Hyper-intriguing Ornament III

The limit is easy to understand. The fraction of a n-dimensional hyperball of radius 1 that is in the n-dimensional hyperball of radius $\frac12$ is $2^{-n}$ , because the exponent on $r$ is $n$ ..

Let us engage the piledriver . Using the radius ratio in this problem, $\frac45$ , the fraction of the volume in hyper-spherical end cap, which number $2 n$ , is:

Using the odd dimensioned hyper-spherical cap formulae, which are fast to compute, the progression towards a limit is easy to see: \begin{array}{rl} 1 & 0.2 \\ 3 & 0.168 \\ 5 & 0.0856 \\ 7 & 0.038192 \\ 9 & 0.0160366 \\ 11 & 0.00650553 \\ 13 & 0.00258142 \\ 15 & 0.00100875 \\ 17 & 0.000389791 \\ 19 & 0.000149336 \\ 21 & 0.0000568287 \\ 23 & 0.0000215085 \\ 25 & \text{8.104170800930324\$\grave{ }\$*\${}^{\wedge}\$-6} \\ 27 & \text{3.042127642435132\$\grave{ }\$*\${}^{\wedge}\$-6} \\ 29 & \text{1.1383156440627037\$\grave{ }\$*\${}^{\wedge}\$-6} \\ 31 & \text{4.24774590586685\$\grave{ }\$*\${}^{\wedge}\$-7} \\ 33 & \text{1.5813248106526474\$\grave{ }\$*\${}^{\wedge}\$-7} \\ 35 & \text{5.874593794653908\$\grave{ }\$*\${}^{\wedge}\$-8} \\ 37 & \text{2.1783826641858435\$\grave{ }\$*\${}^{\wedge}\$-8} \\ 39 & \text{8.06451316612106\$\grave{ }\$*\${}^{\wedge}\$-9} \\ 41 & \text{2.9811522426293965\$\grave{ }\$*\${}^{\wedge}\$-9} \\ 43 & \text{1.1005645447878985\$\grave{ }\$*\${}^{\wedge}\$-9} \\ 45 & \text{4.0581363685967064\$\grave{ }\$*\${}^{\wedge}\$-10} \\ 47 & \text{1.4947362318550126\$\grave{ }\$*\${}^{\wedge}\$-10} \\ 49 & \text{5.5000958468224146\$\grave{ }\$*\${}^{\wedge}\$-11} \\ 51 & \text{2.021992819485542\$\grave{ }\$*\${}^{\wedge}\$-11} \\ 53 & \text{7.427184687963443\$\grave{ }\$*\${}^{\wedge}\$-12} \\ 55 & \text{2.7260391235233023\$\grave{ }\$*\${}^{\wedge}\$-12} \\ 57 & \text{9.998344194441533\$\grave{ }\$*\${}^{\wedge}\$-13} \\ 59 & \text{3.6646664950255834\$\grave{ }\$*\${}^{\wedge}\$-13} \\ 61 & \text{1.342366651319315\$\grave{ }\$*\${}^{\wedge}\$-13} \\ 63 & \text{4.914237412703877\$\grave{ }\$*\${}^{\wedge}\$-14} \\ 65 & \text{1.798065981670408\$\grave{ }\$*\${}^{\wedge}\$-14} \\ 67 & \text{6.575584114225488\$\grave{ }\$*\${}^{\wedge}\$-15} \\ 69 & \text{2.4035637798312907\$\grave{ }\$*\${}^{\wedge}\$-15} \\ 71 & \text{8.781764054282071\$\grave{ }\$*\${}^{\wedge}\$-16} \\ 73 & \text{3.207183691988818\$\grave{ }\$*\${}^{\wedge}\$-16} \\ 75 & \text{1.170825279751455\$\grave{ }\$*\${}^{\wedge}\$-16} \\ 77 & \text{4.272636367879485\$\grave{ }\$*\${}^{\wedge}\$-17} \\ 79 & \text{1.55863354894657\$\grave{ }\$*\${}^{\wedge}\$-17} \\ 81 & \text{5.683872750992984\$\grave{ }\$*\${}^{\wedge}\$-18} \\ 83 & \text{2.0720694479380085\$\grave{ }\$*\${}^{\wedge}\$-18} \\ 85 & \text{7.551457047552192\$\grave{ }\$*\${}^{\wedge}\$-19} \\ 87 & \text{2.7512500232889627\$\grave{ }\$*\${}^{\wedge}\$-19} \\ 89 & \text{1.0020932495292006\$\grave{ }\$*\${}^{\wedge}\$-19} \\ 91 & \text{3.648971790899017\$\grave{ }\$*\${}^{\wedge}\$-20} \\ 93 & \text{1.3283799660362259\$\grave{ }\$*\${}^{\wedge}\$-20} \\ 95 & \text{4.8346872349364785\$\grave{ }\$*\${}^{\wedge}\$-21} \\ 97 & \text{1.7591918852300544\$\grave{ }\$*\${}^{\wedge}\$-21} \\ 99 & \text{6.399720825247548\$\grave{ }\$*\${}^{\wedge}\$-22} \\ \end{array}

