Subject: Real Subsets with the 0-1 Intersection Property
From: lones@lones.mit.edu (Lones A Smith)
Date: 15 Nov 1996 06:55:29 GMT
Let S(x,c) be a closed Borel subset of [0,1], for any x in [0,1] & real c.
Suppose S has the "0-1 intersection property": For any c1 and c2, and for
all x1 <> x2, S(x1,c1) and S(x2,c2) have either zero or one point in common.
CLAIM: {x in [0,1]|union of S(x,c) over all real c has measure >0} is countable
I am quite convinced of this claim, but am spinning my wheels in countable
and continuum sets.
Any help proving it would be greatly appreciated!
Lones
.-. .-. .-. .-. .-. .-.
/ L \ O / N \ E / S \ / S \ M / I \ T / H \
/ `-' `-' `-' `-' `-' `
Lones Smith, Economics Department, M.I.T., E52-252C, Cambridge MA 02139
(617)-253-0914 (work) 253-6915 (fax) lones@lones.mit.edu
Subject: Re: most of you know this few of you care? the surface
From: ale2@psu.edu (ale2)
Date: 15 Nov 1996 08:40:25 GMT
In article <1996Nov14.221903.29402@lafn.org>
ba137@lafn.org (Brian Hutchings) writes:
>
> In a previous article, ale2@psu.edu (ale2) says:
>
> so, *how* is it rounded?... assuming (having no computer loaded
> with *that kind* of software) that it has no discontinuities, at all, and
> no flat areas; eh?
The post started because i thought enough readers (1?, 0, -1?) of
sci.math did not know of the simple surface in question. I saw the
surface in:
"Modern Differential Geometry of Curves and Surfaces" by Alfred Gray.
And i like to plug books i feel are above the ordinary. How one goes
about proving the surface is smooth and in fact is a "rounded cube" is
not my interest.
So... i'm not sure what your getting at...this is not rocket science.
>
> >The equation represents the surface of the rounded cube, but its real
>
> >he: scribbles the equation x^4+y^4+z^4=1 on a drink napkin...
> --
> You *don't* have to be a rocket scientist. (College Career Counselor
> to me, again )
>
> There is no dimension without time. --RBF (Synergetics, 527.01)
Subject: Re: 0.999999999999999999999...=1???
From: "Bob Bridges"
Date: 15 Nov 1996 08:05:38 GMT
> Dieter Dijkstra wrote:
> > Does 0.99... equal 1? I should think so, but can anyone confirm
this?
Gary wrote in article
<328769E6.30A4@pop.pitt.edu>...
> ...If you subtract them and get a nonzero result, then they are
> not equal. (In other words, if you subtract them and get zero, then
> they must be equal.)
If this were alt.folklore.math I'd offer my opininion, which if anyone
cares would have been that .99999... is INFINITELY CLOSE to 1 (but not,
technically, equal to it). But since this is sci.math I'm not entitled
to an opinion, so I'll be careful not to let on that I accidentally had
one.
--
...Bob Bridges, RHBridg@ibm.net
/* When I die, I'd like to go peacefully.
In my sleep.
Like my grandfather.
Not screaming.
Like the passengers in his car. */
Subject: Re: Need help on an stats/prob question
From: David Kastrup
Date: 15 Nov 1996 11:48:58 +0100
Sk8orDie@msn.com (Bob Lee) writes:
> Hi I have a question that I am having trouble with. I'd appreciate any he=
lp.
>
> Suppose the average family income of an area is $10,000.
>
> a) Find and upper bound for the percentage of families with incomes
> over $50,000.
Easy. Assume those families earn $50000 exactly (the minimum).
Assume everybody else earns nothing (the minimum). Should be easy to
get the percentage from that.
> b) Find a better upper bound if it is known that the standard
> deviation of incomes is $8,000.
Easy as well: we know the income to be $10000 on average. And we know
($8000)^2 =3D E (I^2) - ($10000)^2
-> E(I^2) =3D 164000000$^2
> I assume that some kind of distribution must be assumed.
Yes. In this case we again assume all families earning over $50000
earn exactly that. Now how do we compensate for those? If someone is
below $10000 by X, he will contribute -X^2 to the variance, and -X to
the mean. Now we assume that all below $10000 have identical income
again (it remains to be shown that this is the *best* configuration),
then we just need to solve
p.($50000) + (1-p)i =3D $10000
p.(($40000)^2) + (1-p)i^2 =3D 164000000$^2.
