Her: "The hard part is visualizing a 4-d cube......" Me: "Awesome!! I've always wanted to see this, can't wait for some fancy computer graphics which will......." Her: "See you next time!" Me: "........." You can't leave us like that!!!
This is one of the coolest math videos I have ever seen. So many connections between mathematical ideas, simplification of something so interesting and complicated, and good animations. Well done!
We "know" what it's like. We can calculate (more or less) everything about. But most humans can't visualize them. That's where computers come in handy. The general architecture of a computer is 1D. But it can do the maths for "any" dimension you want. It also offers ways of breaking down 4D objects into 3D space using slices or shadows that are then shown on a 2D screen. If you want to you can even solve 4D or 5D Rubik's Cubes (not actually Rubik's) on your computer.
YellowBunny I must disagree; but only about the "computers are 1D" part. Computers do math. That math can be any math we program in to it. Yes, the basic ADD mnemonic is 1D, but many of the SIMD instructions can be either 1D or full matrix operations. Modern GPU matrix operations don't care if you have a 1x32 array of numbers, 2x16 2D points, or a 4x4 list of 4D point. Most displays are limited to 2D representations, but shading tricks our brain into seeing 3D objects. Then there are the various 3D displays; yes they are 2D screens but the brain doesn't care.
The fundamental memory structure of computers is 1D. I'm not familiar with modern GPUs and stuff like that. I mostly know low-level programming. And even if it seems to be higher dimensional the computer is just faking it.
Pascal's triangle has a special place in my heart - I remember accidentally discovering it independently before I knew it existed (in my student years I was looking into new methods of data compression and came up with the exact same pattern while looking at combinations of binary numbers and their cardinalities)
The shapes produced by a diagonal hyperplane passing through a 5D-hypercube are a point, a 5-cell, a rectified 5-cell, and then the same sequence repeats backwards. This can be further generalized to the case of a k-dimensional hypercube. It should be a point, a (k-1) - simplex, a rectified (k-1) - simplex, a birectified (k-1) - simplex, a trirectified (k-1) - simplex, and so on up to a (k-2) - rectified (k-1) - simplex -- which is exactly the same as the (k-1) - simplex, and then the last shape is again a point.
And beyond! Imma just sit back and grab the popcorn now, and wait for angry replies from the folks who can't handle a mathematically incorrect Pixar allusion.
Imagine someone who has never heard of infinite series before and sees the video title "Dissecting Hypercubes with Pascal's Triangle" and they are just like: WHAAAAAAAT!?!?
For those wondering why n choose k appears.The plane we sweep is built in such a way that its equation is x1 +x2 + x3 + ... + xn = k, where we vary k from 0 to n as we "sweep". because all the vertices have either 1 or 0 as coordinates, this equation only has solutions for integer k, and each solution corresponds to choosing k coordinates to be 1 from the n available.
I feel there was something to say about how point plus segment is upright triangle when segment plus point is upside down triangle. The same influence of order of top elements seems to apply to the 6 point figure too.
Might I suggest POV-Ray for help visualising some of these objects? I know support for quaternions (easy 4d, almost like cheating) is built in, and there should be an octonion library available. And if there isn't, I would love the challenge of writing one.
More dissecting hypercubes: I learned "V choose 2 minus S" where V is vertices and S is sides recently, for finding the diagonals in a polygon. Checked and found it generalizes for all dimensions, so there are 16C2-32=88 diagonals in a hypercube.
Two quick notes: First, it is worth saying why the slicing hyperplanes cut out points corresponding to Pascal's triangle. Each stage of the hyperplane as it sweeps along should when it hits a vertex hit every vertex that is the same distance from (0,0...0), and that will correspond to having exactly the same number of 1s in the vertex's coordinates as you can check using the generalized distance formula. Second, since an n-dimensional cube has 2^n vertices, and one's slices must hit every vertex exactly once, one can recover from this the fact that each row of Pascal's triangle sums to a power of 2.
Arrgh! Missed animation opportunity! To get a "feel" for the geometric intersection of objects differing by one dimension, an animation does wonders, as it readily illustrates the "passing through" characteristic independently of the geometric characteristics of the separate intersections themselves. So, for this video, I would have swept the hyperplane continuously along the diagonal, ringing a bell and taking a snapshot whenever one or more vertices of the hypercube intersected the hyperplane. It was an old educational film from the 1950's (IIRC) that literally "opened up" the 4th dimension for me, showing the odd 3D shapes that appear, evolve, then disappear as a 3D hyperplane is swept through the 4D hypercube. It then iterated the process for ever higher dimensions, taking swept "slices of slices" to build a working awareness of higher dimensions using the more familiar 0-3 dimensions. When later, as a hobbyist, I struggled with the notion of string theory's "curled" dimensions, a similar process helped me understand where and how dimensions could "hide" by (crudely) envisioning how they could be "missed" by a swept hyperplane.
