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This sprung from a conversation here. In Stewart's Calculus textbook, he defined $e$ as the unique solution to $\lim\limits_{h\to 0}\frac{x^h-1}{h}=1$. Ahmed asked how do you define $x^h$ is not by $\exp(h\ln(x))$ and I'm not sure.

Does this definition of $e$ even make sense?

Definition here:

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2 Answers2

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The definition of $e$ as the unique number such that $$\lim_{h \to 0}\frac{e^{h} - 1}{h} = 1$$ makes sense, but there are few points which must be established before this definition can be used:

  1. Define the general power $a^{x}$ for all $a > 0$ and all real $x$. One approach is to define it as the limit of $a^{x_{n}}$ where $x_{n}$ is a sequence of rational numbers tending to $x$ (this is not so easy).
  2. Based on the definition of $a^{x}$ above show that the limit $(a^{x} - 1)/x$ as $x \to 0$ exists for all $a > 0$ (this is hard) and hence the limit defines a functions $f(a)$ for $a > 0$.
  3. The function $f(x)$ defined above is continuous, strictly increasing and maps $(0, \infty)$ to $(-\infty, \infty)$ (easy if previous points are established).

From the last point above it follows that there is a unique number $e > 1$ such that $f(e) = 1$. This is the definition of $e$ with which we started. And as can be seen this definition must be preceded by the proof of the results mentioned in three points above. All this is done in my blog post and in my opinion this is the most difficult route to a theory of exponential and logarithmic functions. Easier routes to the theory of exponential and logarithmic functions are covered in this post and next.

Paramanand Singh
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  • arguably it is the most difficult but irritatingly it is also the must intuitively and direct to what we think the definitions "mean". Intuitively $b^n$ is b multiplied by itself n times so "obviously" $b^{n/m}$ is the m-th root of b to the n and as x = lim q then $b^x$ is the limit of $b^q$. I mean "duh" and obviously d(b^x)/dx = $C_bb^x$ so there must be some $e$ where $C_e = 1$ and obviously $e^x$ means $e^x$ and $\ln x$ is just a logarithm. That's obviously what it all "means". It's a pity this is the freaking hardest* approach. – fleablood Apr 09 '16 at 19:54
  • @fleablood: as I say in my blog post "the most intuitive and obvious approach". – Paramanand Singh Apr 09 '16 at 20:45
  • That's a nice blog post btw. – fleablood Apr 09 '16 at 22:21
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$b^x$ can be defined as $\lim_{q\in \mathbb Q \rightarrow x}b^q$. (Isn't it usually so defined?) Or alternatively one can define $e = \lim_{h\in \mathbb Q\rightarrow 0}\frac {x^h - 1}{h}$.

I think it's legit and not circular.

fleablood
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    Of course, this assumes there is a limit. – fleablood Apr 09 '16 at 17:21
  • It works; $b^q$ is defined for rational $q$ in the "usual" way, and you can force the limit to exist using a monotonicity argument. It is essentially the same argument that is used to prove that the only memoryless distributions, i.e. the ones with the property $P(X>t+s|X>t)=P(X>s)$, are the exponential distributions. – Ian Apr 09 '16 at 19:24
  • I think defining new notions of limit is bit complicated and non-standard. The standard calculus texts define limits of functions of real variable and limit of functions of integral variable (sequences). If $x_{n}$ is a sequence of rational numbers with limit $x$ then we can define $a^{x}$ as the limit of sequence $a^{x_{n}}$. The notion of $\lim_{x \in \mathbb{Q} \to a}$ can be made precise by an appropriate definition but it is not very commonly seen. – Paramanand Singh Apr 09 '16 at 19:31
  • This isn't a new definition of limits at all. For every real $x$ there is a rational sequence of {$q_n$}$\rightarrow x$ so we define $b^x$ as $\lim b^{q_n}$. That's all my notation of $\lim_{q\in \mathbb Q\rightarrow x}$ means. We do have to clear up that such a real sequence of {$b^{q_n}$} converges but that's pretty standard and mechanical as Ian points out. (Actually it's a pain in the ass, but never mind...) – fleablood Apr 09 '16 at 19:46
  • The actual difficulty is establishing the uniqueness: why is $\lim_n b^{q_n}$ the same for any given $q_n \to x$? – Ian Apr 09 '16 at 19:54
  • True. It's a pain. But it can be done. need to establish for any $\epsilon > 0$ then there are rational $r < q < t$ such $b^q - \epsilon < b^r < b^q < b^t < b^q + \epsilon$ (assuming $b > 1$; equivalent but "flipped" if 0< b < 1). Then as all $q_n \rightarrow x$ all $b^{q_n}\rightarrow c$ for the same c. No-one's saying it's not a pain. But it is a valid consistent non-circular definition. – fleablood Apr 09 '16 at 20:01
  • @Ian: if $q_{n}, r_{n}$ tend to $x$ then $s_{n} = q_{n} - r_{n} \to 0$ and we need to show that $b^{s_{n}} \to 1$. This is easy. – Paramanand Singh Apr 09 '16 at 20:47
  • @ParamanandSingh Yes, though you don't even need to localize, you use the same monotonicity argument anywhere. – Ian Apr 09 '16 at 21:09