Newsgroup sci.math 152025

Directory

On Thu, 7 Nov 1996, Judson McClendon wrote:
> My previous post got no response, and I wonder if it's because the
> significance of the question wasn't clear?  If the things I have been
> hearing about this issue are true, then this issue should be dynamite! 
	The significance is quite clear to some people: however, for one, 
I am unfortunately not competant enough to help. Don't lose heart: it's 
just a matter of getting the right people's attention for long enough.
-- Niel de Beaudrap

On Thu, 7 Nov 1996, Skoros Laertes wrote:
> Is this right? Is this right?!?!?! Tell me! Tell me! Please, oh 
> please, for the love of God, TTEELL MMEE!!!!!!
> 
> 
> -- Skoros L.
> 
"Is this statement right?" ---> Godel's little known first version of his
                                incompletenesss theorem :)
	In response, yes, your question is right, but is unprovable using TNT.
Niel de Beaudrap
----------------------
jd@cpsc.ucalgary.ca
ps. Do you stutter? :)

Perdon por el error:
La pagina se encuentra en
http://www.geocities.com/CapeCanaveral/8001/

In article b940aace@default, "Robert E Sawyer"  writes:
>(My apologies if this turns out to be a repeat posting -- 
>this is my third try to get it onto the server.)
Appeared here only once. 
>Alan's question may be better posed as follows:
>  
>Suppose X1, X2, ... are iid 0/1 random variables, with pr(Xi=1)=p, 0Let X = .X1X2...; that is, let X be the number whose binary representation
>is given by the infinite sequence X1, X2, ..., so 0<=X<=1.  
I don't exactly know what you mean with this but suppose it is:
Binary		Decimal
0.0		0.0
0.01		0.25
0.1		0.5
0.11		0.75
0.111		0.875
1.0		1.0
and so on. 
>The question:  What is the cumulative distribution function F(t;p) = pr(X<=t;p)? 
>
>Is there a nice form for F(t;p) as a function of both t & p?
Three cases.
1)	p=0 or p=1
Trivial. F(t;p) would be a step function at either 0 or 1, its derivative
a delta function at either 0 or 1.
2)	p=0.5
Trivial too. F(t;p)=t. Why? Just consider the probabilities
pr(X<.1)=pr(X>0.1)=0.5 because p=0.5 for the first toss,
pr(X<0.01)=0.25 because that corresponds to two successive 0 tosses,
and so on. Hence F(t;p)=t  its derivative (wrt to t of course) is
unity everywhere. Please note that 'probability density equal to unity'
does _not_ mean that every sequence has probability unity; the
probability for each individual sequence is the Lebesgue integral
over the subspace of the event space considered, in the case of a single
sequence X this is \int_X^X 1 dx = 0, so that every sequence has
probability density one, but probability zero. It is essential
to see this difference.
3)   	p not in {0,0.5,1}
Tough. Here you end up with what is called a 'binomial multifractal measure'.
It is not a smooth function of anything. You can't take derivatives. See
for example Appendix B and especially the figure on p. 928 in:
@BOOK{Peitgen1992a,
        AUTHOR             = {Heinz-Otto Peitgen and Hartmut J{\"u}rgens and Dietmar Saupe},
        ADDRESS            = {New York},
        PUBLISHER          = {Springer},
        TITLE              = {Chaos and Fractals - New Frontiers of Science},
        YEAR               = {1992},
        ISSN_ISBN          = {0-387-97903-4}
}
>Presumably X has a *continuous* distribution on the interval [0,1] for 0and is uniform for p=1/2.  If F(t;p) is differentiable in t, then the derivative will
>give the "density function". 
As i said, it is a continuous function for p=1/2 only. 
---
-ernst wallenborn.
- Du haelst die Fernbedienung verkehrtherum, Dad!
 - Nicht, wenn es einen Gott gibt, Dumpfbacke...

