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I don't understand what Mathematica is doing to get these solutions. I ask it to solve two equations, the left hand side of both expressions are the same, the right hand sides differ by a ± sign. I really don't think the solutions should be the same, but that is what Mathematica returns, without any comment.

FullSimplify[{
  Solve[1/2 E^(-((2 I l π)/k)) (-(-1 + E^(I ((4 l π)/k + δ))) Sqrt[ρ] + Sqrt[-2 E^(I ((4 l π)/k + δ)) (-2 + ρ) + ρ + E^(2 I ((4 l π)/k + δ)) ρ]) == E^(2 π I m/n), ρ], 
  Solve[1/2 E^(-((2 I l π)/k)) (-(-1 + E^(I ((4 l π)/k + δ))) Sqrt[ρ] + Sqrt[-2 E^(I ((4 l π)/k + δ)) (-2 + ρ) + ρ + E^(2 I ((4 l π)/k + δ)) ρ]) == -E^(2 π I m/n), ρ]
}]

Mathematica returns

{
  {{ρ -> (-1 + Cos[(4 m π)/n - δ])/(-1 + Cos[(4 l π)/k + δ])}}, 
  {{ρ -> (-1 + Cos[(4 m π)/n - δ])/(-1 + Cos[(4 l π)/k + δ])}}
}

Can anyone can shed some light on this, please?

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  • 1
    $\begingroup$ Check for branch cuts ... $\endgroup$ – Dr. belisarius Oct 29 '13 at 3:45
  • $\begingroup$ Yes, that would explain it. $\endgroup$ – Phillip Dukes Oct 29 '13 at 4:05

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