Ramanujan's master theorem is insanely overpowered!!! example using the Fresnel integrals

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  • Опубликовано: 13 июл 2023
  • The one theorem to rule them all.....
    Ramanujan's theorem is one of the most powerful tools in integral calculus and today we're exploring a solution development for the fresnel integrals using this amazing result.

Комментарии • 99

  • @jozsefkimberly3920
    @jozsefkimberly3920 Год назад +169

    I think it would be better for the viewers who don't necessarily study pure maths's at uni to make an introductary video to some special functions and their properties and of course proof why it's working in the first place.

    • @joseph8298
      @joseph8298 Год назад +7

      In addition to this a video covering the best pure maths books to study and/or course material recommendations would be nice, especially if one is wanting to look into higher math for an undergraduate’s

  • @EffySalcedo
    @EffySalcedo Год назад +12

    Ramanujan and Laplace are so intense they are the Double Rainbow.

  • @nickruffmath
    @nickruffmath Год назад +35

    Beautiful approach! Love Ramanujan. No one else like him

    • @EffySalcedo
      @EffySalcedo Год назад +2

      Ramanujan is unique, his mind so rich and beautiful 🥺, whenever I feel claustrophobic, He is a potent natural antidote.

    • @orang1921
      @orang1921 Год назад +4

      bro just spawned in, dropped some absolutely crazy math shit, and dipped

    • @agrimmittal
      @agrimmittal 11 месяцев назад +2

      ​@@orang1921pffft so true about that lmao 😂

  • @maths_505
    @maths_505  Год назад +2

    You can follow me on Instagram for write ups that come in handy for my videos:
    instagram.com/maths.505?igshid=MzRlODBiNWFlZA==
    If you like the videos and would like to support the channel:
    www.patreon.com/Maths505

  • @kazminto9618
    @kazminto9618 Год назад +14

    One ring to rule them all one ring to find them one ring to bring them all and in darkness bind them

    • @atlas132
      @atlas132 Год назад +2

      One Integral to solve them all 😎

  • @MrWael1970
    @MrWael1970 Год назад +4

    Thank you for this fruitful effort. The last step is the integration of cos(x^2) is with respect to x. So, dx in the LEFT hand side is missing. Thanks.

  • @animewarrior7
    @animewarrior7 Год назад

    This helps alot! thank you sir!

  • @zunaidparker
    @zunaidparker Год назад +27

    Next video: drive the Master Theorem itself. This felt like a "shortcut" that required even more complicated assumptions than the method actually applied.

  • @Patapom3
    @Patapom3 Год назад +1

    Amazing!

  • @goubou3872
    @goubou3872 Год назад +15

    I personally prefer the approach of complex analysis, you can take loop that depend on a variable R and the loop is basically a semi circle and just use the residue theorem. You'll just have to majorate 1 or 2 integral and it'll be done.

    • @maths_505
      @maths_505  Год назад +13

      The melin transform and gamma function are both subjects of complex analysis rather than real analysis. So you can still count this video as one for complex analytic methods.

  • @deepjoshi356
    @deepjoshi356 Год назад

    When you are recording screen on galaxy tab, can we do youtube live at the same time?

  • @Maths_3.1415
    @Maths_3.1415 Год назад +11

    I like this one :)

    • @maths_505
      @maths_505  Год назад +10

      And I am officially successful in summoning the ghost of Ramanujan

    • @Maths_3.1415
      @Maths_3.1415 Год назад +7

      ​@@maths_505 lol 😂

  • @ivalle_215
    @ivalle_215 Год назад

    Great video man

  • @hgnb1001
    @hgnb1001 Год назад

    Nice!

  • @pandavroomvroom
    @pandavroomvroom Год назад

    very cool

  • @BorisNVM
    @BorisNVM Год назад +4

    i wouldnt even think this integral converges

    • @yuyan930
      @yuyan930 Год назад

      yeah I don't think it does

    • @laxminarayanbhandari855
      @laxminarayanbhandari855 Год назад +1

      ​​@@yuyan930t literally does. Otherwise he wouldn't have been able to evaluate it. Imagine saying the integral doesn't converge on a video literally solving the integral with valid steps. For a formal proof of convergence, you can just Google search.

