0

I am having problems understanding how to solve second-order linear homogeneous recurrence relations with constant coefficients $b,c$. I have a clear understanding on solving second order linear recurrence relations, but am stumped with this equation as it is given in a form different than what I am used to.

Let the following question be used as an example.

if: $a_0 =0\\a_1 = 1 \\a_2 = 4 \\ a_3 =37$

satisfy the recurrence relation $a_{n+2}+ ba_{n+1}+ ca_n$ where $n\ge 0$ and $b,c$ are constance determine $b,c$ and solve for $a_n$.

In my 'solution' I used the quadratic formula to determine the roots of the equation and then followed with row reduction to for the equation for $a_n$.

$r_1= \frac{(-b + \sqrt{b^2 -4c})}{2}\\ r_2= \frac{(-b - \sqrt{b^2 -4c})}{2}$

which then was used to determine the coeeficients

$ \begin{array}{cc|c} 1 & 1 & 0 \\ r_1 & r_2 & 1 \\ r_1^2 & r_2^2 & 4 \\ r_1^3 & r_2^3 & 37 \\ \end{array}$

$ \begin{array}{cc|c} 1 & 1 & 0 \\ r_1 & r_2 & 1 \\ r_1^2 & r_2^2 & 4 \\ r_1^3 & r_2^3 & 37 \\ \end{array}$

which gave:

$b = (\frac {-1}{r_2-r_1})\\c = (\frac {1}{r_2-r_1})$

resulting in my final answer:

$a_n= b*r_1^n + c*r_2^n$

Please let me know what I have done wrong and any missed steps or errors. I have been struggling with this question for quite some time so the help would be very much welcome!!

NOTE: The solution is $a_n = (\frac{1}{10}(7^n-(-3)^n))$

Miroslav Glamuzina
  • 281

1 Answers1

1

The recurrence relation being (I suppose, since you missed the $=0$) $$a_{n+2}+ b\,a_{n+1}+ c\,a_n=0$$ you properly determined the roots $r_1$ and $r_2$ of the characteristic equation $r^2+br+c=0$.

So, the general solution is $$a_n=k_1\, r_1^n+k_2\, r_2^n$$ and now you need to use the four conditions given to find $k_1,k_2,b,c$.

Using the first and second conditions gives, just as you found, $$k_1 = \frac {-1}{r_2-r_1}=\frac{1}{\sqrt{b^2-4 c}}\qquad k_2 = \frac {1}{r_2-r_1}=\frac{-1}{\sqrt{b^2-4 c}}$$ using $$r_1=\frac{1}{2} \left(\sqrt{b^2-4 c}-b\right)\qquad r_2=\frac{1}{2} \left(-\sqrt{b^2-4 c}-b\right)$$ Now, writing $a_2$ and simplifying as much as possible gives $a_2=-b$, that is to say $b=-4$. Doing the same with $a_3$ gives $a_3=b^2-c$ that is to say $c=-21$.

Using all of the above then gives $b^2-4c=100$.

I am sure that you can take it from here.

Claude Leibovici
  • 260,315