Newsgroup sci.math 157427

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davk@netcom.com (David Kaufman) posted an earnest criticism of the
treatment of units in elementary math textbooks.
To which Michael Stueben posted the bizarre reply:
    Yes, Yes, Yes. Our math texts are terrible. In no math text I
    know of can I find a mention of God's existence. 
Explanation, Mr. Stueben?  Were you serious?  Were you trolling for
flames?  Were you making some enigmatic joke at Mr. Kaufman's expense?
If so, I don't get it.

In article <19970109052100.AAA28472@ladder01.news.aol.com>, qbob@aol.com
(QBOB) wrote:
> In my last two posts I asked about a general, and then specific solution
> (with f'(0)=0 and f(0)=d) of the following differential equation
> 
> f"(x)=-c/f(x) 
Actually you asked for f"(x)=-c/f(x)^2 (as is also clear from your
motivation by Newtonian mechanics), the solution of which is (as I posted
and mailed earlier, Cf.
http://ww2.altavista.digital.com/cgi-bin/news?msg@54960@sci%2emath )
x:=1/sqrt(2*c) * (d^(3/2) * arccos(sqrt(f/d)) + sqrt(f*d*(d-f)))
> Am a correct in assuming that there does not exist a way to
> represent f(x) in terms of x where f(x) is such a natural function?
>
> I hope that I have made a mistake somewhere, please help me, thank you.
Although one can prove that for every x between 0 and
Pi/2*d^(3/2)/sqrt(2*c) there's precisely one f between d and 0 that
satisfies the equation, it has never been solved for f in terms of x using
the ordinary functions of calculus. (To prove that no such solution can
exist would have to rely on the lack of common simplifications of
arccosines and square roots, but that's possibly a hard question with no
short solution, involving transcendental extensions
- the arccos term - of function fields etc.)
Mathematically, the given relation between f and x is good enough to allow
all practical computations: since the intervals of x and f are known, and x
is a strictly decreasing function of f (time x increases as height f
decreases) you can use any number of numerical root-finding algorithms to
determine f given x.
And the derivatives of f wrt to x can be computed as functions of f as
indicated in the verification part of my previous answer.
Cheers, D.C.
--
Dr. Denis Constales - dcons@world.std.com - http://cage.rug.ac.be/~dc/

C. K. Lester wrote:
> 
> In article <32D53060.B3D@fc.hp.com>, Karl Schultz  wrote:
> >C. K. Lester wrote:
> >>
> >> In response to Karl Schultz's prior post,
> >>
> >> >There are no subsets.  The 168-ticket wheel will guarantee a 3-match
> >> >in a 6/49 lotto.
> >> >
> >> >No, the first statement is correct.
> >> >The "wheeled group" is the entire set of 49 numbers.
> >>
> >> So, with a 6/49, buying 168 tickets using the 168-ticket wheel will guarantee
> >> a 3-match...
> >>
> >> So what?
> >
> >Because you were asking about this!!!!!
> 
> NO NO NO... sheesh almighty. I was referring to the "perceived value" of such
> a scheme... as in, "what value is buying 168 tickets for a guaranteed
> three-match?" Maybe I should have said, "Big deal."
The above representation of your question is much better than the
vague "So what?".
The perceived value, IMHO, is as follows.  People like to win.
If they can be sure to walk away with something, then they
might take steps to do that.  The only way to increase your
chances of winning is to play more numbers.  If you are
in the habit of playing 100+ numbers at a time and have
had a long losing streak, you might be inclined to play
the 168-ticket wheel, so that you are sure to have to make
that trip to the counter to claim a prize.  Actually, you
have a 60+% chance of getting 3 wins with 168 tickets,
but that is another story.  So, it is a psychological
thing - sure to get a win.
In the end, you are right.  Big Deal.
The wheel is just a structured way to buy more
tickets, which, in itself will increase chances.
Now, here is a real tough question for wheel experts.
If one plays 168 tickets using the wheel, they are sure
to match 3 at least once.  What does this wheel do to
one's chances to match more than 3???  There was once
a speculation that playing this wheel will reduce the
chances of matching more than 3 on one ticket.  Any
truth to this?

