Hello All! I am having difficulty solving this equation and I hope that there is an elegant solution out there. a11*X1 + a12*X2 + a13*X3 + a14*X1*X2 + a15*X1*X3 + a16*X2*X3 + a17*X1*X2*X3 = 0 a21*X1 + a22*X2 + a23*X3 + a24*X1*X2 + a25*X1*X3 + a26*X2*X3 + a27*X1*X2*X3 = 0 a31*X1 + a32*X2 + a33*X3 + a34*X1*X2 + a35*X1*X3 + a36*X2*X3 + a37*X1*X2*X3 = 0 Best regards and thank you in advance, Tomaz
help solving this equation
Started by ●March 2, 2005
Reply by ●March 2, 20052005-03-02
(X1,X2,X3)=(0,0,0) That will be $100 please. Dirk tomaz wrote:> Hello All! > > I am having difficulty solving this equation and I hope that there isan> elegant solution out there. > > a11*X1 + a12*X2 + a13*X3 + a14*X1*X2 + a15*X1*X3 + a16*X2*X3 +a17*X1*X2*X3> = 0 > > a21*X1 + a22*X2 + a23*X3 + a24*X1*X2 + a25*X1*X3 + a26*X2*X3 +a27*X1*X2*X3> = 0 > > a31*X1 + a32*X2 + a33*X3 + a34*X1*X2 + a35*X1*X3 + a36*X2*X3 +a37*X1*X2*X3> = 0 > > > Best regards and thank you in advance, > > Tomaz
Reply by ●March 2, 20052005-03-02
tomaz wrote:> Hello All! > > I am having difficulty solving this equation and I hope that there is an > elegant solution out there. > > a11*X1 + a12*X2 + a13*X3 + a14*X1*X2 + a15*X1*X3 + a16*X2*X3 + a17*X1*X2*X3 > = 0 > > a21*X1 + a22*X2 + a23*X3 + a24*X1*X2 + a25*X1*X3 + a26*X2*X3 + a27*X1*X2*X3 > = 0 > > a31*X1 + a32*X2 + a33*X3 + a34*X1*X2 + a35*X1*X3 + a36*X2*X3 + a37*X1*X2*X3 > = 0 > > > Best regards and thank you in advance, > > TomazThat's three equations. If the X's are to be solved for in terms of the known or knowable a's, it's possible; otherwise not. What have you tried that doesn't seem to work? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 2, 20052005-03-02
in article 1109777759.628179.12010@g14g2000cwa.googlegroups.com, dbell@niitek.com at dbell@niitek.com wrote on 03/02/2005 10:35:> (X1,X2,X3)=(0,0,0) > > That will be $100 please. > > Dirkwow! what's your going rate, Dirk? you must be billing more than the lobbyists and schisters in DC. or is there a minimum billing even for 30 seconds of time? (all tongue in cheek) :-)> tomaz wrote: >> Hello All! >> >> I am having difficulty solving this equation and I hope that there is > an >> elegant solution out there. >> >> a11*X1 + a12*X2 + a13*X3 + a14*X1*X2 + a15*X1*X3 + a16*X2*X3 + > a17*X1*X2*X3 >> = 0 >> >> a21*X1 + a22*X2 + a23*X3 + a24*X1*X2 + a25*X1*X3 + a26*X2*X3 + > a27*X1*X2*X3 >> = 0 >> >> a31*X1 + a32*X2 + a33*X3 + a34*X1*X2 + a35*X1*X3 + a36*X2*X3 + > a37*X1*X2*X3 >> = 0 >> >> >> Best regards and thank you in advance, >> >> Tomaz >-- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●March 2, 20052005-03-02
r b-j, The charge is based on a highly complex formula based on the number of zeros in the final answer and the amount of time the client has already spent on the problem. I genereously did not include decimal points and the infinite number of succeeding zeros. And, of course, 30 seconds rounds up to an hour. I believe this is all detailed in a government regulation somewhere that can be ordered from NTIS in Springfield, VA. ;-) Dirk
Reply by ●March 2, 20052005-03-02
