Newsgroup sci.math 156778

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Pat Fleckenstein (x58885) wrote:
> 
> In article <32C5616C.195B@math.okstate.edu>,
> David Ullrich   wrote:
> >Milo Gardner wrote:
> >>
> >> Modern computers should not overflow. During the 1970's overflows were
> >> quite common.
> >
> >       But modern computers allow you to pack arbitrarily large
> >integers into 2 bytes??? This is remarkable, where can I get me
> >one of those modern computers?
> 
>     They take advantage of the probabilitic nature of VNLs (very large number
>     (where very large is any number greater than one or less than
>     1/65536)).  It is a well known property of VNLs that the probability
>     that the binary expression of the VNL contains a non-zero digit goes
>     up at a very rapid rate as the VNL becomes large.  At the same time,
>     the probability that the VNL dos not contain a non-zero digit goes
>     down at a very rapid rate.  In fact, it is well known that at the limit,
>     these probabilities proceed to 1 and 0, respectively.
> 
>     Using a complex MINGLE scheme on the proportion of non-zero digits
>     in a VNL (after two's complement inversion, of course), the number
>     is renormalized in MINGLEd probability space as an integer in the
>     range 0 to 127.  The sign bit is then restored and any TWIDDLEing
>     for two's complement is performed.
> 
>     This same scheme is repeated with the proportion of digits that are
>     not non-zero.  These numbers each require a single byte of storage
>     space.
> 
>     It is trivial exercise to show that each pair of MINGLEd bytes
>     corresponds to a unique VNL.  The proof that the MINGLE in
>     probability space results in a distinct pair of bytes given distinct
>     VNLs would require a much deeper knowledge of the MINGLE operation
>     and the arcane mathematics surrounding unitary probabilities than
>     this article is meant to cover.
> 
>     All of this is implemented in hardware so it doesn't take much more
>     than about 4ns.
> 
>     It is hoped that through the use of wavelet transforms, the number
>     of bits required to store an arbitrary number will fall to zero, thus
>     greatly reducing the need for large banks of memory.
> 
>     alter,
Nice one!
MathBlab.  I think I got it first.  Anybody before me?  That's
my first point.  How many points you guys got?
-- 
D.
mentock@mindspring.com
http://www.mindspring.com/~mentock/index.htm

1997 Lehigh University Geometry/Topology Conference 
 			June 19-21, 1997
 			Principal Speakers 
Steve Ferry, SUNY Binghamton: Geometry, controlled topology, and
  			strange spaces. 
Peter Gilkey, University of Oregon: The eta invariant, connective
  			K-theory, equivariant spin bordism, and 
  			metrics of positive scalar curvature.
Robert Gompf, University of Texas at Austin: On the topology of
                        Stein surfaces.
H. Blaine Lawson, SUNY Stony Brook: Curvature, singularities, and
  			Morse theory.
Mark Mahowald, Northwestern University: Curves of higher genus and
  			homotopy theory.
Jon Wolfson, Michigan State University: On a variational problem
  			in symplectic geometry.
  In addition, there will be parallel sessions of 40-minute contributed
  talks, divided roughly into Differential and Complex Geometry,
  Algebraic Topology, and Geometric Topology. Those wishing to give
  a talk should send an abstract to David Johnson or Don Davis by
  April 15. 
  We ask a $30 registration fee of participants who have research
grants.
  On-campus dormitory housing will be provided at a cost of $15 per
  night per room. Single rooms will be provided except to those
  indicating the name of a roommate on their registration form.
  Housing on Wednesday night and/or Saturday night is possible. Please
  submit the registration form by May 1 regardless of whether you
  require housing. 
  Continental breakfast will be provided Friday and Saturday
  mornings, and lunch will be provided Thursday, Friday and Saturday 
  noons. Dinner will be the only meal not provided gratis. On
  Thursday, expeditions to nearby restaurants will be arranged, 
  followed by a party. On Friday there will be a Chinese banquet 
  at a cost of $25.
  More information will be available at the conference Web site as
  it becomes available. Please check back from time to time. You
  can register for the conference on the Web form there, or mail in
  the accompanying registration form.  You can e-mail the form below
  to either of the addresses specified, or s-mail to the address below.
  The URL for the on-line registration form and additional information 
  is:
 	http://www.lehigh.edu/~dlj0/geotop.html
  For more information, contact:
  David L. Johnson or Donald M. Davis
  Department of Mathematics
  14 E. Packer Avenue
  Lehigh University
  Bethlehem, PA 18015-3174
  e-mail: dlj0@lehigh.edu or dmd1@lehigh.edu
-------8<-------8<-------8<-------8<-------8<-------8<-------8<-------
 	1997 Lehigh University Geometry/Topology Conference
 			Registration Form
Name: 
Address: 
Telephone: 
E-mail:
Do you wish to give a talk? 
Title:
Please attach an abstract of your talk, or enclose a DOS-format
floppy disk with a TeX version of your abstract (LaTeX2e preferred),
or e-mail your abstract to the organizers.
Do you need an overhead projector, or a computer with a projector?
If so, specify:
  On-campus dormitory housing will be provided at a cost of $15 per
  night per room. Single rooms will be provided except to those
  indicating the name of a roommate on their registration form.
  If you would prefer to stay in a hotel, there is a new Comfort Suites
  Hotel located 4 blocks from campus. Make your own reservations
  at (610) 882-9700. 
Do you need on-campus housing?
If so, which nights?
Gender (Needed for housing): Male   Female 
Do you wish to share a room with someone?
If so, name:
  On Friday evening there will be a traditional Chinese banquet at
  a local restaurant, at a cost of $25 per person. 
Number attending banquet:
  There is a $30 registration fee only for those supported by external
  research grants. 
If you are supported by such a grant, please check here:
  If you print and mail this form in, please enclose a check, payable 
  to Lehigh University, for room charges, registration fee, and banquet 
  charges, if any, and mail to:
  David L. Johnson or Donald M. Davis
  Department of Mathematics
  14 E. Packer Avenue
  Lehigh University
  Bethlehem, PA 18015-3174
  e-mail: dlj0@lehigh.edu or dmd1@lehigh.edu
-- 
David L. Johnson         		dlj0@lehigh.edu, dlj0@netaxs.com
Department of Mathematics		http://www.lehigh.edu/~dlj0/dlj0.html
Lehigh University
14 E. Packer Avenue			(610) 758-3759
Bethlehem, PA 18015-3174

