A Very Exponential Equation (Homemade!)

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  • Опубликовано: 14 май 2024
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Комментарии • 19

  • @thexavier666
    @thexavier666 2 месяца назад +8

    Because of this channel, my math roots are still strong. Can you also do some 3D integrals?

    • @SyberMath
      @SyberMath  2 месяца назад +1

      I can try 😜

  • @coral_15
    @coral_15 Месяц назад

    Thanks for explaining an interesting problem.
    I have a question about how you made this video.
    What app did you use to make it? I use the app which name is "explain everything" to make my videos, but I am interested in the app you use.
    I am afraid it is not a question about the content of the problem.

  • @barakathaider6333
    @barakathaider6333 2 месяца назад

    👍

  • @Chrisoikmath_
    @Chrisoikmath_ 2 месяца назад +3

    In my view 0^0 is undifined but the limit lim(x^x) = 1 as x trends to 0.

  • @robertibatullin8329
    @robertibatullin8329 2 месяца назад +4

    QA live stream would be great

  • @scottleung9587
    @scottleung9587 2 месяца назад

    Wow, that was so complex only you could understand it - nice job, though!

  • @Stelios.Posantzis
    @Stelios.Posantzis 2 месяца назад

    What's wrong with x being negative infinity?

    • @oenrn
      @oenrn 2 месяца назад

      Infinity is a limit, not a number, so it can't be a solution unless you're talking about limits rather than exact values.

    • @Xnoob545
      @Xnoob545 2 месяца назад

      ​@@oenrn depends on the number system you use
      In hyperreals and surreals it is definitely a number

  • @wonghonkongjames4495
    @wonghonkongjames4495 2 месяца назад

    Good Afternoon, Sir, I've got a suggestion, what if this problem changes to a towering one with infinite "x" powers = 1
    Was there any solution? If it did, what was it?
    I think it would be the latest weirdest towering problem of Maths
    Thank you James 05-23-2024

  • @SidneiMV
    @SidneiMV 2 месяца назад +1

    x^x = 2kπi
    k ≠ 0

  • @GirishManjunathMusic
    @GirishManjunathMusic 2 месяца назад +3

    e↑(x↑x) = 1
    As x↑x ≠ 0:
    e↑(x↑x) = e↑2niπ; n ∈ ℤ∖{0}
    ∴ x↑x = 2niπ
    ln(x)·e↑ln(x) = 2niπ
    taking productlog of both sides:
    ln(x) = W(2niπ)
    x = e↑W(2niπ); n ∈ ℤ ∖{0}.

  • @lawrencejelsma8118
    @lawrencejelsma8118 2 месяца назад +2

    Check Wikipedia. The Lambert W(x) is found in different interval domains of x by Stirling Numbers an abstract iterative recursive algorithm to define what the value is. The Lambert W of a hanging undetermined value of something in the real axis or complex axis in evaluated is useless. The abstract Sterling Number recursive iterative solutions in getting values is worse way to evaluate solutions vs a more defined method (and possibly Newton's Iterative solutions vs Stirling Numbers of form of an iteration constant multiplied with a Stirling Number and to (-1)^m, ln (ln of something)^m/m factorial. I wanted to know what W inverse function of like just 2π was and there are no tables but an abstract reference to Stirling's Numbers of recursive iterations. Wikipedia does a great explanation of iterative series notations for W(x) where x is not between -1/e and +e xa.
    Another level of difficulty in this problem is x^x used as an imaginary 2kπ equivalent of the trigonometric sin and cosine waves repeated solutions of each solution. This is bad to reference your solution is infinite relating to properties of the infinite possibilities of the sin(x) and cosine(x) functions being anything useful also as a solution or set of solutions to try to determine an abstract W(x) result or results.

    • @josepherhardt164
      @josepherhardt164 2 месяца назад

      In the back of my mind is this doubt that the Lambert W() function is in some way, a cop-out.