HDTenEightyP/NetNews huggingface dataset:
From: jbaez@euclid.mit.edu (John C. Baez)
Subject: Re: math and physics degrees
Organization: MIT Department of Mathematics, Cambridge, MA
Date: Mon, 17 Aug 92 19:11:02 GMT

In article <1992Aug17.125911.7989@pellns.alleg.edu> frisinv@alleg.edu(Vincent Frisina) writes:
>   I'm an undergrad in math and physics and eventually want to get a Ph.D.  
>Since my interest shifts from one to the other periodically, I wonder if I  
>can get two Ph.Ds simulataniously without skimping on either. If this is  
>possible, which grad schools are best for it and how long will it take? If  
>not, what's the closest I can get?

If you want to get an academic job (which is mighty hard these days) you
should realize that it's most likely going to be in EITHER a physics or
a math department and they will usually want you to have: 1) a PhD in that
subject, and 2) lots of papers published in journals of that subject.  
It would be a lot of work building up your career in such a way that
your formal qualifications in both math and physics are equal, and the
point is that this work wouldn't stop after you got your PhD.  A much
more typical thing to do is to pick ONE area to be your primary one and
wait until you become famous before you start also claiming to be an
expert in the other area.  All this has to do with the practical issue
of getting jobs, and doesn't have much to do with what you're actually
interested in.  It's a crummy system in many ways, the division of
academia into narrow specialities, but if one is going to succeed in an
academic career one must be aware of this system and have a strategy for
dealing with it.

For example...

I'm a mathematical physicist.  As an undergrad I found that I did a bit
better in math than in physics courses, and I disliked laboratory work.
So I majored in math.  I then got a PhD in math.  But I've liked physics
all along and almost all of my papers are about physics AND math.   For
the most part, though, I've published in math journals (or mathematical
physics journals), which means I usually publish proofs of theorems.  As
a result math departments were reasonably sure I wasn't just a physicist
pretending to be a mathematician, and were willing to hire me and give
me tenure.  Now that I'm slightly more established I am more willing to
play the physicist as well... but when cornered I will always admit that
I am a mathematician, professionally speaking.    

Doing a PhD *simultaneously* in math and physics sounds almost
impossible to me.  Maybe there are some places where it can be done, or
maybe you have twice as much energy and time than other grad
students.  But I suggest that you get a PhD in one or the other, which
needn't prevent you from taking courses in BOTH subjects, as I did.

Get a physicist to explain what physics general exams are like.  I'm
glad I never had to do those.

Newsgroups: sci.math
From: tilo@bernina.ethz.ch (Tilo Levante)
Subject: Eigenvalues and vectors of an infinite matrix
Organization: Swiss Federal Institute of Technology (ETH), Zurich, CH
Date: Thu, 13 Aug 1992 12:02:25 GMT

I have the following matrix and need an algorithm to compute
the eigenvalues and eigenvectors:

         ...    ...    ...    ...
 .....  H + 2I   A*     B*     0      ....
 .....    A    H +  I   A*     B*     ....
 .....    B      A      H      A*     ....
 .....    0      B      A    H -  I   ....
         ...    ...    ...    ...

H is a hermitian nxn matrix
A,B are nxn matrices, A*, B* are the adjoint matrices
I is the nxn unity matrix, 0 is the zero matrix

I know that there are n eigenvalues l1 to ln and that the
other eigenvalues are of the form  li + k for any integer k.

Thanks for any hints

Tilo Levante

Newsgroups: sci.math
From: roach@bikini.wri.com (Kelly Roach)
Subject: Re: u(v^n)w prime puzzle - HINTS
Organization: Wolfram Research, Inc.
Date: Fri, 21 Aug 1992 19:15:53 GMT

In article <a_rubin.714416973@dn66>  writes:
>                                                                     p
> Essentially the same proof shows that if p=uvw is prime, then p | uv w.


     OK.  I think I understand what you are getting at.
For anyone else who didn't get it, I think Arthur Rubin's
solution is "essentially the same" as my hint sheet with
the difference that

     u*10^(a*n+b) + vw
     
     v*Sum[10^(a*i+b),{i,1,n-1}]
 
are used in (2) and p=uvw is used in (4).  The nice feature
to using p=uvw is that p doesn't divide 10^a-1 appearing
in the denominator of the closed form expression for the
geometric Sum.  This is a nice observation.


