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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?

J. M. ain't a mathematician
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OctaviaQ
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  • could you copy some code which generates this? – Andrew Aug 19 '11 at 19:31
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    There are 2 close votes as "off topic," but the faq expressly allows questions related to "software that mathematicians use" (e.g. Mathematica). – anon Aug 19 '11 at 19:31
  • @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. – OctaviaQ Aug 19 '11 at 19:33
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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. – anon Aug 19 '11 at 19:37
  • @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! – OctaviaQ Aug 19 '11 at 19:43
  • I am mildly interested in what generated the hypergeometric expressions... – J. M. ain't a mathematician Aug 21 '11 at 11:55
  • @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. – OctaviaQ Sep 18 '11 at 16:51

1 Answers1

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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 $0^c$.

robjohn
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  • The assumption that $c>1$ should make that irrelevant. –  Aug 19 '11 at 19:21
  • @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]] – OctaviaQ Aug 19 '11 at 19:32
  • @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. – robjohn Aug 19 '11 at 19:33
  • @JandR: what version of Mathematica are you using? I tried Simplify[(1-1)^n,Assumptions->n>0] and got 0 in Mma 8. – robjohn Aug 19 '11 at 19:38
  • @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. – OctaviaQ Aug 19 '11 at 19:45
  • and then there is the question of $0^c$ for complex $c$. – GEdgar Aug 19 '11 at 19:57
  • @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. – OctaviaQ Aug 19 '11 at 19:58
  • @GEdgar: interesting question. What is $0^{iy}$ for $y\in\mathbb{R}$? It sends my mind spinning (around the origin). – robjohn Aug 19 '11 at 21:07
  • @robjohn: What do you mean "usually"? What defines when the definition above is the case, and when it isn't? – OctaviaQ Aug 21 '11 at 17:55
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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 $0^0$. – robjohn Aug 22 '11 at 12:51