Imaginary derivative of x

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  • Опубліковано 9 лют 2025
  • This is the video you've all been waiting for!!! In this video, which is a sequel to my half-derivative of x video, I evaluate the imaginary derivative of x, that is the alpha-th derivative of x, where alpha = i. Although there is no formal definition of the imaginary derivative, I can still calculate it by analogy to what I did with the half-derivative video. Enjoy!

КОМЕНТАРІ • 382

  • @drpeyam
    @drpeyam  7 років тому +217

    Typo: I forgot to put i sinh(pi) in the final answer.
    The answer should be:
    (i-1)/2pi Gamma(i) i sinh(pi) (x cos(ln(x)) + i x sin(ln(x)))
    Which can be written as:
    (i-1)/2pi Gamma(i) sinh(pi) (- x sin(ln(x)) + i x cos(ln(x)))
    Also, in case you’re wondering about e^x, cos, sin:
    Fractional derivatives of exponential and trigonometric functions ua-cam.com/video/k2T0YilPrWw/v-deo.html

    • @christophermusso
      @christophermusso 7 років тому +12

      Dr. Peyam's Show
      Didn't you use the i in front of sinh(pi) to go from (1+i) to (i-1)?

    • @david-yt4oo
      @david-yt4oo 5 років тому +2

      you scared me, so I came to the comment section to see if I was right or wrong

    • @dougr.2398
      @dougr.2398 5 років тому

      We all make mistakes! Thanks for noticing & correcting.... but..... there is perhaps a deeper issue that shows just before 7:39. ....Does simply stating something to be true make it so? (I won’t drag in politics here, but there IS a real-life example or two in the current news)...You claim that the formula derived for real number derivatives is valid for complex numbers. In what way has this been demonstrated, shown or proven?!? [and I’m curious to know if any demonstrated results have important applications and uses). I haven’t finished watching due to other priority tasks, but this is in my « play » list ( = WORK!).

    • @dougr.2398
      @dougr.2398 5 років тому

      P.S. I love how you always thank us for watching first!!! That’s really Really nice of you!!

    • @liahsheep
      @liahsheep 5 років тому +2

      I paused at 17:06 to look for why the sinh(pi) vanished. Thanks for saving my day.

  • @Uni-Coder
    @Uni-Coder 6 років тому +199

    We're ready for quaternions, jth and kth derivatives, and Frobenius theorem

    • @skilz8098
      @skilz8098 5 років тому +15

      Throw some sparse matrices in there with some affine transformations...

    • @naterojas9272
      @naterojas9272 5 років тому +16

      @@skilz8098 Don't for get to add "Artificial Intelligence" into the title for good measure.

    • @dissonanceparadiddle
      @dissonanceparadiddle 4 роки тому +4

      @@naterojas9272 but those two things are in fact relevant to this topic

    • @reinerwilhelms-tricarico344
      @reinerwilhelms-tricarico344 2 роки тому

      Very interesting and well explained, but after about 15 minutes I couldn’t read all the scribbles.

    • @zemoxian
      @zemoxian 2 роки тому

      Why not go straight to Geometric Algebra? Then you get imaginary, quaternions, vectors, and more automatically!

  • @TheJeffSnake
    @TheJeffSnake 5 років тому +43

    This is the right place to learn, to relax, to be amazed, to feel as you are sited in the front row of a master class of mathematics. Please Dr. Peyman, never stop to share with us your knowledge.
    Kind Regards!

    • @azzteke
      @azzteke 10 місяців тому

      Who tf is Peyman??

