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












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Let $B_n$ be the unit ball in $R^n$. We declare $$P_n = left{ xin B_n textrm{ such that }|x_1| < frac{1}{1000}right} .$$
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?










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  • 1




    $begingroup$
    It is doubtful that this integral has a "nice" closed form in terms of $n$.
    $endgroup$
    – Frpzzd
    Jan 5 at 21:04










  • $begingroup$
    Maybe there is an other way to solve it?
    $endgroup$
    – Gabi G
    Jan 5 at 21:04










  • $begingroup$
    Then which volume is bigger: $v(P_n)$ or $v(B_n - P_n)$ ?
    $endgroup$
    – Gabi G
    Jan 5 at 21:08












  • $begingroup$
    @GabiG for whatever reason I misread your question. My apologies. Let me try again....
    $endgroup$
    – Mike
    Jan 5 at 21:40












  • $begingroup$
    So, what I thought about is substituting $x_1 = sin(x_1)$, then it will lead to a reduction formula for the integral I wrote. But I still need some help determining which volume is bgger
    $endgroup$
    – Gabi G
    Jan 5 at 23:20


















1












$begingroup$



Let $B_n$ be the unit ball in $R^n$. We declare $$P_n = left{ xin B_n textrm{ such that }|x_1| < frac{1}{1000}right} .$$
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?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It is doubtful that this integral has a "nice" closed form in terms of $n$.
    $endgroup$
    – Frpzzd
    Jan 5 at 21:04










  • $begingroup$
    Maybe there is an other way to solve it?
    $endgroup$
    – Gabi G
    Jan 5 at 21:04










  • $begingroup$
    Then which volume is bigger: $v(P_n)$ or $v(B_n - P_n)$ ?
    $endgroup$
    – Gabi G
    Jan 5 at 21:08












  • $begingroup$
    @GabiG for whatever reason I misread your question. My apologies. Let me try again....
    $endgroup$
    – Mike
    Jan 5 at 21:40












  • $begingroup$
    So, what I thought about is substituting $x_1 = sin(x_1)$, then it will lead to a reduction formula for the integral I wrote. But I still need some help determining which volume is bgger
    $endgroup$
    – Gabi G
    Jan 5 at 23:20
















1












1








1





$begingroup$



Let $B_n$ be the unit ball in $R^n$. We declare $$P_n = left{ xin B_n textrm{ such that }|x_1| < frac{1}{1000}right} .$$
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?










share|cite|improve this question











$endgroup$





Let $B_n$ be the unit ball in $R^n$. We declare $$P_n = left{ xin B_n textrm{ such that }|x_1| < frac{1}{1000}right} .$$
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?







real-analysis calculus integration multivariable-calculus






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share|cite|improve this question













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edited Jan 6 at 4:33









Travis

62.5k767148




62.5k767148










asked Jan 5 at 21:02









Gabi GGabi G

408110




408110








  • 1




    $begingroup$
    It is doubtful that this integral has a "nice" closed form in terms of $n$.
    $endgroup$
    – Frpzzd
    Jan 5 at 21:04










  • $begingroup$
    Maybe there is an other way to solve it?
    $endgroup$
    – Gabi G
    Jan 5 at 21:04










  • $begingroup$
    Then which volume is bigger: $v(P_n)$ or $v(B_n - P_n)$ ?
    $endgroup$
    – Gabi G
    Jan 5 at 21:08












  • $begingroup$
    @GabiG for whatever reason I misread your question. My apologies. Let me try again....
    $endgroup$
    – Mike
    Jan 5 at 21:40












  • $begingroup$
    So, what I thought about is substituting $x_1 = sin(x_1)$, then it will lead to a reduction formula for the integral I wrote. But I still need some help determining which volume is bgger
    $endgroup$
    – Gabi G
    Jan 5 at 23:20
















  • 1




    $begingroup$
    It is doubtful that this integral has a "nice" closed form in terms of $n$.
    $endgroup$
    – Frpzzd
    Jan 5 at 21:04










