Limit of a vector sequence

define $K = \dfrac{1}{\phi} R$ . then $\begin{bmatrix} p_{n+1} \\ v_{n+1} \end{bmatrix} = \begin{bmatrix} I && I \\ 0 && K \end{bmatrix}\begin{bmatrix} p_{n} \\ v_{n} \end{bmatrix}$ if we let $A= \begin{bmatrix} I && I \\ 0 && K \end{bmatrix}$ , then the nth term in the sequence is $\begin{bmatrix} p_{n} \\ v_{n} \end{bmatrix}=A^n \begin{bmatrix} p_{0} \\ v_{0} \end{bmatrix}$

lemma : $A^n = \begin{bmatrix} I && I+K+K^2+...+K^{n-1} \\ 0 && K^n \end{bmatrix}$ proof (by induction): the base case with n=1 is true. we have $A^{n+1}=A^n A =\begin{bmatrix} I && I+K+K^2+...+K^{n-1} \\ 0 && K^n \end{bmatrix} \begin{bmatrix} I && I \\ 0 && K \end{bmatrix} = \begin{bmatrix} I && I+K+K^2+...+K^{n-1}+K^n\\ 0 && K^{n+1} \end{bmatrix}$ hence proved.

note $\lim_{n\to \infty} K^n = 0$ . Using this we have $\begin{bmatrix} p_{\infty} \\ v_{\infty} \end{bmatrix} = \begin{bmatrix} I && I+K+K^2+... \\ 0 && 0 \end{bmatrix} \begin{bmatrix} p_{0} \\ v_{0} \end{bmatrix}\to p_\infty = p_0+( I+K+K^2+...)v_0$ notice that $R^n$ is periodic, as 4 90 degree rotation is the same as not rotating(i.e identity matrix). infact the powers of n starting at n=1 goes $R,-I,-R,I,R....$ plugging this in we have $p_\infty = p_0+\left(I+\dfrac{1}{\phi}R-\dfrac{1}{\phi^2}I-\dfrac{1}{\phi^3}R+\dfrac{1}{\phi^4}I+.....\right)v_0=p_0 +\dfrac{1}{1+\dfrac{1}{\phi^2}} \left(I+\dfrac{R}{\phi}\right) v_0$ note that geometric series was used. the rest is just plugging in, giving the vector $\boxed{\dfrac{1}{1+\dfrac{1}{\phi^2}} \begin{bmatrix} 1+\dfrac{1}{\phi} \\ 1-\dfrac{1}{\phi} \end{bmatrix}}$

Note to the author:

Generally rotation matrices are written like that:

$R(\theta)=\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$

Indeed your matrix is $R(-\dfrac{\pi}{2})$ but the angle you used is $\dfrac{\pi}{2}$ .

My solution:

Let $w_n=R^n v_0=\phi^n v_n$ .

Using that fact that $R=R(-\dfrac{\pi}{2})$ we see that:

$w_{4n}=(1,1)\\ w_{4n+1}=(1,-1)\\ w_{4n+2}=(-1,-1)\\ w_{4n+3}=(-1,1)$ .

With the relation defining $(p_n)$ we have:

$\underset{n\to +\infty}{\lim} p_n =p=(a,b)=\displaystyle\sum_{n=0}^{+\infty} \dfrac{1}{\phi^n}w_n$

Therefore:

$a=\displaystyle\sum_{n=0}^{+\infty} \dfrac{1}{\phi^{4n}} + \displaystyle\sum_{n=0}^{+\infty} \dfrac{1}{\phi^{4n+1}} -\displaystyle\sum_{n=0}^{+\infty} \dfrac{1}{\phi^{4n+2}} -\displaystyle\sum_{n=0}^{+\infty} \dfrac{1}{\phi^{4n+2}} \\ \text{and,} \\ b=\displaystyle\sum_{n=0}^{+\infty} \dfrac{1}{\phi^{4n}} -\displaystyle\sum_{n=0}^{+\infty} \dfrac{1}{\phi^{4n+1}} -\displaystyle\sum_{n=0}^{+\infty} \dfrac{1}{\phi^{4n+2}} + \displaystyle\sum_{n=0}^{+\infty} \dfrac{1}{\phi^{4n+2}}$

Using the fact that for all integers $k\neq0,\ m$ , if $|x|<1$ then

$\displaystyle\sum_{n=0}^{+\infty} x^{kn+m}=x^m\displaystyle\sum_{n=0}^{+\infty} (x^k)^n=\dfrac{x^m}{1-x^k}$

We get,

$\begin{aligned} a & =\dfrac{ 1+\phi^{-1}-\phi^{-2}-\phi^{-3}}{1-\phi^{-4}}\\ & =\dfrac{\phi^4+\phi^3-\phi^2-\phi}{\phi^4-1} \end{aligned} \\ \text{and,} \\ \begin{aligned} b & =\dfrac{ 1-\phi^{-1}-\phi^{-2}+\phi^{-3}}{1-\phi^{-4}}\\ & = \dfrac{\phi^4-\phi^3-\phi^2+\phi}{\phi^4-1} \end{aligned}$

Hence,

$a+3b=\dfrac{4\phi^4-2\phi^3-4\phi^2+2\phi}{\phi^4-1}$

Using the fact that $\phi^2=\phi+1$ and so $\phi^3=2\phi+1$ as well as $\phi^4=3\phi+2$

$\begin{aligned} a+3b & =\dfrac{12\phi+8-4\phi-2-4\phi-4+2\phi}{ 3\phi+1}\\ & =\dfrac{6\phi+2}{3\phi+1}\\ & =\boxed{2} \end{aligned}$

$R^2=-I$ $\Rightarrow (\frac{1}{\phi}R)^2 = -\frac{1}{\phi^2} I$ $\Rightarrow v_{2k}= (-\frac{1}{\phi^{2}})^kv_{0} \text{ and } v_{2k+1}= (-\frac{1}{\phi^{2}})^kv_{1}$ $\Rightarrow v_{2k} + v_{2k+1} = (-\frac{1}{\phi^{2}})^k(v_{0} + v_{1})$

$p_{2n+1}=\sum_{k=0}^n v_k + \sum_{k=0}^n v_{k+1}$

Now using the fact that for $|r|<1$ , we can write $1+r+r^2+... =\frac{1}{1-r}$ , and putting $r=-\frac{1}{\phi^2}$

$\lim_{n \to \infty} p_{2n+1} =\frac{1}{1+\frac{1}{\phi^2}}(v_0, v_1)=\frac{1}{1+\frac{1}{\phi^2}}(1+\frac{1}{\phi}, 1-\frac{1}{\phi})$

$\lim_{n \to \infty} p_{2n+1}$ $= \frac{\sqrt{5}+3}{\sqrt{5}+5} (\frac{\sqrt{5}+3}{\sqrt{5}+1}, \frac{\sqrt{5}-1}{\sqrt{5}+1})$ $=( \frac{3\sqrt{5}+7}{3\sqrt{5}+5} , \frac{\sqrt{5}+1}{3\sqrt{5}+5})$

$a+3b=\frac{6\sqrt{5}+10}{3\sqrt{5}+5}=\boxed{2}$

Similarly, prove that $\lim_{n \to \infty}p_{2n}$ converges to the same value so that the given limit exists.