Is there a way to prove that the function $e^x$ is its own derivative given only that $$e=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n}$$? I can only seem to find proofs where $e$ being the base such that $e^{x}$ passes through $(0,1)$ with a derivative of one is a given, or that utilize logarithmic differentiation.
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How do you define exponentiation, in particular $e^x$? – Bart Michels Sep 27 '16 at 22:39
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The difficulty lies in defining what $a^x$ should mean when $a>0$ and $x$ is an irrational number. There are ways to do this. The easiest one actually exploits $\exp$ and $\ln$: in this framework we define $a^x$ for $a>0$ to mean $\exp(x \ln(a))$. A significantly harder way to proceed, which does work, uses rational approximation of the exponent (since we can define general rational exponents with positive bases in a straightforward manner). – Ian Sep 27 '16 at 22:40
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Well, nothing wrong with $b^x = \lim_{r\rightarrow x;r \in \mathbb Q} b^r$... – fleablood Sep 27 '16 at 22:41
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Anyway, once you have made sense of the basics, the idea of a direct proof using this definition of $e$ would be to show that $\left ( \lim_{n \to \infty} (1+1/n)^n \right )^x = \lim_{n \to \infty} (1+x/n)^n$, and that you can interchange derivative and limit in the last expression. Neither of these steps is completely trivial. – Ian Sep 27 '16 at 22:43
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Well if we treat x like a constant lim (1 + x/n)^n = e^x, so derive is lim n(1-x/n)^{n-1}*1/n = lim (1-x/n)^{n-1} = e^x. With many little devils in many little details. – fleablood Sep 27 '16 at 22:45
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"\left ( \lim_{n \to \infty} (1+1/n)^n \right )^x = \lim_{n \to \infty} (1+x/n)^n" someone just asked that today. It wasn't that hard. – fleablood Sep 27 '16 at 22:47
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It's not exceptionally hard but it's not trivial either. – Ian Sep 27 '16 at 23:02
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you can show easily that $f(x)= \lim_{n \to \infty} (1+x/n)^n$ fulfills $f(x)^y = f(xy)$ and since $f(1) = e$ it means $f(x) = e^x$. And $f'(x) = f(x)$ follows directly – reuns Sep 27 '16 at 23:15
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Actually, the simplest way to find the derivative of $e^x$ is to first define $ln(x)= \int_1^x \frac{1}{t}dt$. From that it is easy to prove the usual properties of $ln(x)$, that $ln(xy)= ln(x)+ ln(y)$ and $ln(x^a)= aln(x)$. And, of course, that $\frac{dln(x)}{dx}= \frac{1}{x}$ follows from the "fundamental theorem of Calculus".
Then define $e^x$ to be the inverse function to ln(x). Then it is immediate that, with $y= e^x$, $\frac{dy}{dx}= \frac{1}{\frac{dx}{dy}}= \frac{1}{\frac{1}{y}}= y= e^x$.
user247327
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Thoughts: Can we prove $f(x) = \lim_{n\rightarrow \infty} f_n(x)$ then $f'(x) = \lim_{n\rightarrow \infty} f_n'(x)$ when $f_n'$ and $\lim f_n'(x)$ exist?
[I leave this as an exercise to the reader....]
Proving $e^x = \lim_{n\to\infty}(1 + x/n)^{n}$ is relatively easy as $\lim_{n\to\infty}(1 + x/n)^{n}=\lim_{xn\to\infty}(1 + x/n)^{n}=\lim_{n\to\infty}(1 + 1/n)^{xn}= (\lim_{n\to\infty}(1 + 1/n)^{n})^x = e^x$.
So if we can do the first (which might be hard) we have
$(e^x)' = \lim n(1 + x/n)^{n-1}*\frac 1n = \lim (1 + x/n)^{n-1} = (\lim (1+1/n)^{n-1})^x$.
and $e = \lim (1+1/n)^n = \lim(1+ 1/n)^{n-1}(1+1/n) = \lim(1+1/n)^{n-1}\lim (1+1/n) = \lim(1+1/n)^{n-1}*1=\lim(1+1/n)^{n-1}$
So $(e^x)' = (\lim (1+1/n)^{n-1})^x=e^x$.
Martín-Blas Pérez Pinilla
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fleablood
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So $\lim_{n\rightarrow \infty}[\lim_{h\rightarrow 0}\frac{f_n(x + h) - f(x)}h] ?=? \lim_{h\rightarrow 0}[\lim_{n\rightarrow \infty}\frac{f_n(x + h) - f(x)}h]$. Can we say that??? – fleablood Sep 27 '16 at 23:36
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