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I was trying to find $0!$ using the series expansion of $e^x$ and got some odd results.

We know that, $$e^x = \sum_{n =0}^\infty\dfrac{x^n}{n!}$$

Putting $x = 0$ will give us following results.

$$e^0 = \sum_{n=0}^\infty \dfrac{0^n}{n!}\implies1 = \dfrac{0^0}{0!} +\underbrace{ \dfrac{0^1}{1!} + \dfrac{0^2}{2!} + ...\infty}_0$$

So, we are ended up with $1 = \dfrac{0^0}{0!}$

On cross multiplication, we have $0! = 0^0$ (I'm not sure if cross multiplication is a valid step here).

What to do with this? Can we conclude that $0^0 = 1$ because $\lim\limits_{x\to0^+} x^x = 1?$

Is the step of cross multiplication really a valid step above?

Robert Z
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    The definition of $0!$ is $1$, so is the definition of $0^0$, however some authors prefer to keep the latter undefined. I personally disagree with that convention though. – Vivaan Daga Feb 27 '22 at 09:02
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    @SiriusBlack I corrected a few typos. – Robert Z Feb 27 '22 at 09:03
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    Do you want to conclude that $0!=1$, or that $0^0=1$? For the second question, series and polynomials are written under the assumption that $x^0$ stands for the constant $1$ function, so if you manipulate power series you shoud just re-discover it. I can see a few reasons for that choice, but I have never heard of $x^x\to 1$ as $x\to 0^+$ playing a role: in fact, $x\ln x\to 0$ as $x\to 0^+$ is not a particularly comfortable statement to handle with power series. There are reasonable combinatorial arguments in terms of functions $\varnothing \to\varnothing$ for the choice $0!=0^0=1$. – Sassatelli Giulio Feb 27 '22 at 09:08
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    We should really note that the series expansion of $e^x$ only features a "$0!$" factorial because to begin with we know the constant coefficient is $1$, and it is just more elegant to express $1$ as $0!$ to fit it into the series. So, this argument could never work anyway – FShrike Feb 27 '22 at 09:47
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    The hitch is that your "proof" use a convention : $0^0=1$ which cannot be taken for granted. So this method would lead to no more than another convention. Discussion about $0^0$ : https://fr.scribd.com/doc/14709220/Zero-puissance-zero-Zero-to-the-Zero-th-Power – JJacquelin Feb 27 '22 at 10:02
  • Does this answer your question? Zero to the zero power – is $0^0=1$? – Dietrich Burde Feb 27 '22 at 11:59
  • I think the formula assumes that $0!=1$ so as to make the writing of it neat. But I could be wrong. So if I am wrong, recall that the derivative of the exponential is the same function. Then the second term of the derivative of the sum, that is, the term at index $1$, must give $0!=1$. – Artur Wiadrowski Feb 27 '22 at 18:09

1 Answers1

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I would say: facts $(1)$ and $(2)$ are are used in the proof of fact $(3)$. Therefore, it would be "circlular" to deduce $(1)$ or $(2)$ from $(3)$.

$$ x^0 = 1\quad\text{where $x$ is a real variable} \tag1$$

$$ 0! = 1 \tag2$$

$$ e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \tag3$$

I stated $(1)$ in this peculiar way because that formulation is more widely accepted than $$ 0^0=1 \tag{1'}$$

GEdgar
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