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I had an interesting conversation with my teacher today, and we were discussing what the best foot forward for me might be. He asked me to differentiate $e^x$ from first principles, which I did, and then challenged me on:

Begin by defining $e$ as: $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$

He wondered if I found this satisfactory, intuitively speaking. I said: "this is just one of the fundamental and historic definitions of $e$", and he asked me - "why?" And I just couldn't answer beyond repeating: "it's the definition". On that note he then showed a more natural exploration that leads to a discovery of this definition of $e$. He did declare this exploration to be "cheating", which is the focus of my question.

Take $a$ to be a positive real number. Then:$$\begin{align}\lim_{h\to0}\frac{a^{x+h}-a^h}{h}=a^x\lim_{h\to0}\frac{a^h-1}{h}\end{align}$$ And by exploration with graphs, one intuitively sees that the latter term, the derivative of $a^x$ at $0$, clearly should exist, and that there should exist some base $a$ for which this derivative is $1$. Exploring this possibility, define: $$(h_n)_{n\in\Bbb N}=\frac{1}{n}$$ And for each $h_n$, ask which $a$ satisfies: $$\frac{a^{h_n}-1}{h_n}=1$$ And naturally define a sequence $(a_n)$ that solves this equation: $$(a_n)_{n\in\Bbb N}=(1+h_n)^{1/h_n}$$ This leads to the idea that the mystery $a$ for which the derivative of $a^x$ at zero equals $1$ can be found by considering the limits of the sequence $a_n$, as $h_n$ goes to $0$. This finds: $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$ And we call this number $e$. The exploration leads one to suspect that $e$ satisfies $\frac{d}{dx}e^x=1$.

However, he did mention that, in the interests of me studying real analysis, that this approach is "cheating" a little somehow, and that it is not a fully rigorous derivation of that limit. He hinted that I would see why this is shortly after properly studying real analysis.

I have seen a good deal of calculus and analysis around the Internet, in a very erratic way, so I have only some sense of why this approach is unrigorous. However, there are many many people on this forum who have properly studied real analysis, and I am asking for your help in explaining either what's wrong, or how to make it right.

My thoughts:

  1. We have not shown that $a^x$ is sequentially convergent, nor have we shown that $(a_n)$ converges.
  2. Showing that there exists $(a_n),(h_n)$ such that: $$\lim_{n\to\infty}h_n=0,\,\frac{a_n^{h_n}-1}{h_n}=1$$ Is not the same as showing that: $$\lim_{h\to0}\frac{(\lim_{n\to\infty}(a_n))^h-1}{h}=1$$

Any insight into $(1),(2)$ or anything else here would be greatly appreciated - especially any insight into how to make it fully rigorous! I just intuitively feel that $(2)$ is correct, but I cannot know for sure.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Commented Sep 17, 2021 at 17:15

4 Answers 4

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The gaps in the argument that need to be filled with rigor are the following.

  1. First, we need an airtight definition of just what $a^x$ means for at least a real number $a > 0$ and a real number $x$.
  2. Second, we need to rigorously prove that the limit $$\lim_{h \rightarrow 0} \frac{a^h - 1}{h}$$ exists.
  3. Third, we need to prove that the limit $$\lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)^n$$ exists.
  4. Finally, we need to insert the value of this limit, i.e. $e$ into the limit in (2) and show that the limit there is $1$, so that $\frac{d}{dx} e^x = e^x$.

Note here that because we do not have a different expression for the values of the limits above, we have to prove these limits exists as a fact in and of itself without finding their value. You may have done an $\epsilon$-$\delta$ proof before, but every such proof has involved that you already know the limit's value. That bit is important and that, together with (1), are where the real "real analysis" content lies here.

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  • $\begingroup$ I can do 4)! And 1), I think. Suppose I figure out how to show 2) and 3) - is this enough? I was amazed by my teachers exposition of replacing the derivative limit with a sequence in this way, a very natural and heuristic idea, but I wonder how to make that more airtight, if making it more airtight is even necessary. Thank you for your response $\endgroup$
    – FShrike
    Commented Sep 17, 2021 at 6:53
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    $\begingroup$ Your answer says all that I had to say here. +1 for getting at the heart of the issue. $\endgroup$
    – Paramanand Singh
    Commented Sep 17, 2021 at 13:21
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I'm not sure this answer completely to your question but i want to point out that the suspected step seems the following

$$\frac{a^{h_n}-1}{h_n}=1$$

which should be a limit as $n \to \infty$

$$\frac{a^{h_n}-1}{h_n}\to 1$$

or also

$$\frac{a^{h_n}-1}{h_n}= 1+o(1)$$

that is

$$a=\left(1+h_n+o(h_n)\right)^\frac1{h_n} \implies e=\lim_{n\to \infty} \left(1+\frac1n+o\left(\frac1{n}\right)\right)^\frac1{n}$$

which seems to be a more convincing way to obtain the result.

