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When I use Simplify[] or FullSimplify[], specifying domain reals, I get terms with $0^n$ as the coefficient. Isn't that always zero, and thus be simplified out? A example of input/output is below, with the expressions truncated/replaced since they are extremely long.

Input:
Assuming[Reals && c > 5 , FullSimplify[expression]]
Output: $$ \int_0^1 i x... +0^{-1+c} \text{Hypergeometric2F1Regularized}\left[\frac{-1+c}{c},1-c i,2-\frac{1}{c},0^c\right]\ F'[x] \, dx $$

Note, I added the assumption that $c >5$ as overkill to avoid any possible issue with raising 0 to a negative number (or zero), but it didn't help. (Also, note, the $i$ above is a variable, not the imaginary number). Does anyone know why this happens?

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    could you copy some code which generates this?2011-08-19
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    There are 2 close votes as "off topic," but the [faq](http://math.stackexchange.com/faq) expressly allows questions related to "software that mathematicians use" (e.g. Mathematica).2011-08-19
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    @anon: I figured that that was what the "Mathematica" tag was for. If it's unpopular I'm happy to repost in a Mathematica-specific forum.2011-08-19
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    Try inputting `Assuming[Reals && c>5, FullSimplify[0^c]]`. If it doesn't simplify it then you know this is a case of Mathematica not being prepared to simplify a $0^c$ expression even with assumptions.2011-08-19
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    @anon: Thanks for the idea! That input produced "0". After that, I tried running Simplify on the output I got above, and the second run (with the assumptions again) produced the right result. Instead of fiddling with the assumptions, I should have just run it twice. Well, at least the problem is solved! Thanks for your help!2011-08-19
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    I am mildly interested in what generated the hypergeometric expressions...2011-08-21
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    @J.M. oops, I missed this comment somehow. I have forgotten the expression -- I think is was a series of order distributions of a function. I had a lot of them... If I find one, I'll repost.2011-09-18

1 Answers 1

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Usually, especially in the case of binomial identities, $$ 0^n=\left\{\begin{array}{c}1\text{ if }n=0\\0\text{ if }n>0\end{array}\right. $$ is used.

In your particular example, the exponent of $0$ is assumed to be greater than $0$, so I don't see the need for $0^{-1+c}$ or $0^c$. The answer is still correct, but unnecessarily complicated. If your assumption was $c\ge 1$, then $0^{-1+c}$ would be needed, but not $0^c$.

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    The assumption that $c>1$ should make that irrelevant.2011-08-19
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    @robjohn: That is a helpful definition. But, as Ricky points out, why doesn't the assumption then fix this? I even tried $c >5$ to be safe. This is what I input, with the expression cut out for space: Assuming[Reals && c > 5 , FullSimplify[expression]]2011-08-19
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    @Ricky Demer: indeed. However, JandR seemed to think $0^n$ was always zero, so I was explaining why it appears at all. Why it doesn't go away when $c>1$ is assumed seems like a shortcoming in `Simplify`.2011-08-19
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    @JandR: what version of Mathematica are you using? I tried `Simplify[(1-1)^n,Assumptions->n>0]` and got `0` in Mma 8.2011-08-19
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    @robjohn: I am on 7. But, see my comment above. I think that there is something with my expression. Somewhere in the references to other expressions I think I have a "/. something -> 0" so that is probably the cause.2011-08-19
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    and then there is the question of $0^c$ for complex $c$.2011-08-19
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    @robjohn: accepting since this is useful information that may help someone encountering the same confusion, even though it wasn't the root in this particular case.2011-08-19
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    @GEdgar: interesting question. What is $0^{iy}$ for $y\in\mathbb{R}$? It sends my mind spinning (around the origin).2011-08-19
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    @robjohn: What do you mean "usually"? What defines when the definition above is the case, and when it isn't?2011-08-21
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    @JandR: $0^0$ is one of the classic "indeterminate forms" of beginning calculus. For example $\displaystyle\lim_{x\to 0^+}0^x=0$ and $\displaystyle\lim_{x\to 0}x^0=1$. From a set-theoretic point of view, $0^0$ represents the functions from the empty set to the empty set, which is just the empty function, so from that point of view $0^0=1$. See the [sci.math FAQ about $0^0$](http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/).2011-08-22