mm-76
===
This problem was given by my friend.I dont know how to
approach it.Any suggestions are welcomeI pick two numbers,
randomly, and tell you one of them. You are supposedto guess
whether this is the lower or higher one of the two numbers
Ipicked. Can you come up with a method of guessing that does
better thanpicking the response low or high randomly Thaanx
in Advance.Naren.
===
> This problem was given by my friend.>
I dont know how to approach it.> Any suggestions are welcome
I pick two numbers, randomly, and tell you one of them. You
are supposed> to guess whether this is the lower or higher
one of the two numbers I> picked. Can you come up with a
method of guessing that does better than> picking the
response low or high randomly Thaanx in Advance. Naren.If by
number you mean a positive integer and if you randomly tell
me oneof the two integers then I will always say the number
that you told me isthe lower number. Since if you tell me one
of the numbers is X I then knowthat there are X-1 integers
below X and in?itely many integers above X.
===
> This
problem was given by my friend.> I dont know how to approach
it.> Any suggestions are welcome I pick two numbers,
randomly, and tell you one of them. You are supposed> to
guess whether this is the lower or higher one of the two
numbers I> picked. Can you come up with a method of guessing
that does better than> picking the response low or high
randomly Thaanx in Advance.>sure, if the number is positive,
you say high,if the number is negative, you say low.then you
assume the distribution of random numbers is symmetric over
0.Herc
===
|What I noted is the oddity of believing that a
constant factor of a|polynomial, which f^2 is, could factor
off as a function.||It seems unlikely to me that a math
expert would make that mistake.You¶ve only guessed that the
way f^2 divides off is constant. Themistake a mathematician
wouldn¶t make is to just keep sayinghow odd it would be if
their guess were wrong, and pretend that that¶sa logical
argument. Accusing others of oddly believing otherwiseis just
posturing. It¶s not up to others to prove it¶s false; \
it¶s up
to youto prove that it¶s true.Here¶s a related \
example. The
two roots r1(m) and r2(m) of x^2+x+2m=0multiply together to
give 2m, which is divisible by 2. At m=0 one of themis
divisible by 2 and the other one is coprime to 2. But there¶s
nosimple pattern in their common factors with 2:m=1:
r1=(-1+sqrt(-7))/2, r2=(-1-sqrt(-7))/2. 2m=2=r1*r2. But that
meansthat r1 and r2 are themselves the algebraic integer
factors of 2 dividingr1 and r2. Neither r1 nor r2 is
divisible by 2 in the algebraic integers.m=2:
r1=(-1+sqrt(-15))/2, r2=(-1-sqrt(-15))/2. The algebraic
integerfactors that r1 and r2 have in common with 2 are
sqrt(r1) and sqrt(r2).Neither r1 nor r2 is divisible by
2.m=-1: r1=1, r2=-2. The algebraic integer factors that r1
and r2 havein common with 2 are again 1 and 2.Keith
Ramsay
===
> Starting with sequences with equal numbers of
zeros and ones, I think> a new thing to consider are those
sequences having 1/4, 1/8, 1/16,> etcetera ones, and then
also 1/3, 1/5, 2/5, 1/6, 1/7, 2/7, 3/7, 1/8,> 3/8, 1/9,
2/9, 4/9, etcetera many ones, or zeros.> Well, I got into
the case where a quarter of the sequence elements are> ones or
zeros.> n! / ( (n/4)!(3n/4)! 2^n )> I problem here with
getting an identity that converges to a value is> that
(n/4)!(3n/4)! >> (n/2)!(n/2)!. I thought I might be able to
use> Euler¶s formula for Gamma.> ...I want to determine a
relation between (n/4)!(3n/4)! and (n/2)!(n/2)!,I think I
went about it in an overly complicated way.(n/4)!(3n/4)! /
(n/2)!(n/2)! = (n/4)!(3n/4)! / (n/2)!(n/2)!So then, when I
use (n/4)!(3n/4)! in an expression where there wasinstead
(n/2)!(n/2)! that appears to converge, then I should be
ableto multiply it be (n/2)!(n/2)! / (n/4)! (3n/4)! to get
the value ofthe expression when it has (n/2)!^2 instead of
(n/4)!(3n/4)!.The same should hold true for all the cases
like (n/8)!(7n/8)! (whichis greater than (n/4)!(3n/4)! and
also (n/2)!^2) and for(n/3)!(2n/3)!. That is to say, if f(n)
is a simple ratio ofexpressions and that so is g(n) where
each is h(n) divided by theexpression (n/2)!(n/2)! for f(n)
and (n/4)!(3n/4)! for g(n) that f(n)= g(n) * (n/2)!(n/2)! /
((n/4)!(3n/4)!), and that that would be truefor any
expression for rational a/b < 1 that f(n) = g(a, b, n)
*(n/2)!(n/2)! / ( ((a/b)n)! ((1-a/b)n)! ) where f(n) = n!
