The complex number family.

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  • Опубликовано: 22 фев 2022
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Комментарии • 267

  • @tomkerruish2982
    @tomkerruish2982 2 года назад +28

    "...associative..." (*sad octonion noises*)

  • @goodplacetostop2973
    @goodplacetostop2973 2 года назад +29

    17:54 Learning is like rowing upstream: not to advance is to drop back… Have a good day ☀️

  • @ivanklimov7078
    @ivanklimov7078 2 года назад +138

    i love these recent vids on weird algebras and complex numbers, they're quite helpful too since i'm currently studying complex analysis. great job michael!

    • @jsmdnq
      @jsmdnq 2 года назад +2

      I wouldn't call these weird... they are quite natural.

    • @Tim3.14
      @Tim3.14 2 года назад +5

      @@jsmdnq Some of them are natural... Others are rather irrational, and some don't even seem real

    • @rajinfootonchuriquen
      @rajinfootonchuriquen Год назад

      ​@@Tim3.14 lol 😂

  • @freyascats1364
    @freyascats1364 2 года назад +66

    For those looking for more about this topic, there is a very readable book "Complex Numbers in Geometry" by I.M. Yaglom, Academic Press 1968. The naming of "j" numbers varies in the literature. Wikipedia uses "Split-Complex Number". Yaglom uses "Double Number". In the area of physics research, Walter Greiner uses "Pseudo-Complex Number" ("Pseudo-Complex General Relativity", Springer, FIAS Interdisciplinary Science Series, 2016).

    • @no3339
      @no3339 Год назад

      I usually just act like the split complex numbers are just the basis vectors of GA

    • @MarekMadejski
      @MarekMadejski 5 месяцев назад

      There is also one author, M.E. Irizarry-Gelpí, who uses different terms: „perplex numbers” for numbers with j^2=1 and „nilplex numbers” for numbers with ε^2=0. The latter term is unique to this author. I like these terms more than “double numbers” and “dual numbers”, as those are confusing, whereas “nilplex” is immediately associated with ε^2=0.

  • @profkrinkels
    @profkrinkels 2 года назад +35

    10:41 we need to set i = beta / sqrt(abs(beta^2)) so its square will be correct

    • @NicholasPellegrino
      @NicholasPellegrino Год назад

      I got stuck at that too. Hopefully what you said is what he intended. I guess sqrt should be well defined given beta^2 is real.

    • @danielyuan9862
      @danielyuan9862 17 дней назад +1

      @@NicholasPellegrino Yeah that is what he most likely intended, and it seems necessary that beta^2 is real and, more specifically, abs(beta^2)=-beta^2 is a positive real, so that its square root is a well-defined real number.

  • @nosnibor800
    @nosnibor800 2 года назад +61

    Thanks, this is filling in gaps. I am an Electrical Engineer/Systems Engineer and of course very familiar with complex numbers (except we dont use "i" because "i" is instantaneous current ! - we use j instead). I have also come across quaternions which we use to convert 3-space into 4-space when doing dynamic space transformations in say missile control systems (where trouble is encountered when the angles approach 90 degrees using euler angles). I have also seen your tutorial on duel numbers, which I had not come across before. So the overall picture is emerging. The problem with Engineering is we treat maths like a "tool" in order to do calculations. Whereas you mathematicians look at the "toolbox" as a whole. Thanks Mr Penn, keep it coming !

    • @MrLikon7
      @MrLikon7 2 года назад +12

      ah yes, the duel numbers :D
      today: pi vs e!
      fight!

    • @lawrencedoliveiro9104
      @lawrencedoliveiro9104 2 года назад +7

      The “trouble” of which you speak is called “gimbal lock”. We encounter it in CG, too.

    • @nHans
      @nHans 2 года назад +4

      ​@@MrLikon7 Last I heard, Euler brokered a 5-way peace agreement between _e,_ π, _i,_ 0, and 1.

    • @nHans
      @nHans 2 года назад +5

      As a fellow engineer, I feel that the tool-in-a-toolbox approach that we take is as it should be. I wouldn't characterize it as a 'problem.'
      There is way too much knowledge out there for any one person to learn in a lifetime. So we have to be very choosy about what we learn, and carefully manage how much time we spend learning. After all, we do need to leave time for ourselves to solve problems in the real world and earn a living!
      I watch these kinds of videos-math, modern physics, biology etc.-because, yes, I do find them very interesting. But I will not be using them in my work-unless I change my line of work 😜. In fact, I don't even use most of the things I learnt in engineering college-they taught us way too many things, hoping to cover all bases. On the other hand, I've had to unlearn and relearn a lot of other things as my work kept evolving.
      No doubt I could study these topics in detail if I wanted to-but that time would have to come from something else that I'm already doing. Priorities!
      Back in college, if I tried to major in all the courses that I was interested in, I'd never have graduated. As it is, 4 years was plenty! 🤣

    • @Grassmpl
      @Grassmpl 2 года назад

      @@MrLikon7 yes. Epsilon lost to itself. Its square is 0.

