Some other cool examples of metrics are in biology to measure the evolutionary distance between organisms and in linguistics to measure the 'distance' between different words
@@DrTrefor Dunno much about it, but they use an ultrametric with a stronger version of the triangle inequality where every three points forms an isosceles triangle. It's useful because imagine trying to find the least common ancestors of three organisms A, B and C, then you want the 'distance' (i.e. how many years ago their LCA lived) between any pair to be related in a special way. E.g. if A and B's LCA lived 7 mya, and B and C's LCA lived 10 mya, then you don't want the LCA of A and C to be greater than 10 mya.
He helped me solve my homework and then blew my mind at the end of the video for good measure! Maths is so wierd and wonderful. I'm really starting to love it in its own right, rather then just a tool to help me understand physics
As someone who's studying Topology in university right now, this is a nice refresher on metrics. Additionally, with just having finished a chapter on approximation theory in my numerics 2 class, I'm going to add an interesting note: If you look at the chebyshev and the taxi-cab metric, it is pretty easy to find lines on which the best approximation of the origin is not unique. Only with the euclidean distance, such an approximation is always unique, because this metric is defined by a scalar product. This is also generally true, however there are cases where you don't need a metric produced by a scalar product to have unique best approximations to some point in space.
1:03 not really related to the topic, but ironically, the king actually is the weakest piece because it can't capture guarded pieces nor move to squares that are also guarded by other pieces, making it have less attacking and defending potential than even a pawn.
Woaa... It's the first time I realise that Chebyshev metric is actually the King's metric of chess board. What a great explanation professor. You deliver your content as great as always.
Another interesting distance metric is: d(a,b) = max{ | b1 - a1 |, | b2 - a2 |, | (b1 - a1) - (b2 - a2) | } Which gives the distance on a hexagonal grid (with no grid side between the two axes. For a hexagonal grid with a grid side between the two axes, swap the minus for a plus between the two terms in parentheses for the last element.)
Very cool! There are also metric tensors for smooth manifolds where you have something like ds^2 = dθ^2 + sin^2(θ) dφ^2, and to find the distance along a path between two coordinates (θ_1, φ_1), (θ_2, φ_2), you have to integrate ds along that path.
@Scuffed Physics I'm not 100 percent sure about this but a manifold can be a metric space only when it has constant curvature (like a sphere) otherwise the metric tensor varies from point to point then it wont be a metric space since a metric space has a fixed metric
@Scuffed Physics yes but technically that wouldn't be a metric space since the metric itself changes depending on which two points on the manifold you're talking about. a metric space has to have a fixed metric right? lol am i getting the definition of a metric space wrong? also btw a geodesic is by definition the shortest distance between two points on a manifold so "shortest geodesic" is kind of a tautology
@@mastershooter64 You are misunderstanding what a metric tensor is, probably on account of the fact that the terminology is just confusing you. A metric tensor is not a metric. A metric tensor is an inner product. An inner product is a sesquilinear, conjugate-symmetric map that is positive definite. The dot product is an example of an inner product, and the most elementary example. Now, assuming you have an inner product , you can define a norm by saying ||v|| = sqrt(). This turns the manifold into a normed manifold, where the norm is induced by the metric tensor. The metric, in the sense of a metric space, is then defined by d(u, v) = ||u - v||, hence the norm induces a metric on the manifold, and thus the manifold is a metric space. But not every manifold that is a metric space has an inner product. That is where you are getting confused. A manifold with an inner product is called a Riemannian manifold. So the a manifold with a metric tensor is a Riemannian manifold. Such manifolds necessarily do form metric spaces, but the metric tensor is the inner product, not the metric induced by that inner product. I know this is confusing, because a metric, and a metric tensor, are different things, even though the word "metric" features in the name for both mathematical objects. You may think this is poor choice of nomenclature on the mathematician's part. However, the name "metric tensor" makes perfect sense in the context of tensor analysis.
@@angelmendez-rivera351 yes i know a metric tensor is a inner product, and you integrate to get distances on the manifold, no it's not confusing it's pretty intuitive. also it's technically a tensor field over the entire manifold, each point on the manifold has a specific metric tensor on it. lol how can you assume "this is where you are getting confused" the only thing i wasnt sure about was that a metric space can have a metric that depends on where you are on that space, and apparently you can! which is pretty cool
One very important example of metric spaces that are not Euclidean are the L^p spaces in functional analysis. The simplest example is the L^♾ space. We consider functions f : S -> C, where C is the field of complex numbers, and S is a Borel set of C. The supremum norm of f, denoted ||f||_♾, is defined as sup({|f(z)| : z in S}), where sup here denotes the supremum. The space of functions C^S, together with this norm, forms a normed space. The metric d is defined as d(f, g) = ||f - g||_♾. This gives us a way of measuring distances between functions, and the reason this distance notion is so important is because it underlies the notion of uniform convergence. In other words, uniform convergence is just convergence of a sequence, or more generally, a net of functions, with respect to the supremum norm.
Wow! I had no idea that the distances in Chess had their own metric. The king chasing down the pawn example you gave is colloquially known as "the square of the pawn" and it states that the king can catch the pawn if it can step into the "square of the pawn" (a picture of this can explain it more clearly) on its next move. I'm now curious about the knight's movements...
Edit: turns out the knight does make a metric. disregard everything I said The knight’s movements satisfy only the first and second conditions of being a metric, symmetry and “0 distance same point”. It doesn’t satisfy the triangle inequality, as a knight on a1 could move to b3 in 1 move, but to pass through a2, it needs first 3 moves then 2 moves, which don’t add up to 1. It’s a real shame, the knight metric would be so interesting. Is there some way to modify the knight’s movements to make a metric?
@@-minushyphen1two379 any distance function based on counting number of steps should satisfy the triangle inequality, no matter what chess piece is used. A proof by contradiction: Define d(A,B) as the minimum number of steps it would take our chess piece to get from A to B. Assume that the triangle inequality doesn't hold, i.e. we can find some A, B, and C s.t. d(A,C) > d(A,B) + d(B,C). This means that there is a path P from A to C passing through B that is shorter than the shortest path from A to C. But P *is* a path from A to C; it can't be shorter than itself (a contradiction). Our assumption must be false: the function d must satisfy the triangle inequality. It's not hard to see that d satisfies the first condition, and it should satisfy the second as long as every move the chess piece makes is reversible. All this is assuming that the chess piece can get to every square on the chessboard. If it can't get from A to B, then d(A,B) is undefined. So a bishop won't produce a valid distance function.
