if
x=1+ω^x+ω^{2x}
find x
ω is The cube root of unity
It cannot be done algebraicly, because you end up with a variable both normally and exponentiated. 1+ω+ω^2 is easy, it's 0. But each term raised to the power of x means this isn't true.
However, having plotted both the real and imaginary parts of 1+ω^x+ω^{2x}-x between 0 and 2pi, there's a root common to both graphs, x=3. In this case you end up with 1+1+1-3, which is indeed 0.
This is the only solution, as can be seen by taking absolutes of various bits.
However, having plotted both the real and imaginary parts of 1+ω^x+ω^{2x}-x between 0 and 2pi, there's a root common to both graphs, x=3. In this case you end up with 1+1+1-3, which is indeed 0.
This is the only solution, as can be seen by taking absolutes of various bits.
but how
He just explained it to you.
You have omega equal to the "cube root of unity" which is the cube root of 1, which is 1. Regardless of what you choose x to be, you have x = 1 + 1^x + 1^2x, and the only value of x that makes the equation valid is 3. i.e. 3 = 1 + 1 + 1. If you choose x to be, say, 0, you have 0 = 1 + 1 + 1, which obviously isn't true.
You have omega equal to the "cube root of unity" which is the cube root of 1, which is 1. Regardless of what you choose x to be, you have x = 1 + 1^x + 1^2x, and the only value of x that makes the equation valid is 3. i.e. 3 = 1 + 1 + 1. If you choose x to be, say, 0, you have 0 = 1 + 1 + 1, which obviously isn't true.
QUOTE (excaza+Jul 6 2008, 03:08 AM)
You have omega equal to the "cube root of unity" which is the cube root of 1, which is 1
No, the n'the root of unity is cos(2pi.i/n) + i sin( 2pi.in) = exp(2pi.i/n). The other roots of unity are then exp(2pi.i.m/n) for m=2,3,....,n, so there's n of them.
1+ω^x+ω^{2x} = 1 + exp(2pi.x.i/n) + exp(4pi.x.i/n)
Therefore, the real and imaginary parts of 1+ω^x+ω^{2x} - x = 0 are
1 + cos(2pi.x.i/n) + cos(4pi.x.i/n) - x = 0
sin(2pi.x.i/n) + sin(4pi.x.i/n) = 0
You can solve the second one to get x is an integer.
The first one then only have to be checked for integers. Since -x is a monotonic decreasing function and cos(2pi.x.i/n) + cos(4pi.x.i/n) is periodic, with absolute modulus less than 2, you only need to check the region between x=-1 and x=+3 (the offset is due to the +1 term). This can be done by hand because you only need to check 5 values of x.
Job done.
No, the n'the root of unity is cos(2pi.i/n) + i sin( 2pi.in) = exp(2pi.i/n). The other roots of unity are then exp(2pi.i.m/n) for m=2,3,....,n, so there's n of them.
1+ω^x+ω^{2x} = 1 + exp(2pi.x.i/n) + exp(4pi.x.i/n)
Therefore, the real and imaginary parts of 1+ω^x+ω^{2x} - x = 0 are
1 + cos(2pi.x.i/n) + cos(4pi.x.i/n) - x = 0
sin(2pi.x.i/n) + sin(4pi.x.i/n) = 0
You can solve the second one to get x is an integer.
The first one then only have to be checked for integers. Since -x is a monotonic decreasing function and cos(2pi.x.i/n) + cos(4pi.x.i/n) is periodic, with absolute modulus less than 2, you only need to check the region between x=-1 and x=+3 (the offset is due to the +1 term). This can be done by hand because you only need to check 5 values of x.
Job done.
Hmmm, that's what I get for stopping at differential equations. I never knew that about the roots of unity.
AN is wise
AN is wise
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