  • @Alex_Deam
    @Alex_Deam 2 года назад +62

    There's a nice connection between the hyperbola at the end and light cones in special relativity. You can even show that the equivalent of the Cauchy-Riemann equations for split-complex numbers is a wave equation!

    • @neopalm2050
      @neopalm2050 2 года назад +2

      Wait how? I'm not getting that.

    • @rajinfootonchuriquen
      @rajinfootonchuriquen Год назад

      ​@@neopalm2050 it's not hard. I prooved about 3 weeks ago. Now I will see of dual numbers have some kind of CR equations, but just by it's definition it shouldn't.

  • @gudmundurjonsson4357
    @gudmundurjonsson4357 2 года назад +110

    Fun how the final graphs seem to go from a circle in the negative, to a hyperbola in the positive case, and the zeros case is somehow inbetween them. I wonder if you could some connection to the conic sections if you continuously varied what the squared element was from -1 to 1

    • @PhilBoswell
      @PhilBoswell 2 года назад +11

      Since those three conic sections are obtained by intersecting a plane with a double-cone at different angles, I imagine the parameter you're looking for would relate to that angle.

    • @elliottmanley5182
      @elliottmanley5182 2 года назад +9

      The thought that occurred to me is that the hyperbolic trig functions relate to a hyberbola in the same way the standard trigs functions relate to a circle. I wondered if there's a complementary extension of the trig functions that relate to parallel lines. Of course, if there were, it would be completely useless!

    • @angeldude101
      @angeldude101 2 года назад +26

      @@elliottmanley5182 The circular and hyperbolic trig functions are ultimately just the even and odd parts of the imaginary and real exponential functions respectively. So finding a dual cosine and sine would just be to find the even and odd parts of e^αε = 1 + αε + (αε)²/2 + ... Normally you'd go further, but we already hit ε² so everything past it would be 0. So a cosd(α) = 1 and sind(α) = α. Significantly less interesting than sin/cos or sinh/cosh.

    • @KohuGaly
      @KohuGaly 2 года назад +9

      @@angeldude101 _less_ interesting? I'd say the fact that sind and cosd are algebraic instead of transcendental is pretty interesting. Especially from computational standpoint, because it means you can do arithmetic with them with fewer approximations.

    • @NutziHD
      @NutziHD 2 года назад +8

      Visualizing the basis (1,a) of R(a), where a^2 is in R, in a cartesian plane, we can indeed observe that the unit circle, that is all z in R(a) s.t. z•z*=1 (where the * denotes the conjugate) goes from a circle (a^2=-1) to a ellipse (-1

  • @samwinnick4048
    @samwinnick4048 2 года назад +69

    I think there are a few minor mistakes here. In the second case I think you should instead define i=beta/sqrt(-beta^2) instead of beta/|beta^2|, where for a positive real x, sqrt(x) is its unique nonnegative square root. Similarly in the third case I think you should instead define j=beta/sqrt(beta^2) rather than beta/beta^2.

    • @jkid1134
      @jkid1134 2 года назад +10

      I haven't seen this stuff before, but something wasn't quite adding up around there.

    • @anto2593
      @anto2593 2 года назад +9

      i was looking for exactly this comment. Thanks

  • @rivkahlevi6117
    @rivkahlevi6117 2 года назад +105

    "a multiplicative identity, which we will generally call one" 🤣🤣🤣 Gotta love mathematicians.

    • @austinlincoln3414
      @austinlincoln3414 2 года назад +3

      Lmao

    • @hybmnzz2658
      @hybmnzz2658 2 года назад +11

      It's that or "I" or "e"

    • @GeekProdigyGuy
      @GeekProdigyGuy Год назад +8

      that is a useful convention for the abstract general setting - after all we do not usually call the identity matrix 1

  • @pedrocusinato1743
    @pedrocusinato1743 2 года назад +32

    I think i = beta/sqrt(abs(beta²)), the same with j

    • @taeyeonlover
      @taeyeonlover 2 года назад +1

      that made me so confused

    • @schweinmachtbree1013
      @schweinmachtbree1013 2 года назад +4

      ​@@taeyeonlover it's confusing because Michael's notation is confusing; it is better to "write 1/sqrt(abs(beta²)) · beta" to emphasize that the denominator is a scalar multiplication, because if you confuse it for ordinary multiplication then it looks like you can cancel the numerator and denominator ( even more so for "j = beta/sqrt(beta²)" )

    • @taeyeonlover
      @taeyeonlover 2 года назад +1

      @@schweinmachtbree1013 ah no it was just that the sqrt missing threw me off

  • @kruksog
    @kruksog 2 года назад +10

    You're awesome Dr. Penn. I'm a guy with a math degree who otherwise doesn't have much of an inroad to mathematics otherwise these days. Your videos bring me that joy of experiencing math I don't get too often in daily life anymore. Thank you!

