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Why is $$ e^x = \lim_{n \to \infty} \left( 1+ \frac{x}{n} \right)^n $$ also true for an imaginary $x$? Per definition, $$ e = \lim_{n \to \infty} \left( 1+ \frac{1}{n} \right)^n $$ so $$ e^x = \lim_{n \to \infty} \left( 1+ \frac{1}{n} \right)^{nx} $$ and $$ e^x = \lim_{n \to \infty} \left( 1+ \frac{x}{nx} \right)^{nx} $$ I understand now that we can substitute $n$ for $n*x$ as both approach $\infty$, but why does this also work for $x=i$, where $\lim_{n \to \infty} \left(n*i\right) $ would approach $\infty*i$, which is all the way to the top of the complex plane instead of to the right? $$ e^i = \lim_{n \to \infty} \left( 1+ i × \frac{1}{ni} \right)^{ni} $$

(This question came to me at 7:47 in Mathologer's video on $e^{\pi*i}$: https://youtu.be/-dhHrg-KbJ0?t=7m47s)

Ѕᴀᴀᴅ
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Skilly
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  • thank you, fixed. – Skilly Mar 31 '18 at 22:28
  • It would be better if you can tell us what do you mean by the symbol $e^i$. As far as I am concerned $e^i$ is a shorthand for $\cos 1+i\sin 1$. If this is your intended meaning then using very elementary techniques one can prove that $$\lim_{n\to\infty} \left(1+\frac{ix}{n}\right)^n=\cos x+i\sin x$$ for $x\in\mathbb{R} $. – Paramanand Singh Apr 01 '18 at 07:20
  • @ParamanandSingh By $e^i$ I mean plugging in $i$ in the second to last equation in my post. I guess this would've been smarter (generalising $x$ to $ix$): $$ e^{ix} = \lim_{n \to \infty} \left( 1+ \frac{ix}{nix} \right)^{nix} $$ – Skilly Apr 01 '18 at 20:14
  • What do you mean by raising something to exponent $nix$? One must first find a meaning of imaginary exponents before one deals with your question. One of the approaches is to define $e^z$ by $e^z=\lim_{n\to\infty} \left(1+\dfrac{z}{n}\right)^n, z\in\mathbb{C} $ and then define $\log z=w$ if $e^w=z$ for $z, w\in\mathbb {C} $. The symbol $a^b$ for $a, b\in\mathbb{C} $ is then defined as $e^{b\log a} $. – Paramanand Singh Apr 02 '18 at 07:59

4 Answers4

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When one is dealing with the famous limit $$\lim_{n\to\infty} \left(1+\frac{z}{n}\right)^n\tag{1}$$ it is implicitly assumed that $n$ is a positive integer. It can be proved with some effort that the limit above exists for all $z\in\mathbb{C} $ and thus defines a function $f(z) $. Further by definition $e=f(1)$ and again it can be proved using elementary arguments that $f(z_1+z_2)=f(z_1)f(z_2)$ for all $z_1,z_2\in \mathbb {C} $. Now by using algebra it is seen that $f(x) =\{f(1)\}^x=e^x$ for $x\in\mathbb{Q} $. And it therefore makes sense to define the symbol $e^x$ as $f(x) $ even when $x\notin\mathbb{Q} $. Your approach somehow assumes that $e^i$ (or imaginary exponents in general) has a well defined meaning. This assumption needs to be substantiated by proper definitions of imaginary or complex exponents before one can deal with your question.

Also the transition from $e=\lim_{n\to\infty} \left(1+\dfrac{1}{n}\right)^n$ to $e^x=\lim_{n\to\infty} \left(1+\dfrac{1}{n}\right)^{nx}$ involves the concept of raising to power $x$ and the continuity of $g(t) =t^x, t\in\mathbb{R} $. These ideas are non-trivial if $x\notin\mathbb{Q} $ and it is best to work using proper definitions involving the limit $(1)$ as described earlier (see my comments to your question also).

The youtube video you have linked also uses the definition based on limit $(1)$ and the exponent $m$ in video is a positive integer and the guy in video takes $z=i\pi$ and evaluates the expression $(1+i\pi/m)^m$ for $m=1,2,\dots 99$. You can see that imaginary exponents do not come into picture at all.

Paramanand Singh
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  • I understand, thank you. I had thought the definitions were the other way around, $e$ defining $e^x$, because $e$ popped up first in his video and he then combined $nx$ to $m$, which is what confused me. Just accepting that $e^x$ is defined as that limit though works for me, thank you. – Skilly Apr 02 '18 at 18:53
  • There's not any difficulty at all defining $e^z$ for complex $z$ using the series definition. Then your desired formula is an elementary calculation from the binomial theorem--see my answer. – alex-tang Apr 03 '18 at 03:54
  • @alex-tang: the series definition is another alternative. I used the limit definition because the question was based on it. – Paramanand Singh Apr 03 '18 at 04:44
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Yes we can show that in general for $z=x+iy$ we have that $$\lim\limits_{n\to\infty}\left(1+\frac{z}{n}\right)^n=e^z=e^x(\cos{y}+i\sin{y})$$

but not by that limit.

For a proof look here Suppose $z=x+iy$, prove that $\lim\limits_{n\to\infty}(1+\frac{z}{n})^n=e^x(\cos{y}+i\sin{y})$? and to the related OP.

user
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Use the binomial theorem to expand the nth power. As $n\to\infty$, the coefficient of $x^k$ approaches $1/k!$.

alex-tang
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It is valid in a simple generality.. where we are decomposing in binomial series to the log of complex arguments and then re-composing it back.

$$ z= e^{\log \,z} = \lim_{n \to \infty} \left( 1+ \frac{\log \,z} {n} \right)^n $$

Narasimham
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  • Could you explain further? Why is the $i$ in the exponent on the RHS in my last equation simply ignored? – Skilly Mar 31 '18 at 22:53