Solution to $x_{n+2} = x_{n+1} - x_{n}$ does not match initial conditions

Question

I have solved the recursion equation $$x_{n+2} = x_{n+1} - x_{n}$$ and I know the answer is (with a 100% certainty) $$x_n = Ae^{(\frac{in\pi}{3})} + Be^{(-\frac{in\pi}{3})}.$$

Now, as initial values, I am given $$x_0 = 2, \quad x_1 = -3.$$ But solving for $A,B$ with these initial values results in an impossible equation.

With $x_0 = 2$ I get
$$Ae^{i*0} + Be^{i*0} = 2 \implies A+B = 2$$ With $x_1 = -3$ I get $$Ae^{\frac{i\pi}{3}} + Be^{-\frac{i\pi}{3}} = -3$$ $$A\cos{(\frac{\pi}{3})} + iA\sin{(\frac{\pi}{3}})) + B\cos{(-\frac{\pi}{3})} + iB\sin{(-\frac{\pi}{3})} = -3$$ For this to be true, $$(A+B) = \frac{-3}{\cos{\frac{\pi}{3}}} = -6$$ must be true which directly contradicts $A+B=2$.

Where is my reasoning wrong? Again, I am certain $x_n$ and the initial values is correct.

Answer 1

Edit: I failed to compute the roots of the characteristic equation correctly. Your roots are right. I'll leave my comment up, still.

This recursion has characteristic polynomial $\lambda^2-\lambda+1$ (since $x_{n+2}-x_{n+1}+x_n=0$ and the roots are $\lambda_i = \frac{1\pm i\sqrt3}{2}$ as you indicate, so we jump to the conclusion (two distinct roots) that $x_n = A\left(\frac{1+i\sqrt3}{2}\right)^n+B\left(\frac{1-i\sqrt3}{2}\right)^n$. For anyone curious, here is a basic tutorial on the characteristic equation method. As others have pointed out, the coefficients $A$ and $B$ need not be real and that is your mistake.

For those interested, I will flesh out a solution using generating functions whose solutions requires partial fraction at the end. Start with $$x_{n+2}-x_{n+1}+x_n = 0\quad \text{ for all }n\geq 0$$ and multiply both sides by $q^{n+2}$ and sum along all $n\geq 0$: $$0 = \sum_{n\geq 0}\left( x_{n+2}-x_{n+1}+x_n\right)q^{n+2} = (F(q)-x_1q-x_0) - q(F(q)-x_0) + q^2F(q)$$ when you work it all out. Then $$F(q) = \frac{\alpha+\beta q}{1-q+q^2}$$ (ignoring the actual value of the constants, for simplicity) and this should drive home the point that the characteristic equation needs to be solved (so that we can take partial fractions). We are then left with $$F(q) = \frac{A}{1-\lambda_1q} + \frac{B}{1-\lambda_2q}$$ and it is worthwhile to meditate on how these constants got absorbed as they did into new constants.

Answer 2

The condition $$Ae^{i\pi/3}+Be^{-i\pi/3}=-3$$ expands to $$(A+B)\cos{\pi\over3}+(A-B)i\sin{\pi\over3}=-3\ .$$ This is an equation for two complex numbers $A$ and $B$, and can be rewritten as $$\left({1\over2}+i{\sqrt{3}\over2}\right)A+\left({1\over2}-i{\sqrt{3}\over2}\right)B=-3\ .$$ (Your conclusion at this point is simply wrong.) Together with $A+B=2$ you obtain a fine system of two linear equations with complex coefficients.