Find the length of the curve $x^{2k}+y^{2k} =1$

Question

I want to find an expression for length and find the limit $k\rightarrow \infty$

The answer is obviously 8, if we look at the graphs.

Answer 1

A parametrization could be:

$$x(t)=\sqrt[k]{\cos(t)}$$ $$y(t)=\sqrt[k]{\sin(t)}$$

Then, the length is given by $$\int_0^{2\pi}\|\dot \gamma (t)\|dt$$ where $$\gamma (t)=(\sqrt[k]{\cos(t)},\sqrt[k]{\sin(t)})$$

Well, this integral doesn't look easy to calculate :-( But the limit of the figure of equation $x^{2k}+y^{2k}=1$ when $k\to\infty $ is the square of center $O$ that contain the circle of equation $x^2+y^2=1$ (then the limite of this integral is $8$).

Answer 2

First of all $$ x^{2n}+y^{2n}=1\implies y'=-\frac{x^{2n-1}}{y^{2n-1}}\tag{1} $$ Thus, $$ \begin{align} \frac L4 &=\int_0^1(1+y'^2)^{1/2}\,\mathrm{d}x\\ &=\int_0^1\left(1+\left(\frac{x^{2n}}{1-x^{2n}}\right)^{2-1/n}\right)^{1/2}\mathrm{d}x\\ &=\frac1{2n}\int_0^1\left(1+\left(\frac{x}{1-x}\right)^{2-1/n}\right)^{1/2}x^{\frac1{2n}}\frac{\mathrm{d}x}{x}\\ &=\frac1{2n}\int_0^1\color{#00A000}{\left((1-x)^{2-1/n}+x^{2-1/n}\right)^{1/2}}x^{\frac1{2n}-1}(1-x)^{\frac1{2n}-1}\mathrm{d}x\tag{2} \end{align} $$ For $n\ge1$, we have $$ \color{#C00000}{1} \ge\color{#00A000}{\left((1-x)^{2-1/n}+x^{2-1/n}\right)^{1/2}} \ge\sqrt{1-2x(1-x)} \ge\color{#C00000}{1}-\color{#0000FF}{2x(1-x)}\tag{3} $$ To combine $(2)$ and $(3)$ we will use $$ \begin{align} \frac1{2n}\int_0^1\color{#C00000}{1}\,x^{\frac1{2n}-1}(1-x)^{\frac1{2n}-1}\mathrm{d}x &=\frac1{2n}\mathrm{B}\left(\frac1{2n},\frac1{2n}\right)\\ &=\frac1{2n}\frac{\Gamma\left(\frac1{2n}\right)\Gamma\left(\frac1{2n}\right)}{\Gamma\left(\frac1{n}\right)}\\ &=\frac1{2n}\frac{2n\Gamma\left(1+\frac1{2n}\right)2n\Gamma\left(1+\frac1{2n}\right)}{n\Gamma\left(1+\frac1{n}\right)}\\ &=2\frac{\Gamma\left(1+\frac1{2n}\right)\Gamma\left(1+\frac1{2n}\right)}{\Gamma\left(1+\frac1{n}\right)}\\ &=2\left(1+\frac1n\right)\frac{\Gamma\left(1+\frac1{2n}\right)\Gamma\left(1+\frac1{2n}\right)}{\Gamma\left(2+\frac1{n}\right)}\tag{4} \end{align} $$ and $$ \begin{align} \frac1{2n}\int_0^1\color{#0000FF}{2x(1-x)}x^{\frac1{2n}-1}(1-x)^{\frac1{2n}-1}\mathrm{d}x &=\frac1n\mathrm{B}\left(1+\frac1{2n},1+\frac1{2n}\right)\\ &=\frac1n\frac{\Gamma\left(1+\frac1{2n}\right)\Gamma\left(1+\frac1{2n}\right)}{\Gamma\left(2+\frac1{n}\right)}\tag{5} \end{align} $$ Combining $(2)$, $(3)$, $(4)$, and $(5)$, we get $$ \hspace{-1cm}\left(2+\frac1n\right)\frac{\Gamma\left(1+\frac1{2n}\right)\Gamma\left(1+\frac1{2n}\right)}{\Gamma\left(2+\frac1{n}\right)} \le\frac L4 \le\left(2+\frac2n\right)\frac{\Gamma\left(1+\frac1{2n}\right)\Gamma\left(1+\frac1{2n}\right)}{\Gamma\left(2+\frac1{n}\right)}\tag{6} $$ Since $\Gamma(1)=\Gamma(2)=1$, the squeeze theorem says that as $n\to\infty$, $L\to8$.