  • @Jaojao_puzzlesolver
    @Jaojao_puzzlesolver 5 років тому +118

    Thumbnail : *D i x*
    Me : Looks *interesting*

  • @skatelife59
    @skatelife59 6 років тому +69

    Very impressive, but can you do the derivative'th derivative of x

    • @drpeyam
      @drpeyam  6 років тому +27

      Hahaha, good one 😂

    • @mike4ty4
      @mike4ty4 5 років тому +31

      @@drpeyam Interestingly, raising to the power of a differential operator is possible: if D is the differential operator, then you can "formally" find its exponential via
      e^D = 1 + D + D^2/2! + D^3/3! + ...
      where D^n represents n-fold differentiation, and this acts as you'd expect on a function by
      (e^D) f = (1 + D + D^2/2! + D^3/3! + ...)f = f + Df + (D^2 f)/2! + (D^3 f)/3! + ...
      So you _could_ find that the Dth derivative of x should have x^(1-D) as power, which equals x x^(-D) = x e^(-ln(x) D) and the latter can be found using the above series expansion (only will have powers (-1)^n ln(x)^n D^n instead of just D^n in the numerators). Taking the gamma of D, on the other hand ... that I have no idea. But the Dth derivative will be an operator - a very weird one.
      ADD: Actually, e^D has a nice interpretation as the unit translation operator - I just remember: [(e^D) f](x) = f(x + 1) for a suitable f. This has deep significance in quantum mechanics (in theoretical physics), too.

    • @TagRLCS
      @TagRLCS Рік тому

      @@mike4ty4 what the FUCK

    • @datguiser
      @datguiser Рік тому +3

      Now do a matrix-th derivative of x

    • @maxvangulik1988
      @maxvangulik1988 Місяць тому

      @@datguiserD is a matrix

  • @JorgetePanete
    @JorgetePanete 7 років тому +182

    For every ex you've had you have to ask yourself: "Why?", so you can have a y for every x

    • @DiamondSane
      @DiamondSane 5 років тому +4

      I enjoyed this

    • @turolretar
      @turolretar 4 роки тому +4

      all I got is a point at zero

    • @XoPlanetI
      @XoPlanetI 6 місяців тому

      I thought you were talking about ex girl friends

  • @MiroslawHorbal
    @MiroslawHorbal 4 роки тому +9

    Thank you for your videos. Having only learned "vanilla" calculus and using it quite regularly in my day to day life, these videos have been inspiring to remember why I fell in love with mathematics when I was younger.

  • @azmath2059
    @azmath2059 7 років тому +68

    Incredible. Pure maths at it's highest. Just wanted to mention that your presentation has improved remarkably.
    Whiteboard is clear and easy to read, audio is good and your dressed well for the camera.

  • @alanturingtesla
    @alanturingtesla 7 років тому +161

    Peyam is a living legend.

  • @saitaro
    @saitaro 7 років тому +57

    This guy gets better and better.

  • @dysrhythmia
    @dysrhythmia 6 років тому +20

    Hey, I found a way to think of Gamma(i), assuming I did it right. If you plug i into the integral and expand it with Euler's formula, you get two integrals: integral of 1/x*cos(lnx)e^-x and i*1/x*sin(lnx)e^-x. With the u sub: u = lnx, du = 1/x*dx, we get the integral from 0 to infinity of -1/u*cos(u) and -i/u*sin(u). The imaginary part is -pi/2, but the real part diverges. However, evidently the Gamma function integral does not converge absolutely for Re(z)

    • @rarebeeph1783
      @rarebeeph1783 2 роки тому +2

      Plugging (i-1)! into Wolfram Alpha, we get that Gamma(i) is approximately -0.155 + 0.498i. So unfortunately, either something has gone wrong in your calculation, or we're dealing with a multivalued function for which your calculation gives a different branch.

    • @minimo3631
      @minimo3631 22 дні тому

      ​@@rarebeeph1783 The gamma functions only discontinuties are the poles on the negative intergers, and it is holomorphic everywhere else. You can't have a holomorphic (on it's domain of definition) function with different branches unless if it only has isolated singularities, right?

  • @77Fortran
    @77Fortran 4 роки тому +4

    I think Dr Peyam is a great teacher in that his enthusiasm and positivity open the door to the student feeling that they too can learn this cool stuff.

  • @mmukulkhedekar4752
    @mmukulkhedekar4752 7 років тому +42

    wow seems interesting , never seen this before !!!

  • @auroy
    @auroy 5 років тому +9

    When fractional derivative is not confusing enough

  • @insouciantFox
    @insouciantFox Рік тому

    When I saw you break out {tan x}, I got that feeling that only great, beautiful math can give you.
    Oh my lord that's some good stuff right there.

  • @bens4446
    @bens4446 3 роки тому +1

    Fascinating stuff! Also, love your style. Keep 'em coming!

