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Define the function $g_n\left(z\right)=\left(1+\frac{z}{n}\right)^n$ for $\:n\in \mathbb{R^+}$. Then

$$\frac{d}{dz}g_n\left(z\right)=n\left(1+\frac{z}{n}\right)^{n-1}\cdot\frac{1}{n}=\left(1+\frac{z}{n}\right)^{n-1}$$

Define $g_{\infty}\left(z\right)=\lim _{n\rightarrow \infty }g_n\left(x\right)=\lim_{n\rightarrow \infty }\left(1+\frac{z}{n}\right)^n$, then notice that

$$\frac{d}{dz}g_{\infty}\left(z\right)=\frac{d}{dz}\left(\lim _{n\rightarrow \infty }g_n\left(x\right)\right)=\lim _{n\rightarrow \infty }\frac{d}{dz}g_n\left(z\right)$$$$=\lim_{n\to \infty}\left(1+\frac{z}{n}\right)^{n-1}=\lim_{n\to \infty}\frac{\left(1+\frac{z}{n}\right)^n}{1+\frac{z}{n}} =\lim _{n\rightarrow \infty }\left(1+\frac{z}{n}\right)^n=g_\infty\left(z\right)$$

By considering that $g_n\left(0\right)=1\: \forall\:n\in \mathbb{R^+}$, we have the differential equation

$$\frac{d}{dz}g_{\infty}\left(z\right)=g_\infty\left(z\right),\,g_\infty \left(0\right)=1$$

Which also has $e^z$ as a solution. However, the above differential equation has a unique solution, so therefore $$g_\infty \left(z\right)=\lim _{n\rightarrow \infty }\left(1+\frac{z}{n}\right)^n=e^z\,\,\blacksquare$$

I came up with this proof myself, and I think it's quite elegant, but I'm unsure as to the validity of the proof. Something just doesn't seem right to me. Are there any invalid statements or loose ends to this proof? Also, how useful is this as a proof of the limit? You don't need to know theory of differential equations, because you'd recognise easily that $e^z$ is its own derivative. Is there anything in it that might make it inaccessible, assuming you know that $\frac{d}{dz}e^z=e^z$?

Pauly B
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  • You should have a look at http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-2_10.html I have treated the case when $z$ is real. For complex $z$ you will need to work a bit more but idea is similar. – Paramanand Singh Jul 28 '14 at 04:31
  • What is your definition for $e^z$? –  Nov 27 '15 at 16:06

1 Answers1

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In principle, your proof looks pretty good and I might have to use it when I teach calculus! There are definitely important details that would need to be addressed if it were to be 100% rigorous, however. Here are the obvious things I notice:

  • You should be very precise about what assumptions you are making. How are you defining $e^z$? As the unique smooth solution of $\frac{d}{dz}f = f,\; f(0)=1$?

  • When you define $g_\infty$, you should argue why the limit exists and why $g_\infty$ is differentiable. This is probably the most important (and technical) step.

  • You exchanged the order of a limit and a derivative here:

$$ \frac{d}{dz}\left(\lim_{n\rightarrow\infty} g_n\right) = \lim_{n\rightarrow\infty} \left(\frac{d}{dz}g_n\right) $$

You need to explain why this is justified.

icurays1
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  • Thank you for the input!

    For the first point, yes that would be the definition, and it can easily be shown to be unique by the fact that $\frac{d}{dz}(e^{-z} f(z))=e^{-z}(f'(z)-f(z))=0 \implies f(z)=e^z$.

    For the third point, would it be sufficient to say that $n$ and $z$ are independent of each other that justifies the switching?

    That second point is tricky. It can be shown that the sequence converges in a not too bad manner, but showing that $g_\infty (z)$ is differentiable is nasty. I'll work on it, thanks!

     – Pauly B Jul 27 '14 at 19:05
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    For the third point, $n$ and $z$ being independent is unfortunately not enough - the sequence $\frac{d}{dz}g_n(z)$ must converge uniformly. See e.g. here – icurays1 Jul 27 '14 at 19:38