Let $\sum a_nx^n$ be a power series whose radius of convergence is $p\in(0,+\infty)$. If $A, B$ are two real numbers satisfying the relation $a_n+A\cdot a_{n-1}+B\cdot a_{n-2}=0$ for all $n>1$ and $A^2-4B<0$, then find the sum of the power series.
I have tried it but not get the exact sum which is given in the answer. So please help me to solve it. Thanks in advance
The characteristic equation associated with the recurrence relation$$a_n+A\,a_{n-1}+B\,a_{n-2}=0$$is$$t^2+At+B=0$$It has to complex (conjugate) solutions : $r\,e^{\pm i\theta}$ ($r>0$).
If $a_0,a_1$ are also real numbers, then there exist $(\lambda,\mu)\in\mathbb{R^2}$ such that :
$$\forall n\in\mathbb{N},\,a_n=r^n\left(\lambda\cos(n\theta)+\mu\sin(n\theta)\right)$$
Hence, for all $x\in\mathbb{R}$ such that $\vert x\vert<1/r$ :
$$\sum_{n=0}^\infty a_nx^n=\lambda f(x)+\mu g(x)$$
where :
$$f(x)=\sum_{n=0}^\infty(rx)^n\cos(n\theta)=Re\left(\frac{1}{1-rxe^{i\theta}}\right)=Re\left(\frac{1-rxe^{-i\theta}}{(1-rx\cos(\theta))^2+(rx\sin(\theta))^2}\right)$$
that is :
$$f(x)=\frac{1-rx\cos(\theta)}{1-2rx\cos(\theta)+r^2x^2}$$
and, after a similar calculation :
$$g(x)=\frac{rx\sin(\theta)}{1-2rx\cos(\theta)+r^2x^2}$$
The constants $\lambda$ and $\mu$ are obtained by solving the system $\cases{\lambda=a_0\cr r(\lambda\cos(\theta)+\mu\sin(\theta)=a_1}$