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Title: Find real solutions of the system Post by tony123 on Oct 23rd, 2007, 2:52am Find real solutions of the system sin x + 2 sin(x+y+z) = 0, sin y + 3 sin(x+y+z) = 0, sin z + 4 sin(x+y+z) = 0. |
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Title: Re: Find real solutions of the system Post by towr on Oct 23rd, 2007, 3:40am Let's get the easy ones out of the way: x,y,z multiples of pi works.. More solutions would probably entail doing work, so I'll leave that to someone else for now :P |
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Title: Re: Find real solutions of the system Post by pex on Oct 23rd, 2007, 10:16am on 10/23/07 at 03:40:35, towr wrote:
[hideb] S = [sin x + 2 sin(x+y+z)]2 + [sin y + 3 sin(x+y+z)]2 + [sin z + 4 sin(x+y+z)]2. We know that S attains the minimum zero: it cannot get any smaller, because it is a sum of squares; and zero is attained by x=y=z=0, for example. In a minimum, the partial derivatives are zero: dS/dx = 2[sin x + 2 sin(x+y+z)][cos x + 2 cos(x+y+z)] + 50 sin(x+y+z) cos(x+y+z) = 0 dS/dy = 2[sin y + 3 sin(x+y+z)][cos y + 3 cos(x+y+z)] + 40 sin(x+y+z) cos(x+y+z) = 0 dS/dz = 2[sin z + 4 sin(x+y+z)][cos z + 4 cos(x+y+z)] + 26 sin(x+y+z) cos(x+y+z) = 0 But we know that the minimum is zero; as a consequence, all bold factors are zero at each optimum. This clearly leads to the optimum condition sin(x+y+z) cos(x+y+z) = 0. This last condition is satisfied as long as x+y+z is an integer multiple of pi/2. If it is an odd multiple, the first equation of the system becomes (sin x) + 2 = 0, which obviously has no solution. So x+y+z is an even integer multiple of pi/2; that is, an integer multiple of pi. Then, the system of equations simplifies to sin x = 0 sin y = 0 sin z = 0 and all solutions are precisely given by "x, y and z are integer multiples of pi". Second Edit: never mind, I should learn how to take derivatives... :-[ |
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