A Wallis-type product for e.

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  • Опубліковано 19 гру 2024

КОМЕНТАРІ • 75

  • @mostly_mental
    @mostly_mental Рік тому +74

    That's a really pretty result, and the proof is very clear. But how on earth did Pippenger come up with that formula? (Also, really cool to see one of my professors in the wild.)

    • @Alex_Deam
      @Alex_Deam Рік тому +2

      If Pippenger is your professor, you could ask him?

    • @mostly_mental
      @mostly_mental Рік тому +1

      @@Alex_Deam He was when I was in college, but that was a long time ago.

    • @Alex_Deam
      @Alex_Deam Рік тому +1

      @@mostly_mental ah fair

  • @bot24032
    @bot24032 Рік тому +37

    6:30 it should say not 2^(2^(n-2)) terms but 2^(2^(n-2)) total pulled out for those confused

  • @GreenMeansGOF
    @GreenMeansGOF Рік тому +61

    Seeing as Sterling’s Formula is my favorite math result, I definitely want to see the video!

    • @nikolay2263
      @nikolay2263 Рік тому +4

      Agree

    • @allanjmcpherson
      @allanjmcpherson Рік тому +1

      Agreed! Having studied physics, I really like Sterling's approximation. It takes problems in statistical mechanics that would be completely impractical to calculate and turns them into something you can do with a handheld calculator.

  • @elfumaonthetube
    @elfumaonthetube Рік тому +21

    Very interesting. Also, by dividing the Wallis product of pi by the product formula of e it is possible to define a neat product formula for pi/e.

  • @fedorlozben6344
    @fedorlozben6344 Рік тому +1

    23:42
    Michael,yes!
    It is a really good idea about Sterling's approximation.
    I would like to see how to get this useful fact!

  • @stephenhamer8192
    @stephenhamer8192 Рік тому +7

    Awesome exposition. Huge potential for muddle here

  • @jacksonstarky8288
    @jacksonstarky8288 Рік тому +11

    I'd like to know if there's something similar for gamma (the Euler-Mascheroni constant)... and if finding such an infinite product would enable a proof of gamma's irrationality (and possibly also transcendentality).

    • @ardan981
      @ardan981 Рік тому +5

      There's a known wallis-type product for e^gamma but it seems like we don't know of a wallis product for gamma

  • @fredericdutin9076
    @fredericdutin9076 Рік тому +3

    Yes for the Sterling formula video!

  • @sdal4926
    @sdal4926 Рік тому +2

    perfect.well done

  • @hassanalihusseini1717
    @hassanalihusseini1717 Рік тому

    That was tough! Thank you for nice presentation!

  • @dukenukem9770
    @dukenukem9770 Рік тому +4

    I haven’t seen a proof of Sterling’s approximation in years. Please upload one!

  • @cycklist
    @cycklist Рік тому +7

    Astonishing to learn that pi and e are so closely related. What's the intuition behind that?

    • @frankwilhoit
      @frankwilhoit Рік тому +6

      If you take your glasses off, all infinite series are the same.

    • @BrianGriffin83
      @BrianGriffin83 Рік тому +1

      You can see the connection in many simpler formulas, starting with Euler's identity.

  • @jacemandt
    @jacemandt Рік тому +7

    So, if we take finite products that end after each parenthesized set (doubled to offset the 2 in the denominators), they increase to π in one case and increase a little more slowly (because we're including 2^(n-1)-th roots instead of the whole thing) to e in the other case.
    In some sense, that makes e a version of π, when seen as limits, where you're taking care not to increase the products quite as fast for e as you do for π.
    There must be a deeper reason here that e and π are related in this way. Is this a version of their relationship that derives from Euler's formula e^(πi)=-1? Or is this a different relationship entirely? Maybe the answer is to wait for Michael's video on Stirling's formula, since that also expresses a relationship between the constants in terms of factorials?

    • @АндрейДенькевич
      @АндрейДенькевич Рік тому

      You are right. But why powers of 2 are used? Because e is a simplex, that's why, while pi is a sphere.
      2D example. e/pi=2.72/3.14=0.866=sin(60)=1/1.155. This is are "diametr of 2D sphere /length of 1D edge of inscribed 2D simplex".
      3D example. This is are " "diametr of 3D sphere /area of 2D triangle of inscribed 3D simplex".
      Length of 1D edge of inscribed 3D simplex in a unit sphere equals 1.63299932. Use Heron's formula Area= sqrt(3)/4 *1.63299932^2=1.155.