Newsgroups: sci.math
From: wjcastre@magnus.acs.ohio-state.edu (W.Jose Castrellon G.)
Subject: Re: A Non-Cantorian Set Theory question
Organization: The Ohio State University,Math.Dept.(studnt)
Date: Sun, 16 Aug 1992 21:36:57 GMT

In article <1992Aug12.113415.1648@gacvx2.gac.edu> kiran@gacvx2.gac.edu writes:
>Quite a while ago, I read Martin Gardner write in one of his _Mathematical 
>Games_ column in _Scientific American_ that on a plane, a letter like _O_ can 
>be written--allowing smaller O's to be written inside larger O's-- _c_ 
>times where _c_ is the cardinality of the continuum. On the other hand, he
>pointed out, letters like _T_ can only be written aleph-nought times.
>
>A question that has remained in my mind for a long time since is the following
>
>Since we know that non-Cantorian set theories are possible, is there a
>one-dimensional shape which can be written some aleph times where that aleph
>is between aleph-nought and _c_? If so, what would the shape be?
>
>I have once asked my mathematics teacher about it, and he remarked that that
>shape might be a fractal!
>
>Does anyone have any idea on this?

Nice question. If you allow disconnected spaces, the answer is yes:
Let A be a subset of reals of cardinality  K  and call G the set obtained by
taking all finite sums of elements in A and their negatives, then G also has
cardinality K. Now partition the reals into equivalence classes, such that
two elements belong to the same equivalence class iff their difference is an
element of G. If we form a set S by picking one element from each equivalence
class, then the real line is the disjoint union of K-many sets, each congruent
to S  (Just take S_u = {s+u : s belongs to S}, for each u in G). Now by using
Zorn's lemma, S can be constructed such that no more than K-many sets similar
(i.e. homotetic) to  S  can be disjoint in the real line. To get the final sub-
set of the plane, it is enough to give  S some thickness, i.e. make P equal to
the cartesian product of  S  and the interval [0,1], for example (though 
lengthy to include in here, all details are easy to handle, just as for the T).
I believe it is possible to construct a connected set P too, but I havent found
a proof of that yet.

As for well-behaved spaces, I don't see any unreasonable obstacle to prove
that for P path-connected the answer is either one of Aleph_0 or 2^Aleph_0.
Xref: sparky sci.math:10429 soc.culture.sri-lanka:926
From: niranjan@eng.cam.ac.uk (Mahesan Niranjan)
Newsgroups: sci.math,sci.math.stats,soc.culture.sri-lanka
Subject: help(det(toplitz))
Date: 20 Aug 92 10:45:07 GMT
Organization: Cambridge University Engineering Department, UK

Is there a recursive (by dimension) algorithm for computing determinant of
a Toeplitz matrix?

ie if I had matrices...

r0  r0 r1   r0 r1 r2   r0 r1 r2 r3  etc
    r1 r0   r1 r0 r1   r1 r0 r1 r2
	    r2 r1 r0   r2 r1 r0 r1
		       r3 r2 r1 r0

and I want the determinants of all of them, can I do this in a smart way?
(When I go from the 3x3 to the 4x4, I am adding only one new value, r3).

thanx
niranjan@eng.cam.ac.uk
Newsgroups: sci.math
From: bhonsle@bhonsle.iss.nus.sg (Shailendra K Bhonsle)
Subject: Re: u(v^n)w prime puzzle
Organization: Institute of Systems Science, NUS, Singapore
Date: Wed, 19 Aug 1992 02:41:06 GMT