It should also be possible for some chess pieces to create corresponding continuous versions of their metrics on the plane rather than the chessboard, by overlaying an infinite chessboard on the plane whose squares have side length k and then shrinking k, so that the discrete chessboard approximates the continuous plane. The new continuous metric is the limit as k→0 of k*d(A,B). The knight seems to produce a valid continuous metric, with its open ball being an octagon whose horizontal and vertical sides are longer than its diagonal sides. I suspect that the chess pieces that produce valid continuous metrics are the ones that satisfy these two conditions: 1) they produce a valid discrete metric 2) they can't move arbitrarily far in a single step
@@calvincrady well for example a continuous rook metric is well defined as d(a, b) = [a_x ≠ b_x] + [a_y ≠ b_y], which satisfies all rules, it is just not super interesting
In data science there is a machine learning algorithm called Lasso Regression (which prevents overfitting). Anyway it is also called L1 regularization and is depicted visually as a diamond, just like your Taxicab visualization.
6:03 lols 1 way roads where a road A to B curves and road B to A is straight is non metric distance by breaking rule 2 and rule 3. lols if point A that was at X1 and Y3 then moves to point X5 and Y5 and point B is at X5 and Y5 then A and B the distance between them is 0 and by rule 1 they are the same point then if point A moves to X2 and Y4, A and B have not become not the same point and so the distance between them is still 0, a system of dynamic points is non metric, even if you don't want to use time you can make Z location substitute for "then".
I really hope the "intro" part suggests there is more to come :) Don't even care much about the direction, more general? Sure topological spaces are really cool! More in depth? I mean you said you are a calc prof, so that would fit! More specialised with smooth manifolds? If you have the gut to deal with the horrible index mess that differential geometry tends you'd probably make a lot of physicists happy! For real though, even though this video doesn't really discuss anything an analysis 2 (or 1?) lecture wouldn't cover, you animations and presentation style still have me glued to the screen :) Also very nice touch with the motivation on building theories on generalised concepts, from theoretical physics to constructivist, finitist or ZFC-Powerset mathematics, trying to explain why we try to work with the more general version can be hard to do well.
For me I find it is much easier to think about the "different distances" as moves/step between states instead of distance between two points. imagine some state A (for example chessboard arrangement) from which you want to go to state B. Then the metric is just the sum of all steps you need to do in order to get to state B. You can aply this logic easily to various "types of distances". * note that specifuclly the chestboard example could violate the rule d(A, B) = d(B, A) because pawns don't move backwards, but you get the idea.
I think we will need the notion of measure for area instead of metric. Measure generalises 1D Euclidean distance, 2D area in 2D Euclidean space, 3D volume in 3D Euclidean space, all the way to nD hypervolume for nD Euclidean space. Even measure is still compatible for other concepts, such as counting measure that agrees with the concept of cardinality of countable sets.
That is not correct. While distances in Euclidean space are thought of in terms of length, it is more accurate to say that length, in a Euclidean space, is defined in terms of distance. However, this is exclusive to Euclidean space. In actuality, talking in general mathematical terms, distance and length are completely different, unrelated concepts. For length, you want some kind of measure space with topological dimension 1, and use the Lebesgue measure. Metric spaces are a completely unrelated thing. You need not have measurable sets within a metric space, and similarly, you measure space may not have a metric.
@@angelmendez-rivera351 I get your point. But when we talk about area in Euclidean space we should talk about measure instead of metric. That's what I want to emphasise. Let me correct my statement. For R^1, we have to use the notion of length instead of distance if we want it to be correlated to area in R^2, and hence we need the notion of measure instead of metric for the case.
@@rizalpurnawan23 Yes, I agree with you on that point. In fact, that was exactly what I was trying to communicate all along. A measure-theoretic analogue of length in R exists for area in R^2. The notion of distance, though, has no such a thing as a higher-dimensional extension. A distance is just a distance.
What a fantastic job you are doing! Always engaging, inclusive of different entry-levels, and superbly illustrated. My question may be silly or pedantic, but when you talk about open balls of radius 1 for different metrics, your graphs show a clear boundary to the diamond (Manhattan) or the square (Chebyshev), wouldn't the open ball have no boundary? I'm interested in the concept, not finding fault in what is an exceptional presentation.
Absolutely! The thick lines was meant nothing more than to visually distinct, and I didn’t really get into open vs closed. But if I was doing it for pure accuracy I’d draw dashed lines to indicated not including the boundary.
11:21 during editing somehow he realized that he didn't flipped his projection of the video from the camera onto the background. Thought we wouldn't notice, but we did.
That shirt is soooo funny 😂😂😂 Edit:. No one seems to have looked at your shirt yet 2:57 yes that is a distance but we could have walked differently . 6:35 there is a property which says d is always positive.
Is this just a video on its own, or does this belong to a specific course you’re uploading to the channel? I recently read the definition of a metric space in the context of general topology, so I was pleasantly surprised when this notification popped up from you on the topic
I have a question. In the Minkowski metric for spacetime, the distance for light (the spacetime interval) is zero. Doesn't that violate the first condition i.e. if AB =/= 0, A =/= B?
@@feynstein1004 The Minkowski metric, despite the name, is not a metric. Also, the idea of metric in special relativity has nothing to do with metric spaces, but with the metric tensor, which is actually a concept for manifolds, instead.
@@feynstein1004 No, that is incorrect. The Minkowski metric is a metric tensor, not the metric of a metric space. Those are mathematically different concepts.
Thanks for the video! I have a question about PI number in metric spaces: in Euclidean, pi is something beautiful and strange at the same time. Are 'pi' in other metric spaces also so interesting, if we could define pi there at all? Concept of area in that metric spaces feels even weirder.
pi is particularly nice in how it relates perimeter and area. In other metric spaces, such as the two I showed, that relationship between perimeter and area isn't a multiple of pi!
In Euclidean metric pi = 3.14.... but in both in Taxicab and Chebyshev metric pi = 4. In the FLRW metric (the one that space-time obeys at the cosmological scales) pi is a number between 0 and 3.14... depending on the sice of your circle: for small circles (thousands of light years) pi is aproximately 3.14... but for circles with diameters of billions of light years, the value of pi goes closer and closer to zero (these circles have gigantic radiuses but extremele small circunferences, due to the fact that at greater distance the universe was younger and thus smaller than today).