Hi, I am still a newbie, so I am going to present my ugly/elementary solution here for everyone's critique....... Let x2=p*x1, x3=q*x1 where p and q are unknown constants. Subst. them into the orignal equations. Since the trivial case (p=q=0, i.e. x1=x2=x3=0) has been taken care of by Dirk, you can divide everything by x1. Essentially we have reduced the problem to a set of 2nd order equations. Now it looks a lot friendlier than the original 3rd order system: (a17*p*q)*x1^2 + (a14*p+a15*q+a16*p*q)*x1 + (a11+a12*p+a13*q) = 0; (a27*p*q)*x1^2 + (a24*p+a25*q+a26*p*q)*x1 + (a21+a22*p+a23*q) = 0; (a37*p*q)*x1^2 + (a34*p+a35*q+a36*p*q)*x1 + (a31+a32*p+a33*q) = 0; Case 1: p*q=0, (p+q)!=0. Without loss of generality, assume q=0: (a14*p)*x1 + (a11+a12*p) = 0; (a24*p)*x1 + (a21+a22*p) = 0; (a34*p)*x1 + (a31+a32*p) = 0; You should have no problem solving for p and x1 here. (There may or may not be a solution in this case.) Case 2: p*q!=0 By comparing coefficients' ratios, you will get something like: a27*(a14*p+a15*q+a16*p*q) = a17*(a24*p+a25*q+a26*p*q); a37*(a14*p+a15*q+a16*p*q) = a17*(a34*p+a35*q+a36*p*q); a27*(a11+a12*p+a13*q) = a17*(a21+a22*p+a23*q) a37*(a11+a12*p+a13*q) = a17*(a31+a32*p+a33*q) Solve for p and q using the last 2 equations. After you solve p and q, subst. them into the first 2 to see if contradiction occurs. If not, all we have to do now is to solve one quadratic equation in x1. Elegant? No. Methodical? Yes. I hope this helps. Best Regards. KD
Reply by ●March 2, 20052005-03-02
By the way, Jerry, I want to say sorry. I understand that the more experienced guys in comp.dsp want to encourage people to ask questions in a more meaningful way, i.e. by adding their own thoughts and analysis first. That is why you guys seldom answer such a question directly. I ended up solving this problem because at first glance, this looked like a cubic system......and I often wonder when the cubic formula can be simplified...... http://mathworld.wolfram.com/CubicFormula.html
Reply by ●March 2, 20052005-03-02
kd_ei@yahoo.com wrote:> Hi, > > I am still a newbie, so I am going to present my ugly/elementary > solution here for everyone's critique....... > > > Let x2=p*x1, x3=q*x1 where p and q are unknown constants. > > Subst. them into the orignal equations. Since the trivial case (p=q=0, > i.e. x1=x2=x3=0) has been taken care of by Dirk, you can divide > everything by x1. Essentially we have reduced the problem to a set of > 2nd order equations. Now it looks a lot friendlier than the original > 3rd order system: > > (a17*p*q)*x1^2 + (a14*p+a15*q+a16*p*q)*x1 + (a11+a12*p+a13*q) = 0; > (a27*p*q)*x1^2 + (a24*p+a25*q+a26*p*q)*x1 + (a21+a22*p+a23*q) = 0; > (a37*p*q)*x1^2 + (a34*p+a35*q+a36*p*q)*x1 + (a31+a32*p+a33*q) = 0; > > > Case 1: p*q=0, (p+q)!=0. Without loss of generality, assume q=0: > > (a14*p)*x1 + (a11+a12*p) = 0; > (a24*p)*x1 + (a21+a22*p) = 0; > (a34*p)*x1 + (a31+a32*p) = 0; > > You should have no problem solving for p and x1 here. (There may or may > not be a solution in this case.) > > > Case 2: p*q!=0 > > By comparing coefficients' ratios, you will get something like: > > a27*(a14*p+a15*q+a16*p*q) = a17*(a24*p+a25*q+a26*p*q); > a37*(a14*p+a15*q+a16*p*q) = a17*(a34*p+a35*q+a36*p*q); > a27*(a11+a12*p+a13*q) = a17*(a21+a22*p+a23*q) > a37*(a11+a12*p+a13*q) = a17*(a31+a32*p+a33*q) > > Solve for p and q using the last 2 equations. After you solve p and q, > subst. them into the first 2 to see if contradiction occurs. If not, > all we have to do now is to solve one quadratic equation in x1. > > > Elegant? No. Methodical? Yes. I hope this helps. > > > Best Regards.Another way would use matrices, but it leads to the same solution in the end. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 2, 20052005-03-02
