       Re: trouble with pattern matching & manipulating

• To: mathgroup at smc.vnet.net
• Subject: [mg37032] Re: [mg37013] trouble with pattern matching & manipulating
• From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
• Date: Mon, 7 Oct 2002 05:23:58 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```I confess I am not 100% sure what you mean. Would you like to do this
in steps, like you would do it by hand?

In:=
a + b/c + d*(Sqrt[e]/c) == 0;

In:=
Thread[(#1 - a - b/c & )[%], Equal]

Out=
(d*Sqrt[e])/c == -a - b/c

In:=

Out=
d*Sqrt[e] == (-a - b/c)*c

In:=

Out=
d^2*e == (-a - b/c)^2*c^2

In:=
Simplify[%]

Out=
d^2*e == (b + a*c)^2

Of course you can combine all the steps into a single function, but I
think it will be fairly complicated.

My own favourite way to do this sort of thing is:

In:=
Simplify[d^2*e == (d^2*e /. AlgebraicRules[
a + b/c + d*(Sqrt[e]/c) == 0, e])]

Out=
d^2*e == (b + a*c)^2

However, AlgebraicRules has not been documented since version 4. It
should be possible to do this using PolynomialReduce but it seems to
require the sort of skill only Daniel Lichtblau possesses;)

Andrzej Kozlowski
Toyama International University
JAPAN
http://sigma.tuins.ac.jp/~andrzej/

On Sunday, October 6, 2002, at 06:33 PM, Troy Goodson wrote:

> 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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