Newsgroup sci.math 157849

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SAPDMCPC22@ wrote:
: 
: 
: The different infinity cardinalities don't matter at all here.
:  'oo' is (here) the description of a "point" at infinity to get a closured topological 
: real space.
But the set of all real numbers with its (usual) Euclidean
topology is already a closed real space. No additional points
are needed!
d.A.

> ilias kastanas 08-14-90  wrote:
>
>       I just see no reason for empty domains;  contortions, simple vali-
>   dities like  Ax P(x) -> Ex P(x)  fail, etc.  (This was discussed some
>   months ago).
I missed the original discussion. I have found that for a lot of
mathematical modelling in computer science it is more useful to use
many-sorted first-order logic instead of pure first-order logic. For
many-sorted logic, one never simply writes Ax P(x), one instead writes
Ax:T P(x) ("for all x of type (or sort) T, P(x)"). If the mechanisms
available for type-formation are powerful enough, it becomes impossible
to prevent the possibilities of empty types.
Daryl McCullough
CoGenTex, Inc.
Ithaca, NY

James Davenport (masjhd@bath.ac.uk) wrote:
: Yes, there is a sad mathematical fact here.
: The standard real numbers are a subset of the standard complex numbers,
: but this is NOT TRUE of their compactifications (which is really what we
: are talking about here), since R needs a 2-point compactification (+oo and
: -oo  
"Needs"?  There is a standard construction for a compactification
of any locally compact Hausdorff space by adding a single "point
at infinity" and defining a suitable topology. If the space was 
already compact before the construction, then this "one-point
compactification" is in fact (homeomorphic to) the topological
co-product of the space with a singleton.
In the case of R^n (n>0), this construction leads to (a space
hoeomorphic to) the n-sphere S^n. One way to see this is via
stereographic projection.
What about other compactificiations, such as Stone-Czech
or Bohr compactifications?
: whereas C needs a 1-point compactification (projective oo).
: To my mind, this sad mathematical fact means that hardware should concentrate
: on standard R (which can therefore be extended to standard C in the cartesian
: way) and leave the interpretations of infinity to software, since no one
: interpretation will suffice for both R and C (unless we also build C and
: its 1-point compactification in).
: 
In fact, the standard (Alexandrov) compactification mentioned above
when applied to the (real) plane of complex numbers (as distinct from
the complex plane C^2!!!!!) yields the Riemann sphere. Embedding
R in C as the complex numbers invariant under complex conjugation
leads to an embedding of the 1-point compactication of R into the
Riemann sphere as a "longitudinal great circle".
d.A.

Please e-mail me anything you would be willing to contribute to a web
site.

 Note followup to sci.physics; this is not math. 
"Risto Lankinen"  writes:
>
>Doesn't this also mean that the definition of meter is at most as
>accurate as the definition of second?  
 And the means of measuring the speed of light.  
 The shift from the use of a physical artifact (a platinum-iridium 
 bar in Paris) to the present system was made because those 
 measurements (c and a frequency) can be made much more accurately 
 (better than a part in 10^9) than any comparison of two bars 
 of metal.  There is a similar effort underway to replace the use 
 of a lump of metal for the mass unit. 
-- 
 James A. Carr        |  "The half of knowledge is knowing
    http://www.scri.fsu.edu/~jac/       |  where to find knowledge" - Anon. 
 Supercomputer Computations Res. Inst.  |  Motto over the entrance to Dodd 
 Florida State, Tallahassee FL 32306    |  Hall, former library at FSCW. 

In article <32d95617.0@eclipse.wincom.net>, clayton.smith@mustang.com
(Clayton Smith) wrote:
>The problem is as follows.
> 
>  lim   (2^3-1)(3^3-1)(4^3-1) ... (n^3-1)
> x->oo  --------------------------------- = ?
>        (2^3+1)(3^3+1)(4^3+1) ... (n^3+1)
First, split the factors (k^3-1) into factors,
k^3-1 = (k - 1) * (k - y) * (k - y^2),
where y=(-1+I*sqrt(3))/2 is a complex third root of unity, and y^2 too;
similarly,
k^3+1 = (k + 1) * (k + y) * (k + y^2).
Now a product like (2 - a) * (3 - a) * ... (n - a) can be
written in terms of the Gamma function as Gamma(n+1-a)/Gamma(2-a).
Do this for the (k-1), (k-y) etc. factors. Next, take the limit as
n tends to infinity by the Stirling formula, and perform minor wizardry
to kill off the y's in the result (by the formula linking Gamma(x)
with Gamma(1-x), and by Gamma(complex conjugate) = complex conjugate of Gamma.
The result is 2/3.
Maple V r. 4 does all of this this for you:
> product((k^3-1)/(k^3+1),k=2..n);
                                  1/2
2 GAMMA(n) GAMMA(n + 3/2 - 1/2 I 3   )
                           1/2                     1/2
    GAMMA(n + 3/2 + 1/2 I 3   ) GAMMA(3/2 - 1/2 I 3   )
                       1/2    /
    GAMMA(3/2 + 1/2 I 3   )  /  (GAMMA(n + 2)
                            /
                           1/2
    GAMMA(n + 1/2 - 1/2 I 3   )
                           1/2                     1/2
    GAMMA(n + 1/2 + 1/2 I 3   ) GAMMA(5/2 - 1/2 I 3   )
                       1/2
    GAMMA(5/2 + 1/2 I 3   ))
> limit(",n=infinity);
                                               8
                                 - --------------------------
                                            1/2          1/2
                                   (-3 + I 3   ) (3 + I 3   )
> evalc(");
                                             2/3
--
Dr. Denis Constales - dcons@world.std.com - http://cage.rug.ac.be/~dc/

