do normies actually buy this shit?

do normies actually buy this shit?

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  1. 1 month ago
    Anonymous

    Guy single-handedly fricked numberphile https://www.youtube.com/watch?v=YuIIjLr6vUA

    Ramanujan was based though

    • 1 month ago
      Anonymous

      >Ramanujan was based though
      how's the weather in Calcutta?

      • 1 month ago
        Anonymous

        pretty good.

    • 1 month ago
      Anonymous

      what an outdated video

      https://arxiv.org/abs/2401.10981

    • 1 month ago
      Anonymous

      Pretty compelling

    • 1 month ago
      Anonymous

      >debunk video
      >says all the statements made were wrong
      >the very first statement made by himself is wrong
      You can't say that sum of natural numbers is equal to infinity - infinity is not a number.

      • 1 month ago
        Anonymous

        He explains later in the video how when infinity is involved, the meaning of the equals sign will change depending on the context

        • 1 month ago
          Anonymous

          No, he doesn't.
          Series is divergent, therefore has no value.

          Unless you're accepting different more generalized definitions - but then you may as well accept the one where 1+ 2 + 3 + ... = -1/12

          • 1 month ago
            Anonymous

            as you should

            what an outdated video

            https://arxiv.org/abs/2401.10981

  2. 1 month ago
    Anonymous

    Normies do not understand anything about an infinite series anon, and most don't know much about finite ones either.

  3. 1 month ago
    Anonymous

    >muh analytic continuation
    This is what "complex number" ideologists unironically believe.

  4. 1 month ago
    Anonymous

    that an Indian man can be intelligent?

  5. 1 month ago
    Anonymous

    Obvious wrong, but because a mathematician did it, it's right. And it also developed a new sub branch of maths lmao

  6. 1 month ago
    Anonymous

    You are too moronic to understand what is going on there.

    • 1 month ago
      Anonymous

      you are too moronic to realise that the math we have is entirely man made.
      so yes it is susceptible to mistakes.
      there's proof 1=-1,does it make it true? hell no.

      • 1 month ago
        Anonymous

        >there's proof 1=-1,does it make it true? hell no.
        1) no there isnt.
        2) if we couldnt ever distinguish between true and false statements in math, then math would be pointless. you arent allowed to say that some proofs are technically correct but dont count because you dont like them. either state that the proof is erroneous or accept the conclusion.

        • 1 month ago
          Anonymous

          e^(iπ) + 1 = 0
          e^(iπ) + e^(iπ)^2 = 0
          -e^(iπ)^2 = e^(iπ)
          1 = -1

          • 1 month ago
            Anonymous

            e^(iπ)^2 ≠ 1

          • 1 month ago
            Anonymous

            >-1^2 ≠ 1
            lmfao dumb

            >-e^(iπ)^2 = e^(iπ)
            >1 = -1
            I think you're just confusing yourself by excluding parentheses. (e^(iπ))^2 is 1 (which is a fact you use yourself), so both sides of this equation are -1, there's no contradiction.

            -(e^(iπ))^2 = -(e^(iπ)^2) = -1(e^(iπ)^2) = (-(1)e^(iπ)^2) = (-e^(iπ)^2) = -e^(iπ)^2 = 1

            welp, you fricked up the proof. ignoring that, the flaw is that f(x)=x^2 is not an injective function.

            Based. Nobody should be injective or boosted.

          • 1 month ago
            Anonymous

            >>-1^2 ≠ 1
            >lmfao dumb
            Except that you are the dumb one.
            e^iπ^2 = e^(i^2*π^2) = e^(-π^2)

          • 1 month ago
            Anonymous

            >-(e^(iπ))^2 = -(e^(iπ)^2) = -1(e^(iπ)^2) = (-(1)e^(iπ)^2) = (-e^(iπ)^2) = -e^(iπ)^2 = 1
            Again, you're falling victim to your own ambiguous notation. Written properly, your chain of equalities looks like this:
            >-(e^(iπ))^2 = -1 * (e^(iπ))^2 ?= (-1 * e^(iπ))^2 = (-(e^(iπ*~~^2 = 1
            And clearly the ?= step is incorrect since -(a^2) is not equal to (-a)^2 in general. No more (You)s until you're honest about your intended order of operations.

          • 1 month ago
            Anonymous

            -1^2 = -1
            moron

          • 1 month ago
            Anonymous

            >-e^(iπ)^2 = e^(iπ)
            >1 = -1
            I think you're just confusing yourself by excluding parentheses. (e^(iπ))^2 is 1 (which is a fact you use yourself), so both sides of this equation are -1, there's no contradiction.

          • 1 month ago
            Anonymous

            welp, you fricked up the proof. ignoring that, the flaw is that f(x)=x^2 is not an injective function.

  7. 1 month ago
    Anonymous

    Srinivāsa Aiyangār Rāmānujan (ஸ்ரீனிவாஸ ஐயங்கார் ராமானுஜன்) was kino.

  8. 1 month ago
    Anonymous

    It doesn't equal -1/12. If you redefine basic algebra so you can assign non-convergent values to symbols then yes, but all of math collapses into madness if you permit this.

    • 1 month ago
      Anonymous

      But it works

    • 1 month ago
      Anonymous

      way to show your outdated knowledge

      what an outdated video

      https://arxiv.org/abs/2401.10981

  9. 1 month ago
    Anonymous

    Wtf are you even asking? This is no different than the proofs that 1=2 or whatever. It's just your average "using correct math with incorrect assumptions returns contradictory results" thought experiment. I know this is a shitpost thread, but I don't understand what there is to "buy" about this.

    • 1 month ago
      Anonymous

      here is something new for you to learn

      what an outdated video

      https://arxiv.org/abs/2401.10981

  10. 1 month ago
    Anonymous

    you morons do realize that math is made up right?

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