Miguel Lerma wrote:
> 
> In article <32CDD639.2C86@mindspring.com> you wrote:
> [...]
> > No.  The definition says that a meter is 1/299792458 of the distance
> > that light travels in a second in a vacuum.  So, unless you change
> > *this* definition, changes to the definition of a second won't affect
> > it either.
> 
> I did not mean that the definition of "meter" would change,
> I meant that the _length_ of 1 meter would change (if we do
> not change the above definition of "meter"). If you redefine
> "second" making it, say, slightly longer, the meter would also
> become proportionally longer.
I understand what you said.  However, the first time you said it
would *not* change:
> If the meter is defined in terms of the speed of light with specifically
> that constant, then it can't change, since the speed of light will always be
> the speed of light.
-- 
D.
mentock@mindspring.com
http://www.mindspring.com/~mentock/index.htm

In article ,
Jason Reed  wrote:
>I'm attempting to find a function that can be represented by a Taylor
>series and has the property
>
>$$f(e^x)=f(x)+1$$
>
>I think the above falls into the category of Abel functional equations.
>Regardless of that, I reduced the problem to finding the coefficients of
>the power series representation by way of assuming infinitely large
>matrices behave themselves.
[snip]
If you are looking for an entire function to solve the functional 
equation, it does not exist. Reason: the exponential z |-> e^z  has 
infinitely many (complex) fixed points p=e^p. One of them is within 
round-off from  0.3181+1.3372*i, and each of them would force 
    f(p)=f(p)+1.
 Functions of this kind are studied, as you mentioned, in books about 
iterated functions, notably in
Title:          Iterative functional equations / Marek Kuczma, 
BogdanP                   Choczewski, Roman Ger. --
By:             Kuczma, Marek.
                Choczewski, Bogdan.
                Ger, Roman.
Published:      Cambridge ; New York : Cambridge University Press, 1990.
                xix, 552 p. : ill.
Subject(s):  Use s=
                Functional equations
                Functions of real variables 
Series:         Encyclopedia of mathematics and its applications. v. 32
Notes:          Includes bibliographies and index.
ISBN            0521355613
There is a well-behaved function a(x) which counts the iterations of 
  e(x) = e^x - 1
in the sense that a(e(x))=a(x)+1, at least for x>0. It has a logarithmic
singularity as x -> 0+. But then again, some singularity is to be
expected. 
Hope it helps, ZVK (Slavek).

I need to know what the frequency spectrum is of a function divided by 
it's envelope.
The function, amplitude vs. time, will be any waveform that is 
band-limited - there are no frequency components below a or above b hz, 
where b>a.
The envelope would be as calculated using the Hilbert transform, or 
SQRT(I^2 + Q^2).
The output, f(t)/envelope, is to be expressed as a power spectrum 
(=amp^2) vs. frequency, from 0 to 00 hz.
The only modifier to this is that we will later want to lowpass filter 
the envelope, and reduce it's spectrum from the possible 0 to 00 hz to 0 
to c hz, and need to predict the effect on the output spectrum, as 
above.  The most general solution would include c as selectable.
This is a real problem I've had, and I can't solve it, and I haven't 
found anyone who can.  Appreciate some help.
Regards, Howard.

Does anyone know of a simple definition of an "integer"? It is simply
a whole number, correct? Like 1, 2, 239804732098 or what have you?
regards, mark  
YOUR OWN FAX-ON-DEMAND SERVICE A.L.A./US$7.50 MONTH. WITH 
* *NO* * LONG DISTANCE CHARGES. BUSINESS OR PERSONAL USE
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Angel Garcia (bp887@FreeNet.Carleton.CA) writes:
> Richard Mentock (mentock@mindspring.com) writes:
>> Richard Mentock wrote:
>>> 
>>> Angel Garcia wrote:
>>> >
>>> > Robin Chapman (rjc@maths.ex.ac.uk) writes:
>>> 
>>> > > In the recent "The Book of Numbers" (Springer 1996) Conway and Guy
>>> > > give constructions for regular 7, 9 and 13-gons using straightedge,
>>> > > compass and angle trisector. The heptagon construction is amazingly
>>> > > neat.
>>> > >
>>> >      Still has not arrived to U of T. I will look for it as soon as
>>> > it arrives.
>>> >    Yes: adding trisection to ruler&compass; makes many classical problems
>>> > feasible. I wonder which type of approximation or method these recent
>>> > Conway and Guy use for trisection.
>>> 
>>> They present what they call Archimedes's trisection, on p.195.  It
>>> is beautiful.  Assume your ruler has a fixed distance marked upon it.
>>> Draw a circle of that diameter centered at the vertex of the angle to
         OF THAT RADIUS AND NOT "DIAMETER"
!!!
--
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    ***************************