There is a special class of metric spaces called finite-dimensional L^p spaces. These are spaces where the norm of a vector v in R^n is given by ||v|| = [|v0|^p + |v1|^p + ••• + |v(n - 1)|^p]^(1/p), where p = 1 or p > 1, or p -> ♾. The distance function, or metric, is then given by d(v, u) = ||v - u||, where ||•|| is a norm defined as above for some p. Such a norm is called a p-norm. The Euclidean norm is the case where p = 2, the Manhattan or taxicab norm is the case where p = 1, and the Chebyshev norm is the case where p -> ♾. Why am I mentioning p-norms? Because in p-norm spaces, for the metric induced by that norm, the circles defined by that metric do still satisfy the property that the circumference is proportional to its diameter. We can even derive an exact formula for the constant of proportionality. Remember, in this situation, the equation for a circle centered at the origin is |x|^p + |y|^p = r^p, and we only care about centering at the origin, since translation does not affect circumference, in these spaces. Now, if we restrict ourselves only to x > 0, y > 0, that leaves us with a quarter circle, and the circumference will just be 4 times the arclength for this quarter circle. With this in mind, we have that y = (r^p - x^p)^(1/p) for this quarter circle. The formula for arclength, in this case, is given by the integral on [0, r] of [1 + |y'|^p]^(1/p). Notice that y' = 1/p·(r^p - x^p)^(1/p - 1)·(-p)·x^(p - 1) = -(r^p - x^p)^(1/p - 1)·x^(p - 1). Therefore, |y'|^p = (r^p - x^p)^(1 - p)·x^[p·(p - 1)] = [x^p/(r^p - x^p)]^(p - 1). Hence the integrand is {1 + [x^p/(r^p - x^p)]^(p - 1)}^(1/p) = (1 + {(x/r)^p/[1 - (x/r)^p]}^(p - 1)}^(1/p). With the change of variables t = x/r, one has r multiplied by the integral on [0, 1) of {1 + [t^p/(1 - t^p)]^(p - 1)}^(1/p). This 4 times this integral is thus the ratio from circumference to radius, and 2 times this integral is the ratio from circumference to diameter. It is a constant. 2 times the integral is equal to the integral on (-1, 1) of {1 + [|t|^p/(1 - |t|^p)]^(p - 1)}^(1/p), which is the integral corresponding to the upper semicircle, rather than only one quarter circle. For p = 2, the integrand simplifies to 1/sqrt(1 - t^2). For p = 1, the integrand simplifies to 2. So for p = 2, the circumference:radius ratio is π. Actually, π is literally defined by that integral: the symbol π is just an abbreviation for it. Meanwhile, for p = 1, the integral is thus 4. As p -> ♾, it also approaches 4. For an arbitrary metric space, the situation becomes significantly more complicated, and in general, there is no tractable relationship between circumference and radius. For an arbitrary metric space (R^2, d), a circle of radius r centered at (0, 0) is just the set of vectors v such that d(v, 0) = r. The circumference of the upper semicircle is thus equal to some integral of d(v', 0), where v is a parametrization of the circle, and v' its derivative. There is no expectation that this will simplify nicely in terms of the radius.
There needs to be something clarified here. I see many people bringing up the Minkowski metric from special relativity, and the more general concept of the metric tensor fron general relativity, and conflating these concepts with the concepts from the theory of metric spaces. The Minkowski metric is an example of a metric tensor, not of a metric in a metric space. Given a manifold M, a metric tensor is a map M*M -> R that is sesquilinear and conjugate-symmetric. This is completely different from the metric d of a metric space (X, d). These are unrelated concepts, despite the naming scheme. That being said, metric tensors can induce a metric space, if they are positive definite. But the usages of the word "metric" here are different. They are different mathematical concepts.
When I was first studying this stuff, it was with the intention of understanding special and general relativity, and I remember being amused and a bit annoyed by the fact that the "metric" there violates the criteria for a metric space at step one (it's not a positive real function, and this creates the distinction between spacelike, timelike and lightlike intervals).
In the real plane, the Euclidean metric between two vectors u and v is defined by d(u, v) = sqrt[(u0 - v0)^2 + (u1 - v1)^2]. A generalization of this idea gives result to the notion of a p-norm. The Euclidean norm of a vector v in the real plane is given by ||v|| = sqrt(|v0|^2 + |v1|^2). The Euclidean metric is thus defined by d(u, v) = ||u - v||. A p-norm is a generalization of the Euclidean norm. A p-norm is a norm of the form ||v|| = (|v0|^p + |v1|^p)^(1/p), where p in [1, ♾). The case with p = 2 is the Euclidean norm. The case with p = 1 is the Manhattan norm, and p -> ♾ is the Chebyshev norm, also known as the supremum norm, because lim (|v0|^p + |v1|^p)^(1/p) (p -> ♾) = max(|v0|, |v1|). A metric space is then formed by taking d(u, v) = ||u - v|| for each of these norms. Normally, a subscript is added to bar symbols for the norm to indicate which specific p-norm is being worked with. These metric spaces are extremely important in mathematics. One important idea here is that the equation of a circle for the metric space corresponding to a p-norm is |x - h|^p + |y - k|^p = r^p, where (h, k) is the center of the circle, and r its radius. It turns out that you can find a formula for the circumference of such circles. Notice that if one centers the circle at the origin, and restricts oneself to the case with x > 0, y > 0, one can work with the arclength of the upper-right quarter circle instead, and the circumference is 4 times this arclength, and in this case, the equation of a circle can be rewritten as y = (r^p - x^p)^(1/p), where x ranges (0, r). The equation for arclength, in this case, is given by the integral on (0, r) of [1 + |y'|^p]^(1/p), and so |y'| = (r^p - x^p)^(1/p - 1)·x^(p - 1), so |y'|^p = [x^p/(r^p - x^p)]^(p - 1), hence [1 + |y'|^p]^(1/p) = {1 + [x^p/(r^p - x^p)]^(p - 1)}^(1/p) = [1 + {(x/r)^p/[1 - (x/r)^p]}^(p - 1)]^(1/p). Letting t = x/r, with t ranging (0, 1), this results in r multiplied by the integral on (0, 1) of {1 + [t^p/(1 - t^p)]^(p - 1)}^(1/p). This is equal to r/2 multiplied by the integral on (-1, 1) of {1 + [|t|^p/(1 - |t|^p)]^(p - 1)}^(1/p). This is the arclength of the upper right quarter circle, so the circumference is equal to 2·r multiplied by the integral on (-1, 1) of {1 + [|t|^p/(1 - |t|^p)]^(p - 1)}^(1/p). Therefore, the ratio of circumference to diameter of such a circle is just the integral on (-1, 1) of {1 + [|t|^p/(1 - |t|^p)]^(p - 1)}^(1/p). For p = 2, the integrand simplifies to 1/(1 - t^2)^(1/2), and the integral is equal to some real number, which we call π. In fact, this is literally the definition of π. For p = 1, the integand is 2, and so the integral simplifies to 4. For p -> ♾, the integrand also simplifies to 2, and so the integral is also 4. Fun fact: if you graph the integral on Desmos as function of p, one will find that a local minimum value of the integral actually occurs at p = 2, but it is not the global minimum. Also, although (|v0|^p + |v1|^p)^(1/p) is not a norm for p in (0, 1), d(u, v) = |u0 - v0|^p + |u1 - v1|^p actually is a metric. In this case, the equation for the circle is simply |x|^p + |y|^p = 1. When looking at the circumference, the derivation is completely analogous: the integrand for the arclength of the quarter circle is given by 1 + [x^p/(r - x^p)]^(p - 1), and the interval of integration is (0, r^(1/p)). The procedure is now the same, except now, the correct substitution to make is t = x/r^(1/p), giving as result r^(1/p) multiplied by the integral on (0, 1) of 1 + [t^p/(1 - t^p)]^(p - 1). This means that circumference is equal to the integral above, which is a real number independent of radius, multiplied by 4·r^(1/p). Unfortunately, this no longer gives a degree 1 relationship, but this is still a relatively simple expression. For an arbitrary metric space, no nice, general relationship of the sort exists between circumference and radius.
