Hey everyone. If you haven't heard, I'm planning a Q&A video to mark reaching 100k subscribers and I'm still collecting questions for it. Feel free to ask any in reply to this comment! I'm thinking I'll answer some right here right now and save others for the video.

@gametalk3149

8 ай бұрын

How and why is the mandelbrot set so intricate if the equation is "simple". What does the c mean in the mandelbrot set's formula?

@gametalk3149

8 ай бұрын

Why do the Julia sets of the Mandelbrot Set always go to one cycle?

@gametalk3149

8 ай бұрын

Why is the mandelbrot set "nested"; why are there mini copies of the main curve in the mandelbrot set, and why are some of them in spirals; why and the mini copies of the mandelbrot set is spirals. Why aren't they like the serpenski triangle which doesn't have the same type of spiral.

@gametalk3149

8 ай бұрын

Another thing, why does the Julia set of the Mandelbrot set whole, or "dust", disconnected pieces that doesn't have a cycle. Why aren't there Julia sets split in two, or 3.

@evandrofilipe1526

8 ай бұрын

How much geometric algebra have you done? What is your opinion of it so far (if any?)

"just use euler's formula! no, not that one, the other one. no, not that one, the other one. no-"

@umbragon2814

8 ай бұрын

"Discoveries in math are named for the second inventor, because the first is always Euler."

@worldhaseverything

8 ай бұрын

I'm in computer engineering semester 1 and I get 4 euler's formulas in 5 unit

@Treebark1313

8 ай бұрын

gee, maybe we shouldn't name every idea in mathematics after one of three dead European men.

@davidwuhrer6704

8 ай бұрын

​@@Treebark1313It's not their fault they're dead, except Cauchy.

@RichConnerGMN

8 ай бұрын

nice pfp

17:08 Those equations are looking a little sus

@Vsevolodbochkov

3 ай бұрын

When the angle is sus

Among us spotted at 17:13! 2nd row, last theta

The generalized euler's formula looks almost exactly like how you'd do this in projective geometric algebra. u tilt v is effectively the same as u wedge v, producing a bivector that gets factors of sin theta, which produces a rotor that rotates 2 theta radians in the plane of the bivector u wedge v, then since this is automatically normalized you can then add one (equivalent to the identity matrix here) to get the square root of the rotor which rotates theta radians instead of 2 theta.

@user-yb5cn3np5q

8 ай бұрын

Highly underappreciated comment. I was expecting wedge product to show up during the whole video.

@user-yb5cn3np5q

8 ай бұрын

Not to mention exponentiation formula works with AB != BA, it's just going to be a BCH formula.

@KinuTheDragon

8 ай бұрын

I like your funny words, magic man

@davidgillies620

8 ай бұрын

This also splits into the standard q p q^-1 quaternion rotation.

@linuxp00

8 ай бұрын

I expected a connection some vids ago, but he doesn't know GA and instead he is painfully and slowly building LA arguments for what GA already gives us more naturally and directly.

After watching these videos, i have to admit, you are one of my favorite math youtubers, if not my favorite. Congratulations on your journey and i hope you keep making these advanced math videos, because they are amazing!

Here's my comment about geometric algebra Your "tilt" product is the exterior product of vectors in 3D VGA (vanilla geometric algebra, works with linear space of vectors (and other polyvectors too)) which results in a bivector Except exterior product will work even if your vectors aren't orthgonal or aren't normalised (But before expontiating a bivector you will need to normalise it). Bivectors are independent of basis, so no memorizing weird cross product rules etc.

@keris3920

8 ай бұрын

All of this video just describes a small subset of Lie Theory. The rotations they are creating in the video belong the SO3 group, and the "tilt" product is the same as the bracket operator.

I learned about the matrix exponential way back in Differential Equations when deriving the Jordan Normal Form. It's nice to see it make a comeback in something else really cool!

I have been watching many different math KZreadr videos and this is particularly clear, sufficiently abstract and entertaining. Nice job!!!

Please keep going in this series. It is excellent

This video is uploaded at the *perfect* time! I was just learning about this, but it's very hard to learn just using the internet. This is awesome! Thanks a lot! ^w^

goD THIS IS SO INTERESTING !!! thank you for answering some questions i've had for years!

