Solving $ax_{n+1}+bx_n+cx_{n-1}=0$

Question

In a book I found the following:

Consider a recurrence relation of the form $$ ax_{n+1}+bx_n+cx_{n-1}=0 $$ where $a$ and $c$ are both non-zero. Let us try a solution of the form $x_n=\lambda^n$; then $a\lambda^2+b\lambda+c=0$. Denote by $\alpha$ and $\beta$ the roots of this quadratic [...]

Sorry, but I do not understand what the author means with "let $\alpha$ and $\beta$ denote the roots of this quadratic".

What are $\alpha$ and $\beta$? What does he mean?

Answer 1

A quadratic equation in general has two roots (they may not be real numbers, and in the exceptional case there can be a double root).

Since there are two roots and we want to use them, we give them names, calling one root $\alpha$ and the other root $\beta$.

What we know about $\alpha$ and $\beta$ is that $$a\alpha^2+b\alpha+c=0\text{ and }a\beta^2+b\beta+c=0$$

---

Now we can multiply the first of these equations by $A\alpha^{n-1}$ and the second by $B\beta^{n-1}$ and add them to obtain$$a(A\alpha^{n+1}+B\beta^{n+1})+b(A\alpha^n+B\beta^n)+c(A\alpha^{n-1}+B\beta^{n-1})=0$$

And if we then set $x_n=A\alpha^n+B\beta^n$ we see that $$ax_{n+1}+bx_n+cx_{n-1}=0$$

So this choice of $x_n$ provides a solution to the recurrence.

Answer 2

we get the solution $\left\{\left\{x(n)\to c_1 2^{-n} \left(\frac{-\sqrt{b^2-4 a c}-b}{a}\right)^n+c_2 2^{-n} \left(\frac{\sqrt{b^2-4 a c}-b}{a}\right)^n\right\}\right\}$ do you have some initial conditions?

Answer 3

Recurrence relations such as $ax_{n+1}+bx_{n}+cx_{n-1}=0$ are the most general type of recurrence relations that you get in pre-calculus mathematics. The general method to solve these recurrences are that you assume that $x_n$ is a power of a linear polynomial such as $\lambda$. Hence $x_n=\lambda^n$. Now, $$ax_{n+1}+bx_{n}+cx_{n-1}=0$$$$=>a\lambda^{n+1}+b\lambda^{n}+c\lambda^{n-1}=0$$$$=>a\lambda^2+b\lambda+c=0$$$$=>\lambda=\frac {-b\pm \sqrt{b^2-4ac}}{2a}$$ This comes from the quadratic formula also known as Shridharacharya's formula. I(t can be easily found by completing the square in x. Now you can say generally that $\alpha=\frac {-b+\sqrt{b^2-4ac}}{2a}$ and $\beta=\frac {-b-\sqrt{b^2-4ac}}{2a}$. So $x_n=p\alpha^n+q\beta^n=p\left(\frac {-b+\sqrt{b^2-4ac}}{2a}\right)+q\left(\frac {-b- \sqrt{b^2-4ac}}{2a}\right)$. The values of $p$ and $q$ are determined by 2 pre-given values of $x_i$'s.