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The geometric series is usually defined as $\sum_{k=0}^{\infty} a \cdot x^{k}$ where $x$ is on the interval $]-1;1[$, which includes $0$.

My Problem is that substituting $x=0$ for the first Term of the sum gives $a \cdot 0^{0} $. This is an indeterminate form and therefor undefined which means that $x=0$ should not be part of the domain.

I know that the both-sided $lim_{x \rightarrow 0} ( a\cdot x^{0} )$, as well as $\frac{a}{1-x}$ at $x=0$, are equal to $a$. Both shouldn't be reason enough for including $x=0$ in the domain. Under most circumstances having the Geometric Series at $x=0$ equate to $a$ is a good thing, but there must be a better reason for doing so.

Qwox
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    When discussing power series like this, it is assumed (often explicitly) that $0^0 = 1$. While this isn't the convention everywhere in mathematics, it is very much the convention in these circumstances. – Theo Bendit May 17 '22 at 20:48
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    “Really”, the geometric series is $a+a\sum_{k=1}^{\infty}x^k$. The constant value is always there and needed/assumed. We write this as $x^0$ for convenience of putting it as one series, and because $x^0=1$ is a very good convention. So there is no need to worry about any limits, since plugging in $0$ in my first equation gives me just $a+a\cdot0=a$ – FShrike May 17 '22 at 20:48
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    For what it's worth, I view this as an issue associated with a specific way of representing a geometric series (and which can be fixed by defining $0^0=1),$ not something intrinsic about the geometric series itself. – Dave L. Renfro May 17 '22 at 20:52
  • @TheoBendit I have never seen any convention other then $0^0=1$ anywhere outside discussions on m.se about whether $0^0=1$. It is so obviously sensible whether dealing with power series or binomial expansions or empty products or anywhere else you might want to use it. It may make the function $0^x$ discontinuous at $x=0$ but that is unlikely to be an issue given that it is not defined for $x<0$. – Henry May 17 '22 at 22:43

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Generally when dealing with power series (and indeed we do the same when dealing with polynomials), we define $0^0=1$. This is for various reasons, including that we want continuity. Furthermore, we wish for our summation notation to formalize the following notion:

$$\sum_{j=0}^\infty a_j x^j=a_0+a_1x+a_2x^2+\cdots.$$

In particular, we only get this if we define $0^0=1$, as we wish to have

$$a_0x^0=a_0$$

for the first term, for all $x$, including $0$.

Lorago
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Too long for a comment.

Consider $y=x^x$ and make $x=10^{-k}$ and ... compute

$$\left( \begin{array}{cc} k & \left(10^{-k}\right)^{10^{-k}} \\ 0 & 1.0000000000 \\ 1 & 0.7943282347 \\ 2 & 0.9549925860 \\ 3 & 0.9931160484 \\ 4 & 0.9990793900 \\ 5 & 0.9998848774 \\ 6 & 0.9999861846 \\ 7 & 0.9999983882 \\ 8 & 0.9999998158 \\ 9 & 0.9999999793 \\ 10 & 0.9999999977 \end{array} \right)$$

Claude Leibovici
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