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I am a first-year engineering student and attending a course that involves analytic geometry and vector calculus.

While studying for a test I had run into the equation for the surface $\sum: x^4+y^4+z^4 = 1$ and didn't know what shape it had, so I 3D plotted it and found out it looks like a game die, or rather a cube with round edges.

After that I have played with the equation $\sum_{n}:x^{2n}+y^{2n}+z^{2n}=c$ , where: $c\in\mathbb{R},n\in\mathbb{N}$ and noticed that the surface $\sum_{n}$ converges to a cube as $n$ increases, and wondered if there is an explanation or proof of why this happens.

Ido Nir
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    $c=1$ is the easiest to explain with. Each of $x,y,z$ is at most $1$ in absolute value; and if $x$ (say) is less than $1$, then raising it to higher and higher powers makes it smaller and smaller. So for each triple $(x,y,z)$, one of them has to be really close to $1$ (or $-1$) because the other two are contributing less and less as $n$ increases. – Greg Martin Aug 02 '22 at 19:26
  • https://en.wikipedia.org/wiki/Superellipsoid – Dan Uznanski Aug 13 '22 at 10:31

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It is simple to prove so. First of all, this only happens with $c\in\mathbb R^+$. Let's consider first the limit of one power $$\lim_{n\to+\infty}x^{2n}=\begin{cases}+\infty&\text{if }\lvert x\rvert>1\\0&\text{if }\lvert x\rvert<1\end{cases}$$ Due to the fact that $c\in\mathbb R^+\Rightarrow c<+\infty\Rightarrow\lvert x\rvert,\lvert y\rvert,\lvert z\rvert\leq1$, and because $c\neq0$ we have that not all three can be less than one. So at least one coordinate must be $1$ or $-1$, being each one a face of the cube.

There is another prove from the following result. Let $f_n(x_1,...,x_m)=\sqrt[2n]{x_1^{2n}+\cdots+x_m^{2n}}$, then $f_n\xrightarrow{n\to+\infty}\max(\lvert x_1\rvert,...,\lvert x_n\rvert)$. Then, we can write $$x^{2n}+y^{2n}+z^{2n}=c\Rightarrow\sqrt[2n]{x^{2n}+y^{2n}+z^{2n}}=\sqrt[2n]c\xrightarrow{n\to+\infty}\max(\lvert x\rvert,\lvert y\rvert,\lvert z\rvert)=1$$ Wich is the equation for a cube of side 1.

Alba Vika Sánchez Bernal
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