I hadn't noticed the question -----

Question: Is that Amax-Vmax = 2

(we couldn't even write this if dimension weren't dimensionless!)

observation original? Did you use Area = d Volume/dr?

(below) until now. No, I didn't use the Area = ds Volume/dr formula. Just solved separately for Area and Volume, knowing the rough values of the maxima as 2.5 and 0.5: ----- In[1]:= A[x_] := (2 Pi^(x/2) / Gamma[x/2]) / 2^(x-1) In[2]:= FindRoot[A'[x], {x,2.5}, WorkingPrecision -> 105] Out[2]= {x -> 2.476582582306085295207615768858823240301645515180497569319595172\

37241273101141501186216648716028914196677}

In[3]:= V[x_] := (Pi^(x/2) / Gamma[x/2 + 1]) / 2^x In[4]:= FindRoot[V'[x], {x,0.5}, WorkingPrecision -> 105] Out[4]= {x -> 0.476582582306085295207615768858823240301645515180497569319595172\

372412731011415011862166487160289141966772}

----- --Dan

On Dec 26, 2014, at 12:52 PM, Bill Gosper <billgosper@gmail.com> wrote:

Mma solution: In[576]:= D[Log[\[Pi]^(d/2)/2^d/(d/2)!], d] // FullSimplify

Out[576]= 1/2 (EulerGamma - HarmonicNumber[d/2] + Log[\[Pi]/4])

In[577]:= Solve[% == 0]

During evaluation of In[577]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>

Out[577]= {{d -> 2 Root[{-EulerGamma + HarmonicNumber[#1] - Log[\[Pi]/4] &, 0.238291291153042647603807884429}]}}

Contrary to appearances, this is an exact, manipulable quantity:

In[578]:= N[%[[1]], 69]

Out[578]= {d -> 0.476582582306085295207615768858823240301645515180497569319595172372413}

Solving manually instead:

In[579]:= 2*InverseFunction[HarmonicNumber][EulerGamma + Log[\[Pi]/4]]

Out[579]= 2 Root[{-EulerGamma + HarmonicNumber[#1] - Log[\[Pi]/4] &, 0.238291291153042647603807884429}] --rwg (Let's see how many days this delivery takes.)

rwg>This exposes a bug in the terminology "unit ball", which really ought

to mean "unit diameter ball". There is no local content maximum when

the ball is inscribed in a unit cube.<rwg

Is that right? Using diameter = 1 instead of radius = 1,

I'm getting for A(d) / 2^(d-1)

a maximum at d = 2.4765825823060852952076157688588232403016455151805+,

and for V(d) / 2^d

a maximum at d = 0.4765825823060852952076157688588232403016455151805+ .

--Dan

Yikes, I never considered 0<d<1, whereat the "sphere" exceeds the "bounding cube"! --rwg

Question: Is that Amax-Vmax = 2

(we couldn't even write this if dimension weren't dimensionless!)

observation original? Did you use Area = d Volume/dr? --rwg

DanA> Pretending that spheres, balls, and Euclidean spaces can have real dimensions:

* let d_Amax := the real dimension d where the formula

A(d) = 2 pi^(d/2) / Gamma(d/2)

for the (d-1)-dimensional content of the unit (d-1)-sphere in R^d takes its maximum.

-and-

* let d_Vmax := the real dimension d where the formula

V(d) = pi^(d/2) / Gamma(d/2 + 1)

for the d-dimensional content of the unit d-ball in R^d takes its maximum

Then d_Amax = 7.256946404860576780132838388690769236619+ and d_Vmax = 5.256946404860576780132838388690769236619+. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun