| Peter
Gyarmati |
For what values of the parameter t do the simultaneous equations x+y+z+v=0, (xy+yz+zv)+t(xz+xv+yv)=0 have a unique solution? |
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| Brad
Rodgers |
If I'm not making any careless mistakes, it's relatively easy to see that t = 1 /4 (x2 + y2 + z2 + v2 ) satisfies the equation for all (x,y,z,v), so only if t = 1 /4 (x2 + y2 + z2 + v2 ) + f(x,y,z,v), where f(x,y,z,v) = 0 has a unique solution, will the simultaneous equations have a unique solution. Brad |
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| Andre
Rzym |
I had assumed that t was a real number rather than a function of x,y,z,v. Taking t=1, for example, 0=(xy+yz+zv)+t(xz+xv+yv) 0=(xy+yz+zv)+(xz+xv+yv) 0=1/2(x+y+z+v)2 -1/2(x2 + y2 + z2 + v2 ) 0=-1/2(x2 + y2 + z2 + v2 ) Implying the unique solution x=y=z=v=0. The issue is: what about the other values of t? I have a partial solution so far. I'm sure I'm missing something elegant... Andre |
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| Brad
Rodgers |
Oops; regardless of interpretation, my answer is false. I had mentally placed a '+' between the t and the term by which it is being multiplied in the actual problem - I think because they're on separate lines. In any case, I'll try again... Brad |
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| Peter
Gyarmati |
Andre, you must be interested in where this problem comes from. Well I already have one degree as an engineer-teacher, but I was always interested in maths and now I'm attednig this subject at university. This problem and all the others I put up to nrich, were my more difficult homeworks at university. I had no idea for this problem. I read what you wrote, now I'm trying to solve it in different ways, but I think the finish is still far. Thanks for helping me. Peter |
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| Andre
Rzym |
Thanks Peter, I was indeed curious as to where this came from. I've had a few thoughts so far: 1) x=y=z=v=0 is always a solution, so the question is for what t there is any other. 2) One can come up with a simple second solution by construction when t< 0. 3)For t> 0 my thinking is as follows: If (a,b,c,d) is a nonzero solution, then so is (ka,kb,kc,kd). Therefore there must be (at least) a complete line of solutions [including (0,0,0,0)]. Now eliminate x from equation 2, and consider a Taylor expansion about (0,0,0,0). Clearly all first derivatives are zero, so the first nonzero terms look like: (dx,dy,dz,dv)A (dx,dy,dz,dv)T for suitable (symmetric) matrix of derivatives A. So now it is a question of figuring out for which values of A this expression is always positive - because this implies that there cannot be any other solutions. I hope to take another look at it tomorrow. Andre |
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| Andre
Rzym |
Peter, I'm not 100% sure yet, but I think that the critical range of t for a single solution is 1/2(3-sqrt(5)) < t < 1/2(3+sqrt(5)) If t equals the endpoints of the above inequality, we have solutions (x,y,z,v) = (-1,1/2(1-sqrt(5)), 1/2(sqrt(5)-1),1) (x,y,z,v) = (-1,1/2(sqrt(5)+1), -1/2(sqrt(5)+1),1) Does this agree with your thinking? Andre |
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| Peter
Gyarmati |
Andre, Till now I dealt with problem relatively little. I have many such hard problems - see the topic "Elementary geometry" - and I must learn much theory. Sorry, now I see only that your values for (x,y,z,v) are correct. Tomorrow I will concentrate only on this problem. Peter |
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| Andre
Rzym |
Peter, I am now sure that this solution is correct. I can post my (ugly) proof if you wish (I'm sure there must be a neater way), but it roughly goes as follows: 1) Observe x=y=z=v=0 is always a solution. We are looking for values of t for which there is no other solution. 2) Convince oneself that for negative t, there is always a second solution. Therefore t must be nonnegative. 3) If (x,y,z,v) is a solution, so is k(x,y,z,v). Convince oneself that v can be taken to be nonzero and therefore can be taken to be 1 (in the context of this second solution). 4) Now regard the problem as a maximisation problem for which we use Lagrange multipliers. We want to maximise F subject to G=0, by constructing H=F+mG where G=x+y+z+v F=(xy+yz+zv)+t(xz+xv+yv) Requiring that the partial derivatives of H with respect to x,y,z,v, m all equal zero gives: y=(-3+2t)/(-4+t) etc. 5) Substitute the results of (4) into F. This gives F as a function of t. We have computed the stationary value of F over all x,y,z for a given t. It's a cubic in t, with roots -1,1/2(3-sqrt(5)),1/2(3+sqrt(5)). This should allow you to argue that provided t is between the 2nd and 3rd roots, the maximum is negative, and therefore x=y=z=v=0 is the only solution to F=G=0. I am convinced that there must be a neater solution ... Andre |
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| Peter
Gyarmati |
Thanks Andre! Could you send me your full proof? Peter |
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| Andre
Rzym |
Peter, Sorry for the delay. Here is an expansion of some of the points. I can't stress enough that there must be a more elegant solution! 2) Consider (0,1,0,-1). Then F(x,y,z,v) as defined previously will be positive for negative t. However, for sufficiently large a, (a,1-a,0,-1) will be negative. Therefore, since F is a continuous function, there is some a for which F is zero. But for no value of a does this equal (0,0,0,0), therefore there is a second solution of F=0 for t < 0. 4) For H=F +mG as defined above, we require that derivatives with respect to x, y, z, m are all 0. Written out in full: H=(x y+y z+z)+t(x z+x+y)+m(x+y+z+1) hence ¶H/¶x=0=y+t(z+1)+m ¶H/¶y=0=(x+z)+t+m ¶H/¶z=0=(y+1)+t x+m ¶H/¶m = 0=x+y+z+1 Solving: x=(t2-2t+2)/(t2-4t) y=(-2t+3)/(t-4) z=(3t-2)/(t2-4t) m = (-t2+3t-1/(t-4) Andre |
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| Peter
Gyarmati |
Thanks for the solution. If they tell me the solution at the university, I will put it up here, if it is more elegant than yours. Peter |