Beautiful video! (This probably happens a lot) but I was wondering how to do the topic of this (and the previous) video, and thrilled to see such a direct answer on youtube. I was lightly looking into analogues of Euler's formula in higher dimensions a couple months ago, and I think I stumbled upon Rodrigues' function but didn't feel like looking into the context. At the time I was hoping to find a way to generalize a fourier transform to 3d, (mostly to make a present for a friend's birthday, which has long since passed lol) and seeing this video (and the last one in the series) got me very excited. Cheers!

@whatelseison8970

8 ай бұрын

For a 3D analog of Fourier series, see en.wikipedia.org/wiki/Multipole_expansion.

Given that i'm in the midst of learning about robotics, i'm incredibly lucky to have found your content, explanation is the best i've found. Could i be cheeky and ask if you could do anything on the basics of kinematics in the future? The combination of the fluctuating diagrams and amazing explanation in your videos is soo much easier to get into my head than another abstract maths book that only seem to explains things well to other experts. Honestly i truly wish every tutorial was like this one!

great ! morphocular uploaded

very nice video. have just been getting into geometric algebra, couldn't be better timed

Wish videos like this were out when I was first learning about exponential maps and Lie algebras in physics!

Commenting for the algorithm.. looking forward to your next video!

This video is incredible, thank you so much! I'd never heard of this system, it's super interesting. It sounds super similar to Geometric Algebra, to the extent that it would be interesting to see a follow-up short video (or even ablog post! do you have a blog?) making the connexion explicit. (E.g. with wedge products replacing tilt products.) Is this something you might consider doing? At least hashing out what your equations would look like in GA notation? Whatever you do, thanks for all this amazing content, I've learnt so much with your videos! I have a degree in maths but nonetheless, there's always more to learn, and how you write your videos means they're fairly accessible even at very different levels of knowledge. Bravo!!

This video helps me make a relationship between christoffel symbols, tangential acceleration and velocity on a spherical surface.. Thank you so much.

Man, I was struggling to find the general solution to exp(At), where A is a matrix. Now I've seen it, it's so short but elegant!

I'll get back to this to improve my understanding of the topic!

3:10 there is an equilivent linit definition that works

Singular ! Spectacular

23:33 I guess a brief nod is better than nothing huh? I'm partially joking of course but I can't wait to see what you do after this series' final episode? Any insight or not so much? No pressure of course, I love these videos for the most part and I think you should be proud

@morphocular

8 ай бұрын

There's enough demand for Quaternions at least that I'll probably have to do a video on them at some point :) Though I'm not sure when yet, as I need to study up on them more myself before I'll feel prepared enough to make the video.

@evandrofilipe1526

8 ай бұрын

@@morphocular I understand, I only really learned about them briefly when I was doing further maths (I'm not doing that anymore), so I actually don't know that much about formally

I don't know, but it's very beautiful!

Incredible quality of content❤

Another great video (I had to rewatch your earlier one). I've been playing with using quaternions to rotate vectors in 4D space (not just 3D), and I think this provides a great way to do it. Thanks!

OMG IT'S HERE!! Awesome video as always; I love when you cover little-known topics in maths, like wheels and stuff like that. I rarely comment and I just wanted to inform you of my appreciation of your channel.

I have known about 3d number "grid" tho I never understood it, thanks for making this math popular and understandable ❤

I am a simple man, I see a morphocular video, I click.

I stand behind your use of the “mathematician’s” notation for θ and φ. It has numerous advantages: 1) θ being the azimuthal angle means it basically retains its role from polar coordinates. This makes the jump between the two less jarring, and also makes cylindrical coordinates more consistent with polar coordinates. 2) θ represents the sound “th”, which appears in the word “azimuTHal”. φ represents the sound “ph” which is close to the “p” in “Polar” (also in Ancient Greek it was pronounced basically like English “p” so it still works). 3) The azimuthal angle occurs in a longitudinal plane, which can be seen as the horizontal line in the symbol θ. The polar angle is measured with respect to the axis of the sphere, which can be seen as the vertical line in the symbol φ. So, to me, the mathematical convention is just better. It’s closer to polar coords, and allows for easy name-based and symbol-based mnemonics to remember which is which. The only advantage the “physicist’s” definition has (at least that I can come up with) is the fact that it is technically the “standard” one because it was chosen by the ISO. I’m a fan of standards and the ISO in general, but I still consider this one to have been a mistake.