    • @nickruffmath
      @nickruffmath Рік тому +1

      I wonder if it's the grouping with the parentheses being a little misleading for the Wallis example up top. My conceptual understanding of the Wallis product is that an appropriate grouping would be each "group of 4" (two numbers in the numerator and two in the denominator) at that fixed size. Whereas the exponents for e keep doubling the size of the grouping.

    • @nickruffmath
      @nickruffmath Рік тому +1

      But since Pi also shows up in Stirling's formula, maybe they're more related like you're saying

  • @weonlygoupfromhere7369
    @weonlygoupfromhere7369 Рік тому +1

    yes to the Sterling's Approximation video!

  • @worldnotworld
    @worldnotworld 6 місяців тому

    Absolutely astonishing. I don't think I've ever seen anything like a relation between pi and e that is quite so simply "arithmetic" outside of complex analysis. The fact that the change of the exponent from 1 to 1/(2^n) takes us from pi to e (or, more precisely, e/2, but that's still in the "world of e") makes one consider the range of exponentiated "a-sub-n" over "b-sub-n" products in general, for an arbitrary domain of exponents. Call that domain E of exponents "E "-- Do other values for E yield other "important numbers," like but other than e or pi? What about E=1/n? Or maybe E=1/n! ? Or E=e^(-n)? E: Most enticingly: What is the class of functions E=f(x) for E over the reals that has both pi an e in its range for special cases of x? Does it yield phi (golden mean) as well?!

  • @hqTheToaster
    @hqTheToaster Рік тому +4

    Here is another strange equation: Binomial(j,j/e) ~ 1.4027, j ~ 1.2954 , Slope ~ 1.4142
    Anyway, hard to understand, but nice beautiful pattern with exponents. Great video!

  • @NoahPrentice
    @NoahPrentice Рік тому +4

    Would love to see the approximation video!

  • @MortezaSabzian-db1sl
    @MortezaSabzian-db1sl Рік тому +3

    Can you help me solve this problem?
    {A_n+2}+{A_n}={a_n}×{a_n+1}
    Write this recursive sequence in terms of the first and second sentences of the sequence

  • @talastra
    @talastra Рік тому +1

    This was thorough bonkers.

  • @barryzeeberg3672
    @barryzeeberg3672 Рік тому +3

    It is interesting that pi is given by the product of ratios, yet pi is not rational. Presumably this is because it is an infinite product (although this seems to be counter-intuitive)?
    I know that an infinite sum can converge to a limit that is outside of the space of the terms in the sum (if it is not in a Hilbert space), but the product of rational expressions (as in this example) has a numerator and denominator that are both integers, so the product should have the form of a rational expression, no matter how many terms are multiplied?

    • @oliverherskovits7927
      @oliverherskovits7927 Рік тому +5

      The same happens with infinite sums, the sum of rationals is rational, but only a finite sum. In fact every real number is the infinite sum of rationals (it's one of the definitions of the reals as Cauchy sequences). This is indeed counterintuitive, but this is only because we think of infinite sums as a type of sum. In fact it's better to think of infinite sums as just limits (of the partial sums). Then we see that all the strange behaviour happens because limits don't preserve all the properties reals can have (eg limit of positive terms need not be positive)

  • @dariofagotto4047
    @dariofagotto4047 Рік тому +3

    I was waiting for e emerging from the definition and ended up just being dropped as another formula (the approximation) so I'll wait for that video hoping the connection does come up from the ground instead of being taken for granted

    • @nickruffmath
      @nickruffmath Рік тому

      This is a decent video if you're interested: ua-cam.com/video/JsUI40uSOTU/v-deo.htmlsi=WD8HfGKfzFLpJtV_
      The e comes from the fact that the Gamma function is an extension of the factorial to the real and complex numbers. And the Gamma function uses e in its definition. The recovery of the "discreteness" of the factorial comes from using integration by parts on the integral definition of the gamma function, where the unique properties of e^x allow the extraction of the falling power for each step.

  • @endormaster2315
    @endormaster2315 Рік тому

    I would like a video on Sterling's approximation

  • @Zealot0630
    @Zealot0630 Рік тому +2

    I wonder if it is legal to reindex n, because ln(P) is conditional converge, such that rearrange its terms gives different result.

    • @Noct1um
      @Noct1um Рік тому +4

      Shifting the index or reordering a partial(finite!) product before taking a limit is always valid...

  • @The1RandomFool
    @The1RandomFool Рік тому

    That was a big one.

