0
$\begingroup$

I have this equation:

$$\lambda \sin(2 \alpha)+ \sin(2 \alpha \lambda)=0$$

where "$\alpha$" is a known parameter and my desire is to calculate eigenvalues, "$\lambda$". I've tried some newton-raphson and muller codes in Matlab, but it didn't work since I know eigenvalues for some alphas. for example, for $\alpha=\pi/3$, first eigenvalue becomes $0.5122$ and it becomes complex for second one and after. I hope I've explained it clearly.

  • 0
    Neither Newton-Raphson nor Muller work well if you don't have a good initial guess for the eigenvalues you seek...2012-05-04
  • 2
    There must be a mistake somewhere; there are no non-trivial real solutions to this equation for $\alpha=\pi/3$, as you can see from [this plot](http://www.wolframalpha.com/input/?i=plot+{x+sin+%282+pi+%2F+3%29%2C-sin+%282+pi+x+%2F+3%29}+for+x%3D-pi..pi).2012-05-04
  • 0
    @joriki: indeed there is no non-trivial real solution for $0 < \alpha < 1.276782905$ approximately. Perhaps he meant $\alpha = 5\pi/6$, for which there is a solution $\lambda \approx .5122213612$2012-07-16

1 Answers 1