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For $x\in\mathbb R$ we define $$\exp(x) := \sum_{n=0}^\infty \frac{x^n}{n!}. $$ This is the standard definition of the exponential function, e.g. given by Rudin in the introduction to Real and Complex Analysis. It is well-known that this series converges absolutely for all real $x$, so this definition is perfectly acceptable.

I have a rather silly question though. How do we get $$\exp(0)=1 $$ from this definition? It's clear if we use a different definition of $\exp$: $$\lim_{n\to\infty}\left(1+\frac0n\right)^n = \lim_{n\to\infty}1^n=1. $$ But how do we reconcile the expression of the form $0^0$ in $$\exp(0) = \sum_{n=0}^\infty \frac{0^n}{n!} $$

My thoughts: if $x\ne0$, then $x^0=1$, so $\lim_{x\to0}x^0=1$. Hence for any nonnegative integer $N$, $$\lim_{x\to0}\sum_{n=0}^N\frac{x^n}{n!} = 1,$$ and so $$\lim_{N\to\infty}\lim_{x\to0}\sum_{n=0}^N\frac{x^n}{n!}=1. $$ It remains to show that the above is equal to $$\exp(0) = \lim_{x\to0}\exp(x)=\lim_{x\to0}\lim_{N\to\infty}\sum_{n=0}^N\frac{x^n}{n!}. $$ Is the absolute convergence of $\exp(x)$ for $x\ne0$ and the uniform convergence of $\exp$ on any bounded subset of $\mathbb R\setminus\{0\}$ enough to justify the above interchange in limits? If so, post it as an answer, and I will accept it and put this admittedly pedantic question to rest.

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    $\begingroup$ $0^0$ is 1. I don't think this is troubling, but if it troubles you, note that $x^0=1$ for any $x\neq 0$, and extend $x\mapsto x^0$ to $x=0$ by imposing continuity. $\endgroup$ – parsiad Mar 19 '15 at 23:17
  • $\begingroup$ I agree with your argument that $\lim_{x\to0}x^0=1$ (in fact I happened to prove that just now before reading your comment). But the statement $0^0=1$ definitely troubles me. For example, $0^x=0$ for any $x\ne0$, so $\lim_{x\to0}0^x = 0$... $\endgroup$ – Math1000 Mar 19 '15 at 23:34
  • $\begingroup$ @Math1000 I'm replying here becayse there's no need to flood Bernard's message box. Reply: That would be, in a loose sense, changing the definition of $\exp$ you're using. In spite of me preferring what you just suggested, what I actually said was that $\sum\limits_{n=0}^\infty \left(\frac{x^n}{n!}\right):=1+\sum\limits_{n=1}^\infty \left(\frac{x^n}{n!}\right)$. This is OK because the LHS doesn't necessarily have another meaning, due to the $0^0$ problem. If yo''re OK with this, I can ponder posting it as an answer. $\endgroup$ – Git Gud Mar 20 '15 at 14:59
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The standard convention is that when you write $x^n$, then when $n = 0$, this is the constant function $1$, regardless of the value of $x$. This is what people always mean by $x^0$ in a power series $\sum_{n \ge 0} a_n x^n$. This is the unique convention, for example, that makes $x^0$ continuous from the right as a function of $x \in \mathbb{R}_{\ge 0}$.

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  • $\begingroup$ I don't disagree with you, but this doesn't seem like a complete answer because clearly we want to define $\exp$ for $\mathbb R$, not just $\mathbb R_{\geqslant 0}$. $\endgroup$ – Math1000 Mar 20 '15 at 0:22
  • $\begingroup$ @Math1000: sure, and so we extend $x^0$ to be $1$ for all values of $x$. That's the value it should take for all $x \neq 0$ anyway. $\endgroup$ – Qiaochu Yuan Mar 20 '15 at 0:31
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$\exp 0=1+ \dfrac 0{1!}+\dfrac{0^2}{2!}+\dots+\dfrac{0^n}{n!}+\dotsm$

Where is the problem?

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  • $\begingroup$ Where did you get the $1$ from? Simply plugging in $0$ we would have $\frac{0^0}{0!}$ and $0^0$ is an indeterminate form. $\endgroup$ – Math1000 Mar 19 '15 at 23:30
  • $\begingroup$ It is a pure convention: the formula, as such, is valid only for $x\in \mathbf C^*$. Don't forget the first term is equal to $1$, whatever the value of $x\neq 0$. $\endgroup$ – Bernard Mar 19 '15 at 23:36
  • $\begingroup$ Well, Rudin says that "$e^0=\exp(0)=1$, by (1)" (where (1) is the formula given above). I don't see where that conclusion comes from. I agree that I am being a bit pedantic, but I just want to be rigorous here. $\endgroup$ – Math1000 Mar 19 '15 at 23:45
  • $\begingroup$ @Math1000 for $x=0$ the first term would be $0^0$. In this particular case this symbol, $0^0$, is just a convenient way of writing $1$. You don't prove it, it's a definition: $0^0:=1$. However I personally would rather think of $\sum\limits_{n=0}^\infty \left(\frac{x^n}{n!}\right)$ has an abbreviation of $1+\sum\limits_{n=1}^\infty \left(\frac{x^n}{n!}\right)$. I find it less ambiguous and more satisfying. $\endgroup$ – Git Gud Mar 19 '15 at 23:47
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    $\begingroup$ If $x$ is non-integer, the rigorous definition of $a^x$ requires the exponential function be defined. We only can give heuristic arguments for the convention $0^0=1$ here — in particular the function must be continuous at $0$. $\endgroup$ – Bernard Mar 20 '15 at 0:04
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By directly replacing $x$ by $0$ in the sum, the first term is indeterminate $(0^0)$. You have to expand the sum, and then you can evaluate at $x = 0$:

$$ \left.\exp(x)\right|_{x = 0} = \left.\left( 1 + x + \frac{x^2}{2} + \frac{x^3}{3} + \dots \right)\right|_{x = 0} = 1 + 0 + \frac{0^2}{2} + \frac{0^3}{3} + \dots = 1$$

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  • $\begingroup$ Okay, so you are saying that $\lim_{x\to0}\exp(x)=1$? That is what I was trying to show, from the definition, but I got a little bogged down in the details. $\endgroup$ – Math1000 Mar 19 '15 at 23:31

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