  • $begingroup$
    Maybe there is an other way to solve it?
    $endgroup$
    – Gabi G
    Jan 5 at 21:04










  • $begingroup$
    Then which volume is bigger: $v(P_n)$ or $v(B_n - P_n)$ ?
    $endgroup$
    – Gabi G
    Jan 5 at 21:08












  • $begingroup$
    @GabiG for whatever reason I misread your question. My apologies. Let me try again....
    $endgroup$
    – Mike
    Jan 5 at 21:40












  • $begingroup$
    So, what I thought about is substituting $x_1 = sin(x_1)$, then it will lead to a reduction formula for the integral I wrote. But I still need some help determining which volume is bgger
    $endgroup$
    – Gabi G
    Jan 5 at 23:20










1




1




$begingroup$
It is doubtful that this integral has a "nice" closed form in terms of $n$.
$endgroup$
– Frpzzd
Jan 5 at 21:04




$begingroup$
It is doubtful that this integral has a "nice" closed form in terms of $n$.
$endgroup$
– Frpzzd
Jan 5 at 21:04












$begingroup$
Maybe there is an other way to solve it?
$endgroup$
– Gabi G
Jan 5 at 21:04




$begingroup$
Maybe there is an other way to solve it?
$endgroup$
– Gabi G
Jan 5 at 21:04












$begingroup$
Then which volume is bigger: $v(P_n)$ or $v(B_n - P_n)$ ?
$endgroup$
– Gabi G
Jan 5 at 21:08






$begingroup$
Then which volume is bigger: $v(P_n)$ or $v(B_n - P_n)$ ?
$endgroup$
– Gabi G
Jan 5 at 21:08














$begingroup$
@GabiG for whatever reason I misread your question. My apologies. Let me try again....
$endgroup$
– Mike
Jan 5 at 21:40






$begingroup$
@GabiG for whatever reason I misread your question. My apologies. Let me try again....
$endgroup$
– Mike
Jan 5 at 21:40














$begingroup$
So, what I thought about is substituting $x_1 = sin(x_1)$, then it will lead to a reduction formula for the integral I wrote. But I still need some help determining which volume is bgger
$endgroup$
– Gabi G
Jan 5 at 23:20






$begingroup$
So, what I thought about is substituting $x_1 = sin(x_1)$, then it will lead to a reduction formula for the integral I wrote. But I still need some help determining which volume is bgger
$endgroup$
– Gabi G
Jan 5 at 23:20












2 Answers
2






active

oldest

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0












$begingroup$

One can write an explicit formula for the integral in terms of $n$ using a hypergeometric function:
$$operatorname{vol}(P_n) = 2 s cdot {}_2F_1left(frac{1}{2}, frac{1}{2}(-n + 1);frac{3}{2}; s^2right) V_{n - 1}, qquad s := frac{1}{1000}.$$
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 $$int_0^1 (1 - x^2)^{(n - 1) / 2} dx geq int_0^1 left(1 - (n - 1) x^2right) dx = frac{5 - n}{4},$$ 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:



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






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  • $begingroup$
    I don't think this bound on $n$ is right though as $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$, while $B_n = int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$. Even for $n approx 3000$, Vol$(B_{n-1}(1-a^2))$ is close to 1 all the way up to $a approx frac{1}{sqrt{2n}} approx .01 >> .001$ so there is no way $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$ can be at least half of $int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$
    $endgroup$
    – Mike
    Jan 6 at 17:08












  • $begingroup$
    Meanwhile there was a solution already given above that answers the question and is much shorter and more elegant than a direct calculation--which still doesn't answer the question of the volume. Why not just go with that solution instead.
    $endgroup$
    – Mike
    Jan 6 at 17:09