  • @danielgoc2409
    @danielgoc2409 7 років тому +9

    6:53 "Proof by analogy"

  • @TheMauror22
    @TheMauror22 7 років тому +14

    This is insane. I love it.

  • @VideoBrunno9
    @VideoBrunno9 5 років тому

    Sometimes i just open your videos to listen the happiest "all right thanks for watching" ! Its so cool!!

    • @drpeyam
      @drpeyam  5 років тому +1

      Awwwww ❤️

    • @VideoBrunno9
      @VideoBrunno9 5 років тому

      @@drpeyam I cant believe you just answered!! Best wishes from Brazil!! :))

  • @egillandersson1780
    @egillandersson1780 6 років тому +10

    Amazing ! This is new for me. Are these concepts of half-derivative and imaginary-derivative expandable to other functions that polynomial ones ?

    • @drpeyam
      @drpeyam  6 років тому +4

      Yep, see my playlist!

  • @mrbatweed
    @mrbatweed 5 років тому +19

    9 mins to come to the answer, then 13 minutes to rewrite a rewritten formula that you rewrote in order to rewrite it in a rewritten way.

  • @baristha
    @baristha 5 років тому +1

    This is more valuable than a kg of GOLD to me.

  • @mesballo2224
    @mesballo2224 2 роки тому +1

    WOW WOW WOW this is so cool!! Never imagined that :) By the way: if you apply this i-derivative 2 times to x, since i*i = -1 , does this imply you get the -1-derivative of x, that is the integral of x?

    • @drpeyam
      @drpeyam  2 роки тому +1

      Not quite I think you get the 2i derivative of x

    • @mesballo2224
      @mesballo2224 2 роки тому +1

      @@drpeyam ahahahha yeah you are correct!!! Thanks for replaying i was a bit confused ;)

  • @mohaghaderi6077
    @mohaghaderi6077 4 роки тому

    Nice video, thanks Payam jan! Keep the great work up!

  • @RalphDratman
    @RalphDratman 7 років тому +1

    This is the most fun math series ever----thanks so much!

  • @JorgetePanete
    @JorgetePanete 7 років тому +47

    I
    WANT TO
    BELIEVE

    • @davidwright8432
      @davidwright8432 7 років тому +4

      This is the exact beauty of math. No belief needed! Proof does it all. This is sweat of the intellectual brow - not divine revelation! go over the video carefully, write things down, puzzlements included, and don't take 'huh?'for an answer! Good luck!

    • @JorgetePanete
      @JorgetePanete 7 років тому +3

      david wright it was a reference to... nevermind

  • @-doctorwjo
    @-doctorwjo 5 років тому +5

    My braines sanity: "Am I joke to you?"

  • @davidwright8432
    @davidwright8432 7 років тому +4

    Dr Peyam - a delight and pleasure as always! I must say, pulling that derivative out of thin air reminded me of a magician pulling a rabbit out of an 'empty' hat. then, of course, I remembered Oreo, and all was clear!
    Please would you do a sequence on transfinite numbers? I mean, well beyond 'countable and uncountable infinities', Hilert's hotel etc. Sam Sheppard's excellent book 'The Logic of Infinity', Cambridge U. Press, (no flakery here! ) - might give you some notions of the level to pitch your presentations on this. Not post-Postdoc, but past 1st yr undergrad. Thanks!

    • @drpeyam
      @drpeyam  7 років тому +1

      I was gonna do one on Hilbert’s Hotel, but there’s actually an excellent one around already, and I highly recommend you to watch it! ua-cam.com/video/Uj3_KqkI9Zo/v-deo.html

  • @georget8008
    @georget8008 5 років тому +1

    Dr peyam
    Thank you for fixing the angle of the camera with respect to the board from π/6 to π/4!
    :-)
    My question has to do with the generalization you made regarding the A. From integer to real and then to imaginary.
    How do you prove that this generalization is valid?
    And how this generalization is related to the original definition of a derivative which is a limit.
    Thank you
    George

    • @drpeyam
      @drpeyam  5 років тому

      You’re welcome! And probably just by taking limits, since every real number is a limit of rational numbers

