I really liked this video because of all the math this dude did. What a guy.
@Tehom1
Жыл бұрын
Fascinating, especially the last result which I hadn't seen before. I find that an easy way to grasp quaternionic behavior under exponentiation is to see that any single quaternion (other than a purely real one) defines a complex sub-algebra of the quaternions and that sub-algebra behaves the same way as the complex numbers; its purely imaginary part, unitized, plays the same role as the complex number i. A lot of unintuitive stuff got much simpler when I understood that.
@la6mp
Жыл бұрын
This is the first time quaternions made any sense to me, in spite of several attemps. As ususal presented in a clear no-nonsense way. Thanks a lot!❤
@mohammedfarhaan9410
4 ай бұрын
haha same this makes alot more sense then anything ive seen
@adandap
Жыл бұрын
Fun video. BTW, the matrix representation of the quaternions is related to the groups SU(2) and its covering group SO(3), and thus they can be used to represent rotations in three dimensions.
@fjg9657
Жыл бұрын
Nice connection. Thanks!
@felipegabriel9220
Жыл бұрын
wow, never though that i^j would be equal to i*j=k, thats is cool. Nice video as always Michael! :D
@mikeiavelli
Жыл бұрын
Well, ofc you wouldn't think that because it's BY DEFINITION. It's one of the things that make quaternions... quaternions! 😉 Just like i^2 = i*i = -1 is by definition what makes a complex number complex.
@iang0th
Жыл бұрын
@@mikeiavelli The definition says i*j=k, but it's not immediately obvious from the definition that i^j=k, at least not to me.
@fsponj
2 ай бұрын
Wait what about i^^j
@jacksonstarky8288
Жыл бұрын
I was introduced to quaternions by the video done by 3Blue1Brown... and they keep getting stranger and more interesting with every new thing I learn about them. The exponential transformations exposed in this video are an especially brain-bending result.
@monadic_monastic69
Жыл бұрын
I'd suggest looking into the clifford/geometric algebra way of looking at quaternions next. They'll certainly get more interesting, but less strange (an algebra on planes, and there are three planes in 3D space: xy, yz, zx. Oh btw, i, j, k, are basically the normal vectors to these guys if you wanted to think of it that way), and though I like 99.9999% of 3blue1brown's vids, I don't agree with the framing of quaternions as inherently "4-dimensional creatures" (and the clifford/geometric algebra view on this clears this up, also clears up what exponentials mean when you're putting in different objects in there). While at the same time, if you *want* 4D things, you can look at the clifford/geometric algebra generated by a 4D vector space: 'space-time algebra'.
@RadicalCaveman
Жыл бұрын
@@monadic_monastic69 Unless the number of videos produced by 3blue1brown is a multiple of a million, the percentage of them you like cannot be 99.9999.
@CommanderdMtllca
Жыл бұрын
That description though lol
@trueriver1950
Жыл бұрын
The reason we use H for the quarternions is only secondarily because the letter H stands for Hamilton. Firstly: we can't use the now obvs is Q because that's already taken. This is, of course, a rational reason.
@sleepycritical6950
Жыл бұрын
There's a Quarternion version? Damn, you learn something new everyday.
@FractalWoman
3 ай бұрын
I came up with a better, simpler and easier to understand implementation of Euler's formula in quaternion form that can be implemented directly into a computer program to do rotations about an arbitrary axis in 3D space. @FractalWoman "Demystifying Sir William Rowan Hamilton's Quaternions" kzitem.info/news/bejne/146vtHVmiISKmXYsi=62Eao0gdh7O5LPRp A link to the computer code is available in the description.
@jhuyt-
Жыл бұрын
I did not expect spinors to make a surprise appearance at the end. But you are spinning things so I guess I should've
@winteringgoose
Жыл бұрын
What is this video description lol Great vid, I love seeing quaternions at the zoo like this. Wish I saw more in the wild...maybe.
@paulkohl9267
Жыл бұрын
Lovely video, as usual. Great channel to watch. Cheers.
@gab_14
Жыл бұрын
Quaternions written as matrices were the theme of French most difficult math contest (X-ENS, Maths A) last week, to enter the best engineering school.
