Volume of Spheres in Any Dimension (*does it work for fractional dimensions?)

[@EpsilonDeltaMain]

I didn't want to stray too far from the main point of the video, which is about balls and spheres. However, the notion of fractal space(time) and its applications are very interesting in their own right. As a rough analogy, a fabric might be considered two-dimensional, but once you zoom in, it's made of threads, so it behaves more like one-dimension. Once you zoom in even further, the threads start to behave more like a solid rod, so it behaves more like 3-dimension. So the very fabric of spacetime could behave differently at different scales. And fractal/fractional calculus has found many applications not just in theoretical physics but in applied engineering as well. For example, to model heat flow in a porous medium, non-integer order differential equations can be used. As for the genesis of the notion of fractal spacetime, it originated from a technique called dimensional regularization used in quantum field theory (QFT). For example, if we take a 4-dimensional spherical integral of the function f(r) = 1/(1 + r^2)^2 from r = 0 to ∞ (the entire R⁴), the integral diverges. To 'tame' this integral, we take the spherical integral in a dimension less than 4, say 4-ε, then 'sweep some infinities under the rug'. Somehow, the physics works itself out. Some theoretical physicists have conjectured that for this to work, it is possible for the structure of spacetime to be non-integer dimensional.

[@thenationalist8845]

U r doing amazing work!!! 👍

[@giladhr3559]

This topic fascinates me. Do you know where can I read more about it? Or what math is needed to understand it

[@dank.]

​@giladhr3559 there are some links in the description!

[@ozzymandius666]

Absolutely fascinating. I guess as there are different ways of measuring a fractal dimension, then there are different ways of defining a fractal space. As to whether spacetime actually "is" of fractional dimension, I will say only: "All models are wrong. Some models are useful." It seems that fractional dimensioned space models have already become useful in some areas, as you mentioned.

[@ciwix]

0:15 Funnily enough, I am pretty sure it should be the case for every (most) other shapes. for the example of the cube, if we consider 2r = s and substitute it in, we get V=(2r)^3=8r^3 and A=6(2r)^2=24r^2, and we can see how dV/dr = A. as far as i can conclude, it should hold at least all convex n-dimensional objects. this is because we can think about increasing the volume of a convex object like adding a layer on top of it; and the n-dimensional volume of that layer can be described as A*r + E (r:= thickness of the layer, E:=error term, A=(n-1)-dimensional area)). the error E diminishes to zero as r approaches zero. this means that if we add a layer of thickness L to a convex object of radius R, the new volume can be describes as V + \int_{R}^{L-R}Adr, where we integrate in respect to r. now we can just plug in zero for the starting radius and get the volume of a convex object with radius L: which is just the integral of the area in respect to r from 0 to L. the reason it doesn't seem to work for most formulas is because the area and volume must be described in terms of a 'radius', and not anything else; in the cube example, we instead consider everything in respect to the 'diameter', leading to wrong results, because we integrate/differentiate over 2r instead of r. so if you can reduce a formula to just a single radius variable, then everything should work. i know this is not what the video is about, but i thought it was funny to share.

[@magma90]

For a regular polygon, the radius that should be used is the radius of the largest circle that fits inside it

[@JadeVanadiumResearch]

It's dependent on what line segment you use as your reference. The "derivative of the volume" is not a well-defined concept, as you need to choose which line segment you're taking the derivative relative to, and using a different reference length will completely change the answer. For example, if you rewrite the volume & surface area of a sphere in terms of the diameter and then differentiate, then you see dV/dD = A/2, so the equation V' = A magically fails when using a different reference length. Relatedly, as you saw, we get dV/dr = A for the cube when you define r=s/2. This works in general, for any shape with volume & surface area both being finite and nonzero: there always exists a reference length which forces dV/dr = A. This follows due to the fact that volume scales cubically and area scales quadratically. It's possible to rescale the shape so that 3V/A=1, call that the unit shape, and then select any line segment which has length 1 on the unit shape and call that the reference. Where X is the volume of the unit shape, then the surface area of the unit shape is 3X. Consequently, any shape with reference length r will obey V=Xr^3 and A=3Xr^2, and we easily verify dV/dr = A. I think there's an argument to be made, though, that the sphere is the unique shape where the reference length (the length r which forces dV/dr = A to hold) is equivalent to the radius, i.e. half the maximum distance between any two points on the shape. This is related to the well-known fact that the sphere maximizes the ratio V/A for fixed volume V, and the lesser known fact that the sphere also uniquely maximizes the ratio V/r for fixed radius r. I was pretty sure the latter claim is true, but now I'm having trouble finding a reference for it...