@@DrTrefor Also i forgot to thank you for your explanation because you make me realise the importance of the metric abstraction ! I'm looking forward for your future videos!
@@DrTrefor sir can you please upload the video lectures of legendre polynomial, bessel's function I have so many doubts on those topics and I think your lectures are the best suppliment for me to understand those topics with full of visualisation .....
Same thing. Diamond isn't really a colloquial term rather than a rigorous geometric one but usually refers to a rhombus, of which a square is generally regarded as a special case.
You said that all points of distance 1 from a point form an open ball. Surely that is the set of all points of distance *less than* 1. Also, would "genealogical distance" qualify as a metric? eg. the distance from you to your aunt is 3, because you have to go 2 up to your grandparents, then 1 down to your aunt.
Nobody noticed his tshirt? Hippopotamus denoting hypotenuse as the latter is a bit difficult to pronounce and confusing sometimes. I want this t-shirt 🤩
Because the minimum number of single square moves to go between two points on a chessboard is precisely that larger value of the number of ranks separating them and the number of files separating them. If you imagine two points on a chessboard with coordinates (1,1) and (1,5) then (x2-x1, y2-y1) gives (0,4). The formula gives you 4, which is clearly the correct distance.
I'm very sad that you didn't go into the general formula for every metric (called the p-norm), how even Chebychev metric flows out when p=infinity, and how it relates to the superellipse.
Indeed! I’ve found R^+ to be ambiguous, much like the naturals, as to whether it includes zero or not. Regardless, I mean it too. Actually, one need not specify a restriction on the codomain at all, that it is greater than or equal to zero as this follows from the axioms!
@@DrTrefor Yeah, I've seen the set of non-negative real numbers represented as R with a plus sign in the subscript, and this way to represent the positive real numbers, so I thought maybe it was an unintentional mistake
1 day i was thinking of exact same thing and i discovered all of it by my own. I didn't knew this was a concept in maths. Same happened when I discovered sinx function as a infinite series and then i found out it was already done. I was wish I allowed to have mathematical education.
Well, no. That depends largely on the metric space. On a discrete metric space, there are only two possible distances between any two objects: 0, or 1. For this reason, I also like to call it the Boolean metric, even though no mathematician uses this name.
Confusticate and Bebother you and everybody else who brainwashed me into believing the triangle inequality, which left me high and dry when I tried to wrap my mind around the Minkowski metric.
I'm not sure I understand why the notion of distance needs to be symmetric. If you define distance in terms of time, then distances traveled uphill could be longer than downhill. This is quite a common way to measure distances in the real world, so I'm curious why it doesn't qualify as a metric.
Some other cool examples of metrics are in biology to measure the evolutionary distance between organisms and in linguistics to measure the 'distance' between different words
Oh nice! Yes I’ve seen these types of word metrics before but didn’t know the evolution connection
@@DrTrefor Dunno much about it, but they use an ultrametric with a stronger version of the triangle inequality where every three points forms an isosceles triangle. It's useful because imagine trying to find the least common ancestors of three organisms A, B and C, then you want the 'distance' (i.e. how many years ago their LCA lived) between any pair to be related in a special way. E.g. if A and B's LCA lived 7 mya, and B and C's LCA lived 10 mya, then you don't want the LCA of A and C to be greater than 10 mya.
What is this nonsense about "evolutionary distance between organisms"? Don't you know that God created every living thing in seven days?
@@alexisbach That is scientifically false.
@@alexisbach what?
He helped me solve my homework and then blew my mind at the end of the video for good measure!
Maths is so wierd and wonderful. I'm really starting to love it in its own right, rather then just a tool to help me understand physics
As someone who's studying Topology in university right now, this is a nice refresher on metrics. Additionally, with just having finished a chapter on approximation theory in my numerics 2 class, I'm going to add an interesting note:
If you look at the chebyshev and the taxi-cab metric, it is pretty easy to find lines on which the best approximation of the origin is not unique. Only with the euclidean distance, such an approximation is always unique, because this metric is defined by a scalar product. This is also generally true, however there are cases where you don't need a metric produced by a scalar product to have unique best approximations to some point in space.
What do you mean by "the best approximation of the origin"?
1:03 not really related to the topic, but ironically, the king actually is the weakest piece because it can't capture guarded pieces nor move to squares that are also guarded by other pieces, making it have less attacking and defending potential than even a pawn.
That’s what I was thinking lol
Well in endgames you gotta know how to move the king
Woaa... It's the first time I realise that Chebyshev metric is actually the King's metric of chess board. What a great explanation professor. You deliver your content as great as always.
This video is GOLD. Thank you!
Another interesting distance metric is:
d(a,b) = max{ | b1 - a1 |, | b2 - a2 |, | (b1 - a1) - (b2 - a2) | }
Which gives the distance on a hexagonal grid (with no grid side between the two axes. For a hexagonal grid with a grid side between the two axes, swap the minus for a plus between the two terms in parentheses for the last element.)
Very cool! There are also metric tensors for smooth manifolds where you have something like ds^2 = dθ^2 + sin^2(θ) dφ^2, and to find the distance along a path between two coordinates (θ_1, φ_1), (θ_2, φ_2), you have to integrate ds along that path.
technically it's a tensor field lol a tensor field over the entire manifold, so a "metric tensor field" lol riemannian manifolds ftw!!!