  • @panagiotisapostolidis6424
    @panagiotisapostolidis6424 Рік тому

    awesome one

  • @Anonymous-zp4hb
    @Anonymous-zp4hb Рік тому

    That was intense. Pretty sweet result though.

  • @Handelsbilanzdefizit
    @Handelsbilanzdefizit Рік тому +3

    Looks like a product of geometric means.

  • @petterituovinem8412
    @petterituovinem8412 Рік тому +1

    I want to see the sterling video

  • @1991tnh
    @1991tnh Рік тому

    So good

  • @Axenvyy
    @Axenvyy Рік тому +1

    It's almost nauseating to see how similar the expression for e is to Wallis infinite product for pi
    They seem to be completely non related constants🤷‍♀️ and well here we are....

  • @mstarsup
    @mstarsup Рік тому +4

    Very nice video, but it's Stirling, with an "i". :-)

  • @Happy_Abe
    @Happy_Abe Рік тому

    Why are we allowed to replace the terms with their approximations using sterling’s formula?

    • @bluelemon243
      @bluelemon243 Рік тому

      You basically multiply and devide the term with his approximation (so you dont change anything), now , the apporaximation say that in the limit n!/sterling is 1 so you left with sterling alone

    • @Happy_Abe
      @Happy_Abe Рік тому

      @@bluelemon243 ah the fraction is one in the limit I see
      Doing that assumes the original expression itself converges though, which one would have to show

  • @Alan-zf2tt
    @Alan-zf2tt Рік тому +1

    Beautiful and yet aweful at the same time

  • @toddtrimble2555
    @toddtrimble2555 Рік тому +1

    Stirling, not Sterling.

  • @inigovera-fajardousategui3246
    @inigovera-fajardousategui3246 Рік тому +1

    Amazing

  • @davidcroft95
    @davidcroft95 Рік тому

    Plaease upload the Stirling approx video! We always use it in physics and statistics (especially for calculate log(n!)) but noone explain why it's true

  • @ulieggermann4346
    @ulieggermann4346 Рік тому

    Great!

  • @pierreabbat6157
    @pierreabbat6157 Рік тому

    Is this the Futuna product?

  • @henrikschmid
    @henrikschmid Рік тому +2

    I think it's Stirling, not Sterling.

  • @minwithoutintroduction
    @minwithoutintroduction Рік тому

    رحلة شاقة لكن الوصول مريح

  • @skylardeslypere9909
    @skylardeslypere9909 Рік тому +1

    I've got a 'personal' question (unrelated to the video). Are the videos still being edited by the same person as those a few weeks ago? I believe their name was Stephanie? I feel like the editing is less present in the recent videos (not saying that's a bad thing though, I just want to know).

  • @CharIie83
    @CharIie83 Рік тому

    an approximation kinda takes the point out or

  • @kannix386
    @kannix386 Рік тому

    do a pushup everytime he says "two" or "square"

  • @humbledb4jesus
    @humbledb4jesus Рік тому

    it looks like a half decent IQ question: what is the bracketed number raised to the 1/128th power?

  • @paologat
    @paologat Рік тому +5

    Please avoid mixing lower case and upper case N in the same formula. There are so many other letters you could use instead.

    • @D.E.P.-J.
      @D.E.P.-J. Рік тому +2

      It's pretty standard. It works fine as long as you write them differently as Michael does.

    • @BridgeBum
      @BridgeBum Рік тому +3

      @@D.E.P.-J. It is standard, but the problem comes in when he is reading the formulae. I agree with the other poster that I'd prefer to use different letters simply for "production value" clarity.

    • @paologat
      @paologat Рік тому +1

      @@D.E.P.-J. it doesn’t work when you read the formula aloud, unless you specifically say “small n” and “capital n” each time, which Michael didn’t.
      Don’t make formulas harder to read and to check than they need to.

    • @divisix024
      @divisix024 Рік тому +2

      There’s a pretty good reason that math is commonly communicated through writing/reading instead of speaking/listening, Imagine going to a math class and not being shown anything visually, just the lecturer reading out all the expressions and formulae for the entire class.

    • @paologat
      @paologat Рік тому

      @@divisix024 indeed.
      In any case, both when writing and when speaking, I was taught to avoid unnecessary sources of confusion. It’s a form of courtesy to the reader / listener, and it helps the writer / lecturer to avoid mistakes.

  • @fonzi102
    @fonzi102 Рік тому +1

    :)