Denominator of volume of n dimensional ball of radius 1 of dimension 20

If so, then you will discover that the denominator is $\frac{20}{2}!$ and that you can do that computation by hand or on most cheap scientific calculators.

Here is a derivation of the volume formula for dimension n and radius r.

Using Mathematica 11.3. #1 and #2 are the first and second parameters and the & states the formula can be used as functions.

$n_0=1\&$ $\text{answers}=\text{Table}\left[n_i=\text{Evaluate}\left[\text{FullSimplify}\left[\text{Evaluate}\left[\int_{-\text{\$\#\$1}}^{\text{\$\#\$1}} n_{i-1}\left(\sqrt{\text{\$\#\$1}^2-x^2}\right) \, dx\right],\text{\$\#\$1}\in \mathbb{R}\land \text{\$\#\$1}\geq 0\right]\right]\&,\{i,\text{Range}[1,10]\}\right]\text{ giving:}$ $\left\{2 \text{\$\#\$1}\&,\pi \text{\$\#\$1}^2\&,\frac{4 \pi \text{\$\#\$1}^3}{3}\&,\frac{\pi ^2 \text{\$\#\$1}^4}{2}\&,\frac{8 \pi ^2 \text{\$\#\$1}^5}{15}\&,\frac{\pi ^3 \text{\$\#\$1}^6}{6}\&,\frac{16 \pi ^3 \text{\$\#\$1}^7}{105}\&,\frac{\pi ^4 \text{\$\#\$1}^8}{24}\&,\frac{32 \pi ^4 \text{\$\#\$1}^9}{945}\&,\frac{\pi ^5 \text{\$\#\$1}^{10}}{120}\&\right\}$

Then, by examination and a proof by induction, the general formula results: $v=\frac{\pi ^{\frac{\text{\$\#\$1}}{2}} \text{\$\#\$2}^{\text{\$\#\$1}}}{\Gamma \left(\frac{\text{\$\#\$1}}{2}+1\right)}\&;$

The data that lead me to the induction are:

The volume ratios from dimension 2 to 20 are: $2 r,\frac{\pi r}{2},\frac{4 r}{3},\frac{3 \pi r}{8},\frac{16 r}{15},\frac{5 \pi r}{16},\frac{32 r}{35},\frac{35 \pi r}{128},\frac{256 r}{315},\frac{63 \pi r}{256},\frac{512 r}{693},\frac{231 \pi r}{1024},\frac{2048 r}{3003},\frac{429 \pi r}{2048},\frac{4096 r}{6435},\frac{6435 \pi r}{32768},\frac{65536 r}{109395},\frac{12155 \pi r}{65536},\frac{131072 r}{230945},\frac{46189 \pi r}{262144}$

The values of the volume formulae, $\frac{\left(\sqrt{\pi } r\right)^i}{\Gamma \left(\frac{i}{2}+1\right)}$ , are $1,2 r,\pi r^2,\frac{4 \pi r^3}{3},\frac{\pi ^2 r^4}{2},\frac{8 \pi ^2 r^5}{15},\frac{\pi ^3 r^6}{6},\frac{16 \pi ^3 r^7}{105},\frac{\pi ^4 r^8}{24},\frac{32 \pi ^4 r^9}{945},\frac{\pi ^5 r^{10}}{120},\frac{64 \pi ^5 r^{11}}{10395},\frac{\pi ^6 r^{12}}{720},\\\frac{128 \pi ^6 r^{13}}{135135},\frac{\pi ^7 r^{14}}{5040},\frac{256 \pi ^7 r^{15}}{2027025},\frac{\pi ^8 r^{16}}{40320},\frac{512 \pi ^8 r^{17}}{34459425},\frac{\pi ^9 r^{18}}{362880},\frac{1024 \pi ^9 r^{19}}{654729075},\frac{\pi ^{10} r^{20}}{3628800}$

The values of the area formulae, $\frac{\left(\sqrt{\pi }r\right)^{i}}{\Gamma \left(\frac{i}{2}+1\right)}$ are $0,2,2 \pi r,4 \pi r^2,2 \pi ^2 r^3,\frac{8 \pi ^2 r^4}{3},\pi ^3 r^5,\frac{16 \pi ^3 r^6}{15},\frac{\pi ^4 r^7}{3},\frac{32 \pi ^4 r^8}{105},\frac{\pi ^5 r^9}{12},\frac{64 \pi ^5 r^{10}}{945},\frac{\pi ^6 r^{11}}{60},\frac{128 \pi ^6 r^{12}}{10395},\\\frac{\pi ^7 r^{13}}{360},\frac{256 \pi ^7 r^{14}}{135135},\frac{\pi ^8 r^{15}}{2520},\frac{512 \pi ^8 r^{16}}{2027025},\frac{\pi ^9 r^{17}}{20160},\frac{1024 \pi ^9 r^{18}}{34459425},\frac{\pi ^{10} r^{19}}{181440}$

Now, permitting the dimension to be a non-integer real, at what dimension is the volume the largest fraction of a unit n-cube?

$max = n /. Flatten[NSolve[dv[n, 1/2] == 0, n]]\text{ gives }0.476582582306\text{ with a "volume" of }1.03869332805$

The limits of both the volume and area formulae as the dimension goes to $\infty$ is $0$ .

The volume and area formulae for the 0-dimensional and 1-dimensional are somewhat interesting. In the 0 dimensional case, the ball is a point, which is the "universe" and the "universe" has no boundary. In the 1-dimensional case, the ball is a bounded line of length $2r$ with 2 ends.