You have to appreciate PBS's commitment to theses series. I've been a subscriber to Scishow,Vsauce, Numberphile and etc but, I've always felt that I wasn't their targeted audience. They are all good but their incessant take at oversimplifying the content, even dialogues in a attempt to be more palatable to the masses,really demeaned the subject, and failed to capture, due to misunderstanding their audience, the core principles they are trying to convey. After all I wouldn't be watching mathematics on TH-cam,when I could be doing literally anything else, if I didn't deeply enjoy the subject.
I really agree. I first found space time, and that's one of the few science related channels that is understandable to people without a comprehensive math background but still challenges you to learn some real things and represents things pretty close to how they really are. It's strangely refreshing.
If you're looking at slices that aren't hitting the vertices, you can get even more interesting shapes. For example in between the three vertices cases in the good-old 3 dimensional cube, you get a hexagon.
Pascal's triangle can be generalized to higher dimensions, starting with Pascal's pyramid, and in general, Pascal's simplex. *_My question is: Is there a similar geometrical interpretation of higher dimensional Pascal's triangles?_* I tried to think of it myself but failed. However, I know that the outer layer of a Pascal's pyramid is a Pascal's triangle, so a hypothesis would be that Pascal's pyramid describes some geometrical object that looks like a hypercube in three different axes, and then 'something entirely different' along the other axes... I guess that's enough geometry for today...
Yeah, I googled it but couldn't find anything on it. Another thing is, I'm not sure how to interpret the inner layers/walls of a Pascal's pyramid. And they also increase for each successive step... Haha, getting totally confused now! xD
Ermude10 I'm not sure if there's a simple physical analogy, like there is between slicing and pascals triangle, but it should be possible to set up a function from one to the other that can be generalised to 3d space.
The 3rd level of Pascal's pyramid has a 6 in the center. As the plane passes through the cube, between the 2 triangles, its intersection with the cube is a hexagon. Mere coincidence? My almost certainly flawed intuition based on Henry Segerman's 3D shadows of the tesseract tells me that as a cube passes through the tesseract it will intersect it in a figure with 12 vertices, coinciding with the 12's in the 4th level of the pyramid. This is all just musing at this point. I have nothing concrete.
I was thinking about it. I haven't figured it out, but I suspect that at least the second slice is related to edges. The sum of the numbers in the second slice is equal to the number of edges (e) for that n dimensional cube. The second slice can be described as n choose 1 then choose m. (the first being n choose 0 then choose m which is pascal's triangle) For a line, e=1. 1 choose 1 then choose 0 = 1 For a square, e=4, 2 choose 1 then choose 0 = 2, 2 choose 1 then choose 1 = 2. 2+2=4. For a cube, e=12, 3 choose 1 then choose 0=3, 3 choose 1 then choose 1 = 6, 3 choose 1 then choose 2 = 3. 3+6+3 = 12. For the tesseract e=32. 4 c 1 then c 0 = 4, 4 c 1 then c 1 = 12, 4 c 1 then c 2 = 12, 4 c 1 then c 3 = 4. 4+12+12+4 = 32. For the 5D-hypercube e=80. 5c1tc0=5, 5c1tc1=20, 5c1tc2=30, 5c1tc3=20, 5c1tc4=5. 5+20+30+20+5 = 80 At first I thought that it had something to do with sweeping n-m dimensional objects instead of just n-1 but that didn't match the data. I hope that this helps.
Multi-dimensional cubes have another strange property: their diameters get arbitrarily _large_ with increasing dimension. For example, the regular 3D cube with edge length 1 cm looks about the same size (the diagonal is slightly longer but not by much). But a 10,000D cube with edge 1 cm has diameter 1 m! OTOH spheres always have the same diameter (equal to twice the radius) in every dimension.
I think it’s important to consider how visualization works. In a purely mathematical sense, you can create this representation by moving point by point within the space. To make it easy for some of you when you visualizing the “landscape,” consider your viewpoint (dimension), explore (move) in your space and record where you are, and don’t connect all the spaces at once; only the spaces nearby your current viewpoint. And there you have your visualization. When you try and shape higher dimensions, it changes depending on your viewpoint. So don’t focus on all your changes at once because they may not make sense to the human eye
4:20 the number of "1" increases: first of all, there's no 1 (the point), then it increase just to a single one ( {, , } ), then two ( {, , } ) then all of three (the point). I think it's just thanks to the regularity of a cube. Is there a clever correlation with that? I guess: yes, the "Pascal's triangle" stuff and everything else She pointed out.
Here 10:59 you make a triangle where the base has length sqrt(2) and the other two sides are sqrt(1.5). But the result should be an equilateral triangle. You should simply draw a line perpendicular to the old figure, and add a point where the new edges will be as long as the old ones.
Pascals triangle can also be made by repeated convolutions of the vector [1 1]. In a way what she was doing with the hyper-planes is a kind of convolution.
That second triangle and (later) tetrahedron in each row should really be upside down with respect to the first ones. Familiar fact: if you dilate a triangle, it breaks up into triangles and upside-down triangles. Less familiar: if you dilate a tetrahedron, it breaks up into tetrahedra, upside-down tetrahedra, and... octahedra. (Try the tetrahedra with edge-lengths 2.)
