Re: How do you make Mma assume a parameter is real and positive?

*To*: mathgroup at smc.vnet.net*Subject*: [mg7407] Re: [mg7343] How do you make Mma assume a parameter is real and positive?*From*: Richard Finley <trfin at fiona.umsmed.edu>*Date*: Fri, 30 May 1997 01:20:39 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Carl, Sorry, I didn't have time to read your whole background, but for an answer to the basic question: PowerExpand[ expr ] makes the assumption that the arguments are real and positive as you suggested, therefore you can use this ( if you know your arguments do indeed satisfy those constraints) and you will get: PowerExpand [ Sqrt [ a^2 ] ] = a as you desire. Hope that helps...RF At 10:27 PM 5/27/97 -0400, you wrote: >Hi all, > >If I give mma the input Sqrt[3^2] I get back 3, but when I give it >Sqrt[a^2] I don't get back a. This is exactly the behavior I expect. >However, suppose that I want to tell mma that a is a real positive number, >so that I want Sqrt[a^2] to return a. How can I do this? > >Let me give you the background for this question. As a simple example, >suppose I am trying to find the series expansion of > > 2 > b >-------------- > 2 2 >a - Sqrt[a -b ] > >So, I try > >b^2/(a-Sqrt[a^2-b^2])+O[b]^4 > >and mma returns > > 2 > b >---------- + O[b]^4 > 2 >a-Sqrt[a ] > >which is certainly the wrong answer. The correct answer is > > 2 > b >2 a - --- + O[b]^4 > 2 a > >So, I need to tell mma that Sqrt[a^2] is a while it is doing its series >expansion algorithm. Another manifestation of this problem occurs when >taking limits, since mma gives > > 2 > b >Limit[-------------, b->0] ---> 0 > 2 2 > a-Sqrt[a -b ] > >which is also incorrect (it should be 2 a). The usual solution of using >ComplexExpand doesn't help here. My solution for this particular example >was to add new rules for Power: > >Unprotect[Power]; >Power[Power[a,n_],m_]:=Power[a,n m]; >Protect[Power]; > >I would have preferred to add an upvalue to a, but a would be too deeply >buried if I were to try > >a /: Power[Power[a,n_],m_]:=Power[a,n m] > >I am not crazy about the above method, since it slows down the application >of Power everywhere, so I am curious if anybody has a better method. > >Thanks for any suggestions. > >Carl > > > >