@James2210
Жыл бұрын
I really liked this video because of all the math this dude did. what a guy
@sherylbegby
Жыл бұрын
I love the way you present the ideas. Engaging and clear. You asked about the red-brown chalk once. It is a bit hard to see, for example at 2:06. around "example calculations". But these are quibbles. It works for boxes and lines, and the rest of the boards are clear to follow.
@Fire_Axus
2 ай бұрын
your feelings are irrational
@ronaldjorgensen6839
Жыл бұрын
thanks for your persistence
@sayonbasak-x3w
Жыл бұрын
I guess the description is a mathematical secret code
@seneca983
Жыл бұрын
18:45 Why is it log(A)*B rather than e.g. B*log(A)? Isn't that rather arbitrary choice (unless the matrices commute)?
@sciphyskyguy4337
Жыл бұрын
When teaching the addition of many disparate components, I like to underline the components as I deal with them. This establishes a bookkeeping methodology for the students (and me!) that prevents losing track of terms in a long list.
@QP9237
Жыл бұрын
More Quaternions!
@markjk444
Жыл бұрын
I really loved the video it's very good . Thank you sir😊 we will support you
@PetervanderJagt123
Жыл бұрын
i suggest you start with q=a+b1*i+b2*j+b3*k and b=sqrt(b1^2+b2^2+b3^2) then q=a+b*Î with î the specific normalized quarternion here with |î|=1 and î=b1/b*i+b2/b*j+b3/b*k, then e^q becomes very simular to the complex case: q=a+b*î and e^q=e^a*(cos(b)+sin(b)*î) which notation seems much more alike that in the complex situation. Right?
@kummer45
Жыл бұрын
Quaternions, octonions, tetrads, tensors, twistors and vectors are devices for physical description of movement and variation of quantities. This is STRICT computer graphics. Hamilton among many other mathematicians are geniuses.
@vikramkrishnan6414
Жыл бұрын
Suddenly I feel thankful for Heaviside's version of Maxwell's equations
@kevinmartin7760
Жыл бұрын
At 12:38 that seems to be to be a huge leap of faith. I expect that if you, say, expanded e^x to a series, stuck I in it, and simplified, I can see you could get the stated result, but saying it is so just because I^2=-1 seems like an awful stretch.
@jeromemalenfant6622
11 ай бұрын
1:09 The 'where' could/should include ijk = -1. Then the 6 relations below that follow from these 4 relations, which are easier to remember.
@toygartumer7385
3 ай бұрын
I really like this video this guy is awesome CHALK CHALK CHALK BLACK BOARD
@mathunt1130
Жыл бұрын
This comes from Clifford algebra, or as it's known these days, geometric algebra.
@thatwontwork9046
Жыл бұрын
This is exactly what I needed!!
@zwiguy9494
Жыл бұрын
Nice can you do maxwells equations next?!
@Marvelous771
Жыл бұрын
interesting ! thank you very much
@MuffinsAPlenty
Жыл бұрын
The ending is a bit concerning, as you mentioned. We can still define exp(a+BꞮ) = e^a(cos(B)+Ɪsin(B)). From here, let's define the modulus of a quaternion q = a+bi+cj+dk as |q| = sqrt(a^2+b^2+c^2+d^2). Then, letting q = a+bi+cj+dk = a+BꞮ as above, we can define a quaternionic logarithm as log(q) = ln|q| + Ɪarccos(a/|q|), for some choice of arccos(a/|q|) Then we can check that exp(log(q)) = q for all quaternions q. Next, given two quaternions p and q, the next question is how to define q^p. The sneaky part here is that we have two reasonable options: exp(log(q)*p) or exp(p*log(q)). Note that these two do _not_ have to be equal to each other since quaternion multiplication is noncommutative. My proposal, then, is to define two versions of exponentiation - right exponentiation and left exponentiation. At the risk of confusing with tetration, we use the following notations: qᵖ = "q right exponentiated by p" = exp(log(q)*p) ᵖq = "q left exponentiated by p" = exp(p*log(q)) So in your example of i raised to the j, we have to consider a few things: For q = i, we have a = 0, B = 1, Ɪ = i, |q| = 1, so log(i) = ln(1)+i*arccos(0) = iπ/2, let's say. iʲ = "i right exponentiated by j" = exp(log(i)*j) = exp(ijπ/2) = exp(kπ/2) = cos(π/2)+k*sin(π/2) = k ʲi = "i left exponentiated by j" = exp(j*log(i)) = exp(jiπ/2) = exp(−kπ/2) = cos(−π/2)+k*sin(−π/2) = −k This explains the discrepancy in your short where you talk about this. Some interesting things to point out here: if r is a positive real number, then log(r) = ln|r|, which is a real number and commutes with all quaternions. So for any quaternion p and positive real number r, we have rᵖ = ᵖr (i.e., right and left exponentiation of r by q produces the same result). So we haven't introduced an incompatibility between e^p and exp(p) (using a good choice of argument).
