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

2 жыл бұрын

Oh nice! Yes I’ve seen these types of word metrics before but didn’t know the evolution connection

@Alex_Deam

2 жыл бұрын

@@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.

@alexisbach

2 жыл бұрын

What is this nonsense about "evolutionary distance between organisms"? Don't you know that God created every living thing in seven days?

@angelmendez-rivera351

2 жыл бұрын

@@alexisbach That is scientifically false.

@lietpi

2 жыл бұрын

@@alexisbach what?

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.

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

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.

@God-gi9iu

2 жыл бұрын

That’s what I was thinking lol

@d1zputed23

Жыл бұрын

Well in endgames you gotta know how to move the king

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.

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.)

Really well explained video (no surprises there!), thanks for the great content! :)

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.

@DrTrefor

2 жыл бұрын

Ah yes! I know of this metric, but not in that context

@numberandfacts6174

2 жыл бұрын

@@DrTrefor Which value of n 1 to ∞ ∫ ( {x} / (x)^(n+1) ) dx question solution is =( 1/(n+1) ) Where {x} = ( x - floor(x) )

Every one looking for prof like you sir ❤️

Thanks for this video Trefor . It helped me to clear some of my thougts and fill the intuition.

@DrTrefor

2 жыл бұрын

You're very welcome!

This video is GOLD. Thank you!

I love math and your videos make me love it more then ever. Thank you!

@DrTrefor

2 жыл бұрын

I'm so happy to hear that! Math is awesome haha:D

@numberandfacts6174

2 жыл бұрын

@@DrTrefor Which value of n 1 to ∞ ∫ ( {x} / (x)^(n+1) ) dx question solution is =( 1/(n+1) ) Where {x} = ( x - floor(x) )

Hey, just wanted to let you know it is finals season and your calculus 3 videos are saving my life!

@DrTrefor

2 жыл бұрын

Awesome! Good luck!!

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...

@-minushyphen1two379

2 жыл бұрын

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?

@NoNameAtAll2

2 жыл бұрын

@@-minushyphen1two379 d(a1,b3)=1

@calvincrady

2 жыл бұрын

@@-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.

@calvincrady

2 жыл бұрын

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

@Synthetica9

Жыл бұрын

@@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

Sir your teaching process is Always best ...my concept always clear to see your vedio ..mind blowing ❤️

oh, this is a very good video. I was thinking about exactly this topic recently.

Wow, that's really well-explained! Thanks!

@DrTrefor

2 жыл бұрын

Glad you enjoyed it!

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.

@mastershooter64

2 жыл бұрын

technically it's a tensor field lol a tensor field over the entire manifold, so a "metric tensor field" lol riemannian manifolds ftw!!!

@buraianmath

2 жыл бұрын

Infact metric tensor induces a metric space on the manifold. The length between two points would be defined as length of shortest geodesic path between them.

@mastershooter64

2 жыл бұрын

@@buraianmath 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

@buraianmath

2 жыл бұрын

@@mastershooter64 ehm , could you please reread what I said? I meant the metric induced by metric tensor is given as the shortest geodesic distance between two points.

@mastershooter64

2 жыл бұрын

@@buraianmath 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

Beautiful explanation!

@DrTrefor

2 жыл бұрын

Thank you!

Those are awesome examples!

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.

@DrTrefor

2 жыл бұрын

I’m slowly in the background working on what a “calculus on manifolds” series might look like for this channel. Will see!

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.

@DrTrefor

Жыл бұрын

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.

Can you make more videos about metric space it would be very helpful

EXCELLENT EXPLANATION

@DrTrefor

2 жыл бұрын

Glad you think so!

Hello Trefor, Is this the first video of Metric Spaces course (i.e., playlist)?

@DrTrefor

2 жыл бұрын

I don't have it planned as a full course, but likely will do some follow ups:)

Nobody noticed his tshirt? Hippopotamus denoting hypotenuse as the latter is a bit difficult to pronounce and confusing sometimes. I want this t-shirt 🤩

I want that Hyppopotenuse shirt!

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

@DrTrefor

2 жыл бұрын

Not a course, exactly, but I have a sort of short thematic series of cool things in topology and geometry coming out.

@numberandfacts6174

2 жыл бұрын

@@DrTrefor Which value of n 1 to ∞ ∫ ( {x} / (x)^(n+1) ) dx question solution is =( 1/(n+1) ) Where {x} = ( x - floor(x) )

@rizalpurnawan3796

2 жыл бұрын

@@DrTrefor I can't wait to watch that

Great video! Are you starting a course on real analysis?

@DrTrefor

2 жыл бұрын

Sadly not, but I may jump around in some highlights

Wow that's amazing explanation

@DrTrefor

2 жыл бұрын

Glad you liked it!

@emc2884

2 жыл бұрын

@@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 .....

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.

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.

*laughs in pseudo-riemannian manifolds*

Good explanation

@DrTrefor

2 жыл бұрын

thank you!

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).

With Euclidean distances, we easily go from meters to square meters. With metrics (instead of distance), what is the analogous word used for area.

@DrTrefor

2 жыл бұрын

Still area, but what we might mean by area changes.

@rizalpurnawan3796

2 жыл бұрын

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.

@angelmendez-rivera351

2 жыл бұрын

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.

@rizalpurnawan3796

2 жыл бұрын

@@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.

@angelmendez-rivera351

2 жыл бұрын

@@rizalpurnawan3796 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.

Nice video

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?