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Computation of the Length for Large $\boldsymbol{n}$

As $n\to\infty$, the integrand tends to $\left((1-x)^2+x^2\right)^{1/2}x^{-1}(1-x)^{-1}$, which is not integrable on $[0,1]$. This causes convergence problems when trying to evaluate $(2)$ for large $n$. We can use the Beta function to bypass this difficulty.

Break up $(2)$ as $$ \begin{align} \frac L4 &=\frac1{2n}\int_0^1\left((1-x)^{2-\frac1n}+x^{2-\frac1n}\right)^{1/2}x^{\frac1{2n}-1}(1-x)^{\frac1{2n}-1}\mathrm{d}x\\ &=\frac1{2n}\int_0^1\color{#00A000}{\left[\left((1-x)^{2-\frac1n}+x^{2-\frac1n}\right)^{1/2}-1\right]x^{\frac1{2n}-1}(1-x)^{\frac1{2n}-1}}\mathrm{d}x\\ &+\frac1{2n}\mathrm{B}\left(\frac1{2n},\frac1{2n}\right)\tag{7} \end{align} $$ The green integrand in $(7)$ is bounded on $[0,1]$, so $(7)$ behaves much better with computer integration.

Formula $(7)$ can be implemented in Mathematica as

f[n_] := NIntegrate[(Sqrt[x^(2-1/n) + (1-x)^(2-1/n)] - 1) x^(1/2/n-1)(1-x)^(1/2/n-1), {x,0,1}, WorkingPrecision->20]/2/n + Beta[1/2/n, 1/2/n]/2/n

This gives $f[1]=\frac\pi2$ and $f[1000000]=1.99999944669682658902684568$

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Asymptotic Approximation

Consider the integral $$ I(\alpha,\beta)=\frac1{2n}\int_0^1\left((1-x)^{2-\beta}+x^{2-\beta}\right)^{1/2}x^{\alpha-1}(1-x)^{\alpha-1}\mathrm{d}x\tag{8} $$ For $(2)$, we wish to set $\alpha=\frac1{2n}$ and $\beta=\frac1n$. $$ \begin{align} \hspace{-1cm}\frac{\partial}{\partial\beta}I\left(\frac1{2n},0\right) &=-\frac1{4n}\int_0^1\frac{\log(1-x)(1-x)^2+\log(x)x^2}{\left((1-x)^2+x^2\right)^{1/2}}x^{\frac1{2n}-1}(1-x)^{\frac1{2n}-1}\mathrm{d}x\\ &=-\frac1{4n}\int_0^1\frac{\log(1-x)(1-x)^2+\log(x)x^2}{\left((1-x)^2+x^2\right)^{1/2}}x^{-1}(1-x)^{-1}\mathrm{d}x\\ &+O\left(\frac1{n^2}\right)\tag{9} \end{align} $$ Evaluating the integral in $(9)$, we get that $$ I\left(\frac1{2n},\frac1n\right)-I\left(\frac1{2n},0\right)\sim\frac{0.514}{n^2}\tag{10} $$ Thus, we can set $\beta=0$ and only incur a $O\left(\frac1{n^2}\right)$ error. $$ \begin{align} I\left(\frac1{2n},0\right) &=\frac1{2n}\int_0^1\left[\left((1-x)^2+x^2\right)^{1/2}-1\right]x^{\frac1{2n}-1}(1-x)^{\frac1{2n}-1}\mathrm{d}x\\ &+\frac1{2n}\mathrm{B}\left(\frac1{2n},\frac1{2n}\right)\\ &=\frac1{2n}\int_0^1\left[\left((1-x)^2+x^2\right)^{1/2}-1\right]x^{-1}(1-x)^{-1}\mathrm{d}x\\ &+2+O\left(\frac1{n^2}\right)\\ &=2+\frac{\log(2)+\sqrt2\log(\sqrt2-1)}{n}+O\left(\frac1{n^2}\right)\tag{11} \end{align} $$ Putting together $(10)$ and $(11)$, we get $$ \frac L4=2-\frac{\sqrt2\log(\sqrt2+1)-\log(2)}{n}+O\left(\frac1{n^2}\right)\tag{12} $$ where $\sqrt2\log(\sqrt2+1)-\log(2)\approx0.55330329972051571737$