In article <5bf18a$p22@kew.globalnet.co.uk>,
  jimmyp@premier.co.uk wrote:
> 
> Can anyone assist with this number arrangement problem?
> 
> Given n consecutive numbers, 1,2,3,...n the problem is to arrange two
> identical sets of the numbers so that each pair of numbers k enclose
> just k other numbers.
> 
> For example, if n=3, then there are two sets of the numbers 1,2,3 to
> be used, and a solution is:
> 
> 3 1 2 1 3 2
> 
> since there are 3 numbers between the two 3s, 2 numbers between the
> two 2s, etc.
> 
> For n=7 a solution is:
> 
> 2 3 7 2 6 3 5 1 4 1 7 6 5 4
> 
> It is possible to find solutions for n=3 and n=4, but not for n=1,
> n=2, n=5, n=6.
> 
> It is possible to find solutions for n=7 and n=8, but not for n=9 and
> n=10.
> 
> It is possible to find solutions for n=11 and n=12, but not for n=13
> and n=14.
> 
> And so on, with each alternate pair of values for n giving possible
> and impossible solutions, respectively.
> 
> Why is this?
> 
> Also, there is always more than one solution for each value of n, the
> number of solutions increasing as n increases.
> 
> Can anyone find a formula which gives the number of possible solutions
> for a (valid) value of n?
> 
I can show non-existence for n = 3, 4, 7, 8, 11, 12 etc. as follows.
In a given solution let a_j and b_j denote the positions of the first and
second j s for j from 1 up to n. For instance in
	3 1 2 1 3 2
then a_1 = 2, b_1 = 4, a_2 = 3, b_2 = 6, a_3 = 1 and b_3 = 5. In general then
b_j - a_j = j+1. Let A and B be the sums of the a_j s and b_j s respectively.
Then
	A + B = 1 + 2 + ... + 2n = n(2n + 1)
since together the a_j s and b_j s run through all numbers from 1 to 2n. But
	B - A = b_1 - a_1 + b_2 - a_2 + ... + b_n - a_n
	      = 2 + 3 + ... + (n + 1) = n(n + 3)/2 .
But A + B and B - A differ by an even number so they are both even or both odd.
But n(2n+1) has the same parity (evenness/oddness) as n, while n(n+3)/2
is odd whenever n leaves remainder 2 or 3 on division by 4 (and even otherwise).
It follows that if n leaves remainder 1 or 2 on division by 4, then A + B
and B - A have opposite parity, which is impossible.
I can't immediately see a way of finding a solution in the other cases, but
solutions must come in pairs, since reversing the order of each solution
gives a different solution.
Robin J. Chapman                        "... needless to say,
Department of Mathematics                I think there should be
University of Exeter, EX4 4QE, UK        more sex and violence
rjc@maths.exeter.ac.uk                   on television, not less."
http://www.maths.ex.ac.uk/~rjc/rjc.html         J. G. Ballard (1990)
-------------------==== Posted via Deja News ====-----------------------
      http://www.dejanews.com/     Search, Read, Post to Usenet

Jan Zumwalt wrote:
> 
> I'm sorry if I am way off base and intruding. I only caught a small glimpse
> of your discussion but...
> 
> I saw an article about 5 years ago in a scientific magazine written by a
> NASA mathematician paid to consider the problems encountered in near light
> speed travel. The gist of the article included a computer program that
> showed what objects would look like at various speeds.
> 
> The most interesting aspect to me professionally was the observation of
> electrical circuit response at those speeds. For instance at about .75c
> computer response of electronic equipment would become prohibitively slow
> for connections in access of 300ft. Based on your discussion I think this
> Conflicts to one of your opinions but I'm not sure.
> 
-snip-
I don't think the article you read was a very serious one. As stated in
the postulats of theory of relativity, there's no way to distinguish
between any two inertial systems. If the computer is not moving with
respect to an inertial frame of refference, it has to perform the same
as on Earth or anywhere else. No slowing down or whatever.
-- 
\    / | --+--   | | |-\   /\   |-\ |-- /-\ | / \ /
 \  /  |   |     |-| |_/  /--\  | | |-  |   |/   |
  \/   |   |     | | | \ /    \ |_/ |__ \_/ | \  |
vitek@geocities.com

In article , dc@cage.rug.ac.be (Denis
Constales) wrote:
> First, split the factors (k^3-1) into factors,
> 
> k^3-1 = (k - 1) * (k - y) * (k - y^2),
> 
> where y=(-1+I*sqrt(3))/2 is a complex third root of unity, and y^2 too;
> similarly,
> 
> k^3+1 = (k + 1) * (k + y) * (k + y^2).
> 
> Now a product like (2 - a) * (3 - a) * ... (n - a) can be
> written in terms of the Gamma function as Gamma(n+1-a)/Gamma(2-a).
UPDATE on this solution: you don't actually need the Gamma and its
asymptotics here (but you do in more general problems of this type). In
fact, the factorisation rewrites the product as
(2-1)(2-y)(2-y^2)(3-1)(3-y)(3-y^2)...
-------------------------------------
(2+1)(2+y)(2+y^2)(3+1)(3+y)(3+y^2)...
where y is a complex 3rd root of unity; but then y^2=-1-y, so you can
simplify the product to
(2-1)(2-y)(3+y)(3-1)(3-y)(4+y)...
---------------------------------
(2+1)(2+y)(1-y)(3+1)(3+y)(2-y)...
and careful inspection shows that every factor cancels out (you must follow
the (n-1), the (n-y) and the (n+y) strands separately) except for
(2-1)          (3-1)
---------------------------------
     (2+y)(1-y)
and this works out to
                2
---------------------------------
     9/3
i.e. 2/3.
--
Dr. Denis Constales - dcons@world.std.com - http://cage.rug.ac.be/~dc/

