I have a question that I got troubled with and hope for some help.
Let $p_n=(x_n,y_n)\in\mathbb{R}^2$ be a sequence of vectors. It is given that $p_1=(1,0)$ and the following recursions:$$\displaystyle x_{n+1}=\frac{3x_n-\sqrt{3}y_n}{4} \qquad ; \qquad y_{n+1}=\frac{\sqrt{3}x_n+3y_n}{4} $$ Find $\displaystyle \lim_{n\to\infty}p_n$.
I assumed that $\displaystyle \lim_{n\to\infty}x_n=L_1 \ ; \ \lim_{n\to\infty}y_n=L_2$ and solved the system of equations. My assumption was the the sequences converge, but I don't know how to show it. Also, I don't know if my way of finding the limit assuming that it exists is valid.
I will appreciate any kind of help. Thank you!
In vector form we have $$\pmatrix{x_{n+1}\cr y_{n+1}\cr} =\frac14\pmatrix{3&-\sqrt3\cr \sqrt3&3\cr}\pmatrix{x_n\cr y_v\cr}$$ which we can write as $${\bf p}_{n+1}=A{\bf p}_n\ .$$ By recursion, $${\bf p}_{n+1}=A^n{\bf p}_1\ .$$ There are various ways of evaluating the power $A^n$ - look up diagonalisation of matrices if you have not yet studied it - but in this case it is not too hard to prove, for example by induction, that $$A^n=\Bigl(\frac{\sqrt3}{2}\Bigr)^n\pmatrix{\cos(n\pi/6)&-\sin(n\pi/6)\cr \sin(n\pi/6)&\cos(n\pi/6)\cr}\ .$$ Hence $${\bf p}_{n+1}=A^n\pmatrix{1\cr0\cr} =\Bigl(\frac{\sqrt3}{2}\Bigr)^n\pmatrix{\cos(n\pi/6)\cr \sin(n\pi/6)\cr}$$ and you can now easily work out $x_{n+1}$ and $y_{n+1}$ and see what happens when $n\to\infty$.