@Scuffed Physics I'm not 100 percent sure about this but a manifold can be a metric space only when it has constant curvature (like a sphere) otherwise the metric tensor varies from point to point then it wont be a metric space since a metric space has a fixed metric
@Scuffed Physics yes but technically that wouldn't be a metric space since the metric itself changes depending on which two points on the manifold you're talking about. a metric space has to have a fixed metric right? lol am i getting the definition of a metric space wrong?
also btw a geodesic is by definition the shortest distance between two points on a manifold so "shortest geodesic" is kind of a tautology
@@mastershooter64 You are misunderstanding what a metric tensor is, probably on account of the fact that the terminology is just confusing you. A metric tensor is not a metric. A metric tensor is an inner product. An inner product is a sesquilinear, conjugate-symmetric map that is positive definite. The dot product is an example of an inner product, and the most elementary example. Now, assuming you have an inner product , you can define a norm by saying ||v|| = sqrt(). This turns the manifold into a normed manifold, where the norm is induced by the metric tensor. The metric, in the sense of a metric space, is then defined by d(u, v) = ||u - v||, hence the norm induces a metric on the manifold, and thus the manifold is a metric space. But not every manifold that is a metric space has an inner product. That is where you are getting confused. A manifold with an inner product is called a Riemannian manifold. So the a manifold with a metric tensor is a Riemannian manifold. Such manifolds necessarily do form metric spaces, but the metric tensor is the inner product, not the metric induced by that inner product.
I know this is confusing, because a metric, and a metric tensor, are different things, even though the word "metric" features in the name for both mathematical objects. You may think this is poor choice of nomenclature on the mathematician's part. However, the name "metric tensor" makes perfect sense in the context of tensor analysis.
@@angelmendez-rivera351 yes i know a metric tensor is a inner product, and you integrate to get distances on the manifold, no it's not confusing it's pretty intuitive. also it's technically a tensor field over the entire manifold, each point on the manifold has a specific metric tensor on it. lol how can you assume "this is where you are getting confused" the only thing i wasnt sure about was that a metric space can have a metric that depends on where you are on that space, and apparently you can! which is pretty cool
Every one looking for prof like you sir ❤️
One very important example of metric spaces that are not Euclidean are the L^p spaces in functional analysis. The simplest example is the L^♾ space. We consider functions f : S -> C, where C is the field of complex numbers, and S is a Borel set of C. The supremum norm of f, denoted ||f||_♾, is defined as sup({|f(z)| : z in S}), where sup here denotes the supremum. The space of functions C^S, together with this norm, forms a normed space. The metric d is defined as d(f, g) = ||f - g||_♾. This gives us a way of measuring distances between functions, and the reason this distance notion is so important is because it underlies the notion of uniform convergence. In other words, uniform convergence is just convergence of a sequence, or more generally, a net of functions, with respect to the supremum norm.
It's perfectly explained ❤
Love the way it's very easy to understand the point watching this video !
Wow! I had no idea that the distances in Chess had their own metric. The king chasing down the pawn example you gave is colloquially known as "the square of the pawn" and it states that the king can catch the pawn if it can step into the "square of the pawn" (a picture of this can explain it more clearly) on its next move. I'm now curious about the knight's movements...
Edit: turns out the knight does make a metric. disregard everything I said
The knight’s movements satisfy only the first and second conditions of being a metric, symmetry and “0 distance same point”. It doesn’t satisfy the triangle inequality, as a knight on a1 could move to b3 in 1 move, but to pass through a2, it needs first 3 moves then 2 moves, which don’t add up to 1. It’s a real shame, the knight metric would be so interesting. Is there some way to modify the knight’s movements to make a metric?
@@-minushyphen1two379 d(a1,b3)=1
@@-minushyphen1two379 any distance function based on counting number of steps should satisfy the triangle inequality, no matter what chess piece is used. A proof by contradiction:
Define d(A,B) as the minimum number of steps it would take our chess piece to get from A to B. Assume that the triangle inequality doesn't hold, i.e. we can find some A, B, and C s.t. d(A,C) > d(A,B) + d(B,C). This means that there is a path P from A to C passing through B that is shorter than the shortest path from A to C. But P *is* a path from A to C; it can't be shorter than itself (a contradiction). Our assumption must be false: the function d must satisfy the triangle inequality.
It's not hard to see that d satisfies the first condition, and it should satisfy the second as long as every move the chess piece makes is reversible.
All this is assuming that the chess piece can get to every square on the chessboard. If it can't get from A to B, then d(A,B) is undefined. So a bishop won't produce a valid distance function.
It should also be possible for some chess pieces to create corresponding continuous versions of their metrics on the plane rather than the chessboard, by overlaying an infinite chessboard on the plane whose squares have side length k and then shrinking k, so that the discrete chessboard approximates the continuous plane. The new continuous metric is the limit as k→0 of k*d(A,B). The knight seems to produce a valid continuous metric, with its open ball being an octagon whose horizontal and vertical sides are longer than its diagonal sides.
I suspect that the chess pieces that produce valid continuous metrics are the ones that satisfy these two conditions:
1) they produce a valid discrete metric
2) they can't move arbitrarily far in a single step
@@calvincrady well for example a continuous rook metric is well defined as d(a, b) = [a_x ≠ b_x] + [a_y ≠ b_y], which satisfies all rules, it is just not super interesting
In data science there is a machine learning algorithm called Lasso Regression (which prevents overfitting). Anyway it is also called L1 regularization and is depicted visually as a diamond, just like your Taxicab visualization.
Ah yes! I know of this metric, but not in that context
@@DrTrefor Which value of n
1 to ∞ ∫ ( {x} / (x)^(n+1) ) dx question solution is =( 1/(n+1) )
Where {x} = ( x - floor(x) )
I love math and your videos make me love it more then ever.
Thank you!
I'm so happy to hear that! Math is awesome haha:D
@@DrTrefor Which value of n
1 to ∞ ∫ ( {x} / (x)^(n+1) ) dx question solution is =( 1/(n+1) )
Where {x} = ( x - floor(x) )
Really well explained video (no surprises there!), thanks for the great content! :)
Hey, just wanted to let you know it is finals season and your calculus 3 videos are saving my life!
Awesome! Good luck!!
6:03
lols 1 way roads where a road A to B curves and road B to A is straight is non metric distance by breaking rule 2 and rule 3.
lols if point A that was at X1 and Y3 then moves to point X5 and Y5 and point B is at X5 and Y5 then A and B the distance between them is 0 and by rule 1 they are the same point then if point A moves to X2 and Y4, A and B have not become not the same point and so the distance between them is still 0, a system of dynamic points is non metric, even if you don't want to use time you can make Z location substitute for "then".
Thanks for this video Trefor . It helped me to clear some of my thougts and fill the intuition.
You're very welcome!
Sir your teaching process is Always best ...my concept always clear to see your vedio ..mind blowing ❤️
Please make more videos on this .
2:57 it is a distance but we can go differently
I really hope the "intro" part suggests there is more to come :) Don't even care much about the direction, more general? Sure topological spaces are really cool! More in depth? I mean you said you are a calc prof, so that would fit! More specialised with smooth manifolds? If you have the gut to deal with the horrible index mess that differential geometry tends you'd probably make a lot of physicists happy!
For real though, even though this video doesn't really discuss anything an analysis 2 (or 1?) lecture wouldn't cover, you animations and presentation style still have me glued to the screen :) Also very nice touch with the motivation on building theories on generalised concepts, from theoretical physics to constructivist, finitist or ZFC-Powerset mathematics, trying to explain why we try to work with the more general version can be hard to do well.
