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Let $V_n=V(B^n)$ be the volume of the $n$-dimensional unit ball $B^n$. By cross-sectioning $B^n$ along $x_n$-axis, $-1\leq x_n\leq 1$ and by means of similarity of hyper disks we have $$V_n=2\int_0^1(\sqrt{1-x_n^2}\,)^{n-1}V_{n-1}dx_n =B(\tfrac12,\tfrac{n+1}2)V_{n-1}=\frac{\sqrt{\pi}\,\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n+2}2)}V_{n-1}$$ By induction, we obtain the well-known formula $$V_n=\frac{\pi^{\frac n2}}{\Gamma(\tfrac{n+2}{2})}.$$ To find the maximum, I took the derivative and $\frac{dV_n}{dn}=0$ gave the equation $$\psi(\tfrac{n+2}{2})=\ln\pi\tag1$$ where $\psi$ is the Digamma function. WolframAlpha is giving the approximate solution $5.256946$. How could I solve the equation $(1)$ without WA? Is there a closed form for the solution? What is the significance of this number? Thanks for reading.

Bob Dobbs
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1 Answers1

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To find a closed form for a first approximation of the solution, rewrite $$V_n=\frac{\pi ^{\frac n 2}}{\Gamma \left(\frac{n+2}{2}\right)}\qquad \text{as} \qquad\left(\frac n 2\right)!=\frac {\pi^{\left(\frac n 2\right)}} {V_n}$$ Have a look at this old question of mine and use @robjohn's approximation. It leads to $$n\sim 2e\pi\,e^{W(t)}-1\qquad \text{where} \qquad t=-\frac{\log \left(2 \pi ^2 V_n^2\right)}{2 e \pi }$$ $W(t)$ being Lambert function.

Differentiating with respect to $V_n$ $$\frac {dn}{dV_n}=-\frac{2}{V_n (1+W(t))} \quad \implies \quad \frac {dV_n}{dn}=-\frac{1}{2} V_n (1+W(t))$$ $$\frac {dV_n}{dn}=0 \quad \implies \quad 1+W(t)=0\quad \implies \quad t=-\frac 1e$$ $$t=-\frac 1e \quad \implies \quad V_{\text{min}}=\frac{e^{\pi }}{\pi\sqrt{2} }\quad \implies \quad \boxed{\large\color{red}{n\sim 2\pi-1}}$$

The second derivative test shows that this is a maximum of $V_n$. To show it at $V_{\text{min}}$, use the indirect way $$\frac {d^2V_n}{dn^2}=-\frac{\frac{d^2n}{dV_n^2} } {\left(\frac{dn}{dV_n}\right)^3 }=\frac{V_n}{4} \left(1+W(t) \left(W(t)-\frac{1}{e \pi t}+2\right)\right)$$ which, for $V_{\text{min}}$ gives $-\frac{e^{\pi }}{4 \sqrt{2} \pi ^2}$.

To polish the result, perform $\color{red}{\text{one single iteration}}$ of Newton-like method of order $k$; for any $k$ this gives $\color{red}{\text{an explicit formula}}$ for $n_1^{(k)}$. However, we shall look for the zero of function

$$ f(x)=e^{\psi ^{(0)}\left(x\right)}-\pi \qquad \text{where} \qquad x=\frac{n+2}{2}$$

which is very close to linearity.

For Newton method, using $x_0=a$, the formula will be

$$x_1^{(2)}=a-\frac{1-\pi e^{-\psi ^{(0)}(a)}}{\psi ^{(1)}(a)}$$ and for Halley method, it will be

$$x_1^{(3)}=a--\frac{2}{\left(1+\frac{2 \pi }{e^{\psi ^{(0)}(a)}-\pi }\right) \psi ^{(1)}(a)-\frac{\psi ^{(2)}(a)}{\psi ^{(1)}(a)}}$$

Now, using the tight bounds given here we have $$x_{\text{min}}=\frac{1}{\log \left(1+\frac{1}{\pi }\right)}\qquad \text{and}\qquad x_{\text{max}}=\pi+\frac 12$$

Using a secant $$a=\frac {x_{\text{min}}\,f(x_{\text{max}})-x_{\text{max}}\,f(x_{\text{min}}) } {f(x_{\text{max}})- f(x_{\text{min}})}$$ already gives $n=5.256946$ which is exactly Wolfram Alpha result.

Back to $n$, playing with $k$ and converting the $\color{red}{\text{explicit result}}$ to decimals

$$\left( \begin{array}{cc} k & n_{(k)} & \text{method} \\ 2 & \color{red} {5.256946404860576}846027721979484773804672 &\text{Newton} \\ 3 & \color{red} {5.25694640486057678013283}5142847827596458 &\text{Halley} \\ 4 & \color{red} {5.256946404860576780132838388690}927254265 &\text{Householder}\\ 5 & \color{red} {5.2569464048605767801328383886907692366}11 &\text{no name} \\ 6 & \color{red} {5.256946404860576780132838388690769236619} &\text{no name} \\ \end{array} \right)$$

Edit

To obtain a better estimate of $a$, instead of using the secant, build the cubic function $$g(x)=a+b x+c x^2+dx^3$$ such that $$g(x_{\text{min}})=f(x_{\text{min}})\qquad \qquad g(x_{\text{max}})=f(x_{\text{max}}) $$ $$g'(x_{\text{min}})=f'(x_{\text{min}})\qquad \qquad g'(x_{\text{max}})=f'(x_{\text{max}}) $$

Solve the cubic using the trigonometric method and the decimal representation of the $\color{red}{\text{explicit result}}$ is $$a=\color{red} {5.25694640486}44$$ which leads to

$$n_1^{(2)}=\color{red} {5.2569464048605767801328383}98$$ $$n_1^{(3)}=\color{red} {5.2569464048605767801328383886907692366}24$$

Claude Leibovici
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