Deriving $e^{ix} = 1 + ix + \frac{(ix)^2}{2!}+…$

I know how to prove equation:

$$e = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right)^x$$

How can I now derive the series:

$$e^{ix} = 1 + ix + \frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+...$$

Those two seem very similar to me...

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thats not eulers constant.. it's bernoullis constant – user58512 Feb 4 '13 at 21:51

Use the binomial theorem on the following limit:

$$e^z = \lim_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x$$

i.e.,

\begin{align} \left( 1 + \frac{z}{x} \right)^x &= 1 + \frac{x}{1!} \frac{z}{x} + \frac{x (x-1)}{2!} \left (\frac{z}{x} \right )^2 + \frac{x (x-1) (x-2)}{3!} \left (\frac{z}{x} \right )^3 + \ldots \\ &= 1 + \frac{z}{1!} + \frac{z^2}{2!} \left ( 1-\frac{1}{x} \right ) + \frac{z^3}{3!} \left ( 1-\frac{1}{x} \right ) \left ( 1-\frac{2}{x} \right ) + \ldots \\ \end{align}

As $x \rightarrow \infty$, the product terms approach $1$. Therefore

$$\lim_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x = \sum_{k=0}^{\infty} \frac{z^k}{k!} = e^z$$

Now, for $z=i x$,

$$e^{i x} = \sum_{k=0}^{\infty} \frac{(i x)^k}{k!} = 1 + i x + \frac{(i x)^2}{2!} + \ldots$$

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Fist i have to ask you how did u get $e^z = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x$ out of $e = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right)^x$? – 71GA Feb 4 '13 at 21:43
I didn't. I was just answering your question about getting $e^{i x}$, and I was using the simplest way I knew how from a similar limit. I proved to you that the former equation in your comment is true, then I applied $z=i x$ to answer your question. – Ron Gordon Feb 4 '13 at 21:48
Is there any way you could supply a proof for $e^z = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x$ ? – 71GA Feb 4 '13 at 22:13
Isn't that what I did? I am confused. – Ron Gordon Feb 4 '13 at 22:17
@71GA, that limit formula is usually given as the definition of $e^x$. You can also define $f(x)=e^x$ as the function such that $f'(x)=f(x)$ and $f(0)=1$. You can divide the interval $[0,x]$ into n parts, and use step along it using euler integration to estimate $f(x)$. You wind up with that exact formula, and then you take the limit as n goes to infinity. (not too rigorous, but intuitive!) Maybe I should write something up on this... I haven't found it online (which is weird, because I read it online years ago). – NeuroFuzzy Feb 4 '13 at 22:26

If we are allowed to use calculus,

let $$z=1+\frac y{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+\frac{y^4}{4!}+\cdots$$

So, $$\frac{dz}{dy}=0+1+\frac y{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+\frac{y^4}{4!}+\cdots=z$$

$$\implies \frac{dz}z=dy$$

Integrating both sides, $\log z=y+c$

For $y=0,z=1\implies c=\log1-0=0\implies \log z=y\iff z=e^y$

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