I’m slowly in the background working on what a “calculus on manifolds” series might look like for this channel. Will see!
oh, this is a very good video. I was thinking about exactly this topic recently.
Beautiful explanation!
Thank you!
Wow, that's really well-explained! Thanks!
Glad you enjoyed it!
For me I find it is much easier to think about the "different distances" as moves/step between states instead of distance between two points. imagine some state A (for example chessboard arrangement) from which you want to go to state B. Then the metric is just the sum of all steps you need to do in order to get to state B. You can aply this logic easily to various "types of distances".
* note that specifuclly the chestboard example could violate the rule d(A, B) = d(B, A) because pawns don't move backwards, but you get the idea.
With Euclidean distances, we easily go from meters to square meters. With metrics (instead of distance), what is the analogous word used for area.
Still area, but what we might mean by area changes.
I think we will need the notion of measure for area instead of metric. Measure generalises 1D Euclidean distance, 2D area in 2D Euclidean space, 3D volume in 3D Euclidean space, all the way to nD hypervolume for nD Euclidean space.
Even measure is still compatible for other concepts, such as counting measure that agrees with the concept of cardinality of countable sets.
That is not correct. While distances in Euclidean space are thought of in terms of length, it is more accurate to say that length, in a Euclidean space, is defined in terms of distance. However, this is exclusive to Euclidean space. In actuality, talking in general mathematical terms, distance and length are completely different, unrelated concepts. For length, you want some kind of measure space with topological dimension 1, and use the Lebesgue measure. Metric spaces are a completely unrelated thing. You need not have measurable sets within a metric space, and similarly, you measure space may not have a metric.
@@angelmendez-rivera351 I get your point. But when we talk about area in Euclidean space we should talk about measure instead of metric. That's what I want to emphasise.
Let me correct my statement. For R^1, we have to use the notion of length instead of distance if we want it to be correlated to area in R^2, and hence we need the notion of measure instead of metric for the case.
@@rizalpurnawan23 Yes, I agree with you on that point. In fact, that was exactly what I was trying to communicate all along. A measure-theoretic analogue of length in R exists for area in R^2. The notion of distance, though, has no such a thing as a higher-dimensional extension. A distance is just a distance.
What a fantastic job you are doing! Always engaging, inclusive of different entry-levels, and superbly illustrated. My question may be silly or pedantic, but when you talk about open balls of radius 1 for different metrics, your graphs show a clear boundary to the diamond (Manhattan) or the square (Chebyshev), wouldn't the open ball have no boundary? I'm interested in the concept, not finding fault in what is an exceptional presentation.
Absolutely! The thick lines was meant nothing more than to visually distinct, and I didn’t really get into open vs closed. But if I was doing it for pure accuracy I’d draw dashed lines to indicated not including the boundary.
EXCELLENT EXPLANATION
Glad you think so!
11:21 during editing somehow he realized that he didn't flipped his projection of the video from the camera onto the background. Thought we wouldn't notice, but we did.
Ha I didn’t even notice! Silly editing presets:D
Those are awesome examples!
That shirt is soooo funny 😂😂😂
Edit:. No one seems to have looked at your shirt yet
2:57 yes that is a distance but we could have walked differently .
6:35 there is a property which says d is always positive.
haha I love it:D
@@DrTrefor There is a property which says d is always positive.
@@aashsyed1277 yup!
@@aashsyed1277 That one can be derived from the others
@@Alex_Deam ok
Is this just a video on its own, or does this belong to a specific course you’re uploading to the channel? I recently read the definition of a metric space in the context of general topology, so I was pleasantly surprised when this notification popped up from you on the topic
Not a course, exactly, but I have a sort of short thematic series of cool things in topology and geometry coming out.
@@DrTrefor Which value of n
1 to ∞ ∫ ( {x} / (x)^(n+1) ) dx question solution is =( 1/(n+1) )
Where {x} = ( x - floor(x) )
@@DrTrefor I can't wait to watch that
7:39 but a circle is not an open ball? one is the interior and one is the boundary
similar to distance definition are there other definitions for Density ?
Indeed! Density is defined per unit "area", so how you measure distances affect this.
Density is defined in terms of measure theory, so that is mathematically a very different kind of concept.
@@angelmendez-rivera351 thanks for this information
im 16 and your videos are making me really consider doing maths at university!
You totally should!!
Try some real uni examples first. Pure maths is.. well not for everyone. But if you like it, do it!
I have a question. In the Minkowski metric for spacetime, the distance for light (the spacetime interval) is zero. Doesn't that violate the first condition i.e. if AB =/= 0, A =/= B?
I believe the Minkowski metric is know as a "pseudo-Euclidean metric"
@@DrTrefor That sounds interesting. Would you mind elaborating a bit?
@@feynstein1004 The Minkowski metric, despite the name, is not a metric. Also, the idea of metric in special relativity has nothing to do with metric spaces, but with the metric tensor, which is actually a concept for manifolds, instead.
@@angelmendez-rivera351 Ah okay. Thank explains it. Thank you for the reply 🙂
how about the spacetime interval? Is that also a metric?
It can indeed be phrased as a metric!
@@feynstein1004 No, that is incorrect. The Minkowski metric is a metric tensor, not the metric of a metric space. Those are mathematically different concepts.
Can you make more videos about metric space it would be very helpful
Thanks for the video!
I have a question about PI number in metric spaces: in Euclidean, pi is something beautiful and strange at the same time. Are 'pi' in other metric spaces also so interesting, if we could define pi there at all? Concept of area in that metric spaces feels even weirder.
pi is particularly nice in how it relates perimeter and area. In other metric spaces, such as the two I showed, that relationship between perimeter and area isn't a multiple of pi!
In Euclidean metric pi = 3.14.... but in both in Taxicab and Chebyshev metric pi = 4. In the FLRW metric (the one that space-time obeys at the cosmological scales) pi is a number between 0 and 3.14... depending on the sice of your circle: for small circles (thousands of light years) pi is aproximately 3.14... but for circles with diameters of billions of light years, the value of pi goes closer and closer to zero (these circles have gigantic radiuses but extremele small circunferences, due to the fact that at greater distance the universe was younger and thus smaller than today).
