Beautiful maths: Euler's formula
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:
\[\begin{align} e^{\text{ix}} &= cos(x) + i~sin(x) \end{align}\]