This gives us p again, as a nicer bound (I hope).
--
David Kastrup Phone: +49-234-700-5570
Email: dak@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut f=FCr Neuroinformatik, Universit=E4tsstr. 150, 44780 Bochum, Germa=
ny
Subject: VietMath War: Programmer/Analyst needed in war ????
From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium)
Date: 14 Nov 1996 21:28:01 GMT
In article <328A10A3.2610@math.okstate.edu>
David Ullrich writes:
> > Math is indeed the key to the universe. Just don't let academia fool you into
> > thinking they have the only copy.
>
> Certainly not! Just check sci.math - you'll find all sorts of
> extra-academic mathematicians. There's Archimedes Plutonium, for example.
> He's a super-genius; he seems to be the only person in the world who
> realizes that Wiles' proof of Fermat's Last Theorem is wrong. The reason
> is that Wiles doesn't realize that the naturals = Adics. That is, contrary
> to what you and I learned in school, 2 is not a natural number. Natural
> numbers have infinitely many digits extending to the _left_ of the
> decimal point, so for example ...22222.0 is a natural number, while 2
> is not. Perhaps someday there will be another super-genius capable of
> appreciating this.
Thanks for summarizing what I think. I find it very frustrating to
have to correct people who misunderstand what I am doing. Typically
they take a path that I must be 100% wrong in everything that I do.
Then they ask how I could have anything to say about FLT. Then I show
them p-adic counterexamples. Immediately they throw up their arms and
say, aha, here ArchPu is your error. The P-adics are not the counting
numbers.
How do you tell someone who is dense at the starting gate that my
whole push in mathematics is that the Finite Integers are a mirage and
that the p-adics are the Naturals.
I bet there are still professors of physics who believe that
Newtonian Mechanics is correct physics even if you take large slow
moving objects. Their minds simply cannot grasp the fact that Newtonian
mechanics is a fake. They can easily see that gravity is not the flight
of angels pushing Earth off of its linear path and in towards the Sun.
They can see that having "angels" in their theory is messy and that
such a theory is fake. But it is much more difficult for people, even
some students and professors of physics to realize that Newtonian
Mechanics is a fake theory, just as "angels pushing" is a fake theory.
A fake theory may give close approximations but no matter how close it
is a fake theory.
How do I tell the world public that their counting numbers , these
numbers of 0, 1, 2, 3, 4, and so on were as imprecise, as fake as
angels pushing on earth to give gravity. Most people do not see any
difference from those numbers and these p-adic numbers of ....,
...0004, ...0003, ...000002, ...0001, ....000000. Until they see a
p-adic number like ....22222.
I am changing mathematics , math that has been done with counting
numbers for 3 millenium. I have a terrible war on my hands. And I think
that I could make math proofs left and right until I am blue in the
face and for a hundred years and no math professor will heed the call.
But when one physics report announces that p-adics are essential in
physics and that the counting numbers just do not work in the physics
experiment. Well, my day in the sun has come.
Thanks David for your accurate summary of what I am doing. And I
would have thanked you even if you did not have super-genius in your
writing. I find it the case that when people want to attack ad
hominem,it is almost impossible for them to understand what I am about
and it is impossible for them to give an objective unbiased summary of
my work. But when people are open minded and do not take me personally
but try to objectively understand what I am saying, then they can give
an accurate summary of my work.
I needed to repost David's description because I can reuse that
paragraph for the many others who will come flying-on-the-Net,
attacking me , shooting from the hips before asking any questions and
utterly misunderstanding and misrepresenting the idea of Naturals =
p-adics = Infinite Integers
Subject: Lottery probability
From: don.mcdonald@welcom.gen.nz (Don Mcdonald)
Date: Sat, 16 Nov 1996 00:21:48 GMT
-> From: bandy@aplcomm.jhuapl.edu (Mike Bandy F2C )
-> Newsgroups: sci.math (sometimes rec.gambling.lottery.)
-> Subject: Lottery probability
-> Date: 13 Nov 96 17:31:18 GMT
->
->
-> In Maryland there is the lottery game Keno.
