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When a function can be expanded as a Taylor series at $x=x_{0}$ we usually write that:

$$f(x)=\sum_{n=0}^{∞}\frac{f^{(n)}(x_0)(x-x_0)^n}{n!}$$

So my question about writing the Taylor series this way is that when $x=x_{0}$ we get that the partial sum of the series for $n=0$ is $f(x_{0})*0^{0}$ , but isn't $0^{0}$ undefined? So my question is shouldn't we clarify that for $n=0$ the partial sum is $f(x_{0})$?

I know my question does not have any mathematical importance but i had this question for a long time about this notation

Vasilis Georgoulas
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  • The order of evaluation is not commutative. The dummy variable doesn't actually exist since it's a placeholder for writing out all of the terms. So the dummy variable must be evaluated first before the variable for the function. – Ninad Munshi Dec 31 '23 at 21:55
  • @NinadMunshi One must still explain how to simplify $(x - x_0)^0$ when $x$ can be any real number within the radius of convergence. Usually, to simplify a formula without creating special cases, we require that the formula evaluate in that way for all values of the variables, including the inconvenient ones (such as $x = x_0$, in this case). – David K Jan 01 '24 at 00:06
  • Before answering here, see all the other questions on $0^0$ that exist in the forum. – GEdgar Jan 01 '24 at 06:43
  • Related, if not duplicates: https://math.stackexchange.com/q/3539880/42969, https://math.stackexchange.com/q/11150/42969 – Martin R Jan 01 '24 at 09:05

1 Answers1

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$0^0=1$ is the usual setting for it makes formulas like the Taylor series more convenient to write. An extra case for $n=0$ is therefore not necessary.

It also makes sense since $0^0$ is an empty product and $1$ is the neutral element of multiplication. Empty sums are equal to zero, empty products are equal to one. In rare cases, $0^0$ is defined as zero, but this is not the norm and should be explicitly stated if such a convention is used. If not, you may assume that the convention is $0^0=1.$

Marius S.L.
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