0
$\begingroup$

The question is:

Determine the numbers $a_n$ for $n \geq 0$ that satisfy the recurrence relation $a_{n+2} = 4a_{n+1} - 2a_n$ for $n\geq 0$. with Boundary conditions $a_0 = 0$ and $a_1 = 1$.

Now my base step would be to fill in the formula : $$ \begin{align} a_0 :a_2&=4a_1-2a_0 \\ &=4*1 -2*0 =4 \end{align} $$ So we assume it holds for $n = 0$, and thus for $k$. fill in $k+1$: $$ \begin{align} a_{k+3}&=4a_{k+2}-2a_{k+1}\\ \end{align} $$

Which doesnt really help me at all. How do I tackle this? AM I looking at this the wrong way?

$\endgroup$
  • $\begingroup$ yeah it was a typo $\endgroup$ – WiseStrawberry Jun 3 '13 at 15:32
  • $\begingroup$ Do you know about using the roots of the characteristic equation $x^2-4x+2=0$? $\endgroup$ – André Nicolas Jun 3 '13 at 15:32
  • $\begingroup$ yeah. BUt how do I apply that to this? So I guess not. $\endgroup$ – WiseStrawberry Jun 3 '13 at 15:32
  • $\begingroup$ @WiseStrawberry - look up another example like this one and try to emulate what was done there. $\endgroup$ – Hans Engler Jun 3 '13 at 15:35
  • $\begingroup$ You are doing something that looks like induction, which is clearly not what you need here. Take a look at math.stackexchange.com/questions/91242/… to see how these kind of problems should be solved (especially Christian Blatter's answer should help you). $\endgroup$ – dreamer Jun 3 '13 at 15:42
2
$\begingroup$

Since the roots are of the form $2\pm\sqrt{2}$, $a_n=A(2+\sqrt{2})^n+B(2-\sqrt{2})^n$

To see why this works, notice that

$ \left( \begin{array} [cc] a_{n+2} && a_{n+1} \\ a_{n+1} && a_{n} \end{array} \right)=\left( \begin{array} [cc] \\ 4 && -2 \\ 1 && 0 \end{array} \right)\left( \begin{array} [cc] a_{n+1} && a_{n} \\ a_{n} && a_{n-1} \end{array} \right)$ and diagonalise $\left( \begin{array} [cc] \\ 4 && -2 \\ 1 && 0 \end{array} \right)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.