Screaming "geometric algebra" internally while watching this

6:36 : I think it might maybe be good here to (as a next step of how to present this) replace the square of the tilt product, with the negative of the projection, so that we get 1 + (cos(theta) - 1) * (the projection) + sine(theta) * (the tilt product) Or, maybe even, 1 * (1 - (the projection)) + (cos(theta) + (the tilt product) * sine(theta) ) * (the projection) ? So, that way, we have: (The projection onto the orthogonal complement of the plane) + (rotating within the plane) * (projection onto the plane) ? Idk.

Great vid, really touches alot on lie algebra stuff

I had to go back to the previous video to remind myself that you are aware of geometric algebra. Having learned and appreciated the simplicity of it I would really appreciate a follow on video that explains your rotation methods in terms of GA. Freya Holmer did a great talk on the basics called "Why can't you multiply vectors?".

Amazing

Hi Morphocular!! Could you make more videos on rotational mech and circular dynamics from perspective of math (or physics)? Thank you

The exponential follows the Campbell-Baker-Hausdorff formula since it is anticommutative. I wonder if it’s possible to draw any n-dimensional closed curve (practically any shape) using some multidimensional version of Fourier epicycles, the generalized Euler formula and the multidimensional discrete Fourier series with the scalar vector normalized to 1.

@drdca8263

8 ай бұрын

I’ve seen the Baker-Campbell-Hausdorff formula. What do you mean by “because it is antisymmetric”? The commutator is, of course. But I don’t understand what you mean. Edit: also, it is at least possible to take the Fourier transform of the curve by, e.g. taking each coordinate of the vector separately. Though, you don’t really need to consider them separately (picking a basis) in order to do this. You can also just apply the definition of the Fourier transform of a function from the circle to the complex numbers, except with the function you are taking the Fourier transform of being vector valued. The result will be a function from the integers to (complexified) vectors. I guess if you want non-complexified vectors, you take the pairs of vectors associated with pairs of each natural number and its negation, and combine them in the appropriate way in order to go from the exp(i n theta) basis to the sines and cosines basis.. And then uh, well, each of these pairs will give two vectors which... Hm, I guess these will be revolving in an ellipse? This seems surprising to me. v_{n,1} cos(n theta) + v_{n,2} sine(n theta) huh. If this were a circle, with v_{n,1} and v_{n,2} being orthogonal and of equal length, then we could describe this nicely with rotations in a plane as described in this video. But, with the shape being an ellipse rather than a circle... seems rather odd. Weird.

This is super interesting! Would it ever make sense to use this in, say, electrodynamics?

@conando025

8 ай бұрын

Yes actually a lot. You can go down a rabbit whole exploring the idea that the magnetic field isn't a vector field but one of oriented rotations. Though I've mainly seen it through geometric algebra which even though its not the same also contains a similar algebraic structe without relying on matricies

Do you have any lessons on Combinatorics?

Finally

You're very right to stress the non-commutativity of matrix exponentials. A lot of us would have been tricked by that! .. but after a second though, isn't the commutativity that should be considered the special case? As in, in general, exponential is a morphism from an additive to a multiplicative group ... and it just happens that we're mostly used to the commutative group of real number.

Please keep up the quality over quantity.

near the end of the video you criticized the usual spherical coordinates formula for three clutter of its derivative, usually in differential geometry the partial derivatives of the formula would give us basis vectors for the tangent vector space, we'd find the proper linear combination between them by taking the multivariate chain rule of the original formula (without substituting at the start), that naturally gives you a sum of angular speeds multiplied by tangent vectors

oh hello there! 16 minutes since upload

17:00 I see something sus in the expanded derivation

So let say i have a ray in domain repetition, would this help intercept an helix? Like the repeated intersections of the ray on the supporting tube expressed as angular variations?

Can you do this on the 3-sphere?

