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!!!

@shubhamshinde3593

7 жыл бұрын

Well, i guess because visualizing 4-d objects are impossible, even for a computer :'(

@per-axelskogsberg3861

7 жыл бұрын

I also got really excited. A 3d animation might have worked?

@jeffirwin7862

7 жыл бұрын

@Per-Axel Skogsberg, good luck with 3d animations on KZread, an inherently 2d platform.

@per-axelskogsberg3861

7 жыл бұрын

Haha 😂

@samuraimath1864

7 жыл бұрын

See the end of my video. It is not impossible - kzread.info/dash/bejne/famM0NSgYbzWmbw.html

I'm so sad this show ended. It is still wonderful.

I'm so glad I could be the prime subject of this video...

@Donglator

7 жыл бұрын

now time for the fifth dimension! I shall win once again!

@miksurankaviita

7 жыл бұрын

Tesseract A I'm sorry to ruin your hype but she was talking about tesseracts in general and not about you, mr A

@justinward3679

7 жыл бұрын

Let A represent an arbitrary tesseract....

PBS has really upped their game with these webshows

Hypercubes, eh? Well I didn't expect Pascal's Triangle to show up he- NOBODY EXPECTS PASCAL'S TRIANGLE ITS MAIN WEAPON IS SURPRISE

@Ticbow

5 жыл бұрын

Nice Monty python reference

I'm sad you didn't sweep the hyperplane and show the resulting slice as an animation :c

@iwikal

7 жыл бұрын

Shreyas Misra Tricky, yes, but I'm sure it's possible

@YellowBunny

7 жыл бұрын

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.

@brikilian7834

7 жыл бұрын

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.

@Theraot

7 жыл бұрын

Search "Hypercube on a vertex - timeline of cross-sections" by pevogam

@YellowBunny

7 жыл бұрын

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.

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!

I never thought about this before, but... a computational Byte is the set of all vertices of a 8-dimensional hypercube! Mind blown!

That was the best video about tesseracts i have ever seen. This is why I love maths man

@forsaturn4629

5 жыл бұрын

Wrong. Check out this video --> kzread.info/dash/bejne/ZomGuKxsdJvLaM4.html . Better explanation.

Congrats to a well deserved 100k subs! Now, toward infinity!

@dcs_0

7 жыл бұрын

Countable or uncountable infinity?

@nickjimenez9983

7 жыл бұрын

Daniel Shapiro I'm blue

@nickjimenez9983

7 жыл бұрын

Daniel Shapiro sheep cow cow sheep guy man sandwich

@mvmlego1212

7 жыл бұрын

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.

Too bad they didn't finish by animating between the final shapes, like a 4D MRI.

@georgehiggins1320

6 жыл бұрын

I wish they did.

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.

excellent. very clear and i found it easy to understand. thank you.

This is absolutely awesome!

Absolutely brilliant!

In Dungeons and Dragons, those are D4s and D8s. D4s hurt like crazy if you step on them. Great video by the way.

YES!!! I realized this a few years ago and it blew my mind!

Nicely Explained!

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!?!?

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)

@uwuifyingransomware

3 жыл бұрын

That's really cool. I love how pervasive it is! Also, is that the ace flag in your pfp?

Great episode. Thanks

Love these!

Unfortunatly I can't upvote this videos as many times as I would like ^^ Regards from France.

very nice explanation with pascal triangle

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.

@ffggddss

7 жыл бұрын

Exactly right! (You beat me to this.)

@hawkjo

7 жыл бұрын

This is exactly the missing piece of information I was wanting from this video. Thank you.

You guys are awesomely awesome. :)

I can’t get enough of this stuff. 4D is the key.

YOU WILL NEVER HAVE A BETTER VIDEO STRICTLY CONTAINING ONLY MATH THAN THIS ONE!

This is amazing

GODDAMNIT I was hoping to see the hypercube animation! We've all been bamboozled!

Brilliant!

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.

Great video! Keep it up!

Love this show ❤!

Fascinating!

Thank you to PBS Infinite Series for slowing the rate at which my education rots out of my brain.

