1
$\begingroup$

I have a good understanding of the theorems of convergent and divergent sequence and I am able to apply them. But I really don't see how I can solve this thing. We have to use the theorem provided as well in order to solve the question.

Problem

Consider a real sequence $(a_n)$ defined through the formula $$c_2 a_{n+2} + c_1a_{n+1} + x_0a_n = 0 \tag{1}$$ with $c_2, c_1, c_0 \in \Bbb{R}$. The characteristic polynomial of the linear recurrence $(1)$ is defined to be $$p(x) = c_2 x^2 + c_1 x + c_0.$$ By means of the following theorem (do not prove it!)

Theorem If $p(x) = c_2(x - x_1)(x - x_2)$ where $x_1$ and $x_2$ are distinct nonzero complex numbers, then $(1)$ can be solved for $a_n$ and we have $$a_n = B_1x_1^n + B_2x_2^n \quad \forall n \in \Bbb{N} \quad \text{with} \quad B_1, B_2 \in \Bbb{C} \tag{2}$$

  1. find $B_1$ and $B_2$ for the sequence $(a_n)_{n\in\Bbb{N}}$ such that $$a_0 = a, \quad a_1 = b, \quad a_n = \frac{a_{n-1} + a_{n-2}}{2} \quad \forall n \ge 2$$ with $a, b \in \Bbb{R}$.

  2. Moreover, show that the above sequence converges and find its limit.

I tried to re organize the equation of $$a_n=\frac{a_{n-1} +a_{n-2}}{2}$$ to $$a_n-\frac{1}{2}a_{n-1}-\frac{1}{2}a_{n-2}=0$$ but then I didn't know how to take the choice for $x$ or how to proceed after that.

$\endgroup$
1
$\begingroup$

With your reworked equation we have that the characteristic polynomial is given by $$p(x)=x^{2}-\frac{1}{2}x-\frac{1}{2}=(x-1)(x+\frac{1}{2}).$$ By the theorem this implies there exist $B_{1},B_{2}\in\mathbb{C}$ such that $$a_{n}=B_{1}+B_{2}(-2)^{-n}.$$ In particular we have $B_{1}+B_{2}=a$ and $B_{1}-\frac{1}{2}B_{2}=b$. This gives us $B_{1}=\frac{1}{3}(a+2b)$ and $B_{2}=\frac{2}{3}(a-b)$.

Furthermore note that $$\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}B_{1}+B_{2}(-2)^{-n}=B_{1}=\frac{1}{3}(a+2b).$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ from where did you get the (-2)^-n $\endgroup$ – Joe from you Oct 16 '19 at 14:20
  • $\begingroup$ $x_{2}=-\frac{1}{2}=(-2)^{-1}$. $\endgroup$ – Floris Claassens Oct 16 '19 at 15:22

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.