  • @cameronspalding9792
    @cameronspalding9792 Год назад +9

    My favourite way of solving these integrals is using contour integration.

    • @maths_505
      @maths_505  Год назад

      The solution using contour integration is pretty cool and one of the more "unconventional" ways of designing contours. So yes I agree with you that the coolest way is contour integration (though for me the Leibniz rule ties it for top spot)

  • @PaulDean125
    @PaulDean125 Год назад

    What I find interesting is that the two integrals are equal. Equivalently, \int_{0}^{\infty} sin(x^2) \,dx = \int_{0}^{\infty} sin(x^2+\pi/2) \,dx, which by the way nicely corroborates the square root of pi/2. It's surprising when you look at the graphs.

    • @maths_505
      @maths_505  Год назад

      I had the same thought when I first studied these integrals

    • @PaulDean125
      @PaulDean125 Год назад

      @@maths_505 also if the limit is from -infty to +infty, you can do away with the 1/2 and the sqrt(pi/2) is then a straight consequence of the integrals being the same. The integrals being equal is a an equivalent statement to their value being sqrt(pi/2).

  • @zinzhao8231
    @zinzhao8231 11 месяцев назад

    convert to img and Re components, let u = sqrti(x) easy

  • @andrewrivera4029
    @andrewrivera4029 Год назад

    I don’t remember the Mellon transform, it’s sounds like a calc 3 type of equation. I loved these long proofs in engineering school watching bleary eyed in class then banging out problems for the rest of the week.

    • @maths_505
      @maths_505  Год назад +2

      Its Melin not melon 😂

    • @sozeran
      @sozeran 11 месяцев назад +1

      ​@@maths_505it's Mellin, double "L"

    • @maths_505
      @maths_505  11 месяцев назад +1

      The L was taken by the ones who doubted me yo!
      (Yes this is a cover up attempt and thank you for pointing out the mistake)

  • @dominicellis1867
    @dominicellis1867 Год назад +2

    You can use the complex exponential form and solve the subsequent Gaussian integrals.

    • @maths_505
      @maths_505  Год назад +8

      There are a couple of ways to solve those.
      Feynman's trick
      Laplace transform
      This time I wanted to call the ghost of Ramanujan

  • @JorgeLimaJPL
    @JorgeLimaJPL Год назад +3

    Where do I apply such masterpiece? What kind of problems? What kind of industries can be blessed with such knowledge?

  • @zinzhao8231
    @zinzhao8231 11 месяцев назад

    can you do more videos on the melon transform?

    • @maths_505
      @maths_505  11 месяцев назад

      Melin transform bro😂

    • @maths_505
      @maths_505  11 месяцев назад

      If I find some interesting ones then yes sure I'd love to. You can find a table of melin transforms on the internet too

    • @zinzhao8231
      @zinzhao8231 11 месяцев назад

      @@maths_505 ok perfect beautiful, gracias mija

  • @tonebank2000
    @tonebank2000 Год назад

    how does this even make sense? aren't there infinitely many analytic functions that coincide at every integer (just think of adding sin(2piz) to w)? how is this well defined?

    • @morgengabe1
      @morgengabe1 Год назад +3

      Sometimes, mathematics is about finding the best reason to bury your head in some sand and get some calculations done.

  • @allmight801
    @allmight801 Год назад +3

    We have Fourier transform then Laplace transform and now even Mellin transform. Is there some other hidden transformation?

    • @Risu0chan
      @Risu0chan Год назад +4

      Z-transform comes to mind. They are all somewhat related to each other. I think the Z transform is used for sequences and, for engineers, for discrete signals.

    • @allmight801
      @allmight801 Год назад +1

      @@Risu0chan How are they all related to each other?

