Efficiently Solving a System of Equations for Physics Problems

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The discussion focuses on solving a system of equations relevant to a physics problem, specifically involving the equation h+k+l=2n+1 and a complex exponential equation. The initial approach involved factoring out ei(π/2) and simplifying the equation, but the poster faced difficulties progressing from that point. Suggestions were made to express the equations in terms of sine and cosine functions, emphasizing the need to satisfy periodic conditions. Additionally, it was recommended to redefine variables and analyze specific cases for m to gain insights into the general solution. The conversation highlights the complexity of solving such systems and the importance of strategic simplification.
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Homework Statement



Hi everybody, I am trying to find the solution to this system of equations for a bigger physics problem:

h+k+l=2n+1 with n=0,1,2,3...
ei(π/2)(h+k+l)+ei(π/2)(3h+3k+l)+ei(π/2)(3h+k+3l)+ei(π/2)(h+3k+3l)=0

Homework Equations



None

The Attempt at a Solution



Ihave tried to solve it by taking the common factor ei(π/2) out and trying to solve
e(h+k+l)+e(3h+3k+l)+e(3h+k+3l)+e(h+3k+3l)=0 by using h+k+l=0; then, I arrive to 1+e2k+2h+e2h+2l+e2k+2l=0.

From here I don't know how to keep going, so if anybody could point me what to do next it wolud be very helpfull.

Thanks for reading.
 
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Frank Einstein said:
Ihave tried to solve it by taking the common factor ei(π/2) out and trying to solve
That is a factor of ei(π/2)+x, not of ei(π/2)x.
 
Frank Einstein said:

Homework Statement



Hi everybody, I am trying to find the solution to this system of equations for a bigger physics problem:

h+k+l=2n+1 with n=0,1,2,3...
ei(π/2)(h+k+l)+ei(π/2)(3h+3k+l)+ei(π/2)(3h+k+3l)+ei(π/2)(h+3k+3l)=0
To make these equal to zero, you need to enforce conditions on the periodic functions--both real and imaginary.
##\cos( \frac{ \pi(h+k+l)}{2} ) + \cos( \frac{ \pi(3h+3k+l)}{2} )+\cos( \frac{ \pi(3h+k+3l)}{2} )+\cos( \frac{ \pi(h+3k+3l)}{2} ) =0##
and
##\sin( \frac{ \pi(h+k+l)}{2} ) + \sin( \frac{ \pi(3h+3k+l)}{2} )+\sin( \frac{ \pi(3h+k+3l)}{2} )+\sin( \frac{ \pi(h+3k+3l)}{2} ) =0.##

From these, you should be able to say something about the cases where this must be true.
 
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Think of these as :
##\cos( \frac{ \pi(h+k+l)}{2} ) + \cos( \frac{ \pi(h+k+l)}{2} +(h+k)\pi )+\cos( \frac{ \pi(h+k+l)}{2}+ (h+l)\pi )+\cos( \frac{ \pi(h+k+l)}{2}+(k+l)\pi ) =0##
and
##\sin( \frac{ \pi(h+k+l)}{2} ) + \sin( \frac{ \pi(h+k+l)}{2} +(h+k)\pi )+\sin( \frac{ \pi(h+k+l)}{2}+ (h+l)\pi )+\sin( \frac{ \pi(h+k+l)}{2}+(k+l)\pi ) =0.##
 
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Frank Einstein said:

Homework Statement



Hi everybody, I am trying to find the solution to this system of equations for a bigger physics problem:

h+k+l=2n+1 with n=0,1,2,3...
ei(π/2)(h+k+l)+ei(π/2)(3h+3k+l)+ei(π/2)(3h+k+3l)+ei(π/2)(h+3k+3l)=0

Homework Equations



None

The Attempt at a Solution



Ihave tried to solve it by taking the common factor ei(π/2) out and trying to solve
e(h+k+l)+e(3h+3k+l)+e(3h+k+3l)+e(h+3k+3l)=0 by using h+k+l=0; then, I arrive to 1+e2k+2h+e2h+2l+e2k+2l=0.

From here I don't know how to keep going, so if anybody could point me what to do next it wolud be very helpfull.

Thanks for reading.

Please do not use all bold fonts for your solution; it looks like you are yelling at us!
Mod note: Removed excess bold font

Anyway, if you let ##m = 2n + 1##, and if you note that ##h+k+l = m## implies
\frac{i \pi}{2} (3h+3k+l) = \frac{i \pi}{2} (3h + 3k + 3l - 2l) = \frac{i \pi}{2} 3m - i \pi l,
and so forth, you get a much simpler problem, especially if you multiply through by ##e^{-3 m \pi i/2}##.

I would suggest you look at some of the initial cases ##m = 1, 3, 5, \ldots## separately, until you have gained sufficient insight into the nature of the solution for general odd ##m##.
 
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Thank you very much for your anwsers, guess I will be able to proper solve it.
 
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