@MuffinsAPlenty
Жыл бұрын
As a bit of an addendum, in general, exp(p+q) will not be equal to exp(p)*exp(q) since quaternions don't commute. And certainly p^(q+s) will not be equal to p^q * p^s for quaternions p, q, and s (for either left or right exponentiation). As a bit of fun, we can even consider i¹⁺ʲ = exp(log(i)*(1+j)) = exp(iπ/2*(1+j)) = exp(iπ/2+kπ/2) = cos(π/√2) + i*√(2)sin(π/√2)/π + k*√(2)sin(π/√2)/π whereas i¹*iʲ = i*k = −j So these are very different results.
@alikaperdue
Жыл бұрын
Editor note: @5:26 - don't use red chalk for text, just outline. Can't read "exercise". Also, it seems there is a mild glare in the center of your board. Change to position lights or camera to not glare.
@hach1koko
Жыл бұрын
Really interesting, thanks!
@noumanegaou3227
Жыл бұрын
The quatirion field post in exam MathA 2023 in France is beautiful exam to more understanding the quaternions field specialy her matrix represtation
@Meni_Rosenfeld
Жыл бұрын
You can likewise show that j^i = 1/k, which is pretty cool.
@marcosolza3698
Жыл бұрын
Wonderful! Thanks so much 😮
@ere4t4t4rrrrr4
Жыл бұрын
But you didn't show *why*, just because the quaternion I has the property I² = -1, you can apply the Euler formula as if it were a complex number
@williejohnson5172
Жыл бұрын
19:45 Not sketchy at all. In fact you just proved what I have established long ago: ALL QUATERNIONS ARE NATURAL LOGS AND MUST THEREFORE OBEY ALL THE RULES OF EXPONENTS. Thus by the power rule of exponents i^j=i*j=k. You can prove this using DeMoivre's formula. You can also prove it by using his [j^j=e^(-pi/2)]=j*j=-1 if the negative sign preceding pi is i^2 and you treat ^j and i^2 like exponents. In fact you would get e^(i*pi)=-1.
@abdouabdelrazek6037
Жыл бұрын
I would love to thank you a lot for the video. I have a question (extremely curious): Is that method applicable for the higher Hypercomplex Numbers such as Octonions and Sedenions or there will be some differences in the "B" and "I" construction?
@davidwright8432
Жыл бұрын
Generally, every time you go up to a 'next level' of complexity, the algebra changes; in quaternions for instance, commutativity of components under multiplication is lost. I haven't played with Octonions yet, but expect it's a different game.
@GreatestPhysicistOfAllTime
Жыл бұрын
I don't think your derivation of the quaternionic Euler formula is rigorous enough, which is even wrong. Using your result that e^iπ/2=i and so on, we can derive an equation such that ijk=-1=•••=cos(√3/2)π+(i+j+k)/√3*sin(√3/2)π, which is a clear contradiction. You cannot jump to the conclusion just because the square of the large I equals to minus one and suppose it works totally the same to the imaginary unit little i.
@GreatestPhysicistOfAllTime
Жыл бұрын
If you do the calculation e^(a+bi+cj+dk) as e^a•(cosb+isinb)(cosc+jsinb)(cosd+ksind), you will realize that it gives a totally different result to your "Euler's formula". Such a product will not even be definable under the non-commutative multiplication of quaternions.