[@nin10dorox]

Mad respect for standing up against the gamma function!

[@dank.]

YES! THE RISE OF THE PI FUNCTION! I've always wanted an extension of the factorial that actually makes sense. Also, I've somehow never seen that relation between the volume and area - that's really cool!

[@jay_sensz]

0:16 Haven't watched the full video, but that doesn't sound right. It works just fine for a cube if you express volume and surface area in terms of half of the side length t=s/2. Then we have V = 8*t^3, A = 24*t^2, i.e. A = dV/dt. Likewise, the derivative trick doesn't work for spheres if you express the volume and surface area in terms of its diameter D = 2*r, i.e. V = pi/6*D^3, A = pi*D^2, i.e. A ≠ dV/dD In other words, there's nothing special about spheres at all in this regard.

[@Fysiker]

For a tetrahedron with side length a, you don't get the same relation using b=a/2 in your formulas, ie dV/db != S. I worked out that you need c= sqrt(6) a/12 to get dV/dc = S. Maybe there is a nice way to define the side length used in the volume formula of any (sufficiently symmetrical) solid so that the derivative formula works that feels more fundamental than doing out the algebra? In any case, your point about the derivative formula not being special stands, since it does seem like you can always define a length to use in the volume formula so that the derivative gives the surface area (assuming the formula involves just that one length). My formulas: V= sqrt(2) a^3 /12 = 8 sqrt(3) c^3 S= sqrt(3) a^2 = 24 sqrt(3) c^2 c= a*sqrt(6)/12

[@HeavyMetalMouse]

​@@Fysiker You end up needing an analogue for the 'radius' of a polyhedron. For a Square or Cube, this is the distance from center of figure to center of face. That this radius-analogue should hold for other regular polytopes of natural dimension is intuitive by constructing the derivative of the measure of the whole in terms of the measure of the boundary a done in the video for N-Balls and N-Spheres. That is to say, if you have a Regular N-Polytope, and extend its radius-analogue (distance from center of figure to center of (N-1)-face) by an amount 'dr', then the amount of extra area will in limit be the 'surface measure' of the N-Polytope times 'dr'., which creates a directly analogous result to the one in the video. In this manner, we can see why 'side length' is not a good linear measure, as it has no consistent relationship to 'radial measure' in the limit. Thankfully, we can do this without having to go through every specific example of a Regular N-Polytope in higher dimensions separately, as their construction and properties are even more difficult to visualize than N-Balls.

[@jay_sensz]

@@Fysiker You can always define a (reasonably well-behaved) solid 3D shape as some base shape (e.g. sphere with radius 1) scaled by a factor t such that the volume will be V₀*t^3 and the surface area A₀*t^2 where V₀ and A₀ are the volume/area of the base shape. Then you can substitute t = b*u where b is some constant, giving V = V₀*b^3*u^3 and A = A₀*b^2*u^2. Then you set dV/du = 3V₀*b^3*u^2 = A₀*b^2*u^2 and solve for b = A₀/(3V₀). This results in a value of b=1 when A₀ and V₀ are the surface area and volume of a unit sphere or a cube with side length 2. If we take a tetrahedron with side length 1 as the base shape, we get b = 2*sqrt(6) = 12/sqrt(6), which matches what you found.

[@Adam-zt4cn]

​​@@HeavyMetalMouseTo me, it sounds like we invented a generalization of the inscribed n-sphere radius, for shapes which don't even have an inscribed n-sphere. Take the volume and surface formulas of a shape, and express them in the terms of some length _r_ , in such a way, that the surface formula happens to align with the derivative of the volume formula. This quantity _r_ is the generalized inscribed radius. This radius should always be possible to find. Imagine some n-dimensional shape _F_ (which stands for Form, as _S_ was already taken by Surface). This shape has a side length of _a_ , for some arbitrary choice of what a "side" is. The volume of _F_ is _V(F)=b*a^n_ , and its surface is _S(F)=c*a^(n-1)_ , for some constants _b_ and _c_ . Let's express _r_ as some multiple of the side length: _r=k*a_ . After substituting for _r_ , the formulas become: _V(F)=b*k^(-n)*r^n_ _S(F)=c*k^(-n+1)*r^(n-1)_ The deriative of _V(F)_ with respect to _r_ is: (dV/dr)(F)=b*k^(-n)*n*r^(n-1)_ Set the surface equal to the derivative: _S(F)=(dV/dr)(F)_ _c*k^(-n+1)*r^(n-1)=b*k^(-n)*n*r^(n-1)_ _c*k=b*n_ _k=n*b/c_ Therefore, _r=(n*b/c)*a_ . For example, in the case of an equilateral triangle with side length a: _V=(√3/4)*a^2_ _S=3*a_ Therefore, _r=(2*(√3/4)/3)*a_ _r=(√3/6)*a_ which is exactly the radius of the triangle's inscribed circle.

