# Algebra simultaneous equations Question

1. Oct 21, 2016

### Rococo

1. The problem statement, all variables and given/known data
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]

2. Relevant equations

3. 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?

2. Oct 21, 2016

### 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.

3. Oct 27, 2016

### 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 $$sin(0) + cos(k/2)*sin(0) = 0$$

4. Oct 27, 2016

### haruspex

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.

5. Oct 27, 2016

### pat8126

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.

Last edited: Oct 27, 2016
6. Oct 27, 2016

### haruspex

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. gk 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.

7. Oct 27, 2016

### pat8126

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 $$A = Bz^2$$

8. Oct 27, 2016

### haruspex

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

9. Oct 27, 2016

### Staff: Mentor

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