@ArthurRainbow
2 ай бұрын
May I ask what is the definition of exponential? I know many of them, but I admit I don't know how well they extend to quaternion. As a series, I fear that noncommutativity of functions makes me doubt that we can do anything interesting with series anymore. As the unique solution of f'=f with f(0) = 1, I must admit that I have no idea how to prove that such a function exists. I knew the proof of its existence by showing that the series satisfy this property, and analysis showing that this ODE has a unique solution. But it's far from clear that all the mathematics tools created for complex analysis extends to quaternion. So yeah, if we accept that exponential exists and behave similarly on pure imaginary of modulus 1, I can follow you. But it's a huge assumption for me to accept without proof
@amathok6506
Жыл бұрын
Very very interesting video.Million thanks from Morocco.
@xnick_uy
Жыл бұрын
I wonder if we can use the properties of the exponential to write exp(a+bi+cj+dk) = exp(a) exp(bi) exp(cj) exp(dk) = exp(a)(cos b + i sin b)(cos c + j sin c)(cos d + k sin d) ... but this looks very _sketchy_, given that i, j and k do NOT commute. Probably the right way to go about this path is to write the exp as a series and take proper care of the commutation relations.
@angeldude101
Жыл бұрын
You're right in that it's sketchy. In general, the power laws only actually apply when multiplication is commutative. You can pull out the scalar part since it commutes with the rest, but you can't split up the imaginary part.
@kapoioBCS
Жыл бұрын
It is definitely a wrong proof/derivation. The rigors generalized way to deal with something like that is using representation theory
@mikejurney9102
Жыл бұрын
Can jAe^jB((iCe^iD) - (iEe^iF)) be put in the form I*Ge^I*H, where i is complex, j is part of the quaternions, and I is the same as your notation?
@SurfinScientist
Жыл бұрын
Isn't it necessary to prove that e^(BI) = cos B + I sin B, instead of just stating it as a fact, with the argument "because I behaves in a similar way as the complex number i"?
@pwmiles56
Жыл бұрын
You only need that I^2 = -1 which is easily verified. After that the series expansions work in exactly the same way as with e^(Bi)
@SurfinScientist
Жыл бұрын
@@pwmiles56 OK, that makes sense.
@eastherwilson9356
2 ай бұрын
how we can define quaternion j and k without any matrix ? IN case of imaginary number i , i = square root of -1 but in quaternoins they use matrix but matrix is not a operation like square root , matrix is use for telling dimension of any thing , matrix is not a operation .
@Fractured_Scholar
Жыл бұрын
Shouldn't the modulus of the quaternion around 5:30 be under a square root???
@kevinkoslowski2236
Жыл бұрын
Can we have nonstandard quaternions next, please?
@mattcarnevali
Жыл бұрын
I've always treated quaternions like the unit vectors i j k from physics
@APaleDot
Жыл бұрын
You can only represent a translation using standard vectors (via addition), but quaternions represent rotation (via multiplication). You cannot traditionally multiply standard vectors, but if you look into Geometric Algebra, then quaternions can represented as the ratio between vectors (this is also historically how Hamilton conceived of the quaternions). Basically, if you have j (i^-1) then obviously multiplying that by i would give you j, because the i's cancel. In that way j (i^-1) is like the k from quaternions (order still matters in the multiplication). This method of slipping between vectors and quaternions is only valid in 3-dimensional space because the quaternion really represents the _plane_ spanned by i and j, and it just so happens that in 3D, you can represent this plane precisely using a vector normal to the plane, which is the k vector.
@carultch
Жыл бұрын
If you ever wonder why we use the trio of i/j/k of all possible letter combinations, as unit vectors in Physics, it is because Hamilton had wanted quaternions to replace vectors in general. It didn't catch on as Hamilton had hoped, but a lot of his notation did, as well as operations like the dot product and cross product, that were inspired by quaternion math. I prefer x-hat / y-hat / z-hat as the notation for coordinate unit vectors, over i-hat / j-hat/ k-hat.