[@abbasfadhil1715]

can't believe this channel doesn't get as much recognition as it deserves, plz never stop as long as you have loyal fans that never skip a vid

[@antoniusnies-komponistpian2172]

It's a good channel but it's really advanced stuff. I'll soon write my exam in (Real) Analysis 3 in Germany and I have a hard time following along with this. So I guess there are just not that many people who actually understand this.

[@jhacklack]

Really love the shifted gamma function that gets the factorial correct, PI(n) =n! Should be the definition of the gamma function and i have very unkind words to say to whoever defined the gamma function wrong.

[@skilz8098]

When thinking about fractional or partial dimensionalities, yes, it is hard to think of them within a visual geometrical term, yet mathematically they are quite trivial. When we consider full dimensionality such as 0D, 1D, 2D, 3D, ... ND where N is an Integer there are some well-defined properties that we can use to help us understand fractional dimensions. For this demonstration and for simplicity's sake, I'll will primarily focus on 2D and 3D regions of geometrical space although 1D space or linearity is vital within all higher dimensions. Here's a table of comparisons of looking at each ND from various different fields within mathematics. Algebraically based on the exponential and the polynomials: Here I'm not going to concern us with all lower terms or even coefficients, just the highest order term that is analogous to that dimensionality. 0D: f(x) = Arbitrary: single point with no reference, origin, point of reflection, rotation, symmetry. 1D: f(x) = x Linearity: distance 2D: f(x) = x^2 Quadratic: area 3D: f(x) = x^3 Cubic: volume 4D: f(x) = x^4 Hyper: volumetric ND: f(x) = x^N ... Geometrically in conjunction with the above polynomials within an ND coordinate system where N isn't just the exponent or highest degree of the polynomial of that coordinate space but is also the number of Orthonormal Component Vectors where each component vector, i,j,k, ... up to N component vectors are Orthogonal - Perpendicular to each other. // Note: the use of i here is not the imaginary unit i, it is just the i component vector along the typical horizontal axis. 0D: empty set 1D: unitary set, number line 2D: 2D Coordinate space 3D: 4D: ND: Now, when we combine the two fields in contrast to each other we can see that there is a relationship between them such as the following: Within the quadratics for 2D Area: f(x) = x^2 it is analogous to have or exist within at least 2D or greater dimensional coordinate space such as with a coordinate pair: (Xi, Yj). Since a dimensionality is based on the order or the magnitude of the degree of the polynomial and it has a corresponding number of orthogonal component vectors, it is hard to geometrically visual a coordinate space that has a fractional number of unitary vectors. However, this doesn't mean they cannot exist. It's just a limiting factor. What do I mean by this? Well, what does it mean for something to have 1.5 Dimensionality? Well algebraically we can represent this by the polynomial f(x) = x^(3/2). Geometrically we cannot easily do this. Because there is no such vector that is not a right angle to another that maintains perpendicularity, orthogonality or is normal to it. So, we cannot have a coordinate system that has 1.5 orthonormal unit vectors. This is impossible. We cannot represent this polynomial in 1D space either. We can however represent it within 2D space since 2 is > 1.5. Now, there is another trick or technique that we can do to help us with this. Let's consider the polynomial of x^(1/2). This isn't even a full 1D object. It's 1/2. If a full spatial dimensionality is defined to be orthogonal or perpendicular between two vectors such as with the X and Y axes of the XY plane and we know that these are 90 degree or PI/2 rotations giving us 2D space or 2-Degrees of Freedom for rotations. Then what would a fractional dimensionality be with respect to that? If x^1 = 1D with 1 degree of freedom, and x^2 = 2D with 2 degrees of freedom, then x^(3/2) would have 1 full degree of freedom and 1/2 degree of freedom of rotation. With this kind of polynomial or partial polynomial within a 2D coordinate system, we would have full rotation about the horizontal or x axis along the coordinate vector direction, but we wouldn't have access to the rotation along the vertical or coordinate vector direction. What is halfway between the X and Y coordinate axes within 2D planar Euclidean geometry? Well; if going from 1D to 2D is a 90-degree rotation or PI/2 radians, then 1/2 of that would be 90/2 = 45 degrees or PI/2/2 = PI/4 radians. Then within the polynomial we have full rotation of freedom along the x-axis