Ilias Kastanas wrote:
> 
> In article <32D45D77.25C@james.brus.wec.com>,
> deboeck.f  wrote:
> >Hi there,
> >
> >I'm stuck with the following problem.
> >
> >I have a positionner (that's used to move valves) which law of
> >displacement obays to:
> >
> >let s = speed
> >    x = displacement
> >
> >ds/dx = constant say A for 0 < x < x1
> >
> >Boundary conditions
> >s and x = 0 for t = 0
> >s = smax for x = x1
> >
> > S|       ___________ s max
> >  |      /
> >  |     /
> >  |    /
> >  |   /
> >  |  /    A = smax / x1
> >  | /
> >  |/
> >  +------+----------------
> >  0      x1               x
> >
> >I try to find both s and x versus time t.
> >
> >General solution
> >----------------
> >I tried to introduce dt in the equation:
> >
> >(ds/dt)*(dt/dx) = A or (ds/dt)*1/s = A
> >
> >A general solution is s(t) = C1 * [exp(A*t) +C2]
> >
> >Particular solution
> >-------------------
> >
> >The trick is to define the constants C1 and C2.
> >
> >I can write that s(t=0) = 0 = C1 * [exp(A*0) +C2] = C1 * [1+C2]
> >
> >As C1 is expected to be related to smax (thus C1 not null), C2 = -1 and
> >
> >s(t) = C1* [exp(A*t) - 1]
> >
> >
> >Now what about C1? I know that s = smax for x = x1. If I find the time T
> >for which x = x1, I can put T in s(t) and try to solve s(T) = smax so I
> >find C1. Unfortunately, I don't find the solution.
> >
> >Indeed, x(t) = integral (s(t) * dt)) = C1 * [(1/A) *exp(A*t) - t]
> >between t = 0 to t
> >
> >For t = T I have x(T) = x1 = C1 * [(1/A) *(exp(A*T) - 1) - T] = C1 *
> >f(T)
> >
> >There is no analytic way to extract T (and a solution exits only for 0 <
> >T < 1/A) but I try to replace C1 with the expression above in the speed
> >equation, which is evaluated for t = T and for which I know s(T) = smax
> >
> >s(T) = x1/f(T) * [exp(A*T) -1] = smax
> >
> >
> >The problem is that I find T = 0 which is obviouly wrong. So, where did
> >I fail?
> >
> >A possible root of confusion is that I have 2 constants C1 and C2 and I
> >have at least 3 boundary conditions:
> >2 for t = 0 where both x and s = 0
> >1 for s = smax for x = x1
> 
>         Yes; and if s were 0 at t=0, it would stay zero!
> 
>    Integrating, x (not s) = C1 (e^At + C2); x(0) = 0 gives  x = C1 (e^At - 1).
>    Then  s = A C1 e^At.   Find the t = T when x = x1 and apply s = s_max:
> 
>         s = (s_max - A x1) e^At,   x = (1/A) (s_max - A x1) e^At.
> 
>    So  s_max - A x1 had better be > 0, not = 0 !  It is  s  at t = 0.
> 
>    For T we find  A T = - log(1 -  A x1/s_max )  or, with  q = A x1/s_max,
> 
>         T  =  (x1 /s_max) (1 + 1/2 q  +  1/3 q^2  + ... )
> 
>                                                         Ilias
Ilias,
Thanks for answering.
I bypassed the difficulty by assuming non-zero positon and speed at time
0 by asssuming s0= smax/100 and x0=xmax/100. Then I solved (with the
exopnential we both found) the problem analytically for s and x > s0,
x0.
Although this seems to be an easy physical problem, it gave headaches to
my collegues here. The root could be that t=s=x=0 is a singularity (or a
pole if using Laplace Transform) and could not be estimated
mathematically, unless one considers to introduce the acceleration
(non-nul, otherwise nothing happens!) in the game. By considering the
physics (introducing the acceleration), one finds a Dirac function at t
= 0 for position when using Laplace functions. This problem exists (I
think) also if another displacement law than linear is used.
Thanks again
Frederic

In article <32d957c8.0@eclipse.wincom.net>, clayton.smith@mustang.com
(Clayton Smith) wrote:
> I have a question regarding the uses of calculus.  Is it possible to
> determine the surface area of a function in two parameters over a
> given range?  For example, would it be possible to find the surface
> area of the paraboloid given by:
> 
> y = f(x,z) = x^2 + z^2
> 
> over the range 0<=x<=1, 0<=z<=1.  If this can be done, what is the
> general procedure I would use to determine this or any other surface
> area? 
You integrate over that x,z range the square root of 1 plus the squared
partial x derivative plus the squared partial z derivative, i.e.
sqrt ( 1 + (df/dx)^2 + (df/dz)^2 ),
read curly d for the partial derivative here.
The proof relies on determining the direction of the normal vector to the
surface and thereby its angle with the vertical, cf. any calculus textbook,
or e.g. the pale green Schaum book on Calculus, etc. (Mentioning that one
because it's there I first encountered the formula.)
--
Dr. Denis Constales - dcons@world.std.com - http://cage.rug.ac.be/~dc/

Nick Johnson-Hill (nickjh@globalnet.co.uk) wrote:
> : Is it possible to have factorials of non-integers ? I was always taught
> : that the answer is NO but I found the following on a Mathematics web
> : page:
> 
> : (1/2)! = Square root of Pi
Bad page... (1/2)! = Gamma(3/2) = sqrt(Pi)/2.
The Gamma function 
                       infinity
                      /
                     |           (-t)  (x - 1)
         GAMMA(x) =  |          e     t        dt
                     |
                    /
                      0
is a continuous version of (x-1)!, and the only smooth function to be 
1. logarithmically strictly convex for x>0 (i.e. the second derivative of
its log is strictly positive over x>0),
2. to satisfy f(x+1)=x*f(x) for all x>0, and
3. f(1)=1.
--
Dr. Denis Constales - dcons@world.std.com - http://cage.rug.ac.be/~dc/

kfoster@rainbow.rmii.com (Kurt Foster) wrote:
} 
}   "I am a HAL Nine Thousand computer Production Number 3.  I became
} operational at the HAL Plant in Urbana, Illinois, on January 12, 1997."
} --  "2001 a space odyssey" -- a novel by arthur C. Clarke
borism@interlog.com (Boris Mohar) writes:
>
>  And if you shift right the letters HAL you will get his dady's name.
 Anyone who knows computers knows there are not many from that 
 company on the list of super-fast ones, particularly in the 60s. 
 If you recall the number scheme used by CDC, wherein the Cray 1 
 would have been the 8600, you are a bit closer, since that would 
 have put a 9000 series in the 1990s.  However, the model machine 
 was the Illiac IV (note the place of origin), which once sung 
 "Daisy".  [Thanks, CNN, for that factoid.]
-- 
 James A. Carr        |  "The half of knowledge is knowing
    http://www.scri.fsu.edu/~jac/       |  where to find knowledge" - Anon. 
 Supercomputer Computations Res. Inst.  |  Motto over the entrance to Dodd 
 Florida State, Tallahassee FL 32306    |  Hall, former library at FSCW. 

