- RudiI am sure Gregoire will respond to this, but thought I would put my tuppence worth in and wish everyone a Merry Christmas at the same time.As I understand it, this mailing list was established to foster conversation between members. If you start judging the questions by what YOU know, we will all lose out.The concept that a function with an infinite base can have a finite value is, after all, counter-intuitive. Until, at least, you realise that (in practice) most of that is an infinitely long axis multiplied by an infinitely small height. The solution I was taught as a fledgling was that integrating e-to-the-x-squared can only be done by calculus sleight-of-hand (it's a mathematician's trick). As an engineer, I personally don't give a fetid dingos kidney, because it works!Keep up the questions, Digby, you keep us all awake, interested and on our toes. The questions out of left field are the ones which make us re-assess all those things we take for granted.Good wishes to all for the winter solstice. Where I am, the shortest day is around 5 hours long and dark with it!Isobelhttp://www.kriging.com/whereisshe.html
wrote:*Rudi Dutter <R.Dutter@...>*Dear members of the list,

Is it the goal of the list to discuss mathematical basic calculus? Everything

can be read e.g. in

Abramowitz and Stegun (eds.) (1964): Handbook of Mathematical Functions.

National Bureau of Standards.

Best regards,

Rudi Dutter

John.Butler@... wrote:

>

> Hello list,

>

> Here is the 'trick' I came across in diffusion studies, which might not not

> fully satisfy the pure mathematicians in the list, but I can see the

> picture, so I like it. This goes as follows :-

>

> Start with the standard statement that the integral from minus to plus

> infinity of e raised to power -1/2((x-m)^2)/(s^2) dx = ssqrt(2p)

> First step substitute y = (x-m)/s into the above relationship.

>

> Now we want to show that A= (is equivalent to) the transformed integral

> from minus infinity to plus infinity of e raised to the power -1/2 y^2 dy =

> sqrt(2p)

> If we consider A^2, then we can write this as the product of two integrals

> of the above form, one in x and the other in y. This is then combined as

> the double integral (with the same limits) of e raised to the power -1

> /2(x^2 +y^2) dx dy. This double integral is the volume under the bell

> shaped surface e^(-1/2(x^2 + y^2)).

>

> Now x^2 + y^2 = the square of the distance of point x, y from the origin,

> denote this by r^2. The area of a ring, distance r from the origin, and

> width dr, = 2prdr.

>

> The volume of a cylindrical shell with this annulus as its base, and with

> height, e^(-1/2(r^2)), so that it just touches the bell shaped surface is

> (e^(-1/2(r^2))) 2prdr

>

> Therefore A^2 = 2p times the integral from zero to infinity of (r times

> e^(-1/2(r^2)) dr . On integration this gives A^2 = 2p[-e^(-1/2(r^2))]

> evaluated from zero to infinity. Substituting in for the limits - gives

> when inifinity (or try a large number) is substituted for r, the

> expression yields zero, when zero is substituted for r, then the expression

> yields minus 1, which is subtracted from the upper limit, so it becomes

> plus 1, giving A^2 = 2p. We wanted to find A, so we take the square root

> of this giving, A = sqrt(2p).

>

> Sorry I didn't have the symbolic representation, but if you write it out it

> you should be able to folllow the above line of reasoning.

>

> In some versions of the above 'trick' you might see that they define the

> 'square' base area with co-ordinates x and y, but then replace x & y more

> formally with polar co-ordinates r and q giving x = r cos q and y = r sin

> q, They then replace the element of area (dx, dy) by (r dr dq), and the

> range of integration has to be altered from the square OABC (O = origin, C

> = distance along x axis, A = distance along y axis, B = opposite corner to

> O) to the quadrant OAC, which results in an error denoted by e. This is the

> space between the outer 'box' and the curve joining x (C) and y (A). In

> this case the limits of integration are :- outer limit from 0 to R, inner

> limit from 0 to p/2. The expression to be evaluated changes to

> (e^(- r^2))r dr dq, plus the error term e. The volume represented by e has

> a base area which is less than 1/2 R^2, and a maximum height of exp (-

> R^2). Thus

>

> e < 1/2 R^2 exp( - R^2), thus as R goes to infinity, e goes to 0. This

> might satisfy Digby?

>

> Thus A^2 = 1/4p - 1/4 p exp(- R^2) + e

>

> Therefore, as R goes to infinity, A^2 goes to 1/4p. In this latter case

> the algebra had been simplified at the start by removing the normalising

> factor, 2/(sqrt(p)), which is used to make erf ( inifinity ) = 1.

>

> I hope this helps.

>

> John Butler

>

> Gerald van den Boogaart on 20/12/2005 09:54:17

>

> Please respond to boogaart@...