@@DrTrefor i fail linear algebra rip
@@DrTrefor Which value of n
1 to ∞ ∫ ( {x} / (x)^(n+1) ) dx question solution is =( 1/(n+1) )
Where {x} = ( x - floor(x) )
There is a special class of metric spaces called finite-dimensional L^p spaces. These are spaces where the norm of a vector v in R^n is given by ||v|| = [|v0|^p + |v1|^p + ••• + |v(n - 1)|^p]^(1/p), where p = 1 or p > 1, or p -> ♾. The distance function, or metric, is then given by d(v, u) = ||v - u||, where ||•|| is a norm defined as above for some p. Such a norm is called a p-norm. The Euclidean norm is the case where p = 2, the Manhattan or taxicab norm is the case where p = 1, and the Chebyshev norm is the case where p -> ♾. Why am I mentioning p-norms? Because in p-norm spaces, for the metric induced by that norm, the circles defined by that metric do still satisfy the property that the circumference is proportional to its diameter. We can even derive an exact formula for the constant of proportionality. Remember, in this situation, the equation for a circle centered at the origin is |x|^p + |y|^p = r^p, and we only care about centering at the origin, since translation does not affect circumference, in these spaces. Now, if we restrict ourselves only to x > 0, y > 0, that leaves us with a quarter circle, and the circumference will just be 4 times the arclength for this quarter circle. With this in mind, we have that y = (r^p - x^p)^(1/p) for this quarter circle. The formula for arclength, in this case, is given by the integral on [0, r] of [1 + |y'|^p]^(1/p). Notice that y' = 1/p·(r^p - x^p)^(1/p - 1)·(-p)·x^(p - 1) = -(r^p - x^p)^(1/p - 1)·x^(p - 1). Therefore, |y'|^p = (r^p - x^p)^(1 - p)·x^[p·(p - 1)] = [x^p/(r^p - x^p)]^(p - 1). Hence the integrand is {1 + [x^p/(r^p - x^p)]^(p - 1)}^(1/p) = (1 + {(x/r)^p/[1 - (x/r)^p]}^(p - 1)}^(1/p). With the change of variables t = x/r, one has r multiplied by the integral on [0, 1) of {1 + [t^p/(1 - t^p)]^(p - 1)}^(1/p). This 4 times this integral is thus the ratio from circumference to radius, and 2 times this integral is the ratio from circumference to diameter. It is a constant. 2 times the integral is equal to the integral on (-1, 1) of {1 + [|t|^p/(1 - |t|^p)]^(p - 1)}^(1/p), which is the integral corresponding to the upper semicircle, rather than only one quarter circle. For p = 2, the integrand simplifies to 1/sqrt(1 - t^2). For p = 1, the integrand simplifies to 2. So for p = 2, the circumference:radius ratio is π. Actually, π is literally defined by that integral: the symbol π is just an abbreviation for it. Meanwhile, for p = 1, the integral is thus 4. As p -> ♾, it also approaches 4.
For an arbitrary metric space, the situation becomes significantly more complicated, and in general, there is no tractable relationship between circumference and radius. For an arbitrary metric space (R^2, d), a circle of radius r centered at (0, 0) is just the set of vectors v such that d(v, 0) = r. The circumference of the upper semicircle is thus equal to some integral of d(v', 0), where v is a parametrization of the circle, and v' its derivative. There is no expectation that this will simplify nicely in terms of the radius.
There needs to be something clarified here. I see many people bringing up the Minkowski metric from special relativity, and the more general concept of the metric tensor fron general relativity, and conflating these concepts with the concepts from the theory of metric spaces. The Minkowski metric is an example of a metric tensor, not of a metric in a metric space. Given a manifold M, a metric tensor is a map M*M -> R that is sesquilinear and conjugate-symmetric. This is completely different from the metric d of a metric space (X, d). These are unrelated concepts, despite the naming scheme. That being said, metric tensors can induce a metric space, if they are positive definite. But the usages of the word "metric" here are different. They are different mathematical concepts.
When I was first studying this stuff, it was with the intention of understanding special and general relativity, and I remember being amused and a bit annoyed by the fact that the "metric" there violates the criteria for a metric space at step one (it's not a positive real function, and this creates the distinction between spacelike, timelike and lightlike intervals).
In the real plane, the Euclidean metric between two vectors u and v is defined by d(u, v) = sqrt[(u0 - v0)^2 + (u1 - v1)^2]. A generalization of this idea gives result to the notion of a p-norm. The Euclidean norm of a vector v in the real plane is given by ||v|| = sqrt(|v0|^2 + |v1|^2). The Euclidean metric is thus defined by d(u, v) = ||u - v||. A p-norm is a generalization of the Euclidean norm. A p-norm is a norm of the form ||v|| = (|v0|^p + |v1|^p)^(1/p), where p in [1, ♾). The case with p = 2 is the Euclidean norm. The case with p = 1 is the Manhattan norm, and p -> ♾ is the Chebyshev norm, also known as the supremum norm, because lim (|v0|^p + |v1|^p)^(1/p) (p -> ♾) = max(|v0|, |v1|). A metric space is then formed by taking d(u, v) = ||u - v|| for each of these norms. Normally, a subscript is added to bar symbols for the norm to indicate which specific p-norm is being worked with. These metric spaces are extremely important in mathematics.
One important idea here is that the equation of a circle for the metric space corresponding to a p-norm is |x - h|^p + |y - k|^p = r^p, where (h, k) is the center of the circle, and r its radius. It turns out that you can find a formula for the circumference of such circles. Notice that if one centers the circle at the origin, and restricts oneself to the case with x > 0, y > 0, one can work with the arclength of the upper-right quarter circle instead, and the circumference is 4 times this arclength, and in this case, the equation of a circle can be rewritten as y = (r^p - x^p)^(1/p), where x ranges (0, r). The equation for arclength, in this case, is given by the integral on (0, r) of [1 + |y'|^p]^(1/p), and so |y'| = (r^p - x^p)^(1/p - 1)·x^(p - 1), so |y'|^p = [x^p/(r^p - x^p)]^(p - 1), hence [1 + |y'|^p]^(1/p) = {1 + [x^p/(r^p - x^p)]^(p - 1)}^(1/p) = [1 + {(x/r)^p/[1 - (x/r)^p]}^(p - 1)]^(1/p). Letting t = x/r, with t ranging (0, 1), this results in r multiplied by the integral on (0, 1) of {1 + [t^p/(1 - t^p)]^(p - 1)}^(1/p). This is equal to r/2 multiplied by the integral on (-1, 1) of {1 + [|t|^p/(1 - |t|^p)]^(p - 1)}^(1/p). This is the arclength of the upper right quarter circle, so the circumference is equal to 2·r multiplied by the integral on (-1, 1) of {1 + [|t|^p/(1 - |t|^p)]^(p - 1)}^(1/p). Therefore, the ratio of circumference to diameter of such a circle is just the integral on (-1, 1) of {1 + [|t|^p/(1 - |t|^p)]^(p - 1)}^(1/p). For p = 2, the integrand simplifies to 1/(1 - t^2)^(1/2), and the integral is equal to some real number, which we call π. In fact, this is literally the definition of π. For p = 1, the integand is 2, and so the integral simplifies to 4. For p -> ♾, the integrand also simplifies to 2, and so the integral is also 4. Fun fact: if you graph the integral on Desmos as function of p, one will find that a local minimum value of the integral actually occurs at p = 2, but it is not the global minimum.