Really cool. Thanks for explaining so clearly. Please use an editor so you don't misuse words such as "comprise" (it's "composed of" or "comprises" but not both) and "infamous", and I promise not to mix up vertex and vortex.
The last digits of even powers of 2 are: 2 4 8 6 /cyclebacktostart 2 (2) 4 (4) 8 (8) 16 (6) 32 (/cyclebacktostart) Plus, Pascal's triangle to me best describes powers of 11.
What I love about this is that since all the resulting intersections are regular polytopes, and Pascal's triangle tells you how to construct them, Pascal's triangle literally generates regular polytopes. So when Wikipedia says "In five and more dimensions, there are exactly three regular polytopes", Pascal's triangle begs to differ. en.m.wikipedia.org/wiki/Regular_polytope#Higher-dimensional_polytopes
Bertie Blue Aren't they though? All the cross sections presented here were regular. All the side lengths are equal, and all the faces are equal. I suppose it's difficult figuring out what a tetrahedron plus an octahedron looks like. But, there is a way. After you cut the tesseract, the piece between the tetrahedron and the octahedron is the shape we're after.
When dragging the 3D cube through 4D what point of reference on the 3D cube are you using to match the two together. With the first number on the 4th row of pascals triangle, it is a point, which means that only one point from the 3D cube and the 4D cube are touching. Because of this it can be assumed the point of reference on the 3D cube is a side face facing towards the middle of the 4D cube, but if this point of reference is consistent though the entire procedure then the last vertices would not be achievable unless for the last step, the point of reference is changed to the opposite face of the 3D cube, which to me seems wrong because of inconsistent step length and segmentation. could you clarify? As long as changing the point of reference on the 3D cube is allowed i get it, if not this seems like a flaw. Also if the point of reference is using the center of the 3D cube then it would need to change size as it passed through the 4D cube for the initial and final vertices to be achievable. sorry if im missing something here. Maybe Pascals pyramid could be utilized here?
A bit nit picky, but when we place down our hyperplane triangles for 3d on Pascal's Triangle, shouldn't we flip the orientation to correspond to the direction of the slices? So one triangle should point up while the other should point down. I would assume the same thing is applicalical for the 4d slices and would also be flipped.
Let's see. Correct me if I'm wrong, the cross section is always in form of regular polytopes right? And there's only 3 polytopes in dimension 5 and above, so how do we identified which one to use? is it the cross polytopes or the simplex polytopes?
Ralph Strocchia I forgot that you "only have two favorites and so will only pick the same two over and over" and literally thought you meant "if asked to choose favorite two, you will pick all of the puppies" lmao
The same way that a 2D (planar) figure can be formed from points that occupy 3D space. The key is that they all lie on that "hyperplane" in 4D (the one we're using to "slice" through the tesseract), which is itself, a 3D space; so any figure entirely within it, has to have 3 or fewer dimensions.
Just to help explain more, you can take a 2d XY plane and all the shapes there and say they have a Z coordinate of 0. Of course, now you could lift or lower them "off" their home 2d plane, changing their Z coordinate, giving them a whole new 2d plane. You could also twist them, changing all the coordinates. Similarly, in a 3d space, all objects would have a W coordinate of 0. But then you can lift or lower them "out", changing the W coordinate, putting them into new 3d spaces. Then you can twist and turn them, changing all the coordinates, yet still retaining a 3d object floating in 4d space. Has your brain melted yet? This is about the point where mine does.
The more I watch, the higher on a logarithmic scale, my non understanding status moves, till it ultimately engulfs my entire limited universe in a black hole.😅
First thanks, as always, for this awesome video. Please could you mak a series on Hilbert problems and millenium problems. I can't find any decent video on youtube that treats any of the problems like you do. I think it would be a great series and fun to watch!
Hey I have a quick question, for the octahedron at 11:50 why do we only connect each point on the lower plane to two others on the upper plane and not all points on the upper plane? I'm guessing it's simplified since the missing lines just go through the middle of the shape anyway and don't change the shape when it's filled in... but for higher dimensions it would be useful to know otherwise I cant tell what point connects to what, especially because I cant really visualize it xD. Amazing vid by the way :), love pbs!!
Notice the number of vertices with sum of coordinates in each cube forms a Pascal triangle Dimensions sum of coordiantes-number of such coordinates 1D 0-1 2D 0-1 1-2 2-1 3D 0-1 1-3 2-3 3-1 4D 0-1 1-4 2-6 3-4 4-1
i have a big problem the 5d cube slice has a 10 vertex cross section. there's no shapes in 4d that has exactly 10 vertexes that is Pechora. are we missing a 7th shape
With creating the third (/second if you're counting from 0) 'slice' of the tesseract, why are the triangles 'reversed', instead of having them directly above each other (meaning, one is just the other, only moved up/down). I see that it produces a more regular shape that way, but what's the more rigorous reason behind it? Is this because the way the two triangles (that were slices of the 3d cube) were made in 'reverse order': the first from a point and a line, and the second from a line and a point?