-> It is played as followed:
->
-> There are 80 numbers
-> The game randomly picks 20 from the 80
-> You pick a number of 'spots' - ranging from 1 to 10.
-> You get a payback on the number of spots that
-> match the selected numbers.
->
-> Please give the formula for calculating
-> the probability of, say, choosing
-> 3 spots (you select your birthday, for example)
-> and you match 2 of the 3.
->
-> --
-> Mike Bandy
-> bandy@aplcomm.jhuapl.edu
-> Johns Hopkins University
-> Applied Physics Laboratory
REM > Donmcd.lotto.keno.txt20dd80, 5/11/94, 0300.
Bas V Progm DonMcD.Lotto.Keno.Ken20dd80, 5/11/94, 0300.
/
Keno Lotto, NEW Zealand 31/10/1994.
Draw 20 nos. from 1 to 80.
Select max 10 Group. 0 to 10 Match.
/
group= 1 .. (denom= 80)
match= 1 odds 1: 4 pay$= 2 return 50 cents.
.. odds 1: 4, bank favour factor= 2 contin.
/
group= 2 .. (denom= 3160)
match= 2 odds 1: 16.63 pay$= 10 return 60.1 cents.
.. odds 1: 16.6315789, bank favour factor= 1.66 contin.
/
group= 3 .. (denom= 82160)
match= 2 odds 1: 7.21 pay$= 2 return 27.8 cents.
match= 3 odds 1: 72.07 pay$= 20 return 55.5 cents.
.. odds 1: 6.55183413, bank favour factor= 1.8 contin.
/
group= 4 .. (denom= 1581580)
match= 3 odds 1: 23.12 pay$= 3 return 13 cents.
match= 4 odds 1: 326.44 pay$= 150 return 58.9 cents.
.. odds 1: 21.5930098, bank favour factor= 1.7 contin.
/
group= 5 .. (denom= 24040016)
match= 3 odds 1: 11.91 pay$= 2 return 16.8 cents.
match= 4 odds 1: 82.7 pay$= 20 return 41 cents.
match= 5 odds 1: 1550.57 pay$= 300 return 60.3 cents.
.. odds 1: 10.3442232, bank favour factor= 1.66 contin.
/
group= 6 .. (denom= 300500200)
match= 3 odds 1: 7.7 pay$= 2 return 26 cents.
match= 4 odds 1: 35.04 pay$= 3 return 34.5 cents.
match= 5 odds 1: 323.04 pay$= 40 return 46.9 cents.
match= 6 odds 1: 7752.84 pay$= 1000 return 59.8 cents.
.. odds 1: 6.18880476, bank favour factor= 1.67 contin.
/
group= 7 .. (denom= 3.1767164E9)
match= 4 odds 1: 19.16 pay$= 2 return 10.4 cents.
match= 5 odds 1: 115.76 pay$= 4 return 13.9 cents.
match= 6 odds 1: 1365.98 pay$= 500 return 50.5 cents.
match= 7 odds 1: 40979.31 pay$= 4000 return 60.3 cents.
.. odds 1: 16.2374695, bank favour factor= 1.66 contin.
/
group= 8 .. (denom= 2.89875372E10)
match= 4 odds 1: 12.27 pay$= 2 return 16.3 cents.
match= 5 odds 1: 54.64 pay$= 4 return 23.6 cents.
match= 6 odds 1: 422.53 pay$= 100 return 47.3 cents.
match= 7 odds 1: 6232.27 pay$= 500 return 55.3 cents.
match= 8 odds 1: 230114.61 pay$= 10000 return 59.7 cents.
.. odds 1: 9.77156031, bank favour factor= 1.68 contin.
/
group= 9 .. (denom= 2.31900297E11)
match= 4 odds 1: 8.76 pay$= 2 return 22.8 cents.
match= 5 odds 1: 30.67 pay$= 3 return 32.6 cents.
match= 6 odds 1: 174.84 pay$= 10 return 38.3 cents.
match= 7 odds 1: 1690.11 pay$= 200 return 50.2 cents.
match= 8 odds 1: 30681.95 pay$= 2000 return 56.7 cents.
match= 9 odds 1: 1380687.65 pay$= 50000 return 60.3 cents.
.. odds 1: 6.53376058, bank favour factor= 1.66 contin.
