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When I first studied power series in high school, the teacher gave the following general definition:

\begin{equation} f(x)=\sum_{n=0}^{\infty}a_n (x-x_0)^n \end{equation}

He then proceeded to explain that when both $x=0$ and $x_0 =0$, the first term of the power series is defined to simply be $a_0$. How does one justify $0^0 = 1$ in this case?

giobrach
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  • Since the exponents are integers, $w^n$ is a product of $n$ factors $w$. If $n = 0$, it is an empty product. The empty product has the value $1$. – Daniel Fischer Apr 24 '16 at 22:50
  • Because the exponent $0$ is an integer and there's no problem whatsoever in defining $a^0=1$ *for any* $a$ (including $a=0$) when dealing with integer exponents. Powers with arbitrary real exponents are only defined for positive bases. – egreg Apr 24 '16 at 23:19

2 Answers2

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There is no real need to justify it. It is just a convention that simplifies notation and makes it uniform.

Martin Argerami
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To justify $0^0=1$ one can refer to the definition of products via $$x^n=\prod_{k=1}^nx.$$ Since we have a lower upper bound than the lower bound for $n=0$ the product is defined as $1$, since it is the empty product. This interpretation is the one I have seen throughout calculus and analysis, especially to justify things like the binomial theorem.

In contrast to $\displaystyle\lim_{x\to 0+0}x^x=1$ we have $\displaystyle\lim_{x\to 0+0}0^x=0$ so be aware of that case as well.

Christian Ivicevic
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