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The graph of the function $x^{n}+y^{n}=r^{n}$ for certain large values of $n$ looks suspiciously like a square.

See this page from wolframalpha. Have any results been proven regarding this observation? What do we call this figure anyway?

kodyv
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    To expand a bit on Gadi's answer below: One name would be the $r$-sphere of the $L^{p}$-norm on $\mathbb{R}^2$ (something inside me refuses to use $n$ instead of $p$). You can prove that $\lim_{p \to \infty} |x|{p} = |x|{\infty}$ and the $r$-sphere of the $L^{\infty}$-norm is cube of side length $2r$. The phenomenon is essentially the same as the one that occurs when looking at the graph of $x^n$ on $[0,2]$, which approaches a right angle for large $n$. – t.b. May 22 '11 at 11:42
  • Let me say an answer using basic maths. Without a loss of generality, assume $r=1$; a different value of $r$ simply scales the picture. More importantly, If $n$ is very large and $x$ or $y$ are smaller than one, then $x^n$ and $y^n$ are much smaller than one - they're virtually zero, unless $x$ or $y$ are extremely close to one. For example, $0.9^{100}$ is still close to zero. So if $x^n+y^n=1$, then either $x$ or $y$ or both have to be extremely close to 1 from below so that the high power contributes almost one. – Luboš Motl May 22 '11 at 11:53
  • In other words, for large even $n$, $x^n+y^n=1$ is essentially equivalent to $x\approx \pm 1$ or $y\approx \pm 1$ (with the other variable allowed to be essentially anything), which are the equations of the square. Note that neither $x$ nor $y$ can exceed one, so that's why you only get the 4 line segments. – Luboš Motl May 22 '11 at 11:55
  • @Luboš Motl: That's quite good enough to post as an answer IMHO. – ShreevatsaR May 22 '11 at 11:56
  • J’ai trouvé une merveilleuse réponse à cette question, mais la marge est trop étroite pour la contenir. :) – Wok May 22 '11 at 17:18

4 Answers4

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You have just rediscovered the max-norm.

More precisely, you have noted that as $p$ becomes large, the unit circle in the $l_p$ norm looks similar and similar to the one of the $l_\infty$ norm.

Community
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Gadi A
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Sometimes it's called a superellipse - see, e.g., http://en.wikipedia.org/wiki/Superellipse

Gerry Myerson
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By symmetry, you can consider the equation $y^n+x^n=r^n$ for $0 \leq x \leq r$. Rewrite as $$ y(x) = \sqrt[n]{{r^n - x^n }} = \sqrt[n]{{r^n - r^n \bigg(\frac{x}{r}\bigg)^n }} = r\sqrt[n]{{1 - \bigg(\frac{x}{r}\bigg)^n }}, $$ for $0 \leq x \leq r$. This shows that $y$ is strictly decreasing from $r$ to $0$ as $x$ varies from $0$ to $r$, respectively, and that the sequence of functions $y(x) = y_n (x)$ converges pointwise, as $n \to \infty$, to the function $f$ defined by $f(x)=r$ if $0 \leq x < r$ and $f(r)=0$; moreover, the convergence to $f$ is uniform for $x \in [0,a]$, for any $0 < a <r$ (but not for $x \in [0,r]$, since $y(r)=0$). This accounts for the square shape.

Shai Covo
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For the super circle defined by $$x^{n}+y^{n}=r^{n}$$ use the parametrization $$x(t)=\pm r \cos ^{\frac{2}{n}}(t)\qquad \text{and} \qquad y(t)=\pm r \sin ^{\frac{2}{n}}(t)$$ for $0 \leq t \leq \frac \pi 2$.

Its area is given in terms of the gamma function by $$A_n=4r^2\, \frac{\Big[\Gamma \left(1+\frac{1}{n}\right)\Big]^2}{\Gamma \left(1+\frac{2}{n}\right)}$$

If you consider large values of $n$

$$A_n=4r^2\left(1-\frac{\pi ^2}{6 n^2}\right)+O\left(\frac{1}{n^3}\right)$$ Using, as you did, $n=24$, $A_{24}=3.98924 r^2$

Claude Leibovici
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