Re: trouble with pattern matching & manipulating

*To*: mathgroup at smc.vnet.net*Subject*: [mg37040] Re: trouble with pattern matching & manipulating*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Mon, 7 Oct 2002 05:24:24 -0400 (EDT)*References*: <anp0p6$qvg$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Troy, True, interactive manipulation can be difficult. However, here is one way to do what you want. We have to do the same thing to both sides of the equation. (# - D*Sqrt[K]/C)&/@(A+B/C+D*Sqrt[K]/C\[Equal]0 A + B/C == -((D*Sqrt[K])/C) Together/@% (B + A*C)/C == -((D*Sqrt[K])/C) #C&/@% B + A*C == (-D)*Sqrt[K] #^2&/@% (B + A*C)^2 == D^2*K NOTES. Here is how #C&/@ (lhs ==rhs) works: #C&/@ (lhs ==rhs) --> #C&[lhs]==#C&[rhs] --> lhs C == rhs C --> ... f/@( expr) is special for for Map[f, expr] expr& is special for Function[expr] Please look up Map and Function in the Help Browser. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Troy Goodson" <Troy.D.Goodson at jpl.nasa.gov> wrote in message news:anp0p6$qvg$1 at smc.vnet.net... > I'm a newbie and, of course, the first thing I want to do is apparently > one of the most complicated... > > I have an expression that looks like this: > > A + B/C + D*Sqrt[E]/C = 0 > > A,B,C,D, & E are all polynomials in x > I want it to look like this > > (D^2)*E = (A*C + B)^2 > > At that point, I'll have polynomials in x on both sides. Finally, I > want the equation to be written out with terms grouped by powers of x, > but I think I can do that part :) > > I'll be very grateful to anyone who can give me some pointers. Or, at > least point me to some tutorial in the Mathematica documentation. I've > been looking over the documentation and I found Appendix A.5 in The > Mathematica Book, but that doesn't help me. I _need_ some examples. I > did find a couple of well-written posts in this newsgroup, but not quite > close enough to what I want. > > Thanks! > > Troy. > > =-=-=-=-=-=-=-=-=-= > > FYI, here's the expression I'm working with. > > > denom = Sqrt[(B^2 - r^2)^2 + 4*(r^2)*(b^2)] > cnu = (2*b^2 - B^2 + r^2)/denom > snu = -2*b*Sqrt[B^2 - b^2]/denom > sif = 2*r*b/denom > cif = (r^2 - B^2)/denom > > pdr = -Cos[ds]*Sin[q]*(snu*cif + > cnu*sif) - Sin[ds]*(cnu*cif - snu*sif) > > 0 == -(B^2 - b^2)*V^2/(r^2) + (((B*V)^2)/( > r^2) - 2*w*b*V*Cos[q]*Cos[ds] + (w* > r)^2 - (w*r*pdr)^2)*(Cos[qr])^2 > > Although I said it's a polynomial in x, it's really a polynomial in "b" > that I'm after. >