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Consider the sequence in $\mathbb{R}^{2}$ given by
$$
x_{n}=\left(\frac{1}{n} \cos \left(\frac{\pi n}{2}\right), \frac{1}{n} \sin \left(\frac{\pi n}{2}\right)\right)
$$
Recall that we can equip $\mathbb{R}^{2}$ with several metrics, including the following:
The $L^{2}$ metric, given by
$$
d_{2}\left(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right)=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}
$$
The $L^{1}$ metric, given by
$$
d_{1}\left(\left(x_{1}, y_{1}\right)\right)=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right| .
$$
The $L^{\infty}$ metric, given by $d_{\infty}\left(\left(x_{1}, y_{1}\right)\right)=\max \left\{\left|x_{1}-x_{2}\right|,\left|y_{1}-y_{2}\right|\right\}$
The discrete metric, given by
$$
d_{d}\left(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right)=\left\{\begin{array}{ll}
1 & \text { if }\left(x_{1}, y_{1}\right) \neq\left(x_{2}, y_{2}\right) \\
0 & \text { if }\left(x_{1}, y_{1}\right)=\left(x_{2}, y_{2}\right)
\end{array}\right.
$$

(a) Compute the $L^{2}$ distance between $x_{n}$ and $(0,0)$. That is, compute $d_{2}\left(x_{n},(0,0)\right)$. Your answer should be written in terms of $n$.
(b) Use (a) to show that $x_{n} \rightarrow(0,0)$ with the $L^{2}$ metric.
(c) Compute the $L^{1}$ distance between $x_{n}$ and $(0,0)$. That is, compute $d_{1}\left(x_{n},(0,0)\right)$. Your answer should be written in terms of $n$.
(d) Use (c) to show that $x_{n} \rightarrow(0,0)$ with the $L^{1}$ metric.
(e) Compute the $L^{\infty}$ distance between $x_{n}$ and $(0,0)$. That is, compute $d_{\infty}\left(x_{n},(0,0)\right)$. Your answer should be written in terms of $n$.
(f) Use (d) to show that $x_{n} \rightarrow(0,0)$ in the $L^{\infty}$ metric.
(g) Show that $x_{n}$ does not converge to $(0,0)$ in the discrete
in Mathematics by Bronze Status (8,688 points) | 12 views

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(a) We compute the $L^{2}$ distance as
$$
d_{2}\left(x_{n},(0,0)\right)=\sqrt{\left(\frac{1}{n} \cos \left(\frac{n \pi}{2}\right)\right)+\left(\frac{1}{n} \sin \left(\frac{n \pi}{2}\right)\right)^{2}}=\frac{1}{n}
$$
(b) Since the distance is $1 / n$ and $1 / n \rightarrow 0$ in $\mathbb{R}$, then $x_{n} \rightarrow(0,0)$.
(c) We compute the $L^{1}$ distance as
$$
\begin{array}{l}
d_{1}\left(x_{n},(0,0)\right)=\left|\frac{1}{n} \cos \left(\frac{n \pi}{2}\right)-0\right|+\left|\frac{1}{n} \sin \left(\frac{n \pi}{2}\right)-0\right|= \\
\left|\frac{1}{n} \cos \left(\frac{n \pi}{2}\right)\right|+\left|\frac{1}{n} \sin \left(\frac{n \pi}{2}\right)\right|= \\
\left|\frac{1}{n}\right|\left(\left|\cos \left(\frac{n \pi}{2}\right)\right|+\left|\sin \left(\frac{n \pi}{2}\right)\right|\right)
\end{array}
$$
Notice that $\cos (n \pi / 2)$ and $\sin (n \pi / 2)$ take on the values $-1,0$, or $1 .$ In particular, when one is 0, the other is $\pm 1$ and vice versa. Thus, in absolute values, exactly one is 0 and one is $1 .$ So, our distance reduces to
$$
d_{1}\left(x_{n},(0,0)\right)=1 / n .
$$
(d) From (c), $d_{1}\left(x_{n},(0,0)\right)=1 / n \rightarrow 0 .$ Thus, $x_{n} \rightarrow(0,0)$ in the $L^{1}$ metric.

(e) We compute the $L^{\infty}$ distance as
$$
\begin{array}{c}
d_{\infty}\left(x_{n},(0,0)\right)=\max \left\{\left|\frac{1}{n} \cos \left(\frac{n \pi}{2}\right)-0\right|,\left|\frac{1}{n} \sin \left(\frac{n \pi}{2}\right)-0\right|\right\}= \\
\max \left\{\left|\frac{1}{n} \cos \left(\frac{n \pi}{2}\right)\right|,\left|\frac{1}{n} \sin \left(\frac{n \pi}{2}\right)\right|\right\}= \\
\left|\frac{1}{n}\right| \max \left\{\left|\cos \left(\frac{n \pi}{2}\right)\right|,\left|\sin \left(\frac{n \pi}{2}\right)\right|\right\}
\end{array}
$$
As mentioned above, the two terms are either 0 or 1 at opposite times. Thus, the maximum is equal to
1. So, the distance is given by
$$
d_{\infty}\left(x_{n},(0,0)\right)=\frac{1}{n} \text { . }
$$

(f) From (e), we have that $d_{\infty}\left(x_{n},(0,0)\right)=\frac{1}{n} \rightarrow 0 .$ Thus, $x_{n} \rightarrow(0,0)$.
(g) Notice that for all $n,|\cos (n \pi / 2)| \neq|\sin (n \pi / 2)| .$ Thus, in particular, they are never both 0 at the same time. So, they are never equal to $(0,0) .$ So, $d_{d}\left(x_{n},(0,0)\right)=1$ for all $n$. This does not converge to 0, so $x_{n} \not A(0,0)$ (or any other point).
by Bronze Status (8,688 points)

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