rjc@maths.ex.ac.uk wrote:
> I can show non-existence for n = 3, 4, 7, 8, 11, 12 etc. as follows.
Very nice! but shouldn't that be "non-existence for 1,2,5,6,9,10 etc.",
instead?
-- 
D.
mentock@mindspring.com
http://www.mindspring.com/~mentock/index.htm

Patrick De Geest (Patrick.DeGeest@ping.be) writes:
>  wrote:
>> I'm always looking for all sorts of information about palindromic
>> numbers. I devoted a whole website to these fascinating and beautiful
> -----
>> Record_nr(?) for palindromic squares with NONpalindromic base (sometimes 
>> called 'root')
>>
>>	  ( 306.950.094.269.977.057 ) ^ 2 =
>>         94.218.360.372.347.802.120.874.327.306.381.249 [length 35]
>>
>>         This one looks great !. How did you find it ?: just by trial-error
>> testing or is it there a program which shortens the way ?.
> 
> Certainly not by trial-error testing ! I wrote a MASM assembly language
> program which now still runs in a DOS box on my WIN95 pc. It all started
--------
> fascinating and appealing than 'palindromic subsets' of some irrational
> numbers ? Or do you seek other perhaps more practical things rather than
> the beauty of palidromic numbers ?
> 
    By no means !: palindromes are beautiful no matter what; but SIMPLICITY
is the climax of beauty. Thus if one can get 14 or 20 digit palindrome
out of the SIMPLE square root of a 3 or 4 digit integer then such combination
would be much more appealing and beautiful, no ?. The DIFFICULTY in such
search is that your intelligent program (and probably ANY other program)
would fail to shorten the search... correct me, please, if you know better:
thus the SUPREME beauty of something SIMPLE AND UNPREDICTABLE would concurr
with palindromism to create really astonishing conundrums.
Nontheless your findings so far MUST be kept on record, of course, as
important intermediates between plain conundrums and analytic
quasi-conundrums like that one of Ramanujan related to 163^(1/2).
--
Angel, secretary of Universitas Americae (UNIAM). His proof of ETI at
Cydonia and complete Index of new "TETET-96: Faces on Mars.." by Prof.
Dr. D.G. Lahoz (leader on ETI and Cosmogony) can be studied at URL:
     http://www.ncf.carleton.ca/~bp887    ***************************

Michael A. Stueben wrote: 
> Yeah, there is a bunch of joke proof methods listed somewhere: 
> 1. Proof by hearsay: "Prof. Rubin told me it was true."
>  
> ---
I've heard it said that if one can get three Geometers to
nod their heads at your proof then it is correct.

Peter Diehr wrote:
> > Black Holes in the GR sense remain hypothetical.
> 
> You haven't been following the news very closely, have you?
There is still no positive, undeniable evidence that a black hole exists.
I think it's safe to say that most physicists are pretty sure there exist,
an we have some convincing candidates, but there isn't quite the degree of
certainty yet that would warrant your objection.
-- 
                             Erik Max Francis | max@alcyone.com
                              Alcyone Systems | http://www.alcyone.com/max/
                         San Jose, California | 37 20 07 N 121 53 38 W
                                 &tSftDotIotE; | R^4: the 4th R is respect
     "You must surely know if man made heaven | Then made made hell"

I'm trying to figure out the equation for a cone that is not
symetrical about a cartensian axis (i.e. rotated about one or more axes)
Every book I've seen has equations for symetrical cones.  Can anyone
give me some ideas how this is done?
Thanks,  David

Sergio Marra wrote:
> 
> I am trying to output some numbers into a file, but the command Print seems to
> work only to the screen.
> I do something like
> 
> f[i_]=If[expression, Print[i], 0]
> Do[ f[x], {x, 0, 10^7,1 } ]
> 
> I tried with a redirection after Do or even after if but does not work.
> 
> Any idea? Thanks in advance
Have you used the command Put?
-- 
D.
mentock@mindspring.com
http://www.mindspring.com/~mentock/index.htm

Paul Schlyter (pausch@electra.saaf.se) wrote:
: In article ,
: David Kastrup   wrote:
:  
: > Which is why oo is *not* a real number.  All expressions involving
: > real numbers are either undefined, or equal exactly one real number.
:  
: You mean expressions like:
:  
: sqrt(4)         =  +2 or -2
:  
: sqrt(sqrt(16))  =  +2, -2, +2i or -2i
:  
: arctan(1)       =  pi/4 + n*pi/2    where n is any integer
I guess he meant _algebraic_ expressions. By defining a non-algebraic
function, you  can of course map real numbers to anything you like.
However a function value, whatever it is, must be uniquely defined.
Therefore you should write the value of the above "multivalued" functions
as  sqrt(4) = {-2,2} etc. (i.e. the value is uniquely defined by a set),
in order to be precise.  
Of course such a precise notation becomes soon very awkward. Therefore
humans tend to the sloppier notation. Nonetheless it is just a sloppy
notation and not the real thing. Computer algebra systems _must_ deal with
the real thing to produce correct results (e.g. if you are familiar
with Maple, look at the semantics of the "RootOf"-function).
-- 
Ulrich Lange                       Dept. of Chemical Engineering
                                   University of Alberta
lange@gpu.srv.ualberta.ca          Edmonton, Alberta, T6G 2G6, Canada