Gary Hampson wrote:
> 
> In article <32C90B5C.64B1@public.ibercaja.es>, benigno
>  writes
> >Hi,
> >       I need to handle Big matrix of around 800 x 800 to implement
> >       some optimization algorithms, I would use C++ libraries if
> >       possible to use on BorlandC++ 4.5, but if there is any shareware
> 
> Unless the matrix has some structure which allows many short cuts and
> reduced storage (eg Toeplitz), then just get on with coding it. 800*800
> is not that big (unless of course in the optimisation you need to
> evaluate A.x many times, or you have some time critical conditions.
> --
> Gary Hampson
If you're multiplying matrices of that size (800 x 800), you might
want to investigate the Strassen matrix multiplication algorithm.
One (old) reference for this algorithm is "Fundamentals of Computer
Algorithms," Horowitz and Sahni, Computer Science Press, 1978, 
Section 3.7, pp. 137-140
This was a fun GWU grad school homework assignment some years ago.  I
probably have UCSD Pascal code that demonstrates the algorithm
somewhere if you're interested.
efg
-- 
Earl F. Glynn          EarlGlynn@WorldNet.att.net
EFG Software              913/859-9557  Voice/Fax
   Scientific/Engineering/Medical Applications
             Overland Park, KS  USA

In article <32CDAE1E.D39@scf.usc.edu>,
Ryan Cormney   wrote:
>
>I apologize for using the word hate and blurb. I was not being fair.
	Hey, it's rap! :-)  Don't worry.
>I have realized from reading more poems lately, I like having to look up
>words in a poem. As you can tell I'm a novice at this game, however I do
>personally think I have potential as a poet.
	I am sure. (I write for long years, longer than
	I care to remember, without any abilities, but I still
	believe or hope that I have potential). And the
	social poetic environment (Internet) is an incredible
	advantage not present in the past.
> If you'rr uninterested in
>helping me out that is fine,
	I already did, twice. My remark was written in good
	faith. Then I explained it in a greater detail.
	You may disagree. But it will not hurt to ponder about.
> but I would like you to take one last look
>at a poem I just wrote(Generic Hieratic) and tell me what you think.
	First dwell on "Inside". Maybe I am wrong, maybe your
	poem was fine. I don't think so, but... If I were right
	then was it just question of technique or of the very concept
	of the poem? And if you insist on the concept but agree
	about its generic character (I see neither of the
	lovers in that poem) then in what direction are you going
	to push that poem to make it unique.  At the moment it
	seems to me that your text is a common denominator
	of the experience of every non-virgin.  The role of the poetic
	eye is to see things in a different, unique, unobvious way
	(I don't mean weird or superficial--there is place in poetry
	for these two too, even if I personally am not too big of
	a fan of them).
>Thanks RBC
	Good luck,
		Wlod

In article <5a8t4g$8t1@news00.btx.dtag.de>,
Christian Wieschebrink  wrote:
>    There are no positive integers p,q, such that 
>    p^3 - q^2 = 1 
...is true. There aren't even any rational solutions except p=1, q=0.
(It's an elliptic curve of rank zero and torsion of order 2).

e9209h60@gamma.ntu.ac.sg wrote:
>I am interested in knowing about algorithms for converting a
>covariant tensor to a contravariant one and vice versa. I know that
>for tensors of rank n and indices i_1, i_2, ..., i_n = 1..2,
>this can be achieved by inner product of the tensor with the
>permutation tensor, e, defined by,
>              _
>             |   1   if i=1, j=2
>      e_ij = {  -1   if i=2, j=1
>             |   0   if i=j
>              -
>In a general case when i_1, i_2, ..., i_n = 1..r,
>how can we achieve this transformation from covariant to contravariant
>tensor and vice versa ?
>
>Thanks
>
>Satish
>
I don't know where to begin.  I think you have several ideas mixed up.
 First, the e_ij you defined is the 2-dimensional permutation symbol.  It 
is not a tensor (except with respect to a quite narrow group of 
transformations.)  Second, for higher dimensional spaces, the definition 
is more general, usually stated in terms of odd and even permutations of 
the index values.  Third, to get the tensor you must multiply be the 
square root of the determinant of the metric tensor or its reciprocal, 
depending on whether you want the contravariant or covariant version.  
Third, while an inner product with any tensors will produce a tensor of 
the type revealed by the surviving indices, it will generally not be the 
one you started with, especially if you use the permutation tensor.  
Fourth, to obtain the _associated_ tensor of the opposite type, raise or 
lower the indices with the metric tensor.
Harry

This corrects my post of Dec. 30, 1996 on the thread, "Hypergeometric
Probability Function."
A reference to "sci.math.stat" should read "sci.stat.math" -- I regret the
error.
Patrick T. Wahl
( no institutional affiliation )

Angel Garcia wrote:
> 
> Angel Garcia (bp887@FreeNet.Carleton.CA) writes:
> > Richard Mentock (mentock@mindspring.com) writes:
> >> Richard Mentock wrote:
> >>>
> >>> Angel Garcia wrote:
> >>> >
> >>> > Robin Chapman (rjc@maths.ex.ac.uk) writes:
> >>>
> >>> > > In the recent "The Book of Numbers" (Springer 1996) Conway and Guy
> >>> > > give constructions for regular 7, 9 and 13-gons using straightedge,
> >>> > > compass and angle trisector. The heptagon construction is amazingly
> >>> > > neat.
> >>> > >
> >>> >      Still has not arrived to U of T. I will look for it as soon as
> >>> > it arrives.
> >>> >    Yes: adding trisection to ruler&compass; makes many classical problems
> >>> > feasible. I wonder which type of approximation or method these recent
> >>> > Conway and Guy use for trisection.
> >>>
> >>> They present what they call Archimedes's trisection, on p.195.  It
> >>> is beautiful.  Assume your ruler has a fixed distance marked upon it.
> >>> Draw a circle of that diameter centered at the vertex of the angle to
> 
>          OF THAT RADIUS AND NOT "DIAMETER"
Yes, you are right.  Sorry.
-- 
D.
mentock@mindspring.com
http://www.mindspring.com/~mentock/index.htm