  • @cpiantes
    @cpiantes 2 года назад +6

    Can you do an episode on Clifford algebras?

  • @AleksyGrabovski
    @AleksyGrabovski 2 года назад +4

    Your videos reminded me how I love abstract algebra.

  • @wladwladsnotmyrealname9082
    @wladwladsnotmyrealname9082 2 года назад +7

    One arguably nice property of the double numbers (which are defined to be R(j)) is that linear algebra over them is non-trivial. A matrix over the double numbers is a pair of matrices over the real numbers (A,B^T). Addition happens componentwise, and multiplication is sort-of componentwise: (A,B^T) * (C,D^T) = (AC,(DB)^T). The final algebraic operation to define is the "conjugate-transpose" operation: (A,B^T)^* = (B,A^T). Then you have that the Hermitian matrices are pairs (A,A^T) which are a non-trivial embedding of the square matrices, the unitary matrices are pairs (P,(P^-1)^T) which are a non-trivial embedding of the invertible matrices, the "spectral theorem" is just the Jordan canonical form for real matrices, the Cholesky decomposition is effectively the LU decomposition for real matries, the normal matrices are pairs of commuting real matrices, and so on. There is a non-trivial analogue of the Singular Value Decomposition as well.

    • @wladwladsnotmyrealname9082
      @wladwladsnotmyrealname9082 2 года назад +3

      Hopefully, this won't be perceived as a rant by most.
      I think that the analogue of complex analysis over the double numbers is not as nice as their linear algebra. This is because linear algebra uses the conjugation operation (a + bj)^* = a - bj. Linear algebra over the double numbers depends strongly on this operation. Complex analysis studies the analytic functions, which do not include conjugation. I think that because of the isomorphism R(j) = RxR, a holomorphic function over the double numbers is just an arbitrary pair of differentiable real functions, so double-number analysis is effectively real analysis AFAICT.

  • @TheDannyAwesome
    @TheDannyAwesome 2 года назад +7

    I also like to consider these number systems as the ring of polynomials over the real numbers quotiented out by a quadratic polynomial. In the case of the dual numbers, x^2, in the case of the complex numbers, x^2+1, and in the case of the split complex numbers, x^2-1.

    • @adityajain16701
      @adityajain16701 2 года назад

      Those are indeed all isomorphic to their respective 2D R-algebras

    • @angeldude101
      @angeldude101 2 года назад +3

      1 + -1 = 0, j² + i² = ε²

    • @schweinmachtbree1013
      @schweinmachtbree1013 2 года назад

      @@angeldude101 nice!

    • @thewhitefalcon8539
      @thewhitefalcon8539 4 месяца назад

      This raises the question of other quotient rings such as real polynomials/x³+x²+1. Maybe they end up just being complex numbers with a weird choice of I or maybe not

  • @neonsilver1936
    @neonsilver1936 Год назад +1

    I've watched 2 of your videos now, and I can already tell that I'm going to learn a LOT from you. Thank you for showing me that I need to improve my mathematics.

  • @camperbbq3026
    @camperbbq3026 2 года назад +2

    Please do more content on the split complex numbers! I love your videos!

  • @dtrimm1
    @dtrimm1 2 года назад

    Great videos - I’m loving them, and learning a ton - thanks for doing them!

  • @SiiKiiN
    @SiiKiiN 2 года назад +1

    I like these style of videos which helps build a landscape for motivation behind algebras

  • @nmmm2000
    @nmmm2000 2 года назад +8

    Woaw :) Definitely need to render Mandelbrot in Split complex numbers too...

    • @thewhitefalcon8539
      @thewhitefalcon8539 4 месяца назад

      There's a fractal called the Burning Ship Fractal that you get if you mess up the Mandelbrot equations slightly.

    • @nmmm2000
      @nmmm2000 4 месяца назад

      @@thewhitefalcon8539 burning ship utilize standard complex numbers, but get absolute value (modul) from the complex part (is not that easy but this is the basic idea)

  • @m1323fj
    @m1323fj 2 года назад

    Really nice video. I thoroughly enjoyed it and would like to see more content like this.

