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Han de Bruijn

Posted: Tue Apr 29, 2008 9:14 am

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Continuous Geometric Means

Definition of the Geometric Mean at:

http://en.wikipedia.org/wiki/Geometric_mean

The relationship with the arithmetic mean of logarithms at that page
can be used to derive an expression in case the medium is continuous:

GM = exp( integral(t=0..L) ln( f(t) ) dt / L )

We couldn't find many references on the Internet (Google) concerning
the _continuous_ version of the geometric mean. Curious about it ..

Example.

For a straight line segment, we substitute:

f(t) = (A.t - B)^2 + (C.t - D)^2 for 0 <= t <= 1 ;

Where A = x_2 - x_1 ; B = x - x_1 ; C = y_2 - y_1 ; D = y - y_1 ;

Meaning that f(t) is the squared distance between a point of the line
segment and a point (x,y) in the plane. Evaluate this a little bit:

f(t) = (A^2 + C^2) t^2 - 2.(A.B + C.D).t + (C^2 + D^2)

And make dimensionless: f(t) = t^2 - 2.a + b^2 .

Where: a = (A.B + C.D)/(A^2 + C^2) and b^2 = (C^2 + D^2)/(A^2 + C^2)

The discriminant of the quadratic f(t) turns out to be: a^2 - b^2 =

D = - (A.D - B.C)^2 / (A^2 + C^2) ; it is negative or zero. The case
D = 0 should be handled as a special case, where: f(t) = (t - a)^2 .

So we have to calculate:

E(x,y) = exp( integral(t=0..1) ln( t^2 - 2.a + b^2 ) dt )

A computer algebra system (MAPLE) has been employed for this purpose:

http://groups.google.nl/group/sci.math/browse_frm/thread/6540aee9c6a068ff

Quote:

collect(collect(int(ln(t^2-2*a*t+b^2),t=0..1,continuous),arctan),ln);

And the special case:

Quote:

collect(collect(int(ln((t-a)^2),t=0..1,continuous),arctan),ln);

The result has been programmed in Delphi Pascal. Here are some details:

var
A,B,C,D,N,I,S,E,F : double;
begin
A := two.x - one.x; C := two.y - one.y;
B := dit.x - one.x; D := dit.y - one.y;
N := sqr(A) + sqr(C); { assume nonzero }
S := (sqr(B) + sqr(D))/N; { b^2 }
I := (A*B + C*D)/N; { a }
D := sqr(A*D-B*C)/N; { discriminant }
E := 0;
if (D > 0) then
begin
D := sqrt(D);
F := (1-I)*ln(abs(1+S-2*I)) + I*ln(S)
- 2*D*(arctan((I-1)/D) - arctan(I/D)) - 2;
E := exp(F/2);
end;
if (D = 0) and (I <> 1) and (I <> 0) then
begin
F := ln(sqr(1-I)) - 2 - 2*I*(ln(abs(1-I)) - ln(abs(I)));
E := exp(F/2);
end;
if (D = 0) and ((I = 0) or (I = 1)) then E := exp(-1);
integraal := E;

And when visualized it looks as follows:

http://hdebruijn.soo.dto.tudelft.nl/jaar2008/segment.jpg

Similar attempt for a circle has resulted in the integral published at:

http://groups.google.nl/group/sci.math/browse_frm/thread/b1634000e5babd75

Quote:

exp(int(ln((a+cos(t))^2+(b+sin(t))^2),t=0..2*Pi)/(2*Pi));

It is suspected (Conjecture !) that the outcome is simply: (a^2 + b^2) .

Anyone to confirm or deny ?

Han de Bruijn

Angus Rodgers

Posted: Tue Apr 29, 2008 9:59 am

Guest

On Tue, 29 Apr 2008 16:14:27 +0200, Han de Bruijn
<Han.deBruijn@DTO.TUDelft.NL> wrote:

Quote:

Similar attempt for a circle has resulted in the integral published at:

http://groups.google.nl/group/sci.math/browse_frm/thread/b1634000e5babd75

exp(int(ln((a+cos(t))^2+(b+sin(t))^2),t=0..2*Pi)/(2*Pi));

It is suspected (Conjecture !) that the outcome is simply: (a^2 + b^2) .

Anyone to confirm or deny ?

When a = b = 0, the value of the expression is:

exp(int(ln(1),t=0..2*pi)/(2*pi)

= exp(0) = 1

!= 0 = a^2 + b^2

--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril

Angus Rodgers

Posted: Tue Apr 29, 2008 10:02 am

Guest

On Tue, 29 Apr 2008 15:59:41 +0100, I wrote:

Quote:

exp(int(ln(1),t=0..2*pi)/(2*pi)

That should have another right-hand parenthesis at the end.

--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril

G. A. Edgar

Posted: Tue Apr 29, 2008 10:18 am

Guest

Quote:

Hardy, Littlewood, Polya, INEQUALITIES, Section 6.7

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

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