In article <1992Aug18.171532.14274@wri.com>, roach@bikini.wri.com (Kelly Roach) writes:
|> 
|> 
|> 
|>      Prove or disprove:  There are three non-empty
|>      strings of digits u,v,w such that all the
|>      numbers in
|>           L = {u(v^n)w | n is a natural number}
|>             = {uw, uvw, uvvw, uvvvw, uvvvvw, ...}
|>      are prime numbers.
|> 
|> 
|>      Time to say a few more words about my u(v^n)w prime
|> puzzle which I posted yesterday.  I am definitely not looking
|> for any solutions involving u=v="0".  Ordinary syntax only
|> please.  No leading zeros in u.
|>      Some interesting patterns:
|> 
|>      u="3",v="3",w="1"
|>      31, 331, 3331, 33331, 333331, 3333331, 33333331
|> 
|>      u="1",v="36",w="1"
|>      11, 1361, 136361, 13636361, 1363636361, 136363636361
|> 
|>      u="17",v="57",w="09"
|>      1709, 175709, 17575709, 1757575709, 175757575709,
|>      17575757575709, 1757575757575709, 175757575757575709
|> 
|> Below each line giving u,v,w values appear a lot of prime
|> numbers.  These patterns do eventually fail:
|> 
|>      333333331 = 17*19607843
|>      13636363636361 = 17*1321*5693*106661
|>      17575757575757575709 = 232433*75616446785773
|> 
|> The question is, are there any {u,v,w} examples which do
|> not fail?  That always give prime numbers?
|>      I know the solution to this puzzle.  I think the
|> solution can be understood fairly easily by anyone that
|> has had a first course in number theory.  See if you
|> can discover it.
|> 
|> 
|> 
|> 
|> 
|> 

-- 

Hi,
I think the answer is no, it is not possible to have such a set of u v and w.
Since I am not a hardcore mathematician (never done anything in number theory 
formally), let me first give me the intuition and then the proof:

1. Intuition: If it were possible then it will not be such a big deal to 
discover the biggest prime number etc. (just keep on repeating "v") provided 
one is able to find u,v, & w.

2. proof:

The approach is to take a integer x (let's say prime integer) and prove that by repeating 
v one gets a string u(v^m)w (for positive integer m) which is divisible by x.

Let's take x=9 (because it has nice properties for base 10 numbers).

then for a number "uv"(ie. 10u+v) 
     uv=u+v (mod 9)


In our problem definition:

let u+w=m(mod 9)

Now it is possible to find a n for given v such that

                     v^n==n.v== 9-m (mod 9) ==> simple number theory 


Hence for this n u(v^n)w= u+w+n.v==0(mod 9)



So for any such u,v,w there is at least one number which is divisible by 9.


So by contradiction we prove that it is not possible to have such triples.


-----------------------------

Is this proof satisfactory??

Shailendra
(bhonsle@iss.nus.sg)
Inst. of Systems Science, Singapore  

Newsgroups: sci.math
From: umatf071@unibi.uni-bielefeld.de (0105)
Subject: Re: A Word-Problem
Date: Thu, 13 Aug 92 13:57:04 GMT
Organization: Universitaet Bielefeld

This is the problem the word-problem comes from:

                   1
                  1 1
                 2 2 3
                3 2 3 3
               3 3 1 1 2
              1 2 2 1 2 2
             1 1 2 3 3 1 1
            2 3 3 1 3 2 1 3
           2 2 3 1 1 2 2 3 3
                                                            o
This figure shows that you can tile a triangle T9 with T2: o o

QUESTION: which T(n) can be tiled with T2?

SOLUTION:
 The number of points of T(n) is ( n+1 \choose 2 ) = t(n).
 3 | t(n)  iff  n = 0, 2 (mod 3).

 CASE I: if n = 0, 2, 9, 11 (mod 12) then T(n) is tileable.
   From T9 you can get T11 and T12 adding:

      1 2 2 1 2 2 1 2 2 1
     1 1 2 1 1 2 1 1 2 1 1

   or
      1 1 2 2 3 3 1 1 2 2
     2 1 3 2 1 3 2 1 3 2 1
    2 2 3 3 1 1 2 2 3 3 1 1

   With T11, T12, and k times

     1 2 2 1 2 2 1 2 2 1 2 2
    1 1 2 1 1 2 1 1 2 1 1 2

   you can construct T(n+12) from T(n).