  • $begingroup$
    In general, for general positive large $K$ letting $P^K_{xi} = {x in B_n; |x_i| le frac{1}{K} }$, this set has half the volume iff $n geq Y$ for some $Y in theta(K^2)$ and not $theta(K)$. Intuitively $B_{n-1}(sqrt{1-a^2})dx =(sqrt{1-a^2})^{n-1}$ has to be about $frac{1}{2}$ by the time $a$ is $frac{1}{K}$, and this will only happen if $n$ is that large.
    $endgroup$
    – Mike
    Jan 6 at 17:30












  • $begingroup$
    Typo in my last comment: *Intuitively Vol$(B_{n-1}(sqrt{1-a^2}) )$ has to be half Vol$(B_{n-1})$ by the time $a$ is $frac{1}{K}$; as Vol$(B_{n-1}(sqrt{1-a^2}) )$ $= (sqrt{1-a^2})^{n-1}$ Vol$(B_n)$ this will happen only if $n$ is as large as $theta(K^2)$.
    $endgroup$
    – Mike
    Jan 6 at 17:37












  • $begingroup$
    The solution doesn't claim that the bound $n > -2 log 2 / log(1 - s))$ is sufficient to guarantee that $operatorname{vol}(P_n) > operatorname{vol}(P_n)$ (it's not). Rather, it claims that bound is sufficient to guarantee the inequality that comes immediately afterward, and that latter inequality is use to establish the asymptotics of $operatorname{vol}(P_n)$. Consulting a CAS shows that the smallest $n$ where $operatorname{vol}(P_n)$ first exceeds $operatorname{vol}(B_n - P_n)$ is $approx 4.55 cdot 10^5$.
    $endgroup$
    – Travis
    Jan 6 at 21:25





















0












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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 $Q_{ni} = B_nsetminus P_{ni} = {x in B_n; |x_i| > frac{1}{1000} }$. 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$(Q_{ni}) le 1000000times$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.





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



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



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






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    2 Answers
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    2 Answers
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    active

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    active

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    0












    $begingroup$

    One can write an explicit formula for the integral in terms of $n$ using a hypergeometric function:
    $$operatorname{vol}(P_n) = 2 s cdot {}_2F_1left(frac{1}{2}, frac{1}{2}(-n + 1);frac{3}{2}; s^2right) V_{n - 1}, qquad s := frac{1}{1000}.$$
    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 $$int_0^1 (1 - x^2)^{(n - 1) / 2} dx geq int_0^1 left(1 - (n - 1) x^2right) dx = frac{5 - n}{4},$$ 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:



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






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I don't think this bound on $n$ is right though as $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$, while $B_n = int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$. Even for $n approx 3000$, Vol$(B_{n-1}(1-a^2))$ is close to 1 all the way up to $a approx frac{1}{sqrt{2n}} approx .01 >> .001$ so there is no way $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$ can be at least half of $int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$
      $endgroup$
      – Mike
      Jan 6 at 17:08












    • $begingroup$
      Meanwhile there was a solution already given above that answers the question and is much shorter and more elegant than a direct calculation--which still doesn't answer the question of the volume. Why not just go with that solution instead.
      $endgroup$
      – Mike
      Jan 6 at 17:09












    • $begingroup$
      In general, for general positive large $K$ letting $P^K_{xi} = {x in B_n; |x_i| le frac{1}{K} }$, this set has half the volume iff $n geq Y$ for some $Y in theta(K^2)$ and not $theta(K)$. Intuitively $B_{n-1}(sqrt{1-a^2})dx =(sqrt{1-a^2})^{n-1}$ has to be about $frac{1}{2}$ by the time $a$ is $frac{1}{K}$, and this will only happen if $n$ is that large.
      $endgroup$
      – Mike
      Jan 6 at 17:30












    • $begingroup$
      Typo in my last comment: *Intuitively Vol$(B_{n-1}(sqrt{1-a^2}) )$ has to be half Vol$(B_{n-1})$ by the time $a$ is $frac{1}{K}$; as Vol$(B_{n-1}(sqrt{1-a^2}) )$ $= (sqrt{1-a^2})^{n-1}$ Vol$(B_n)$ this will happen only if $n$ is as large as $theta(K^2)$.
      $endgroup$
      – Mike
      Jan 6 at 17:37