15:37 shouldn’t U and V be matrix-to-scalar functions, or do I misunderstand what you’re getting at here?

@morphocular

8 ай бұрын

U and V are meant to be functions that take a scalar as input and return a matrix as output ("scalar-to-matrix"). Their derivative is then relatively easy to define, as it just comes down to taking the derivative of each entry in the matrix. If it was the other way around, it'd be more complicated to define, as you'd be taking the derivative of a scalar with respect to a variable matrix.

would you consider this method of rotation to have any merit in computer graphics, or are quaternions the way to go?

@bloom945

8 ай бұрын

I always forget to mention, but I loved the video of course! Your videos are always such high quality, it's seriously impressive

@morphocular

8 ай бұрын

I haven't done enough work with computer graphics myself to really know for sure, but I believe for sheer computational speed and stability, matrices tend not to be the best choice for representing rotations, and my impression is quaternions are kind of the gold standard for rotation algorithms much of the time. But to me, the merit of this generalized Euler formula approach with matrices is making 3D+ rotations easier to analyze (hence the example with spherical coordinates in the video).

@APaleDot

8 ай бұрын

Quaternions are definitely simpler for 3D rotations. This could be seen as a generalization of quaternions, so it ends up being a bit more complicated. For instance, if you actually grind through the algebra for the sandwich product between a unit quaternion and a 3D vector (represented as a quaternion with no real part), you get a formula that is almost identical to the one shown in the video. Specifically, if 'n' is a unit vector representing the axis of rotation, the resulting product has three terms: an (n · v)n term containing the component of the vector that lies along the axis of rotation an (n × v) term containing the projection of v onto the plane of rotation, then rotated 90° and an (n × v) × n term containing the projection of v onto the plane of rotation The full formula is qvq* = (n · v)n + cosθ ((n × v) × n) + sinθ (n × v) Comparing to the formula in the video, we see the projection onto the plane is handled by -cosθ ². Given that ² projects onto the plane and then rotates 180°, its negation is the same projection given by ((n × v) × n). The sinθ term clearly corresponds with no extra work, so all that we're left with is the I + ² matrix. This matrix takes the original vector and subtracts its projection on the plane ( -² is the projection as previously stated). In other words, it removes the components of the vector that lie in the plane, leaving only the components which are orthogonal to the plane. In 3D, this is the projection onto the axis of rotation, same as (n · v)n. So it really is the same mathematics underneath, but quaternions are very compact and easy to lerp and stuff.

@keris3920

8 ай бұрын

If you are both rotating and translating, you should have a look at the SE3 Lie Group, which combines the SO3 Lie Group used in this video with a translation vector.

This presentation is so good for Euler's Formula. But min. 12:13 is not clear. Please check it. Thank you so much.

What is that letter on minute 23:20?

@17thstellation

8 ай бұрын

It's the Sinhala letter ඩ (ḍa), also widely memed due to its resemblance to an amogus

I dont know math but fire music

ok, I'm 2 minutes in and this screams "Geometric algebra" (or Clifford algebra if you prefer). Let's see if I'm right...

Hi Morph! Has Thanksgiving come early? Yes. Yes it has. Also, sine of amogus.

everyone like and comment ,btw great vid

Wait, but doesn't the second rotation flatten it to the XY plane?

this "tilt product" sounds a lot like the geometric product between 2 vectors in GA it does the same thing except it creates a rotor instead of a matrix

@2fifty533

8 ай бұрын

actually nevermind, it's more like the outer product

@keris3920

8 ай бұрын

​@@2fifty533it's the bracket operator seen in Lie Theory. Have a look at the SO3 Lie group and associated Lie Algebra, it's actually pretty incredible that they managed to rediscover Lie Algebras in this way.

17:13 when the sine is sus..?

Why do this when we have bivectors?

"How to rotate from one vector to the other" You mean the Geometric Mean of the two vectors? :P

Why do you use non standard symbols like instead of standard symbols like ( and ) ?

@morphocular

8 ай бұрын

It's partly because it conveys important meaning in this case. is meant to represent something more and different than an ordinary cross product, and so the angle brackets serve to make it clear it is a different operation.