    • @Risu0chan
      @Risu0chan Год назад +4

      by a change of variable and/or a transform of the function:
      Laplace L{f} (is) = Fourier F{f} (s)
      Mellin M{f} (s) = L{f(e^-x} (s) = F{f(e^-x)} (-is)
      I'm not familiar with the Z transform, but the relationship is explained on Wikipedia.

    • @morgengabe1
      @morgengabe1 Год назад +2

      In computer science there are more. They call them "correspondences" but don't quite formalize them into "transforms" the way mathematicians do (i think because they have so many languages/notations to choose from). One example is between Church's Lambda calculus and Turing's machines. You could also argue that each of these has a correspondence to complete Godel languages; all support derivations (programs) through conditional execution and arbitrary memory.
      Curry-howard correspondence is an other big one. NOt an expert but suspect curry's work, in particular, has strong connections to Galois theory.
      They have varying proximities to "more conventional" areas of mathematics.
      I think the common underlying theme is that transforms/correspondences relate languages that can detect each other's errors. In that way they're quite like isomorphisms between "versions" of mathematics.

  • @quidam3810
    @quidam3810 11 месяцев назад

    Awesome ! I feel like a 5 year old discovering grown-ups toys !!

  • @farhadazadi
    @farhadazadi Год назад +1

    Hey may I ask what app you are using?

  • @NurHadi-qf9kl
    @NurHadi-qf9kl Год назад

    .x^2=z, 2x dx=dz

  • @jonsmith8579
    @jonsmith8579 Год назад +4

    are there any underpowered theorems?

  • @hassanbabajantabar3202
    @hassanbabajantabar3202 Год назад +5

    After watching this solution you have more appreciation for Laplace

    • @maths_505
      @maths_505  Год назад +15

      Laplace vs Ramanujan would be a nice integration war. I'll make a video on that soon.

    • @joshuahoover7700
      @joshuahoover7700 Год назад +1

      @@maths_505 Feynman would dwarf either of them tbh

    • @maths_505
      @maths_505  Год назад +1

      @@joshuahoover7700 the integration GOAT 🔥🔥🔥

  • @Mayk_thegoat
    @Mayk_thegoat Год назад

    Sir am ur fan and i need the inverse Laplace of 1 over [(square root of s)+1]
    Plzzz

    • @johnbussio7742
      @johnbussio7742 Год назад +4

      This isn’t google

    • @Mayk_thegoat
      @Mayk_thegoat Год назад +2

      @@johnbussio7742 you need to back up before you get smacked up

    • @maths_505
      @maths_505  Год назад +1

      I think its t^(-1/2)e^(-x)/sqrt(pi)

    • @Mayk_thegoat
      @Mayk_thegoat Год назад

      @@maths_505 sir
      I meant that 1 is not under the sqrt
      that means 1/sqrt(s)+1

    • @morgengabe1
      @morgengabe1 Год назад +1

      ​@@Mayk_thegoatfunny, but he has a point.

  • @fudgenuggets405
    @fudgenuggets405 Год назад

    But what if c-a-t really spelled, "dog"?

  • @joshuahoover7700
    @joshuahoover7700 Год назад

    when the power series doesn't exist

  • @sogga_fan
    @sogga_fan Год назад

    nice video u sound like andrew tate

  • @djUStylor
    @djUStylor Год назад +2

    I think you made a mistake. When you replace the Us back to Xs, it should be x.e^(-ix^2). Or are you calling a completely different variable x?

    • @maths_505
      @maths_505  Год назад +5

      Its just a rename. Not a substitution. The variable in case of a definite integral is just a dummy variable so you can rename it anytime. It doesn't change the functions or limits involved so it doesn't matter what you call it.

    • @benhetland576
      @benhetland576 Год назад +1

      @@maths_505 I disagree... If it's just a rename as you claim, then this latter x is not the same as the first one, which you already declared equal to x^2. But if it is the latter x = u then it becomes confusing since you suddently have two different expressions for the I= where both include an x, but not the _same_ x. Conversely, if it is not really a rename then it is no longer the value of I being expressed. You do need to change the limits (both times you substitute), but incidentally x=0 u=0 and also x->inf => u=x^2 -> infinity. That might explain why one can "get away with" the "rename"...