    • $begingroup$
      The solution doesn't claim that the bound $n > -2 log 2 / log(1 - s))$ is sufficient to guarantee that $operatorname{vol}(P_n) > operatorname{vol}(P_n)$ (it's not). Rather, it claims that bound is sufficient to guarantee the inequality that comes immediately afterward, and that latter inequality is use to establish the asymptotics of $operatorname{vol}(P_n)$. Consulting a CAS shows that the smallest $n$ where $operatorname{vol}(P_n)$ first exceeds $operatorname{vol}(B_n - P_n)$ is $approx 4.55 cdot 10^5$.
      $endgroup$
      – Travis
      Jan 6 at 21:25


















    0












    $begingroup$

    One can write an explicit formula for the integral in terms of $n$ using a hypergeometric function:
    $$operatorname{vol}(P_n) = 2 s cdot {}_2F_1left(frac{1}{2}, frac{1}{2}(-n + 1);frac{3}{2}; s^2right) V_{n - 1}, qquad s := frac{1}{1000}.$$
    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 $$int_0^1 (1 - x^2)^{(n - 1) / 2} dx geq int_0^1 left(1 - (n - 1) x^2right) dx = frac{5 - n}{4},$$ 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:



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






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I don't think this bound on $n$ is right though as $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$, while $B_n = int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$. Even for $n approx 3000$, Vol$(B_{n-1}(1-a^2))$ is close to 1 all the way up to $a approx frac{1}{sqrt{2n}} approx .01 >> .001$ so there is no way $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$ can be at least half of $int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$
      $endgroup$
      – Mike
      Jan 6 at 17:08












    • $begingroup$
      Meanwhile there was a solution already given above that answers the question and is much shorter and more elegant than a direct calculation--which still doesn't answer the question of the volume. Why not just go with that solution instead.
      $endgroup$
      – Mike
      Jan 6 at 17:09












    • $begingroup$
      In general, for general positive large $K$ letting $P^K_{xi} = {x in B_n; |x_i| le frac{1}{K} }$, this set has half the volume iff $n geq Y$ for some $Y in theta(K^2)$ and not $theta(K)$. Intuitively $B_{n-1}(sqrt{1-a^2})dx =(sqrt{1-a^2})^{n-1}$ has to be about $frac{1}{2}$ by the time $a$ is $frac{1}{K}$, and this will only happen if $n$ is that large.
      $endgroup$
      – Mike
      Jan 6 at 17:30












    • $begingroup$
      Typo in my last comment: *Intuitively Vol$(B_{n-1}(sqrt{1-a^2}) )$ has to be half Vol$(B_{n-1})$ by the time $a$ is $frac{1}{K}$; as Vol$(B_{n-1}(sqrt{1-a^2}) )$ $= (sqrt{1-a^2})^{n-1}$ Vol$(B_n)$ this will happen only if $n$ is as large as $theta(K^2)$.
      $endgroup$
      – Mike
      Jan 6 at 17:37












    • $begingroup$
      The solution doesn't claim that the bound $n > -2 log 2 / log(1 - s))$ is sufficient to guarantee that $operatorname{vol}(P_n) > operatorname{vol}(P_n)$ (it's not). Rather, it claims that bound is sufficient to guarantee the inequality that comes immediately afterward, and that latter inequality is use to establish the asymptotics of $operatorname{vol}(P_n)$. Consulting a CAS shows that the smallest $n$ where $operatorname{vol}(P_n)$ first exceeds $operatorname{vol}(B_n - P_n)$ is $approx 4.55 cdot 10^5$.
      $endgroup$
      – Travis
      Jan 6 at 21:25
















