From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math,sci.math.num-analysis Subject: Re: Trigonometric problem Date: 1 Apr 1995 07:20:55 GMT In article <19950331.083811.682463.NETNEWS@cc1.kuleuven.ac.be>, Nicolas Point wrote: > I have this set of Equations : > > Cos(a) +Cos(b) -Cos(c) = -k > Cos(5a)+Cos(5b)-Cos(5c) = 0 > Cos(7a)+Cos(7b)-Cos(7c) = 0 > > The problem is to find the domain of k for which a solution {a,b,c} respects > the following condition : > > 0 < a,b,c < Pi/2 I don't have a full answer but this should get you pretty far. 0. Replace c by pi-c to put more symmetry in the equations; note that then we need c> Pi/2. 1. Express all equations in terms of a1=Cos(a), b1=Cos(b), and c1=Cos(c). For example, Cos(5a)=16 a1^5 - 20 a1^3 + 5 a1. Then the problem is to find the k which admits a solution in which -1 < c1 < 0 < a1,b1 < 1 2. Express all equations in terms of s1=-(a1+b1+c1), s2=(a1 b1 + a1 c1 + b1 c1) and s3 = -(a1 b1 c1). For example, (a1^5+b1^5+c1^5) = -s1^5 + 5 s2 s1^3 - 5 s3 s1^2 - 5 s2^2 s1 + 5 s2 s3. The first equation is simply s1 = k. The second is linear in s3, so we may solve this equation for s3 : s3 = -(k/20)*( (16*k^4-20*k^2+5) + (-80*k^2+60)*s2 + 80*s2^2 ) / (4*k^2-3-4*s2) When substituting into the remaining equation, we find that the original set of equations is equivalent to a single cubic in s2: 0 = (-1575+9800*k^2+3072*k^10-15360*k^8-24080*k^4+28160*k^6)+ (-20*(4*k^2-3)*(256*k^6-784*k^4+700*k^2-175))*s2+ (89600*k^2+46080*k^6-19600-116480*k^4)*s2^2+ (44800*k^2-11200-35840*k^4)*s2^3 We can solve this equation symbolically and substitute into the previous equation to get expressions for s1, s2, and s3. (Any of the real solutions for s2 may be used: we want the full solution set.) Then a1, b1, c1 may be recovered as the roots of the equation X^3 + s1 X^2 + s2 X + s3 = 0. 3. Analyze the role of k as follows. You want to know when two roots of this last equation lie in (0, 1), for example. You'll have to handle separately the case when 0 or 1 is a root, but in general you can determine the number of roots in an interval using a Sturm sequence. There is a chain of polynomials which I don't quite recall (starts with the cubic, ends with its discriminant) with the property that the number of roots in the interval is N(0) - N(1), where N(x) is the number of changes in sign in the sequence {f1(x), ..., f4(x)}. By writing down the appropriate fi, you then obtain a collection of inequalities on s1, s2, s3 which must be satisfied for all three roots of the cubic to lie in [0, 1]. The result of step 2 is to provide formulas for the si in terms of k; substituting them in, we obtain a collection of inequalities on functions of k which describe the region you need. (There will be several cases to consider corresponding to the different choices of solutions for s2, and the choices of locations for the sign changes. You want the union of all these solution sets for k.) The ultimate answer is some subset of (-3, 3), of course, so the final presentation of the answer will just amount to some number crunching. dave &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Date: Fri, 7 Apr 95 17:07:51 +0200 To: rusin@math.niu.edu (Dave Rusin) From: point@mathro.fpms.ac.be (Nicolas Point) Subject: Re: Trigonometric problem >Out of curiosity -- did you complete the problem? I posted a partial >solution but did not return to it. Dear dave, First of all : Thanks for spending some time on our problem and for posting a solution (however partial it may be). Unfortunately, I haven't seen it on any newsgroups! In Belgium, the are some "newsgroup restructuration" for the moment and it is possible (?) that I missed it by switching between two newservers. Can you send it directly by mail (point@mathro.fpms.ac.be)? Anyway, all I have for the moment are numerical solutions (see the appended mail for example). It is not elegant solution but... I'll keep you informed if I find anything. Thanks again, Nicolas. =========================================================================== Nicolas Point point@mathro.fpms.ac.be Faculte Polytechnique de Mons Rue de Houdain, 9 Tel: +32 65 374686 B-7000 Mons : +32 65 374141 Belgium Fax: +32 65 374689 =========================================================================== Enclosed Mail: ------------------------------------------------------------ Date: Mon, 3 Apr 1995 16:20:49 +0100 (CET) To: point@mathro.fpms.ac.be Subject: quiz posted in sci.math.num-analysis Dear Nicholas, I spent half an hour this sunny sunday to study your problem. No easy way for a theorical approach (manipulation on Tchebichev polynomials ?). Numerically with a constraint minimisation by simplex method, I found three valid domains for k : [ -1.8546 (a=0.09,b=0.52,c=1.56) , -1.8459 ( 0.14, 0.41, 1.51) ] [ -0.9323 ( 0.43, 0.02, 0.23) , -0.2777 ( 1.42, 0.96, 1.11) ] [ 0.2266 ( 0.92, 1.43, 0.20) , 0.5100 ( 1.57, 1.08, 0.18) ] Hope this help (Iam not sure !!!) ps: I would be happy to hear for a theorical solution by email or post to the News group ! BOETTNER Jean Claude LCSR-CNRS 1c Avenue de la Recherche Scientifique F-45071 Orleans Cedex 02 Tel : 38.51.54.95 Fax : 38.51.76.70 ------------------------------------------------------------------------------ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Date: Fri, 7 Apr 95 12:59:08 CDT From: rusin (Dave Rusin) To: point@mathro.fpms.ac.be, rusin@math.niu.edu Subject: Re: Trigonometric problem Sorry you didn't get to see this. [copy of my post sent -- djr]