  • @eschudy
    @eschudy 2 года назад

    Very cool!! More please!

  • @stephenmorton9789
    @stephenmorton9789 Год назад

    Great stuff, thank you

  • @unflexian
    @unflexian 2 года назад +4

    You're making me love abstract algebra:D

  • @starsun7455
    @starsun7455 2 года назад +1

    I've been studying hyperbolic for several months, so it was really good that Professor Penn dealt with 'split complex numbers' called hyperbolic numbers.

  • @Skyb0rg
    @Skyb0rg 2 года назад +3

    You should explain Clifford algebras! Many of those "families" can co-exist in those algebras, and some examples (such as R_3; R_3,2; and R_3,0,1) model 3D spaces really well.

  • @stevenwilson5556
    @stevenwilson5556 2 года назад +1

    wow so cool. great way to tie this together

  • @AJ-et3vf
    @AJ-et3vf Год назад

    Great video. Thank you

  • @zemoxian
    @zemoxian 2 года назад +4

    Reminds me of the various geometric algebras. The multivectors have similar bases and can even reproduce quaternions and Minkowski spacetime.

  • @Ricocossa1
    @Ricocossa1 2 года назад +3

    Circles in R(j) and R(\epsilon) are like lines of constant proper time in Mikowskian and Gallilean space-times.

    • @radadadadee
      @radadadadee 25 дней назад

      for a stationary observer?

  • @copernic7511
    @copernic7511 Год назад +1

    That merch is great! No way I'm forgetting that formula now.

  • @NutziHD
    @NutziHD 2 года назад

    Awesome video!!

  • @joel.9543
    @joel.9543 2 года назад +2

    Amazing!

  • @jantarantowicz1306
    @jantarantowicz1306 2 года назад

    as always, great video!

  • @neopalm2050
    @neopalm2050 2 года назад +2

    I predict the clifford algebras and perhaps the weyl algebras at some point in the near-ish future

  • @Zeitgeist9000
    @Zeitgeist9000 2 года назад +12

    Love this area of math, thanks for the cool content!

    • @shalvagang951
      @shalvagang951 2 года назад +1

      which area of math is this I would really love to know

    • @Zeitgeist9000
      @Zeitgeist9000 2 года назад +1

      I was wondering the same thing

    • @shalvagang951
      @shalvagang951 2 года назад +1

      @@Zeitgeist9000 I have searched for it and I got the result something about hypercomplex number in group representation theory

    • @shalvagang951
      @shalvagang951 2 года назад +1

      @@Zeitgeist9000 I just wanna ask you one thing that anytime you hear about any new math topic that has recently been discovered what would be your thoughts about it pls answer

    • @shalvagang951
      @shalvagang951 2 года назад +1

      like you wanna learn it or just "ahh heck with it I am not doing this"

  • @sciencewithali4916
    @sciencewithali4916 2 года назад

    Amazing topic 👏 thank you very much. You second RUclips channel truly is very ins8ghtful and helpful

  • @nbooth
    @nbooth 2 года назад +4

    Excellent, thank you. Any chance you'd do a video/series on multivectors from geometric algebra?

  • @wilderuhl3450
    @wilderuhl3450 2 года назад

    Fascinating

  • @sebastiandierks7919
    @sebastiandierks7919 2 года назад +4

    Maybe a follow-up video on Grassmann numbers, i.e. the the exterior algebra of a vector space, and their differentiation and integration (Berezin integral) would be interesting, as they are a generalisation of the dual numbers! Would love to see that video :) Basically, I don't understand how you integrate over a Grassmann variable. In the context of this video, how would you integrate over the dual number epsilon? It's just a single basis element of the algebra!?

  • @General12th
    @General12th 2 года назад +2

    This is a really neat video!
    I don't think I would have been able to follow along if I was just reading through a textbook, even if the book had all the same arguments laid out in the same order. I wonder what makes lectures different and easier for me to digest.

    • @xrhsthsuserxrhsths
      @xrhsthsuserxrhsths 2 года назад

      I had this problem too, I still do to some extent, but it gets better. It also depends on how familiar is the subject you read.

  • @alexey_burkov
    @alexey_burkov 2 года назад +2

    omg, that preview picture 😂

  • @apteropith
    @apteropith Год назад +1

    these can all be understood as planar rotational algebras for vector spaces with particular metrics, as well
    with the right choice of setup you can get some fairly impressive use out of them (and their higher dimensional analogues)
    such as representing arbitrary translations as dual-number rotations around a paraboloid, which allows the composing of translations with (regular) rotations much more effectively
    ... i don't have much opportunity to make use of this knowledge myself, but it is a subject that fascinates me

  • @travisporco
    @travisporco 2 года назад

    fascinating stuff. Would love to see more applications of the dual numbers

    • @angeldude101
      @angeldude101 2 года назад +1

      Dual numbers can be used for automatic differentiation since the "imaginary" term acts like the derivative in many ways. While it's not specifically the dual numbers, a basis squaring to 0 is also useful in representing translations of objects, much like complex numbers are good at rotating, and split-complex numbers are good at hyperbolic boosts (more niche than the other two, but very useful in relativity).

