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So, I know there is a post that already asks this question, but the question I’m asking is if my proof ,up to now, is correct, and how to finish it. I have not studied sin and cos, so please refrain from using those in your answer.

So, if we break it down, $e^{i\pi}=i^2=-1$. If we replaced the values in this equation by, say, 2, 3 and x, then we get $2^{3x}=3^2=9$ To get 9, we need to find $\frac {log_2(9)}{3}$, which is equal to about $1.056641667$. If we replace $x$ by this new number, then we get $2^{3\times1.056641667}$, which is roughly equal to $9$. Now, going back to $e^{i\pi}$, we can apply the same strategy. We need to find $\frac{log_e(\pi)}{i}$. $log_e(\pi)$ is easy, it’s about 1.144729886. I’m just wondering how to get $\frac {1.144729886}{i}$.

Rhys Hughes
  • 13,103

1 Answers1

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It's very difficult to formulate a proof of this without the use of $\sin$ and $\cos$, so I will do my best to explain every usage of it.

The Taylor Expansion of $e^y=1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+...$

Applying $y=ix$ we get $e^{ix}=1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+...$

Here is where $\sin$ and $\cos$ come into play. Their Taylor Expansions are: $$\cos(x)=1-\frac{x^2}{2x}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$$ $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$

This means that $e^{ix}=\cos(x)+i\sin(x)$ by inspection.

$$\to e^{i\pi}=\cos(\pi)+i\sin(\pi)$$ Looking here you will see that $\cos(\pi)=-1$ and $\sin(\pi)=0$.

Hence $e^{i\pi}=-1+0i=-1$.

Rhys Hughes
  • 13,103