4
$\begingroup$

Prove that $$\left( \frac{nS_n}{S_{n-1}}\right)^{1/n} \le 2$$ where $S_i$ is the volume of the $i$th dimensional unit ball and $n\ge 2$.

I think we can use the fact that an $n$-dimensional unit ball is contained in a hypercube whose edge measures $2$ units, and the ball contains a hypercube whose edge measures $\sqrt{2}$ units. So we have $\sqrt{2}^n\le S_n \le 2^n$ and thus $$\left( \frac{nS_n}{S_{n-1}}\right)^{1/n} \le \left( \frac{n2^n}{\sqrt{2}^{n-1}} \right)^{1/n} = n^{1/n}2^{\frac{n+1}{2n}} = n^{1/n}\sqrt{2}\sqrt{2}^{1/n}.$$

So if we take $n^{1/n} \le 2$ and $\sqrt{2}^{1/n}\le \sqrt{2}^{1/2}$, then an upper bound to my problem is $2^{1+3/4}$ which isn't good enough.

Do we need another approach or do we get better bounds on $n^{1/n}$?

$\endgroup$
  • $\begingroup$ Since $3^{1 / 3} \sqrt(2) \sqrt{2}^{1 / n} > 2$, no estimate on $n^{1 / n}$ will work. $\endgroup$ – Travis Willse Jan 26 '18 at 12:55
  • $\begingroup$ Your argument is very nice, but it only works if $n^{\frac{1}{n}}2^{\frac{n+1}{2n}} \le 2 \Leftrightarrow n \le 2^{\frac{n-1}{2}}$, which is true for all $n \ge 7$, since the function on the right is exponential. So if you can find a better bound for lower $n$'s it's fine, but it might require an explicit formula?? $\endgroup$ – 57Jimmy Jan 26 '18 at 12:55
4
$\begingroup$

It is enough to show that $S_n\leq 2\,S_{n-1}$, which is trivial, since

$$\{(x_1,\ldots,x_{n-1}):x_1^2+\ldots+x_{n-1}^2\leq 1\}\times[-1,1]\supset \{(x_1,\ldots,x_{n}):x_1^2+\ldots+x_{n}^2\leq 1\}.$$ Over $n\geq 2$, $(2n)^{1/n}$ attains its maximum, $2$, at $n=2$.

$\endgroup$
2
$\begingroup$

Since the volume of a $n$-dimensional unit ball is (see volume of an $n$- ball)$$ S_n = \frac{π^{\frac{n}{2}}}{Γ\left( n + \frac{1}{2} \right)}, $$ then\begin{align*} \left( \frac{nS_n}{S_{n - 1}} \right)^{\frac{1}{n}} \leqslant 2 &\Longleftrightarrow \frac{n \sqrt{π} \cdot Γ\left( n - \frac{1}{2} \right)}{Γ\left( n + \frac{1}{2} \right)} \leqslant 2^n\\ &\Longleftrightarrow \frac{n \sqrt{π}}{n - \frac{1}{2}} \leqslant 2^n. \end{align*} Because$$ 2^n \geqslant 4 > 2 \sqrt{π} \geqslant \frac{n \sqrt{π}}{n - \frac{1}{2}}, $$ then $\displaystyle \left( \frac{nS_n}{S_{n - 1}} \right)^{\frac{1}{n}} \leqslant 2$.

$\endgroup$
  • $\begingroup$ This looks like "cheating". I don't think you're supposed to assume that formula to begin with. $\endgroup$ – Deepak Jan 26 '18 at 14:02
  • 1
    $\begingroup$ @Deepak Nowhere in the question prohibits use of explicit formulae for $S_n$. $\endgroup$ – Saad Jan 26 '18 at 14:07
  • $\begingroup$ The use of the gamma function is fine. I was trying to find a more intuitive explanation but I came across this result in a paper where they casually slipped this in. They could have used the gamma function. $\endgroup$ – Picasso Jan 26 '18 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.