    • @drewmandan
      @drewmandan 2 года назад

      Probably has applications for things involving perturbations. Maybe even stochastic calculus.

  • @CTJ2619
    @CTJ2619 2 года назад +3

    I like learning about the quaterians

  • @drewmandan
    @drewmandan 2 года назад +1

    This looks so much like general relativity and closed/open/flat space! Cool

  • @user-fi6if8gx3g
    @user-fi6if8gx3g 2 года назад +1

    A lovely family indeed.

  • @Grassmpl
    @Grassmpl 2 года назад +1

    You should never use round parenthesis when the ring extension does not form a field. Use square parenthesis.
    R[epsilon], R[j], R(i) this last one forms a field

  • @Khwartz
    @Khwartz 2 года назад

    Excellent! Very Clear I think thanks of a Very Nice Progression in the Lecture :) (y) (y)

  • @spudhead169
    @spudhead169 Год назад

    I think you spilled something down your hoodie there Mike.

  • @Dalroc
    @Dalroc 11 месяцев назад

    Love it that Michael can't differentiate between his a's and his u's.
    Makes me feel better about my own a's and u's!... and my z's and 2's. and my I's and 1's..

  • @CesarMaglione
    @CesarMaglione 2 года назад +2

    Buena pizarra Michel 👍

  • @aweebthatlovesmath4220
    @aweebthatlovesmath4220 2 года назад +1

    Numbers are cool!

  • @Nickle314
    @Nickle314 2 года назад +5

    On the quaternions, the more human understandable approach is bivectors from a geometric algebra, and how they multiply.

    • @angeldude101
      @angeldude101 2 года назад +1

      It's definitely much easier to see (e1e2)(e2e3) = e1e2e2e3 = e1e3 = -e3e1 than to try and remember ij = k.

    • @Nickle314
      @Nickle314 2 года назад +3

      @@angeldude101 Agreed. GA to me has been a revelation. Covers large parts of Tensor calculus, Quaternions, Complex numbers, Dirac algebra .... Then you have the physical equations.
      ∇F=J/(ϵ0c)+MI
      One equation, the whole lot.
      On complex numbers, I think there is a case for teaching the GA approach first at a very basic level, to show that things can square to -1 in a real sense.

    • @Jono98806
      @Jono98806 2 года назад

      Quaternions are actually a subset of the geometric algebra (as are complex numbers and split complex numbers).

    • @Noam_.Menashe
      @Noam_.Menashe 2 года назад

      It's also a bit like the cross product.

    • @angeldude101
      @angeldude101 2 года назад +3

      @@Noam_.Menashe But the main difference is that the cross product is evil and needs to die in a fire because it's completely unintuitive both geometrically and algebraically. The wedge product by contrast is very intuitive both geometrically and algebraically.

  • @_P_a_o_l_o_
    @_P_a_o_l_o_ 2 года назад +3

    Wonderful content! Will you consider making a video on quaternions?

    • @elliottmanley5182
      @elliottmanley5182 2 года назад

      If so, how about taking a similar "family" approach and explaining how constraints of addition and multiplication drop away as you move from complex to quaternions to octonions to sedenions?

    • @elliottmanley5182
      @elliottmanley5182 2 года назад

      A bit more background reading today: I started to get a handle on Clifford algebras and then got lost completely when I got to Bott Periodicity.

    • @cpiantes
      @cpiantes 2 года назад

      @@elliottmanley5182 8N = N8

  • @user-hh5bx8xe5o
    @user-hh5bx8xe5o 2 года назад

    Geometrically, the multiplication of the algebra is a composition of a dilatation and a rotation for C, a reflection with y=x for R(j) and a projection along x=0 for R(epsilon)

  • @brattok
    @brattok Год назад

    In 13:45 we could also say that R(j) is not a field because polynomial x²-1 has more than 2 roots

  • @bobdowling6932
    @bobdowling6932 2 года назад +12

    Typo at 11:05-ish and again a little later. I think you are missing a square root in your normalizations to get i and j. The denominator needs to be sqrt(-beta^2) and sqrt(beta^2) respectively.

    • @badlydrawnturtle8484
      @badlydrawnturtle8484 2 года назад +2

      Thank you. That makes sense. I had to pause the video and look through the comments for an explanation when that bit didn't follow.

