       RE: trouble with pattern matching & manipulating

• To: mathgroup at smc.vnet.net
• Subject: [mg37047] RE: [mg37013] trouble with pattern matching & manipulating
• From: "DrBob" <drbob at bigfoot.com>
• Date: Mon, 7 Oct 2002 05:25:00 -0400 (EDT)
• Reply-to: <drbob at bigfoot.com>
• Sender: owner-wri-mathgroup at wolfram.com

```Try this:

A + B/C + D*(Sqrt[E]/C) == 0
(#1 - %[[1,{1, 2}]] & ) /@ %
(C*#1 & ) /@ %
(#1^2 & ) /@ %
Simplify[%]

Also, be aware that E is the natural logarithm base, reserved for that
purpose.

DrBob

-----Original Message-----
From: Troy Goodson [mailto:Troy.D.Goodson at jpl.nasa.gov]
To: mathgroup at smc.vnet.net
Subject: [mg37047] [mg37013] trouble with pattern matching & manipulating

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.

```

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