/
group= 10 .. (denom= 1.64649211E12)
match= 0 odds 1: 21.84 pay$= 3 return 13.7 cents.
match= 5 odds 1: 19.44 pay$= 2 return 24 cents.
match= 6 odds 1: 87.11 pay$= 5 return 29.8 cents.
match= 7 odds 1: 620.68 pay$= 50 return 37.8 cents.
match= 8 odds 1: 7384.47 pay$= 500 return 44.6 cents.
match= 9 odds 1: 163381.37 pay$= 20000 return 56.8 cents.
match= 10 odds 1: 8911711.18 pay$= 250000 return 59.6 cents.
.. odds 1: 9.05382312, bank favour factor= 1.68 contin.
Prog Ken20dd80 e n d. Stopped *
--------------
Program BBC BasV "donmcd.lotto.keno.20/80", 02.11.96, 21h
Calculate odds of winning lower tier Lottery prizes.
By don.mcdonald@welcom.gen.nz.
Hit -- Kk.eno or Ll.otto:
(Standard NZ Lotteries Commission shortcuts,)
OR any other key, general / other lotteries.
6/49, 5/45, etc.
Keno 20/80
Enter combo or how many nos you chose,
(Keno ticket group) - 0=quit ?10
Enter minimum correct = ?0
Balls = 80
lucky / unlucky = 20 / 60
combo (group) = 10
minimum correct = 0
t =(Balls, combo) = 1.64649211E12
c (lucky, c)x (unlucky,combo-c);
nr; s; odds = 1:t/nr | cumul 1:t/s
20 (20,20)x(60,-10) 0 0
...
number sum
11 (20,11)x(60,-1) 0 0
10 (20,10)x(60,0) 184756 184756 8911711.18 8911711.18
9 (20,9)x(60,1) 10077600 10262356 163381.372 160439.972
8 (20,8)x(60,2) 222966900 233229256 7384.46877 7059.54364
7 (20,7)x(60,3) 2.6527344E9 2.88596366E9 620.677332 570.517271
6 (20,6)x(60,4) 1.89007326E10 2.17866963E10 87.112608 75.5732806
5 (20,5)x(60,5) 8.4675282E10 1.06461978E11 19.4447786 15.4655412
4 (20,4)x(60,6) 2.42559402E11 3.4902138E11 6.78799543 4.71745344
3 (20,3)x(60,7) 4.40275889E11 7.89297269E11 3.73968267 2.08602281
2 (20,2)x(60,8) 4.86137961E11 1.27543523E12 3.38688242 1.2909257
1 (20,1)x(60,9) 2.95662853E11 1.57109808E12 5.56881628 1.04798811
0 (20,0)x(60,10) 7.53940276E10 1.64649211E12 21.8384952 1
RUN. or Escape
*
don.mcdonald@welcom.gen.nz (Wellington, New Zealand.)
Subject: Re: BETROTHED numbers.
From: fredh@ix.netcom.com (Fred W. Helenius)
Date: Fri, 15 Nov 1996 10:46:29 GMT
jlame@math.ohio-state.edu (John Lame) wrote:
>In article <56bkcq$foe@cantuc.canterbury.ac.nz>,
>Bill Taylor wrote:
>[snip]
>>My query is, are there any "affianced sequences", of length k > 2.
>>
>>i.e. n2 = f(n1), n3 = f(n2) ... n_k = f(n_(k-1)) & n1 = f(n_k);
>>
>>where f(n) = sigma(n)-n-1, (as above).
>>
>The only affianced sequences involving any natural numbers <= 10000 have
>length k=2. These are:
>{48,75} {140,195} {1050,1925} {1575,1648}
>{2024, 2295} {5775,6128} {8892, 16587} {9504, 20735}
>All other numbers <= 10000 eventually collapse to zero or lead
>to one of the above 2-cycles.
I find that there are 95 such pairs with smaller term < 2^27, but
still no longer cycles. Email me if you'd like to see all 95.
A short cycle with an odd number of terms would seem to be unlikely,
since one of the terms, say n, would have to have sigma(n) be odd,
which occurs only if n is a square or twice a square.