  • @beatadalhagen
    @beatadalhagen 2 года назад +1

    Will there be a follow-up involving combining these?

  • @Grassmpl
    @Grassmpl 2 года назад

    An associative algebra is a ring and a real vector space, with the compatibility of scalar multiplication.

  • @rv706
    @rv706 Год назад +2

    Please denote the dual numbers by R[ε], not R(ε), cause it's just a ring not a field; generally the notation with round parentheses is for field extensions, as opposed to ring (or algebra) extensions.

  • @JakobVirgil
    @JakobVirgil 2 года назад +1

    I have noticed that if I want to use dual and complex numbers together that it sort of forces me into quartics but I am having a hard time expressing why. Could you shed some light on this?

  • @radupopescu9977
    @radupopescu9977 2 года назад +1

    In fact, complex number and split complex numbers, are subalgebras of bicomplex numbers.

  • @deltalima6703
    @deltalima6703 2 года назад +1

    Cohl Furey over at perimeter institute did a series relating octonions to QM and GR. Perimeter is legit, as I am sure most people know, having been a phd and nobel prize factory for awhile. It might be worthwhile adding octonions in as well, they are not as useless as you would suspect.

    • @deprivedoftrance
      @deprivedoftrance 2 года назад

      Now I might need to go re-watch them as it's been a while.
      Those were good ones.

  • @pajrc1234
    @pajrc1234 Год назад +1

    So when I saw the video on the dual numbers I tried writing down a bunch of relations between it and i and the matrix stuff. I discovered that e^(+-b epsilon)=1+-b*epsilon--the same curve that represents the "unit circle." I figure the same is true with j. Also, (ep*i)^2+(i*ep)^2=1 and (ep*i)+(i*ep)=-1. It's weird because these are not commutative. I'll have to add j to the mix to find other other things. I'm going to see if f(x+b*ep)=f(x)+b*ep*f'(x) works even when b is complex, and complex algebra treats dual numbers the same way

  • @gabrielfrank5142
    @gabrielfrank5142 Год назад

    Some questions:
    1. Is the fact that the dual and split cases are not a field is related (i.e. iff) to the fact there unit circles are not continuous?
    2. Can we define other types of numbers based on arbitrary "unit circle" curve geometry?

  • @Raptormind1
    @Raptormind1 2 года назад

    I feel like I'm missing something. at 5:50, x+y(alpha) is just shorthand for x(1)+y(alpha) where 1 is the multiplicative identity of the original vector field, right? So at 7:31, isn't 4a+b^2 shorthand for 4a+b^2 times the 1 from the vector field? So shouldn't beta squared be a real number times the 1 from the vector field, not just a real number? What am I missing?

  • @skillerror951
    @skillerror951 2 года назад +2

    Looks like curvature of space

  • @tothm129
    @tothm129 2 года назад +1

    can I give you a closed curve and then say its a unit circle and then you work backwards

  • @nrrgrdn
    @nrrgrdn 2 года назад +1

    Thanks, why don't we hear of this in a standard abstract algebra course

  • @littlecousin5630
    @littlecousin5630 Год назад

    10:44
    Doesn’t that yield i^2=B^2/|B^2|^2=|B^2|(-1)/|B^2|^2=-1/|B^2|. How do we get to -1 from there?

  • @przemekmajewski1
    @przemekmajewski1 2 года назад +1

    Nice video! I have a remark: You've been talking about associative algebras, then suddenly you've switched to "normed algebras", IMO it was too quick a jump. The notion of an adjoint and how to introduce norm are important stuff

    • @przemekmajewski1
      @przemekmajewski1 2 года назад

      In fact in your very nice proof of the fact how many "2D algs" there are you've shown, that any such algebra is in fact a normed *-algebra, as you nicely constructed the element "beta" that squared to a central element.

  • @catherinebernard3282
    @catherinebernard3282 Год назад

    Really silly thing to point out, but if you start at 17:23 or so you can hear the chalk break, and if you go frame by frame you can actually see it fall... which for some reason I found amusing.

  • @Grassmpl
    @Grassmpl 2 года назад +1

    Only in C does the modulus induce a metric.

  • @brian554xx
    @brian554xx Год назад

    Just a stray thought that I hope will be picked up by someone with more talent than I have: Is there a way to use dual numbers in string theory for the collapsed dimensions? Is there such a thing as a hybrid of quaternions and duals? I don't see any conflict. Can one make something with multiple epsilons?
    Perhaps this has been tried. If so, could you direct me to search terms that will help me find it?