calvitti@kevin.ces.cwru.edu wrote:
> 
> given that the solution of a linear system dxdt = A.x in R^n is
> 
> (*)     x(t) = exp(A.t).x(0)
> 
> is the time "t" to reach x(t) from x(0) well defined for all A? seems
> that t would be given by the matrix natural log: nl(A). at least this
> works for scalar systems and i imagine that if A can be diagonalized,
> then it should be just as easy in higher dimensions.
> 
> intuitively, the time between two points in the state space - provided
> they are connected by a flowline - is well defined; it could be found
> by numerical integration for instance. however, i don't know how this
> is related to the matrix logarithm.
> 
> for example if A = {{-1,0},{0,-2}} then the (uncoupled) solutions are:
> 
>         x(t) = exp(-t).x(0)
>         y(t) = exp(-2t).y(0)
> 
> where {x(0), y(0)}, {x(t), y(t)} corresponding to the initial and
> final points in R^2 are given. the eigenvalues of A are {-1,-2}.
> however, doesn't the matrix log of a matrix with negative eigenvalues
> have complex entries? how is this matrix log computed to begin with?
> (matlab has a function to compute it numerically)
> 
> can someone email/post some insight or point to the literature?
> thanks for the info,
You have to be careful: Even though tA might have
negative eigenvalues, exp(tA) will not.  Provided
you take the right branch of the logarithm
log(exp(tA)) = tA.
The easiest way to see that the theoretical
solution of dx/dt= Ax with A constant is x(t) =
exp(tA)x(0) is to expand exp(tA) as a MacLaurin
series: exp(tA) = I + tA + (1/2)(tA)^2 + ...
Differentiate this w.r.t. t, and you'll see
immediately that dx/dt= Ax.
Finally, the solution of functions of matrices is
actually quite interesting - and not at all
trivial.  You might, for example be tempted to
compute the matrix exponential by the Taylor's
series.  This turns out to be a bad idea. For
some of the standard techniques, refer to Golub
and Van Loan's classic "Matrix Computations" (now
in third edition!)  
In my opinion, though, the best technique to
compute functions of matrices is the
Schur-Frechet algorithm by C. Kenney and
A.J. Laub.  In fact, the first paper on this
subject was concerned with the computation of the
Matrix Logarithm (Technical Report number
CCEC-95-0714 "A Schur-Fréchet Algorithm for
Computing the Logarithm of a Matrix" C. Kenney
and A.J. Laub, July 14th, 1995.) I'm not sure if
it has been published yet, but you can get a
reprint by snail mail from UCSB.  See the web
page at URL:
http://www-ccec.ece.ucsb.edu/pubs/techrpts/index.html
for further information.
I hope this helps, Cheers, John
-------------------------------------------------
 Dr. J.J. Hench  
 Dept. of Mathematics, Univ. of Reading, England   
 Institute of Informatics and Automation, Prague
-------------------------------------------------

Hello,
I post this problem just by curiosity, since it does not
refer to my actual research problems.
It is known that for instance x^2+x+41
exhibits a great density of prime numbers in its
values.
I was wondering whether this property extends to
the corresponding homogeneous 2-ary polynomial:
x^2+xy+41y^2.
For instance on couples x,y such that gcd(x,y)=1
and -100 <= x,y <= 100, the binary form exhibits
61.47 percent density.
Posing this problem this way raises another
problem, which is close to the problem of
idoneal numbers: if N is expressible, in a
unique way on the form x^2+xy+41y^2 with
gcd(x,y)=1, does it implies that N is a
prime number or the square of a prime
number. The property of unique representation
is decidable for each N, since the minimum of
the form is 161 x^2 when x is fixed.
The next step would be to wonder how this
property extend to others binary (or more)
forms.
Maybe some of our number theorists have
something to say about this problem.

Adam Logan (alogan@fas2.fas.harvard.edu) wrote:
: 
: Let R --> S be an injective homomorphism of rings (or, more generally, let
: S be a faithful R-module).  If the tensor product of S with itself over R,
: with respect to the given homomorphism, is flat over R, does it follow that
: S is flat over R?
: 
: (The consensus among people I've asked seems to be "no", but no one has come
: up with a counterexample.)
Well, I'd rather suggest "Yes".
Let's try to prove it.
Flatness of S means that S\tensor_R is an exact functor.
If S\tensor S is flat, but S itself is not flat, there must be a short
exact sequence 0 \to A \to B \to C \to 0 of R-modules such that
taking tensor products with (S\tensor S) yields another exact sequence,
but A\tensor S \to B\tensor S is not injective. 
Thus there must exist an element w\in A\tensor S which is mapped onto zero.
Now, A\tensor S\tensor S \to B\tensor S\tensor S is injective,
hence w\tensor s must be zero for all s \in S.
In the case where S is a ring (and not only a faithful module) this can't
happen: The multiplication of S yields a map \phi: A\tensor S\tensor S
\to A\tensor S such that \phi(w\tensor 1)=w, hence w\tensor 1 can't be zero
for non-zero w.
Thus the assumption "S\tensor S is flat and S is not" leads
to a contradiction.
Joerg