... 
The analysis may be easier if one assumes a *parabolic curve* 
deflection.  This also gives a slightly more gradual curve. 
Viz. maximum curvature at vertex. 
The result is again,  h = sqrt(yard * furlong) 
                = 3*sqrt(220)  = sqrt(1980)   , feet. 
                same  as interim solution -quoted below. 
This is a friendly and true  relationship between imperial units..!# 
"Mile plus foot" gives sqrt "yard*furlong." 
                       (3 feet,  220 yards, 660 feet, 10 chains, 
                                 one-eighth mile.) 
y = a*x^2 
y' = dy/dx = 2 a x. 
Arc length 
     ds = sqrt(1+y'^2)dx 
        = sqrt(1+ (2ax)^2) dx 
        ~ (1+2a^2.x^2).dx 
        s = [ x + 2/3.a^2.x^3] 
5281/2 = 5280/2 + 2/3.a^2.(5280/2)^3 
a^2    =  3/4.(2/5280)^3 
deviation = a.(5280/2)^2 
...       =  sqrt(3*5280/8) 
          =  sqrt(3*660.)  =  sqrt (yard * furlong.) 
          =  3*sqrt(220) 
          ~  3*15.-    feet. 
          =  45 - 1/2 - 1/356. Ft. 
Don.McDonald@welcom.gen.nz 
: From: kfoster@rainbow.rmii.com (Kurt Foster) 
: Subject: SPOILER, "Mile plus a foot" rail problem####### 
: Date: 5 Nov 1996 19:06:28 GMT 
: Message-ID: <55o37k$c7e@rainbow.rmii.com> 
>  A steel rail, just over 1 mile long, is rigidly fixed at one end, and 
>clamped at a point exactly one mile away, so exactly 1 mile of the rail 
>forms a straight line segment between the fixed end and the clamped point. 
>  Now, the clamp is loosened, and exactly one foot of additional rail is 
>forced between the clamp and the fixed end; the clamp is then 
>re-tightened. 
>  The section of rail between the clamp and the fixed end then bulges 
>out to one side in what you may assume is a circular arc whose greatest 
>deflection from the original straight line is right in the center.  The 
>question is, how much is that deflection? [Note: 1 mile = 5280 feet.] 
: (V) h = sqrt(3Ls/8), approximately, if s/L is small. 
:  
:      Plugging L = 5280, s = 1 into (IV) and (V) give 
:  
: A = sqrt(1/880) = 0.033709992 radians, approximately, and 
:  
: h = sqrt(1980) or 44.49719 feet, approximately.  ..[don.mcdon # agreed] 
:  
:      By using Newton's method to solve (I) to double precision accuracy, 
: ...and the displacement is 
:    h = 44.4984550191008  feet. 
--- 

Does anyone know where I can find a proof of Pascal's Mystic Hexegon
Theorem on the web (or ftp, newsgroups, etc.)?

i want to find the limit value of 'x*ln(sinx)' at x=pi
can anyone tell me how to solve the problem
please help me , thank you very much 

On 9 Nov 1996, WAPPLER FRANK wrote:
[snipped to eliminate spaces twixt lines]
> susieq wrote:
> > In article <55u4od$4mr@newton.pacific.net.sg>, u07j4@altron.com.sg says...
> > 
> > >maths is great!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
> 
> >                ---------------------------
> >        physics is more *fun*!!!!! (i *like* getting physical, don't you?)
> >        what's so great about math? it can't define a proton.
> >                                        love,
> 
> That's what >>you think<<, susieq.                           Frank  W ~@) R
> 
	Boy, leave it to an inconsequential posting to let the rivalries 
show. Why, without even knowing what it was, I could post a similar 
one-liner about autodynamics, and everyone here would line up like little 
toy soldiers into their battle formations. Let's be civil!
"Can't we all just get along?" -- A Reasoning Being
Niel de Beaudrap
----------------------
jd@cpsc.ucalgary.ca

On Sun, 10 Nov 1996, Special Agent Orange wrote:
> 
> 
>      3/2                 3    1/2             1/2 
> (-1)        =      [ (-1)   ]          =  (-1)           =      i
> 
>      3/2                   1/2   3           3
> (-1)        =      [ (-1)         ]      =   i          =       - i
> 
> The correct one is......
> 
> 
                                                             i pi
	I guess the difference is the difference between   e       and
  3i pi                   
e       . Both are equal to -1, but using the equation
    i theta   n       i n theta        i pi              3/2    i 3pi/2
 (e         )     = e          , -1 = e       yields (-1)    = e
                                                             = -i,
                                       3i pi            3/2     i 9pi/2
                          while  -1 = e      yields (-1)     = e
                                                                i pi/2
                                                             = e
                                                             = i.
	I suspect that this means that, like 0/0, -1 may actually be 
indeterminate. However, one can avoid the problem by going with the 
convention that the number is represented in the, er, "Eulerian" (?) form 
with the smallest value for theta. At least, I suspect such a convention 
exists: This explains why root 1 = 1, root -1 = i (as opposed to -i), etc.
This way, (-1)^(3/2) would be -i.
So, did I pass?
Niel de Beaudrap
----------------------
jd@cpsc.ucalgary.ca

