Re: What we get from (0.0*x), (0.0^x) and similar stuff

*To*: mathgroup at smc.vnet.net*Subject*: [mg58821] Re: What we get from (0.0*x), (0.0^x) and similar stuff*From*: "David W. Cantrell" <DWCantrell at sigmaxi.org>*Date*: Wed, 20 Jul 2005 00:29:20 -0400 (EDT)*References*: <dbidbr$bk3$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

ted.ersek at tqci.net wrote: > I think some of the code I sent in my last message on this subject was a > bad idea. My current thoughts on this subject are summarized below. > ------------------------------ > > I like what I get from the next line. If a user wants to ignore the fact > that certain values will give (0^0) or (0*Infinity) they can add > definitions to do what they want. If a user wants (0^x) to return > If[x==0,Indeterminate,1] they can add a definition to do that. If you were thinking about my previous response when you said that, please note that I was clearly talking about x^0, rather than 0^x, when I mentioned If[x==0, Indeterminate, 1]. > In[1]:= > Clear[x,y]; > {0^x,0.0^x,0.0*x*y} > > Out[2]= > {0^x, 0.x, 0. x y} Surely you meant {0^x, 0.^x, 0. x y}. > ------------------------------ > The next line should return the approximate number 0. Should it? Given only that az is approximately zero and -Infinity < x < Infinity, the product az*x could potentially be large. I suggest that 0. x is the correct thing to return. > In[3]:= > FullSimplify[ 0.0*x, -Infinity < x < Infinity ] > > Out[3]= > 0. x > > ------------------------------- > (0^-2) and (0.0^-2) return ComplexInfinity. Shouldn't the next line > return {ComplexInfinity, ComplexInfinity}. Yes and no, resp., IMO. See below. > In[4]:= > FullSimplify[ {0^x,0.0^x}, -Infinity < x < 0 ] > > Out[4]= > {Infinity, Indeterminate} Using version 5.1.0.0 for Windows, I get In[10]:= FullSimplify[{0^x, 0.0^x}, -Infinity < x < 0] Out[10]= {0^x, Indeterminate} so I don't know how you got Infinity for 0^x when -Infinity < x < 0. But in any event, Infinity is incorrect here. As you suggested earlier, the best result would be ComplexInfinity. Returning 0^x unchanged, as my version did, while not the best here, is at least not wrong. Messier is 0.0^x when -Infinity < x < 0. Indeterminate is the answer given by both our versions, and I believe it is the correct answer as well. Given only that az is approximately zero and -Infinity < x < 0, the expression az^x could potentially be, say, close to 1 (taking |x| to be very small). Thus, since |az^x| need not be large, we should not return ComplexInfinity as you had suggested. But we would have a different situation if the exponent were not allowed to be close to 0. For example, given -Infinity < x < -1, it might seem reasonable if 0.0^x simplified to ComplexInfinity, although that is not what Mathematica does: In[11]:= FullSimplify[0.0^x, -Infinity < x < -1] Out[11]= Indeterminate > ------------------------------- > Sqrt[0.0] and (0.0^2.3) return the approximate number 0. Shouldn't the > next line also return approximate zero. No. The reason is the same as that above, where I explained why FullSimplify[0.0^x, -Infinity < x < 0] should not return ComplexInfinity. But again we would have a different situation if the exponent were not allowed to be close to 0. For example, given 1 < x < Infinity, it might seem reasonable if 0.0^x simplified to 0., although that is not what Mathematica does. > In[5]:= > FullSimplify[ 0.0^x, 0 < x < Infinity] > > Out[5]= > Indeterminate Regards, David Cantrell

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