There's an easy way to visualize the slices. Imagine taking a square and slicing it by the hyper plane (a line in this case). The intersections are a point, a line, and a point. But what happened to the square? It got cut into a point, a triangle, another triangle, and a point. Hmm, isn't that familiar? Let's step up a dimension. Take a cube and slice it by the hyperplane. The pieces are a point, a tetrahedron, an octahedron, you get the idea. This is no coincidence: the orthogonal projections of nth dimensional cubes are n-1 dimensional cubes, and the orthogonal projections of the hyperplane intersections are the pieces after the slice on the projection.
When sweeping the 2D plane along the diagonal of the 3D cube, why should it intersect (1,0,0), (0,1,0), and (0,0,1) at the same time? Is there a way to show this, and show that it applies to all n-dimensional cubes as well?
Question: is it not possible to predict the shapes you produce just from symmetry considerations? All of those shapes in all of the dimensions you showed were the most symmetric distribution of vertices available, no?
One observation... When finding the shape of a 2D slice through the 3D cube, a 2D plane is placed above another 2D plane with the point from the first column adding to the line of the second column to create the 2D shape of the slice at that point (a triangle). Shouldn't this really be a 1D plane (a line) with a point placed above another 1D plane (another line) with a line segment to create the 2D slice? This matches better with the idea that when constructing the 3D slices of the 4D cube, that each shape is placed on a 2D plane, one above the other which was shown in the video.
What four dimentional shape is the slice made by a four dimentional hyperplane through the five dimentional hypercube through with the 5C2 and 5C3 verticies. Its the geometric addition of a regular tetrahedron and a regular octahedron but no regular 4d polytopes have tem verticies.
Cheers, I looked it up on wikipedia; "can be positioned on a hyperplane in 5-space as permutations of (0,0,0,1,1) or (0,0,1,1,1)" so it seems to be the right one. I'm assuming the (0,0,0,0,1) and (0,1,1,1,1) polytopes are regular 5-cells.
I mean, the fact that (1,0,0,0) (0,1,0,0) etc. makes a regular tetrahedron is pretty obvious because they are all the same distance from each other (namely, root 2), and the tetrahedron is the only 3d shape like that
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Her: "The hard part is visualizing a 4-d cube......"
Me: "Awesome!! I've always wanted to see this, can't wait for some fancy computer graphics which will......."
Her: "See you next time!"
Me: "........."
You can't leave us like that!!!
Well, i guess because visualizing 4-d objects are impossible, even for a computer :'(
I also got really excited. A 3d animation might have worked?
@Per-Axel Skogsberg, good luck with 3d animations on TH-cam, an inherently 2d platform.
Haha 😂
See the end of my video. It is not impossible - th-cam.com/video/KuXnrg1YpiY/w-d-xo.html
I'm so sad this show ended. It is still wonderful.
PBS has really upped their game with these webshows
I'm so glad I could be the prime subject of this video...
now time for the fifth dimension! I shall win once again!
Tesseract A I'm sorry to ruin your hype but she was talking about tesseracts in general and not about you, mr A
Let A represent an arbitrary tesseract....
Hypercubes, eh? Well I didn't expect Pascal's Triangle to show up he-
NOBODY EXPECTS PASCAL'S TRIANGLE
ITS MAIN WEAPON IS SURPRISE
Nice Monty python reference
I never thought about this before, but... a computational Byte is the set of all vertices of a 8-dimensional hypercube! Mind blown!
This is one of the coolest math videos I have ever seen. So many connections between mathematical ideas, simplification of something so interesting and complicated, and good animations. Well done!
That was the best video about tesseracts i have ever seen. This is why I love maths man
Wrong. Check out this video --> th-cam.com/video/4URVJ3D8e8k/w-d-xo.html . Better explanation.
I'm sad you didn't sweep the hyperplane and show the resulting slice as an animation :c
Shreyas Misra Tricky, yes, but I'm sure it's possible
We "know" what it's like. We can calculate (more or less) everything about. But most humans can't visualize them. That's where computers come in handy. The general architecture of a computer is 1D. But it can do the maths for "any" dimension you want. It also offers ways of breaking down 4D objects into 3D space using slices or shadows that are then shown on a 2D screen. If you want to you can even solve 4D or 5D Rubik's Cubes (not actually Rubik's) on your computer.
YellowBunny I must disagree; but only about the "computers are 1D" part. Computers do math. That math can be any math we program in to it. Yes, the basic ADD mnemonic is 1D, but many of the SIMD instructions can be either 1D or full matrix operations. Modern GPU matrix operations don't care if you have a 1x32 array of numbers, 2x16 2D points, or a 4x4 list of 4D point.
Most displays are limited to 2D representations, but shading tricks our brain into seeing 3D objects. Then there are the various 3D displays; yes they are 2D screens but the brain doesn't care.
Search "Hypercube on a vertex - timeline of cross-sections" by pevogam
The fundamental memory structure of computers is 1D. I'm not familiar with modern GPUs and stuff like that. I mostly know low-level programming. And even if it seems to be higher dimensional the computer is just faking it.