/((n/2)!(n/2)! 2^n) = g(n,1,2) and g(n,a,b) = n! / ( (a/b)n!
(1-a/b)n!2^n ). Thus I was surprised when that did not appear
to be the casefor the small ?ite values of n that I
calculated.I guess it does, I may have had some other error
in my program, Ithink I already tried it.f(4)=
1.500000000000000000000000000000000000000000000000000000000000
f(16)=
1.571044921875000000000000000000000000000000000000000000000000
f(64)=
1.589548059967470344452933339596256701042875647544860839843750
f(256)=
1.594211517956484839624950947759399703028478483734198325900710
f(1024)=
1.595379577150690797583016460816556042629587428727203086922211
f(4096)=
1.595671726561313225912210533040091324648758050191856710414061
f(16384)=
1.595744772287097827906598649182402930052438187131273212971257
I still need to understand the proof of why that value in the
limit asn->oo is sqrt(8/Pi), as I haven¶t seen one yet, but it
appears that itis very simple to generate the expressions for
the expected number ofsequences with a given fraction of ones
or zeros.n! / ((n/2)!^2 2^n) ~= sqrt(8)/ sqrt(Pi*n)n! /
((a/b)n! (1-a/b)n! ! 2^n) ~= ( sqrt(8) (a/b)n! (1-a/b)n! )
/(n/2)!^2 sqrt(Pi * n)With this and Stirling¶s formula it is
possible to determine n^n, e^n,(a/b)n!, 2^n, and n in terms
of each other and Pi.Another thread today on sci.math is
discussing probabilities thatequal zero. Pick a number
between zero and one, I¶ll probably betit¶s one half, \
with
probability at least in?itesimally greater thanone half.
After all, the average value of all of the reals betweenzero
and one is one half. Of any characteristic of the amount of
onesand zeros of any in?ite string of coin tosses, it is
more probablethat it has nearly 1/2 ones and 1/2 zeros than
any other fraction. Then again, on pick a number between zero
and n for unknown n, Ipick four.Could in?ite monkeys each
typing at random in?itely generate theworks of Shakespeare?
Easily, the work of Shakespeare is a ?itestring. How many
cigarettes would they smoke?The probability of selecting a
given sequence at random from the 2^npossible sequences is 1/
2^n. That evaluates to zero for in?ite nin terms of x, y, or
z unrelated to n, but 2^n / 2^n evaluates toequal to 1 for
any n.Ross
===
>> handled by professor Chapman. Let m=[K:Q].
We can represent multiplication> I¶m not, and I never have
been, a professor.In the U.S., wouldn¶t your equivalent
position be professor? Or atleast associate professor?I never
understood the British system. How many levels are there
anyways?Anyway, my point is that in the States, the normal
form of address forsomeone in your position would be
professor, so it¶s natural people aregoing to keep doing
this. Are you going to keep correcting everybody whoaddresses
you that way? I think this is the second time I¶ve seen you
dothis recently.
===
NUMEROLOGY 5if 2 is 100, 50 is 135, 1000
is 145what is 5 billion?my guess is 733 or 469, at least
300unless its only 216this is where I¶m trying to get a ?ure
for the highest IQ, (2 year old post)2 is 100 means if you are
average (one in two) then your IQ is 100.If you are one in 50
(top 2% - mensa) then IQ is 135.with no knowledge about the
extremities of the bell curve or IQnotice I am spot on here
with 216, as most ?ures for highest IQare just over 200 to
220, which ties in with the gist of the post that I canjust
guess things correctly.Although here is my best demo of
guessing where I beat 81 to 1 odds
live.http://tinyurl.com/j9vr Mmmmm-mmmm! Wholesome humble
pie! I could try lying to save face but I have to admit it:
Four out of four right.Herc
===
> Is there an algorithm to
determine the minumum number of generators for a> ?ite group
? Clearly, for a cyclic group it is 1, for a symmetric>
permutation group or a dihedral group it is 2 but what about
any ?ite> group ? I asked John Conway the inverse question -
Can every group begenerated from a set of generators whose
orders multiply to give theorder of the group? and he assured
me that they could, at least up tothe ?st Janko group. This
puts an upper limit on the number ofgenerators needed. S4
needs all 4 {3x2x2x2=24). GAP can provide a setof generators
via
RelatorsOfFpGroup(Image(IsomorphismFpGroup(SmallGroup(m,n))))
,but this may not be minimal.Roger Beresford.Make things as
simple as possible, but no simpler (A. Einstein.)
