Are the Metrics \(d_1\), \(d_2\), and \(d_{\infty}\) Strongly Equivalent?

  • MHB
  • Thread starter mathmari
  • Start date
  • Tags
    Equivalence
In summary, the conversation discusses the metrices $d_1, d_2, d_{\infty}$ in $\mathbb{R}^2$ and their unit balls. It is shown that these metrices are strongly equivalent, as proven through various inequalities and visualized with graphs. The conversation also addresses the graphical representation of the unit ball for $d_{\infty}$.
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! :eek:

Consider the following metrices in $\mathbb{R}^2$. For $x,y\in \mathbb{R}^2$ let \begin{align*}&d_1(x,y)=|x_1-y_1|+|x_2-y_2| \\ &d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} \\ &d_{\infty}(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\}\end{align*}

Draw the unit ball $B_i(0,1)=\{y\in X\mid d_i(0,y)<1\}$ in each metric.

Show that the metrices $d_1, d_2, d_{\infty}$ are strongly equivalent. Could you give me a hint for the first part?

As for the second part:

Without loss of generality, we assume that $\max \{|x_1-y_1|,|x_2-y_2|\}=|x_2-y_2|$.

\begin{align*}d_{\infty}(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\}=|x_2-y_2|\leq |x_1-y_1|+|x_2-y_2|=d_1(x,y)\end{align*}

\begin{align*}d_1(x,y)&=|x_1-y_1|+|x_2-y_2|\leq \max \{|x_1-y_1|,|x_2-y_2|\}+\max \{|x_1-y_1|,|x_2-y_2|\}\\ & =2\max \{|x_1-y_1|,|x_2-y_2|\}=2d_{\infty}(x,y)\end{align*}

So we get \begin{equation*}d_{\infty}(x,y)\leq d_1(x,y)\leq 2d_{\infty}(x,y)\end{equation*} Does this mean that $d_{\infty}(x,y)$ and $d_1(x,y)$ are strongly equivalent? (Wondering)

\begin{align*}d_{\infty}(x,y)&=\max \{|x_1-y_1|,|x_2-y_2|\} =|x_2-y_2|=\sqrt{|x_2-y_2|^2}=\sqrt{(x_2-y_2)^2}\\ & \leq \sqrt{(x_1-y_1)^2+(x_2-y_2)^2}=d_2(x,y)\end{align*}

\begin{align*}d_2(x,y)&=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}\leq \sqrt{(x_1-y_1)^2}+\sqrt{(x_2-y_2)^2}\\ & =|x_1-y_1|+|x_2-y_2| \leq \max \{|x_1-y_1|,|x_2-y_2|\}+\max \{|x_1-y_1|,|x_2-y_2|\}\\ & =2\max \{|x_1-y_1|,|x_2-y_2|\} = 2d_{\infty}(x,y)\end{align*}

So we get \begin{equation*}d_{\infty}(x,y)\leq d_2(x,y)\leq 2d_{\infty}(x,y)\end{equation*} Does this mean that $d_{\infty}(x,y)$ and $d_2(x,y)$ are strongly equivalent? (Wondering) Do we have to show that also for $d_1$ and $d_2$ ? Or does this follow from the above? (Wondering)
 
Physics news on Phys.org
  • #2
mathmari said:
Hey! :eek:

Consider the following metrices in $\mathbb{R}^2$. For $x,y\in \mathbb{R}^2$ let \begin{align*}&d_1(x,y)=|x_1-y_1|+|x_2-y_2| \\ &d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} \\ &d_{\infty}(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\}\end{align*}

Draw the unit ball $B_i(0,1)=\{y\in X\mid d_i(0,y)<1\}$ in each metric.

Show that the metrices $d_1, d_2, d_{\infty}$ are strongly equivalent. Could you give me a hint for the first part?

We have for instance $d_1(0,y) = |0-y_1|+|0-y_2| = |y_1|+|y_2|$ don't we? (Wondering)

So $B_i(0,1)=\{y\in X\mid d_i(0,y)<1\} = \{y\in X\mid |y_1|+|y_2|<1\} = \{(x,y)\in \mathbb R^2\mid |x|+|y|<1\}$.

We can draw it like:
[DESMOS]advanced: {"version":7,"graph":{"squareAxes":false,"viewport":{"xmin":-1.2,"ymin":-1.2,"xmax":1.2,"ymax":1.2}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"\\left|x\\right|+\\left|y\\right|<1"}]}}[/DESMOS]
mathmari said:
As for the second part:

Without loss of generality, we assume that $\max \{|x_1-y_1|,|x_2-y_2|\}=|x_2-y_2|$.

\begin{align*}d_{\infty}(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\}=|x_2-y_2|\leq |x_1-y_1|+|x_2-y_2|=d_1(x,y)\end{align*}

\begin{align*}d_1(x,y)&=|x_1-y_1|+|x_2-y_2|\leq \max \{|x_1-y_1|,|x_2-y_2|\}+\max \{|x_1-y_1|,|x_2-y_2|\}\\ & =2\max \{|x_1-y_1|,|x_2-y_2|\}=2d_{\infty}(x,y)\end{align*}

So we get \begin{equation*}d_{\infty}(x,y)\leq d_1(x,y)\leq 2d_{\infty}(x,y)\end{equation*} Does this mean that $d_{\infty}(x,y)$ and $d_1(x,y)$ are strongly equivalent? (Wondering)

\begin{align*}d_{\infty}(x,y)&=\max \{|x_1-y_1|,|x_2-y_2|\} =|x_2-y_2|=\sqrt{|x_2-y_2|^2}=\sqrt{(x_2-y_2)^2}\\ & \leq \sqrt{(x_1-y_1)^2+(x_2-y_2)^2}=d_2(x,y)\end{align*}

\begin{align*}d_2(x,y)&=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}\leq \sqrt{(x_1-y_1)^2}+\sqrt{(x_2-y_2)^2}\\ & =|x_1-y_1|+|x_2-y_2| \leq \max \{|x_1-y_1|,|x_2-y_2|\}+\max \{|x_1-y_1|,|x_2-y_2|\}\\ & =2\max \{|x_1-y_1|,|x_2-y_2|\} = 2d_{\infty}(x,y)\end{align*}

So we get \begin{equation*}d_{\infty}(x,y)\leq d_2(x,y)\leq 2d_{\infty}(x,y)\end{equation*} Does this mean that $d_{\infty}(x,y)$ and $d_2(x,y)$ are strongly equivalent?

