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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? –  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. –  Jand 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! –  Jand Aug 19 '11 at 19:43

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up vote 5 down vote accepted

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. –  Ricky Demer 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]] –  Jand 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
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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

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