I have a bunch of small (4-10 points) data sets (x vs y), gathered from
experimental data.  All can be very accurately approximated by a cubic. 
However, the curvature of the lines vary--some are very linear.  What I
need is an accurate measure of the curvature of the lines, so I can
compare them to each other.  (What I want to do is see if any other
external variables effect the curvature of the x-y plot.)
Thanks,
-Paul

ibokor (ibokor@metz.une.edu.au) wrote:
: SAPDMCPC22@ wrote:
: : 
: : The different infinity cardinalities don't matter at all here.
: :  'oo' is (here) the description of a "point" at infinity to get a closured topological 
: : real space.
: 
: But the set of all real numbers with its (usual) Euclidean
: topology is already a closed real space. No additional points
: are needed!
What he meant is "compact space". BTW: The word "real" in "closed real
space" makes no sense.
-- 
Ulrich Lange                       Dept. of Chemical Engineering
                                   University of Alberta
lange@gpu.srv.ualberta.ca          Edmonton, Alberta, T6G 2G6, Canada

Martijn Dekker wrote:
> 
> pausch@electra.saaf.se (Paul Schlyter) wrote:
> 
> :In article ,
> :David Kastrup   wrote:
> :
> :> Which is why oo is *not* a real number.  All expressions involving
> :> real numbers are either undefined, or equal exactly one real number.
> :
> :You mean expressions like:
> :
> :sqrt(4)         =  +2 or -2
> :
> :sqrt(sqrt(16))  =  +2, -2, +2i or -2i
> :
> :arctan(1)       =  pi/4 + n*pi/2    where n is any integer
> :
> :?????
> 
> indeed. sqrt(4) is not a real number unless you fix one particular
> branch (i.e. choosing sign). sqrt is not a one-valued function.
sqrt(4) = 2 
4^(1/2) = +/-2
> --
> Martijn Dekker
> UvA - Math department Universiteit van Amsterdam
> PFF - Computer Games
> http://turing.fwi.uva.nl/~mdekker/pff/  (Xcogitate, Templar)
Wilbert Dijkhof

David Debes wants "a cone that is not symetrical around a cartensian axis.
 ... Can anybody give me some idea how this is done?"
First, gently, we say, "symmetrical" and "Cartesian."
Now, I expect you want a symmetrical cone ( rather than lumpy ) whose
center line is not on a Cartesian axis.  Instead, it is shifted or tilted
or both.
There are two basic ways to change Cartesian coordinates.  Shifting is
easiest, while rotation gets complicated without matrices.  Rather than
introduce matrices, I will do this step-by-step.  ( If you want to be
efficient, computer graphics texts can show you how to do everything in
one step, but the equations are messy. )
Let's agree that capital letters are for the new coordinates and
lower-case letters are for the original coordinates.  A shift that moves
the old origin to point (a,b,c) is:
X = x + a             also given as:         x = X - a
Y = y + b                                         y = Y - b
Z = z + c                                          z = Z - c
You substitute the new expression for each x, y and z in the old formula. 
To be concrete, let's make (a,b,c) the point (1,2,3) and say the old
formula was:
z^2 = x^2 + y^2           ( a cone surface growing up the z-axis )
so the new formula is:
(Z-3)^2 = (X-1)^2 + (Y-2)^2
Rotations can be decomposed into rotation around the x-axis, then the
y-axis, and finally the z-axis.  You can shift before or after rotating,
or both if you want a real mess.  I'll stick to a clockwise rotation of 30
(degrees) around the z-axis.
X = x * cosine(30) - y * sine(30)
Y = x * sine(30)   + y * cosine(30)
Z = z  
And then you substitute into your equation as before.
I hope this is the information you need.
Patrick T. Wahl
( no institutional affiliation )

Jeffrey Rubin wrote:
> 
> In trying to complete an exercise (for my own study) in Rudin's Functional
> Analysis, I need to show the following:
> 
> Let K be an arc on the unit circle, U is the open unit disc (in C) and
> D (well, it should be called \bar{U}) is the closed unit disc.  Let
> f be a continuous (complex-valued) function on K.  Then, given epsilon,
> there is a function, g, which is holomorphic in U and continuous on D
> such that the restriction of g to K is within epsilon of f on all of K.
	First you need to suppose that K is a proper subarc of the unit
circle - I'm not certain whether the circle itself is an "arc", but the
result is certainly false in that case. (Exercise: If K is the unit
circle and f(exp(it)) = exp(-it) then you're sunk.)
	Supposing that, the result is immediate from Mergelyan's theorem
(which you can find in Real & Complex Analysis, among other places.)
	But in fact this is much less deep than Mergelyan's theorem. If 
you could show that there exists a function holomorphic in a neighborhood 
of K which approximates f within epsilon on K then you'd be done by Runge's
theorem - in fact you'd get polynomials approximating f on K from
Runge's theorem. A cheap way to get a function holomorphic in a
neighborhood of K which is within epsilon of f on K would be 
Fourier series:
	You can extend f to a continuous function on the entire circle
any way you want, and then approximate that by a trigonometric
polynomial. Now you have numbers a_(-N), ... a_N  such that the sum
from -N to N of a_n*exp(int) is within epsilon of f(exp(it)) for
exp(it) in K. But that trigonometric polynomial is in fact the restriction
to the circle of a function holmorphic in a neighborhood of the circle
(in fact in the plane minus the origin), since exp(it) is the restriction
of z to the circle and exp(-it) is the restriction to the circle of 1/z .
-- 
David Ullrich
?his ?s ?avid ?llrich's ?ig ?ile
(Someone undeleted it for me...)