Great video as always. Can you please talk about fractals and perhaps the Mandelbrot set?

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.

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.

wow! mind blowing !

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.

This is madness!

That was so tight

If you had shown hos those hyperplanes combined for a tesseract you would have nailed the video. Anyway, great episode!

what a great tool. thank you. I might see if i can figure out the 3 dimensional shadow of a 5 dimensional cube

at 5:05 i just start smiling like stupid, this picture just makes me so happy lol

amazing channel

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!

It's like a powerpoint. Use the motion of video as the main visual tool to communicate change, patterns, similarity and differences.

Interesting, helps me imagine a little bit more the 4D Hyperspace

My brain hurts so good! :)

Well, that was interesting !

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.

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.

Thank-you!

You blew my mind again! :) Too bad you couldn't show the tetrahedon and the octahedron in the tesseract. :(

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 KZread,when I could be doing literally anything else, if I didn't deeply enjoy the subject.

@semicharmedkindofguy3088

7 жыл бұрын

You might enjoy 3blue1brown and mathologer.

@johnmorrell3187

7 жыл бұрын

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.

@SlashCrash_Studios

3 жыл бұрын

@Definitely a George Soros funded bot Shut up. You have no power here this comment section _IS_ nerds

I immediately like the host. I was so afraid of it being some douchy guy, but she seems so cool?

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...

@DeserdiVerimas

7 жыл бұрын

Interesting thought! Really intrigued to know the answer now...

@Ermude10

7 жыл бұрын

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

@DeserdiVerimas

7 жыл бұрын

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.

@stevethecatcouch6532

7 жыл бұрын

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.

@abramthiessen8749

7 жыл бұрын

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.

Awesome

I understand thanks!

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.

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!!

damn! so close yet so far. imagination part is hard. ironic tho how everything starts with 1 and ends with 1.

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.

"what's a hyperplane?" a spaceship.

Is there any graph theoretic additive rule for those vertices addition for the higher dimension ? (as it is not clear from the video )

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.

awesome

awesome episode! but you really need to draw the second triangle upside down on the summery screen.

that's very interesting

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]

Great !

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

@Holobrine

7 жыл бұрын

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.

what shapes do the slices make as dimensions approach infinity?

Hypercubes.....awesomesauce! It always blew my mind that the surface area of the hypersphere maximises around 7.26 dimensions. Is there an elegant manner to understand why this is (beyond the derivation from math world)? Also, why is the Schrodinger Equation in the background of the Q&A portion? That's physics! =P

Do a video on Gödel's incompleteness theorems, or Set Theory!

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.)

My current math professor said he had a professor in college who was born blind and had no problem visualizing objects over three dimensions.

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.

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

I understand this one!!

Pascal's triangle is also known by earlier Indian mathematician Pingla as Meru Prastara

pbs is so great

Huh, that’s funny. I was just thinking about this a few days ago.

5:53 False. If you have 5 puppies and need to choose your favorite 2, there is only one way to do so.

@cortster12

7 жыл бұрын

Let them battle it out until only two are left, thus leaving the choice to chance?

@mvmlego1212

7 жыл бұрын

+Ralph -- Technically, yes, but it's pretty clear what she meant.

@ralphinoful

7 жыл бұрын

I just found it funny, and wondered if anyone else caught the slight misuse of language.

@pokestep

7 жыл бұрын

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

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.

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.

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.

I have trouble with dimensions higher than 3. But I do love PT.

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.

#MindBlown...!!!!!!! this was awsome...|!

Very nice video!!! What about the perfect numbers? I guess it could gives better answer about coordinated in any dimensions. The choice must be LnCn, k where k have to be 2, the first prime number, using n as the perfect numbers.

Can we a video about - set, class and collection as described in set theory especially taking into account the associated paradoxes with it...I would love to help if I could

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?

Ah ah this joke at the end. But thanks for the insight in the 4D case

I love your brain.

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.

@rodrigon.almeida8093

7 жыл бұрын

The rectified 5-cell has 10 vertices! It's a semirregular polytope though

@yujiokitani4492

7 жыл бұрын

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.