@alexanderst.7993
Жыл бұрын
I legit wanna know where the inspiration for the descriptions come from :D
@MichaelPennMath
Жыл бұрын
My ADHD brain is allow to run wild and I just type stream of consciousness. That's the truth. -Stephanie MP Editor
@Fine_Mouche
Жыл бұрын
why it's not i⁴=-1, j⁴=-1, k⁴=-1 ? I ask that because in complex, i is like a 90° rotation, so here, it could be 45°, (just i and j would be 60° so i³=-1, j³=-1) Or maybe it could be i²=-1, j³=-1, k⁴=-1. (but surely no more ij=k, jk=i, ki=j, but maybe anothers emerging proprieties) Oh ! we could imagine a infinite serie Sum_n=0,n=inf of x_n+i_n where (i_n)^n = -1 With a sub set of that where at a range, all x_n = 0 (to be polynomial-like)
@wolliwolfsen291
Жыл бұрын
Recall at 18:20, why is a+b*i+c*j+d*k equal to this matrix? The Rest is clear, you can prepare i by setting a=c=d=0 and b=1, j by setting a=b=d=0 and c=1, using exponent rules, matrix multiplication along with A^B=(e^(logA))^B to get i^j=k
@kapoioBCS
Жыл бұрын
It is not equal, it is just different representation called the complex matrix representation..
@schweinmachtbree1013
Жыл бұрын
Michael shouldn't have used an equals sign, he should have used something like a ⇿ (I mean a two-headed arrow, like ←→ as one symbol, but when I put the default two-headed arrow ↔ it comes out as an emoji XD)
@gabenugget114
Жыл бұрын
r u ok? im always here 4u
@daniellosh1015
Жыл бұрын
The more interesting result is j^i = -k
@ChuffingNorah
Жыл бұрын
Hey Prof - you forgot ijk = -1 Just Saying! 😅🤣😂
@UdssRAP
Жыл бұрын
I mean he gave us ij=k which is equivalent to ijk=-1 if you assume k^2=-1
@vaheakli4551
Жыл бұрын
19:04 shouldn't it be -1 in left corner? edit: yep, 20 seconds later it should 🤗
@FrankHarwald
2 ай бұрын
Question: what are is (i+j)^(i+j) & (i+j+k)^(i+j+k) ?
@siyuanchen659
Жыл бұрын
It's very helpful for me to understand the exponential map of quaternions. I just realized this video was talking about en.wikipedia.org/wiki/Exponential_map_(Lie_theory)#Examples ("In the quaternions H" section), which corresponds to 13:40 if assign a=0. Thanks!
@enysuntra1347
Жыл бұрын
1:15 Wrong Audio. It is read "kj=i", correct (and on blackbord) jk=i. Can you correct the audio for visually impaired/ppl just listening to the creation, too?
@enysuntra1347
Жыл бұрын
Obvious check: if multiplication is in alphabetical séquence, wrapping at k->i (modulus), it's positive. If it's backward, i wrapping to k, it's negative.
Before looking at what you did, my guess is for the cosine of the angle to each axis.
@jacksonrocks4259
Жыл бұрын
It might be just the way it’s drawn, but do quaternions have anything to do with galois theory?
@kapoioBCS
Жыл бұрын
This derivation is sketchy, since by transitions to H (Hamilton Quaternions) we have lost the commutativity of the ring H
@mykolapokhylets3105
Жыл бұрын
If I try to compute e^(ix+jy) directly I get a zero coefficient for k. But if try to compute (e^ix)(e^jy) I end up with a term of i*j*sin(x)*sin(y) = k*sin(x)*sin(y), which looks like a contradiction to me.
@kapoioBCS
Жыл бұрын
e^(ix+iy) ≠ e^ix e^iy since the ring of quaternions is not commutative
@@forcelifeforce oh we so pedantic! Then e^ is making zero sense and it should be replaced by the exp map etc etc
@deltalima6703
Жыл бұрын
This is as dodgy as string theory. Smh
@oshaakau
Жыл бұрын
ngl this thumb kinda sucks
@kapoioBCS
Жыл бұрын
I am waiting to see the octonions video with the nonassocietivity hell 😂
@HarperChisari
Ай бұрын
Just wait for the sedenions or Bi-Sedenions!
@manucitomx
Жыл бұрын
What a gnarly way to start the week. Thank you, professor.
@tylerdancey6085
Жыл бұрын
I really liked this video because of all the math this dude did. what a guy
@whendiditfall
Жыл бұрын
It is not clear to me why i^j should be k instead of -k. Why do we say that (e^(i*pi/2))^j is e^(ij*pi/2) instead of e^(ji*pi/2)?