===
>> Is
there an algorithm to determine the minumum number of
generators for a>> ?ite group ? Clearly, for a cyclic group
it is 1, for a symmetric>> permutation group or a dihedral
group it is 2 but what about any ?ite>> group ? I asked John
Conway the inverse question - Can every group be>generated
from a set of generators whose orders multiply to give
the>order of the group? and he assured me that they could, at
least up to>the ?st Janko group. This puts an upper limit on
the number of>generators needed.Yes, but this upper limit
would be much too large in general. >S4 needs all 4
{3x2x2x2=24).I am not sure what you mean by that. S4 is a
2-generator group.>GAP can provide a set>of generators via
RelatorsOfFpGroup(Image(IsomorphismFpGroup(SmallGroup(m,n))))
,>but this may not be minimal.But the question was to ?d a
minimal generating set. If you just wanta generating set,
take the whole of G.Derek Holt.
===
The Principal Of Induction
which is:a statement P is true for P=1,and true for P=k+1 if k
was true,then P is trueIs it the axiom or theorem??B4E
===
>
The Principal Of Induction which is: a statement P is true
for P=1,> and true for P=k+1 if k was true,> then P is true
Is it the axiom or theorem??>In my text it says it lies at a
different level than the axioms,together with this rule.If F
and F->G then Gthe rule of induction is like the rule of
inference, they are added to the axioms.But I don¶t see any
formal reason for its special case, at this point it
lookslikely as a step to complete formalism, but we are short
on categorising manyother proof mechanisms.Herc
===
> I can¶t
seem to construct one. Is there such a mapping? (R is the
real>> numbers)>Maybe this will help. I will show a bijection
from the interval (0, 1) to>the interior of a unit cube. Let a
point in (0 , 1) be .a_1a_2a_3.... This>corresponds to the
unique point (.a_1a_4a_6..., .a_2a_5a_8...,>>a_2a_5a_8...) in
the interior of the unique cube. Now given the
point>(.a_1a_2a_3a_4...., .b_1b_2b_3..., .c_1c_2... ) in the
interior of the unit>cube correspond to the unique point
(.a_1b_1c_1a_2b_2c_2...) in the>interval ( 0, 1). Of course
you must talk about numbers like .399999..... Is>it
.39999.... or is it .4000...??>Steve>A space-?ling curve
should do the trick here as well, no?-Davis
===
> N^n injects
into N? How is this possible? Does this mean that Z^n also>
injects into Z?> This seems ridiculous to me, but I will have
to think about it more after I> read all of the>
responses.[...]N^2 injects into N as follows:(a,b) |-> 2^a *
3^b .For larger n, one can use any n distinct primes.This
bears a resemblance to Godel numbering.David Bernier
===
You
are playing the following game. There are 3 players armed
with a gun.Player #1 hits his target 1/3 of the time. Player
#2s hits her target halfof the times and player #3 never
misses. 1st player #1 shoots, then player#2 shoots if she is
still alive, then player #3 shoots if she is still aliveand
then back to #1,etc. You are player #1. What is your best
strategy? Wedo assume that all shooters make the optimal
decision for themselves and ifyou are shot you die, ie you
are out of the game.Steve
===
Steven> You are playing the
following game. There are 3 players armed with agun.> Player
#1 hits his target 1/3 of the time. Player #2s hits her
target half> of the times and player #3 never misses. 1st
player #1 shoots, then player> #2 shoots if she is still
alive, then player #3 shoots if she is stillalive> and then
back to #1,etc. You are player #1. What is your best
strategy? We> do assume that all shooters make the optimal
decision for themselves andif> you are shot you die, ie you
are out of the game.> SteveSuch problems are taken quite
seriously in some circles :) but there can¶tbe any point in
shooting at player #2. Or am I missing something?Most of us
have probably seen some form of the Xtown
MathematicsExamination, where Xtown is the name of some rough
neighbourhood nearby.Question 1) If Billy sells 20 grams of
cocaine per day, and makes $30 pergram, then how many days
will he need to pay back the loan shark if ...Question 2) If
an AK-47 holds 30 rounds per clip, and ?es 4 rounds
persecond...You get the
idea.Larry
===
VnkYa.28002$O42.10475@news02.roc.ny...> You are
playing the following game. There are 3 players armed with
agun.> Player #1 hits his target 1/3 of the time. Player #2s
hits her target half> of the times and player #3 never
misses. 1st player #1 shoots, then player> #2 shoots if she
is still alive, then player #3 shoots if she is stillalive>