Wieschebrink@t-online.de (Christian Wieschebrink) wrote:
>Does anyone know a proof of the statement below?
>I can't find the answer myself.
>    There are no positive integers p,q, such that 
>    p^3 - q^2 = 1 
>    is true.
>Of course the conjecture might be wrong. What is
>the smallest solution (p,q) then?
Outline of proof (due to Euler)  :
If p,q are solutions, then p^3=q^2+1=(q+i)(q-i) 
(where i^2= -1).  The arithmetic of Gaussian integers
(complex numbers of the form a+bi, where a,b are integers)
then requires
   q+i = k(a+bi)^3  where a,b are integers and 
k=1 or -1 or i or -i.  Taking complex conjugates of both sides,
    q-i = k'(a-bi)^3   where k'=1 or -1 or -i or +i (respectively).
Subtracting,
      2i = k(a+bi)^3 - k'(a-bi)^3
"and an easy calculation show that this is impossible."
 (I saw this in American Mathematical Monthly, v103, #7, 
Aug-Sept 1996 in the article "Catalan's Conjecture" 
by Paulo Ribenboim )
G Hunsberger
Dept of Mathematics
Hartwick College
Oneonta, NY

In article <19970103140800.JAA15072@ladder01.news.aol.com>,
  wrote:
>
>In article  <560tlm$nih@gap.cco.caltech.edu>
>  ikastan@alumnae.caltech.edu (Ilias Kastanas) wrote:
>:
>:In article <19961108163000.LAA25851@ladder01.news.aol.com>,
>:  wrote:
>:>In article  <55rq1t$963@nuke.csu.net>
>:>Ilias Kastanas wrote:
>:>>
>:>>    So in Mr. Tleko's world there are no analytic functions at all ....
>:>       There are. e^z is analytic in the whole plane.
>:	Good, but if  e^z = R + iI then according to your "formula",
>:
>:	e^z = sqrt(R^2 + I^2) * (cos(atan(I/R)) + i*sin(atan(I/R)) .
>:
>:   So then, pursuing your "argument", the multiplicity of arctan means
>that  e^z is not analytic!
>        The formula for e^z you wrote should read:
>
>        e^z=e^(x+iy)=(e^x)*(e^(iy))=(e^x)*(cos(y)+isin(y))
>
>        where        R=(e^x)*cos(y)   and  I=(e^x)*sin(y).
>
>        There is no arctan involved and e^z is an analytic function.
>
>        Please make a note of it.
	Quite unexpected, months after the previous posts!
	Mr. Leko, I'm glad you realized there is no need for arctan.  Yes,
   e^z is analytic (phew!);  so is  z, with  R = x, I = y,  right?
							Ilias

I recommend the following wonderful book by Rademacher:
      Author: Rademacher, Hans, 1892-1969.
       Title: Lectures on elementary number theory 
              by Hans Rademacher.
 Publication: Huntington, N.Y. : R.E. Krieger, 1977, c1964.
    Material: ix, 146 p. ; 24 cm.
        Note: Reprint of the ed. published by Blaisdell Pub. Co., New
              York, in series: A Blaisdell book in the pure and
              applied sciences, Introduction to higher mathematics.
        Note: Includes index.
     Subject: Number theory.
--------------------------------------------------------------------------
   Bob Silverman (numtheor@tiac.net) wrote:
   : martind@veblen.acns.carleton.edu (Daniel T. Martin) wrote:
   : >I'm looking for a source that will give Dirichlet's proof that arithmetic
   : >sequences with the initial value and increment coprime contain an infinite 
   : >number of primes.
   : >All the references I've found so far avoid the proof (usually by saying that
   : >analytical number theory is beyond the scope of the book).
   : >Does anyone know of a (English-language) source that actually gives a proof?
   : H. Davenport  "Multiplicative Number Theory".  Springer
   :
	 I'll second that.  Also "Number Theory" by Borevich and Shafarevich
   (Academic Press); there's a version of the result also in Tom Apostol's
   "Introduction to Analytic Number Theory" (Springer Undergraduate Texts).
	 Any number of books on algebraic number theory or classfield theory also
   give proofs.
--
Ray Mines          
ray@nmsu.edu
Mathematical Sciences
New Mexico State University
Las Cruces, New Mexico 88003

Michel Hack asked about bounds on the size of maximum partial quotients in
the continued fraction expansion of certain rational numbers.  He studies
(5^a)/(2^b) when this is near to 1.
Short answer:  For practical continued fractions, look at "Knuth Volume
2."
I consulted D.E.Knuth, _Seminumerical Algorithms_, Volume 2 of _The Art of
Computer Programming_ (second edition, 1981) and found references in the
solutions to his exercises.  See 4.5.3, # 24 and # 35.  (pp. 361-362,
604-606)
A.Kinchin, Compos. Math. 1 (1935), 361-382 "proved that the sum of the
first n partial quotients ... of a real number X will be asymptotically n
lg n, for almost all X. ... the behavior is different for rational X."  
(Knuth uses lg(x) to denote the base-two logarithm.)
A.C. Yao and D.E.Knuth, Proc. Nat. Acad. Sci. 72 (1975), 4720-4722
apparently addresses the case of rational X.
I'm curious:  Knuth covers a wealth of information, including the formulas
for 2X and 1/X when you know X as a continued fraction.  Naively, I
suggest this course:
1.  The continued fraction for 1/(5^a) is trivial.
2.  Multiply by 2 using the Hurwitz formulas cited by Knuth.
3.  Repeat step 2 until you have (2^b)/(5^a).
4.  Use the 1/X formula to get (5^a)/(2^b).
I suspect you will learn more by studying what happens to the individual
partial quotients as you repeatedly apply step 2.  It may turn out to
relate to the binary representation of powers of 5.  (This may be
nonsense, but it's free.)
Hope this helps.
Patrick T. Wahl
( no institutional affiliation )

On Fri, 03 Jan 1997 12:26:14 -0500, Richard Mentock
 wrote:
>sfly wrote:
>> Consider the nontrivil case ,0<=x,y<=1
>> so  let x=1-a,y=1-b
>> then x^y+y^x=(1-a)^(1-b)+(1-b)^(1-a)=(1-a)/((1-a)^b)+(1-b)/((1-b)^a)
>> and (1-a)^b<=1-ab ,(1-b)^a<=1-ab
>This works for a=b=.5, for instance, but not for a=b=.1
>Try again!
su3g4196t87!!
no,I'm no wrong! 
and
tiju......Try again yourself....!
>> so  x^y+y^x>=(2-a-b)/(1-ab)=1+(1-a)(1-b)/(1-ab)>=1 ##
>> look like easy.....