  • @tomctutor
    @tomctutor 2 роки тому

    Ok there is a simple formula for F(D)e^(ax) where D = d/dx operator of course ( *The D-Op Theorem* in fact used a lot in solving differential equations )so before I state the relevance here, I give quick simple example of the power of this theorem:
    Eg, solve y' ' -5y' +6y =e^(4x)
    then [D^2 -5D +6D]y = e^(4x)
    soln y = [1/(D-2)(D-3)] e^(4x) = F(D)e^(4x) where a = 4
    y = [1/(4-2)(4-3)] e^(4x) = (1/2)e^(4x) the particular integral
    complete solution need to add homogeneous [D^2 -5D +6D]y_h = 0 the traditional method with y_h = Ae^(2x)+Be^(3x) of course.
    Now that out the way we need D^(i)x = D^(i)e^(lnx) = D^(i) e^(u) using u = ln x, unfortunately we need to redefine D for new variable u
    which I believe is D_x = {(u-1)e^(u)}D_u (this part I used d/dx = (d/du)(du/dx) chain rule = (xlnx - x)d/du but not 100% certain here)
    so D_x^(i) = (d/dx)^(i) x = {(u-1)e^(u)}^(i)}D_u^(i) e^(u)
    = {(u-1)e^(u)}^(i)}^(i)1^(i)
    = i(x)^(i+1)ln(x/e)
    which if correct should be equivalent to Dr Peyam's derivation.
    But who am i definitely not Pimi thats for sure.

  • @bikalbaral1394
    @bikalbaral1394 5 років тому

    Happy to watch informative video from a cheerful maths teacher :)

  • @garyhuntress6871
    @garyhuntress6871 5 років тому +7

    Do fractional derivatives have any usefulness when analyzing physical systems?

    • @isaiahzepf5842
      @isaiahzepf5842 5 років тому +4

      I saw a video a week or two ago where it was used as an alternate way to solve the tautochrone problem.

  • @IronMaidenEE
    @IronMaidenEE 6 років тому +1

    Since your formula for the Ath derivative of x^N is proved by induction, it means it holds for all a in integers. I don't think you can generalise it just like that for complex numbers as well, because it's a different domain. Correct me if I'm wrong.

  • @t.n.t1229
    @t.n.t1229 3 роки тому

    it's beautiful, love it :3 but i think you should put camera closer at the final answer, it's a little bit blurred

  • @hectordomingotroncoso3047
    @hectordomingotroncoso3047 Місяць тому

    Muy impresionante,My dear profesor!!!!!!

    • @drpeyam
      @drpeyam  Місяць тому

      Gracias!! 😄

  • @christosvoskresye
    @christosvoskresye 5 років тому +5

    To be convincing, this would need to work for functions that are not simple power laws.

  • @pythagorasaurusrex9853
    @pythagorasaurusrex9853 5 років тому +1

    "aye pi aye"... aye aye aye... :) Weird shit, but mind blowing. Never thought about a derivate this way. I always learn something new :)

  • @power2survive
    @power2survive 6 років тому +2

    Hey Dr. Peyam
    Two questions
    Is there any Interpretation of imaginary differentiation?
    Would you like to do a video about fractional differential equations?

  • @Topstudentonyoutube
    @Topstudentonyoutube 10 місяців тому

    Hi Drpeyam, may you please tell me what branch or research paper you got this from. If I can know more about this branch, I will be able to develop a formula that has the potential to solve the Riemann hypothesis

  • @VeanZann
    @VeanZann 4 роки тому +1

    Awesome, I love that passion! :D

  • @davidchan8732
    @davidchan8732 4 роки тому

    thank u, the illustration is realy down to every detail

  • @MrAksupriatna
    @MrAksupriatna 3 роки тому

    Awesome presentation!!......Have you also done for quaternion order derivative?

  • @sachitvarshney1495
    @sachitvarshney1495 5 років тому

    You discovered new Maths ..
    You are Euler in the present world

  • @112BALAGE112
    @112BALAGE112 7 років тому +1

    Where does the "Fact" at 20:20 come from? I couldn't find anything like it. I tried to check numerically and it turned out to be false.

