Find the volume of $P_n = {x \in B_n \textrm{ s.t. } |x_1| < \frac{1}{1000}}$ and the volume of $B_n-P_n$, and determine which is bigger

Question

Let $B_n$ be the unit ball in $R^n$. We declare $$P_n = \left\{ x\in B_n \textrm{ such that }|x_1| < \frac{1}{1000}\right\} .$$ I want to calculate the volume of $P_n$ and $B_n - P_n$ and determine which is bigger.

I tried to use Fubini's theorem here and found $$P_n = V_{n-1}\int_\frac{-1}{1000}^\frac{1}{1000} \left(\sqrt{1-x_1^2}\right)^{n-1}dx_1 ,$$ where $V_{n-1}$ is the volume of the unit ball in $R^{n-1}$. I got to this answer since the volume of a ball in $R^n$ with a radius $r$ is $V_n r^n$.

However here I get stuck since I don't to solve this integral. I couldn't really solve it even with the help of Wolfram Alpha.

Am I doing something wrong?

Answer 1

For large enough $n$, the set $P_{ni} = \{x \in B_n; |x_i| \le \frac{1}{1000} \}$ is bigger.

Indeed, for each $i$, let us write as $Q_{ni} = B_n\setminus P_{ni} = \{x \in B_n; |x_i| > \frac{1}{1000} \}$. So it suffices the show the following inequality: Vol$(Q_{n1})$ $<\frac{1}{2} \times$Vol$(B_n)$ for $n$ sufficiently large. We do this next.

Then by symmetry each $Q_{ni}$ has the same volume, and of course $\cup_n Q_i \subset B_n$. However, each $x \in B_n$ is in at most $1000^2 =1000000$ of the $Q_{ni}$s [make sure you see why] which implies the following inequality: $\sum_n$ Vol$(Q_{ni}) \le 1000000\times$Vol$(B_n)$, which, as all the $Q_{ni}$s have the same volume, in turn implies the following string of inequalities: Vol$(Q_{n1}) \le \frac{1000000}{n} \times $Vol$(B_n)$ $<\frac{1}{2} \times$Vol$(B_n)$ for $n$ sufficiently large, which yields precisely what you want to show.

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ETA on the other hand, for general (large) positive $K$, let $P^K_{ni}$ be the set $\{x \in B_n; |x_i| \le \frac{1}{K} \}$. Then the inequality Vol$(P^K_{ni}) < \frac{1}{2} \times$ Vol$(B_n)$ only if $n \ge \theta(K^2)$. Indeed, let us set $a' = \frac{4}{K}$. Then for all $a < a'$, we note that Vol$(B_{n-1}(\sqrt{1-a^2})) \geq (\sqrt{1-a^2})^{n-1}$Vol$(B_{n-1})$ $\geq \frac{1}{2}$Vol$(B_{n-1})$ for $n < \frac{K^2}{8}$. This implies

$$\text{vol}(B_n) \ge \int^{a'}_{-a'} \text{Vol}\left(B_{n-1}\left(\sqrt{1-x^2}\right)\right) dx$$

$$ > 2 \int^{\frac{1}{K}}_{-\frac{1}{K}} \text{Vol}\left(B_{n-1}\left(\sqrt{1-x^2}\right)\right) dx \ \doteq \ 2 \text{Vol}(P^K_{ni}). $$

Answer 2

One can write an explicit formula for the integral in terms of $n$ using a hypergeometric function: $$\operatorname{vol}(P_n) = 2 s \cdot {}_2F_1\left(\frac{1}{2}, \frac{1}{2}(-n + 1);\frac{3}{2}; s^2\right) V_{n - 1}, \qquad s := \frac{1}{1000}.$$ Unless you have a good deal of intuition for hypergeometric functions, though---I don't---this probably doesn't illuminate the point of the problem much, to say nothing of its second part.

On the other hand, applying Fubini's Theorem in the same way you did but this time to an integral for $V_n$ gives $$V_n = V_{n - 1} \int_{-1}^1 (1 - x^2)^{(n - 1) / 2} dx,$$ so (after rewriting the integrals using symmetry) we're comparing $$\operatorname{vol}(P_n) = 2 V_{n - 1} \int_0^s (1 - x^2)^{(n - 1) / 2} dx \qquad \textrm{and} \qquad \operatorname{vol}(B_n - P_n) = 2 V_{n - 1} \int_s^1 (1 - x^2)^{(n - 1) / 2} dx ,$$ or just as well, the integrals $$\int_0^s (1 - x^2)^{(n - 1) / 2} dx \qquad \textrm{and} \qquad \int_s^1 (1 - x^2)^{(n - 1) / 2} dx .$$

Since $(1 - x^2)^{(n - 1) / 2} \leq 1$, the first integral satisfies $$\int_0^s (1 - x^2)^{(n - 1) / 2} \leq s .$$ On the other hand, we have $$\int_0^1 (1 - x^2)^{(n - 1) / 2} dx \geq \int_0^1 \left(1 - (n - 1) x^2\right) dx = \frac{5 - n}{4},$$ so the second integral is $$\int_s^1 (1 - x^2)^{(n - 1) / 2} dx > \frac{5 - n}{4} - s,$$ and hence:

$$\textbf{For small $n$ we have } \operatorname{vol}(B_n - P_n) > \operatorname{vol}(P_n) \textbf{.}$$

On the other hand, the second integral satisfies $$\int_s^1 (1 - x^2)^{(n - 1) / 2} dx \leq (1 - s) (1 - s^2)^{(n - 1) / 2} \leq (1 - s^2)^{(n - 1) / 2},$$ whereas for large $n$ (explicitly, $n > -2 \log 2 / \log(1 - s)$), a naive comparison for the first integral gives $$\int_0^s (1 - x^2)^{(n - 1) / 2} dx \geq \frac{1}{2} \sqrt{1 - 4^{- 1 / n}} .$$ Expanding the r.h.s. in a series at $\infty$ gives $\frac{1}{2} \sqrt{1 - 4^{- 1 / n}} = \sqrt{\frac{\log 2}{2}} n^{-1 / 2} + O(n^{-3 / 2})$. In particular, $\int_0^s (1 - x^2)^{(n - 1) / 2} dx$ decays much more slowly in $n$ than $(1 - s^2)^{(n - 1) / 2}$, so:

$$\textbf{For large $n$ we have } \operatorname{vol}(B_n - P_n) < \operatorname{vol}(P_n) \textbf{.}$$