Amiel Ferman (amielfu@mail.tlm.openu.ac.il) wrote:
: b(n) = the sum of b(k) * b(n-1-k)  
:              for k = 0 .. (n - 1)
: 
: (that's b(0)*b(n-1) + b(1)*b(n-2) + .. + b(n-1)*b(0))
: 
: 
: b(n) = ( 1/(n+1) ) * ( 2n )
:                      (  n )
These are called "Catalan numbers".  You could just substitute the
known answer into the recursion.

Consider N delivery people.  They each have several (scheduled) deliveries to
make each day.  Suppose they are at location z1(i), i = 1, ..., N, at time T
and the locations of the time (T+1) deliveries are at z2(i), i = 1, ..., N.
The problem is to minimise
        D(p) = (sum over i = 1 to N) distance(z1(i), z2(p(i))
over all permutations p (with distance taken in the plane, say).  This can be
generalised by replacing the distance function with a more general weighting
function (hence taking account of actual travel times, say).
Is this a known problem?  Are there any decent algorithms?
Thanks, Wayne
(Please send e-mail as well as posting.)

In article ,
Dennis Yelle  wrote:
:In article <32CD9764.670475AA@nwu.edu> Jeffrey Ely  writes:
:>Does there exist a function on a well-behaved domain
:>(say an interval of the real line) which is discontinuous
:>at every point?
:
:Yes.
:
:I am tempted to end this post here, but....
:
:f(x) = 1 iff x is rational
:f(x) = 0 iff x is not rational.
:
 Now try something harder: a function defined on a discrete set and 
discontinuous everywhere. Deadline: April 1, 1997 :-)=
Seriously: There is even a subset of R (Borel measurable), whose
characteristic function is discontinuous everywhere on R, and so badly so
that no re-definition on a set of measure zero can produce a point of
continuity. Just construct a Cantor-like set of locally positive measure
in every interval [n,n+1] (n integer), squeeze recursively a scaled copy
of it into every interval of the complement, and unite all these sets. 
 Worse yet, there is a function (Borel measurable again) defined on R
which maps every non-empty open interval onto all of R.
Good luck, ZVK (Slavek).

                             SEARCH FOR TECHNICAL EXPERTS
Product and/or process development requires specific information, knowledge
and/or experience. Specific information/knowledge/experience is mostly not easy
to track down, even not in centres for technology. One reason is that specific
knowledge/experience is mostly linked to individuals, the experts on a
specific topic.
In order to further the search for specific knowledge I am setting up a
database that refers to experts on all kinds of technological topics.
Companies that look for specific knowledge/experience can use the database
to get in touch with the needed expert.
If you are an expert and you like to be included in the database, please
send me (by email) the following information:
KEYWORDS describing your expertise:
* Field of technology, e.g. chemistry: 1 keyword.
* Application in terms of product/process, e.g. thermocouples: 1 - 3 keyword(s).
* Application in terms of industry/activity, e.g. refinery: 1 - 3 keyword(s).
* Description of your specific expertise: preferably 3 keywords, e.g.
  degradation, carbonmonoxide, misinterpretation; if not possible: a small text
  is allowable.
 Some Rules:
 -For the selection of keywords you may use your own terminology.
 -A keyword may consist of more than one word.
 -Use the above description for each separate expertise you offer.
 -If you feel the use of keywords is too restrictive for a good description,
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PERSONAL details:
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*      Email address and/or facsimile number.
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Confidentiality: My name is J.H. Heerings (Dieren, The Netherlands). I am
writing from a personal interest and as an individual (no company is involved).
The above information will not be used for mailing lists or otherwise; only for
the abovementioned database. I will contact you at the moment the database will
start to run and is accessible to industry.
The information should be emailed to: heerings@worldaccess.nl

could some nice person give me the formula(s) for how to 
determine the probabilities for how likely it would be 
to guess the correct card in a standard Zeno (???) deck of 
ESP cards...
deck consists of 25 cards, 5 kinds of cards
star, cirlcle, square, cross, wavy lines.
and a second problem:
i have a program on my 48sx that displays a jiggling dot on the 
screen, what is the probability, over a fixed amount of time, that 
it will deviate from the center of the screen...???
( the program allows the user to hit various keys which un(?)-
predictably fudge the random number generator )
and the third problem:
i have another program on my 42 and 48 ( HP calculators )
that play a game very much like mastermind
but with 'bug-brain' the user tries to guess a 6 digit 
number, such that no two digits are the same
and after each guess the program tells the user how 
many digits, both position & value, are correct
with zero blind luck allowed, how many guess', given the 
most perfect guessing strategy, should be nesessary to guess all
6 digits...???
( i can get all 6 in 6 to 8 guess' (usually))
thanx!
sproogles...        babynous

