Why is the plot of $|x|^N + |y|^N =2$ almost a square when $N$ is large?

Question

Is there any reason , that justifies the fact that the plot of

$$ |x|^N + |y|^N =2 $$

for really big $ N > 100000 $ is almost a square ?

https://www.wolframalpha.com/input?i=+plot++%7Cx%7C%5E456+%2B+%7Cy%7C%5E456++%3D1&lang=es for example

Answer 1

The equation is equivalent to $(|x|^N + |y|^N)^{1/N} =2^{1/N}$. As $N\to\infty$, we have $(|x|^N + |y|^N)^{1/N}\to\max\{|x|,|y|\}$ (see this post for a proof) and $2^{1/N}\to1$. Therefore, as $N$ becomes large, the plot of $|x|^N+|y|^N=2$ can be approximated by the plot of $$\max\{|x|,|y|\}=1,$$ which is exactly a square.

Answer 2

For the more general case of $$x^{n}+y^{n}=r^{n}$$ 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$.

In particular the inside area is given by $$A_n=4r^2\, \frac{\Big(\Gamma \left(1+\frac{1}{n}\right)\Big)^2}{\Gamma \left(1+\frac{2}{n}\right)}$$ I $n$ is large, using Stirling approximation $$A_n \sim 4r^2\left(1-\frac{\pi ^2}{6 n^2}+O\left(\frac{1}{n^3}\right)\right)$$

If $r=2^{\frac{1}{n}}$, this gives $$A_n=4+\frac{8 \log (2)}{n}-\frac{2 \left(\pi ^2-12 \log ^2(2)\right)}{3 n^2}+O\left(\frac{1}{n^3}\right)$$

For $n=10^4$ the approximation gives $4.00055449038333$ while the exact value is $4.00055449038560$