and then back to #1,etc. You are player #1. What is your best
strategy? We> do assume that all shooters make the optimal
decision for themselves andif> you are shot you die, ie you
are out of the game.This is a well-known problem.The best
strategy is to pass one¶s turn, or to shoot at the
sky.--Julien Santini
===
I¶ll be back.===>>Yes, that is what
I want to say.>>Take a very big n, for example, n =
e^10^10^100:>>if you pick an integer at random between 1 and
n, the probability that it is>>prime is almost 0, much less
than the probability that you won the biggest>>monthly
lottery prize in USA 10^80 times in a row buying only one
ticket>>each month.I think that¶s why it¶s big news \
when they
?d huge primes of many>(thousands?) of digits. Because you
can¶t just pick those at random>and have any reasonable hope
of success.Not really. The largest prime found so far has
around 4 million digits,which is far, far, smaller than the
example above. In the millions ofdigits range, the odds of
picking a prime at random are about the sameas some of the
easier lotteries (and an order of magnitude better if
yousieve out small prime multiples).The rarity of primes only
increases linearly with the number of digits,whereas the
running time of the primality testing algorithm
increasespolynomially. If you¶re going to attribute the
dif?ulty to any onething it should be the computational cost
of the algorithm. -- Erick
===
> If the percentage of primes as
a fraction of integers was not 0, > it would be an interesting
constant.And if my Aunt had balls...Still, there are a couple
or three things you might or might not know.1) There IS a
constant limiting density (the type you like) of SQUARE-FREE
numbers - no squared-prime factors (or cubed etc). It is
6/pi^2 .2) There IS a constant limiting probability that TWO
numbers will be CO-prime. if they are randomly chosen in
almost any sensible way. It is the same!3) Given a large
prime chosen at random, the probabilities that it is of type
4n+1 or 4n-1 are half each. Ditto for 6n+1 and 6n-1.There are
many other results of these types. All good clean
fun.-
- Bill Taylor
W.Taylor@math.canterbury.ac.nz---
--- Watch out for
my de?itive series of books on Prime Numbers. Just coming
into print now... volume I: The Even
Primes---
---
===
 This problem of baselines is why
economists use logarithms,> and why they sometimes construct
indexes that are based> on the logs of values -- instead of
the natural prices.>AIUI the reason economist use log sis
because the procesesses they aremodeling have exponential
drift. Eg they tend to grow propotionally to theircurrent
size. It is easier for economists to think in terms of
thisinstantaneous growth than the absolute amount of growth.
Forgive me if thisis what you meant.However I think with most
(serious) currencies the stochastic elementoutweighs the
exponential so they tend to just graph values.> If you don¶t
get more speci? help, I suggest that you> browse the basic
textbooks on economics in a library.> Index and logarithm
might be key-words.> [ snip, rest] --> Rich Ulrich,
wpilib@pitt.edu> http://www.pitt.edu/~wpilib/index.html>
Taxes are the price we pay for civilization. Justice
Holmes.
===
> I think that the curves formed by two surfaces
rolling (without sliding > or twisting) should be the
geodesic curves on the two surfaces, but I am > not sure. Is
this right? are there any references on this topic? This is
not true.Consider, for example, the sphere rolling on the
plane. The positionand orientation of the sphere can be
described by a point in themanifold M = R^2 x SO(3) (R^2
representing the location of the pointof contact, SO(3)
representing the rotation of the sphere). Let u(t)be the
curve in M that represents the path traced by the point
ofcontact.To capture the non-slip and non-twist conditions,
you need to look atthe lift of u into the tangent bundle TM =
R^2 x R^2 x SO(3) x so(3).The no-twist condition reallly means
that the sphere is not allowed torotate around the vertical
axis; this means that the so(3)-componentof the velocity
vector u¶(t) looks something like [ 0 0 -a2] [ 0 0 a1] [ a2
-a1 0] .Now, let u(t) = (x,R) and u¶(t) = (x,v,R,A) . Then
the no-slipcondition is v = r [ a1 ; a2 ]where r is the
radius of the sphere. You can use Chow¶s theorem toshow that
any point in M can be reached by a no-slip and no-spin
path,which is certainly not true of geodesic motion.Note that
the no-slip and no-spin conditions are linear constraints
onthe tangent vector; that is, an admissable tangent vector
must lie ina certain linear sub-space of the tangent space.