Tasos Serghides wrote:
> 
> Consider the sequence:    1, 4, 7, 10, ....
> 
> Is, "1" the first term or is "4" the first term?  The text that I am 
> using refers to "1" as the "a (subknot)" number and for "4" as the "a 
> (subone)" number.  That is if "4" is the subone number of the sequence 
> then it is the first number of the sequence - I guess this is a case 
> where the language we speak and the notation we use, do not match.
> 
> In this convention, and I assume that this is a convention question, 
> what is the first number of the sequence?  This of cource has 
> implications when you are talking about the nth number of the
> sequence.
1 is the first term.  The natural (counting) numbers begin with 1.  Your
text has made a poor choice of notation unless they're leading up to
something where that notation is useful.
Sometimes it's quite helpful to begin a sequence with a zeroth term.  A
good example is the series a0 + a1x + a2x^2 + a3x^3 + ... + anx^n (an
nth degree polynomial).  Here the a(i) coefficient corresponds to the
ith power of x, and a(0) is the coefficient of x^0.
Another place where a zero term might make sense is in computer arrays. 
In hexadecimal notation, a 256-element array might well be indexed from
X(00) to X(FF) where 0, ..., 9, A, ..., F represent the decimal numbers
0 through 15.
But unless there's a reason for doing something like this, it's more
natural to start counting from 1.
James

In article <32cd4e71.49410189@snews.zippo.com>, Mark  wrote:
>Does anyone know of a simple definition of an "integer"? It is simply
>a whole number, correct? Like 1, 2, 239804732098 or what have you?
The question is very definitely NOT as simple as one would think.
There are no elementary ways of defining an integer.  Now to make this
clear, an elementary method is one using only the properties of an
ordered field, with no usage of sets or having a known set of integers 
around. 
One could say that x is an integer if sin(\pi*x) = 0, but this is
not what is being wanted.  The number 1 is defined as the unit for
multiplication.  There are many non-elementary ways of defining 
integer, or the set of integers. 
One can define the SET of all positive integers as the smallest set
which contains 1, and whenever it contains x, contains x+1.
Alternatively, one can define a positive number y to be an integer
if the smallest SET which has y as an element, and whenever u is
an element, u-1 is an element, has 0 as an element.  Or that the
set has a negative element, and any positive element v satisfies
v <= v*v.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu         Phone: (317)494-6054   FAX: (317)494-0558

Hayo Hase is approximating the volume "beneath" a region on the surface of
a sphere and continuing down to the center..  He wants to pack together
some tetrahedrons with a common vertex at the sphere's center and with
adjacent triangles made from points on the surface of the sphere.
The volume desired is in proportion to the volume of the whole sphere as
the area of the surface region is the the surface of the whole sphere. 
Why not forget about the tetrahedrons and just think about surface area?
Using the same triangulation of the surface, use a classic  formula to
find the area of each spherical triangle.  The difficulties in Hase's
original method will not arise, no matter how triangles are chosen to
cover the given region.  The only difficulty is choosing points that
closely approximate the boundary.  Every triangulation is equally good
since all spherical triangle edges are confined to the surface.
Patrick T. Wahl
( no institutional affiliation )

get a life!

In article <32CBE758.505F@winternet.com>, American Sharecom
 writes
>Would like to subscribe to SCI.MATH
>
Looks like you just did!
AFAIK this is unmoderated and folks just come and go.
Enjoy!
-- 
Richard H Gould 
rhgould@gocomp.demon.co.uk

Brent Hetherwick wrote:
> 
> What are some reasonable conditions to put on A, B, and C so that
> if A x B is isomorphic to A x C, then B is necessarily isomorphic to C,
> if x denotes the tensor product operator?
> 
> --
> $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666
>                        hetherwi@math.wisc.edu
> $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666
If A is faithfully flat, the following condition holds (almost by 
definition): For every morphism B -> C, the corresponding morphism 
A x B -> A x C is an isomorphism if and only if B -> C is an 
isomorphism. It is therefore sufficient that A is faithfully flat.
To give more explicit conditions, it is necessary to know what kind
of objects A,B and C are. For instance, if they are vectorspaces, no
condition is needed - as every vectorspace is faithfully flat. 
Eivind Eriksen

Jacob  writes:
> 
>  Prove that:    13 divides (17^47 + 2^12)^14 -4
                   n     
Suppose           a  = a + 1     (mod p)
             (p-1)/2
and         a        = -1        (mod p)
then multiplying the 1st equation by the 2nd^m, m odd, gives
          n+m(p-1)/2
         a           = -(a + 1)  (mod p)
Choosing p=13, a=2, n=4, m=15 we have
   94     4+15*6
  2   =  2           = -3      (mod 13)
So   2^94 = 4^47 = 17^47 = -3  (mod 13)
and                 2^12 =  1  (mod 13)  by FlT (Fermat's little Theorem)
thus   (17^47 + 2^12)^14 = (-3+1)^14  (mod 13)
                         = (-2)^2     (mod 13)  since FlT => (-2)^12 = 1
                         = 4          (mod 13)
As an alternative to the first equation above, one could instead use
a = (1/a + 1)^2,  i.e.  a^3 = (a + 1)^2,   for a = 4   (mod 13)
as in John McGowan's post. One can generalize the problem to 
other values of a and p by choosing a = some solution of the
first equation (in either form) that is simultaneously a
primitive root modulo p (hence satisfying the second equation).
-Bill Dubuque

You wrote : I'm glad you realized there is no need for arctan.
   Not for e^z which is analytic (has no zeros). Try to write the formula
   for f(z) = z or any other polynomial. They are not analytic, right?
tleko@aol.com 