@rehanchopdar617
2 ай бұрын
Because multiplication is not commutative it is anti commutative that is just (iπ/2)j that is ijπ/2 but if we were to write jiπ/2 we need to add a -ve signed as well so the exponent can be written as -jiπ/2
@alnfsyh9403
2 ай бұрын
@rehanchopdar617 I'd like to ask for clarification on why the multiplication of the exponents would lead only for i to be multiplied by j and not j to be multiplied by i I understand that the Hamiltonians aren't commutative. I am just asking why the exponents would multipy by that order.
@rehanchopdar617
2 ай бұрын
@@alnfsyh9403 how do you interpreted (a^m)^n. I would say a^(m×n) like I said multiplication is anti commutative m×n=-n×m there are lots of ways to prove the exponent and multiplication anti commutativity, you can search them up
@lorisschirar6680
Жыл бұрын
Both of those formulae are special cases of the Clifford algebra formulation of 3D rotations :)
@protocol6
Жыл бұрын
Consistency with the order of the imaginary and its coefficient might be a good idea, especially when you start getting into quaternions. I mean, if you try to go the route of complexifying a complex number, (a + bi) + (c + di)j = a + bi + cj + dk is fine but (a + ib) + j(c + id) = a + ib + jc - kd is slightly problematic. You could do (a + ib) + j(c - id) = a + ib + jc + kd, of course, which is a nice callback to the particular complex matrix form you use; though I prefer a different form. Anyway, the signatures are pretty arbitrary here (as sign is, generally) so long as you are consistent, throughout. I'd just stick with trailing imaginaries, personally, but then you run into a formatting issue with Euler's formula. If you want to use the leading i in Euler's and the conjugate imaginary component in the matrix it's probably best to stick with leading everywhere to avoid problems.
@scottmiller2591
Жыл бұрын
Fun video. Lord Kelvin declared that quaternions, “though beautifully ingenious, have been an unmixed evil to those who have touched them in any way, including Clerk Maxwell” 👿
@Thalesfreitas96
Жыл бұрын
What are the connection between the quaternions and the Pauli matrices?
@pwmiles56
Жыл бұрын
The Pauli matrices are generators of the Lie group of the unitary 2x2 matrices with determinant +1, i.e. SU(2). The key relation is U = exp(b sigma_x + c sigma_y + d sigma_z) in which sigma_x = [0 1] [1 0] sigma_y = [0 -i] [i 0] sigma_z = [1 0] [0 -1] (i.e. the Pauli matrices) b, c and d are real coefficients U is an arbitrary member of SU(2) The relation to quaternions is by putting sigma_x = i, sigma_y = j, sigma_z = k U = q, an arbitrary quaternion of modulus 1 Michael's result for a=0 is q = exp(bi + cj + dk) = cos(B) + I sin(B) with I = (bi + cj + dk)/B, B = sqrt(b^2+c^2+d^2)
@MooImABunny
Жыл бұрын
when I learned that Pauli matrices have such a similar Euler's identity I was really surprised. But if you add the identity you get exactly the quaternion algebra, so knowing that, the quaternion exponentiation makes perfect sense
@peersvensson9253
Жыл бұрын
As a physicist I feel like you should discuss the connection to SU(2). It's quite simple to show the version of Euler's formula in that context as well.
@ScienceTalkwithJimMassa
Жыл бұрын
This was a very interesting, fun presentation. Thank you professor))
@MrGranddy
Жыл бұрын
Hi Michael, there is a major simplification when applying the Power Rule for Exponents to the quaternions, you said exp(i * pi /2) ^ j is exp(i * j * pi / 2) yet I think j being on the right side of i is not so obvious. Why is this the case? Since normally this rule was applied to commutative numbers, now with quaternions there must be a special explanation of this specific order. Edit: Probably this is related to the problem you mention in the short video.
@ntesla66
Жыл бұрын
I loved this! Would love to see this extended to the Clifford representation.
@zemoxian
Жыл бұрын
Whenever I see a video about complex numbers I automatically wonder about quaternions. Whenever there’s quaternions I automatically wonder about geometric algebras. Fun fact about Euclidean (Clifford) geometric algebra is that the basis vectors square to 1. Products of non-parallel vectors create bivectors. Unit bivectors square to -1 so they act like i in Euler’s formula. In fact, the scalars plus bivectors form a sub algebra that’s equivalent to the quaternions!