 CASE II: if n = 3, 5, 6, 8 (mod 12) then T(n) is not tileable.
   This is the hard case. You can look at the equivalent triomino
   puzzle. In this case T5 would look like:

              o                The allowed L-triominoes are:
            o o                      o         o o
          o o o   = T5             o o   and   o
        o o o o
      o o o o o                (only 180-degree rotation)

   Now let's try group theory:
   You can read about this technique in the article:
     Dmitry V. Fomin, Getting it together with "polyominoes",
     quantum 2:2 (Nov/Dec 1991) 20-23, 61
   So you get the group: G = < a, b| aabb=baba, bbaa=abab >.
   If a^n b^n <> (ba)^n for n = 3, 5, 6, 8 (mod 12) then is T(n) not
   tileable. Is a^n b^n = (ba)^n for some n = 3, 5, 6, 8 (mod 12) then
   a^3 b^3 = (ba)^3. This you can see be the same technique as in CASE I.
   You have the additional possibility of reduction in the group. If you
   show a^3 b^3 <> (ba)^3 then CASE II is solved. David Sibley found a
   homomorphism, which shows this inequality. So CASE II is proofed.

Annotation:
  Noam Elkies is right, that the homomorphis of David Sibley shows that
  it is implossible to tile an odd rectangle with the L-triomino (only
  180-degree rotation allowed) too. In this case you have to show that
  a^3 b <> b a^3.

Torsten Sillke
From: wilson@web.ctron.com
Newsgroups: sci.math
Subject: Fundamental Theorems
Date: 12 Aug 92 14:46:42 GMT
Reply-To: wilson@web.ctron.com ()
Organization: Cabletron Systems INc.
Originator: wilson@web


    My hastily compiled votes on various fundamental theorems:


    Fundamental Theorem of Geometry:

	The Independence of the Parallel Postulate:

	The parallel postulate is independent of the axioms of absolute
	geometry.

	Try to beat this one for historical importance as well as its 
	bearing on mathematical thought.

    Fundamental Theorem of Probability:

	The Central Limit Theorem:

	The distribution of the mean of a sample drawn from any
	distribution of positive variance tends towards a normal
	distribution as the size of the sample grows large.

	Puts the normal distribution in its place.

    Fundamental Theorem of Complex Numbers:

	The Fundamental Theorem of Algebra:

	All roots of a polynomial with complex coefficients are complex.

	Complex numbers arose historically as solutions to irreducible
	polynomials over the reals.  That the complex numbers split
	such polynomials, and that no further splitting of complex
	polynomials is necessary, is the trademark of the complex field.

    Fundamental Theorem of Linear Algebra:

	The Primary Decomposition Theorem:

	Given linear operator T on finite dimensional vector space V, V
	may be expressed as the direct sum of invariant subspaces of T.

    Fundamental Theorem of Group Theory:

	Cayley's Theorem:

	Every group is isomorphic to a subgroup of the permutation group
	of some set.

    Fundamental Theorem of Statistics:

	No Entry.

    Fundamental Theorem of Trigonometry:

	Euler's Theorem:

	exp(x+iy) = exp(x)(cos y + i sin y)

	Explains circular functions in terms of the exponential function.

    Fundamental Theorem of Set Theory:

	No Entry.  (I incline, however, towards elevating Russell's
	paradox to the status of the Fundamental Theorem of Inconsistent
	Set Theory).

    Fundamental Theorem of Logic:

	Godel's incompleteness theorem:

	Any sufficiently rich theory is undecideable.

	In short, Principia Mathematica will have to wait.

    Fundamental Theorem of Mathematics:

	The principle of mathematical induction:

	Let P(x) be a proposition about natural number x.  If
	    1.  P(0)
	    2.  For all x, P(x) ==> P(x+1)
	Then P(x) is true for all numbers.

	This principle pervades mathematics.  To understand, appreciate,
	and use mathematical induction is the entry ticket to mathematics.

	(Bob Silverman points out that mathematical induction is an
	axiom.  I believe that this makes it no less a theorem).


