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?

He means, "let $\alpha$ and $\beta$ be the two only (eventually) distinct complex numbers such that $a\alpha ^2+b\alpha +c=0$ and $a\beta ^2+b\beta +c=0$". — Git Gud, Oct 25 '14 at 13:47

Mark Bennet · Accepted Answer · 2014-10-25T14:01:15.633

3

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.

edited Oct 25 '14 at 14:01

answered Oct 25 '14 at 13:47

Mark Bennet

100,194

So root of a quadratic equation is only another word for "solution of the quadratic equation? – mathfemi Oct 25 '14 at 13:50
@mathfemi Indeed - you will find the solutions of the polynomial equation $p(x)=0$ very often referred to as the roots of the polynomial $p(x)$. – Mark Bennet Oct 25 '14 at 13:51
Ok. Then it was only a missunderstanding. And why is $y_n=A\alpha^n+B\beta^n$ then a solution? – mathfemi Oct 25 '14 at 13:54
@mathfemi I've added a section to the answer. – Mark Bennet Oct 25 '14 at 14:01

score 0 · Answer 2 · answered Oct 25 '14 at 13:57

0

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?

answered Oct 25 '14 at 13:57

Dr. Sonnhard Graubner

95,283

score 0 · Answer 3 · answered Oct 25 '14 at 14:27

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.

Thank you. Now I search the general solution of $h_i=ph_{i+1}+qh_{i-}$ where $0<p=1-q<1$. Can you help me? — mathfemi, Oct 25 '14 at 14:42
I have that the common solution is $h_i=A\alpha^i+B\beta^i$ with $\alpha=\frac{1+\sqrt{1-4pq}}{2p}, \beta=\frac{1-\sqrt{1-4pq}}{2p}$. But the solution says that for $p\neq q$ the general solution is $h_i=A+B\left(\frac{q}{p}\right)^i$. — mathfemi, Oct 25 '14 at 14:57

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

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