@Apollorion
Жыл бұрын
@@zemoxian Even when the vectors have more than 3 dimensions?
@person1082
Жыл бұрын
if you feed vectors into it you get hyperbolic rotations
@monadic_monastic69
Жыл бұрын
@@Apollorion clifford/geometric algebra works independent of the dimension of the vector space that you feed it. It just requires a vector space *and* the inner product structure you feed it (i.e. if some of the vectors square to -1, or even 0). Even a generalization of the cross product called the 'wedge product' works in a geometric algebra generated by a vector space of more than 3 dimensions. It comes down to the way the cross working based off what the orthogonal complement *is* of the vector you're feeding it (and the cross product fails, because it demands to output just a vector while there are more than 1 orthogonal vectors now to a given vector in 4D. The wedge product which works on multivectors doesn't have this limitation) EDIT: One more thing I want to add is a lot of people will mention wedge-products as not having a visualization to them. I would be wary of statements like these, they can be represented as the plane spanned by those two vectors you're 'wedging' (but this plane, or circle, or whatever also has an orientation to it. So e1 ^ e2 has the opposite orientation to e2 ^ e1). They may have very good algebraic explanations, but not the geometric explanation handy, and that's totally ok! (the vice-versa also can happen, and you should be wary of that too when/if you run into it).
@Apollorion
Жыл бұрын
@@monadic_monastic69 Thank you for your response. I realise I do not know enough about the clifford/geometric algebra yet.
@ghkthILAY
Жыл бұрын
I think the contradiction showed in the short video can also be seen here. when you take i^j=(e^{i*pi/2})^j pulling j inside of the exponential isn't well defined I think, would it necessarily be multiplication from the left or should it be from the right? because of the commutation relations i and j have it can be either e^{k*pi/2} or e^{-k*pi/2}.
@ghkthILAY
Жыл бұрын
also, as usual great video- I love those videos of yours where you push poke known math in unusual ways to see what comes out
@Nickle314
Жыл бұрын
We're nudging closer to the Clifford Algebra and Geometric Algebra
@DavidFMayerPhD
Жыл бұрын
How much of Complex analysis can be extended to the Quaternions? I have often wondered about this. Do you have any sources or references? Using i as quaternion and complex in same equation is VERY confusing.
@Noam_.Menashe
Жыл бұрын
It's somewhat related to multivariate complex analysis. But I'm not an expert and I don't know how the loss of commutativity changes the analysis.
@michaelaristidou2605
Жыл бұрын
Not much, because they are non commutative. For example, the Foundamental Theorem of Algebra fails!
@DavidFMayerPhD
Жыл бұрын
@@michaelaristidou2605 Thanks.
@schweinmachtbree1013
Жыл бұрын
The analogue of complex analysis for quaternions is called quaternionic analysis.
@DavidFMayerPhD
Жыл бұрын
@@schweinmachtbree1013 Is it good for anything?
@-.......-
Жыл бұрын
I really liked this video because of all the math this dude did. what a guy :)
@xizar0rg
Жыл бұрын
The title for this video is quite clear and descriptive. The title card is also quite clear and descriptive. Thank you.
@MessedUpSystem
Жыл бұрын
Some years ago I calculated a general formula for a quaternion raised to a quaternion power. Quite messy and arduos, but was fun and the result is quite neat, a scaled rotation along the axis of both original quations followed by a rotation about the axis of their cross product. (Everytime I say "the axis of" I mean the axis defined by the purely imaginary part, and the cross product being within the imaginary parts also) Edit: of course, this was all abusing notation and glossing over the sketchiness of using exponentials on a non-commutative algebra, but I was on my first year of college so chill I didn't have the knowledge to take that much into consideration xD
@cxpKSip
Жыл бұрын
In a way, e^{i\theta} is an abuse of notation.
@friendlyone2706
Жыл бұрын
@@cxpKSip notations are made to be abused stretched to the ultimate - that's how all ideas are tested and great ones are created. Keep it up!
@scalex1882
Жыл бұрын
Exponentiation of a matrix is just applying the series representation of the exponential to the matrix. It is so amazing to see the concept of exponentials generalize this way... Same for the logarithm. Thank you for making the video!!