-- 
David W. Wilson (wilson@ctron.com)

Disclaimer: "Truth is just truth...You can't have opinions about truth."
- Peter Schikele, introduction to P.D.Q. Bach's oratorio "The Seasonings."
From: jk87377@cc.tut.fi (Juhana Kouhia)
Newsgroups: sci.math,sci.psychology
Subject: Re: mathematician's marrying ages.
Date: 19 Aug 92 18:52:15 GMT
Organization: Tampere University of Technology


In article <86536@netnews.upenn.edu> zorrilla@cattell.psych.upenn.edu
(Eric Zorrilla) writes:
>
>>It is researched that those who are not married does have more
>>mental, say, problems.
>
>In almost every study I know of, this is simply a correlation.  How
>many people do you know of that want to get married to someone with a
>"mental" problem? (co-dependents and masochists excluded  :)   )

Well, my English is not perfect in psychology field.
I didn't mean that they have mental problems in sense of lunatic/mad.
Also, I didn't mean that they are allready mad in their young age.

>>Lonely people usually die more early than
>>non-lonely people.
>
>Same issue as above

My sentences above above and above means the same effect, actually.
Let me explane again.

>(though there is slightly better research in psychoneuro-
>immunology suggesting an effect of social support in humans and
>controlled studies of changes in "social environment" [e.g., housing
>status, gender-housed status, etc.] in animals (though interpreting
>any of this as being "lonely" is beyond my view)

I have read from the newspapers about animals in zoos, how they
are lonely (alone) and actually die for their loneliness (also after
their partner has died).
I have also read about how some elder people die soon after
the dead of their wife/husband.

So, the researchs I have read about tells that
people married lives longer and are more healty than
people not married.
(And the case is not simply as: "In almost every study I know of, this is
simply a correlation.  How many people do you know of that want to get
married to someone with a "mental" problem?" as suggested.)

Having many friends is not the same than having a caring living
partner.


Juhana Kouhia
Newsgroups: sci.math
From: roach@bikini.wri.com (Kelly Roach)
Subject: Re: u(v^n)w prime puzzle - HINTS
Organization: Wolfram Research, Inc.
Date: Thu, 20 Aug 1992 23:30:27 GMT

In article <a_rubin.714331997@dn66>  writes:
> uvw can't divide 10^a-1; it's larger!


     OK, I agree.  How is this useful?  Shailendra K Bhonsle
assumed p=uw all the way back in step 1.  Repairing step 1
to use a different p seems far from obvious.


Newsgroups: sci.math
From: israel@unixg.ubc.ca (Robert B. Israel)
Subject: Re: Help - non-integral powers of a matrix?
Keywords: matrix
Organization: University of British Columbia, Vancouver, B.C., Canada
Date: Thu, 13 Aug 1992 06:20:59 GMT

In article <1992Aug12.231708.3644@galois.mit.edu> jbaez@nevanlinna.mit.edu
(John C. Baez) writes:
                                                                                
>In article <a_rubin.713653963@dn66> a_rubin@dsg4.dse.beckman.com (Arthur 
>Rubin) writes:                      
(referring to the series log A = sum_{k=1}^infinity (-1)^(k-1) (A-I)^k/k)
>>log "obviously" converges if all eigenvalues of A are strictly within 1 of 1,  
                                                                                
>To clarify, perhaps, let me add that this is not only "obvious", it's           
>true, at least if A is diagonalizable.  If A is a                               
>not-necessarily-diagonalizable matrix log A is defined if ||A - 1|| < 1;        
>I don't think the eigenvalue condition above is sufficient.                     

It is sufficient.  On the other hand, |z - 1| <= 1 with z <> 1 (which I put
forward in another post) is NOT enough.

We can assume without loss of generality that A has only one eigenvalue r.
Then A = rI + N where N is nilpotent (say N^n = 0).  Expanding all the terms
of our series in powers of N, it is enough to get convergence for the 
coefficients of each N^j, 0 <= j < n.  The series for the coefficient of N^j 
is
    sum_{k = j}^infinity (-1)^(k-1) binom(k,j)/k (r-1)^(k-j)
= ((r-1)^(-j)/j!) sum_{k = j}^infinity (-1)^(k-1) (k-1)(k-2)...(k-j+1) (r-1)^k
and this converges if |r - 1| < 1.

