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The limit definition by Euler is given as

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

and thus for large $N$ we have

$e^z = (1 + \frac{z}{N})^N$.

Now my question is: how "fast" does this approximation converge? I.e., how do I have to choose $N$ in order for my error to be approximately zero? I know that I have to choose $N$ dependent on $z$ since for larger $z$, I also have to choose larger $N$. But how exactly does this relation look like? Is it enough for $N$ to be a polynomial in $z$?

anon
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    Welcome to MSE. Who says that $e^z=\left(1+\frac zN\right)^N$ for large $N$? This is false. – Another User Jun 20 '22 at 17:48
  • $N$ depends on $z$ and on what you want $\left\lvert e^z-\left(1+\frac zN\right)^N\right\rvert$ to be. – Sassatelli Giulio Jun 20 '22 at 17:50
  • @AnotherUser Hey, can you maybe expand on that? Why is it false? – anon Jun 20 '22 at 20:19
  • Because, for instance $\left(1+\frac zN\right)^N=0$ when $z=-N$, whereas $\exp$ has no zeros. – Another User Jun 20 '22 at 20:32
  • @anon There are a few reasons why $e^z=\left(1+\frac zN\right)^N$ doesn't hold. One is that the statement "$f(z)=\lim_{n\to\infty}f_n(z)$ implies $f(z)=f_N(z)$ for large $N$" is not a thing. If we want to look at some details of this case: since you were shown that for all $N$ there is some $z_N\in\Bbb C$ such that $e^{z_N}\ne\left(1+\frac{z_N}N\right)^N$, the identity theorem and corollaries tell you that for all $n$ the set $S_n=\left{z\in\Bbb R,:, e^z=\left(1+\frac zn\right)^n\right}$ is closed and discrete. It follows that $\bigcup_{n\in\Bbb N}S_n$ is countable and with empty interior. – Sassatelli Giulio Jun 21 '22 at 07:49
  • @SassatelliGiulio, @ AnotherUser thank you both for your answers, that clarifies it! – anon Jun 21 '22 at 07:52
  • @AnotherUser But wait, what if I choose N dependently on x? So, if x is a binary number with n bits and I choose N as 2^n. Does it work then? – anon Jun 21 '22 at 13:36
  • No. If $x\in(0,\infty)$, then $e^x$ is greater than $\left(1+\frac xN\right)^N$, for every $N\in\mathbb{N}$. – Another User Jun 21 '22 at 13:38

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