Pascal's triangle has a special place in my heart - I remember accidentally discovering it independently before I knew it existed (in my student years I was looking into new methods of data compression and came up with the exact same pattern while looking at combinations of binary numbers and their cardinalities)
That's really cool. I love how pervasive it is! Also, is that the ace flag in your pfp?
The shapes produced by a diagonal hyperplane passing through a 5D-hypercube are a point, a 5-cell, a rectified 5-cell, and then the same sequence repeats backwards.
This can be further generalized to the case of a k-dimensional hypercube. It should be a point, a (k-1) - simplex, a rectified (k-1) - simplex, a birectified (k-1) - simplex, a trirectified (k-1) - simplex, and so on up to a (k-2) - rectified (k-1) - simplex -- which is exactly the same as the (k-1) - simplex, and then the last shape is again a point.
Congrats to a well deserved 100k subs! Now, toward infinity!
Countable or uncountable infinity?
Daniel Shapiro
I'm blue
Daniel Shapiro
sheep cow cow sheep guy man sandwich
And beyond!
Imma just sit back and grab the popcorn now, and wait for angry replies from the folks who can't handle a mathematically incorrect Pixar allusion.
Imagine someone who has never heard of infinite series before and sees the video title "Dissecting Hypercubes with Pascal's Triangle" and they are just like: WHAAAAAAAT!?!?
For those wondering why n choose k appears.The plane we sweep is built in such a way that its equation is x1 +x2 + x3 + ... + xn = k, where we vary k from 0 to n as we "sweep". because all the vertices have either 1 or 0 as coordinates, this equation only has solutions for integer k, and each solution corresponds to choosing k coordinates to be 1 from the n available.
Exactly right! (You beat me to this.)
This is exactly the missing piece of information I was wanting from this video. Thank you.
Too bad they didn't finish by animating between the final shapes, like a 4D MRI.
I wish they did.
In Dungeons and Dragons, those are D4s and D8s. D4s hurt like crazy if you step on them.
Great video by the way.
I feel there was something to say about how point plus segment is upright triangle when segment plus point is upside down triangle. The same influence of order of top elements seems to apply to the 6 point figure too.
YOU WILL NEVER HAVE A BETTER VIDEO STRICTLY CONTAINING ONLY MATH THAN THIS ONE!
I can’t get enough of this stuff. 4D is the key.
excellent. very clear and i found it easy to understand. thank you.
YES!!! I realized this a few years ago and it blew my mind!
Might I suggest POV-Ray for help visualising some of these objects? I know support for quaternions (easy 4d, almost like cheating) is built in, and there should be an octonion library available.
And if there isn't, I would love the challenge of writing one.
GODDAMNIT I was hoping to see the hypercube animation! We've all been bamboozled!
More dissecting hypercubes: I learned "V choose 2 minus S" where V is vertices and S is sides recently, for finding the diagonals in a polygon. Checked and found it generalizes for all dimensions, so there are 16C2-32=88 diagonals in a hypercube.
very nice explanation with pascal triangle
This is absolutely awesome!
Absolutely brilliant!
Thank you to PBS Infinite Series for slowing the rate at which my education rots out of my brain.
Two quick notes: First, it is worth saying why the slicing hyperplanes cut out points corresponding to Pascal's triangle. Each stage of the hyperplane as it sweeps along should when it hits a vertex hit every vertex that is the same distance from (0,0...0), and that will correspond to having exactly the same number of 1s in the vertex's coordinates as you can check using the generalized distance formula.
Second, since an n-dimensional cube has 2^n vertices, and one's slices must hit every vertex exactly once, one can recover from this the fact that each row of Pascal's triangle sums to a power of 2.
Unfortunatly I can't upvote this videos as many times as I would like ^^
Regards from France.
If you had shown hos those hyperplanes combined for a tesseract you would have nailed the video. Anyway, great episode!
Nicely Explained!
I absolutely love this channel and Kelsey might be the mathematician that could explain math to all those math hating students. Got to admire her.
Arrgh! Missed animation opportunity!
To get a "feel" for the geometric intersection of objects differing by one dimension, an animation does wonders, as it readily illustrates the "passing through" characteristic independently of the geometric characteristics of the separate intersections themselves.
So, for this video, I would have swept the hyperplane continuously along the diagonal, ringing a bell and taking a snapshot whenever one or more vertices of the hypercube intersected the hyperplane.
It was an old educational film from the 1950's (IIRC) that literally "opened up" the 4th dimension for me, showing the odd 3D shapes that appear, evolve, then disappear as a 3D hyperplane is swept through the 4D hypercube. It then iterated the process for ever higher dimensions, taking swept "slices of slices" to build a working awareness of higher dimensions using the more familiar 0-3 dimensions.
When later, as a hobbyist, I struggled with the notion of string theory's "curled" dimensions, a similar process helped me understand where and how dimensions could "hide" by (crudely) envisioning how they could be "missed" by a swept hyperplane.
You have to appreciate PBS's commitment to theses series.
I've been a subscriber to Scishow,Vsauce, Numberphile and etc but, I've always felt that I wasn't their targeted audience.
