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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...

Joseph Quinsey
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71GA
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2 Answers2

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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 $z=i x$,

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

Ron Gordon
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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 04 '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 04 '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 04 '13 at 22:13
  • Isn't that what I did? I am confused. – Ron Gordon Feb 04 '13 at 22:17
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    @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). –  Feb 04 '13 at 22:26
  • @rigordonma You start with the equation $e^z = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x$ which i have no reason to believe in because you have not prooven it and to me (a newbie) it has no background. What you have shown is how to evolve the mentioned equation into a series $1 + i x + \frac{(i x)^2}{2!} + \ldots$. – 71GA Feb 04 '13 at 22:38
  • @71GA: let me see, perhaps you mean prove it from $e^z = \lim_{x \rightarrow \infty} \left [ \left( 1 + \frac{1}{x} \right)^x \right ]^z= \lim_{x \rightarrow \infty} \left [ \left( 1 + \frac{1}{x} \right)^z \right ]^x = = \lim_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x $ because of the binomial theorem. – Ron Gordon Feb 04 '13 at 22:51
  • How do you justify the interchange of the limit and the summation? – Antonio Vargas Feb 04 '13 at 23:20
  • @AntonioVargas: I don't follow you. You mean how do I justify taking the limit of the individual terms in my sum? Each individual term in the sum has a limit that is well-defined, so the limit of the sum is the sum of the individual limits of the terms. – Ron Gordon Feb 05 '13 at 01:10
  • That is not true in general. Usually you need to invoke the Lebesgue dominated convergence theorem to say something like that, but I don't see how to use it here. – Antonio Vargas Feb 05 '13 at 01:40
  • See, for example, this question. – Antonio Vargas Feb 05 '13 at 01:58
  • @AntonioVargas: OK, thanks for pointing ths out, although I'm not exactly sure how much use this is to the OP. – Ron Gordon Feb 05 '13 at 02:06
  • @NeuroFuzzy If you would write somthing about that it would be great! – 71GA Feb 05 '13 at 17:45
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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 $y=0,z=1\implies c=\log1-0=0\implies \log z=y\iff z=e^y$

lab bhattacharjee
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