Using FullSimplify to check hand algebra

*To*: mathgroup at smc.vnet.net*Subject*: [mg69277] Using FullSimplify to check hand algebra*From*: "Goyder, Hugh " <h.g.d.goyder at cranfield.ac.uk>*Date*: Tue, 5 Sep 2006 05:30:42 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

The output from some computer algebra gives me ee1 = (-(1/Sqrt[2]))*Sqrt[(1/(KK*r^2))*(w*((-(2 + KK + 2*nu))*w + Sqrt[4*KK^2 + (KK - 2*(1 + nu))^2*w^2]))] I also have the following assumptions and can use FullSimplify. ass = {w > 0, KK > 0, nu > 0, r > 0} s1 = FullSimplify[ee1, ass] FullSimplify does not give me a nice form. I don't mind this because I realise that my nice form may not be the same as that given by Mathematica. Using hand algebra I converted the expression to a nice form for my application s2 = (-(w/r))*Sqrt[-(1/2 + (1 + nu)/KK) + Sqrt[1/w^2 + (1/2 - (1 + nu)/KK)^2]] (Which when evaluated annoyingly puts the r on the denominator of the whole expression not just under the w), I now wish to check that I have not made any slips so I try the following hoping to get the answer True. However, I don't and on checking I don't find any slips. So why do I not get the answer True? FullSimplify[s1 == s2, ass] I note that I do get the answer True if I use the following but I think this is less strict than the previous test. (Under what circumstances is this the same as the previous test?) FullSimplify[s1/s2 == 1, ass] Similarly using PowerExpand does not help. (I don't wish to use PowerExpand because complex solutions are a possibility.) FullSimplify[PowerExpand[s1] == PowerExpand[s2], ass] So the general question is when can I use FullSimplify to check my hand algebra? Hugh Goyder Cranfield University

**Follow-Ups**:**Re: Using FullSimplify to check hand algebra***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

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