--
Fred W. Helenius
Subject: Re: Vietmath War: Wiles FLT lecture at Cambridge
From: JC
Date: Fri, 15 Nov 1996 10:08:33 +0000
ghidrah wrote:
>
> > The mathematics literature even up to this date, is horribly lacking
> > in any elementary discussions of p-adics, what they are, how to
> > multiply and divide with them. There strange characteristics. Why this
> > lack? The answer is that noone but me ever thought they were anything
> > more than a extension. I am the first to realize that they are the
> > Naturals themselves, and that the Finite Integers were a field of
> > ghosts, or angels that fit on the end of a needle.
> >
> Jean Pierre Serre has a book called "A Course in Arithmetic" where he
> speaks of many of the elementary properties of p-adic integers. This book,
> I think, is something of a standard in the subject.
AP has also written several, er, 'papers' on the p-adics, where he makes
little or no sense. These papers, I think, are somewhat non-standard in
the subject.
Subject: Re: Name of a fraction
From: fleck@dirk (Pat Fleckenstein (x58885))
Date: Thu, 14 Nov 1996 18:10:59 GMT
Bob Silverman wrote:
>>Bob Silverman wrote:
>
>>> No other name is needed. I aim this concept at elementary teachers who, for some
>>> reason, think it necessary to divide fractions into 'proper' and 'improper'.
>>> This distinction is meaningless.
>
>Please explain why dividing the rationals into two sets, one with absolute value
>less than or equal to 1, and one with absolute value greater than one should be
>meaningful. Tell us where the meaning lies and why it is important. And please
>explain what a "numerical purist" is. A concept in mathematics is either useful
>or it isn't. Where is the use here?
I agree that elementary teachers making the distinction between
'proper' and 'improper' fractions is a bogus thing. (Additionally,
I think the words 'proper' and 'improper' show a bias that isn't
really present, in much the same way that 'real' and 'imaginary'
show an artificial bias.)
But, numbers inside the unit sphere and numbers outside the unit
sphere.... I'd absolutely appreciate some decent words to quickly
make that distinction. Given x and y reals, x**y has very different
behaviors is x is in [0,1] than if x is in (1,infinity). And, it
has very different behaviors if y is on (0,1] than if y is in
(1,infinity).
We have 'positive' and 'negative' to refer to numbers on either
side of zero. Certainly, '1' is as important to the group as
'0' is.
alter,
pat
Subject: please help: non polynomial function
From: s84213@vcldec7.polito.it (Stoppa Igor)
Date: 15 Nov 1996 11:53:55 GMT
Hi !
We are two students in electronical engineering(third year) at
Politecnico di Torino, Italy and we are in need of help for a problem in
numerical analysis.
How can you rappresent non polynomial function (not exprimable using a
sum of powers of x) on a PC ? (Expecially sin(x), cos(x), and exp(x) )
We tought about Mc Laurin's serie whit several enanchements, such as
reducing every angle to the range { -pi/4 , pi/4 }, but our teacher told
us that there is a best way, which gives a smaller error when abs(x) is
relatively far from 0 (near pi/4).
Can anyone help us? We are in a great hurry since we have to terminate
this job in a few days and we would like to add the results of theese
better algorithms, comparing them with those from the simpler algorithms
we used.
Thanks a lot and please forgive us for the bad english.
Guido & Igor
P.S. : If you can help, send the answer not only to the newsgroup,
but also through e-mail at guidov@net4u.it because otherwise we
couldn't read it until monday.
Subject: Re: 0.999999999999999999999...=1???
From: David Kastrup
Date: 15 Nov 1996 14:24:15 +0100
"Bob Bridges" writes:
> > Dieter Dijkstra wrote:
> > > Does 0.99... equal 1? I should think so, but can anyone confirm
> this?
>
> Gary wrote in article
> <328769E6.30A4@pop.pitt.edu>...
> > ...If you subtract them and get a nonzero result, then they are
> > not equal. (In other words, if you subtract them and get zero, then
> > they must be equal.)
>
> If this were alt.folklore.math I'd offer my opininion, which if anyone
> cares would have been that .99999... is INFINITELY CLOSE to 1 (but not,
> technically, equal to it).
Well, once we agree on that we mean real numbers, *then* we have some
laws. One consequence is that I have a number smaller or equal to any
imaginable positive number, it must be zero (all real numbers can be
defined by converging intervals). Any interval converging to zero
measure in the reals is defining exactly one number. That's a basic
property of the definition of reals. Once we accept that, we have to
consider 1 as equal to 0.9999..., as the difference is smaller than
any conceivable positive number.