On Sun, 10 Nov 1996, Dan Razvan Ghica wrote:
> On 7 Nov 1996, Archimedes Plutonium wrote:
> 
> >  
> > > Please, what are "p-adics"?
> > 
> >  Each of them are Infinite Integers. Around 1901 Kurt Hensel in Germany
> > extended the integers through a series operation.
> > 
> >  The Finite Integer such as 1 is supposedly finite, nothing to the
> > right or left of it.
> > 
> >   Infinite Integers all of them have an endless string of digits to the
> > leftward, thus 1 is .....000000001  or 231 is .....00000231 but not
> > every Infinite Integer repeats in zeros, for instance the Infinite
> > Integer
> >    ....9999999999998 is equivalent to -2
> [...]
> 
> I don't really see how the way we write a number is important. If
> "...9998" and "-2" are equivalent I would naively go for the "-2" notation
> just because it lacks the confusing "..." at the left end. Just like I
> prefer "2.0" to "1.999...". But I might be wrong. 
> 
> It would be interesting if Archimedes Plutonium would steer his postings
> away from anti-mathematical-establishment conspiration-theory-esque
> rantings and tell us more about these mysterious p-adics, their fine
> properties and their potential impact on life from mathematics and physics
> to, say, accounting. 
> 
> I need more background before I start fighting the VietMath war.
> 
> Regards,
> DRG
	Now why didn't I think of this? Arch, would it be possible to 
please talk more about p-adics and less about how they will cause the 
collapse of mathematics? Serious request, here: I don't mean to be snide 
or sarcastic. Tell us about them.
Niel de Beaudrap
----------------------
jd@cpsc.ucalgary.ca

On Mon, 11 Nov 1996, Dieter Dijkstra wrote:
> Does 0.99... equal 1? I should think so, but can anyone confirm this?
> 	Dieter
Yes.
	Two ways to think about it:
Cheap way: for any digit a, the number 0.aaaaaaaaaaaaaaaaaaaaaaaa... = a/9
           for a = 9, this would mean 0.9999999999... = 9/9 = 1.
Better way: 0.9999999999999999999999999... is basically 1-n, where n is 
            a very, very, very, very, small number. We say that n is
            /infinitesimally/ small, which means that it is so small, we
            cannot measure it. For all practical purposes, n = 0. So,
            0.99999999999999999999... = 1-n = 1-0 = 1.
	In truth, in base ten (normal decimal numbers), the fraction 0.99999... 
dosen't actually exist. Anything that would evaluate to it, like the two
examples above, end up evaluating to 1 in the end.
Hope this helps.
Niel de Beaudrap
----------------------
jd@cpsc.ucalgary.ca

Aaron Birenboim (aaron@swcp.com) wrote:
: I have come across a simple, finite series problem...  I'm not sure there
: is an answer.
: I want to know if there might be a simple formula equivalent to :
:      1*1 + 2*2 + 3*3 + 4*4 + ... + n*n
: I'm looking for a simple answer like n*(n-1)/2 = 1+2+3+4+5+...+n
: --
: Aaron Birenboim   |         aaron@ptree.abq.nm.us            | Albuquerque, NM
: http://www.swcp.com/~aaron         RESUME> http://www.swcp.com/~aaron/res.html
: PearTree Consulting (WWW, UNIX, Scientific Computing,...)
This sum is 1/6*n*(n+1)*(2n +1)
The sum of third powers from 1 to n =
1/4*n^2*(n+1)^2 = (1+2+....+n)^2.
etcetera. The sum of k-th powers is from 1 to n
is always a polynomial of degree k+1 in n.
Hope this helps.
Henno Brandsma

