Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2002
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2002

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: trouble with pattern matching & manipulating

  • To: mathgroup at smc.vnet.net
  • Subject: [mg37073] Re: trouble with pattern matching & manipulating
  • From: Troy Goodson <Troy.D.Goodson at jpl.nasa.gov>
  • Date: Tue, 8 Oct 2002 07:17:34 -0400 (EDT)
  • Organization: JPL/Caltech
  • References: <anp0p6$qvg$1@smc.vnet.net> <anrkga$3v5$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In <anrkga$3v5$1 at smc.vnet.net> Allan Hayes wrote:
> 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)


I think I have to apologize for the lack of clarity in my original post.  
I had tried to word it carefully, but I deceived myself.  I should have 
said:

I have an expression that can be put into this form:
     A + B/C + D*Sqrt[K]/C = 0
     A,B,C,D, & K are all polynomials in x
I need to get it into that form and, in the end, I want it to look like 
this
(D^2)*K = (A*C + B)^2

I think I gave the impression that I have polynomials A,B,C,D, & K at my 
fingertips.  I don't.   The expression I have is given at the end of 
this message.  

I'm still trying to digest the respones I've garned so far.  In the 
meantime, I decided to post this clarification.

> 
> "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[K]/C = 0
>>
>> A,B,C,D, & K are all polynomials in x
>> I want it to look like this
>>
>> (D^2)*K = (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.
>>
> 
> 
> 
> 


  • Prev by Date: Re: Re: Accuracy and Precision
  • Next by Date: Re: factoring quartic over radicals
  • Previous by thread: RE: trouble with pattern matching & manipulating
  • Next by thread: Rounding numbers