Strange Spheres (extra footage) - Numberphile 

A circle with a circumference of Tau*(One radius)+-i.

A lot, but not enough

Sebastián Espejo Loyaga so you mean about a parker lot.

The answer is yes.

Seven.

I wouldn't call it a lemon. I would call it a Parker Circle.

a higher dimensional Parker Sphere

+1234Daan4321 is a lemon a parker orange?

kourii That can't work you see. Parker oranges are supposed to be blue.

It was a "Parker Explanation"

@@natfailsyoutube8163 But a Parker square is a failed attempt at a square, while he didnt even try to explain it (because you shouldnt worry about it)

Including the line "It is trivial to see that..."

In which number of dimensions? It can't be in 7d because when d=7, 4^d=16348, so 7 is too small, but when d=8, 4^d=65536, so 8 is too big.

I looked at the numbers for the even numbered dimensions up to 32. The volumes of the spheres were outstripping the volumes of the boxes. The ratios for even numbered dimensions from 2 to 32 are approximately 0.308 0.749 1.9805 5.564 16.335 49.589 154.575 492.294 1596.137 5253.993 17521.223 59097.318 201331.945 692018.789 2397635.534

Steve's Mathy Stuff thanks for doing that, those are surprising results

I just realized that I confused myself by reading too many comments. Some people are talking about 2 x 2 x 2 x ... cubes. I should have made my calculations using a 4 x 4 x 4 x ... cube. I'm off by a factor of 2^k. The actual ratios are 0.03368825528 0.0192765711 0.01170117025 0.007736233366 0.005433539366 0.00398795459 0.003026657236 0.002358623777 0.001877953681 0.001522195063 0.001252649559 0.001044346271 0.0008806186624 0.0007500199415 0.0006444927204 0.0005582430245

The formula for the "volume" of the central sphere in 2k-D is (pi^k/k!)*(sqrt(2k) -1)^2k. The "volume" of the cube is 4^2k. So the ratio is [pi^k * (sqrt(2k) -1)^2k]/ [k!*4^2k]. I can't find any clever way to show the limit is 0 as k increases.

en.wikipedia.org/wiki/Volume_of_an_n-ball#Low_dimensions

I'd personally never given much thought to how you'd even mathematically define "spiky", but the definition used by Matt (proportion of shape removed by removing a certain portion of a linear dimension) feels intuitive enough - if you think about a star (a shape most people would consider spiky), you can see that it would give a much lower value than a circle (or even a sphere). When I watched the original video, the reason why "spiky" felt reasonable to me as descriptor for higher dimensional shapes is that a sphere in any dimension is based on the surface of points equidistant from some other point, and the conversion from the rectangular coordinates to the distance from the centre is Pythagoras' Theorem, the square root of the sum of squares. The more dimensions you have to split the distance to the centre between, the more extreme points you have where nearly all of the distance is invested into one of the dimensions, while the majority of points that share the distance about evenly between all of the dimensions are even closer to the centre point than usual because there are more dimensions to split the distance between. So, in a circle, there are four extreme points (two for each dimension), and a reasonably large amount of the circle exists near each of those points because there are only two dimensions to split the distance between. In a sphere, there are now six extreme points, and a reduced amount of the sphere exists near each of those points because distance is being split between three dimensions. While higher dimensions are a pain to visualise, and as a purely recreational mathematician I haven't actually studied higher dimensional shapes, it's intuitive enough to me that extrapolating that relationship would create the expectation that as you add more dimensions, and even more extreme points, you'd have less and less of the higher dimensional sphere near the extreme points, which is basically the definition of spiky.

I still have a couple of problems with the description of "spiky". First, spikiness implies a lack of convexity. Imagine a star shape: if you draw a line segment from a point in one spike to a point in another spike, there are points on the segment outside the star. In other words, the star is not convex. Hyperspheres in any number of dimensions are convex. And how many spikes does a sphere have, anyway? 3? 6? Where are the spikes located? Doesn't the presence of spikes violate the rotational symmetry? When you rotate a sphere, don't the spikes rotate with it? If there are, say, six spikes on a sphere, wouldn't you have to rotate a sphere 90 degrees to rotate one spike into another spike? But a sphere should have continuous rotational symmetry. I'm still having a really hard time visualizing these spikes.

