8 thoughts on “Taking math to a whole new level”

  1. But that’s not even an equation.

    I this case it should be sin(sqrt(x^2+y^2))/sqrt(x^2+y^2) instead of sin(sqrt(x^2+y^2+z^2))/sqrt(x^2+y^2+z^2).

  2. MM is correct, u = f(x,y,z) would produce a 4D plot. The paper plot is instead z = f(x,y) and should not include the z^2 part in the equation.

  3. You’re forgetting that we can’t completely see the left part of the expression. It’s entirely possible that the expression we see is equal to a constant, making this curve a cross-section of hyperspace

    1. While it is a possibility that the graph was intended as a cross-section of hyperspace, I do believe that it is more likely that the author made a mistake when writing the formula.
      Without knowing what might be hidden on the left side of the expression, it does look like the fraction is purposefully placed in the centre of the available writing area. Considering that the author did make an effort to make a really nice 3D graph (it’s well made, the edges seems to be coloured for height and the paper is placed so that the pre-printed lines match); I claim that a constant placed in the hidden area to the left of the expression would not fit. It would make the placement of the expression rather awkward and make the rest of the effort put down to be more or less in vain – I mean, why make such a nice graph and then not take an effort to place the expression nicely?.
      Sure, a counter argument could be “if the z^2 part is a mistake, why be so sloppy as to not double check the expression?”; but frankly, I do believe it’s a more likely mistake rather than placing the formula 5 mm too far too the left when the pre printed lines on the paper matches on a <1 mm scale.

  4. The z^2 terms are clearly erroneous. If we remove them then we would get a curve that looks similar to the one created.

    Consider instead, the case where we have a constant on the left:
    k = sin(sqrt(x^2+y^2+z^2))/(sqrt(x^2+y^2+z^2))

    Then we have sin(t) = kt, x^2+y^2+z^2 = t^2. If |k| >= 1 then we have just t = 0, which gives the point (0,0,0). If 0 < |k| < 1 then we have a finite number of discrete points t, each of which corresponds to a sphere, centred at (0,0,0) of radius t, and for |k| = 0 we have an infinite number of spheres centred at (0,0,0). Regardless of the constant, we certainly wouldn't get a single surface like the one in the image, but a number of concentric spheres…

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