In article <5bc184$q32$1@dartvax.dartmouth.edu>, Archimedes Plutonium
 writes
>   If the US had been a parliamentary form of government where all
>politicians are elected and not these cabinets that linger from one
>administration to another and really run the government. Then,
>hypothetically, is it  highly likely that the Vietnam War would have
>never occurred? Or if it had, would not a parliamentary form of
>government gotten the US out quicker? One can argue that the US Vietnam
>War was chiefly the result of foolish advisors to the president.
>
Interestingly enough, while the US was busy getting bogged down in
Vietnam, the UK was engaged in fighting in Borneo, in remarkably similar
political situations. The UK military position wasn't as good as that of
the US; the Borneo border was massively longer than that which the
Americans had to deal with, and the terrain very much harder.
Nonetheless, the UK was successful.
One can debate why this should be; however, there was no great "anti-
war" debate in the UK. I suspect that this was in part because of
different attitudes.
>  Perhaps this is a great research inquiry as to see which form of
>democracy is superior-- the US or the UK parliamentary.
>
>  In a parliamentary system, the likelihood of foolish advisors doing
>so much damage is minimized, I suspect.
>
If only. You don't live in Britain, do you?
>  Same thing in mathematics, where math is run by the old geezers who
>control the math journals. They print and publish the pipsqueak little
>progress. And they do their utmost best to keep out anything that is
>big, new and exciting and important.  In fact, they mostly publish that
>which furthers their own self interests or
>you-rub-my-hand-I-rub-your-hand.
>
>  The clowns that got the US into Vietnam are the same sort of
>intellectual clowns that control the mathematics publishing journals
>and who hate an idea such as    Naturals = P-adics = Infinite Integers.
Ah, paranoia. Isn't it wonderful.
-- 
Alison Brooks  
O-

Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) wrote:
>   If the US had been a parliamentary form of government where all
>politicians are elected and not these cabinets that linger from one
>administration to another and really run the government. Then,
>hypothetically, is it  highly likely that the Vietnam War would have
>never occurred? Or if it had, would not a parliamentary form of
>government gotten the US out quicker? One can argue that the US Vietnam
>War was chiefly the result of foolish advisors to the president.
>
>  Perhaps this is a great research inquiry as to see which form of
>democracy is superior-- the US or the UK parliamentary.
>
>  In a parliamentary system, the likelihood of foolish advisors doing
>so much damage is minimized, I suspect.
>
While I would be the first to argue that there are serious problems with
what passes for a democratic system of government in the US, I would have
to point out that while it is true that the entire government in the UK
changes on a fairly regular basis, much of the UK's foriegn policy is
dictated by career diplomats. Unlike the US system where, as I understand
it, the President can appoint anybody to be an ambassador, all diplomats 
in the UK are
civil servants, who have actually received training for the job that they
do. Ambassadors especially would have had many years of experience before
reaching their posts. Although the government has some influence in who
gets promoted and who doesn't, there are no such things as political 
appointees. John Major (or Tony Blair if you are reading this a week or
two from now ;)  ) cannot say that he wants X random person from outside
the service to be ambassador to Mongolia. He can choose one person from a
list of suitable people, but thats about it.
The point I'm trying to make is that the civil service is far more
powerful in the UK than in the US, and this leads to a certain amount of
stability/stagnation from government to government, particularly in
foreign affairs and trade relations (if indeed there is a distinction
between the two), and this compensates for the more frequent change at
the helm, so to speak.
In a parlimentry democracy, there seems to be a need for a conservative
civil service to keep the whole show on the road from year to year; the
politicians do all the talking and the civil servants really run the
country. Although it can be argued that the Vietnam conflict was prolonged
by bad advice, where did this advice come from. Was it from ambassadors
and civil servants, or was it from the military/industrial complex as
Mr. Stone would argue? If the later is the case would a parlimentry system
really have made a difference?
As for comparing the two systems, the US is not a true democracy. Because
of the Electoral College system, the President is not actually elected
by the citizens themselves. The electoral college of a state votes
for the President based on the recomendations of the people of the state,
but that can by no means be called one person, one vote. A citizen votes
to mandate a representative to vote a certain way, but that citizen
doesn't directly vote to put the president in the White House. CNN also
has alot to answer for, while people in the West Coast and Alaska and
Hawaii are still voting, CNN was announcing results from the East Coast
and declaring an outcome, which obviously discouraged voters in the West
from going to the poles. In the UK and here in Ireland, no results can
be announced until all polling stations have closed (yes even though we
are very small, outlying islands vote on different days). Yes there are
exit polls conducted, but these only give an indication, not a result.
But I digress, 'cause CNN wouldn't have been a major factor in the
Vietnam conflict. 
The UK system of Parlimentry Democracy isn't all that democratic either,
based on a 'first past the post' system. Each constituancy has but one
represntative, and the person with the most votes wins. This means that 
even if the majority of voters voted against a candidate, they can still
be elected if they get more votes than anyone else individually. It is
therefore almost impossible for candidates from the smaler parties, the
Greens, the Liberals, Natural Law or Monster Raving Looney Party, to
ever get anyone elected. It also has a detrimental effect on candidates
running as individuals, rather than on a party ticket.
Here in Ireland, as in other true democracies ;) we operate on a
Proportional Representation system. This works by having more than one
Representative per constituancy, usually between 3 and 5. The total
valid poll is calculated and a quota is tallied, I think its 50% +1,
but I'm not sure. When voting, people vote for the candidates in order
of prefernce, for example, Bill Clinton first, Ross Perot second and
Bob Dole third. They then tally all the first prefernces. If anybody
has reached the magical 50%+1 quota on the first count, they are elected.
What happens then is the distribution of their surplus. All their second
preferences are given a certain value, and added onto the results from
the first round of those who still remain. This goes on until all the
vacancies have been filled. If no-one reaches the quota during any given
round, they eliminate the person with the lowest tally, and distribute
their other prefrences between the remaining candidates. Yes, it is
a complicated system, and counts usually go on well into the night and
next day, particularly if, as they usually do, an eliminated candidate
asks for a recount. What this does mean, however, is that there are a
wide range of people elecetd from many different parties, or none at all,
and the Dail, our parliment, is actually representative of the way people
actually voted. Yes this means that Coalition Governments are the norm,
but this means that the views of the majority of the population are
actually reprented in Government. At the moment our Government is a
coalition of three parties, centre-right, centre-left, and left, the
last government was right and centre-left. Mix and match is the name
of the game.
But back to the point, a UK style parlimentry government would not, in
my opinion, have finished in Vietnam any earlier than the US did in OTL,
however, a parlimentry government based upon a PR electoral system would
have, as it would have been more representative of the views of the people
and would have represented the strong anti-war movement prevelant at the
time, rather than the intrests of the military/industrial complex.
Just my 2.3 Euros worth
Dave

would  appreciate  help  with  this  MI  question .
Consider  the  Fibonnaci  sequence  1,1,2,3,5,8,....
which  may  be  defined  as  t(1) = t(2) = 1 , 
 and    t(n+2) = t(n+1) + t(n)
Prove   by  MI 
t(2n)  =  t(n){t(n+1)  +t(n-1)} 
  stan