In article <32D4D3B9.386C@ipvvis.unipv.it>, Enrico Petracchi
 wrote:
> Which iterative method I have to use to solve nonsimmetric sparse linear
> system ??
> I'm looking for a free available package to solve it. 
> C is better, but fortran is ok too. 
You might try GMRES; there's an implementation of it in the Meschach matrix
package in C language. Cf. http://www.netlib.no/netlib/c/meschach/readme
--
Dr. Denis Constales - dcons@world.std.com - http://cage.rug.ac.be/~dc/

On 6 Jan 1997, John Edser wrote:
> Does anybody know if ZERO and NOUGHT are different?
> 
> It seems to me that zero is an invisible line
> from which numbers become positive or negative.
> 
> Nought means nothing.
You should ask a linguist.  In regard to the group of integers to which 
you refer, there exists an additive identity, 0, such that x+0=0+x=x
so "nothing" <==> this 'line' you mention.

Hi netters,
I just have a question to a possible algorithm to solve the following problem:
Given a set of points in a 3-D room, is there any (and only one) flat, and the
sum of distances from these points to the flat is the minimum? If yes, how to
get the flat?
Can anyone over there give me algorithm or hint? Please send email to
mackpei@cs.tu-berlin.de
Thanks in advance!

Clark Cooper wrote:
> 
> sockeye wrote:
> >...
> >
> > "A particle starts at the origin at t = 0 and moves along the s-axis
> > in such a way that its velocity at position s is ds/dt = [cos(Pi*s)]^2
> > (i.e., cosine squared of Pi times s). How long will it take the
> > particle to reach s = 1/4?"
> >
> > ...
> >         Thanks loads,
> >
> >                 =Eric
> 
> You need to integrate OVER S, which means that you need to get
> the expression in s on the same side as the `ds'.
> 
> ds/dt  = [cos(PI*s)]^2 =>
> dt = ds/[cos(PI*s)]^2
> 
> Integrate the right with respect to s and the left with respect
> to time and you will get:
> 
>        sin(pi s) - 1      cos(pi s)
>        ------------- - ---------------
>         2 cos(pi s)    2 sin(pi s) - 2
> t =    -------------------------------
>                      pi
> 
Hmmm, is this the output of some symbolic program (e.g., Mathematica)?
If one writes dt = ds/[cos(PI*s)]^2 as dt = [sec(PI*s)]^2 ds and
integrates both sides, the right side is an "elementary" integration
that my students were supposed to recognize; the result of integrating
and applying the condition (s=0 when t=0) is  t = [tan(PI*s)]/PI
Ron Winther

I am looking for a program to compute the integral homology of
a finite simplicial complex.  Does anyone know where I can
find such a program?  Thanks in advance.
-- 
David Radcliffe				radcliff@alpha2.csd.uwm.edu

In article <5b5a80$lj4@sun168.rz.ruhr-uni-bochum.de>,
  winkejbd@rz.ruhr-uni-bochum.de (Joerg Winkelmann) wrote:
> 
> Adam Logan (alogan@fas2.fas.harvard.edu) wrote:
> : 
> : Let R --> S be an injective homomorphism of rings (or, more generally, let
> : S be a faithful R-module).  If the tensor product of S with itself over R,
> : with respect to the given homomorphism, is flat over R, does it follow that
> : S is flat over R?
> : 
> : (The consensus among people I've asked seems to be "no", but no one has come
> : up with a counterexample.)
> Well, I'd rather suggest "Yes".
> Let's try to prove it.
> Flatness of S means that S\tensor_R is an exact functor.
> If S\tensor S is flat, but S itself is not flat, there must be a short
> exact sequence 0 \to A \to B \to C \to 0 of R-modules such that
> taking tensor products with (S\tensor S) yields another exact sequence,
> but A\tensor S \to B\tensor S is not injective. 
> Thus there must exist an element w\in A\tensor S which is mapped onto zero.
> Now, A\tensor S\tensor S \to B\tensor S\tensor S is injective,
> hence w\tensor s must be zero for all s \in S.
> In the case where S is a ring (and not only a faithful module) this can't
> happen: The multiplication of S yields a map \phi: A\tensor S\tensor S
> \to A\tensor S such that \phi(w\tensor 1)=w, hence w\tensor 1 can't be zero
> for non-zero w.
> Thus the assumption "S\tensor S is flat and S is not" leads
> to a contradiction.
This is a very nice argument. But one can simplify it even more. I'll denote the
tensor product by o. Let m: S o S --> S be the multiplication map, and define
f: S --> S o S by f(s) = s o 1. Then m(f(s)) = s which shows that the surjection
m splits, and so S o S has a direct summand isomorphic to S. Thus if S o S is
flat, then so is S.
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

