       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))*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

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

```

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