    0












    0








    0





    $begingroup$

    One can write an explicit formula for the integral in terms of $n$ using a hypergeometric function:
    $$operatorname{vol}(P_n) = 2 s cdot {}_2F_1left(frac{1}{2}, frac{1}{2}(-n + 1);frac{3}{2}; s^2right) V_{n - 1}, qquad s := frac{1}{1000}.$$
    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 $$int_0^1 (1 - x^2)^{(n - 1) / 2} dx geq int_0^1 left(1 - (n - 1) x^2right) dx = frac{5 - n}{4},$$ 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:



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






    share|cite|improve this answer









    $endgroup$



    One can write an explicit formula for the integral in terms of $n$ using a hypergeometric function:
    $$operatorname{vol}(P_n) = 2 s cdot {}_2F_1left(frac{1}{2}, frac{1}{2}(-n + 1);frac{3}{2}; s^2right) V_{n - 1}, qquad s := frac{1}{1000}.$$
    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 $$int_0^1 (1 - x^2)^{(n - 1) / 2} dx geq int_0^1 left(1 - (n - 1) x^2right) dx = frac{5 - n}{4},$$ 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:



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







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 6 at 4:28









    TravisTravis

    62.5k767148




    62.5k767148












    • $begingroup$
      I don't think this bound on $n$ is right though as $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$, while $B_n = int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$. Even for $n approx 3000$, Vol$(B_{n-1}(1-a^2))$ is close to 1 all the way up to $a approx frac{1}{sqrt{2n}} approx .01 >> .001$ so there is no way $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$ can be at least half of $int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$
      $endgroup$
      – Mike
      Jan 6 at 17:08












    • $begingroup$
      Meanwhile there was a solution already given above that answers the question and is much shorter and more elegant than a direct calculation--which still doesn't answer the question of the volume. Why not just go with that solution instead.
      $endgroup$
      – Mike
      Jan 6 at 17:09












    • $begingroup$
      In general, for general positive large $K$ letting $P^K_{xi} = {x in B_n; |x_i| le frac{1}{K} }$, this set has half the volume iff $n geq Y$ for some $Y in theta(K^2)$ and not $theta(K)$. Intuitively $B_{n-1}(sqrt{1-a^2})dx =(sqrt{1-a^2})^{n-1}$ has to be about $frac{1}{2}$ by the time $a$ is $frac{1}{K}$, and this will only happen if $n$ is that large.
      $endgroup$
      – Mike
      Jan 6 at 17:30












    • $begingroup$
      Typo in my last comment: *Intuitively Vol$(B_{n-1}(sqrt{1-a^2}) )$ has to be half Vol$(B_{n-1})$ by the time $a$ is $frac{1}{K}$; as Vol$(B_{n-1}(sqrt{1-a^2}) )$ $= (sqrt{1-a^2})^{n-1}$ Vol$(B_n)$ this will happen only if $n$ is as large as $theta(K^2)$.
      $endgroup$
      – Mike
      Jan 6 at 17:37












    • $begingroup$
      The solution doesn't claim that the bound $n > -2 log 2 / log(1 - s))$ is sufficient to guarantee that $operatorname{vol}(P_n) > operatorname{vol}(P_n)$ (it's not). Rather, it claims that bound is sufficient to guarantee the inequality that comes immediately afterward, and that latter inequality is use to establish the asymptotics of $operatorname{vol}(P_n)$. Consulting a CAS shows that the smallest $n$ where $operatorname{vol}(P_n)$ first exceeds $operatorname{vol}(B_n - P_n)$ is $approx 4.55 cdot 10^5$.
      $endgroup$
      – Travis
      Jan 6 at 21:25




















    • $begingroup$
      I don't think this bound on $n$ is right though as $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$, while $B_n = int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$. Even for $n approx 3000$, Vol$(B_{n-1}(1-a^2))$ is close to 1 all the way up to $a approx frac{1}{sqrt{2n}} approx .01 >> .001$ so there is no way $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$ can be at least half of $int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$
      $endgroup$
      – Mike
      Jan 6 at 17:08