In article <32D3A3DA.2E4B@math.chalmers.se>,
Jan Stevens   wrote:
>Michael A. Stueben wrote:
>> 
>> I was trying to list the different kinds of proof for my H.S.
>> precalculus students. So I gave direct, indirect, math
>> induction and proof by contraposition. Fine. But later I
>> thought what about this: proof by example (offer am
>> illustration of a situation or give directions for a
>> construction)? Or proof by verification
>> (substitute and complete a calculation). Then proof by cases
>> could be a proof made up of a mixture of all types of proof.
>
>
>
>You missed proof by intimidation.
Also Proof by Least Astonishment.
("I'd be surprised if it weren't true.")
--Andrew.
-- 
"I shall drink beer and eat bread in the House of Life."

In article <70D6wCA0Ep0yIwcx@tharris.demon.co.uk>, Rebecca Harris
 writes
>In article , David Kaufman 
>writes
>>                What Is Ethical Truth?
>>
>>
>>Introduction:
>>------------
>>
>>       Is a holy person (who never tells a lie) lying, if they
>>hide a person being chased by a killer (when confronted by 
>>the would be murderer) say, "The person ran that way"?
>>
>>       From a science or math prospective, which usually 
>>ignores the ethical consequences of revealing certain 
>>truths, the holy person told a propositional falsehood. 
>>
>>       However, from an ethical prospective, the holy person 
>>told the ethical truth because Truth in its human dimension 
>>also includes not harming others. Truth creates harmony, 
>>peace and joy. 
>>
>>
>>Is God All Powerful?
>>-------------------
>>
>>       One area that has caused me much mental suffering is 
>>the belief that I had because my spiritual teachers told me 
>>that God is all powerful and all loving.
>>
>>       However, I still believe God is all loving, but I now 
>>believe God is not all powerful, yet I still believe my 
>>spiritual teachers told the ethical Truth as described above
>>in the introduction.
>>
>>       How can an all powerful God allow legs to be blown off 
>>by land mines and all the other horrors happening right now 
>>as you read these lines?
>>
>>       To a would be logical person, if someone has the power 
>>to act to correct wrongs before their eyes and doesn't, then
>>that person is accountable for the harm they allow to 
>>happen. 
>>
>>       So the only conclusion about an all knowing and all 
>>loving God is that God can't be all powerful. Otherwise, it 
>>would lead a logical person to either ignore the existence 
>>of God or to hate God for Mankind's ongoing physical 
>>suffering.
>>
>>
>>How To Explain Why Holy People Say God Is All Powerful:
>>------------------------------------------------------
>>
>>       Breaking away from accepted beliefs is not only 
>>difficult for true believers (as I am), but requires setting
>>up an alternate plausible (non-contradictory) system of 
>>ideas to explain, "Why Do Holy People Say God Is All 
>>Powerful?" 
>>
>>       To help others, who might also be struggling with the 
>>question of how not to hate an all powerful God, I offer the
>>following 2 scenarios for our current condition.
>>
>>       On scenario (that is used to explain out current 
>>situation) is that the Devil created the world and that an 
>>all loving God to lesson Mankind's suffering made a deal 
>>with the Devil to gain access to the world through God's 
>>Saints. 
>>
>>       However, the deal with the Devil requires that all 
>>God's Saints claim God is all powerful and is the Creator of
>>all that exists. Otherwise, the Devil regains the power to 
>>create even greater harm to Mankind.
>>
>>       The other scenario imagines that God did create the 
>>universe--perhaps our galaxy the Milky Way--and has a grand 
>>plan to stop emerging suffering in the other galaxies. 
>>
>>       To stop enormous suffering elsewhere requires our 
>>current training in this galaxy to become fit for the tasks 
>>ahead to serve the universe by eventually creating peace and
>>joy for all sentient beings.
>>
>>       I hope the above ideas help those who are currently 
>>struggling with the concept of God as I did. My Best Wishes.
>>
>>------------------------------------------------------------
>>A brief note of value:
>>---------------------
>>
>>       Getting teachers to used numbered words properly in 
>>solving constant rate math and science problems could lesson
>>student and teacher suffering enormously, I believe.
>>
>>       If you are interested in promoting constant rate 
>>solutions using numbered words in the K-12 school system (or
>>elsewhere) as outlined briefly in my current post titled, 
>>"Numbered Word Forms To Algebra Equations." in newsgroup
>>k12.chat.science, please e-mail me your interests on this 
>>matter.  Thanks.
>>
>>-----------------------------------------------------------
>>    C by David Kaufman,                   Jan. 5, 1997
>>    Remember: Appreciate Each Moment's Opportunities To
>>          BE Good, Do Good, Be One, And Go Jolly.
>>
>>Note: Please feel free to share this 2 page article with 
>>      others without charge.
>
>What is was the point in writing all that "stuff" about god???
>I am an athieist(probably wrong spelling)But I believe that everyone is
>allowed their own opinion......So why preach about "the wonderful and
>powerful god"?
R33BOX, you are probably not athiest, but agnostic. But I agree with you. These
people think they can change somebody's life just by talking to them...
-- 
Goddess
The girl who cried "MONSTER!" and got her brother....
E-mail : goddess@segl.demon.co.uk
Homepage: http:/www.segl.demon.co.uk/frances