@Tehom1
Жыл бұрын
Yes. My favorite example of generalized exponentiation is the math of squeezed light, such as what LIGO uses. It's the sinh and cosh of the sum/ difference of the creation and annihilation operators, the quantum mechanical operators that mean "add a particle" and "remove one". Since sinh and cosh are just sums of exponentials, we are exponentiating to the power of "add a particle" etc and it actually works!
@scalex1882
Жыл бұрын
@@Tehom1this sounds awesome, is there a KZitem resource you can point to??
@Tehom1
Жыл бұрын
@@scalex1882 I read this in a scientific paper and I'm afraid I don't know of a youtube video about it, sorry.
@apteropith
Жыл бұрын
regarding the end, there: it feels as though we'd need to define a separate left-sided exponentiation to account for the lack of commutativity, at a bare minimum
@riccardofroz
Жыл бұрын
I actually finally understand quaterions now.
@thefunpolice
11 ай бұрын
God I love mathematics. It's so lovely to play with ideas in this way.
@FractalWoman
3 ай бұрын
At the 19:42 mark, shouldn't the lower left "i" be "-i" ? In other words, shouldn't the lower left of your matrix "-c+di actually be -(c+di) = -c-di. When you do this, then you get the matrix [0 c+di], [-c-di 0] and when you let c=0 you get [0 i], [-i 0] = k. I can see this clearly when I use the 4x4 matrix implementation of quaternions directly which I show in my video: @FractalWoman "Demystifying Sir William Rowan Hamilton's Quaternions" kzitem.info/news/bejne/146vtHVmiISKmXYsi=62Eao0gdh7O5LPRp
@Happy_Abe
Жыл бұрын
What’s the short video that shows why this is sketchy?
@MichaelPennMath
Жыл бұрын
It’ll be out tomorrow. On KZitem Shorts. -Stephanie MP editor
@Happy_Abe
Жыл бұрын
@@MichaelPennMath looking forward!
@Jooolse
Жыл бұрын
@@MichaelPennMath So satisfying to find the answer to the question you were about to ask :)
@elgefe5442
Жыл бұрын
i^j = ±k depending on how you apply rule for the exponential of an exponential (right vs left).
@Jeathetius
Жыл бұрын
This is true of matrix exponentiation as well-it gives rise to two kinds of exponentiation-left and right-which for commutative multiplication are the same. What is less obvious to me is what the principal branch of the quaternion logarithm should be. If you have the Log of -1, which quaternion do you use? Any purely imaginary quaternion with modulus 1 will work.
@kono152
Жыл бұрын
THE DESCRIPTION LMAO
@RGAstrofotografia
Жыл бұрын
Do you know the Dihedrons? kzitem.info/news/bejne/zad-mnaCgXmkeaw What should be the Euler formula for them?
@schweinmachtbree1013
Жыл бұрын
for the exercise at 5:27, writing q for a + bi + cj + dk and q*/|q|^2 for (a - bi - cj - dk)/(a^2 + b^2 + c^2 + d^2), one needs to check _both_ q(q*/|q|^2) = 1 and (q*/|q|^2)q = 1, since quaternions are noncommutative - we have |q| ≠ 0 if and only if q ≠ 0 so q*/|q|^2 gives the inverse of any nonzero quaternion, and one can clear the denominators in the checks, meaning it is equivalent to check that qq* = |q|^2 = q*q.
@lajont
Жыл бұрын
You do not really need to look at both if you find that one of them is |q|². If you found that qq*=|q|², you could note that q*q=(qq*)*=(|q|²)*=|q|², and say that you found the answer since |q|² is a real number. If you would find that q*q≠|q|², this would be a non-issue since therefore q^(-1)≠q*/|q|² as noted above.
@1.4142
Жыл бұрын
Gently press
@friendlyone2706
Жыл бұрын
My first introduction to quaternions was glancing through an applied electrical engineering text. Fascinating something so theoretical is so essential to describe electrical behavior.
@AkamiChannel
7 ай бұрын
The unit quaternion algebra is isomorphic to su(2), which is the Lie algebra that describes the angular momentum of fermionic particles (in other words, the so-called 1/2 spin of the elementary matter particles such as the electron, quark, and neutrino). And the "vector algebra" (dot product and cross product) which is used to write the classical laws of EM were historically extracted from the quaternion algebra to begin with. It is indeed fascinating.
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