When we are talking about something different, like infinite number
strings or so (p-adics are somewhat like that), then both can denote
different entities. When we are talking about identities of notation,
than both are different as well.
When we are talking about real numbers, those represented by the above
expressions are equal.
> But since this is sci.math I'm not entitled
> to an opinion, so I'll be careful not to let on that I accidentally had
> one.
Good thing you didn't. I won't as well.
--
David Kastrup Phone: +49-234-700-5570
Email: dak@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut f=FCr Neuroinformatik, Universit=E4tsstr. 150, 44780 Bochum, Germa=
ny
Subject: Re: Laws of Flight
From: carl@aoainc.com (Carl Witthoft)
Date: 15 Nov 1996 15:48:38 GMT
In article <56dkhr$rcc@niamh.indigo.ie> gerryq@indigo.ie (Gerry Quinn) writes:
>
>(ii) Suppose a large dome was held at high pressure so as to create the
>maximum safe atmospheric density ( 10-20 kg/m3 I suppose ). Would
>human-powered flight be possible inside such a dome?
>
It occurs to me to compare this to a "dome" full of water or some
other fluid of a density comparable to that of the human body.
Then you could "fly."
Before people start screaming, just consider this: penguins fly under
water, in the sense that they flap their wings in a manner similar
to birds in the air. Conversely, extremely small insects "fly" thru
air with a wing motion reminiscent of animals paddling or treading
water.
So I guess a more interesting question is: at what fluid density can
a human, with or without surface area enhancements, produce dynamic
lift?
--
Carl Witthoft @ Adaptive Optics Associates
carl@aoainc.com 54 CambridgePark Drive, Cambridge,MA 02140 617-864-0201
"Eight ever; nine never."
Subject: Re: The anchored string revisited, but now in 3D ?
From: ale2@psu.edu (ale2)
Date: 15 Nov 1996 13:37:31 GMT
In article <56c3ou$sng@r02n01.cac.psu.edu>
ale2@psu.edu (ale2) writes:
> If i plug Iain Mains book twenty more times he said he might get me a
> free copy (not).
(Make that nineteen more times)
>
> In Iain Main's book "Vibration and Waves in Physics" he considers the
> physics of the anchored string. Now what gives me goosebumps about this
> system is that it has the same frequency wavelength relationship as a
> massive quanta in one dimension!
You also forgot to mention what about the above also gives you
goosebumps. The example which Iain Main uses is of a string under
tension, anchored at two points, and with a bunch of springs attached
to the string which give it an additional sideways restoring force. All
the springs in his example all lie in a single plane (that plane also
contains the string). Now for the goosebumps. If the string lies in say
the z axis, the string will have two fundamentally different modes of
vibration:
1) the string moves in the plane containing the springs ("massive"
modes),
2) the string moves in the plane perpendicular to the plane containing
the springs ("massless" modes)!
3) almost forgot the longitudinal modes of vibration (also "massless")
>
> (children and mental midgets are easily impressed)
>
> A while back i asked the readers of sci.physics to come up with a
> physical system which has the same frequency wavelength relationship as
> a massive quanta in 3 dimensions. Well either no one cared about my
> question or they did not see it (i've been kilefiled?) ?
>
> I think i have something now that works ?
Yes, i think the below might work, but your three-dimensional system of
inductors and capacitors no longer has massless modes like Iain Mains
example with the anchored string. Please try a little harder!
>
> Consider an infinite 3 dimensional system of masses and springs such
> that each mass has six springs attached to it in a symmetric fashion,
> and all the masses are hooked together by the springs and form a cubic
> array. This is the system one considers as a simple model of vibrations
> in solids?
>
> Now transform the above system:
>
> 1) replace each spring with an inductor and capacitor in series,
>
> 2) each mass is replaced with one end of a capacitor and the other end
> of the capacitor is grounded.
>
> Now perturb the system at some small region (apply an oscillating
> voltage at a point where one of the masses once was) for a long time
> with less than some critical frequency and energy is not absorbed after
> steady state is reached, but increase the frequency above the critical
> value and energy propagates out of the small region?