whuang@ugcs.caltech.edu (Wei-Hwa Huang) wrote:
> marnix@worldonline.nl (Marnix Klooster) writes:
> > * How much agreement is there among mathematicians about beauty
> >and elegance of theorems and proofs?  Where does this agreement
> >come from?
> 
> Usually there is agreement, but for certain cases there isn't.
> Consider the following problem:
> 
> "A rectangle with integral sides has an area numerically equal to 
> its perimeter.  What are the possible dimensions of this rectangle?"
To summarize the three solutions:
 1. enumerates the possibilities;
 2. uses a geometrical trick (covering with unit squares);
 3. uses an algebraical trick (factorization).
> We can probably agree that Solution 1 is not very elegant.
Yes.  It works, but it gives no insight.
> Now, which is more elegant, solution 2 or solution 3?  Both are
> similar.  People with a more geometrical bent might prefer solution 2;
> people with a good grounding in number theory might prefer solution 3.
> Solution 1 has an algorithmic bent;
> Solution 2 has an intriguing solving method;
> Solution 3 has an "aha" when you add the 4 to both sides.
I think solution 2 has an "aha" too -- where does the idea of
filling with unit squares come from?  And, at least to me, the
"aha" in solution 3 is not so big.  As soon as on reaches
    ab - 2a - 2b = 0
it is very tempting to try a factorization of the left hand side.
So the crucial step is not adding 4, but factorizing.  And this
step is suggested by the shape of the formula.
> Personally, I prefer Solution 3, since the method solves seemingly unrelated
> problems like "I have two pairs of integers.  The product of one pair
> is the sum of the other, and vice versa.  What are my numbers?"
In my experience algebraic (calculational) proofs usually are
easier to generalize than geometric ones.  But then again, I
haven't done much geometry.
> > * Is a more elegant or beautiful theorem or proof `better' than
> >a clumsy or ugly one?  Why?
> 
> Pragmatically, the entire point of a proof is to convince some reader.
> To take a simile, it's a lot like writing a mystery novel; a bad one
> tends to make the reader bored, and has a detective who seems to be
> following a script, trying all combinations as he goes, while a good
> one is gripping, has a detective who has flashes of insight that the
> reader would never have thought of, and leaves the reader convinced
> at the end.
A good analogy!  The difference is, of course, that the mystery
novel author often conceals necessary facts, or at least makes
them hard to discover for the reader.  A mathematician should
make these as explicit as possible.
> Is it a proof if it's so long and inelegant that no one reads it?  
> It's a good philosophical question, and that's been a subject of debate
> among mathematicians recently (cf. proof of the Four Color Theorem).
Formally, of course, a proof is correct only if it shows how to
derive the theorem from the axioms and inference rules that one
is using.  This level of formality is usually not enlightening.
But any proof should be presented in a way that convinces the
reader that a translation into a fully formal proof is possible.
And this is why formula manipulation proofs are important: they
make this translation easier, while still being easy to
understand.  Hence they are more convincing.
> > * What is the use of formula manipulation in finding and
> >presenting proofs?  Does it help or does it obstruct?  Generally,
> >is a formula-manipulation proof better (or more elegant) than a
> >mostly-text one, or is it worse (or clumsier)?
> 
> Again, taste differs.  See Solutions 2 and 3 above; the
> formula vs. text is certainly in full force there.
> 
> In general, mathematicians like formula manipulation better,
> since it is more verifiable.
I certainly like formula manipulation better.  But I suspect that
many mathematicians still prefer text proofs.
>                               However, text proofs are
> usually easier to understand.  
Hmm.  Obviously, taste differs.  I find solution 3 easier to
understand.  In general, text proofs seem to require too much
interpretation, while (if done right) formula manipulation proofs
can be understood without interpreting the symbols.  This makes
formula manipulation my favourite.
> The "elegance" is also different; which is more elegant, 
> figuring out to tile squares inside or figuiring out 
> that adding 4 factors the expression?
This is true.  Despite the fact that a geometrical proof often
does not generalize, still such proofs can be elegant.
One part of this question is unanswered: How do you *find* a
proof?  In my experience, concentrating on formula manipulation
makes it easier to find proofs.
> Wei-Hwa Huang
Groetjes,
 <><
Marnix
--
Marnix Klooster        |  If you reply to this post,
marnix@worldonline.nl  |  please send me an e-mail copy.

Dan Kotlow  wrote:
> I think brevity, at least relative brevity, would be an important factor
> to most mathematicians.  So would economy of ideas.  IMHO, the vanG
> proof is so awful that it calls into question whether the author of
> the cited book can be a mathematician.  Not only does it substitute a
> whole lot of words for what can be proved in two lines of high school
> algebra, but it introduces a nonstandard system of representing numbers
> and contains an unnecessary descent argument. 
I don't agree that Van Gasteren's proof is awful, but I admit
that the alternative proofs are nicer.  Still, calling someone a
non-mathematician because he or she gives an awful proof isn't
really appropriate.  There are a number of text books (written by
professional mathematicians, I presume) out there that contain
ugly proofs where more elegant ones exist.
Calling someone who does things the inelegant way a
non-mathematician doesn't seem to be productive.  Rather, point
him or her the right way.  (As you all have done by giving me
alternative proofs.  Thanks!)
Incidentally, in "On the Shape of Mathematical Arguments" Van
Gasteren gives a lot of advice on how to design notation and
present proofs.  Most of it is good advice, too.  (At least, it
helped me in writing clearer proofs.)  Looking back, perhaps Van
Gasteren has ignored her own advice in the proof that I posted.
Groetjes,
 <><
Marnix
--
Marnix Klooster        |  If you reply to this post,
marnix@worldonline.nl  |  please send me an e-mail copy.

