The dunk already happened as you can see but here's the link if you wanna go marvel at the real thing: https://twitter.com/renatokara/status/1412484734949675013?s=19

  • Fakename_Bill [he/him]
    ·
    3 years ago

    It's been so long since I took calculus. Can you just cancel the diagonal dx, or is the whole point of the joke that you can't?

    • 0karin728 [any]
      ·
      3 years ago

      For literally every conceivable situation that anyone who isn't a professional mathematician or physicist would ever encounter, yes you absolutely can treat dy/dx as a fraction.

      Because it basically is a fraction, either the limit of a fraction as both parts go to zero, or a fraction of two infinitesimals (numbers between 0 and the smallest or positive or negative real number). A lot of mathematicians get sad when you use infinitesimals but it's fine.

      • Pezevenk [he/him]
        ·
        3 years ago

        For literally every conceivable situation that anyone who isn’t a professional mathematician or physicist would ever encounter, yes you absolutely can treat dy/dx as a fraction.

        Not really.

        df/dx=df/dt.

        If you pretend they're fractions you will find dx/dt=1 which is wrong in general. For instance, let's say f(x)=3, x(t)=sint+t.

        There is a lot of confusion that can be caused in instances like that.

        EDIT: I suppose in this case you could say df is 0 so you can't do that, but there is other confusing stuff that can happen if you don't pay attention to what the derivatives represent. For instance, you may have df/dx(0)=df/dt(0) in which case it is a really bad idea to treat them as fractions.

        • 1267 [he/him]
          ·
          3 years ago

          Fuck, I have killed a lot of brain cells.

        • 0karin728 [any]
          ·
          3 years ago

          The df terms in df/dx and df/dt represent fundamentally different things tho, so you couldn't just cancel them like that even if you're thinking of it as a fraction. The df term in df/dt is some function of t (say g(t)dt, if you think of dt as an arbitrarily small incriment in t) and in df/dx it's some function of x (say h(x)dx)).

          This turns df/dx =df/dt into (g(t)dt)/dt) = (h(x)dx)/dx, which reduces to g(t)=h(x), which is fine and doesn't cause any contradictions.

          • Pezevenk [he/him]
            ·
            3 years ago

            The df terms in df/dx and df/dt represent fundamentally different things tho

            That is why you shouldn't think of them as fractions lol

            EDIT: What I mean is that when you look at the notation and treat it as a simple fraction, the dfs look like they're the same thing.

    • Pezevenk [he/him]
      ·
      3 years ago

      You typically can't do that if you want to be rigorous in math. It's more complicated.