    • @drpeyam
      @drpeyam  7 років тому +2

      There’s a video about that coming on Monday. And it’s possible that the minus sign is a plus sign, that’s why numerically it might be false

  • @artbymou361
    @artbymou361 7 років тому +29

    sir,what is the derivative of x with respect to fractional part of x

    • @drpeyam
      @drpeyam  7 років тому +20

      Wow, beautiful question! But it’s still the same answer but with alpha = {x}. I doubt that the gamma part can be simplified, but the x part becomes x^(1-{x})

    • @etienneparcollet727
      @etienneparcollet727 7 років тому +3

      I think the question wasn't (d/dx)^{x}*x but more of dx/d{x} which, I presume would be 1.

    • @dlevi67
      @dlevi67 7 років тому +1

      Except at integers where it's not differentiable (I think).

    • @etienneparcollet727
      @etienneparcollet727 7 років тому +1

      It is because you can make a continuation:
      it's dx coming from left and from right, for every integer.

    • @handhdhd6522
      @handhdhd6522 5 років тому +1

      Angel Mendez-Rivera he meant (d/d{x}) x

  • @XoPlanetI
    @XoPlanetI 6 місяців тому +1

    If you have doctorate in Math, why do you need a paper in your hand?

    • @drpeyam
      @drpeyam  5 місяців тому

      What does one have to do with the other?

  • @tz233
    @tz233 6 років тому +3

    Uh, Dr. Peyam....I think you just broke calculus ;)

  • @stevewhisnant
    @stevewhisnant 7 років тому +19

    Didn't he loose a factor of sinh(pi) from the gamma function along the way?

    • @GreenMeansGOF
      @GreenMeansGOF 7 років тому +3

      Yes, he forgot to write sinh(π).

    • @drpeyam
      @drpeyam  7 років тому +4

      I did, my bad!

  • @leonardromano1491
    @leonardromano1491 7 років тому +1

    To all those people asking where the sinh(π) went:
    Isn't it obvious that he was working in units where sinh(π)=1 in this certain part?

    • @drpeyam
      @drpeyam  7 років тому

      Leonard Romano Hehehe, that’s a great way of putting it :)

  • @daisnour396
    @daisnour396 4 роки тому

    Thanks Dr. Peyam, very interested.What is the interest to compute the imaginary derivative in our real Life ?

  • @alejandrojoseurielessalced423

    Best regards, I have a question, where can I find information or text to delve deeper into the fractional derivative of complex order, that is, when z has a real and imaginary part other than zero, it would also be good if you uploaded a video explaining this case. thank you

  • @Metalhammer1993
    @Metalhammer1993 5 років тому +7

    imaginary derivatives: the kind of derivatives year eleven students come up with on the exam after half a year of not doing their exercises?^^

  • @tianyizhou775
    @tianyizhou775 5 років тому

    Very enjoyable tutorial! Thank you for the video

  • @digvijaygadhavi7418
    @digvijaygadhavi7418 6 років тому +1

    i am a high school student but still i can understand it because of Dr.peyam sir's great explanations.

    • @mike4ty4
      @mike4ty4 6 років тому

      Do you want to go to be a Mathematician when you get older?

    • @digvijaygadhavi7418
      @digvijaygadhavi7418 6 років тому

      @@mike4ty4 yes. I want to become a mathematician and i also want to do PHD in mathematics

    • @mike4ty4
      @mike4ty4 6 років тому

      @@digvijaygadhavi7418 Are you keeping your Grades Up? Also, have you not been gotten down by Insulters and Bullies? You make sure that if there are any present to understand they are chock full of BS all the time, every time, no matter what. I have to say that so that just in case it is happening to someone else, I can try to stop from happening to them what happened to me in my own teen years. If there aren't, or you already know, good for you. But if there are, you better take that all in. And even if not, maybe there's someone else who needs it.

    • @digvijaygadhavi7418
      @digvijaygadhavi7418 6 років тому

      @@mike4ty4now I am in std.12 in india, and I am preparing for iit-jee engineering entrance exam of india. In my institute I do top in regular exam of physics,chemistry, mathematics and i got 1st rank everytime in mathematics but not in other subjects

    • @mike4ty4
      @mike4ty4 6 років тому

      @@digvijaygadhavi7418 Good on the maths. I am not familiar with these exams though or what you need to get on them to be admitted to University, as I am not Indian.