This collection ofsub-spaces forms a sub-bundle of the
tangent bundle, often called adistribution. (Not to be
confused with the distributions inanalysis such as the
so-called Dirac delta function.)This topic is studied
extensively in both differential geometry andcontrol
theory/mechanics. Keywords: non-holonomic
constraints;Frobenius theorem; Pfaf?n systems;
distributions; derived ?howtheorem.Kevin.
===
> [snip]>
See the section ``Delta > 0 is Square¶¶ in> Solving \
the
equation ax^2 + bxy + cy^2 + dx + ey + f = 0 - PDF File> at
http://hometown.aol.com/jpr2718/> Either methods in that
section or elsewhere in the ?e should cover what you> ask,
depending on what you mean by ``this form of \
equation,¶¶ and
what you mean> by `` very large.¶¶> As another poster
noted, your equation has no solutions. As written, the lhs>
is odd and the rhs is even.By ?very large¶ I mean that
cof?ients and variables may have severalhundred digits.But
if factoring is required... I guess I¶m out of luck,
right?
===
>My question now is how we can>integrate over a
sphere using spherical coordinates when they are>badly
behaved at several points, they don¶t constitute an atlas
so>under this coordinate regime the sphere isn¶t a manifold,
what is it,>why do integrals still work.One approach is to
use a partition of unity. We can ?d twocoordinate charts and
functions f and g that are nonzero onlyon each chart
separately, but where f+g = 1 everywhere. Thenthe integral of
a function u can be computed as the integralof f*u + g*u,
where the integral of f*u can be evaluated using theone
coordinate chart, and the integral of g*u can be
evaluatedusing the other one.For other manifolds more
coordinate charts may be needed of course.Keith Ramsay
===
|I
presume that Petry wants to subordinate the free enquiry of
mathematics|to the commericial/governmental/military
sponsored|interests of the scientists. Luckily he has no
power or in?.I¶m a little bit surprised at how popular
the point of view is, that sayswhat science and mathematics
need is to be more ?mly under thethumb of those
sponsors.Keith Ramsay
===
|But is there any reason to think
that constructive mathematics|*helps* in applied
mathematics?A constructivist once told me that it did. Of
course, one can shrug thisoff as bias, but I don¶t have any
particular reason to think that he wasclaiming this because
he held constructivist opinions, rather than that heheld
constructivist opinions because he¶d discovered such things
as it¶shaving advantages in applicability. He also
investigated this kind of thingmore thoroughly than most
people. It¶s always a little tricky because thepeople who do
the most investigation tend to be self-selected for
havingcome to believe that what they¶re investigating is good
stuff, but I haven¶tseen good reason to think he¶s \
mistaken
about this.By comparison, I can only offer a relatively
off-the-cuff impression. Myhunch is that what is good about
it for applications (whether it actuallymakes it better than
classical mathematics or not) springs from the samesource as
all its other kinds of goodness, rather than because of
anextra advantage toward application. I say this partly
because the latternotion seems to be a popular target for
criticism, as if what we¶re doingwhen we weigh constructive
and classical mathematics against each other istrying to
decide whether the advantage in applicability is enough
tooutweigh losses thought to exist in other areas.|I don¶t
think it really makes|much difference. Constructively, you
can try to prove that (for|example) a differential equation
has a solution, and the proof|itself will provide an
algorithm for computing the solution.|Classically, one would
break it into two different tasks:|(1) Deciding whether the
equation has a solution, and|(2) Finding an algorithm for
computing the solution. If|the classical mathematician
manages to complete task 2,|then hasn¶t he accomplished the
same practical goal as|the constructive
mathematician?Backtracking to determine whether the end
result is computable doesn¶tstrike me as conducive to
complete illumination of what is really goingon.