Yep. (Nod)

mathmari said:
Do we have to show that also for $d_1$ and $d_2$ ? Or does this follow from the above?

Strong equivalence of metrics is an equivalence relationship, which implies indeed that $d_1$ and $d_2$ are strongly equivalent by the transitivity property. (Nod)

Alternatively we can show it using the inequalities you found above:
$$\frac 12 d_1(x,y) \le d_\infty(x,y) \le d_2(x,y) \le 2 d_\infty(x,y) \le 2 d_1(x,y)$$
So:
$$\frac 12 d_1(x,y) \le d_2(x,y) \le 2 d_1(x,y)$$
Therefore $d_1$ and $d_2$ are strongly equivalent. (Nerd)

We can also illustrate it graphically:
[DESMOS]advanced: {"version":7,"graph":{"squareAxes":false,"viewport":{"xmin":-1.6,"ymin":-1.6,"xmax":1.6,"ymax":1.6}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"\\left|x\\right|+\\left|y\\right|<1"},{"type":"expression","id":"2","color":"#c74440","latex":"\\sqrt{x^2+y^2}<1"},{"type":"expression","id":"3","color":"#2d70b3","latex":"\\left|x\\right|+\\left|y\\right|<\\sqrt{2}"}]}}[/DESMOS]
(Cool)
 
  • #3
Klaas van Aarsen said:
We have for instance $d_1(0,y) = |0-y_1|+|0-y_2| = |y_1|+|y_2|$ don't we? (Wondering)

So $B_i(0,1)=\{y\in X\mid d_i(0,y)<1\} = \{y\in X\mid |y_1|+|y_2|<1\} = \{(x,y)\in \mathbb R^2\mid |x|+|y|<1\}$.

How do we draw the last one?

We have $d_{\infty}(0,y) = \max\{|0-y_1|, |0-y_2|\} = \max\{|y_1|, |y_2|\}$. So we get $B_{\infty}(0,1)=\{y\in X\mid d_{\infty}(0,y)<1\} = \{y\in X\mid \max\{|y_1|, |y_2|\}<1\} = \{(x,y)\in \mathbb R^2\mid \max\{|y_1|, |y_2|\}<1\}$.

But which is the graph of that? (Wondering)
 
  • #4
mathmari said:
How do we draw the last one?

We have $d_{\infty}(0,y) = \max\{|0-y_1|, |0-y_2|\} = \max\{|y_1|, |y_2|\}$. So we get $B_{\infty}(0,1)=\{y\in X\mid d_{\infty}(0,y)<1\} = \{y\in X\mid \max\{|y_1|, |y_2|\}<1\} = \{(x,y)\in \mathbb R^2\mid \max\{|y_1|, |y_2|\}<1\}$.

But which is the graph of that?

[DESMOS]advanced: {"version":7,"graph":{"squareAxes":false,"viewport":{"xmin":-1.2,"ymin":-1.2,"xmax":1.2,"ymax":1.2}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"\\max\\left(\\left|x\\right|,\\left|y\\right|\\right)<1"}]}}[/DESMOS]
(Emo)
 
  • #5
Ahh yes (Tmi)

Thank you so much! (Blush)
 

Related to Are the Metrics \(d_1\), \(d_2\), and \(d_{\infty}\) Strongly Equivalent?

1. What is the concept of strong equivalence of metrices?

The concept of strong equivalence of metrices refers to the idea that two different metrices can produce the same results or conclusions when applied to a set of data. This means that the two metrices are interchangeable and can be used interchangeably in a given situation.

2. How is strong equivalence of metrices different from weak equivalence of metrices?

The main difference between strong and weak equivalence of metrices is the level of agreement between the two metrices. Strong equivalence means that the two metrices produce exactly the same results, while weak equivalence means that there is some degree of variation or difference in the results produced by the two metrices.

3. What are some examples of strong equivalence of metrices?

One example of strong equivalence of metrices is when two different methods of measuring temperature, such as using a thermometer and a thermal camera, produce the same temperature readings. Another example is when two different questionnaires designed to measure the same construct yield similar results.

4. How is strong equivalence of metrices determined?

Strong equivalence of metrices is determined by comparing the results produced by the two metrices on the same set of data. If the results are identical, then the metrices are considered to be strongly equivalent. This can also be confirmed through statistical analysis and testing.

5. Why is strong equivalence of metrices important in scientific research?

Strong equivalence of metrices is important in scientific research because it allows for the validation and replication of results. If two different metrices produce the same results, it increases the confidence in the findings and supports the generalizability of the results. It also allows for the comparison of results from different studies that use different metrices, making it easier to build upon existing research.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
71
Replies
3
Views
964
Replies
12
Views
2K
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
621
  • Linear and Abstract Algebra
Replies
34
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
331
  • Linear and Abstract Algebra
Replies
4
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
Replies
4
Views
307
Back
Top