For the record, the first 9 hyper-spherical cap formulae are: $\begin{array}{rl} 1 & h \\ 2 & \tan ^{-1}\left(\frac{h-r}{\sqrt{-h (h-2 r)}}\right) r^2+\frac{\pi r^2}{2}+(h-r) \sqrt{-h (h-2 r)} \\ 3 & -\frac{1}{3} h^2 \pi (h-3 r) \\ 4 & \frac{1}{12} \pi \left(6 \tan ^{-1}\left(\frac{h-r}{\sqrt{-h (h-2 r)}}\right) r^4+3 \pi r^4-2 \left(\sqrt{-h (h-2 r)} r (h+3 r)-6 \sqrt{-h^5 (h-2 r)}\right) r-4 \sqrt{-h^7 (h-2 r)}\right) \\ 5 & \frac{1}{30} h^3 \pi ^2 \left(3 h^2-15 r h+20 r^2\right) \\ 6 & \frac{1}{180} \pi ^2 \left(30 \tan ^{-1}\left(\frac{h-r}{\sqrt{-h (h-2 r)}}\right) r^6+15 \pi r^6+2 (h-r) \left(8 h^4-32 r h^3+22 r^2 h^2+20 r^3 h+15 r^4\right) \sqrt{-h (h-2 r)}\right) \\ 7 & \frac{1}{210} h^4 \pi ^3 \left(-5 h^3+35 r h^2-84 r^2 h+70 r^3\right) \\ 8 & \frac{\pi ^3 \left(210 \tan ^{-1}\left(\frac{h-r}{\sqrt{-h (h-2 r)}}\right) r^8+105 \pi r^8-2 \sqrt{-h (h-2 r)} \left(48 h^7-336 r h^6+808 r^2 h^5-680 r^3 h^4+6 r^4 h^3+14 r^5 h^2+35 r^6 h+105 r^7\right)\right)}{5040} \\ 9 & \frac{h^5 \pi ^4 \left(35 h^4-315 r h^3+1080 r^2 h^2-1680 r^3 h+1008 r^4\right)}{7560} \\ \end{array}$

Another charming problem on a Sunday morning, Comrade! Thank you!

Let me try to explain this in the simplest terms I can think of, without resorting to fancy formulas. In what follows, let $b$ be the volume of the unit ball in $\mathbb{R}^{n-1}$ . For example, for $n=3$ , we have $b=\pi$ , the area of the unit disk in $\mathbb{R}^2$ .

We can inscribe a cylinder of volume $6\times 4^{n-1}b$ into our hyperball by letting $x_1^2+...+x_{n-1}^2\leq 16$ and $|x_n| \leq 3$ . Likewise we can inscribe each of the $2n$ caps (for example, $x_n\geq 4)$ into a cylinder of volume $3^{n-1}b$ since $x_1^2+...+x_{n-1}^2\leq 9$ . Thus $\frac{V_{caps}}{V_{total}} \leq \frac{2n\times 3^{n-1}}{6\times 4^{n-1}}$ , approaching 0 as $n \rightarrow \infty$ . Thus the percentage of the volume of the hyperball residing within the hypercube will approach $\boxed{100}\text{\%}$ .

If you are confused be my remarks (and by the cylinders in particular), consider the case $n=3.$