Aesthetically this is perhaps the main thing about what
littleconstructive mathematics I¶ve been able to do that I
?d most appealing:it seems much more like a complete laying
bare of the nature of amathematical situation, not by heroic
efforts at incorporatingcomputability considerations
afterward, but simply in the natural courseof development of
the area.|The algorithms might turn out to be different,
because|the constructive mathematician will come up with an
algorithm|that is provably correct *constructively*, while
the classical|mathematician only needs to prove that it is
correct *classically*.The algorithms might turn out to be
different, of course, but thedifference between ?ding a
constructive proof and a classical proofof its correctness is
perhaps different from what you think. For a fairlybroad class
of results of this kind, any classical proof provides one
witha constructive proof in an automatic way: if for each m,
there exists ann such that P(m,n) where m and n are integers
and P is primitive recursivehas a classical proof, then it
has a constructive proof. The slogan used(by certain
logicians) is that classical and constructive mathematics
have the same provably recursive functions. So the difference
hinges on moresubtle differences between the two standpoints.I
think an example I mentioned not so long ago might serve again
here.In the book of Pour El and Richards, they consider
different senses ofcomputability for solutions of a wave
equation. In constructivemathematics two of them correspond
naturally to two ways of de?ingL^2(R^3). The usual de?ition
in classical mathematics is that they arethe square integrable
functions on R^3, modulo the ones that are zeroexcept on a set
of measure zero. Alternatively it can be de?ed as
thecompletion in the L^2 norm of a space of smooth test
functions.During a constructive development, the two
de?itions are kept distinctbecause they aren¶t
constructively equivalent. You are naturally led to seethe
differences between them. Doing a classical development seems
to me tobe a little misleading, because you prove along the
way that the twode?itions are the same (i.e.,
nonconstructively equivalent) and thusare liable not to
respect the distinction. Then it reappears in a moreconfusing
guise when you try to work out some computability theory
relatedto them. I think the people who¶ve cited it as an
example of potentiallyuncomputable physics, for example, have
essentially been conned byclassical mathematics¶ own
misleading appearance.So one possible difference in how a
mathematician working from a classicalstandpoint would
develop an algorithm, and how a mathematician working froma
constructive standpoint would, is that the terms in which the
classicalmathematician is considering the problem are to the
constructivemathematician a bit like chimeras, combinations
of animals from distinctspecies thought of as if they were
simply individual animals. A proof mightin its course might
cultivate an element of L^2 (a sheep) and then make offwith
it as a square integrable function modulo equality almost
everywhere(a goat) without being aware of what kind of
information is being thrownout in doing so.Keith
Ramsay
===
[snip]> I don¶t think that any mathematicians are
saying that ZFC> is the *only* way to think about real
numbers. They are> just saying that it is an extremely
powerful way to think> about them.But it isn¶t. It is the
best way to avoid any furtherthoughts or deeper analysis into
the nature of real numbers.It¶s _the_ way to free math of
philosophical considerations.Originally, mathematicians
thought that one¶s philosophicalassumptions couldn¶t \
possibly
in? the math. This view isstill popular in math. But it
isn¶t true (intuitionism is one example,but perhaps more will
be found, when we start looking).When it comes to foundations,
things simply can be much moresubtle and interesting than is
suggested by FOL/ZF. And thatinsight is overwhelmingly
absent.> Are there actual examples of results that> could not
be obtained using ZFC?Sure. But mainstream maths can¶t
appreciate these as¶actual results¶.But let me ask \
another
question:Is every result provable from ZFC a sensible result
of mathematics?(I¶m not saying useful, or applicable:
?sensible¶.)For example, what do we ever do with the set
P(P(P(P(P(P( N )))))) ?Then why do large sets play such a
large role in ZFC?It seems to me that maths could do with
much _less_ than ZFC, andit would require quite some
(worthwhile) effort to ?d out how.>BTW, i know that
intuitionism often leads to irritation>on the side of
classical mathematicians.>They think that intuitionists
sort of want to replace>classical math with something
else.>But did non-Euclidean geometry ?replace¶ Euclidean
geometry?>No, it just made an end to its exclusive
monopoly.>And that only made math richer, not poorer.
Right? I don¶t have any problem> with including intuitionist
mathematics. I only have a> problem with David¶s wanting to
*exclude* classical> mathematics.It is questionable whether
the current foundations of mathare really representing
?classical mathematics¶. To me, ZFCis assuming far too
much.Another problem with the current foundations is, that
inpractice it is an alibi to stop thinking about
foundations.And a third problem is:ZFC makes us all sound
like mediaeval wizards from the past,talking in their own
language, about weird fantasy objects,only known by the
in-crowd, after years of study.And, be fair: can anyone
become a mathematician, nowadays,without being turned into a
ZFC-talker?Herman Jurjus
===
|And, be fair: can anyone become
a mathematician, nowadays,|without being turned into a
ZFC-talker?As often as people on usenet refer to ZFC, I think
the typicalmathematician knows very little about it, nor does
it makemuch difference to them.Keith Ramsay
===
|And, be fair:
can anyone become a mathematician, nowadays,>|without being
turned into a ZFC-talker?As often as people on usenet refer
to ZFC, I think the typical>mathematician knows very little
about it, nor does it make>much difference to them.Yes, I
think you are right.It occurred to me recently that I have
almost no instinct for what canbe proved without using AC.For
example, can you prove that there is no group G with |Aut(G)|
= 3(a problem that arose recently on sci.math) without using
AC ?You reduce easily to an in?ite abelian group G in which
all elements haveorder 2 or, equivalently, an in?ite vector
space over the ?ld of order2. Without AC, can you say
anthing at all about Aut(G) ?Derek Holt.