"Nathan Crowder"  wrote:
>How many real solutions does the equation sin(x)=(x/100) have?
>Keep in mind this was a high school ciphering problem on my math team.
>There must be a trick or something.  My friends and i have wrote programs
>in pascal to do it but we only had 1.5 minutes and no computer. HELP?
You just need to graph sinx and x/100 and realize that x/100 is
positive to the right of zero and that x/100 = 1 at x=100 and that is
the last 'moment' that sinx can intersect x/100  also every 'hump'
that intersects x/100 intersects it twice except at x=0 and x=100 and
x=-100.  You should beable to guess that it is something like
integer_value[100/(2*pi)] = approximate double hits of a hump on the
postive side
so 2*integer_value[100/(2*pi)]  + {-1, 0, or 1}
Tim

  I don't know if this will help, but if we let A' and (I-B)' stand for the inverses 
if they exist, then:
  (A -BA)x = kx  becomes  (I-B)y = kA'y or  Ax = k(I-B)'x where y = Ax.
 Now if either A or (I-B) is positive or negative definite, you at least have a 
recasting of the original problem into a well understood one.  It might also
point to Lanzcos-type solution algorithms.
Regards,  Chuck Crawford

Hi,
Cliff Pickover's latest book may interest you,
because it contains loads of mathematical ideas.
THE LOOM OF GOD: MATHEMATICAL TAPESTRIES AT THE END OF TIME.
(Plenum Publishing).  Even though it's not out until Spring, the
publisher is accepting order now at www.plenum.com.
Here is the books' Table of Contents:
1  Are Numbers Gods?
2  The End of The World
3  Pentagonal Numbers
4  Doomsday: Friday 13, November, A.D. 2026
5  666,666, Gnomons, and Oblong Numbers
6  St. Augustine Numbers
7  Perfection
8  Turks and Christians
9  The Ars Magna of Ramon Lull
10 Death Stars, a Prelude to August 21, 2126
11 Stonehenge
12 Urantia and 5,342,482,337,666
13 Fractals and God
14 Behold the Fractal Quipu
15 The Eye of God
16 Number Caves
17 Numerical Gargoyles
18 Astronomical Computers in Canchal de Mahoma
19 Kabala
20 Mathematical Proofs of God's Existence
21 Eschaton Now
22 Epilogue
Postscript 1. Goedel's Mathematical Proof of God's Existence
Postscript 2. Mathematicians Who Were Religious
Postscript 3. Author's Musings
Smorgasbord for Computer Junkies
Notes                                                                          
References
About the Author
Here are some blurbs from the back of the jacket.
                        THE LOOM OF GOD
            Mathematical Tapestries at the Edge of Time
                         (Plenum, 1997)
"As far as I know, Clifford Pickover is the first
mathematician to write a book about areas where math and
theology overlap.  Are there mathematical proofs of God?
Who are the great mathematicians who believed in a deity?
Does numerology lead anywhere when applied to sacred
literature?  Pickover covers these and many other off-trail
topics with his usual verve, humor, and clarity.  And along
the way the reader will learn a great deal of serious
mathematics."
- Martin Gardner
"Pickover has done it again, with a marvelously
entertaining, historical romp through the unexpected
connections between mathematics and mysticism."
- Paul Hoffman, President/Editer-in-Chief, Discover magazine
"Without peer as an idea machine, Cliff Pickover proves
equally adept at writing, The Loom of God is a well-crafted
piece of mathematical science fiction."
- Charles Aschbacher, Book Review Editor, Journal of Recreational
Mathematics   
"Pickover's lively, provocative travel guide takes readers
into the fascinating realm of mystic math, from perfectly
strange numbers to fractured geometries and other curious
nooks and crannies of ancient worlds and modern times."
- Ivars Peterson, Science News,
Author of "The Mathematical Tourist: Snapshots of Modern Mathematics"
"Chock full of mathematical treats, The Loom of God takes
you on a trip which explores ideas in a totally non-
threatening, enjoyable format.  Entertaining and informative
adventure of Pickover's fictional characters -- Theano and
Mr. Plex -- bring to life such things as:  the golden mean,
spirals in hyperspace, the Inca quipus, string theory, the
wild and diverse world of numbers.  A must for the
I-hate-math person as well as the mathematical explorer."
- Theoni Pappas, author of The Joy of Mathematics
"If you ever doubted that science and religion have
commonality, this is the book for you.  In The Loom of God,
Cliff Pickover, in his irrepressible style, frolics through
a forest of mathematical curiosities and historical tidbits,
all skillfully woven into a futuristic fantasy, leaving you
to wonder where he learned all that."
- Julien C. Sprott, Professor of Physics, University of Wisconsin-Madison
Cliff Pickover has a home page at
http://sprott.physics.wisc.edu/pickover/home.htm
that has images of the book cover as well as some other
artwork from the book.   
Regards, Charon                                                                
---------------------------------------------------------
Get Your *Web-Based* Free Email at http://www.hotmail.com
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Hello. I've a little problem:
I must invert this function: y = (x^2) + x     for x >= 0
Is there anybody that could help me?
Thanx.
   +---------------------------------------+
   /   - - ---> C y b e r M a n <--- - -   /
   /    _______ Milano -- ITALY _______    /
   /->>> e-mail:  cyberman@zerocity.it <<<-/
   /          PGP KeyID: 487C6475          /
   +---------------------------------------+
        - ----< Key fingerprint >---- -
8F 50 99 35 BF A4 83 6C 68 11 C9 31 DC A7 65 F2
      <----------------@----------------->

http://www.oei.es/sipconc.htm

Ok, this may sound mildly stupid, or like I just plain have too much
time on my hands, but it actually seems to have some validity in my
eyes.  
Most people have experienced looking through a camera that is hooked
up to a TV displaying what the camera sees, then looking at the TV and
you get really funky images.  Well, I have found maybe 2 pages in a
book dealing with just the basics of it and its connection to fractals
and iteration (I mean just think about it, camera is like inputing a
value in a function and the TV is the output).  It goes on to explain
it somewhat and a basic setup for experimentation with it.  I've
noticed images seem to make the most sense when there is a 1:1 mapping
ratio between the TV and camera (camera is zoomed in exactly on TV
screen), that way there is no dilation of any sort of the image.  I
have been experimenting a little bit with the angle of the camera and
the period of the image produced.  If anybody has any sort of idea
what I'm talking about and knows of where I might find further
information on the subject e-mail jhalcrow@mindspring.com