  • @i_deepeshmeena
    @i_deepeshmeena 7 років тому +12

    your content is much more advanced and good as compared to *bprp* and fapable maths keep going :)

    • @drpeyam
      @drpeyam  7 років тому +7

      Thanks so much!!! They do have some pretty advanced stuff too, though! :)

  • @g0rgth3b0rg
    @g0rgth3b0rg 7 років тому +5

    I think the equation looks nicer using Gamma(i).

  • @ZipplyZane
    @ZipplyZane 4 роки тому

    It seems to me that you could check this definition by checking to see if D^-i(D^i{x^n)) = D^1(x^n). Though with how complicated the answer to one of those is, I'm not sure how well you could get everything to cancel out.

  • @cmilkau
    @cmilkau 5 років тому +1

    Does this satisfy D^a = e^(a log D), treating D as linear operator? Can you even take the log of D? It seems positive semidefinite but it's not index 0 and I can't recall the exact conditions.

  • @amanasci2481
    @amanasci2481 4 роки тому +1

    When you love mathematics too much and you start imagining things 😂

  • @Peter_1986
    @Peter_1986 5 років тому +1

    And here I was thinking that _real_-valued fractional derivatives were crazy.

  • @IoT_
    @IoT_ 4 роки тому

    What books do you use to prepare the information about fractional analysis?

    • @drpeyam
      @drpeyam  4 роки тому

      Brezis functional analysis

  • @DaanSnqn
    @DaanSnqn 6 років тому

    You really have the gangsta way of doing calculus

  • @kehindeajibade5278
    @kehindeajibade5278 4 роки тому

    Interesting. Don't you think that when you find the alpha derivative of x^5. There should be a condition that alpha must be less or equal to 5? Is that necessary?

  • @rybaplcaki7267
    @rybaplcaki7267 7 років тому +11

    17:00 What's happend with this sinh???

    • @drpeyam
      @drpeyam  7 років тому +7

      Ryba Plcaki My bad, it’s a typo

  • @skeletonrowdie1768
    @skeletonrowdie1768 6 років тому

    i love this!! you are so creative!

  • @soup1649
    @soup1649 3 роки тому

    i'd love to see a proof of the gamma(i) definition!

  • @perlindholm4129
    @perlindholm4129 5 років тому

    Nice. Idea - What if every function has a crack? Defined by an input that makes it impossible to express it as a machine learning model() function.

  • @hishan.farfan
    @hishan.farfan 7 років тому +37

    good lord! what is happening in there

  • @stydras3380
    @stydras3380 6 років тому +1

    Love the idea of Dⁱ :) But I don't think the integral of Γ(i) converges. If I remember correctly, the integral representation of Γ(s) is only convergent for Re(s)>0.

  • @Rundas69420
    @Rundas69420 7 років тому +1

    After warching this video I'm completely convinced that you consumed some substances I named my channel after xD.

  • @SteamPunkLV
    @SteamPunkLV 7 років тому +5

    these are going to be 20 really good minutes :)

  • @user-jn4qk3qi4g
    @user-jn4qk3qi4g 4 роки тому +1

    amazing video! When I saw this, the I thought we can just take the square root of its integral as it’s the square root of its -1st derivative. How wrong I was....

  • @KenhelExcallius
    @KenhelExcallius 12 днів тому

    I’ve thought about generalising this to include matrices and my attempts have been questionable
    I wonder what’s your opinion on this

  • @warface_cheaters_caught
    @warface_cheaters_caught 7 місяців тому

    it was stunning af

  • @Wolf-if1bt
    @Wolf-if1bt 2 роки тому

    Could we get the same result by using Fourier transform ? Given the fact that derivation is linear and that deriving sin(x) substracts pi/2 to the phase, I can guess that i-th derivative of sin(wt) is
    (w^i)*sin(wt-i*pi/2). And thus we should sum these sin functions to get de i-th derivative of any periodic function. Of course this doesn't work for x (aperiodic)

  • @Gamma_Digamma
    @Gamma_Digamma 5 років тому

    Imaginary derivatives?
    Great, now we're dreaming math like Ramanujan

  • @VerSalieri
    @VerSalieri 7 років тому

    So, derivative degrades the function by a power of 1(x^n becomes x^n-1], half derivative degrades it by a power of half (x becomes square root of x), and now imaginary derivative comes up with ln(x) (which is far more degradation than square root of x)... no idea what to do with this one, I’ll just step back and admire it from afar.