"Christopher Anderson"  writes:
> 
> While in an interview I was asked why when a value of x is placed into
> x^3-x it is divisible by 6. ... How do I show this using algebra?
More generally, for any squarefree integer m, and any integer x,
      phi(m) + 1
    x            = x  (mod m)                                [1]
Recall m is squarefree means that each prime occurs to power at
most one in m, say  m = p q ... s, and in this special case the 
Euler totient function is simply  phi(m) = (p-1) (q-1) ... (s-1).
In your case we have m=6=2*3  so  phi(m)=1*2  and  x^3=x (mod 6).
To prove [1] above we first prove that it holds (mod p) for
any prime p dividing m:  clearly [1] holds if x = 0 (mod p);
otherwise [1] follows from Fermat's Little Theorem (FlT) as
follows:
     phi(m) + 1      p-1 q-1 ... s-1 + 1    
   x             =  x
                     p-1 q-1 ... s-1
                 = (x   )            * x
                                          p-1
                 =  x             since  x   = 1 (mod p)  by FlT
Thus since  x^(phi(m)+1)-x  is divisible by the distinct primes 
p,q,...,s  it must be divisible by their product  m = p q ... s 
(a special case of the Chinese Remainder Theorem).
Note that the above proof fails if m is not squarefree since
x^(p^(n-1)*(p-1)) = 1 (mod p^n) need not hold if p divides x,
e.g. consider x=p=n=2: 2^2 != 1 (mod 4); thus x^3 != x (mod 4).
In general, for any finite group G of size n, all elements x in G
satisfy  x^n = 1. Fermat's Little Theorem is a very special case.
These are basic results that can be found in all elementary texts
on algebra, groups, number theory, or finite fields, etc.
-Bill Dubuque

ASARI Hirotsugu wrote:
> I'm no physict or engineer, but it seems that there are lot more variables
> involved in the problem than just the rate (of what?) and the angle.  In
> other words:  No, I can't figure out.
You might be able to find the response I gave to the question, look for
it and you should completely understand the solution!
---------------------------------------------------------------------

In article <3287499b.0@news.rz.uni-passau.de>,
 schweigh@fmi.uni-passau.de (Markus Schweighofer) writes:
|> Can anybody give me an example of a finite normal field extension
|> E/K (i.e. E is a splitting of a polynomial over K) such that
|> E is not a splitting field of an irreducible polynomial over K.
|> Please mail it also to me (schweigh@fmi.uni-passau.de). Thanks! 
Off the top of my head and therefore probably not bugfree.  Find and
repair any errors before submitting this as homework ;)  :
K = F_p(X,Y)  rational function field in two variables over `the' finite
              field with p elements, p prime
E = K(X^{1/p},Y^{1/p})
              i.e. the splitting field of
              (T^p-X)(T^p-Y)  =  (T - X^{1/p})^p (T - Y^{1/p})^p
              \___in K[T]__/     \__________ in E[T] __________/
Now if F is any irreducible polynomial of degree > 1 in K[T] which has
a root in E, that root is a rational function in X^{1/p} and Y^{1/p}
with coefficients in F_p, so its p-th power is an element of K (and is
not a p-th power in K).  It follows that all subfields of E/K which are
splitting fields of irreducible polynomials over K are of degree p over
K, whereas [E:K] = p^2.
[Inspired by van der Waerden, Algebra 1, 8th ed., sect.46.  It is obvious
that E/K can't be separable;  a separable normal extension is always the
splitting field of the (irreducible) minimal polynomial of a primitive
element.  So the above is about the simplest example of an inseparable
extension with _no_ primitive element.  Unless I've made a mistake some-
where, that is.]
Enjoy, Gerhard
-- 
* Gerhard Niklasch  *** Some or all of the con-
* http://hasse.mathematik.tu-muenchen.de/~nikl/ ******* tents of the above mes-
* sage may, in certain countries, be legally considered unsuitable for consump-
* tion by children under the age of 18.  Me transmitte sursum, Caledoni...  :^/