or i vector direction, but we do not have full rotation about the y-axis or j. What we do have rotation about is the line generated by the rotation from the x-axis by 45 degrees. This 45-degree rotation gives us the line y = x that has a slope of 1, and a y-intercept of 0. We could represent this as being within 2D space of but only having access to where j/2 represents the fact that the second degree of rotation is not a full orthonormal vector with respect to i. This means that i and j/2 are 45 degrees from each other and not 90-degree right angles. This is one way that we can perceive partial dimensional spaces. For any N dimensional space where N is a + integer > 1 then there exists N orthogonal or perpendicular coordinate axes. When we have a dimensionality that is partial, such as with RD space where R is any + real number > 0. Then that partial dimensionality, fractional or decimal portion is a divisor of 90 Degrees or PI/2. In other words: 1/2 dimensionality is to divide theta or perpendicularity by 2. 1/3 dimensionality is to divide it by 3 and so on. This is probably the easiest or best way to fully understand this. Now as to try and graph such a polynomial itself is not trivial. We cannot easily represent it by an N-vector component with N coordinate pairs where N is an + integer value > 0. We can't easily represent a fractional coordinate. What we can do is to take our fractional exponent of the polynomial and round up to the nearest integer value. For example: x^(3/2) would require at least x^2 with to represent, and x^(7/8) would require at least x^1 with to represent. If we have something like x^(17/5) then it would require at least x^4 with to represent since 5 goes into 17, three times and there is a fractional portion left over that requires the 4th dimension. And here the portion that is left over is (2/5) this means we would have to take 90 degrees or PI/2 and multiply it by 2/5. giving us 2PI/2*5 = PI/5. So here the 4th dimensional space coordinate vector would look something like this: where Wl/5 is PI/5 or 180 degrees / 5. We don't have full orthogonality within the 4th dimension. Here we have full access to the first 3 dimensions but the last degree of freedom the rotation isn't orthogonal to the others. The line of rotation is fractional or partial to it. Within 3D graphics programming we use 4D matrices, quaternions, and such to do rotations and before we can render the frame buffer to the screen, we always have to take into account the Z-depth buffer for projection reasons and we also have to account for the w/divide. This is kind of a similar concept, but not exactly the same. When we get into the mathematical concepts of partial dimensions which I haven't even covered or mentioned complex values is when we start to dive into complex structures such a fractal geometry such as seen with the Julia and Mandelbrot Sets. This phenomenon does exist. When we look at say a riverbank or a shoreline, how much detail is there when we begin to zoom in and out and what kind of shape as well as what is their perimeter at each level of detail? Just food for thought!

[@mcpecommander5327]

Intro is wrong. It is the case for all shapes, so long as you parameterize it correctly. E.g.: Parameterize a cube in terms of its inner radius, which is half the side length. We get that V=s^3=(2r)^3=8r^3. Also, A=6s^2=6(2r)^2=24r^2. These match up, with the derivative of volume equaling the surface area. This makes sense, if you calculate volume as the integral of all the cubic shells propegating outward from the center of the cube, the surface area is how much dV will be added if you expand out in all directions. Thats why you have to parameterize it so that you are expanding in all directions equally, rather than in only half the directions, as when you are increasing side length.

[@MeshremMath]

It depends what measure on a shape you use as x, to whether or not the volume is the integral of the surface area.

[@logician1234]

Honestly, I don't think I learned much from the video. You just said "we can imagine higher-dimensional balls" and threw random formulas without derivation. You didn't explain why this doesn't work for fractional dimension, you just pointed out some random paper without any explanation. After that you went on a random tangent and even that wasn't motivated or explained. I understand that the math may be more involved and that you didn't want to scare off your viewers, but what is the point of tackling these kinds of topics then? I know that what I wrote may sound harsh to you, but I see potential in your content, and that you know your stuff. I like your channel and the topics you choose. I believe that with a little more effort you could make perfect math videos.

