An Infinite Irrational Tower

Поділитися
Вставка
  • Опубліковано 15 лис 2024

КОМЕНТАРІ • 24

  • @ianchristian7949
    @ianchristian7949 7 місяців тому +21

    Lambert's W function always feels like cheating.

    • @moeberry8226
      @moeberry8226 7 місяців тому +4

      By that logic then the logarithm is also cheating. The point is we know how to calculate and get approximate values but we don’t have all day to keep approximating better and better estimates. For example 2^x=3 gives a number greater than 1 but less than 2 for x and we can keep getting better estimates but telling a computer or calculator log base 2 of 3 is much faster than wasting our time.

    • @SyberMath
      @SyberMath  7 місяців тому +2

      good point

  • @Mehrdad_Basiry-fj4rl
    @Mehrdad_Basiry-fj4rl 7 місяців тому

    Learn alot... thanks...❤❤❤.

  • @spelunkerd
    @spelunkerd 7 місяців тому

    I raised both sides of the first equation to 1/x to get x on one side. Then with a little algebra came to lnx = -W(ln(sqrt2)). As usual, your approach was more elegant, but it is reassuring to show that the two methods converge to the same result.

  • @scottleung9587
    @scottleung9587 7 місяців тому

    Nice job!

  • @sandem4592
    @sandem4592 7 місяців тому

    This was a very nice equation to solve. However I think also an interesting result is that this also gives us a way to compute W(x), when -1/e ≤ x ≤ e (not sure about convergence, mostly educated guesses here) with W(x) = x * (e^-x)^(e^-x)^(e^-x)... I don't think it's awfully useful, but hey at least there's a new way.

  • @SIB1963
    @SIB1963 7 місяців тому +2

    The problem with using Lambert's W function is that it's a magical solution that you don't solve analytically, but with Wolfram Alpha or something like that. As long as I have the magic box, I can get the solution. But if I have access to Wolfram Alpha, I can probably plug in the original equation and get the answer anyway.
    In short, Lambert's W function is a cool thing, but it doesn't seem to help me understand the math any better. It's just an algorithm in my bag of tricks that yields an answer without giving me any gut-level insight into what I'm doing.

    • @markosth09
      @markosth09 7 місяців тому +1

      By your logic the logarithm is also magic. You can’t always get a nice answer. For the equation 2^x=3, x is some number greater than 1 and less than 2, which is transcendental. The point is that these function have nice properties and we know methods to approximate these functions very well (Taylor Series, Numerical Integration, Padé Approximations, Newton’s method, etc), and we can tell a computer to approximate these functions instead of wasting our time. In fact, you don’t need more than a calculator to approximate the Lambert W function with sufficient precision.

  • @GeoffryGifari
    @GeoffryGifari 7 місяців тому +1

    Does it make sense to have a mathematical object like that which "extends to both sides"?
    ..... y^y^y^y ......
    so there isn't really a "bottom" term?

    • @GeoffryGifari
      @GeoffryGifari 7 місяців тому

      could this be used in the study of recursions?

  • @GeoffryGifari
    @GeoffryGifari 7 місяців тому +4

    0.76666665?
    so that infinite square root power tower is rational?
    interesting

    • @pelledanasten1615
      @pelledanasten1615 7 місяців тому

      Nah

    • @TedHopp
      @TedHopp 7 місяців тому +1

      Very doubtful that it's rational. The video makes clear that the computed value is only approximate.

    • @GeoffryGifari
      @GeoffryGifari 7 місяців тому

      @@TedHopp but is it impossible?

    • @TedHopp
      @TedHopp 7 місяців тому +1

      @@GeoffryGifari I wouldn't say impossible. It might be rational (although I would bet against it). I was just pointing out that the video made no claim that 0.766665 was an exact value.

  • @Chris_387
    @Chris_387 7 місяців тому

    That's a solution to the equation x²=2^x

    • @duytuanlengo5038
      @duytuanlengo5038 7 місяців тому

      Definitely it isn't. The equation x² = 2^x have 2 positive solutions x=2 & x=4, 1 negative solution x=-e^-W(ln2/2).
      So the value of the tower in this case is the additive inverse of the negative solution above.

    • @Chris_387
      @Chris_387 7 місяців тому

      @@duytuanlengo5038 it is the same just negative

  • @giuseppemalaguti435
    @giuseppemalaguti435 7 місяців тому +1

    e^(-W(ln2/2))=0,76666

    • @esnofru
      @esnofru 7 місяців тому

      e^(-W(-ln(sqrt(2)/2))

  • @kingaka699
    @kingaka699 7 місяців тому

    First to comment?