a beautiful gamma function identity -- Legendre's duplication formula.

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  • Опубліковано 19 гру 2024

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  • @Jacob.Peyser
    @Jacob.Peyser Рік тому +37

    This is by far the best derivation to date of the connection between the Gamma and Beta functions on UA-cam.

  • @FrankHarwald
    @FrankHarwald Рік тому +15

    Here's something I discovered about one or two years ago: Legendre's Duplication formula also works pretty much the same over finite fields - just replace the gamma functions with corresponding factorials & the sqrt(pi) will become another constant (solely dependent on the cardinality of the finite field). Proof by complete induction.
    (It can be further generalized to work over arbitrary rings but that's rarely useful unless none of the factorials contain the ring's center, which they almost always do except for very small values so the equation becomes a trivial 0 = 0.)

    • @drdca8263
      @drdca8263 Рік тому

      That sounds very cool! However, I’m unsure quite what you would mean by Gamma(n+(1/2)) in this case.
      Do you mean like, viewing (1/2) as the element which is the reciprocal of 2 in your field? I guess that is the thing that would make sense... ok, I guess that is what you mean.
      By the center of the ring, do you mean like, some zero divisors or something? When you mentioned general rings, I initially was thinking commutative rings, but in commutative rings, I would consider all elements to be in the center, as they commute with everything?
      Oh, also, do you have this defined for finite field of prime power cardinality, or just for prime cardinality? I don’t see how you would define the factorial for e.g. F_9 . Is there a standard way to define that?
      This kind of thing, taking analogies of functions that appear in complex analysis, and looking at analogies in finite fields, is something I rather like the idea of, so I would be happy to hear more about what you’ve done on this!

  • @QmcometdudeShardMaster
    @QmcometdudeShardMaster Рік тому +22

    I would absolutely love to see some videos on finding Gamma of values that are more complicated denominators. I've actually spent quite some time trying to find good ways to work with them on my own first and then with various online sources but I know any video made about it here would definitely frame it much more nicely. Plus, they're cool in general.

    • @Noam_.Menashe
      @Noam_.Menashe Рік тому +3

      There aren't really any good values with bigger denominators, but there are if you multiply some gamma functions.

  • @tasosgiot7038
    @tasosgiot7038 Рік тому +1

    After watching this video I am afraid of the following words
    BUT NOW WE ARE GONE DO A CHANGE OF VARIABLES

  • @xizar0rg
    @xizar0rg Рік тому +17

    How much of finding the right change of variables is genius and how much is brute force? I know a lot of them are "easy" to see just from experience, but so many feel as though they're pulled out of thin air. I imagine some of those come from cheating (re-deriving known results to make things tidier), but that would just make the original derivations all the more mind-boggling.

    • @Alan-zf2tt
      @Alan-zf2tt Рік тому +4

      My guess: by hard work - trial and error

    • @jaopredoramires
      @jaopredoramires Рік тому +3

      I was about to comment exactly this. Also wondering how did people come up with such clever change of variables and how much of it came from experience and malice

    • @Alan-zf2tt
      @Alan-zf2tt Рік тому +2

      @@jaopredoramires my guesses: analytical approaches , trial and improvement and, increasingly observation and subconscious reasoning? Maybe it is the stuff of PhD researchers?

    • @ConManAU
      @ConManAU Рік тому +2

      It’s about 50% genius and 50% brute force, but most of the genius comes from having spent a long time brute forcing other integrals until you learn what tricks are more likely to work.

    • @Alan-zf2tt
      @Alan-zf2tt Рік тому

      @@ConManAU I wonder if there is another aspect. The sub-conscious mind is apparently more powerful than the conscious part.
      It sort of explains why some people see a solution then trundle though groundwork to find it it was true (or false or only part way there as the case might be)

  • @goodplacetostop2973
    @goodplacetostop2973 Рік тому +9

    18:29

  • @jacksonstarky8288
    @jacksonstarky8288 Рік тому +2

    Yes to more of this. The Gamma and Beta functions have been subjects of interest to me ever since learning about (first) the factorial generalization and (second) the zeta function in its various forms.

  • @nickbierman7008
    @nickbierman7008 Рік тому

    That derivation was a joy to watch. And the final result is beautiful!

  • @hausmaun1800
    @hausmaun1800 Рік тому +4

    actually, factorials can symplify to the (2n-1)!!, so that the coefficient before pi would be (2n-1)!!/2^n

  • @matematicagoiania9341
    @matematicagoiania9341 8 місяців тому

    Yes, I would like to see and learn about it Professor; congratulations for your awesome math classes!👏👏👏👏

  • @md2perpe
    @md2perpe Рік тому +7

    Isn't the change to polar coordinates overkill? Just merge the two integrals and then change coordinates by u = st and v = s(1-t). Everything will fall out from there.

