Algebra simultaneous equations Question

[Rococo]

Homework Statement

Let $g_k = 2cos(k/2)$ and $z=e^{ip(N+1)}$ where N is an integer.

There are two simultaneous equations:

$E^2 = (g_k + e^{ip})(g_k + e^{-ip}) = 1 + g_k^2 + 2g_k cos(p)$ [1]

$(1+z^2)E^2 = (g_k + e^{-ip})^2 z^2 + (g_k + e^{ip})^2$[2]

Eliminate $E^2$ to show that:

$sin[pN] + g_k sin[p(N+1)] = 0$[3]

Homework Equations

The Attempt at a Solution

I've tried substitution of [1] into [2] and then equating the real parts/imaginary parts of the equation. Also have tried equating coefficients of terms like $e^{i2p(N+1)}$ but couldn't get anywhere with it.

Subsituting [1] into [2] and expanding all complex exponentials into sines and cosines, equating real parts gives me:

$1 + cos(2p(N+1)) = cos(2pN) + cos(2p) + 2sin(2p(N+1))g_k sin(p)$

And equating imaginary parts gives me:

$sin(2p(N+1)) = sin(2pN) - 2g_k sin(p)cos(2p(N+1)) + sin(2p) + 2g_k sin(p)$

But I can't seem to get equation [3] from these. Also, equating the coefficients of $e^{i2p}$ I get:

$e^{i2pn} = -2i e^{i2pN} g_k sin(p) + 1$

But again I can't rearrange it to equation [3] after equating real/imaginary parts. Does anyone have any insight as to how to do this?

 

[Simon Bridge]

Looking at: $E^2 = 1+g_k^2 + 2g_k\cos p$

Looks like the cosine rule in trig doesn't it? Is there a clue by considering the matter geometrically?
Solving for $E^2$ in [2] and subbing into [1] gives:
$$\frac{(g_k+e^{-ip})^2z^2 + (g_k+e^{ip})^2}{1+z^2} = 1+g_k^2 + 2g_k\cos p$$... this looks like a combination of real and imaginary terms on the LHS and only reals on the RHS. Is this how you are thinking? The imaginary terms have to cancel out?

Note: put a backslash before the trig function to get it to format correctly like $\sin \theta$ instead of $sin\theta$.
You may want to adopt a shorthand like $s=\sin p$ and $S=\sin(N+1)p$ to ease writing.

 

[pat8126]

This may be completely wrong, but I think there's a quick workaround to the problem. Even in the highly unlikely case that the answer is correct, the reasoning is probably wrong regardless.

Let A = $$(g_k + e^{ip})$$
Let B = $$(g_k + e^{-ip})$$

Therefore $$E^2 = AB$$

If $$(1 + z^2)E^2 = (g_k + e^{-ip})^2 z^2 + (g_k + e^{ip})^2$$

then $$AB + ABz^2 = B^2z^2 + A^2$$

Just looking at both sides of the equation, it appears that

$$AB = A^2$$
$$ABz^2 = B^2z^2$$

As such, $$A = B$$
Therefore, $$e^{ip} = e^{-ip}$$

which means $$p = 0$$

With that in mind $$sin(0) + cos(k/2)*sin(0) = 0$$

 

[haruspex]

Just looking at both sides of the equation, it appears that

It was looking useful up to that point, but your next step seems to be treating the equation in A, B and z as true for all z with constants A, B. There is no basis for that.

Anyway, it is just as well that you went wrong. This is a homework forum. The idea is to provide hints and point out errors, not lay out complete solutions.

 

[pat8126]

It was looking useful up to that point, but your next step seems to be treating the equation in A, B and z as true for all z with constants A, B. There is no basis for that.

I knew it was too good to be true - way too simple. Back to the long form of calculation.

By the way, what's the right answer? Does the angle p = 0 or is it irrelevant? Unless that's the key to the problem, which would not be allowed until after the original poster solved it.

 

[haruspex]

I knew it was too good to be true - way too simple. Back to the long form of calculation.

By the way, what's the right answer? Does the angle p = 0 or is it irrelevant? Unless that's the key to the problem, which would not be allowed until after the original poster solved it.

As I posted, your working down to there looked good. Use it to express z in terms of A and B. Simplify, then substitute in what A, B and z represent.
The problem is a bit strange... k plays no role. g_k is just any real in the range -2 to +2, and it turns out that even that is not necessary. Maybe it is just inherited from some original context.

 

[pat8126]

$$AB + ABz^2 = B^2z^2 + A^2$$

Thanks haruspex, good thoughts! With some algebra, the equation simplifies even further to:

$$AB - A^2 = B^2z^2 - ABz^2$$
$$A(B-A) = Bz^2(B - A)$$

At this point, even if I solve it, the original poster must show the answer first.

I wouldn't be able to divide by (B - A) if A = B, since that would be division by zero. In the event A did not equal B, then $$A = Bz^2$$

 

[haruspex]

if A = B, since that would be division by zero

Right, but in that case p=0 and the equation to be proved is trivially true anyway.

 

[Mark44]

The OP hasn't chimed in since last Friday, so let's hold off any more discussion until he comes back...