"My" proof for Euler's identity

Few years back, I was sitting in the classroom during my M.E. getting bored. Then I remembered of one of the most beautiful equations - Euler's identity:
\[e^{i\pi}+1=0\]
So I decided to prove it using Taylor series.
Yes, you can ask why? The answer is - because I was bored and because I can...

So, first a reminder - Taylor series definition:
\[
\sum_{n=0}^{\infty}\frac{f^{(n)}(a)\cdot(x-a)^n}{n!}
\]
Now, Taylor series for some basic functions (\(a=0\)):
\(

\begin{align*}
e^x &= e^0+\frac{e^0x}{1!}+\frac{e^0x^2}{2!}+\frac{e^0x^3}{3!}+\dots \\
&= 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots
\end{align*}
\)

\(
\begin{align*}sin(x) &= 0+\frac{cos(0)x}{1!}+\frac{-sin(0)x^2}{2!}+\frac{-cos(0)x^3}{3!}+\dots \\
&= \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots
\end{align*}
\)

\(
\begin{align*}cos(x) &= 1+\frac{-sin(0)x}{1!}+\frac{-cos(0)x^2}{2!}+\frac{sin(0)x^3}{3!}+\dots \\
&= 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots
\end{align*}
\)

Let's start. If we assign \(x=ix\) in \(e^x\) we get:
\(
\begin{align*}
e^{ix} &= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \dots \\
&= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \dots \\
\end{align*}
\)

If we reorder what we got, we get:
\(
\begin{align*}
e^{ix} &= \underbrace{1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots}_\text{cos(x)} + i \cdot (\underbrace{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots}_\text{sin(x)}) \\
&= cos(x) +i \cdot sin(x) \end{align*}
\)

Let's assign \( x = \pi \):
\(
\begin{align*}
e^{i\pi} &= cos(\pi) +i \cdot sin(\pi) \\
&= -1 + i \cdot 0 \\ &= -1 \end{align*}
\)

So, finally we got what we wished for:
\(
e^{i\pi} + 1 = 0
\)


OK, now - for all of you saying "of course - the identity is just a special case for Euler's formula: \(e^{ix}=cos(x) +i \cdot sin(x)\)" - I know and I "proved" it (well, it's not a real proof and it's not really mine to claim...). Oh, and I also wanted to write some \(\LaTeX\)...
So that was me goofing around out of boredom..


The \(\LaTeX\) in this post is powered by MathJax