>
Subject: Re: Need help on an stats/prob question
From: Gus Gassmann
Date: Fri, 15 Nov 1996 10:33:11 -0800
David Kastrup wrote:
>
> Sk8orDie@msn.com (Bob Lee) writes:
>
> > Hi I have a question that I am having trouble with. I'd appreciate any help.
> >
> > Suppose the average family income of an area is $10,000.
> >
> > a) Find and upper bound for the percentage of families with incomes
> > over $50,000.
>
> Easy. Assume those families earn $50000 exactly (the minimum).
> Assume everybody else earns nothing (the minimum). Should be easy to
> get the percentage from that.
>
> > b) Find a better upper bound if it is known that the standard
> > deviation of incomes is $8,000.
>
> Easy as well: we know the income to be $10000 on average. And we know
> ($8000)^2 = E (I^2) - ($10000)^2
> -> E(I^2) = 164000000$^2
>
> > I assume that some kind of distribution must be assumed.
>
> Yes. In this case we again assume all families earning over $50000
> earn exactly that. Now how do we compensate for those? If someone is
> below $10000 by X, he will contribute -X^2 to the variance, and -X to
> the mean. Now we assume that all below $10000 have identical income
> again (it remains to be shown that this is the *best* configuration),
> then we just need to solve
>
> p.($50000) + (1-p)i = $10000
> p.(($40000)^2) + (1-p)i^2 = 164000000$^2.
>
> This gives us p again, as a nicer bound (I hope).
This is needlessly complicated. If you know the standard deviation, you
can use Chebychev's theorem.
-------------------------------------------------------
gus gassmann (Horand.Gassmann@dal.ca)
School of Business Administration, Dalhousie University
Halifax, Nova Scotia, Canada , B3H 1Z5
ph. (902) 494-1844
fax (902) 494-1107
http://ttg.sba.dal.ca/sba/profs/hgassmann/hgassman.html
Subject: Re: Polaner Hypothesis. Im not quite conviced but lets look again.
From: JC
Date: Fri, 15 Nov 1996 14:16:47 +0000
Michael Gary Kramer wrote:
>
> 1961 Elementary calculus. San Diego State College.
>
> My teacher trundles out this business and we talk about it.
>
> Please excuse my crude constructions as I am as mathematically naive as I
> sound. This has stuck in my mind for many years though. Who the hell
> Polaner is beats me.
>
> Polaner Hypothesis:
>
> Before you sits a very large! piece of paper . to the left and right of a
> decimal point extend without limit an infinite sequence of random
> integers selected from (0 to 9).
>
> beneath this sequence is another random expansion extending to the left
> and right of a decimal point.
>
> random is whatever fulfills best definitions of it (im not sure here what
> to say) and its definition asserts among other things that every finite
> sequence of numbers must exist within these random expansions.
>
> remove the decimal points.
>
> Hypothesised: "IT IS POSSIBLE to translate one expansion in such a way
> that ALL integers of one are identical to the expansion beneath. ALL
> without limit."
>
> Polaner asserts that IF IT IS NOT POSSIBLE then there must exist a finite
> sequence of one expansion which will not correspond in the expressed
> manner with at least one sequence of the other expansion. As this
> circumstance violates the very definition of the construction the
> assertion that the two infinite expansions can be placed into 1 to 1
> correspondence in the stated manner is proven by reductio ad absurdum.
>
> This is fun stuff for freshmen, which 35 years later Im afraid I still
> am. But help me be a sophomore.
>
> Is it provable that such a translation is not possible?
>
> Does this ,as my teacher cheerfully and mystifyingly suggested, imply
> that all such infinite sequences of integers must be the same but out of
> joint with one another merely by translation?
>
> Sincerely
>
> Michael G. Kramer
>
I've never heard of this hypothesis, but it sounds extremely
dodgy. Since you don't define random it's difficult to
disprove such an assertion, but there are various reasons
to dismiss it. If all random sequences are identical
via a translation then there are only countably many
random sequences, a heavily counter-intuitive result.
If you are asserting that, if every finite sequence in one
expression corresponds to a finite sequence from the other,
then they are equal modulo a translation, then this assertion
is *definitely* false. We could do this by a stepwise
construction, with steps 1,2,3,4... At each step we fill
in a new finite patch of our sequences so that both
will contain all finite subsequences. We also place
unequal values at crucial points to rule out the possibility
of each translation map in turn.
JC
