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A255135 Dimensions where the volume of an L^p unit ball is maximized.

%I A255135 #33 Nov 20 2022 18:31:10
%S A255135 1,5,16,41,102,242,558,1263,2817,6214,13583,29471,63548,136305,291019,
%T A255135 618849,1311314,2769847,5834119,12257072,25691785,53738815,112188059,
%U A255135 233796875,486435094,1010552580,2096469429,4343666482
%N A255135 Dimensions where the volume of an L^p unit ball is maximized.
%C A255135 The volume of an n-dimensional L^p-ball of radius R is vol(R, p, n) = ((2^n)*(gamma((1/p)+1))^n /gamma((n/p)+1)*R^n for all real p > 0. For the Hilbert space L^2 we recover the familiar Euclidean volume of a unit n-ball (Pi^(n/2))/gamma((n/2)+1) which attains a maximum value at dimension n=5. The present sequence describes the same phenomenon for positive integer values of p.
%C A255135 From _Robert L. Diersing_, May 26 2020: (Start)
%C A255135 We can find very tight bounds for a(n):
%C A255135 define c := 2*gamma(1 + 1/p), then
%C A255135   floor(p*(c^p - 1/2) - 1/2) <= a(n) <= ceiling(p*(c^p - 1/2)).
%C A255135 First we treat V(n, p, 1) as a real-valued smooth function w.r.t. n, then we can maximize it. Since log(x) is an increasing function, the critical points of log(V(n, p)) will be the same as the critical points for V(n, p).
%C A255135   (d/dn)log(V(n, p)) = log(c) - 1/(p(digamma((n+p)/p))).
%C A255135   (d^2/dn^2)log(V(n, p)) = -polygamma(1, (n+p)/p))/p^2 < 0 for all n > 0, p > 0.
%C A255135 This is because polygamma(1, x) = Sum_{k>=1} 1/(x + k)^2 > 0 for all x > 0.
%C A255135 Therefore (d/dx)V(n, p) is a decreasing function w.r.t. n, which means there is only one local extremum. We can bound that local extremum similar to the way that we find bounds for the case where p=2 and the radius varies: see Wikipedia link.
%C A255135 Let's call this extremum n_.
%C A255135   dV/dn = log(c) - 1/(p*(digamma((n_+p)/p))) = 0;
%C A255135   p*log(c) = digamma((n_ + p) / p);
%C A255135   p*(inverse_digamma(plog(c)) - 1) = n_.
%C A255135 We can see from [2] that
%C A255135   log(y - 1/2) < digammma(y) < log(y + e^-m - 1),
%C A255135 where m is the Euler-Mascheroni constant.
%C A255135 Thus
%C A255135   log((n_+ p)/p - 1/2) < p*log(c) < log((n_ + p)/p + e^-m - 1);
%C A255135   c^p - e^-m + 1 < (n_ + p)/p < c^p + 1/2;
%C A255135   p*(c^p - e^-m) < n_ < p*(c^p - 1/2).
%C A255135 We can now improve on these bounds; define:
%C A255135   n` = p*(c^p - 1/2).
%C A255135 We can prove
%C A255135   n` - n_ < 1/2;
%C A255135   p*(c^p - 1/2 - inverse_digamma(p*log(c))) < p/c^p < 1/2;
%C A255135   c^p + 1/2 - 1/c^p < inverse_digamma(log(c^p)), say x := c^p;
%C A255135   x + 1/2 - 1/x < inverse_digamma(log(x));
%C A255135   digamma(x + 1/2 - 1/x) < log(x)
%C A255135 (inverse_digamma is an increasing function).
%C A255135 We see from [1] that digamma(1/log(1 + 1/x)) < log(x), therefore we only need to show
%C A255135   digamma(x + 1/2 - 1/x) < digamma(1/log(1 + 1/x)),
%C A255135 and since inverse_digamma is an increasing function, it suffices to show
%C A255135   x + 1/2 - 1/x < 1/log(1 + 1/x),
%C A255135 which is true for x > 0, and can be seen from an inequality solver such as Wolfram Alpha: https://www.wolframalpha.com/input/?i=x+%2B+%C2%BD+-+1%2Fx+%3C+1%2Flog%281+%2B+1%2Fx%29.
%C A255135 thus
%C A255135   n` - 1/2 < n_ < n`.
%C A255135 Now we have to confirm that n_ is the global maximum by checking the boundaries.
%C A255135   V(p, 0) = c^p < n_;
%C A255135   Limit_{n->oo} V(p, n) = lim_{n->oo} (2*sqrt(2*Pi/p))^n / (sqrt(2*Pi*n/p) n^n) = 0 < n_.
%C A255135 Therefore n_ is the global maximum, and thus we can solve for a(n).
%C A255135   floor(n` - 1/2) <= a(n) <= ceiling(n`).
%C A255135 So to find a(n) we only need to iterate over at most three values. (End)
%H A255135 Necdet Batir, <a href="https://arxiv.org/abs/1705.06547">Inequalities for the Inverses of the Polygamma Functions</a>, arXiv:1705.06547 [math.CA], 2017.
%H A255135 N. Elezovic, C. Giordano and J. Pecaric, <a href="https://dx.doi.org/10.7153/mia-03-26">The best bounds in Gautschi's inequality</a>, Math. Inequal. Appl. 3 (2000), 239-252.
%H A255135 Wikipedia, <a href="http://en.wikipedia.org/wiki/Volume_of_an_n-ball#Balls_in_Lp_norms">Balls in L^p norms</a>
%F A255135 From _Robert L. Diersing_, May 26 2020: (Start)
%F A255135 Define c = 2*gamma(1 + 1/p), and n` = p*(c^p - 1/2)
%F A255135 for x in [floor(n` - 1/2), floor(n` + 1/2), floor(n` + 3/2)]
%F A255135 a(n) = the value x that gives the largest volume. (End)
%e A255135 a(2) = 5 because the sequence vol(R=1, p=2, n) increases monotonically up to n=5 and then decreases monotonically.
%e A255135 The approximate values for vol(1,p,a(p)) are 2, 5.263789015, 50.05958637, 5970.510613, p = 1..4. - _Wolfdieter Lang_, Mar 20 2015
%t A255135 f[p_] = Round[FindArgMax[((2^n)*(Gamma[1+(1/p)])^n)/Gamma[1+(n/p)], n]]
%K A255135 nonn
%O A255135 1,2
%A A255135 _Dimitris Cardaris_, Feb 14 2015
%E A255135 Edited by _Wolfdieter Lang_, Mar 20 2015
%E A255135 More terms from _Robert L. Diersing_, May 26 2020