This looks like it could be a very useful tool! So - this tilt-product is your own invention (or discovery)? Could be very useful indeed. I guess, one could also use it generalize curl to any dimension? How would that look like? nabla-tilt-vector, I guess? ....yes - I know - there's also the exterior derivative available as generalization of curl - but being able to express such things via matrices makes it easier for me to think about it

The formula around 23:20 looking pretty sus ngl

17:10 Is this ????? AMONGUS ??!!!

11:16 Don't worry I won't *Proceeds to pull out a gun*

23:08 I disagree with the character, watching the terms shuffle around and morph into other terms to produce a formula is simply amazing

17:15 something about this expression is… suspicious

Why did you botfarm boost this?

But what is going on with that theta at 23:24?? It's on crack or something.

Quaternions

Remember galosian product ei*ej=-e(i+j mod 3)

Please give me any references you have on this.

3:00 Having the I term makes this ugly. The correct term in the series is (A^0)/(0!). Yes, this can simplify to I, but kept in that form, it prevents that (1-cos) in the subsequent grouping of the terms.

I saw that amoogus at 17:06, sus...

24:25 is sus

Anyone else notice the amogus person in the spherical coordinates matrix formula derivative?

17:07 sus

5:47 acting like i and sincos

10th?

17:10 suddenly i found amogus

Apologies for being critical, but I think it might be useful. Some of us have real difficulty if there is a backing music track. It makes me not want to listen to a video like this. I doubt very much that having backing music will make people more interested/more likely to "like" and so on, so I suggest not having one in future.

Wait, but that's just Geometric Calculus with extra steps.

@LimeHunter7

8 ай бұрын

Yes, the tilt product is definitely doing something like a wedge product here, but the correspondence isn't obvious to me

@naturallyinterested7569

8 ай бұрын

@@LimeHunter7 In my (very limited) understanding of both he uses the same algebraic structure as bivectors but embedded as a subspace in matrix algebra (what I mean by extra steps) so he can do matrix calculus on the problem without first developing GC, but then needs to "apply" his structure to a basis vector, doing all the fine-and-gritty matrix multiplication instead of having the geometric product of multivectors directly.

@angeldude101

8 ай бұрын

​@@LimeHunter7The tilt product literally gives the matrix representation of a bivector. The reason it behaves differently is because it's being applied in 1-sided transformations on "ordinary vectors," whereas applying a transformation in geometric algebra is always done to other transformations (since all geometric objects in GA can also be used as transformations), which requires a double sided application. Doing only a single product only _composes_ the objects' associated transformations.

17:15 amogus

Mock Theta

You thought you could hide your amogus from me?

@svz5990

8 ай бұрын

I think they just rendered the theta wrong

@CyborusYT

8 ай бұрын

@@svz5990 no, I think that was intentional

17:11 I don't think I'm safe.

I made the quantum computing very easily🎉🎉🎉because👨‍🔬 indian🇮🇳 are living with it from⌚ millions of years🇮🇳🇮🇳🇮🇳🇮🇳

Have a look at Lie theory, if you haven't already. Your tilt product operator already has a name, and so does your algorithm.

math

I guess it's the Lie Group.

Great video. My only critique is that your pronunciation of "t" as [t] in words like "derivative" and "velocity" sounds kind of forced and unnatural.

Actually complex exponentials already lose a property compared to real exponentials. If a and b are complex: (e^a)^b is not generally equal to (e^b)^a, because both cannot generally be rewritten as e^(a*b). ( See en.wikipedia.org/wiki/Exponentiation#cite_note-Clausen1827-33 ) Matrix exponentials just "lose" an extra property.

NUMBEEEEERS

Are we STILL Learning from NEANDERTHALS CONSTANTS..FFS

main takeaway from this video: ඞ

@geekjokes8458

7 ай бұрын

it also reminded me of my analytical mechanics class where we took the time derivative of the "monster matrix" lol after class i asked the professor about a more "intuitive" way to look at it (i wasnt _just_ talking about that rotation matrix, but anyway), and although he's part of the mathematical physics department, he didn't talk about this connection (to be fair, there a lot more to talk about, i dont think he would have had time) maybe he didn't think I'd be up for it...