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  • $\begingroup$ Thank you for this response. It seems to me that this addresses some points. Do you have any insight into other unrigorous aspects? For example, your RHS on the final equation, the $e$ limit, we need to know that the limit converges (I suppose I could show this for myself, but the next point I'm hazy on): how valid is it to replace the derivative limit with the limit of these sequences? $\endgroup$
    – FShrike
    Commented Sep 16, 2021 at 20:35
  • $\begingroup$ @FShrike Yes this seems the big issue in the argument. I fix a step for the little-o notation. $\endgroup$
    – user
    Commented Sep 16, 2021 at 20:51
  • $\begingroup$ Does $o(1)$ mean that as $n\to\infty$ it goes to $0$? I think $o(h_n)$ worked as well, because as $n\to\infty$, $h_n\to0$ and if something is dominated by $h_n\to0$ then that something also goes to zero very quickly. Have I made a mistake? $\endgroup$
    – FShrike
    Commented Sep 16, 2021 at 20:53
  • $\begingroup$ @FShrike For the last limit we can of course show that limit exists. $\endgroup$
    – user
    Commented Sep 16, 2021 at 20:53
  • $\begingroup$ @FShrike $o(1)$ indicates something that goes to zero as $n\to \infty$ and by this the equality $\frac {a^{h_n}-1}{h_n}= 1+o(1)$ is equivalent to $\frac{a^{h_n}-1}{h_n}\to 1$. $\endgroup$
    – user
    Commented Sep 16, 2021 at 20:55
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I believe another aproach would be more rigorous and that might be the reason why your teacher is implying this is not. Here you assumed that the form of the function is $a^x$.

Another approach is to use analytical functions, which are functions that can be given almost anywhere by their Taylor series. If you assume that there exists a function $\exp(x)$ which is it's own derivative, then it must necessarily be analytical, since $\forall x \in \mathbb{R}$ its Taylor series is convergent.

From this it follows that: $$\exp(x; x_0) = \sum_{k=0}^{\infty} \exp^{(k)}(x_0)\frac{(x-x_0)^k}{k!} = \exp(x_0) \sum_{k=0}^{\infty} \frac{(x-x_0)^k}{k!}$$ Without loss of generality, we can choose $\exp(0) = 1$ and use $x_0=0$, and we end up with the following definition of $\exp(x)$: $$\exp(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!} $$ $$e := \exp(1)$$

Then you can show that $$\exp(1) = \sum_{k=0}^{\infty} \frac{1}{k!} = \big(1 + h_n)^{\frac{1}{h_n}}$$ for every $h_n, \lim_{n\to\infty} h_n = 0$

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    $\begingroup$ Yes! So this is the kind of reasoning I was giving to him. He just wanted me to approach the question from more basic principles, without invoking the heavy-duty theorems of real and complex analysis. Thank you for your answer, but I'm really looking for insights into the particular exploration he showed me $\endgroup$
    – FShrike
    Commented Sep 16, 2021 at 17:05
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Maybe you are being asked to think about this in terms of sequences of functions. You can define the sequence of functions $$f_n: (0,\infty)\to\mathbb R,\; f_n(a) = \frac{a^{1/n}-1}{1/n}$$ which has limit $$f(a) = \lim_{n\to\infty} f_n(a) = \lim_{n\to\infty} \frac{a^{1/n}-1}{1/n}.$$ From your post, we have that $$\forall n\geq 1,\; f_n(a_n) = 1$$ and want to determine whether this implies $$f(\lim_n a_n) = 1.$$ One way you might proceed to think about this is to define the inverses $$f_n^{-1}(x) = \left(1+\frac{x}{n}\right)^n.$$ Our question may be rephrased as $$\text{Does } \lim_{n\to\infty} f_n^{-1}(1) = f^{-1}(1)?$$ Well, according to this other stackexchange post Convergence of a sequence of functions and their inverses, the answer is yes, if you can show that $f^{-1}$ is continuous.

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  • $\begingroup$ Great! I suspected this in the: “my own thoughts” section. Suppose I show we have sequential continuity: is this enough to say that $\lim_{h\to0}\frac{a^h-1}{h}=\lim_{n\to\infty}f_n(a_n)=1$? $\endgroup$
    – FShrike
    Commented Sep 17, 2021 at 6:57
  • $\begingroup$ I think you need something kind of strong, like the above, i.e. show that $f^{-1}$ is well defined and continuous. Alternatively, I think you can also prove the result if you can show the $\{f_n\}_n$ are equicontinuous at $a$: $\forall \epsilon > 0$, there exists $\delta>0$ s.t. $\forall n > N_\epsilon,\; |f_n(a') - f_n(a)| < \epsilon$ if $|a'-a|<\delta$. Using this, you can show that $|f_n(a_n) - f(a)| \to 0$ as desired: $|f_n(a_n) - f(a)|\leq |f_n(a_n) - f_n(a)| + |f_n(a) - f(a)| \to 0$. The right term goes to zero by definition; the left goes to zero by equicontinuity. $\endgroup$
    – ffffffyyyy
    Commented Sep 17, 2021 at 17:03

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