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Can anybody help me solve this transcendental equation by maple ?

Posted on 2008-03-17 09:23 By andy.zhou.nuaa (Maple Rating 1 17)
Categories:

   The equation like that:

  a*(x/sqrt(x^2-1)+2*x/(Pi*sqrt(x^2-1))-1-2/(Pi*x))=1

when i used solve command in Maple, the result is :

   RootOf((-Pi^2-2*Pi^2*a)*_Z^6+(2*Pi^2+4*Pi^2*a+a^2*Pi^2)*_Z^4+(4*a*Pi+4*a^2*Pi)*_Z^3+(-a^2*Pi^2-2*Pi^2*a-Pi^2)*_Z^2+(-4*a*Pi-4*a^2*Pi)*_Z-4*a^2)

  this result is not what i want, i need exactly solution. 

 Can anyone help me?

 Thanks very much for your help!!

 

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acer's picture

wallpaper

Comment on 2008-03-17 09:48 from acer (Maple Rating 4 1356)

In Maple 11.02,

eq:=a*(x/sqrt(x^2-1)+2*x/(Pi*sqrt(x^2-1))-1-2/(Pi*x))=1;
_EnvExplicit:=true;
sol:=[solve(eq,x)];
nops(sol);

acer

hi, your method seems doesn't work in maple.

Comment on 2008-03-17 10:05 from andy.zhou.nuaa (Maple Rating 1 17)

Dera acer:

  I hve used your method just now. but it still doesn't work.

 some wrong tips display:

    sol:=[solve(eq,x)];
       "  Warning, solutions may have been lost "

      Note: My maple version is maple 11.0.

Scott03's picture

Try updating

Comment on 2008-03-17 10:09 from Scott03 (Maple Rating 4 468)

Since Maple updates have many fixes and other updates in them, it is possible that this may be fixed in one of the versions between 11.00 and 11.02 and that may be why it works for Acer and not you.  You can download the update of Maple here (http://www.maplesoft.com/support/downloads/index.aspx).

 

Scott

acer's picture

about `a`

Comment on 2008-03-17 10:30 from acer (Maple Rating 4 1356)

You might also help maple out, by informing it of some extra information about `a`. For example,

sol:=[solve(eq,x,Explicit)] assuming a>0;

(I am not 100% sure, but the Warning may be related to Maple's not knowing anything about `a`. It might also actually be innocuous here.) Or perhaps you want a solution where x>1, or where x>Pi/2, etc, which could also be supplied as additional assumptions. Or, you could take your original RootOf() result, which represents the roots of a 4th degree polynomial, and try hitting that with allvalues().  It's just a question, but if you have no such extra information then how would you choose amongst the 4 explicit roots?

acer

thanks, but still wrong information.

Comment on 2008-03-17 10:40 from andy.zhou.nuaa (Maple Rating 1 17)

Dear acer:

   can u give me the result in your maple software?

    thanks very much for ur help! 

                                                                                                           andy

 

Doug Meade's picture

show us, please

Comment on 2008-03-17 11:01 from Doug Meade (Maple Rating 4 715)

Can you please share with us EXACTLY what you are trying? Either upload your Maple file (worksheet or document) to MaplePrimes or cut-and-paste the input (and output) from your session. Then, we should be able to provide a solution that will definitely work for you.

Doug

---------------------------------------------------------------------
Douglas B. Meade  <><
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu       
Phone:  (803) 777-6183         URL:    http://www.math.sc.edu/~meade/
acer's picture

sorry

Comment on 2008-03-17 11:06 from acer (Maple Rating 4 1356) 
> length(sol);
                                   11401819

So, no, I won't paste it. It's pages and pages and pages in length. Here's another way that you might obtain it, with some smarter management of the common subexpressions. Take that RootOf result, which I think was,

RootOf((2*a*Pi^2+Pi^2-4*a^2*Pi-4*a^2)*_Z^4+(4*a*Pi+4*a^2*Pi)*_Z^3+(-Pi^2-2*a*
Pi^2-a^2*Pi^2+4*a^2)*_Z^2+(-4*a*Pi-4*a^2*Pi)*_Z-4*a^2)

Now solve the general quartic, for explicit roots, and substitute in those coefficients. Test at some random value of `a`.

eq:=a*(x/sqrt(x^2-1)+2*x/(Pi*sqrt(x^2-1))-1-2/(Pi*x))=1;
solg:=[solve(C0*x^4+C1*x^3+C2*x^2+C3*x+C4,x,Explicit)]:
newsol:=subs({C0=(2*a*Pi^2+Pi^2-4*a^2*Pi-4*a^2),
 C1=(4*a*Pi+4*a^2*Pi),C2=(-Pi^2-2*a*Pi^2-a^2*Pi^2+4*a^2),
 C3=(-4*a*Pi-4*a^2*Pi),C4=-4*a^2},solg);
seq(evalf(subs(x=newsol[i],a=3,lhs(eq)-rhs(eq))),i=1..nops(newsol));

If I'm not mistaken, that newsol is a list of 4 elements, each of which satisfies `eq` the original eq.

acer

Robert Israel's picture

Equation

Comment on 2008-03-17 12:25 from Robert Israel (Maple Rating 4 1431)

First of all, it's an algebraic equation, not a transcendental one.  The only thing transcendental in it is the number Pi, but that doesn't make the equation transcendental.

