|x+y| ≦ |x| + |y|
The first 2 cases are easy since they are clear cut.
Case 1: x ≧ 0 and y ≧ 0
|x+y| = x+y, |x| = x, |y| = y and we can finish the rest of the proof
Case 2: x < 0 and y < 0
|x+y| = -(x+y), |x| = -x, |y| = -y and we can finish the rest of the proof
The last 2 cases are not that easy since they are not clear cut
Case 3: x ≧ 0 and y < 0
This is not so clear cut because x+y could be pos or neg
Case 4: x < 0 and y ≧ 0
This is not so clear cut either because x+y could be pos or neg
Or perhaps I just have to live with having 6 cases and prove them separately?
I’m trying to show the triangle inequality of the following distance measure but I am stuck. Can anyone help please?
Let x and y are two n-vector of proportions ( i.e. the sum of elements of x (and y) is one). The distance between them is defined as:
d(x , y) = [ sum_j [ ln (x_j / g(x) ) - ln (y_j / g(y) ) ]^2 ]^1/2,
where g(.) is geometric mean. My try is as follows:
Proof: Let z be any n-vector of proportion. then by definition of d we have
d(x , z) = [ sum_j [ ln (x_j / g(x) ) - ln (z_j / g(z) ) ]^2 ]^1/2
= [ sum_j [ ln (x_j / g(x) ) - ln (y_j / g(y) ) + ln (y_j / g(y) )- ln( z_j / g(z) )} ]^2 ]^1/2
= [ sum_j [ {ln (x_j / g(x) ) - ln (y_j / g(y) ) } + { ln (y_j / g(y) )- ln( z_j / g(z) )} ]^2 } ]^1/2
After opening square and applying summation
d(x, z) = [ d^2 (x, y) + d^2 (y, z) + 2 sum_j (a_j b_j) ]^1/2 , (Eq. 1)
where
a_j = { d (x_j / g(x) ) - ln (y_j / g(y) ) }
and
b_j = { ln (y_j / g(y) )- ln( z_j / g(z) ) }.
How could I conclude from Eq. 1 that
d(x, z) > = d (x, y) + d ( y, z) ??
Any help will be appreciated.
Let x and y be two vectors of proportions and m is the mean of x and y. If the m is projected onto both x and y, the the projection of m onto x is p_x(m) and the projection of m onto y is p_y(m). Thus, the distance between x and y is defined as: d(x, y) = [ SqE(p_x(m) - m) + SqE(p_y(m)-m) ]^(1/2),
where SqE is the squared Euclidean distance between the projections and m.
If there is any other point z, then how could I prove that
d(x, z) is less than or equal to [d(x, y) + d(y, z )]?
I have tried to use Chauchy-Schwartz inequality but it did not work.
Any idea or help please? Thanks
Let x1 and x2 be two m dimensional vector on the standard simplex.i.e. both x1 and x2 are vector of proportions. The distance between them is defined as
d(x1, x2) =
[sum_{j= 1,...,m} [(x1_j / a1)- p_j ]^2 ] ^ (1/2) + [sum_{j= 1,...,m} [(x2_j / a2)-p_j]^2 ] ^ (1/2).
where p_j = (x1_j + x2_j)/2, a1 and a2 are the values that minimised d(x1,x2). Show that
[ d(x1, x2) + d(x2, x3) ] is greater than or equal to d(x1, x3).
I have shown numerically that inequality holds but having problems to prove it mathematically. I have tried Cauchy-Schwarz inequality, but thy didn’t work due to the complex form of the distance and included optimal values.
Would anyone mind helping me with the prove of the triangular inequality for the above d? Any help would be highly appreciated. Thank you
I am given the problem: Prove that the function on [;X \times X;] given by [;d(x,y)=[d_{1}(x_{1},y_{1})^{2}+\cdot{}\cdot{}\cdot{}+d_{n}(x_{n},y_{n})^{2}]^{1/2};] is a metric. My attempt goes like this:
(1) [;d(x,y) \geq 0;] because it is the sum of metrics, which are definite positive
(2) [;d(x,y) = [d_{1}(x_{1},y_{1})^{2} + d_{2}(x_{2},y_{2})^{2}]^{1/2} = [d_{2}(x_{1},y_{1})^{2} + d_{1}(x_{2},y_{2})^{2}]^{1/2}=[d_{2}(y_{1},x_{1})^{2} + d_{1}(y_{2},x_{2})^{2}]^{1/2} = d(y,x);]
(3) [;d(x,y)=0 \implies d_{1}(x_{1},y_{1})^{2}=0 \textrm{ and } d_{2}(x_{2},y_{2})^{2}=0;] which only holds for [;x=(x_{1},x_{2})=(y_{1},y_{2})=y;]
(4) Not sure where to start. I've seen proofs for [;\mathbb{R}^{2};] using the Cauchy Inequality, but they have been hard for me to follow and I'm not sure if they are applicable to the more general case of [;X \times X;]. How would you go about proving functions defined over [;X \times X;] follow the triangle inequality?