In article <567rpd$nh@ultra0.rdrc.rpi.edu>,
Jeffrey A. Young  wrote:
>In article <561do5$2ui@gap.cco.caltech.edu>,
>Ilias Kastanas  wrote:
>>In article <5607lo$rvj@ultra0.rdrc.rpi.edu>,
>>Jeffrey A. Young  wrote:
>>>In article <55tvo6$un@sun001.spd.dsccc.com>,
>>>Mike McCarty  wrote:
>>>>In article <55lpte$osm@ultra0.rdrc.rpi.edu>,
>>>>Jeffrey A. Young  wrote:
>>
>>...
>>
>>>Not at all.  Every event (sequence) has probability zero.
>>>Only the individual events you can specify are impossible.
>>>
>>>>You are also pretty fuzzy about "fully specify". What does that mean?
>>>
>>>Well substitute "identify explicitly as an individual event before the trial
>>>occurs" then.  Obviously there are uncountably many events in the population
>>>and only a countable number that anyone could specify individually
>>>beforehand.
>>>
>>>>If I specify the class of events which can occur, then I can certainly
>>>>predict what the outcome of the trial will be. It will be one of the events.
>>>
>>>But you cannot specify all the events in the class individually.
>>>That is the difference.
>>
>>
>>	This notion of 'impossible' does not seem to make sense.  You are
>>   granting the feasibility of infinitely many trials.  In any single trial
>>   both H and T are possible outcomes.  So how can any infinite sequence of
>>   H's and T's be "impossible"?  
>
>How can you claim to know anything about what happens at infinity?
	You made such a claim in the first place.
	A good part of mathematics studies infinity.
>>What difference does it make whether it can be "fully specified" or not?
>
>It was intended as one possible counter-argument.
	To what?  Sorry if I missed an argument of yours, but I didn't see
   any; just the assertion that some  sequences are impossible.
>>	Any one of  H, HH, HHH ... is surely possible; what does it mean,
>>   then, to say "HHHH... is impossible"?  Same for any infinite sequence;
>>   all its finite initial segments are possible.  Stating a priori that even
>>   one particular infinite sequence is "impossible" contradicts the assumption
>>   about single trials; it sounds like a conspiracy theory, or animism.
>
>Uhhh, I don't see anything about "infinity" in the assumption
>about single trials.  Aren't you using other assumptions as well?
	The one I stated above, that infinitely many single trials are possi-
   ble.  You apparently agree with that.  So why is it impossible for all to
   come out H?
	In another posting you stated your claim is _consistent_.  That is a
   retreat from saying it is true (parallel postulate, etc)... but I'm afraid
   it isn't even that.  It is fine to define a uniform probability on some
   subset of [0, 1], and have everything else a priori impossible.  But when
   you consider infinite sequences of H and T it is plain logical error to
   assert that some are not there.
							Ilias

jmccarty@sun1307.spd.dsccc.com (Mike McCarty) wrote:
>In article <01bbcf0b$ca153d00$310296c2@club-internet.club-internet.fr>,
>gdm  wrote:
>)If x is very big and e very small, I want to write
>)Log(x+e)=Log(x)(1+a*e+be^2+c*e^3+....)
>)what are the values of a, b, c ?
>)thanks, 
>)Gilles de Montety, Paris, France
>
>
>Looks suspiciously like a homework problem. But still,
>
>Log(x+e) = Log[x(1+e/x)] = Log(x)+Log(1+e/x) = 
>Log(x)+[e/x - (e/x)^2/2 + (e/x)^3/e - ...]
= Log(x) ( 1+  e/xlnx  - e^2/2x^2lnx  .... )
Thanks goodness! That agrees with my power series I got by repeated
differentiation, although your method is a lot quicker.
Simon

Dean Povey wrote:
> 
> dean@psy.uq.oz.au (Dean Povey) writes:
> Here is an extract:
> =====>
> SR and AD Comparison
> 
> The general relativity equation for advance of the Mercury perihelion is:
> 
>                  6 pi GM
>           T  =  ----------           [Pardon my ascii, DGP]
>               c^2 r (1 - e^2)
> 
> Where e = eccentricity, c = light speed and G, M, and r have the usual
> meaning in this paper.
> 
> This equation yields, in a century:
> 
>                   42.4" for Mercury
>                      8" for Venus
>                      4" for Earth
>                      1" for Mars
> 
> In AD gravitation, the perihelion advance for each planet is
> proportional to the square root of the division of the solar mass by
> the orbital radius power 3.
radius? by this you mean *mean distance*, right?
> 
>              Tp = sqrt(M / r^3)      [ditto: DGP]
> 
> If the Mercury value is taken as 43", the values for the other planets are:
Please tell me you're not required to take Mercury's observed advance
as input.  If so, the oft-quoted claim that AD predicts Mercury's 
perihelion advance is a load.  And I don't mean a load of bananas.
Cheers,
Todd
------------------------------------------------------------------
Todd K. Pedlar   -  Northwestern University - FNAL E835
Nuclear & Particle Physics Group
------------------------------------------------------------------
Phone:  (847) 491-8630  (708) 840-8048  Fax: (847) 491-8627
------------------------------------------------------------------
WWW:	http://numep1.phys.nwu.edu/tkp.html
------------------------------------------------------------------