Also, although (|v0|^p + |v1|^p)^(1/p) is not a norm for p in (0, 1), d(u, v) = |u0 - v0|^p + |u1 - v1|^p actually is a metric. In this case, the equation for the circle is simply |x|^p + |y|^p = 1. When looking at the circumference, the derivation is completely analogous: the integrand for the arclength of the quarter circle is given by 1 + [x^p/(r - x^p)]^(p - 1), and the interval of integration is (0, r^(1/p)). The procedure is now the same, except now, the correct substitution to make is t = x/r^(1/p), giving as result r^(1/p) multiplied by the integral on (0, 1) of 1 + [t^p/(1 - t^p)]^(p - 1). This means that circumference is equal to the integral above, which is a real number independent of radius, multiplied by 4·r^(1/p). Unfortunately, this no longer gives a degree 1 relationship, but this is still a relatively simple expression.
For an arbitrary metric space, no nice, general relationship of the sort exists between circumference and radius.
I had a problem recently that really made me struggle, until I realized I was measuring distance wrong
Is it possible to create a series about topological spaces?
Indeed! I’m not exactly doing a series, but my background is topology, and I’ve got a number of cool topology videos in the works
@@DrTrefor Also i forgot to thank you for your explanation because you make me realise the importance of the metric abstraction ! I'm looking forward for your future videos!
Wow that's amazing explanation
Glad you liked it!
@@DrTrefor sir can you please upload the video lectures of legendre polynomial, bessel's function I have so many doubts on those topics and I think your lectures are the best suppliment for me to understand those topics with full of visualisation .....
8:27 actually it is a square. In a diamond the angles are not straight.
Same thing. Diamond isn't really a colloquial term rather than a rigorous geometric one but usually refers to a rhombus, of which a square is generally regarded as a special case.
*laughs in pseudo-riemannian manifolds*
You said that all points of distance 1 from a point form an open ball. Surely that is the set of all points of distance *less than* 1.
Also, would "genealogical distance" qualify as a metric? eg. the distance from you to your aunt is 3, because you have to go 2 up to your grandparents, then 1 down to your aunt.
Yes indeed on both accounts.
@@DrTrefor I look forward to saying to my cousins "By my metric, you don't make my will".
Great video! Are you starting a course on real analysis?
Sadly not, but I may jump around in some highlights
Is this useful before I start general relativity
I mean will it give me basics?
Yes I think it is helpful for the metrics you will see in GR
Hello Trefor,
Is this the first video of Metric Spaces course (i.e., playlist)?
I don't have it planned as a full course, but likely will do some follow ups:)
So the Chebyshev metric is the one used by D&D 5e? 🤔
I want that Hyppopotenuse shirt!
Which value of n
1 to ∞ ∫ ( {x} / (x)^(n+1) ) dx question solution is =( 1/(n+1) )
Where {x} = ( x - floor(x) )
Once all the pawns are gone, is the number of moves to go from chessboard state A to state B, a distance metric?
Nobody noticed his tshirt? Hippopotamus denoting hypotenuse as the latter is a bit difficult to pronounce and confusing sometimes.
I want this t-shirt 🤩
Please make a series on vector space
I have one! Check out my linear algebra playlist
I'm still trying to make sense of this formula on chebyshev distance. Why do we need a max there?
Because the minimum number of single square moves to go between two points on a chessboard is precisely that larger value of the number of ranks separating them and the number of files separating them. If you imagine two points on a chessboard with coordinates (1,1) and (1,5) then (x2-x1, y2-y1) gives (0,4). The formula gives you 4, which is clearly the correct distance.
I'm very sad that you didn't go into the general formula for every metric (called the p-norm), how even Chebychev metric flows out when p=infinity, and how it relates to the superellipse.
Let me just say I have more videos planned:D
Scary Derivations of “The Metric Tensor”
Good explanation
thank you!
In the definition, I think the co domain should be non-negative real numbers not just positive
Indeed! I’ve found R^+ to be ambiguous, much like the naturals, as to whether it includes zero or not. Regardless, I mean it too. Actually, one need not specify a restriction on the codomain at all, that it is greater than or equal to zero as this follows from the axioms!
@@DrTrefor Yeah, I've seen the set of non-negative real numbers represented as R with a plus sign in the subscript, and this way to represent the positive real numbers, so I thought maybe it was an unintentional mistake
What about the "1" metric: D(p1, p2) = 0 if p1=p2, otherwise 1
Love this “discrete” one!
Respected sir may u start tutorial of statistics inference part.
1 day i was thinking of exact same thing and i discovered all of it by my own. I didn't knew this was a concept in maths. Same happened when I discovered sinx function as a infinite series and then i found out it was already done. I was wish I allowed to have mathematical education.
Nice video
My favorite metric is d(a,b)=1 if a!=b, else d(a,b)=0 iff a=b
Distance metrics are used also in clustering algorithms.
Should have talked about the discrete metric to really mess with people's minds
Upvotes 1.3% of views, comments 0.2% of views - look, fellow viewers, it's a good video so feed the algorithm.
All praise the algorithm!
Wouldn't Feynman ask what are ALL possible distances between any two points, and the answer would be, infinite?
In many contexts that would be true!
Well, no. That depends largely on the metric space. On a discrete metric space, there are only two possible distances between any two objects: 0, or 1. For this reason, I also like to call it the Boolean metric, even though no mathematician uses this name.
Confusticate and Bebother you and everybody else who brainwashed me into believing the triangle inequality, which left me high and dry when I tried to wrap my mind around the Minkowski metric.
The Minkowski metric is a metric tensor, not the metric of a metric space.
I'm not sure I understand why the notion of distance needs to be symmetric. If you define distance in terms of time, then distances traveled uphill could be longer than downhill.
This is quite a common way to measure distances in the real world, so I'm curious why it doesn't qualify as a metric.
Colloquially, but not in any rigorous setting. Two points don't get further apart because you walk slower.
Make video on group theory
I plan to actually!
@@DrTrefor thanks you are life saver
I love math
me too haha!
Ily❤❤❤
Exercise: is sin(A-B) a distance? :)
I'm gonna do an open ball of distance 2
You can come if you are either my friend, or a friend of a friend of mine
(topology joke)
Quadrance
That was dope !!!!!!! ¯\_(ツ)_/¯
❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤
But… you didn’t prove any results about metric spaces.
I would have watched this if you weren't so patronising , bordering on oleaginous.
Merch link 404s, link should have /collections/ not /pages/ (Try checking logged out)
Thank you!