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How do I find explicit formula for $(x_n,y_n)$ and show that the sequence tends to converge to $(0,0)$?

In $\mathbb{R}^2$, the sequence $(x_n,y_n)$, $n\in \mathbb{N}_0$ is recursively defined: $\begin{pmatrix}x_{n+1}\\ y_{n+1}\end{pmatrix}=\left (\begin{matrix}0 & 1\\ 1/4 &0 \end{matrix} \right )\begin{pmatrix}x_n\\ y_n\end{pmatrix}$.

So for $(x_0,y_0)=(4,2)$. I have found some follow members:$(x_1,y_1)=(2,1),(x_2,y_2)=(1,1/2),(x_3,y_3)=(1/2,1/4),(x_4,y_4)=(1/4,1/8)...$ At this point I see that $x_{n+1}=x_n/2$. But how do I write it correctly? How to show that it converges in $(0,0)$?

Laja
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2 Answers2

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We have $x_1=4$ and $x_{n+1}=x_n \div 2$. This means to get to the next term, we divide by $2$. Therefore, if we want to go three terms forward, we divide by $2$ three times, or divide by $2^3=8$. If we want to go five terms forward, we divide by $2$ five times, or divide by $2^5=32$. We have $x_0$ and want to go to $x_n$, so we want to go $n$ terms forward. This means we need to divide by $2$ for $n$ times, or divide by $2^n$, so we have: $$x_n=\frac{x_0}{2^n}=\frac{4}{2^n}$$

Similarly, with $y_n$, we have: $$y_n=\frac{y_0}{2^n}=\frac{2}{2^n}$$

Hopefully, you can see how both of these sequences converge to $0$ as $n \to \infty$ (hint: look at denominator), meaning we have: $$\lim_{n \to \infty} (x_n, y_n)=(0, 0)$$

Noble Mushtak
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$ \begin{pmatrix}x_{n+1}\\ y_{n+1}\end{pmatrix} = \left (\begin{matrix}0 & 1\\ 1/4 &0 \end{matrix} \right )\begin{pmatrix}x_n\\ y_n\end{pmatrix} = \left (\begin{matrix}0 & 1\\ 1/4 &0 \end{matrix} \right )\left (\begin{matrix}0 & 1\\ 1/4 &0 \end{matrix} \right )\begin{pmatrix}x_{n-1}\\ y_{n-1}\end{pmatrix} = \dots = \left (\begin{matrix}0 & 1\\ 1/4 &0 \end{matrix} \right )^{n+1}\begin{pmatrix}x_0\\ y_0\end{pmatrix} $

and you need to power the matrix. Compute few powers ($n=1,2,3,4$), you'll notice that for even exponent, it is a multiplicity of $\left (\begin{matrix}1&0\\ 0&1 \end{matrix}\right)$.

user340508
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