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My textbook explains that the power series: $\sum_{n=0}^{\infty} x^{n}/n!$ converges for $x=0$ because the terms of the series get the value 0.

My problem with this argument is the first term, which is $0^{0}$. But this is undefined? Someone who can explain this?

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The first term is taken to be $1$ for all $x$. A less compact way of writing the series is $$1 + \sum_{n=1}^{\infty}\frac{x^n}{n!},$$ since it is the Taylor series of $e^x$, so the constant term is $f(0) = e^0 = 1$. – Arturo Magidin May 27 '12 at 22:07
This convention is nearly universally observed to prevent the awkwardness Arturo has described. – ncmathsadist May 27 '12 at 22:11
Great! I found the errata of the book and it was a mistake. – characters May 27 '12 at 22:12
possible duplicate of Zero to the zero power - Is $0^0=1$? – Nate Eldredge Jun 8 '13 at 20:18
up vote 7 down vote accepted

In the context described in the question, it is a convention that $0^0 = 1$.

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In a power series, $x^n$ is not exponentiation operation on real numbers. Instead, it uses a different exponentiation operation; e.g. repeated multiplication suffices for this purpose. Therefore, $x^0=1$ identically.

Of course, most people don't like to pay attention to this level of detail in syntax, so they just treat $x^0=1$ when $x=0$ as a convention.

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There are calculus books that say that 0^0 is undefined. The reason for this is tradition; long ago, before continuous functions were well understood, Gauss placed 0^0 in a table of "indeterminate forms", a concept that becomes obsolete once you know the relation between limits and continuous functions.

There are numerous places in mathematics where 0^0 is implicitly assumed to be 1. So if you want consistency, then 0^0 must be defined as 1.

Some people say that sometimes 0 is better and sometimes 1 is better, but this is not true, the value 0 is never useful, and the value 1 never leads to contradictions.

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Welcome to MSE. I feel compelled to point out that your usage of "never" seems at odds with the customary mathematical definition. How does $$\lim_{x\to 0} (e^{-k/x^{2}})^{x^{2}} = e^{-k}, k \geq 0 \text{ arbitrary},$$fit into your linguistic scheme? – Andrew D. Hwang Apr 17 '14 at 19:05

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