Consider a power series (complex in the general case): $$\sum_{n \geq 0} \mathrm{a_n (z-z_0)^n}$$ It converges at least for $\mathrm{z=z_0}$ and the sum function $F(z)$ in $z_0$ is $a_0$, that is

$$\mathrm{F(z)}=\sum_{n \geq 0} \mathrm{a_n (z-z_0)^n \,\,\,\, and \,\,\,\, F(z_0)=a_0}$$

Why is $F(z_0)=a_0$? Shouldn't is be equal to $0$ since the series in $z=z_0$ is the series $\sum_{n \geq 0} 0$ ?

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    $\begingroup$ but before plugging the value is $a_0+a_1(z-z_0)+a_2(z-z_0)^2 \ldots$ $\endgroup$ – Phicar Oct 20 '16 at 15:08
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    $\begingroup$ But $0^0 = 1$ (except when it's undefined or defined differently, but in the context of power series, it's $1$). $\endgroup$ – Daniel Fischer Oct 20 '16 at 15:09
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    $\begingroup$ $z_0^0=1$. In the context of power series even "$0^0=1$", which is a convention justified by how nice things work with it. $\endgroup$ – user378947 Oct 20 '16 at 15:12

You'll see why if we just write out the first few terms.

If we have a a function $F(z)=\sum_{n=0}^{\infty}a_n(z-z_0)=a_0+a_1(z-z_0)+a_2(z-z_0)+....$. Then as you can see no matter what you choose z to be $a_0$ is unaffected.

Lets choose $z=z_0$




No, because $\lim_{x\rightarrow 0} x^0=1$. The first term, the $n=0$ term, is nonzero.

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    $\begingroup$ This has nothing to do with limits. $0^0=1$ is simply the definition that makes polynomials and power series work. $\endgroup$ – Henning Makholm Oct 20 '16 at 15:17
  • $\begingroup$ It has everything to do with limits. $0^0$ is undefined, to make sense of $F(0)$ in the question you need to evaluate $\lim_{x\rightarrow 0}F(x)$. $\endgroup$ – Wouter Oct 20 '16 at 15:19
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    $\begingroup$ $0^0$ is only "undefined" if you willfully ignore its definition -- which has nothing to do with limits. The only limit in the question is the limit of partial sums -- there is no limit where $x$ moves anywhere. $\endgroup$ – Henning Makholm Oct 20 '16 at 15:20
  • $\begingroup$ In fact, when people say $0^0=1$ what they mean is $\lim_{x\rightarrow 0}x^0=1$. They surely don't mean $\lim_{x\rightarrow 0}0^x=1$, for that would be false. $\endgroup$ – Wouter Oct 20 '16 at 15:20
  • $\begingroup$ When I say $0^0=1$ I mean that $0^0=1$. I don't mean $\lim_{x\to 0}x^0$ and I don't mean $\lim_{x\to 0}0^x$, because $0^0$ has nothing to do with limits. Exponentiation $R\times\mathbb N\to R$ can be defined for any ring $R$ (or even a monoid with no addition), no matter whether it even has a topology, and $a^0=1$ is the base case of that definition. $\endgroup$ – Henning Makholm Oct 20 '16 at 15:22

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