===
This problem has
nothing to do with school, work, or anything like that. It¶s
just a problem that occurred to me one day, and has been
driving me nuts. Perhaps somebody here can provide me the
solution, or at least give me some pointers.Here¶s the
problem.Consider a game with the following rules: You are
given an initial score and a time limit. We¶ll denote the
time remaining by t, and your score at that time by x[t].
Your score changes over time as a random walk with a fudge
factor that attracts it to zero: x[t-dt] = x[t]*e^(-dt) +
Z*sqrt((1-e^(-2*dt))/2)where e is the bast of natural
logarithms and Z is a standard normal random value (mean
zero, variance 1).(Note that, for dt<<1, x[t-dt]-x[t] is
approximately -x[t]*dt + Z*sqrt(dt).)You may stop the game at
any time. If you stop the game, your ?al score is the score
you have when you stop it. If you do not stop the game, your
?al score is the score you have when time expires.QUESTION
#1: What is the optimal strategy? (I trust that the meaning
of strategy and optimal are clear. I can post de?itions if
necessary.)QUESTION #2: Given an initial score x0 and time
t0, and assuming that you play the optimal strategy, what is
the expected value of the game?
===
> currently, the politic
does to the people what they can¶t do back.> I know this,
100,000 people in Townville know this and will> report to you
that the government tortures adam of the bible> The population
of Townsville is closer to 130,000 - which brings upthe
question, what else characterizes the 30,000 or so people
ofTownsville that do not know this thing you assert (namely,
that thepolitic does to the people what they can¶t do back)?>
> the politic does for the people what they can¶t do for
themselves,> as far as I am concerned, this means the same
as> Seperate Church and State!> So, the politic does to the
people what they can¶t do back, and alsothe politic does for
the people what they can¶t do themselves. Andthese are not
necessarily related, I suppose, anyway I think I¶mfollowing
you so far, although the signi?ance of what you are tryingto
say is failing me.> Let Adam decide when he wants to be
?med, stop the Truman Show.> Adam is here for the future of
humanity, not your personal training.> My personal training?
Wait - what do you know of me? I¶m not oneof the 130,000 some
residents of Townsville. But I accept yourstipulation that
there is an Adam, presumably a resident ofTownsville, and
that he is here for the future of humanity. You arenot
explicit about what being here for the future of humanity
means,but I¶ll assume Adam means to do humanity some good.> I
give you all this chance to do for yourselves, if the people>
can take religion then the government will be YOUR audience.>
If taking religion means realising that religion plays an
importantrole to a subset of humanity, I¶m not sure I
understand how thisrealization leads to the condition that
then the government becomes myaudience, or what the
signi?ance of this is.> If the people can¶t do for
themselves the seperation of church> and state then the
government is not going to let go of a free> nuclear blast
shield that is Adams current immortality, and> their free
in?ite knowledge they spy on all his life.> Mr. Cooper, are
you currently keeping up with your medication? Canyou
recommend any way for the average reader to be able to tell,
fromyour posts, which ones were posted while you are stable
and medicated,and which ones are done whilst in the dark grip
of a serious mentalillness?> If the people can¶t seperate
church and state, then you are> forever to live a life that
the politic will do to you anything> that you can¶t do back.
