Starting with power series and building on derivation, factorials, exponential function, trigonometry and complex numbers we’ll get to Euler’s formula; some of the most beautiful mathematics I’ve seen coming together.

Power series

Let’s start with powers of $x$ :

$\begin{align} x^0~~~x^1~~~x^2~~~x^3~~~... \end{align}$

Obviously:

$\begin{align} x^0 = 1\\ x^1 = x \end{align}$

So here’s usual way of writing powers of $x$ :

$\begin{align} 1~~~x~~~x^2~~~x^3~~~... \end{align}$

We can create a function where we multiply each power of $x$ by some coefficient and we add up the infinite series:

$\begin{align} f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... \end{align}$

Derivations of a power series

So now let’s look at the derivations for $f(x)$ :

$\begin{align} f'(x) &= a_1 + 2~a_2 x + 3~a_3 x^2 + 4~a_4 x^3 + ...\\ f''(x) &= 2~a_2 + 2~3~a_3 x + 3~4~a_4 x^2 + 4~5~a_5 x^3 +...\\ f'''(x) &= 2~3~a_3 + 2~3~4~a_4 x + 3~4~5~a_5 x^2 + ...\\ ... \end{align}$

If we choose $x = 0$ then all the terms except the constant at the front go away:

$\begin{align} f(0) &= a_0\\ f'(0) &= a_1\\ f''(0) &= 2~a_2 = 2!~a_2\\ f'''(0) &= 2~3~a_3 = 3!~a_3\\ ... \end{align}$

And in general because $2~3~4~...~n$ is $n~factorial$ written as $n!$ :

$\begin{align} n^{\text{th}}~derivative~of~f(0) &= n!~a_n \end{align}$

One useful fact is that for two functions $g$ and $h$ :

$\begin{align} \left\{\exists x_0~|~n^{\text{th}}~derivative~of~g(x_0) = n^{\text{th}}~derivative~of~h(x_0)~,~\forall n >= 0\right\} \end{align}$

then $g$ and $h$ are equal for all $x$ .

So then we could pick some function and then choose the coefficients $a_0, a_1, a_2, a_3, ...$ so that we have equality with that function for $x = 0$ .

Exponential function as a power series

In particular let’s pick $e^x$ for which we know that all its derivatives are equal (i.e. $e^x$ )

$\begin{align} (e^x)' &= e^x\\ (e^x)'' &= e^x\\ (e^x)''' &= e^x\\ ... \end{align}$

We also know that $e^0 = 1$ :

$\begin{align} e^0 &= 1\\ (e^0)' = e^0 &= 1\\ (e^0)'' = e^0 &= 1\\ (e^0)''' = e^0 &= 1\\ ... \end{align}$

To express $e^x$ as a power series we need to choose the coefficients $a_0, a_1, a_2, a_3, ...$ for our power series so that:

$\begin{align} 1 &= a_0\\ 1 &= a_1\\ 1 &= 2!~a_2\\ 1 &= 3!~a_3\\ ... \end{align}$

which leads to:

$\begin{align} e^x = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + ... \end{align}$

Sine as a power series

Let’s pick another function: $sin(x)$ . The derivatives of $sin(x)$ are:

$\begin{align} (sin(x))' &= cos(x)\\ (sin(x))'' = (cos(x))' &= -sin(x)\\ (sin(x))''' = (-sin(x))' &= -cos(x)\\ (sin(x))'''' = (-cos(x))' &= sin(x)\\ ... \end{align}$

For $x = 0$ the values are:

$\begin{align} sin(0) &= 0\\ (sin(0))' = cos(0) &= 1\\ (sin(0))'' = -sin(0) &= 0\\ (sin(0))''' = -cos(0) &= -1\\ (sin(0))'''' = sin(0) &= 0\\ ... \end{align}$

To express $sin(x)$ as a power series we need to choose the coefficients $a_0, a_1, a_2, a_3, ...$ for our power series so that:

$\begin{align} 0 &= a_0\\ 1 &= a_1\\ 0 &= 2!~a_2\\ -1 &= 3!~a_3\\ 0 &= 4!~a_3\\ ... \end{align}$

which leads to:

$\begin{align} sin(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + ... \end{align}$

Cosine as a power series

The relative of $sin(x)$ is $cos(x)$ . Its derivatives are:

$\begin{align} (cos(x))' &= -sin(x)\\ (cos(x))'' = (-sin(x))' &= -cos(x)\\ (cos(x))''' = (-cos(x))' &= sin(x)\\ (cos(x))'''' = (sin(x))' &= cos(x)\\ ... \end{align}$

For $x = 0$ the values are:

$\begin{align} cos(0) &= 1\\ (cos(0))' = -sin(0) &= 0\\ (cos(0))'' = -cos(0) &= -1\\ (cos(0))''' = sin(0) &= 0\\ (cos(0))'''' = cos(0) &= 1\\ ... \end{align}$

To express $cos(x)$ as a power series we need to choose the coefficients $a_0, a_1, a_2, a_3, ...$ for our power series so that:

$\begin{align} 1 &= a_0\\ 0 &= a_1\\ -1 &= 2!~a_2\\ 0 &= 3!~a_3\\ 1 &= 4!~a_3\\ ... \end{align}$

which leads to:

$\begin{align} cos(x) = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6 + ... \end{align}$

Euler’s formula

Remember that in complex numbers we have the imaginary number for which its powers are:

$\begin{align} i^0 &= 1\\ i^1 &= i\\ i^2 &= -1\\ i^3 &= -i\\ i^4 &= 1\\ ... \end{align}$

So going back to $e^x$ , what if we use an (complex) imaginary $ix$ we get:

$\begin{align} e^{\text{ix}} &= 1 + ix + \frac{1}{2!}(ix)^2 + \frac{1}{3!}(ix)^3 + ...\\ &= 1 + ix - \frac{1}{2!}x^2 - i\frac{1}{3!}x^3 + ...\\ &= (1 - \frac{1}{2!}x^2 + ...) + i( x - \frac{1}{3!}x^3 + ...) \end{align}$

which lead to Euler’s formula:

$\begin{align} e^{\text{ix}} &= cos(x) + i~sin(x) \end{align}$

References

Gilbert Strang: Power Series/Euler’s Great Formula | MIT Highlights of Calculus