    • @Mavhawk64
      @Mavhawk64 Год назад +8

      @@benhetland576 That's what he said in the lecture. He mentioned that because the bounds remain the same, he can just rewrite a new function that is equivalent to the old function using different variables. That's like saying f(x) = x^2 and g(t) = t^2 (on same domain/range, etc). We can then say f(x) = g(x) = x^2 or f(t) = g(t) = t^2. There's no difference. When doing the integral, both integrals evaluate to the same outcome with the provided bounds, and thus we can do a change of dummy variables to get it in the standard terms of x again. Of course, @maths_505 is doing this in an 11 minute video instead of writing a whole LaTeX paper with all of the steps mapped out, he just glosses over the simple fact, and children in the comments section have to argue, even though they will learn how simple of a concept an exchange of dummy variables is in Abstract Math in First Year.

    • @Mavhawk64
      @Mavhawk64 Год назад +1

      In his case, he is saying I(u=x) = I(x) = *integral with respect to x*

    • @frenchimp
      @frenchimp Год назад +4

      @@benhetland576 You need to understand the meaning of a mute or dummy variable. In this case u and x are dummy variable, and they can be renamed as you see fit.

  • @thefirstsurvivor7240
    @thefirstsurvivor7240 Год назад

    After watching im starting to think that 1 + 1 doesnt equal 2

  • @hotdog77189
    @hotdog77189 Год назад

    I clicked because of the clickbait title and video.

  • @jkid1134
    @jkid1134 Год назад +2

    You kinda have to prove the theorem though

    • @maths_505
      @maths_505  Год назад +4

      Bruh
      That needs a video of it's own
      But yeah that's a nice idea although I think Michael Penn has a proof video on this.

    • @jkid1134
      @jkid1134 Год назад +4

      @@maths_505 it's just that there's a big difference between a video where I have to believe you and a video where I don't. My math journey is all from first principles, I have verified it all with my own eyes, yknow? I'll look for the Michael Penn video.

    • @lih3391
      @lih3391 Год назад +1

      @@jkid1134that's great, but theres always too much to prove, since mathematical knowledge is so vast. That michael penn video is very nice tho.

    • @morgengabe1
      @morgengabe1 Год назад +1

      ​@@lih3391maybe we just need shorter proofs. I think that when you approach a problem with tools it may accept but does not need, the solution is always harder.

  • @shakaibsafvi97
    @shakaibsafvi97 Год назад +1

    I'm having a headache.....
    Too much complexity for my brain....

    • @morgengabe1
      @morgengabe1 Год назад +2

      Loool, i feel you.
      it's a little more straight forward if you start at about 4 minutes. He explains the master theorem, which seems to define an other way to compute a mellin transform. Hopefully he does a derivation soon. Would be a lot easier for people who don't do the calculus every day to follow.

  • @formationcovadishydrauliqu6255

    Ramajan said hhh sum of 1_1,,,,= 0.5 I think Ramajan stupid

    • @shubhankarmahashabde1029
      @shubhankarmahashabde1029 3 месяца назад

      and i think you are stupid

    • @aravindakannank.s.
      @aravindakannank.s. 3 месяца назад

      for an alternating series of 1 and 0 which is diverging due to the assumption of constant ( converging) series he does get the answer of 1/2 .
      u can't simply say he is stupid.

    • @aravindakannank.s.
      @aravindakannank.s. 3 месяца назад +1

      by assuming the series of 1-1+1-1+1-1+...... converging
      he got a value of 1/2
      how in the world u can say it's stupid.
      it's like a probability of any one answer that is either 1 or 0

  • @gourabsen5886
    @gourabsen5886 Год назад +1

    Anyone suggest me how to remove this kind of worthless videos permanently...

    • @050138
      @050138 Год назад +3

      😂 such worthless videos can never be permanently removed as long as there are worthless people who click on it, watch it and post worthless comments 🤣