Hi. My name is Heather and I am a junior in high school.  I could use
some verification of my work on 2 project problems I am doing.
Question 1:
Graph f(x)= x, g(x)=sin x, and h(x)= x(sin x).  Compare the magnitudes
of f(x), g(x), and h(x) when x is "close to zero."  Write a short
paragraph explaining why the limit of h(x) is 0 when x approaches 0.
	I graphed the three funtions on my graphing calculator.  All three
share the point (0,0), but approach it from different positions. Since
h(x) is the product of f(x) and g(x), and both of the latter functions
have a magnitude of 0 at 0, then does it follow that (since 0x0 = 0),
h(0) = 0?  Does this also show that the limit of h(x) as x --> 0 is 0?  
Question 2:
The number of units in inventory in a small company is 
N(t) = 25(2[[((t+2)/2)]]-t), t is in the interval [0,12]
Note: [[ ]] = greatest integer function
	The real number t is the time in months.  Sketch the graph of the
function and discuss its continuity.  How often must thsi company
replenish its inventory?
	I thought about it and wanted to know if the inventory needs to be
replaced when N(t)=0?  Also, do you have any tips on how to graph this
function?
Thank you for your time and help.  Since I check my email more than this
newsgroup, please email any responses to:  timbidou@injersey.com 
Thanks again,
Heather

I would guess that the reason is to finally get to a 
divergenceless tensor. That is necessary to deal 
with persistent rather than transient events. This
may not be correct since my GR course was back
in 1966. 
Larry
Jeff Cronkhite  wrote in article
<5am7t2$i8c@catapult.gatech.edu>...
> I have a question from back in my GR class which I have never gotten
> around to finding an answer to, and wonder if the newsgroup could be
> of help.
> 
> After deriving the Riemann curvature tensor, one typically proceeds by
> contracting on two of the indices to arrive at the Ricci tensor, and
> then using the Bianchi identities to define the Einstein tensor.  The
> mathematics is straightforward enough, but I have never seen a good
> explanation for the motivation behind this procedure.  What does
> contracting the indices of the Riemann tensor do for us (or put
> another way, what is the interpretation of the Ricci tensor)?
> 
> Any help or good references would be greatly appreciated.
> Thanks,
> 
> jmc
> 
> 

david.bowdley@cableinet.co.uk (David Bowdley) wrote:
>As far as I knew the definition of the metre (or meter) is defined in
>SI Units as 1650763.73 times the wavelength in a vacum of the light
>emitted by a certain isotope of Krypton or something like that. I have
>got this definition from a number of books including a reference book
>to scientific data published by The Open University here in the UK.
>
>Best Regards
>David Bowdley
>Bilston, England
Oooops Was it really me who said the things above. I knew as I sent
the Usenet message that I mightbe wrong and then five minutes after
posting I read a more up to date book and there it was the truth about
the meter. Sorry folks I'll just crawl back into my shell and leave
this thread to those who know what they are talking about.
HAPPY NEW YEAR
Best Regards
David Bowdley - embarrased physicist :-)
Bilston, England
========================================================
e-mail:   david.bowdley@cableinet.co.uk
http://www.cableinet.co.uk/users/david.bowdley/index.htm
========================================================

Jeff Cronkhite wrote:
> 
> I have a question from back in my GR class which I have never gotten
> around to finding an answer to, and wonder if the newsgroup could be
> of help.
> 
> After deriving the Riemann curvature tensor, one typically proceeds by
> contracting on two of the indices to arrive at the Ricci tensor, and
> then using the Bianchi identities to define the Einstein tensor.  The
> mathematics is straightforward enough, but I have never seen a good
> explanation for the motivation behind this procedure.  What does
> contracting the indices of the Riemann tensor do for us (or put
> another way, what is the interpretation of the Ricci tensor)?
> 
> Any help or good references would be greatly appreciated.
> Thanks,
> 
> jmc
Jeff,
	I'll take a stab at this one. My understanding is that
the Reimann tensor is both non-covariant *and* divergent. 
This is a "bad" thing. Hence, it is contracted to form the Ricci
tensor, which *is* covariant, but still divergent).
(The tensors must all be *covariant* in order to be combined with
the Stress-Energy-Momentum tensor (T(u_v) in the field equations). 
	Now, however, the Ricci tensor remains divergent (it will not
conserve energy). So another term "R" was introduced by a great
intuitive leap by Einstein. The application of "R" results in
a tensor that non-divergent, the Einstein tensor:
	G(u_v) = R(u_v) - 0.5 g(u_v) R
	Finally, then, is the Field Equation itself:
	R(u_v) - 0.5 g(u_v) R = -8 (pi) G T(u_v)
One must be careful, though, because the Ricci tensor can yield a
zero curvature while the Reimann tensor can yield a non-zero
curvature. Hmmmmmm. I also hope that there are no typo's in 
the above equations. If so, I'm sure you can "fill-in".
	I hope this doesn't cause even more confusion. Anyway......
	Bye for now,
		Tim Tasto

In article <32CC4934.6567@sqruhs.ruhs.uwm.edu> donniet@sqruhs.ruhs.uwm.edu writes:
 > that is true.  lim 1/x as x->0 does not exist and is considered to be
 > either -oo or +oo.
Or simply not distinguish -oo and +oo.  To distinguish them may lead to
severe problems when you are switching to complex numbers.  The fact that
at one stage IEEE switched in their definition of floating-point numbers
from a projective infinity to a dual infinity leads to severe problems
stating basic properties about complex f-p math.  That is, unless you
switch to a polar representation of complex quantities, but polar complex
math is numerically much less stable than an euclidian equivalent.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/

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Richard Miao (RMiao@worldnet.att.net) wrote:
: i have been unable to integrate some functions that seemed easy enough but 
: the answers i came up with were incorrect. Can somebody show me how to do 
: them. Most importantly, i need to know the steps involved. I could easily 
: do them with a calculator but i need to know how to do it by hand. I have 
: tried u-sub but it does not seem to work. If anyone could help, i'd 
: greatly appeciate it.
: 
: (3x^2)
: -------
: (x^2+1)
: 
: xe^x - xe^-x
Hint: 
Use integration by parts for the second one (u=x v'=e^x-e^(-x)) 
The fastest way to solve the first is probably based on recognizing that:
 3x^2      3x^2 + 3       3                3
------- =  --------- - -------  =  3 -  -------
 x^2+1      x^2 + 1    x^2 + 1          x^2 + 1
The first term gives you 3x. For the second, substitute x=tan y. 
-- 
Ulrich Lange                       Dept. of Chemical Engineering
                                   University of Alberta
lange@gpu.srv.ualberta.ca          Edmonton, Alberta, T6G 2G6, Canada