0

I was reading brilliant wiki on recurrence relations. It says, if the characterstic polynomial has complex roots. Say $$r=2e^{\pm i\theta}\\\text{where }\theta=\arctan(\sqrt{15})\\\text{for relation }x_n=x_{n-1}-4x_{n-2}$$ Then the solution are given by $$x_n=\alpha2^n\cos{n\theta}+\beta2^n\sin{n\theta}$$ I don't understand why it shouldn't be $$x_n=\alpha2^ne^{in\theta}+\beta2^ne^{-in\theta}$$ or how the two solution are same. Please help.

Anvit
  • 3,379
  • why it shouldn't be It could be that, depending on $x_0,x_1$. But if those are real, then you wouldn't expect $x_n$ to ever be a non-real complex. Take a closer look at how $,\alpha,\beta,$ depend on $x_0,x_1$. – dxiv Apr 29 '18 at 06:27
  • @dxiv I think the two solutions are one and the same just that $\alpha$ and $\beta$ are different in both. Not sure how though – Anvit Apr 29 '18 at 06:30
  • 1
    Right, those are not the same $\alpha,\beta$. Write down the formulas for $x_0,x_1$ and it should become obvious. – dxiv Apr 29 '18 at 06:31

0 Answers0