  • @elliottmanley5182
    @elliottmanley5182 2 года назад +2

    Really nice survey that made these (previously, to me) very abstract concepts accessible.
    Wondering if there's an algebra that joins more than one of these imaginary bases: e.g. a four-dimensional R(i,j,epsilon)

    • @elihowitt4107
      @elihowitt4107 2 года назад +1

      Interesting if this mixture yields anything

    • @angeldude101
      @angeldude101 2 года назад +2

      Geometric algebra lets you specify the signature of the algebra as how many bases square to 1, -1, and 0. You can have G_1,1,1(R) which has {1, i, j, ε, ij, iε, jε, ijε}. In practice you don't need this since two positive bases already have an imaginary product, and it's possible to make a dual basis vector from a positive and a negative basis.

    • @opfromthestart3645
      @opfromthestart3645 2 года назад

      I believe that this 4d structure would just be the set of all 2x2 matrices. If you represent i,j, and epsilon as matrices it makes sense

    • @opfromthestart3645
      @opfromthestart3645 2 года назад

      You could also make them as 8x8 matrices (technically in 3 ways, but all will act the same). You can also make them into 4×4 matrices in three distinct ways, following each pair, which won't commute with the other one in the pair, but they should both commute with the one left out.

  • @timseytiger9280
    @timseytiger9280 Год назад +1

    Nice, how about redefining the absolute value so we always get the "same" picture?

  • @user-qu2oz2ut2h
    @user-qu2oz2ut2h Год назад +1

    hmm
    dual and split-complex Mandelbrot set analogs... Sweeps of "i" from minus infinity up to infinity...

  • @marcoottina654
    @marcoottina654 Год назад

    13:20 is R(j) just the negated complex of C ?
    17:55 also, it seems that:
    -) C is something like "two dimensions swirling all together in a coupled, attractive way, with the "balancing" fact that the radius always stays the same (1)"
    -) R(j) are two "divergent" dimensions, like a "swirling where the interaction leads to divergence, like particle with the same electrical charge"
    -) R(epsilon) seems two completely unrelated, never-interacting "dimensions", which closely resembles the very concept of "parallelism"

  • @Fine_Mouche
    @Fine_Mouche 2 года назад

    17:54 : "ok that's a good place to start" So it will have a next episode of this series about "hypercomplexe family" ?

  • @arnbrandy
    @arnbrandy 2 года назад +1

    IIUC he worked out the modulus of dual and split-complex numbers by multiplying them by their conjugate, very much like with complex numbers.
    I don't understand why would we use the same definition here. I understand the motivation behind using this definition for complex numbers, it gives the distance from the origin. Why don't we define modulus differently for the other algebras so we get the same property.
    I found it fascinating how the modulus definition leads to different conic curves but I wonder why would it be especially important.

    • @angeldude101
      @angeldude101 2 года назад

      If you were to just square a split-complex number, then you'd get something like (x+yj)^2 = x^2 + 2xyj + y^2; not quite what we're looking for. The conjugate is needed to eliminate the middle term of the expansion and get a pure difference of squares.

  • @MisterPenguin42
    @MisterPenguin42 29 дней назад

    What should I have studied before I can begin to understand this video?

  • @mattiasselin4955
    @mattiasselin4955 2 года назад

    Nice

  • @christressler3857
    @christressler3857 2 года назад

    Using ß for beta..
    I don't think you want,
    i = ß/|ß^2|, unless by |•| you mean something like the *norm*, iow a square root value.
    It might've made more sense to put,
    i = ß/|ß| which looks more like a unit vector than your notation does.

  • @mvsh
    @mvsh 2 года назад

    I expected the "circle" in the dual-numbers plane to be a parabola after seeing the circle and hyperbola because of the conic sections… Is there anything that will give us it?

  • @HoSza1
    @HoSza1 Год назад

    Well after digging around a bit, it looaks like the defining properties of H are i²=j²=k²=ijk=-1 and the other three "rules" in the video are in fact theorems that are easy consequences of these four. I wonder what happens if we define ijk=1 (or zero) instead, is the resulting structure still useful in some ways?

    • @James2210
      @James2210 11 месяцев назад

      Split quaternions? Only j^2 and k^2 are 1

  • @antoniusnies-komponistpian2172

    Though j just has the properties that 1 and -1 have in common. j is a kind of Schrödinger's 1 that you give a letter just because you don't know if it's positive or negative 😅

  • @EMAngel2718
    @EMAngel2718 2 года назад +1

    I wonder if the very similar correlation between basis sign and shape to that of spatial curvature has any meaning

    • @EMAngel2718
      @EMAngel2718 2 года назад

      I just realized it's actually the reverse but still

  • @geelaw
    @geelaw 2 года назад

    It's a bit confusing why something in A in the form of x*1+0*alpha is considered to be in R (as you mentioned 1 means the identity in A) --- it would be instructive to first show that R is naturally embedded into A so that x in R can be regarded as x*1 in A.