Xue-Jun PEI wrote:
> 
> Hi netters,
> 
> I just have a question to a possible algorithm to solve the following problem:
> 
> Given a set of points in a 3-D room, is there any (and only one) flat, and the
> sum of distances from these points to the flat is the minimum? If yes, how to
> get the flat?
> 
> Can anyone over there give me algorithm or hint? Please send email to
> 
> mackpei@cs.tu-berlin.de
> 
> Thanks in advance!
Unfortunately, there may not be a unique solution.  Take, for instance,
the case where you have eight points at the vertices of a cube.
Clearly, by symmetry there won't be a unique plane which minimizes
the sum of the distances from the points.
In cases where there is a unique plane which minimizes the sum of
the distances one can indeed find the equation of the plane, but
it's not--at least at first glance--a simple task.  Let your points
be (xi, yi, zi) and the desired plane be given by Ax + By + Cz + D = 0.
The distance from the point to the plane is well known:
     di = abs(A*xi + B*yi + C*zi + D) / sqrt(A^2 + B^2 + C^2)
so you want to minimize the \sum{i = 1, n}(di) or perhaps the
sum of the squares of those distances di.  In either case, an
approach would be to find the partial derivatives of the sum
with respect to A, B, C, and D, set each of the derivatives to zero
and solve the resulting system for A, B, C, D.  This, I'll warn you in 
advance, is not a task for the faint of heart, since the resulting
four equations aren't linear in A, B, C, D.
Hope this helps,
Rick
-----------------------------------------------------
Rick Decker                   rdecker@hamilton.edu
Department of Comp. Sci.      315-859-4785
Hamilton College
Clinton, NY  13323            =  !=  ==  (!)
-----------------------------------------------------

John Edser  wrote:
>Could somebody do my simple problem?
>
>FB=(((Phi)^x-(-Phi^-x))/sqrt(5))^gg
>
>what then does:
>
> gg =
	Well, first let me say that this looks like the closed
form for the Fibonacci numbers.  I assume your Phi is the 
honest-to-God Phi=[1+sqrt(5)]/2.  In that case, your formula
will yield the xTH Fibonacci number when you plug in an integer
x, provided gg=1.  Perhaps that is what you're ultimately looking 
for.
	If not, then we have ln(FB)=gg*[ln(Phi^x-(-Phi)^-x)-ln(5)/2].
Divideth thee both sides by thy string of confusing symbols 
on the right hand, and thy value for gg will be got.  
  .,-:::::   :::.         ....:::::: @niu.edu -- http://www.math.niu.edu/~caj/
,;;;'````'   ;;`;;     ;;;;;;;;;````
[[[         ,[[ '[[,   ''`  `[[.     "I'd like a large order of FiboNachos."
$$$        c$$$cc$$$c ,,,    `$$      "Okay sir, that'll cost as much as a
`88bo,__,o, 888   888,888boood88    small order and a medium order combined."
  "YUMMMMMP"YMM   ""` "MMMMMMMM"  _____________________________________________