    • $begingroup$
      Meanwhile there was a solution already given above that answers the question and is much shorter and more elegant than a direct calculation--which still doesn't answer the question of the volume. Why not just go with that solution instead.
      $endgroup$
      – Mike
      Jan 6 at 17:09












    • $begingroup$
      In general, for general positive large $K$ letting $P^K_{xi} = {x in B_n; |x_i| le frac{1}{K} }$, this set has half the volume iff $n geq Y$ for some $Y in theta(K^2)$ and not $theta(K)$. Intuitively $B_{n-1}(sqrt{1-a^2})dx =(sqrt{1-a^2})^{n-1}$ has to be about $frac{1}{2}$ by the time $a$ is $frac{1}{K}$, and this will only happen if $n$ is that large.
      $endgroup$
      – Mike
      Jan 6 at 17:30












    • $begingroup$
      Typo in my last comment: *Intuitively Vol$(B_{n-1}(sqrt{1-a^2}) )$ has to be half Vol$(B_{n-1})$ by the time $a$ is $frac{1}{K}$; as Vol$(B_{n-1}(sqrt{1-a^2}) )$ $= (sqrt{1-a^2})^{n-1}$ Vol$(B_n)$ this will happen only if $n$ is as large as $theta(K^2)$.
      $endgroup$
      – Mike
      Jan 6 at 17:37












    • $begingroup$
      The solution doesn't claim that the bound $n > -2 log 2 / log(1 - s))$ is sufficient to guarantee that $operatorname{vol}(P_n) > operatorname{vol}(P_n)$ (it's not). Rather, it claims that bound is sufficient to guarantee the inequality that comes immediately afterward, and that latter inequality is use to establish the asymptotics of $operatorname{vol}(P_n)$. Consulting a CAS shows that the smallest $n$ where $operatorname{vol}(P_n)$ first exceeds $operatorname{vol}(B_n - P_n)$ is $approx 4.55 cdot 10^5$.
      $endgroup$
      – Travis
      Jan 6 at 21:25


















    $begingroup$
    I don't think this bound on $n$ is right though as $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$, while $B_n = int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$. Even for $n approx 3000$, Vol$(B_{n-1}(1-a^2))$ is close to 1 all the way up to $a approx frac{1}{sqrt{2n}} approx .01 >> .001$ so there is no way $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$ can be at least half of $int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$
    $endgroup$
    – Mike
    Jan 6 at 17:08






    $begingroup$
    I don't think this bound on $n$ is right though as $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$, while $B_n = int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$. Even for $n approx 3000$, Vol$(B_{n-1}(1-a^2))$ is close to 1 all the way up to $a approx frac{1}{sqrt{2n}} approx .01 >> .001$ so there is no way $P_n = int_{-.001}^{.001}$Vol$B_{n-1}(sqrt{1-x^2})dx$ can be at least half of $int_{-1}^{1}$Vol$B_{n-1}(sqrt{1-x^2})dx$
    $endgroup$
    – Mike
    Jan 6 at 17:08














    $begingroup$
    Meanwhile there was a solution already given above that answers the question and is much shorter and more elegant than a direct calculation--which still doesn't answer the question of the volume. Why not just go with that solution instead.
    $endgroup$
    – Mike
    Jan 6 at 17:09






    $begingroup$
    Meanwhile there was a solution already given above that answers the question and is much shorter and more elegant than a direct calculation--which still doesn't answer the question of the volume. Why not just go with that solution instead.
    $endgroup$
    – Mike
    Jan 6 at 17:09














    $begingroup$
    In general, for general positive large $K$ letting $P^K_{xi} = {x in B_n; |x_i| le frac{1}{K} }$, this set has half the volume iff $n geq Y$ for some $Y in theta(K^2)$ and not $theta(K)$. Intuitively $B_{n-1}(sqrt{1-a^2})dx =(sqrt{1-a^2})^{n-1}$ has to be about $frac{1}{2}$ by the time $a$ is $frac{1}{K}$, and this will only happen if $n$ is that large.
    $endgroup$
    – Mike
    Jan 6 at 17:30






