frac{m^1}{1!} + \frac{n^1}{1!}) + (\frac{m^2}{2!} + \frac{2}{2}\frac{m^1 n^1}{1! 1!} + \frac{n^2}{2!}) + (\frac{m^3}{3!} + \frac{3}{3}\frac{m^2 n^1}{2! 1!} + \frac{3}{3}\frac{m^1 n^2}{1! 2!} + \frac{n^3}{3!}) ... \\ &= \sum_{k=0}^{\infty} \frac{(m + n)^k}{k!} \\ &= f(m + n) \end{aligned}

Also similar to an exponent, we can show for our function $$f(x)$$, that $$f(0) = 1$$.

##### Identity of f(0)
Show that $$f(0) = 1$$ for our function: $$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$

Solution:

Substituting $$x=0$$,

\begin{aligned} f(0) &= \sum_{k=0}^{\infty} \frac{0^k}{k!} \\ &= \frac{0^0}{0!} + \frac{0^1}{1!} + \frac{0^2}{2!} + ... \\ &= 1 + 0 + 0 + ... \\ &= 1 \end{aligned}

Done!

Given that our function $$f(x)$$ behaves just like an exponent should behave, it must be equivalent to some number to the power of $$x$$. We can find that number by working out $$f(1)$$.

##### Find f(1)
Show that $$f(1) \approx 2.717$$

Solution:

Evaluating,

\begin{aligned} f(1) &= \sum_{k=0}^{\infty} \frac{1!}{k!} \\ &= \sum_{k=0}^{\infty} \frac{1}{k!} \\ &= \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + ... \\ &= 1 + 1 + 0.5 + 0.1667 + 0.0417 + 0.0083... \\ &= 2.717 \end{aligned}

The exact value (if we summed to infinity) is assigned a special name in maths, e.

So,

$e^1 = f(1) = \sum_{k=0}^{\infty} \frac{1}{k!} \approx 2.71828...$

Now any value of our function can be evaluated simply as a power of $$e$$:

$f(3) = e^3 = 2.71828... ^3 = 20.08534...$

## Derivatives of exponentials

What if we need to work with other exponents of $$e$$, such as $$e^{3x}$$? How do we work out the derivative - how fast $$e^{3x}$$ is changing as $$x$$ changes? Your memory may (or may not) tell you that:

$\frac{d}{dx} e^{3x} = 3e^{3x}$

or more generally, if $$g$$ is another function of $$x$$, that

$\frac{d}{dx} e^{g} = \frac{dg}{dx}e^{g}$

This is all we need for our tooling here, though we can also see why this is the case given the chain rule for derivatives which says that:

$\frac{d}{dx}f(g) = \frac{dg}{dx} \times \frac{d}{d g}f(g)$

together with the special property of our function $$f(x) = e^x$$ that the derivative at any point is equal to the function itself, $$\frac{d}{dx}f(x) = f(x)$$.

## Euler's formula

Now we have enough background to appreciate the beauty of Euler's formula. Euler, like Roger Cotes before him, noticed that if he evaluated this exponent function with a special type of value, an "imaginary" value whose square is negative, the result is a combination of the trigonometric functions $$\cos$$ and $$sin$$. That is, if we define $$i$$ such that $$i^2 = -1$$, and evaluate $$e^{ix}$$, it conveniently splits into two sums, one of the odd powers and one of the even powers:

\begin{aligned} e^{ix} &= \sum_{k=0}^{\infty} \frac{(ix)^k}{k!} \\ &= \frac{(ix)^0}{0!} + \frac{(ix)^1}{1!} + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} ... \\ &= (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} ...) + i(\frac{x^1}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} ...) \\ &= \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{2k!} + i \sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!} \\ &= \cos(x) + i\sin(x) \end{aligned}

That is, the two separate summations are the actual definitions of $$\cos(x)$$ and $$\sin(x)$$.

Notice that, with our introduction to Geometric Algebra, we did not need to invent an imaginary unit $$i$$ here, as we have already seen that the product of basis vectors $$\mathbf{xy}$$ has the same property that $$(\mathbf{xy})^2 = -1$$, but we will pick up this thread in Euler's formula and Geometric Algebra next.

## Derivatives of cos and sin

It's worth noting that we can calculate the derivative of $$e^{ix}$$ and as a result work out the standard derivations of $$\cos$$ and $$sin$$ that we may have memorized in school or otherwise:

\begin{aligned} \frac{d}{dx}e^{ix} &= ie^{ix} \\ &= i\cos(x) + i^2\sin(x) \\ &= -\sin(x) + i\cos(x) \end{aligned}

and so comparing the real parts of $$e^{ix}$$ and $$\frac{d}{dx}e^{ix}$$ show that

$\frac{d}{dx}\cos(x) = -\sin(x)$

while comparing the imaginary parts show that

$\frac{d}{dx}\sin(x) = \cos(x)$