  • @nosnibor800
    @nosnibor800 5 років тому

    Very nice, but I wonder where you would find a complex derivative? I always think of j as being a rotation operator (sorry i'm an engineer). So "rate of rotation" in a complex (Hilbert) space ?? Doesn't this come up in quantum mechanics ?? Forgive my guess work I'm only a humble electrical engineer, but I sometimes worship at the church of mathematics.

    • @drpeyam
      @drpeyam  5 років тому

      Super interesting questions!!! I’m not really sure, but there should be applications somewhere

    • @nosnibor800
      @nosnibor800 5 років тому

      @@drpeyam In vector calculus we have the gradient, divergence and curl operators, in a normal 3D vector space. These operators use integer derivatives that we are used to. But what about in higher dimensional complex spaces. What are the equivalent operators? Is the j derivative a curl in a complex space ?

  • @serraihacanmatheux2494
    @serraihacanmatheux2494 4 роки тому

    Do you recommend any books on Fractional Calculus?
    For the beginner

  • @zeeek3348
    @zeeek3348 7 років тому +1

    Dr Peyam, what we can do with the fractional part of tanx or another fractional part? Its just and concept?Actually im studying Pure Mathematic but im starting, anyway, amazing video as always

    • @zeeek3348
      @zeeek3348 7 років тому

      It’s okay to say that the fractional part of X its X - the greatest integer of X?

    • @drpeyam
      @drpeyam  7 років тому +1

      Correct, the frac part of x is x minus the integer part (floor) of x. So it’s basically as important as the floor of x, except what’s nice is that it’s always between 0 and 1.

  • @stydras3380
    @stydras3380 7 років тому

    wow... thats awesome :0 how would we integrate with respect to i now? :0 and how could we generalize that...
    Awesome concept and execution!
    Also... did you lose sinh(π) when simplifying or did I miss sth?

    • @drpeyam
      @drpeyam  7 років тому +1

      Integrating with respect to i is differentiation with respect to -i, so just use the formulas with -i :) And yep, I forgot about that factor

    • @stydras3380
      @stydras3380 7 років тому

      Dr. Peyam's Show oh wow! thank you!!!

  • @harisimer
    @harisimer 7 років тому +5

    rip sinus hyperbolicus, it became as meaningless as 1 in multiplication

    • @drpeyam
      @drpeyam  7 років тому

      harisimer My bad!

  • @robkim55
    @robkim55 5 років тому

    I cannot read what is on the black /whiteboard ; is it possible to put the camera near the board?

  • @elimarburger1659
    @elimarburger1659 4 роки тому

    Me trying to play hangman:
    Advanced level math: gamma function

  • @markusheler2762
    @markusheler2762 3 роки тому

    Is it possible to put the formula around 7:00 in it's generalized glory for complex a,b to D^b (x^a)=Gamma(b+1)/Gamma(b+-1-a)x^(b-a)?

  • @gustavorc25
    @gustavorc25 3 роки тому

    Nice, but i have a question:
    What is: D^i ! (x) = ?

  • @theoleblanc9761
    @theoleblanc9761 6 років тому

    I have a question, your définition formula for the derivitive only works for α

  • @galSyehoshua
    @galSyehoshua Рік тому

    Does it also have a motivation?

  • @mihalistsouk1432
    @mihalistsouk1432 7 років тому

    Just subscribed, you rock!

    • @drpeyam
      @drpeyam  7 років тому

      Thank you!!! :D

  • @simplyyummy92
    @simplyyummy92 7 років тому

    That's Crazzyyy! Can you prove Gamma(i) or i! next!?!?!?!?!?! Also maybe !i (sub fractorial) or i!! (double clucktorial)

    • @drpeyam
      @drpeyam  7 років тому

      What do you mean by prove Gamma(i)? There’s nothing to prove!

