Given that $a(0) = 1, a(1) = -1$, $a(n)= -4a(n-2)$ for all n>= 2, Solve the recurrence relation.
I know that the characteristic polynomial is $x^2+4=0$, using the quadratic equation we get the two roots as +- $\sqrt{-16}/2$ which i simplified to $+-2i$
The solution to the recurrence relation will be in the form
$a(n) = A(0+-2i)^n+B(0+-2i)^n$
how can I solve this recurrence? My prof used De Moivres Theorem to solve something similar but I do not understand the theorem
For a general method of solution, you can see Theorem 4.2, part(3) and Example 4.4(c)
The characteristic equation is $r^2 + 4 = 0$ with solutions
$r = \pm ~i~2$
which means
$\rho = 2$ and $\theta = \frac{\pi}{2}$
Hence the general solution is
$a_n = 2^n \left( A \cos {\frac{n \pi}{2}} + B \sin {\frac{n \pi}{2}}\right)$
Putting $n = 0$ in the above solution we have
$a_0 = A = 1$
Putting $n = 1$ in the above solution we have
$a_1 = 2^1 \left( A \cos {\frac{1 \pi}{2}} + B \sin {\frac{1 \pi}{2}}\right) = -1$
Notice that $\cos {\frac{\pi}{2}} = 0$
or, $a_1 = 2B = -1 \implies B = -\frac{1}{2}$
Finally
$a_n = 2^n \left( \cos {\frac{n \pi}{2}} -\frac{1}{2} \sin {\frac{n \pi}{2}}\right)$
Please let me know if you could follow all the steps.