God is on your side, be on his.> phone any of 100,000
people in Townsville Australia to> verify, just like the
following 4 posts, that the truman show> is running live
there, and the truman is constantly tortured> by a satelite
that penetrates buildings and a team of narcisistic> agenst
that work around the clock to persecute Adam. ME>
aus.tv know The Truman Show is true>
_____________________________> I¶m from Townsville and
YOU ARE the Truman!> http://tinyurl.com/iky5> I was in
Townsville over the weekend, and I heard him.> Very spooky!>
http://tinyurl.com/iky8>phone someone in Townsville,
half of you must know someone there,>every day I go out
people say THERES THE TRUMAN> I¶m in Townsville. We¶re \
sick
of you.> http://tinyurl.com/iky9>
http://tinyurl.com/iky4> You rule Truman!>
===
>
received didn¶t help me enough to ?ure out a proof. I am not
a> student, so please don¶t feel that you are helping me
cheat on> homework or something. I am using this group
precisely because I> don¶t have the advantage of a
professor to ask questions of.> The question:>
Let P(A) denote the power set for some set A.> A subset B
of P(A) is called an antichain if no element> of B is a
subset of any other element of B. With N denoting the set>
of natural numbers, prove that P(N) contains an uncountable
antichain.> The hints I recieved said to look for an
uncountable antichain in> either P(Q) or P(QxQ), where Q is
the set of rational numbers. Since> Q and QxQ have the same
cardinality as N, this should solve the> problem. As most
of these types of exercises use a diagonal argument,> I
expect that one is probably required. So, what is the answer?
> There is a Theorem to the effect that there is a subset of
P(N) of the> power of the continuum such that each element of
A is in?ite and if X> and Y belong to A then X / Y is ?ite
(and consequently neither is> contained in the other).> For
each real r, 0 < r < 1 let > U(r) = { 2^n*(?*r) +1) : n =
1, 2, ...}. > A = {U(r) : 0 < r < 1, r real}.
===
>> Well, that
was too easy because it¶s already upper triangular.>> But if
that 0 in the second column wasn¶t there, what you¶d \
do...>So
this example is too special, that¶s why it can be done in one
step,>instead of Strang¶s three steps? You said: But if that
0 in the second>column wasn¶t there, ...>Which zero? U1?In
the previous sentence, I said because it¶s upper
triangular.I¶m referring to the 0 in the second column that
makes A1 upper triangular.>So can you explain more clearly
when can I use the simpli?d GUESSING,>when I need to use
Strang¶s method?OK, let¶s try another example. [ 3 -2 \
8 ] A =
[ -4 6 6 ] [ 1 -4 11 ]This has an eigenvalue 5 with normalized
eigenvector [ 2/3 ] [ 2/3 ] [ 1/3 ]which we complete to the
unitary matrix [ 1/2 ] [ 5 ] [2/3  0 ] [ 3 ] [ ] [ 1/2
1/2 ] U1 = [ 4 5 5 ] [2/3 - --  ] [ 15 5 ] [ ] [ 1/2
1/2] [ 2 5 2 5 ] [1/3 - -- - --] [ 15 5 ]Then [ 1/2
1/2] [ 11 5 74 5 ] [5 - --- - ---] [ 5 15 ] U1^(-1) A
U1 = [ ] [ -14 ] [0 29/5 --- ] [ 15 ] [ ] [0 -18/5 46/5 ]Note
the ?st column has 0¶s below the diagonal, but the second
columndoesn¶t, so it¶s not upper triangular yet.Take \
the
lower right 2 x 2 submatrix [ -14 ] [29/5 --- ] A2 = [ 15 ] [
] [-18/5 46/5]This has an eigenvalue 10 with normalized
eigenvector [ 1/2 ] [ 2 85 ] [ --- ] [ 85 ] [ ] [ 1/2] [
9 85 ] [- ---] [ 85 ]Complete this to a 2 x 2 unitary
matrix [ 1/2 1/2] [ 2 85 9 85 ] [ --- ---] [ 85 85 ]
U2 = [ ] [ 1/2 1/2] [ 9 85 2 85 ] [- --- ---] [ 85 85
]which is the lower right 2 x 2 submatrix of the 3 x 3 unitary
matrix [1 0 0 ] [ ] [ 1/2 1/2] [ 2 85 9 85 ] [0 ---
---] V2 = [ 85 85 ] [ ] [ 1/2 1/2] [ 9 85 2 85 ] [0 -
--- ---] [ 85 85 ]Note that U2^(-1) A2 U2 is upper
triangular. Then you want [ 1/2 1/2 1/2 1/2 ] [ 2 5 85 3 5 85
] [2/3   ] [ 255 85 ] [ ] [ 1/2 1/2
1/2 1/2] U = U1 V2 = [ 7 5 85 2 5 85 ] [2/3 -  -
] [ 255 85 ] [ ] [ 1/2 1/2 1/2 1/2] [ 2 5 85 2 5
85 ] [1/3  - ] [ 51 85 ]and lo! [ 1/2
1/2 1/2 1/2] [ 8 5 85 89 5 85 ] [5  -
-] [ 17 255 ] U^(-1) A U = [ ] [0 10 8/3 ] [ ] [0
0 5 ]Robert Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2