David Ullrich (ullrich@math.okstate.edu) wrote:
<>
Thank you David.  As in the past, you have come through for me with a
clear direct answer to my question.  I really appreciate you volunteering
so much of your time to help people out on the net (sci.math in particular)
and I hope you will continue to do so.
-- Jeff

Cyberman wrote:
> I must invert this function: y = (x^2) + x     for x >= 0
> 
> Is there anybody that could help me?
Solve the equation x^2+x-y=0 for x. This gives
x=(-1+sqrt(1+4y))/2 or x=(-1-sqrt(1+4y))/2. Of course, since
the function x -> x^2 + x is not one-to-one, so the whole problem
is not very well defined (though it is typical in this newsgroup
to bicker about whether sqrt(4) is +/-2 or just 2), but you might
say that an inverse function is
  y -> (-1+sqrt(1+4*y))/2
     David A. Madore
    (david.madore@ens.fr,
     http://www.eleves.ens.fr:8080/home/madore/index.html.en)

In article <32ddcd24.2071802@news.inet.it>, cyberman@zerocity.it (Cyberman) writes:
|> Hello. I've a little problem:
|> 
|> I must invert this function: y = (x^2) + x     for x >= 0
|> 
|> Is there anybody that could help me?
|>
  Since this sounds pretty much like homework, I'll say only the following:
  Use the Quadratic Formula to solve x^2 _ x - y = 0 for x as a 
  two-valued function of y. Then note that only _one_ of the roots
  is nonnegative (at least, this is true for y >= -1/4 -- but we're
  already assuming y >= 0, since that follows from the similar
  assumption about x).
|> Thanx.
|> 
|> 
|>    +---------------------------------------+
|>    /   - - ---> C y b e r M a n <--- - -   /
|>    /    _______ Milano -- ITALY _______    /
|>    /->>> e-mail:  cyberman@zerocity.it <<<-/
|>    /          PGP KeyID: 487C6475          /
|>    +---------------------------------------+
|>         - ----< Key fingerprint >---- -
|> 8F 50 99 35 BF A4 83 6C 68 11 C9 31 DC A7 65 F2
|>       <----------------@----------------->
-- 
 Ed Hook                              |       Coppula eam, se non posit
 Computer Sciences Corporation        |         acceptera jocularum.
 NASA Langley Research Center         | Me? Speak for my employer?...<*snort*>
 Internet: hook@cscsun3.larc.nasa.gov |        ... Get a _clue_ !!! ...

In article ,
Alison Brooks   wrote:
>In article <5bc184$q32$1@dartvax.dartmouth.edu>, Archimedes Plutonium
> writes
>>   If the US had been a parliamentary form of government where all
>>politicians are elected and not these cabinets that linger from one
>>administration to another and really run the government. Then,
>>hypothetically, is it  highly likely that the Vietnam War would have
>>never occurred? Or if it had, would not a parliamentary form of
>>government gotten the US out quicker? One can argue that the US Vietnam
>>War was chiefly the result of foolish advisors to the president.
>>
>
>Interestingly enough, while the US was busy getting bogged down in
>Vietnam, the UK was engaged in fighting in Borneo, in remarkably similar
>political situations. The UK military position wasn't as good as that of
>the US; the Borneo border was massively longer than that which the
>Americans had to deal with, and the terrain very much harder.
>
>Nonetheless, the UK was successful.
>
>One can debate why this should be; however, there was no great "anti-
>war" debate in the UK. I suspect that this was in part because of
>different attitudes.
>
I suspect the UK was successful because Borneo is not next-door to China.
The UK did not have the restrictions on elminating the insurgency's main 
base of operations (N. Vietnam for the US) that the US did.  
>>  Perhaps this is a great research inquiry as to see which form of
>>democracy is superior-- the US or the UK parliamentary.
>>
>>  In a parliamentary system, the likelihood of foolish advisors doing
>>so much damage is minimized, I suspect.
>>
>
>If only. You don't live in Britain, do you?
>
A PM can pull all sorts of stunts on his own in Britain.  Aside from
Borneo, there was an effective undeclared war between Britain and the
early Bolshevik government for a while.  Britishers can probably come up
with numerous examples of the parliamentary system failing to prevent 
Vietnams.  
And then there's the fictional example of PM Urquhart, who starts a war 
in Cyprus to build his memory, the phrase "Frances, this could be our 
Falklands" dripping from his wife's lips.....
-- 
"A story to me is like a line on which I hang little baubles of incident"-Al
Capp

I am in need of a proof by contradiction that the set of prime numbers is
infinite.
Starts by assuming that the set of primes is finite and the largest prime
is P.
Let x = P! and  let y = x + 1. From here is where I can't get things
straight.
I think we want to find the lowest number  that will divide  x and also
show that because of our assumprion it also divides x which can't be
possible since x and y are consecuctive integers. This  is our
contradiction that show our original assumption is wrong and the set in
infinite.
Any help in the details would be appriecated.
Daryl
Lively,Ontario

The JakeWeb Math Challenge is stll continueing. Check out
www.jmartin.home.ml.org to enter. Solve a problem and have your name
listed in my hall of fame!
Good Luck!!!
-- 
Jacob Martin
jacobmartin@geocities.com or try jake@scientist.com
http://jmartin.home.ml.org