    $begingroup$
    In general, for general positive large $K$ letting $P^K_{xi} = {x in B_n; |x_i| le frac{1}{K} }$, this set has half the volume iff $n geq Y$ for some $Y in theta(K^2)$ and not $theta(K)$. Intuitively $B_{n-1}(sqrt{1-a^2})dx =(sqrt{1-a^2})^{n-1}$ has to be about $frac{1}{2}$ by the time $a$ is $frac{1}{K}$, and this will only happen if $n$ is that large.
    $endgroup$
    – Mike
    Jan 6 at 17:30














    $begingroup$
    Typo in my last comment: *Intuitively Vol$(B_{n-1}(sqrt{1-a^2}) )$ has to be half Vol$(B_{n-1})$ by the time $a$ is $frac{1}{K}$; as Vol$(B_{n-1}(sqrt{1-a^2}) )$ $= (sqrt{1-a^2})^{n-1}$ Vol$(B_n)$ this will happen only if $n$ is as large as $theta(K^2)$.
    $endgroup$
    – Mike
    Jan 6 at 17:37






    $begingroup$
    Typo in my last comment: *Intuitively Vol$(B_{n-1}(sqrt{1-a^2}) )$ has to be half Vol$(B_{n-1})$ by the time $a$ is $frac{1}{K}$; as Vol$(B_{n-1}(sqrt{1-a^2}) )$ $= (sqrt{1-a^2})^{n-1}$ Vol$(B_n)$ this will happen only if $n$ is as large as $theta(K^2)$.
    $endgroup$
    – Mike
    Jan 6 at 17:37














    $begingroup$
    The solution doesn't claim that the bound $n > -2 log 2 / log(1 - s))$ is sufficient to guarantee that $operatorname{vol}(P_n) > operatorname{vol}(P_n)$ (it's not). Rather, it claims that bound is sufficient to guarantee the inequality that comes immediately afterward, and that latter inequality is use to establish the asymptotics of $operatorname{vol}(P_n)$. Consulting a CAS shows that the smallest $n$ where $operatorname{vol}(P_n)$ first exceeds $operatorname{vol}(B_n - P_n)$ is $approx 4.55 cdot 10^5$.
    $endgroup$
    – Travis
    Jan 6 at 21:25






    $begingroup$
    The solution doesn't claim that the bound $n > -2 log 2 / log(1 - s))$ is sufficient to guarantee that $operatorname{vol}(P_n) > operatorname{vol}(P_n)$ (it's not). Rather, it claims that bound is sufficient to guarantee the inequality that comes immediately afterward, and that latter inequality is use to establish the asymptotics of $operatorname{vol}(P_n)$. Consulting a CAS shows that the smallest $n$ where $operatorname{vol}(P_n)$ first exceeds $operatorname{vol}(B_n - P_n)$ is $approx 4.55 cdot 10^5$.
    $endgroup$
    – Travis
    Jan 6 at 21:25













    0












    $begingroup$

    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 $Q_{ni} = B_nsetminus P_{ni} = {x in B_n; |x_i| > frac{1}{1000} }$. 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$(Q_{ni}) le 1000000times$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.





    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



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



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






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      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 $Q_{ni} = B_nsetminus P_{ni} = {x in B_n; |x_i| > frac{1}{1000} }$. 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$(Q_{ni}) le 1000000times$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.





      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



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



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






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $Q_{ni} = B_nsetminus P_{ni} = {x in B_n; |x_i| > frac{1}{1000} }$. 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$(Q_{ni}) le 1000000times$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.





        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



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



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






        share|cite|improve this answer











        $endgroup$



        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 $Q_{ni} = B_nsetminus P_{ni} = {x in B_n; |x_i| > frac{1}{1000} }$. 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$(Q_{ni}) le 1000000times$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.





        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



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



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







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 6 at 19:42

























        answered Jan 6 at 3:02









        MikeMike

        4,284412




        4,284412






























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