  • @HoSza1
    @HoSza1 Год назад

    About quaternions: I understand that i²=j²=k²=-1 rules may naturally come as an extension of the complex numbers, but where do the last 3 rules come from? I was possible to derive jk=i and ki=j from the other rules, but I had to use ij=k, and I don't know if this last one can be derived too, or it is a free choice itself.

    • @dantebroggi3734
      @dantebroggi3734 Год назад

      I believe the `jk=i` form rules are generally derived from a rule `ijk = -1`.

  • @anonymous-vs8oo
    @anonymous-vs8oo 2 года назад +1

    @MichealPenn Sir
    I am in grade 10(midschool) should I learn from other books than syllabus. Actually I like to do things which are easier and all that. But I want to to do everything which can be learned. So should I do what I love or should I do only fron syllabus?
    Please answer anyone.

    • @nHans
      @nHans 2 года назад

      Well, I suppose it depends on which country you're in.
      If you're in India, then-based on my own experience-I strongly suggest that you master the syllabus for your board exams before you do anything else. Passing class 10 and 12 with high marks is of paramount importance in the Indian education system. That, and preparing for engineering or medical entrance exams.
      I don't know enough about other countries to advise you. If you have the time, energy, and resources to study further-you're a very lucky person, and by all means, do so. Although, even in that case, I'll suggest that you don't spend all your time studying academic topics. Balance it with physical sports, hobbies, and social activities with friends-I mean, with real people, and involving in-person interactions.

    • @bobdowling6932
      @bobdowling6932 2 года назад

      I don’t know what your syllabus contains, but if you don’t have complex numbers on your syllabus they would be my recommendation. So much of algebra “fits” better in the complex context. I can still remember the buzz of realizing “where the solutions went” for quadratics when the discriminant (the bit in the square root in the classical formula) goes negative. You also get to understand what’s really going on with the Mandelbrot set and the corresponding Julia sets. Euler’s equation makes trigonometric identities easy. It’s the one thing that gives back most, IMHO.

  • @DirkDanckaert
    @DirkDanckaert 2 года назад

    Small error in the normalisation op beta: i,j := beta/|beta| and ot beta/|beta|^2

  • @VideoFusco
    @VideoFusco 2 года назад

    Case 3 is also known as hyperbolic numbers

  • @djsmeguk
    @djsmeguk 2 года назад +2

    This must have some sort of relationship to the conics.

  • @weonlygoupfromhere7369
    @weonlygoupfromhere7369 2 года назад

    My brain is too small for this

  • @williamwright1440
    @williamwright1440 Год назад

    Has anyone seen any papers/videos/textbooks on algebras where the products of the basis vectors with themselves are not necessarily real? E.g. an algebra over the reals where the basis vectors are 1 (the real number) and x (not a real number, or anything in particular, for that matter), with 1*1=1 (of course), 1*x=x, x*1=x, and x*x=x.

  • @Number_Cruncher
    @Number_Cruncher 2 года назад +5

    Does the norm, that was used to calculate the modulus, fulfil the axiom of positive definiteness in each of the the three cases? I doubt, that it does. Therefore, I'm hesitant to accept these objects being generalized circles.

    • @Theo0x89
      @Theo0x89 2 года назад +4

      The norm is like a field norm, which doesn't have the axiom of positive definiteness.

  • @HoSza1
    @HoSza1 Год назад

    According to wikipedia split complex numbers can describe the points of the 2d Minkowsky space just like cemplex numbers describe the "normal" 2d plane. I suppose split numbers are used in relativity then, but did Einstein use them in his original work or is this an insight from those who came after to interpret his work and find more modern way to write down the original formulas?

    • @MattMcIrvin
      @MattMcIrvin Год назад

      It's a later insight. Minkowski represented special-relativistic spacetime by putting a factor of i from the complex numbers on the time coordinate, but defining the norm in a "Euclidean" way otherwise, without taking complex conjugates, so that a minus sign appeared in the calculation. In one space and one time dimension that's equivalent to the split-complex numbers, just written slightly differently.

  • @Kris_not_Chris
    @Kris_not_Chris 2 года назад

    does this mean there's a connection between the hyperbolic functions and split complex numbers like there is between the trigonometric functions and complex numbers?

    • @angeldude101
      @angeldude101 2 года назад +1

      Yup! e^ix = cos(x) + isin(x), e^jx = cosh(x) + jsinh(x). Dual numbers have an odd variant: e^iε = 1 + εx.