I felt the same way about the odd description of 'spikiness', since it seemed to imply a lack of convexity, but when he describes the meaning in this video, I get it a little better - perhaps a better description would be they're 'more tightly curved' - since the total curvature along a multi-dimensional 'surface' depends on the curvatures of all the different dimensions of it - a surface of a sphere is 'more tightly curved' for the same radius than a circle is, because there are two directions for it to curve in, and curvature is intuitively multiplicative. Now, comparing curvature in 2 dimensions to curvature in 3 dimensions is kind of like comparing apples to oranges, but it is a reasonable way to get a sort of intuitive handle on where the difference comes from. Every point is more tightly curved, but there's a LOT more room for points to exist at. The sphere doesn't have some discrete number of 'spikes', it's just that every single point on the sphere is 'sharper' in that sense than the sphere of the next lower dimension.

He should've said it is more pointy, not spiky.

That explanation certainly makes more sense than "spikiness", but if the radius of a hypersphere is 1, the Gaussian curvature will be 1 in any number of dimensions. If it is less than 1, the curvature will be greater than 1 and it will grow with the number of dimensions. If it is greater than 1, the curvature will be less than one, and it will get smaller as the number of dimensions increase.

It's freakier than that. Pi is raised to increasing powers every _other_ dimension. en.wikipedia.org/wiki/Volume_of_an_n-ball#Low_dimensions The proportion that a sphere fills a box is only 100% for 1 dimension. It not only decreases, it accelerates from there (the denominator is essentially factorial): 1 - 100% 2 - 79% (pi/4) 3 - 53% (pi/6) 4 - 31% (pi^2/32) 5 - 16% (pi^2/60) 6 - 8% (pi^3/64) 7 - 4% (pi^3/840) 8 - 1.6% (pi^4/6144) 9 - 0.6% (pi^4/15120) 10 - 0.25% (pi^5/122880)

It is odd because the radius of the sphere is growing bigger and bigger eventually getting out of the box not that the space that it takes up is getting bigger

@@supersmashbghemming6445 One year later: The radius is growing bigger and bigger indeed, but the possible directions that the spheres need to cover is growing too, every higher dimension is a new direction so in order for the n-sphere to touch every direction the n-box has, the n-sphere itself is covering less n-volume. It's easy to understand in our 3 dimensionality, a sphere covers less volume inside a box than a circle because the 3d box has more directions than the 2d box, our 3d sphere is not getting spikier, its extending to 2 new directions, up and down. To understand how geometrical objects work in higher dimensions you analize the sequence of the formulas that rules our dimension in comparison to the lower ones. Don't try to use actual third dimensional formulas, they don't make sense in any place other than in our 3d world.

Limited precision of an 8-byte real...

It would fit, the 11D sphere is still only 2 units across in every direction, while the 11D box (I'm assuming all sides are the same length) is 4 units wide, and the hypotenuse across the 10D opening would be ~12.65 units across.

I see That makes sense As much sense as I can make of 11 dimensions, at the very least Thank you

The radius does, but so too does the distance from the centre of the box to the corners. The centre of the box is always distance 2 to the centre of a face, but the corners are sqrt(d*2^2)=sqrt(4d)=2*sqrt(d) away from the centre, so the ball/sphere/round thing (with radius sqrt(d)-1) will always be short of the corners. In fact, as you go to higher dimensions, the corners get further away from the ball (2*sqrt(d)-[sqrt(d)-1] = sqrt(d)+1 ). This means the ball escapes further and further out of the box, but gets further and further away from being totally free!

