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Message no. 1
From: shadowrn@*********.com (Hahns Shin)
Subject: Extrapolation (was Re: [OT] Earthdawn Questions)
Date: Sat Feb 2 14:40:01 2002
> > >When you include the open-ended rule, you can't get an exact
average. You can
> > >only extrapolate... :-)
> >
> > Sorry to disappoint you, but you can. Let's get a dice with d
sides. To get
> > the average of the results with this dice, open-ended style, you
just have
> > to mulitply its usual average (d+1)/2 by d/(d-1). This last factor
is the
> > sum of the (converging) numercial serie 1+1/d+1/d²+1/d^3....
>
> Isn't that extrapolating?
No, actually, it isn't. Because it is a converging sequence, when
taken to a limit of infinity it will sum to a finite number. Thus, you
CAN get an exact average (of course, sometimes this is like
calculating pi to the nth digit). It's sort of like the converging
sequence you get from Zeno's Tortoise and Achilles paradox (the
debunking of this paradox is a popular way to introduce middle school
students to the fact that infinite converging sequences can sum to a
finite number). Extrapolation, by definition, estimates a value that
is outside a given known range by using the values within the given
known range. An example would be predicting the world population 2
years from now (easy) or predicting the Dow-Jones industrial average
tomorrow (hard). But this is nitpicking... I think the word you want
is estimation, rather than extrapolation. However, there are many
infinite sequences that sum to a rational number or even integers,
thus not being an estimation at all. Of course, a sufficiently
powerful estimation can functionally act as the final result (e.g. pi
is 3.14159 for the purposes of most high school science classes), and
such is the way with calculating the averages for Open Ended Earthdawn
tests (phew!).

Link to proving and debunking Zeno's paradox:
http://www.mathpages.com/rr/s3-07/3-07.htm

Hahns Shin, MS II
Budding cybersurgeon
"Fairy tales do not tell children the dragons exist. Children already
know that dragons exist. Fairy tales tell children the dragons can be
killed."
-G. K. Chesterton
Message no. 2
From: shadowrn@*********.com (Achille Autran)
Subject: Extrapolation (was Re: [OT] Earthdawn Questions)
Date: Sun Feb 3 22:15:01 2002
>From: "Hahns Shin" <Hahns_Shin@*******.com>
>
> > > >When you include the open-ended rule, you can't get an exact
>average. You can
> > > >only extrapolate... :-)
> > >
> > > Sorry to disappoint you, but you can. Let's get a dice with d
>sides. To get
> > > the average of the results with this dice, open-ended style, you
>just have
> > > to mulitply its usual average (d+1)/2 by d/(d-1). This last factor
>is the
> > > sum of the (converging) numercial serie 1+1/d+1/d²+1/d^3....
> >
> > Isn't that extrapolating?
>No, actually, it isn't. Because it is a converging sequence, when
>taken to a limit of infinity it will sum to a finite number. Thus, you
>CAN get an exact average (of course, sometimes this is like
>calculating pi to the nth digit)...

<POWA SNIPPA>

Yeah, what he said. The key word here is "converging," and
Hahns' discourse was understated in my "the (/converging/) numerical serie".

/Estimating/ the sum of series, or the limit of sequences, is another kind
of fun. Sometimes we merely now that a sum or a limit exists (it converges)
and we can compute approximations, sometimes we are able to get an easy
formula for it. Here is a proof for the calculation of the sum at hand
(foreword: READER, YES, YOU, READER, YOU ARE NOT, REPEAT, NOT COMPELLED TO
READ THE FOLLOWING NOTES BROUGHT TO YOU BY THE SICKLY COMPLETION IMPULSE OF
THE ALTOGETHER SICK MATHEMATICIAN WHO WROTE THESE LINES. NOW, CONSIDER
YOURSELF WARNED. YOU ARE ABOUT TO READ MATHEMATICS. WE DECLINE ALL
RESPONSIBILITIES CONCERNING POSSIBLE DAMAGES, WETHER NEUROLOGICAL OR CAUSED
TO SURROUNDING MATERIALS BY EPILEPTIC SEIZURES. WE WON'T COVER PERSONNAL
THERAPY AND DIVORCE EXPENSES IF YOUR WIFE OR HUSBAND AND CHILDREN LEAVE YOU
IN CASE OF UNFORTUNATE SUBJECT BROUGHT UP DURING THE FAMILIAL EVENING MEAL.
DON'T EVEN THINK ABOUT PAYBACK. WE HAVE PROOVES THAT WOULD DRIVE INSANE THE
MOST HARDENED WOODCHUCK OR AIRBORNE CYPRINUS CARPIO. READ AT YOUR OWN RISK.)

Let S(n) be the partial sum:
S(n) = 1 + 1/d + 1/d^2 + ... + 1/d^n
Obviously, we have S(n+1) = S(n) + 1/d^(n+1)
But we also have:
S(n+1) - (1/d)*S(n) = [1+1/d+...+1/d^n+1/d^(n+1)] - [1/d+...+1/d^(n+1)] = 1
Using the first equation to replace S(n+1), we get:
(1-1/d)*S(n) + 1/d^(n+1) = 1
i.e.
S(n) = [1 - 1/d^(n+1)] / [1 - 1/d]
Since when n tends toward infinity, 1/d^(n+1) tends toward zero, S(n) as a
sequence has a limit and this limit is:
S = 1/(1 - 1/d) = d/(d-1)
QED
T'was easy, isn't it? *honest, broad smile*

BTW, Zeno's paradox series is exactly the same as our open-ended dice
serie, with d being the ratio between the speeds of Achilles and the turtle.

/Extrapolation/ is something totally different. E.g., when you use the
Bezier curve tool with a drawing program, fitting a curve to a handful of
control points and tangents, the computer is using extrapolation methods.
In fact, it extrapolates a whole curve out of a few points and derivatives.

>Hahns Shin, MS II
>Budding cybersurgeon with l337 Mathz Skillz! ;-)

Molloy

Further Reading

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