I have a graph G=(V,E) which doesn't satisfy triangle inequality, I want to solve the TSP question. Then I have an algorithm to solve this question that needs to satisfy triangle inequality.
How to convert a graph that doesn't satisfy triangle inequality to one that satisfies triangle inequality?
Who can give me some suggestions, Thanks!
.
Hello everyone, I have to say that math learning is not easy at all . But math is really magic to all of us. I'm learning something about triangle now and I think this theorem is the basic knowledge you should know. This video is easy to understand and I hope everyone can learn something from this video. Have a nice day to you !
Hi all,
In my graduate math course, we've recently been introduced to metric spaces. One of the things we're doing is proving that something constitutes a distance. While proving that d(x,y) = 0 iff x =y, d(x,y)=d(y,x) and d(x,y) > 0.
However, one thing that I'm struggling with is proving that the triangle inequality holds. While stating the triangle inequality is fine, I usually don't know how to proceed to actually "prove" it.
Whenever I look at a proof of it it appears to be more like a "matter of fact" statement rather than a proof. Has anyone else experienced this or does anyone have links/resources to get a better understanding of this?
Thanks!
So I have this question that I have been stuck on for like an hour. I have no clue where and how the triangle inequality comes into play. I would really appreciate a hint or direction. I can see that it is true by drawing but other than that I am at a loss.
We were given a hint that |a-b|=|a-x-(b-x)| which is all well and good, but what the hell is x and how does this hint help me at all.
|a-b|=|a-x-(b-x)|<=|a-x|+|b-x|?
https://imgur.com/gallery/fWop6T8
Thanks in advance!
Why is it that you can take the triangle inequality directly on an infinite sum, but first must do it on the finite sum and then take the limit to infinity?
I’m currently brushing up on my math skills for a entrance type of exam. I took a practice test and the question was,
“Which of the following could be the perimeter of a triangle with two sides that measure 13 and 5?”
I figured using Pythagorean’s theory would be the easiest way to solve it but it told me the answer was “26.5” and that I should use triangle inequality theory instead. Can someone explain if the question is right or wrong and why?
Given nondegenerate triangle ABC with sides a = BC, b = AC, c = AB, we must show that a+b > c, a+c > b, and b+c > a. WLOG, let a >= b >= c. Then it is obvious that a+b > c and a+c > b since all lengths are positive. Therefore, we must only show that b+c > a.
Now we utilize the fact that shortest distance between two points in a 2-D plane is a straight line. The shortest path from point B to point C is via line BC which corresponds to side a. Any other path must be longer. Therefore, b+c must strictly be greater than a because it represents a path that is not a line from point B to point C.
Would this be a complete proof? I recently took high school geometry and a proof was never presented in class.
Any feedback is greatly appreciated :)
America, because it's not a metric space.
Let x, y, and z are m dimensional vectors of proportions such that z is the mid point of x and y. If the distance between x and y is defined as
d(x, y) = [ sum_j [ (x_j / a1) - z_j ]^2 ]^(1/2) +
[ sum_j [ (y_j / a2) - z_j ]^2 ]^(1/2),
where
a1 = sum_j (x_j)^2 / sum_j (x_j z_j) and
a2 = sum_j (y_j)^2 / sum_j (y_j z_j),
then show that [d(x, z) + d(z, y)] is greater than d(x,y) when x and y are not equal.
I have tried this by using Chauchy-Schwartz inequality but I am having problem. Could anyone please help me? Thanks
Hi! I’m not sure whether it’s an appropriate subreddit for this question or not. If not then apology. So this question is:
Let x1 and x2 be two m dimensional vector of proportions. The distance between them is defined as
d(x1, x2) =
sum_{i = 1,2} sum_{j= 1,...,m} [ [ ( x_ij / a_i ) - pi_j ] ^ 2 ] ^ (1/2),
where a_i and p_j are the values that minimised d(x1, x2) subject to the constraint ((sum_{j=1,...,m}p_j)=1). Show that the triangular inequality holds for d.
I have shown numerically that inequality holds but having problems to prove it mathematically. Would anyone mind helping me with the prove of the triangular inequality for the above d? Any help would be highly appreciated. Thank you
Let x1 and x2 be two m dimensional vector of proportions. The distance between them is defined as
d(x1, x2) =
[sum_{i = 1,2} sum_{j= 1,...,m} [ [ ( x_ij / a_i ) - pi_j ] ^ 2 ] ^ (1/2),
where a_i and p_j are the values that minimised d(x1, x2) subject to the constraint ((sum_{j=1,...,m}p_j)=1). Show that the triangular inequality holds for d.
I have shown numerically that inequality holds but having problems to prove it mathematically. Would anyone mind helping me with the prove of the triangular inequality for the above d? Any help would be highly appreciated. Thank you
Hello everyone, I have to say that math learning is not easy at all . But math is really magic to all of us. I'm learning something about triangle now and I think this theorem is the basic knowledge you should know. This video is easy to understand and I hope everyone can learn something from this video. Have a nice day to you !