[@user-fl5nv7oh3z]

As I started from a lower level, I feel I could learn a lot. And as I can not follow, this will trigger to look for other sources...

[@VideoFusco]

For the cube or the square it doesn't work because s is the length of the side, and the side varies from both ends when a cube is subjected to a homothetic transformation (which is implicit in a derivation operation), while r, the radius of a circle or a sphere, varies from both ends. If you express the volume of the cube, or the area of ​​the square, by the half-side d (which, like the radius of a sphere or a circle, is also the distance of each face of the cube, or side of the square, from the center of symmetry of the figure) the thing continues to work: the area of ​​a square as a function of d is 4d^2 whose derivative is 8d which is precisely the perimeter, the volume of a cube as a function of d is 8d^3 whose derivative is 24d^2 which is precisely the total surface area.

[@jaytravis2487]

I loved that large portion of the video he devoted to the fractal dimensions mentioned in the title.

[@alexsere3061]

Hey, I loved the video, but I expected it to be a derivation of the formula, which I had already seen around. This felt more like an introduction. Which is okay, but I felt like there was potential for more

[@VaradMahashabde]

The derivative relationship does hold for a cube, the derivative should be in the terms of the corresponding "radius", half side length or distance from the center to the face, instead of the "diameter" i.e. the side length. If s=2h, the volume is 8h^3 and the surface area is 24h^2, it's derivative

[@poetrywithbennit]

Really well explained. Thanks

[@KazmirRunik]

0:15 I have a nit with this little bit. It DOES work for cubes, but you have to model it using spherical coordinates and integrate over that surface. Here, you change from volume & area with respect to radius R for the sphere to volume & area with respect to side length S for the cube. That's a sleight of hand; R & S are not analogous metrics!

[@HJ-gg6ju]

The pacing of this video is excellent.

[@geraltofrivia9424]

Great content

[@roneyandrade6287]

I really need to read that paper, i want to know why anh how the volume of a fractal dimension sphere is such and such

[@artemetra3262]

babe wake up new epsilondelta upload 🗣️🔥

[@1.4142]

What about plugging in complex numbers?

[@user-pr6ed3ri2k]

Square in fractal dimensions What is the second derivative or the first integral's geometrical meaning Intuitive in the sense that the lowest possible change in surface area is the derivative (this makes no sense but whatever)

[@faisalsheikh7846]

Awesome❤

[@yorailevi6747]

I have a question about the transition at 1:09 where you say "Insignificantly small" I have been hearing this line of reasoning quite often and although I understand it enough to say "sure" I also know there are edge cases where It cannot apply (I don't remember right now). When can one actually neglect the higher order epsilon values? I have heard that this exact issue has been troubling math before modern calc was formalized... so why all of a sudden do we reason without limits? how can we reason about it with limits and still get the same answer?

[@azmah1999]

To reason with limits just keep all the terms and take the limit dr to 0. You'll see that dv/dr = a term constant w.r.t. dr plus other terms with positive powers of dr. This give you the same result as if you only considered the linear term in the first place.

[@dank.]

I haven't yet seen any cases where it doesn't work, but you can generally show it to br true by appealing to properties of the wedge product. Differentials aren't really multiplied together, but rather joined by this wedge product (ex. dx² is shorthand for dx^dx, where ^ denotes the wedge product). You can think of the wedge product as computing the area of a parallelogram where the sides are the components of the wedge operation. If the components are the same, then the parallelogram collapses into a line and thus has no area. It's also seemingly just taken as an axiom in the defenition of differential forms, although it says it has something to do with poincare's lemma, but I don't know enough math to understand what's going on there tbh. You can find the axioms under the "defenition" tab. en.m.wikipedia.org/wiki/Exterior_derivative I also found this but I don't know how helpful this will be math.stackexchange.com/questions/1778665/why-is-the-wedge-product-of-a-1-form-and-itself-0

[@zopsinan5842]

What happens if we don't call the higher powers of dr insignificant. Is there a branch of math that explores those kinds of situations?

[@HaramGuys]

Calculus of finite differences

[@MooImABunny]

yooooo a Π function user Γ is such a silly definition, why Legendre, why did you do this to us?? Reject historical errors, embrace sensibility

[@Kraflyn]

what? that's all? :D

[@RobertMStahl]

hbar = h / 2pi