• LifeInMultipleChoice@lemmy.world
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    14 hours ago

    I read it as surface area, thus being the amount of space on the sphere itself.

    A=4πr2 is the formula if I remember correctly, so I just figure only radius can be altered to match infinity.

    Maybe someone will tell me I’m missing something

    • thebestaquaman@lemmy.world
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      12 hours ago

      I believe the surface area of an n-dimensional hypersphere is (n - 1) pi r^{n - 1}. In that case (I may have some factors wrong here, just going off memory), an infinite-dimensional hypersphere has infinite surface area as long as it has non-zero radius.

      • loppy@fedia.io
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        7 hours ago

        It does indeed scale with r^(n-1), but your factors are not close at all. It involves the gamma function, which in this case can be expanded into various factorials and also a factor of sqrt(pi) when n is odd. According to Wikipedia, the expression is 2pi(n/2)r(n-1)/Gamma(n/2).

        • thebestaquaman@lemmy.world
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          2 hours ago

          Edit: ignore the below, I forgot my pi-factor in the gamma function for half-integers…

          Edit 2: Since you’re right, my missing gamma-factor completely changes this. An infinite-dimensional hypersphere will have zero surface area for any (finite?) radius.

          Original dum-dum:

          While I’m completely open that my factor is likely wrong here, the expression you provided is definitely wrong in the 3D case (I’m assuming the r superscript on the pi was a typo), since it doesn’t give n = 3 => A = 4 pi r^2.