They are all good but their incessant take at oversimplifying the content, even dialogues in a attempt to be more palatable to the masses,really demeaned the subject, and failed to capture, due to misunderstanding their audience, the core principles they are trying to convey.
After all I wouldn't be watching mathematics on TH-cam,when I could be doing literally anything else, if I didn't deeply enjoy the subject.
You might enjoy 3blue1brown and mathologer.
I really agree. I first found space time, and that's one of the few science related channels that is understandable to people without a comprehensive math background but still challenges you to learn some real things and represents things pretty close to how they really are. It's strangely refreshing.
@Definitely a George Soros funded bot
Shut up. You have no power here this comment section _IS_ nerds
If you're looking at slices that aren't hitting the vertices, you can get even more interesting shapes. For example in between the three vertices cases in the good-old 3 dimensional cube, you get a hexagon.
Great episode. Thanks
Pascal's triangle can be generalized to higher dimensions, starting with Pascal's pyramid, and in general, Pascal's simplex.
*_My question is: Is there a similar geometrical interpretation of higher dimensional Pascal's triangles?_*
I tried to think of it myself but failed. However, I know that the outer layer of a Pascal's pyramid is a Pascal's triangle, so a hypothesis would be that Pascal's pyramid describes some geometrical object that looks like a hypercube in three different axes, and then 'something entirely different' along the other axes...
I guess that's enough geometry for today...
Interesting thought! Really intrigued to know the answer now...
Yeah, I googled it but couldn't find anything on it. Another thing is, I'm not sure how to interpret the inner layers/walls of a Pascal's pyramid. And they also increase for each successive step... Haha, getting totally confused now! xD
Ermude10 I'm not sure if there's a simple physical analogy, like there is between slicing and pascals triangle, but it should be possible to set up a function from one to the other that can be generalised to 3d space.
The 3rd level of Pascal's pyramid has a 6 in the center. As the plane passes through the cube, between the 2 triangles, its intersection with the cube is a hexagon. Mere coincidence?
My almost certainly flawed intuition based on Henry Segerman's 3D shadows of the tesseract tells me that as a cube passes through the tesseract it will intersect it in a figure with 12 vertices, coinciding with the 12's in the 4th level of the pyramid.
This is all just musing at this point. I have nothing concrete.
I was thinking about it. I haven't figured it out, but I suspect that at least the second slice is related to edges.
The sum of the numbers in the second slice is equal to the number of edges (e) for that n dimensional cube.
The second slice can be described as n choose 1 then choose m. (the first being n choose 0 then choose m which is pascal's triangle)
For a line, e=1. 1 choose 1 then choose 0 = 1
For a square, e=4, 2 choose 1 then choose 0 = 2, 2 choose 1 then choose 1 = 2. 2+2=4.
For a cube, e=12, 3 choose 1 then choose 0=3, 3 choose 1 then choose 1 = 6, 3 choose 1 then choose 2 = 3. 3+6+3 = 12.
For the tesseract e=32. 4 c 1 then c 0 = 4, 4 c 1 then c 1 = 12, 4 c 1 then c 2 = 12, 4 c 1 then c 3 = 4. 4+12+12+4 = 32.
For the 5D-hypercube e=80. 5c1tc0=5, 5c1tc1=20, 5c1tc2=30, 5c1tc3=20, 5c1tc4=5. 5+20+30+20+5 = 80
At first I thought that it had something to do with sweeping n-m dimensional objects instead of just n-1 but that didn't match the data.
I hope that this helps.
at 5:05 i just start smiling like stupid, this picture just makes me so happy lol
Great video as always. Can you please talk about fractals and perhaps the Mandelbrot set?
Multi-dimensional cubes have another strange property: their diameters get arbitrarily _large_ with increasing dimension. For example, the regular 3D cube with edge length 1 cm looks about the same size (the diagonal is slightly longer but not by much). But a 10,000D cube with edge 1 cm has diameter 1 m! OTOH spheres always have the same diameter (equal to twice the radius) in every dimension.
I immediately like the host. I was so afraid of it being some douchy guy, but she seems so cool?
awesome episode! but you really need to draw the second triangle upside down on the summery screen.
I think it’s important to consider how visualization works. In
a purely mathematical sense, you can create this representation by moving point
by point within the space. To make it easy for some of you when you visualizing
the “landscape,” consider your viewpoint (dimension), explore (move) in your
space and record where you are, and don’t connect all the spaces at once; only
the spaces nearby your current viewpoint. And there you have your
visualization. When you try and shape higher dimensions, it changes depending
on your viewpoint. So don’t focus on all your changes at once because they may
not make sense to the human eye
That was so tight
4:20
the number of "1" increases: first of all, there's no 1 (the point), then it increase just to a single one ( {, , } ), then two ( {, , } ) then all of three (the point). I think it's just thanks to the regularity of a cube. Is there a clever correlation with that? I guess: yes, the "Pascal's triangle" stuff and everything else She pointed out.
Here 10:59 you make a triangle where the base has length sqrt(2) and the other two sides are sqrt(1.5). But the result should be an equilateral triangle. You should simply draw a line perpendicular to the old figure, and add a point where the new edges will be as long as the old ones.