Let's make a clear distinction between "log A is defined" (i.e. there is
some (highly non-unique) B such that exp(B) = A) and "the above series
converges".  For log A to exist, all you need is that A is nonsingular.
Moreover, you can get log A without any infinite series.  

If, as above, A = rI + N with N^n = 0 and r <> 0, then 
    log A = log r + log(I + N/r)
          = log r + sum_{k=1}^{n-1} (-1)^(k-1) (N/r)^k /k 
In general, for any nonsingular A, decompose the space of column vectors 
into eigenspaces corresponding to each eigenvalue; on each eigenspace, A has 
the above form.  Of course, finding the eigenvalues is difficult and 
numerically unstable in general, so this is more of a theoretical than a 
practical solution. 


-- 
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             or israel@unixg.ubc.ca
University of British Columbia
Vancouver, BC, Canada V6T 1Y4
From: fc03@ns1.cc.lehigh.edu (Frederick W. Chapman)
Newsgroups: sci.math
Subject: Re: mathematician's marrying ages.
Date: 22 Aug 92 00:29:41 GMT
Organization: Lehigh University

In article <1992Aug19.185215.303@cc.tut.fi>, jk87377@cc.tut.fi (Juhana Kouhia) 
writes:

[...]

>Having many friends is not the same than having a caring living
>partner.
>
>
>Juhana Kouhia


HEY!  STOP NAGGING!   :-)

-- 

o ------------------------------------------------------------------------- o
|  Frederick W. Chapman, User Services, Computing Center, Lehigh University |
|    Campus Phone:  8-3218     Preferred E-mail Address:  fc03@Lehigh.Edu   | 
|         "The day after yesterday is the second-to-last day before         |
|               the rest of your life the day after tomorrow."              |
o ------------------------------------------------------------------------- o
From: wsadjw@rw7.urc.tue.nl (Jan Willem Nienhuys)
Newsgroups: sci.math
Subject: Re: Young Gauss and the arith. series (was: Bell's _Men of Mathematics_)
Date: 12 Aug 92 15:38:16 GMT
Reply-To: wsadjw@urc.tue.nl
Organization: Eindhoven University of Technology, The Netherlands

In article <1992Aug12.103511.4051@kth.se> TordM@VanD.PhySto.SE writes:
>
> as I read it in a volume called "Sigma", which is about 6 books about the
>mathematical history, the teacher asked the children to add the numbers
>from 1 to 100, and Gauss wrote the answer down, and handed it to his teacher
>and said "da liegt es!". 

"Ligget sie" (in his dialect).  The story goes on that the slate had to be
put face down, and all the others piled on top in order of arrival.
All the time little Gauss was sitting quietly waiting intil the last
pupil was ready.  In those days it seems, corporal punishment was quite
common for boys who had the answer wrong, and acted as smart alecks.
So as all the slates were turned one by one (with many wrong answers),
the tension must have mounting . . .

The story, by the way, used to be told by Gauss himself, no written records
by witnesses exist.

JWM
From: andrew@rentec.com (Andrew Mullhaupt)
Newsgroups: sci.math
Subject: Re: Help - non-integral power of a matrix?
Keywords: matrix
Date: 13 Aug 92 14:08:06 GMT
Organization: Renaissance Technologies Corp., Setauket, NY.

In article <Aug.10.15.45.34.1992.26563@clam.rutgers.edu> gonzalez@clam.rutgers.edu (Ralph Gonzalez) writes:
>
>Hi.  Does anyone know of an algorithm to find a non-integral
>power of a matrix, e.g. A^.5 or A^1.3? Thus, A^2 is the same
>as AxA and A^0 is the identity.

Yes. _Run_ do not walk, to the nearest copy of _Matrix Computations_ by
Golub and van Loan. Read carefully the part about the matrix exponential.
Do not fail to notice the ominous title of one of the more important
references on this subject: "Nineteen Dubious Ways to Compute the Matrix
Exponential", and thus chastened, implement one of the recommended algorithms
(Shur decomposition or Squaring, scaling and Pade).

I've had good success with the squaring, scaling & Pade approach.

Later,
Andrew Mullhaupt


P.S. this should be in a FAQ, right?
Xref: sparky sci.math:10507 sci.physics:13315