kd_ei@yahoo.com wrote:> By the way, Jerry, I want to say sorry. I understand that the more > experienced guys in comp.dsp want to encourage people to ask questions > in a more meaningful way, i.e. by adding their own thoughts and > analysis first. That is why you guys seldom answer such a question > directly.A short answer like that from me is partly didactic and partly lazy. I usually see no reason to do the grunt work if I can help by pointing someone in the right direction. Better for him, easier for me. > I ended up solving this problem because at first glance, this> looked like a cubic system......and I often wonder when the cubic > formula can be simplified......I was going to say yes, but ...> http://mathworld.wolfram.com/CubicFormula.html... I see you found the answer. I need to look it up whenever I use it, but I know where to find it. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 2, 20052005-03-02
Jerry Avins wrote:> kd_ei@yahoo.com wrote: > >> Hi, >> >> I am still a newbie, so I am going to present my ugly/elementary >> solution here for everyone's critique....... >> >> >> Let x2=p*x1, x3=q*x1 where p and q are unknown constants. >> >> Subst. them into the orignal equations. Since the trivial case (p=q=0, >> i.e. x1=x2=x3=0) has been taken care of by Dirk, you can divide >> everything by x1. Essentially we have reduced the problem to a set of >> 2nd order equations. Now it looks a lot friendlier than the original >> 3rd order system: >> >> (a17*p*q)*x1^2 + (a14*p+a15*q+a16*p*q)*x1 + (a11+a12*p+a13*q) = 0; >> (a27*p*q)*x1^2 + (a24*p+a25*q+a26*p*q)*x1 + (a21+a22*p+a23*q) = 0; >> (a37*p*q)*x1^2 + (a34*p+a35*q+a36*p*q)*x1 + (a31+a32*p+a33*q) = 0; >> >> >> Case 1: p*q=0, (p+q)!=0. Without loss of generality, assume q=0: >> >> (a14*p)*x1 + (a11+a12*p) = 0; >> (a24*p)*x1 + (a21+a22*p) = 0; >> (a34*p)*x1 + (a31+a32*p) = 0; >> >> You should have no problem solving for p and x1 here. (There may or may >> not be a solution in this case.) >> >> >> Case 2: p*q!=0 >> >> By comparing coefficients' ratios, you will get something like: >> >> a27*(a14*p+a15*q+a16*p*q) = a17*(a24*p+a25*q+a26*p*q); >> a37*(a14*p+a15*q+a16*p*q) = a17*(a34*p+a35*q+a36*p*q); >> a27*(a11+a12*p+a13*q) = a17*(a21+a22*p+a23*q) >> a37*(a11+a12*p+a13*q) = a17*(a31+a32*p+a33*q) >> >> Solve for p and q using the last 2 equations. After you solve p and q, >> subst. them into the first 2 to see if contradiction occurs. If not, >> all we have to do now is to solve one quadratic equation in x1. >> >> >> Elegant? No. Methodical? Yes. I hope this helps. >> >> >> Best Regards. > > > Another way would use matrices, but it leads to the same solution in the > end. > > JerryBut would a matrix representation been clearer? I looked at the system of equations and thought, "That should mean something." But I couldn't *recognize* what. just my 2 cents