4 years later but... What is taking a 2D slice out of a 9D object, think about what it would mean to take such a slice out of a 4D hypersphere, it would actually equal to taking a line segment 1D of a 3D object, since the 4D object is not merely composed of lines and faces, but also whole 3D structures. (e.g. a cube's outer shell is made out of 6 squares shifted into 3D; a hypercube would be made out of 8 cubes! shifted into 4D, it's outer shell is composed of whole 3D objetcs) Therefore, taking out a 2D segment out of a 9D structure, would be comparatively removing nothing, since it is 7 dimensions far from its own, just as 0D (a point, no directions) is merely 3 dimensions far from ours.

Lucas Balaminut thinking in 4d is impossible. Understanding it can only be done with lower dimensional slices and shadows. Giving up higher dimensional global understanding makes it easier to slice and dice 4d object to analyze parts in lower dimension. It is also easier two follow the numbers than thinking about it geometrically

Probably.

(Yes, there is. Easiest way to work it out is probably with integration in 3D, but I don't have the whiteboard pens for that atm!)

In 1 dimension, there's only one possible shape (a line), so the closet equivalent to 4 circles in a 4x4 box would be 2 2-unit long lines within a 4 unit long segment. That would just mean the two lines would take up the entirety of the box, and there would be no space between them. This actually works with the algorithm described. Square root of 1 - 1 = 0.

A 2D *circle* is the outside/boundary of a 2D *disk* . A 3D *sphere* is the outside/boundary of a 3D *ball* .

A solid circle could be called a disc/disk?

I would rather put the other way round: A disc/disk could be called a solid circle. Since that would start with something which is defined exactly, and then attach a alternative description to it. But without further nitpicking: yes, that's right.

Someone is going to get a maths question on the volume of a sphere and write 0 because boundary has no thickness or volume...

Hmm.. That's a tricky one. Maybe the sphere itself has zero volume, but it still contains a non-zero volume inside?

linkviii Not thin enough. :P

thats a super nice point!!!!!!!!!

achu11th the definition of a dimension is that, each dimension is at one right angle to each other.

Ipad Air so explain to me how I should imagine this problem in more than 4D (which I can get behind). 9D for me requires a 9 D cube to solve the problem and even then it is hard to imagine it for me. So that's why I asked to begin with. I couldn't see a possibility to make a pythagoras at these high levels of dimensions.

Well, each new direction is separated by 90 degrees from previous ones by definition, that is just how maths works when working with higher dimensions, and that is how mathematicians defined them. You could possibly assume something different but then you have to state it clearly so that nobody is confused, and if somebody doesn't do that, you can safely assume that they mean standard definition of a dimension. As for second part, yes, with higher dimensions you have more "volume" inside cube, but increase in middle spheres "volume" is not what is so amazing about this demonstration, but the fact that this sphere in higher dimensions has radius simingly bigger than box containing it. EDIT: with each new dimension you use the result from pithagoras theorem from previous step, i.e. when you step from 2nd to 3rd dimension you use result of pithagoras in 2nd dimension ( sqrt(2) ) and 1, which gives you sqrt(3). Then when you step up to 4th dimension you use sqrt(3) and 1 which gives you sqrt(4). Then in 5th dimension you use sqrt(4) and 1, and so on, and so on.

Preedx2 the second part helped me realise the thinking error, I was thinking about. That's why I said "or I am not concentrating enough?". As for the first part, it is the question of how you can proove it to be an exactly 90 degree angle. I know, that it is mathematical definitions involved saying, that it just is 90 degree each step to another dimension, but it still is a valuable question. What if it wasn't 90 degree dimension hopps and therefore would rely on another formula, which would make the spheres sphere again? Or as I stated, what if the new space allowed you to somehow fix the issue with the packaging? As for the radius getting bigger than the containment, I would say, it is unexplainable for me right now (still keeping it a sphere and inside the box). Maybe it is some sort of relativ distance as in physics, where a meter can be smaller than meter in the blackhole, because of gravitation. Maybe n-dimension, can provide a distorted lenght of what it actually is? Or as I said, maybe pythagoras stops working after 3D.

Yes?

To eat those oranges with the sticky tape still on :P