>

> To: "Digby Millikan"

> cc: "AI Geostats mailing list"

>

> Subject: Re: [ai-geostats] Cumulative gaussian distribution

>

> Dear Digby Millikan,

>

> William Harper told us that the area under the Gauss density

> f(x)=exp(-x^2 /2)/(sqrt(2pi))

> is 1 and we all know that he is right.

>

> Anyway I'll try to answer your consern, how that could be true.

>

> Indeed to those who might smile on Digbys consern, I have to say that this

> is

> a nontrivial finding of higher mathematical calculus, which probably non of

> the non mathematicians in the mailing list could prove. On one hand you

> need

> Lebesgue integration theory or at least improper Riemann integrategrals to

> even define, what the area under an infinite curve might be. Second the

> actual value is to my knowledge most easy obtained as a result of the

> residual theorem from function theory. Or has anyone a simple idea where

> the

> pi comes in and don't forget that the integral function of the curve does

> not

> have a closed form. Its just that we learned that the area is one, and we

> are

> good belivers.

>

> > I was wondering how the area under a gaussian distribution curve is 1 if

> > the tails go off to infinite?, the area must be infinite unless some

> unless

> some

>

> Your concern might be reduced to the concern, how a sum of infinitly many

> summands might stay finite, since the area under the tails is the sum of

> the

> areas in the intervalls from i to i+1 for all integer number i. I will give

> a

> simple example of such sum of infinitly many summands adding to 1.

>

> The i-th summand should be a_i = 0.5^i.

>

> We start to add

> 0 +0.5 = 0.5

> 0.5 +0.25 = 0.75

> 0.75 +0.125 = 0.875

> ...

>

> In each line we added the half of that what was missing to 1

>

> 0 +0.5 = 0.5 ; (1-0 )/2=0.5

> 0.5 +0.25 = 0.75 ; (1-0.5 )/2=0.25

> 0.75 +0.125 = 0.875 ; (1-0.75)/2=0.125

> ...

>

> It is easy to accept that we never get a value bigger than 1 because we

> always

> only add the half of what was missing to 1. On the other hand the distance

> to

> 1 is cut down to a half in every step and thus finally the distance to 1

> drops under any e > 0.

>

> The abstraction that in this case the values of the sequence converges to 1

> and that than the value of the infinite sum is 1, is one of the great and

> early historic achievments of calculus long before derivatives and

> integration and is linked to the initial definition of real numbers (which

> are actually defined as equivalence classes of (Cauchy)-sequences of

> rational

> numbers).

>

> A historic joke on that problem of finite sequences is the saying about the

> greek hero Achill who was known as a fast runner and a turtel. It goes as

> follows: A turle is running (or better say slowly crawling) 10 meter in

> front

> of Achill. Can he catch the turtle. The historic argument was: No, because

> in

> the time Achill needs to run the 10meter (say 1sec, he was really fast) the

> turtle advanced a little say 1 Meter, than in the time Achill needs to run

> this meter the turle advances again, and so on for ever again, such that

> Achill never reaches the turtle, since we get an infinite sum of timeslices

> before he reaches the turtle. However all of us know that after 2 seconds

> Achill has passed the turtle by 8 Meter since he got 20 Meters and the

> turle

> only 2. So we all know that Achill catches the turle. Only our brains play

> a

> trick on us in making us believe that an infinite sum of times must be

> infinite hindering Achill to catch the turle by taking him infinite time to

> run after it.

>

> Merry Chrismas,

> Gerald v.d. Boogaart

>

> Am Dienstag, 20. Dezember 2005 00:03 schrieb Digby Millikan:

> > Dear Madam/Sir,

> >

> >

> >

> > I was wondering how the area under a gaussian distribution curve is 1 if

> > the tails

> >

> > go off to infinite?, the area must be infinite unless some unless some

> > approximation

> >

> > is made?

> >

> >

> >

> > Digby

>

> --

> -------------------------------------------------

> Prof. Dr. K. Gerald v.d. Boogaart

> Professor als Juniorprofessor fuer Statistik

> http://www.math-inf.uni-greifswald.de/statistik/

>

> office: Franz-Mehring-Str. 48, 1.Etage rechts

> e-mail: Gerald.Boogaart@...

> phone: 00+49 (0)3834/86-4621

> fax: 00+49 (0)89-1488-293932 (Faxmail)

> fax: 00+49 (0)3834/86-4615 (Institut)

>

> paper-mail:

> Ernst-Moritz-Arndt-Universitaet Greifswald

> Institut f?r Mathematik und Informatik

> Jahnstr. 15a

> 17487 Greifswald

> Germany

> --------------------------------------------------

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~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=

From: Prof. Dr. Rudolf Dutter

Dept. of Statistics and Probability Theory

Vienna University of Technology,

Wiedner Hauptstr. 8-10

A-1040 Vienna, Austria

Tel. +43 1 58801/10730

FAX +43 1 58801/10799

E-Mail: R.Dutter@...

Internet: http://www.statistik.tuwien.ac.at/public/dutt/

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