We know that
$$ e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},x\in \mathbb R, $$
which can be written out $$ e^x=\frac {x^0}{0!}+\frac {x^1}{1!}+\frac {x^2}{2!}+\cdots, $$
but $0^0$ isn't well defined.
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So is your question about the series at zero? – Sean Roberson Feb 09 '20 at 06:56
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3In this case we take $0^0=1.$ – Allawonder Feb 09 '20 at 07:02
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2One usually defines $0^0$ as an exact quantity to be $1$ whether in the setting of analysis or the discrete situation. But it is important to keep in mind that $f(x,y)=x^y$ is not continuous at $(0,0)$ with this definition (or any other definition) of $0^0$. – Ian Feb 09 '20 at 07:26
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1https://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1 – Hans Lundmark Feb 09 '20 at 08:48
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In the general formulation of Taylor series centered at $x=0$, $$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \ldots$$ (where the equality above holds on analytic functions -- don't worry about these technicalities at this time). Thus, for $f(x) = e^x$, we get $$e^x = e^0 + e^0 x + \frac{e^0}{2!} x^2 + \ldots = 1 + x + \frac{1}{2!} x^2 + \ldots$$ You'll notice that your $0^0$ doesn't appear here; that's because this formula is really what the Taylor series of $e^x$ whereas you are right in saying that the collapsed formula in summation notation is technically undefined at $x=0$. Just think of summation notation as a shorthand (or "notational trick") for this formula, where this formula is valid for all $x$. Often times in math, mathematicians do away with details in their notation that makes it "technically incorrect", for the purpose of communicating something at a high/concise level.
D.R.
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