  • @SuhelHasnein
    @SuhelHasnein Рік тому +3

    Let's all appreciate for the hard work done here so far,Thank you sir 🙏

  • @kylesendgikoski4231
    @kylesendgikoski4231 Рік тому

    YES!!!!.....please make a video about generic fractional factorials and possibly discuss, roughly, why there are a variety of approaches.
    Pretty sure I sent you an email inquiry about fractional derivatives a while back. Hope to see more of this.
    Great content like always!

  • @kilianklaiber6367
    @kilianklaiber6367 Рік тому

    So this is an excercize in change of variables. After changing the variable of integration a dozen times, you suddenly arrive at this amazing result. Feels like magic.

  • @faithnlove
    @faithnlove Рік тому

    Very pleasing to watch. My favorite so far.

  • @GreenMeansGOF
    @GreenMeansGOF Рік тому +1

    Yes. Show us the generalizations of these types of results.

  • @Calcprof
    @Calcprof Рік тому

    It should be remarked that this formula is really easy for z a positive integer. Use Γ(z) = (z-1)!, Γ(z+1) = z Γ(z), and Γ(1/2) = sqrt(π). Now just break up (2n-1)! into even and odd factors (2 4 6 8 ...) (1 3 5 ...)

  • @Packerfan130
    @Packerfan130 Рік тому

    how are you able to write a product of two integrals as a double integral at 3:24?

  • @cm_63_France
    @cm_63_France Рік тому

    ok, great! these gamma and beta functions are treasures!

  • @emanuellandeholm5657
    @emanuellandeholm5657 Рік тому +2

    10:23 surely dt = dx/2?
    Edit: spoke too soon! Nice problem tho. :)

  • @manucitomx
    @manucitomx Рік тому

    Thank you, professor

  • @gp-ht7ug
    @gp-ht7ug Рік тому

    Yes please more videos
    And maybe an explanation of the Beta function

  • @HershO.
    @HershO. Рік тому

    maybe you could do a video for a proof of eulers reflection formula.

  • @normanstevens4924
    @normanstevens4924 Рік тому +4

    Am I missing something? Surely this is just a trivial application of Γ(x + 1) = xΓ(x) and the value of Γ(1/2).

    • @DeJay7
      @DeJay7 Рік тому +1

      Using this formula you would have to either find the value of every integer + 1/2 consecutively up to what you want or derive this formula which maybe could be done inductively? In any case, yes this was overly complicated but ultimately just informative, it explores the Β(z,w) function

    • @drdca8263
      @drdca8263 Рік тому

      For the final example maybe, but I think this also shows that the duplication formula holds for arbitrary (positive?) z?

  • @TomFarrell-p9z
    @TomFarrell-p9z Рік тому +2

    Sure, I'd like to see a video about other denominators in the argument of the Gamma function. At the risk of discouraging mathematicians who delight in the math for it's own sake 🙂, the Gamma and Beta functions have lots of useful applications for engineers designing radar detection systems and designing communications systems descriminating between waveforms in the presence of noise. I've use these results without knowing how they were derived, so this is very interesting!

    • @ChefSalad
      @ChefSalad Рік тому

      Unfortunately, I don't think you get anything nice out of the Gamma function with denominators other than 1 and 2. Or so I've been told.

    • @kylesendgikoski4231
      @kylesendgikoski4231 Рік тому

      ​​@@ChefSaladI saw a video a few months back on general fractional factorials.....I think. But I can't find it now.....2 minutes later I now know I was thinking about fractional derivatives, equally interesting I think.

  • @alexbush9250
    @alexbush9250 Рік тому

    When I put n=0 in I get 0 but it should be sqrt(pi)

  • @bot24032
    @bot24032 Рік тому +2

    while this is interesting, isn't it easier to use Γ(x+1)=xΓ(x) if we know Γ(½)? i don't think that the proof in the video is the best way to find a formula for half-factorials

    • @drdca8263
      @drdca8263 Рік тому

      It seems to me that the duplication formula should apply for non-integer x as well?

    • @bot24032
      @bot24032 Рік тому +1

      @@drdca8263 yes, and for that it is great. i think the title said smth anout a formula for half-integer factorials initially but has now been changed

  • @Mohamed.Soltan1991
    @Mohamed.Soltan1991 Рік тому

    Wonderful 🌹🌹

  • @khaledchatah3425
    @khaledchatah3425 Рік тому

    "lets zee"