@DrTrefor

2 жыл бұрын

I believe the Minkowski metric is know as a "pseudo-Euclidean metric"

@feynstein1004

2 жыл бұрын

@@DrTrefor That sounds interesting. Would you mind elaborating a bit?

@angelmendez-rivera351

2 жыл бұрын

@@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

2 жыл бұрын

@@angelmendez-rivera351 Ah okay. Thank explains it. Thank you for the reply 🙂

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.

@DrTrefor

2 жыл бұрын

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!

@mikip3242

2 жыл бұрын

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).

@masternobody1896

2 жыл бұрын

@@DrTrefor i fail linear algebra rip

@numberandfacts6174

2 жыл бұрын

@@DrTrefor Which value of n 1 to ∞ ∫ ( {x} / (x)^(n+1) ) dx question solution is =( 1/(n+1) ) Where {x} = ( x - floor(x) )

@angelmendez-rivera351

2 жыл бұрын

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.

Respected sir may u start tutorial of statistics inference part.

Is it possible to create a series about topological spaces?

@DrTrefor

2 жыл бұрын

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

@user-xv6cg1vf3w

2 жыл бұрын

@@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!

similar to distance definition are there other definitions for Density ?

@DrTrefor

2 жыл бұрын

Indeed! Density is defined per unit "area", so how you measure distances affect this.

@angelmendez-rivera351

2 жыл бұрын

Density is defined in terms of measure theory, so that is mathematically a very different kind of concept.

@meguellatiyounes8659

2 жыл бұрын

@@angelmendez-rivera351 thanks for this information

how about the spacetime interval? Is that also a metric?

@DrTrefor

2 жыл бұрын

It can indeed be phrased as a metric!

@angelmendez-rivera351

2 жыл бұрын

@@feynstein1004 No, that is incorrect. The Minkowski metric is a metric tensor, not the metric of a metric space. Those are mathematically different concepts.

Please make a series on vector space

@DrTrefor

2 жыл бұрын

I have one! Check out my linear algebra playlist

im 16 and your videos are making me really consider doing maths at university!

@DrTrefor

2 жыл бұрын

You totally should!!

@maxmusterman3371

2 жыл бұрын

Try some real uni examples first. Pure maths is.. well not for everyone. But if you like it, do it!

Is this useful before I start general relativity I mean will it give me basics?

@DrTrefor

2 жыл бұрын

Yes I think it is helpful for the metrics you will see in GR

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.

@DrTrefor

2 жыл бұрын

Ha I didn’t even notice! Silly editing presets:D

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".

Once all the pawns are gone, is the number of moves to go from chessboard state A to state B, a distance metric?

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.

@DrTrefor

2 жыл бұрын

Let me just say I have more videos planned:D

I'm still trying to make sense of this formula on chebyshev distance. Why do we need a max there?

@leeprice133

6 ай бұрын

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.

Scary Derivations of “The Metric Tensor”

So the Chebyshev metric is the one used by D&D 5e? 🤔

Which value of n 1 to ∞ ∫ ( {x} / (x)^(n+1) ) dx question solution is =( 1/(n+1) ) Where {x} = ( x - floor(x) )

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.

@DrTrefor

2 жыл бұрын

haha I love it:D

@aashsyed1277

2 жыл бұрын

@@DrTrefor There is a property which says d is always positive.

@DrTrefor

2 жыл бұрын

@@aashsyed1277 yup!

@Alex_Deam

2 жыл бұрын

@@aashsyed1277 That one can be derived from the others

@aashsyed1277

2 жыл бұрын

@@Alex_Deam ok

In the definition, I think the co domain should be non-negative real numbers not just positive

@DrTrefor

2 жыл бұрын

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!

@anshumanagrawal346

2 жыл бұрын

@@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

Should have talked about the discrete metric to really mess with people's minds

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.

@DrTrefor

2 жыл бұрын

Yes indeed on both accounts.

@godfreypigott

2 жыл бұрын

@@DrTrefor I look forward to saying to my cousins "By my metric, you don't make my will".

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.

@angelmendez-rivera351

2 жыл бұрын

The Minkowski metric is a metric tensor, not the metric of a metric space.

What about the "1" metric: D(p1, p2) = 0 if p1=p2, otherwise 1

@DrTrefor

2 жыл бұрын

Love this “discrete” one!

My favorite metric is d(a,b)=1 if a!=b, else d(a,b)=0 iff a=b

Ily❤❤❤

7:39 but a circle is not an open ball? one is the interior and one is the boundary

Make video on group theory

@DrTrefor

2 жыл бұрын

I plan to actually!

@motherisape

2 жыл бұрын

@@DrTrefor thanks you are life saver

Wouldn't Feynman ask what are ALL possible distances between any two points, and the answer would be, infinite?

@DrTrefor

2 жыл бұрын

In many contexts that would be true!

@angelmendez-rivera351

2 жыл бұрын

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.

8:27 actually it is a square. In a diamond the angles are not straight.

@leeprice133

6 ай бұрын

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.

Quadrance

Distance metrics are used also in clustering algorithms.

That was dope !!!!!!! ¯\_(ツ)_/¯

Exercise: is sin(A-B) a distance? :)

I love math

@DrTrefor

2 жыл бұрын

me too haha!

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.

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.

@leeprice133

6 ай бұрын

Colloquially, but not in any rigorous setting. Two points don't get further apart because you walk slower.

Upvotes 1.3% of views, comments 0.2% of views - look, fellow viewers, it's a good video so feed the algorithm.

@DrTrefor

2 жыл бұрын

All praise the algorithm!

❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤

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)

@DrTrefor

Жыл бұрын

Thank you!