MaplePrimes - answers and comments on Question, How to calculate the triple integral

24/5 ?

Axel Vogt — Wed, 17 Aug 2011 18:12:10 Z

see the attached file hirnyk_max_squar.mws

one way

pagan — Wed, 17 Aug 2011 22:34:44 Z

Easiest (to me) is to get rid of the `max` altogether. I don't know how to do that conceptual part in Maple. (I looked at the 3d case as a plot, and considered the symmetry...)

5! * 2^5 * Int(Int(Int(Int(Int(x^2,v=0..w),w=0..z),z=0..y),y=0..x),x=0..1):

CodeTools:-Usage( value(%) );
memory used=1.12MiB, alloc change=0.75MiB, cpu time=31.00ms, real time=31.00ms

                              160
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                               7

I don't think you should need Maple for this step:

simplify( max(x^2,y^2,z^2,w^2) ) assuming w>0, w<z, z<y, y<x, x<1;

                                2
                               x

The hard bit is figuring out the multiplicative factor, on account of the symmetries.

sketch

Axel Vogt — Thu, 18 Aug 2011 12:32:03 Z

Here is a sketch, how I try to understand pagan's solution:

If we agree to reduce to the (open) unit hypercube and that max( ()^2 ) = max(..)^2
then I think one can argue as follows:

The set, where any of the coordinates may coincide, is low dimensional (analytic of
dim <= n-1) and thus does not contribute to the integral.

Denote the complement by U.

Let be U0 := points with 0 < x1 < x2 ... < xn < 1. The symmetric group Sn operates by
permutation on the coordinates. Any of those maps U0 to U. Any point in U has just one
ordering of its coordinates, thus is an image of a point in U0 by a unique permutation.

Thus U is the disjoined union of images of U0 under the symmetric group.

And the integral is the sum over those images.

Now we have to convince ourselfs, that this works reasonable on the integrals.

The permutations do not change volume. And on the function max(...) they fit, since
max is invariant under permutations. Therefore the integrals in the sum have all the
same value, that over U0.

There are as many summands as elements of Sn, hence n!-times that over U0.

The integral over U0 is just as already written down by pagan for n=5, 0< v<w<z<y<x < 1
as it is seen by 'resolving the range' on paper.

That is a bit of handwaving, but probably can be made sound.

Exact value & higher dimensions

Markiyan Hirnyk — Wed, 17 Aug 2011 18:33:57 Z

This is a good work, many thanks from me to you. As far as I understand it, you have calculated the integral numerically. However I expect the exact solution with Maple.
The command int(max(x^2, y^2, z^2), [x = -1 .. 1, y = -1 .. 1, z = -1 .. 1]) is spinning. Trying to caIculate this with Maple 13, I didn't succeed to solve some inequalities. Also how about higher dimensions?

symbolical

Axel Vogt — Wed, 17 Aug 2011 18:59:52 Z

@Markiyan Hirnyk

Checking the results is numerically, the rest is symbolically.

My thoughts:

By symmetry reduce to the unit cubes, there are 2^dimension of those.
On them one has max(x^2,t^2) = max(x,t)^2 and max(a,b,c) = max(a, max(b,c)).

Now take assumptions, write as 'piecewise' and integrate iteratively.

Have not done that for higher dimensions, since the numerical check is painful,
but the 'recipe' should work.

Vote up

Markiyan Hirnyk — Wed, 17 Aug 2011 19:13:23 Z

@Axel Vogt Indeed, you are right. You are a real expert in integrals! In the general case the expected answer is 2^n*n/(n+2). Maybe, I 'll try to do that in the case n=4.

PS. I have added your answer to my favorites so my vote up has vanished. In any case you got points.