This is madness!
You blew my mind again! :) Too bad you couldn't show the tetrahedon and the octahedron in the tesseract. :(
Love these!
Fascinating!
Pascals triangle can also be made by repeated convolutions of the vector [1 1]. In a way what she was doing with the hyper-planes is a kind of convolution.
That second triangle and (later) tetrahedron in each row should really be upside down with respect to the first ones.
Familiar fact: if you dilate a triangle, it breaks up into triangles and upside-down triangles.
Less familiar: if you dilate a tetrahedron, it breaks up into tetrahedra, upside-down tetrahedra, and... octahedra. (Try the tetrahedra with edge-lengths 2.)
Brilliant!
wow! mind blowing !
This is amazing
what shapes do the slices make as dimensions approach infinity?
Really cool. Thanks for explaining so clearly. Please use an editor so you don't misuse words such as "comprise" (it's "composed of" or "comprises" but not both) and "infamous", and I promise not to mix up vertex and vortex.
The last digits of even powers of 2 are:
2 4 8 6 /cyclebacktostart
2 (2) 4 (4) 8 (8) 16 (6) 32 (/cyclebacktostart)
Plus, Pascal's triangle to me best describes powers of 11.
Is there any graph theoretic additive rule for those vertices addition for the higher dimension ? (as it is not clear from the video )
What I love about this is that since all the resulting intersections are regular polytopes, and Pascal's triangle tells you how to construct them, Pascal's triangle literally generates regular polytopes. So when Wikipedia says "In five and more dimensions, there are exactly three regular polytopes", Pascal's triangle begs to differ.
en.m.wikipedia.org/wiki/Regular_polytope#Higher-dimensional_polytopes
Bertie Blue Aren't they though? All the cross sections presented here were regular. All the side lengths are equal, and all the faces are equal. I suppose it's difficult figuring out what a tetrahedron plus an octahedron looks like. But, there is a way. After you cut the tesseract, the piece between the tetrahedron and the octahedron is the shape we're after.
You guys are awesomely awesome. :)
Love this show ❤!
It's like a powerpoint. Use the motion of video as the main visual tool to communicate change, patterns, similarity and differences.
When dragging the 3D cube through 4D what point of reference on the 3D cube are you using to match the two together. With the first number on the 4th row of pascals triangle, it is a point, which means that only one point from the 3D cube and the 4D cube are touching. Because of this it can be assumed the point of reference on the 3D cube is a side face facing towards the middle of the 4D cube, but if this point of reference is consistent though the entire procedure then the last vertices would not be achievable unless for the last step, the point of reference is changed to the opposite face of the 3D cube, which to me seems wrong because of inconsistent step length and segmentation. could you clarify? As long as changing the point of reference on the 3D cube is allowed i get it, if not this seems like a flaw. Also if the point of reference is using the center of the 3D cube then it would need to change size as it passed through the 4D cube for the initial and final vertices to be achievable. sorry if im missing something here. Maybe Pascals pyramid could be utilized here?
A bit nit picky, but when we place down our hyperplane triangles for 3d on Pascal's Triangle, shouldn't we flip the orientation to correspond to the direction of the slices? So one triangle should point up while the other should point down. I would assume the same thing is applicalical for the 4d slices and would also be flipped.
damn! so close yet so far. imagination part is hard.
ironic tho how everything starts with 1 and ends with 1.
Interesting, helps me imagine a little bit more the 4D Hyperspace
what a great tool. thank you. I might see if i can figure out the 3 dimensional shadow of a 5 dimensional cube
Let's see. Correct me if I'm wrong, the cross section is always in form of regular polytopes right? And there's only 3 polytopes in dimension 5 and above, so how do we identified which one to use? is it the cross polytopes or the simplex polytopes?
5:53 False. If you have 5 puppies and need to choose your favorite 2, there is only one way to do so.
Let them battle it out until only two are left, thus leaving the choice to chance?
+Ralph -- Technically, yes, but it's pretty clear what she meant.
I just found it funny, and wondered if anyone else caught the slight misuse of language.
Ralph Strocchia I forgot that you "only have two favorites and so will only pick the same two over and over" and literally thought you meant "if asked to choose favorite two, you will pick all of the puppies" lmao
I have a question.
How can a 3D solid (e.g. the tetrahedron) be formed from points which are described by 4D co-ordinates?
Thanks in advance :)
The same way that a 2D (planar) figure can be formed from points that occupy 3D space.
The key is that they all lie on that "hyperplane" in 4D (the one we're using to "slice" through the tesseract), which is itself, a 3D space; so any figure entirely within it, has to have 3 or fewer dimensions.
Thank you - that makes sense to me now :)
Just to help explain more, you can take a 2d XY plane and all the shapes there and say they have a Z coordinate of 0. Of course, now you could lift or lower them "off" their home 2d plane, changing their Z coordinate, giving them a whole new 2d plane. You could also twist them, changing all the coordinates. Similarly, in a 3d space, all objects would have a W coordinate of 0. But then you can lift or lower them "out", changing the W coordinate, putting them into new 3d spaces. Then you can twist and turn them, changing all the coordinates, yet still retaining a 3d object floating in 4d space. Has your brain melted yet? This is about the point where mine does.
The more I watch, the higher on a logarithmic scale, my non understanding status moves, till it ultimately engulfs my entire limited universe in a black hole.😅
Thank-you!
Great video! Keep it up!
First thanks, as always, for this awesome video. Please could you mak a series on Hilbert problems and millenium problems. I can't find any decent video on youtube that treats any of the problems like you do. I think it would be a great series and fun to watch!
Hey I have a quick question, for the octahedron at 11:50 why do we only connect each point on the lower plane to two others on the upper plane and not all points on the upper plane?
I'm guessing it's simplified since the missing lines just go through the middle of the shape anyway and don't change the shape when it's filled in... but for higher dimensions it would be useful to know otherwise I cant tell what point connects to what, especially because I cant really visualize it xD.
Amazing vid by the way :), love pbs!!
Notice the number of vertices with sum of coordinates in each cube forms a Pascal triangle
Dimensions sum of coordiantes-number of such coordinates
1D 0-1
2D 0-1 1-2 2-1
3D 0-1 1-3 2-3 3-1
4D 0-1 1-4 2-6 3-4 4-1
i have a big problem the 5d cube slice has a 10 vertex cross section. there's no shapes in 4d that has exactly 10 vertexes that is Pechora. are we missing a 7th shape
Try dungeons and dragons dice. 10 sider...
Seeing that higher dimensions only have 3 regular hyper-hedron, what would you look for next in the shapes of (n choose k) for k = [2,n-2]
With creating the third (/second if you're counting from 0) 'slice' of the tesseract, why are the triangles 'reversed', instead of having them directly above each other (meaning, one is just the other, only moved up/down). I see that it produces a more regular shape that way, but what's the more rigorous reason behind it? Is this because the way the two triangles (that were slices of the 3d cube) were made in 'reverse order': the first from a point and a line, and the second from a line and a point?
There's an easy way to visualize the slices. Imagine taking a square and slicing it by the hyper plane (a line in this case). The intersections are a point, a line, and a point. But what happened to the square? It got cut into a point, a triangle, another triangle, and a point. Hmm, isn't that familiar? Let's step up a dimension. Take a cube and slice it by the hyperplane. The pieces are a point, a tetrahedron, an octahedron, you get the idea. This is no coincidence: the orthogonal projections of nth dimensional cubes are n-1 dimensional cubes, and the orthogonal projections of the hyperplane intersections are the pieces after the slice on the projection.
When sweeping the 2D plane along the diagonal of the 3D cube, why should it intersect (1,0,0), (0,1,0), and (0,0,1) at the same time? Is there a way to show this, and show that it applies to all n-dimensional cubes as well?
What's the relation between dissecting hypercubes diagonally by the hyperplane and platonic shapes and polytopes for higher dimensions
Question: is it not possible to predict the shapes you produce just from symmetry considerations? All of those shapes in all of the dimensions you showed were the most symmetric distribution of vertices available, no?
My current math professor said he had a professor in college who was born blind and had no problem visualizing objects over three dimensions.
One observation... When finding the shape of a 2D slice through the 3D cube, a 2D plane is placed above another 2D plane with the point from the first column adding to the line of the second column to create the 2D shape of the slice at that point (a triangle). Shouldn't this really be a 1D plane (a line) with a point placed above another 1D plane (another line) with a line segment to create the 2D slice? This matches better with the idea that when constructing the 3D slices of the 4D cube, that each shape is placed on a 2D plane, one above the other which was shown in the video.
What four dimentional shape is the slice made by a four dimentional hyperplane through the five dimentional hypercube through with the 5C2 and 5C3 verticies. Its the geometric addition of a regular tetrahedron and a regular octahedron but no regular 4d polytopes have tem verticies.
The rectified 5-cell has 10 vertices! It's a semirregular polytope though
Cheers, I looked it up on wikipedia; "can be positioned on a hyperplane in 5-space as permutations of (0,0,0,1,1) or (0,0,1,1,1)" so it seems to be the right one. I'm assuming the (0,0,0,0,1) and (0,1,1,1,1) polytopes are regular 5-cells.
Pascal's triangle is also known by earlier Indian mathematician Pingla as Meru Prastara
I mean, the fact that (1,0,0,0) (0,1,0,0) etc. makes a regular tetrahedron is pretty obvious because they are all the same distance from each other (namely, root 2), and the tetrahedron is the only 3d shape like that
Well, that was interesting !
"what's a hyperplane?" a spaceship.
What's the background music that starts at 3:33?
Does this form an infinite group under the operation of 'geometric addition' you explained? Would be quite interesting!
Can you do Arrow's impossibility theorem? Would be interesting to know what kind(s)/how math proves such statement.
amazing channel
How does this apply to the negative portion of Pascal's triangle?
I have trouble with dimensions higher than 3. But I do love PT.
I understand thanks!
Could a cube (generally) be described as a unit subspace?