The RootOf solution is almost certainly more informative than the solution using radicals, and better for numerical computation.

If you write 1/sqrt(x^2-1) = t and multiply by Pi*x, your equation becomes

a*t*(Pi+2)*x^2 - (1+a)*Pi*x - 2*a = 0

The resultant of a*t*(Pi+2)*x^2 - (1+a)*Pi*x - 2*a and t^2*(x^2-1)-1 with respect to t
is

P=(Pi+2*a*Pi+2*a)*(-2*a+Pi)*x^4+4*a*Pi*(1+a)*x^3-(a*Pi+2*a+Pi)*(a*Pi-2*a+Pi)*x^2-4*a*Pi*(1+a)*x-4*a^2

According to the theory of resultants, any x for which there exists t such that both equations are satisfied must be a root of P.  So this is where the RootOf comes from. 

The one question that is tricky is whether t will be 1/sqrt(x^2-1) or -1/sqrt(x^2-1).  For example, if a = -15*Pi/(40*Pi+74) one of the roots of P is 5/4, but this is not a solution of your equation because the corresponding t is -1/sqrt(x^2-1) = -4/3 rather than 1/sqrt(x^2-1).  It seems to me that for 0 < a < Pi/2  there will be a negative root of P that doesn't satisfy your equation, while for -Pi/(2*(Pi+1) < a < 0 there will be a positive root of P that doesn't satisfy your equation.

I don't know why there is the "Solutions may be lost" warning, but it could be related to that (so it's not really a matter of solutions being lost, but rather of bogus solutions being found).

 

acer's picture

nice analysis

Comment on 2008-03-17 14:10 from acer (Maple Rating 4 1356)

This is a nice analysis. (Somehow I figured that the OP might not be interested in such forensics, but then I didn't have nearly as nice and clear a process as you've shown. I also goofed and typo'd x>Pi/2 instead of a>Pi/2, and so which wouldn't have helped.)

But now it raises a question. How much would be involved in training `solve` to be so smart (for a restricted class of problems, no doubt)? Even if I set _EnvConditionalSolutions to be 'true' I can't get `solve` to return a nice piecewise solution. It seems that `int` has made more progress in recent releases of Maple than has `solve`, in this sort of respect.

Hopefully the OP will also pay attention to your comment about RootOf and subsequent numeric evaluation. The same goes for quite a bit of subsequent symbolic computation too, of course (which is part of why we have RootOf). I should have been more clear about why I called an explicit solution wallpaper.

acer

Sorry, the Equations isn't right.

Comment on 2008-03-18 08:23 from andy.zhou.nuaa (Maple Rating 1 17)

 This is the right Equation:

 eqns := a*(x/sqrt(x^2-1)+2*x/(Pi*(x^2-1))-1-2/(Pi*x)) = 1

 it have the same problem with the first equation!

 i am very sorry for waste someone's time!!

Robert Israel's picture

Similar story

Comment on 2008-03-18 11:29 from Robert Israel (Maple Rating 4 1431) 
> solve(eqns,x);

Warning, solutions may have been lost

RootOf((2*a*Pi^2+Pi^2)*_Z^6+(-a^2*Pi^2-4*a*Pi^2-2*Pi^2)*_Z^4+(-4*a*Pi-4*a^2*Pi)*_Z^3+(Pi^2+a^2*Pi^2+2*a*Pi^2)*_Z^2+(4*a*Pi+4*a^2*Pi)*_Z+4*a^2)

> P := subs(_Z=x,op(%));
  galois(subs(Pi=pi,P), x);

"6T16", {"S(6)"}, "-", 720, {"(1 6)", "(2 6)", "(3 6)", "(4 6)", "(5 6)"}

In this case we have a polynomial of degree 6 that, in general, is not solvable by radicals.   So the option of an "exact" solution does not exist.

Again, there are branches of the RootOf solution that are not solutions of your equation: note that P has degree 2 in a, but eqns is of degree 1 in a.  For almost all x with |x| > 1 there will be two real values of a satisfying P, but since eqns is only of degree 1 in a, only one of these can be a solution of eqns.

 

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