    • @simplyyummy92
      @simplyyummy92 7 років тому

      Dr. Peyam's Show ohh I thought there was more? 19:09 :)
      Also i got that i! Is numerically equal to: 0.498015668118... - 0.154949828302...i. Also another amazing result is (i^i)! is =Gamma(exp(-π/2)+1) which is a REAL number. I guess it makes sense since i^i is real! Awesome stuff πam :) Love it!

    • @drpeyam
      @drpeyam  7 років тому +1

      That’s because i^i is real :) Use i = exp(pi/2 i) in the base

  • @dataweaver
    @dataweaver 2 роки тому

    Wouldn't it be cleaner to use the Pi function? It's exactly like the Gamma function, except that Pi(n)=n! when n is a natural number.

    • @drpeyam
      @drpeyam  2 роки тому

      Gamma is more common

  • @MrRyanroberson1
    @MrRyanroberson1 7 років тому

    Maybe using the series expansion of sine one could go on to define those

  • @aaronsmith6632
    @aaronsmith6632 4 роки тому +1

    I reinvented fractional derivatives myself in college, and I was always curious about D^i. Thank you!!!

  • @linusschwan6299
    @linusschwan6299 7 років тому +1

    Cool video, got most of what you said, but what does sinh(x) mean?

    • @juauke
      @juauke 7 років тому +4

      Linus Schwan hyperbolic sine
      You can find more about it here :D : en.m.wikipedia.org/wiki/Hyperbolic_function

  • @alimoharam4362
    @alimoharam4362 4 роки тому

    does the imaginary derivative mean the fractional integral ? since the integral is a derivative of the (-1) order or (inverse function)

  • @ZipplyZane
    @ZipplyZane 7 років тому

    Is there any equation expressible with elementary functions where the i-th derivative produces a result that is also expressible with elementary functions?
    Or any real function where the i-th derivative is also a real function?

  • @kharnakcrux2650
    @kharnakcrux2650 2 роки тому

    I think this is what they call Umbral techniques. shady dark techniques that can still give some insight.
    This is what i tinkered with years ago... arriving at exactly this.
    I saw the complex order as just being "rotations".
    So, i apply it to X^p in general.... something strange happens when p=integer..... VS non-integer powers.
    what i call a "Derivative spectrum", just collapses to Zero.
    But non-integer: the Derivatives go on & on to lower and lower degrees.
    -- Integrals become VERY ugly. I had an odd hunch...... that the subtle reason why integrals are such beasts... has to do with the poles Gamma has on the negative side, while it's a nice curve on the right.
    it's beyond my why. But i had another hunch that i could test whether any function's integral is non-elementary.....
    i applied fractional derivatives to X^X. couldn't get far with that ;0

  • @FedericoYulita
    @FedericoYulita 7 років тому

    I have to ask: Did you make this up or is this something that has been done before?

    • @drpeyam
      @drpeyam  7 років тому +3

      I made it up :)

    • @sachitvarshney1495
      @sachitvarshney1495 5 років тому

      @@drpeyam I think Dr Peyam you must release these results to AMS ..

  • @mike4ty4
    @mike4ty4 6 років тому +3

    That formula for Gamma(i) using an integral of a chopped up tangent function - how the heck was that developed?!
    Also You seem like a Sweet person I would like to Hug and Meehhmmhhrr, Okay? :) Mehhr.

    • @drpeyam
      @drpeyam  6 років тому +3

      Awwwww *hug*

    • @mike4ty4
      @mike4ty4 6 років тому

      @@drpeyam :) mehhr. Thanks.

  • @beardymonger
    @beardymonger 5 років тому +1

    Beautiful, thank you for your work!
    One small nitpick about Gamma: I've read, in a great book by the great author and expositor H. M. Eswards "Riemann's Zeta Function", that the whole Gamma(x) = (x-1)! (i.e. that *shift by 1* ) was a blunder by the great Legendre. Gauss used Pi(x) which is *not* shifted by 1. So for obscure reasons we keep needlessly adding "+1" or "-1" to our formulas :-) Same with Pi that sould have been 2Pi btw.