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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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    People sometimes believe that $0^0$ can't be defined because they feel that a discontinuity is the same thing as a contradiction. In case you're wondering in which contexts you can replace $0^0$ by $1$, there is an easy to remember rule for that: It's always OK.2017-02-16

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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, Cauchy 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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    Setting $0^0 = 1$ never leads to contradictions if you follow the rules of mathematics (if you believe that a discontinuity is the same thing as a contradiction then all bets are off).2017-02-16