Re: Re: Simplifying constants...bug?

*To*: mathgroup at smc.vnet.net*Subject*: [mg18610] Re: [mg18559] Re: [mg18489] Simplifying constants...bug?*From*: "Andrzej Kozlowski" <andrzej at tuins.ac.jp>*Date*: Tue, 13 Jul 1999 01:01:33 -0400*Sender*: owner-wri-mathgroup at wolfram.com

You can always understand such things by looking at FullForm. In the first case you have In[1]:= FullForm[(a + c*d)/(b + c*d)] Out[1]//FullForm= Times[Plus[a, Times[c, d]], Power[Plus[b, Times[c, d]], -1]] There are two instances of Times[c,d] and both are replaced by -z. In the second case you have: In[2]:= FullForm[(a + c*d)/(c*d)] Out[2]//FullForm= Times[Power[c, -1], Power[d, -1], Plus[a, Times[c, d]]] There is only one instance of TImes[c,d] and it is replaced by -z. So everything is as it should be. -- Andrzej Kozlowski Toyama International University JAPAN http://sigma.tuins.ac.jp http://eri2.tuins.ac.jp ---------- >From: calvitti at boes.ces.cwru.edu To: mathgroup at smc.vnet.net >To: mathgroup at smc.vnet.net >Subject: [mg18610] [mg18559] Re: [mg18489] Simplifying constants...bug? >Date: Sat, Jul 10, 1999, 3:18 PM > > > here's some intersting behavior in 3.0: > > (a + c*d)/(b + c*d) //. c*d -> -z > > gives (as expected): > > (a - z)/(b - z) > > yet > > (a + c*d)/(c*d) //. c*d -> -z > > no longer simplifes the denominator: > > (a+b-z)/(c*d) > > anyone know why? > > +------------------------------------------+ > alan c > systems and control engineering > case western reserve university > > calvitti at alpha.ces.cwru.edu > +------------------------------------------+ > > > > "Ersek, Ted R" <ErsekTR at navair.navy.mil> writes: > >> Morten G. Dyndgaard wrote: >> ----------------------------- >> I am working with some annoyingly long equations that I want to simplify >> by including one set of parameters in a constant (or function) A, and >> another in B and so on. >> >> You can easily do: >> A = b+c+d+e >> >> but what I want to do is the reverse: >> b+c+d+e = A >> ----------------------------- >> >> This should help. >> In[1]:= >> expr=a+b+b1+c+c1+d+d1+e; >> >> >> In[2]:= >> expr/.a+b+c+d+e->A >> >> Out[2]= >> A+b1+c1+d1 >> >> >> Regards, >> Ted Ersek