# Dimension that maximizes surface area and volume unit $n$-sphere

Let $$S_n(R)$$ and $$V_n(R)$$ denote the surface area and volume of an $$n$$-sphere with radius $$R$$ respectively.

It is well known that $$S_n(R) = \frac{n \pi^{\frac{n}{2}}R^{n-1}}{\Gamma\left(\frac{n}{2}+1\right)}\quad\text{and}\quad V_n(R) = \frac{\pi^{\frac{n}{2}}R^{n}}{\Gamma\left(\frac{n}{2}+1\right)}$$ where $$\Gamma(z)$$ is the gamma function: $$\Gamma(z) = \int_0^\infty e^{-x} x^{z-1}\, dx.$$

Question: Let $$v$$ and $$s$$ be positive integers such that $$S_s(1)$$ and $$V_v(1)$$ are the maximum. What is $$v-s?$$

If I am not mistaken, I saw somewhere that $$s = 7$$ and $$v = 5$$ (I think) but I couldn't find that source anymore.

While trying to prove the result, I realize that I am not able to do so.

Any hint is appreciated.

Hint: The only tricky bit is the $$\Gamma$$-function. For $$n$$ even you need the $$\Gamma$$-function at integer values, so you can just use $$\Gamma(k)=(k-1)!$$ For $$n$$ odd you need $$\Gamma(k+\frac{1}{2})$$ for $$k$$ an integer. Here you can use the Legendre duplication formula, which will work out to $$\Gamma(k+\frac{1}{2})=\frac{2^{1-2k}\sqrt{\pi}(2k-1)!}{(k-1)!}$$ Now you can just compute the first few terms and see at what point they start decreasing again. $$s=7$$ and $$v=5$$ looks roughly correct but I haven't done the computation.

Edit: I will write out the solution for the $$S_{2k}(1)$$, the other $$4$$ cases work similar. $$S_{2k}(1)=\frac{2k\pi^{2k/2}}{\Gamma(2k/2+1)}=\frac{2k\pi^{k}}{k!}=2\frac{\pi^k}{(k-1)!}$$ So if we go from $$k$$ to $$k+1$$, the volume gets multiplied by $$\frac{\pi}{k}$$, so until $$k=3$$ this is bigger than $$1$$, for $$k\ge 4$$ this is smaller. The maximum occurs at $$k=4$$. Hence the highest surface area of an even dimensional sphere is at $$k=4$$ or $$n=8$$.

• I got the part for even $n$ and odd $n.$ What I don't get is that how do we know that $V_n(1)$ and $S_n(1)$ will start decreasing after some point and never increase again? – Idonknow Nov 1 '19 at 0:41

Remember that $$\Gamma(x+1) = x \Gamma(x).$$ You can deal with the odd and even dimensions separately.

Now if you start with the ratio $$V_n = \frac{\pi^{n/2}R^{n}}{\Gamma\left(\frac{n}{2}+1\right)}$$ for some particular value of $$n,$$ and you add two the dimension repeatedly, you are multiplying the denominator by $$x,$$ $$x+1,$$ $$x+2, \ldots$$ (where $$x = \frac n2 + 1$$) whereas the numerator is multiplied by $$\pi$$ each time. First do this sufficiently many times so that the factor $$x+k$$ in the denominator is greater than $$\pi,$$ and then a sandwich with $$0 \leq V_i \leq \text{constant} \times \left(\frac\pi{x+k}\right)^i$$ shows that $$V_i \to 0.$$

For $$S_n = nV_n$$ you might add two to the dimension enough times so that $$n + k > 2\pi$$ and $$(n+2)/n \leq 2,$$ and then you have $$0 \leq S_i \leq \text{constant} \times \left(\frac{2\pi}{x+k}\right)^i.$$

So each of the formulas has a maximum over all odd dimensions and a maximum over all even dimensions, and whichever is greater is the maximum over all dimensions.