We invite you to access educational enrichment programs, including free 
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Patrick De Geest (Patrick.DeGeest@ping.be) writes:
> 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 ?.
     I would be interested in palindromes which are impredictable: like
the 20 first digits of the square root of something (reciprocal of your
example above); generally irrational numbers showing palindromism.
--
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    ***************************

John  wrote:
Jean-Christophe Janodet wrote:
> 
> It's easy to explain why the sum of two negative
> numbers is negative, using the example of a barometer.
> Does anybody have a similar example to justify that
> the product of two negative integers is positive ?
> Or else, what are the theoretical reasons ?
> 
> Sincerely, Jean-Christophe.
A non-mathematical friend once asked me the same question, and this
was the argument I came up with---
We agree that zero times any number is zero and a number plus its
negative is  zero-- so, for example,
0 * (-3) = ( 2 + (-2)) *(- 3) = 2*(-3) + (-2)*(-3) = 0
Therefore, 2*(-3) must be the negative of (-2)*(-3) since they sum to
zero. 
 A similar argument
0 * 3 = ( 2 + (-2)) * 3 = 2*3 + (-2)*3= 0
shows that (- 2)*3 must be the negative of 2*3 since they sum to zero.
Putting the two conclusions together, since 2*3 is positive, 2*(-3)
must be negative; and since (- 2)*3 = 2*(-3)  must be negative,
(-2)*(-3) must be postive..
It -does- seem unnecessarily complicated..
Matt Feinstein
mfein@aplcomm.jhuapl.edu

Jean-Christophe Janodet wrote:
> 
> It's easy to explain why the sum of two negative
> numbers is negative, using the example of a barometer.
> Does anybody have a similar example to justify that
> the product of two negative integers is positive ?
> Or else, what are the theoretical reasons ?
> 
> Sincerely, Jean-Christophe.
Hi Jean-Christophe,
This difficulty is similar to the "1/0" flurry now on this site.
The explanation I found is:
"... To include the new symbols -1, -2, -3, ... in an enlarged
arithmetic which embraces both positive and negative integers, we must,
of course, define operations with them in such a way  that the original
rules of arithmetical operations are preserved. For example (-1)*(-1) =
1, which we set up to govern the multiplication of negative integers is
a consequence of our desire to preserve the distributive law...."
This was not an easily found rule and once found it was not easily
explained. Euler had trouble with it so why not you and I?
It appears that (-1)*(-1) = 1 is sort of an axiomatic afterthought.
This was all gleaned from _What is Mathematics?_ by Richard Courant and
Herbert Robbins, Oxford Univ. Press, 1941 (14th printing 1969), page 55.
So, unless someone out there has clearer paths to enlightenment, it's
what I believe.
Regards, 
John

Angel Garcia  wrote:
>
>Richard Mentock (mentock@mindspring.com) writes:
>
>> Hey, no stopping there.  Let's see the list of misprints!
>    
>       Yes: one cannot say 'there is ONE missprint or typo' without saying
>where. because this easily could become a brutal 'slandering' if that
>accusation is not substantiated.
	And one must be especially careful when assuming something
to be a typo in a mathematics book or article.  Think of the frequency
with which people come stomping into this group insisting that 
Marilyn Vos Savant is the biggest idiot on earth, because she printed
a puzzle solution in her column that didn't seem intuitively correct?
I admit that her book on the FLT has tarnished her reputation among
mathematicians, but long before that people were too quick to chalk
up alarming (but correct) result as a mistake on her part.
	On the other hand, I should point out, you do a great service
to the community here on sci.math by reporting errors.  If it's a few
typos and nothing more, then we will be prepared when we buy it.  
If a book is laughably bad, many people who might have paid real
green money for it won't, and will be very thankful.  A few months ago, 
I pointed out (alleged?) that Mike Schneider's book, _A Beginner's 
Guide_to_Constructing_the_Universe_, had so many errors both 
mathematical and non (and I don't mean typos, but real, honest-to-Ghod, 
"where the Hell did he get that from?" errors) that I would give a 
prize to whoever found the most.  I've been compiling a list of all
the mismathematics in that book over the course of this last semester,
and may post it if it will help someone out there avoid a purchase
she/he would later regret.  If anyone else has a copy of that infernal
thing, let me know:  perhaps you've seen some I haven't.
 ,oooooooo8    o     ooooo@math.niu.edu  -==-  http://www.math.niu.edu/~caj/
o888'   `88  ,888.    888   
888         ,8'`88.   888 "Hey, you got your chocolate in my cod liver oil!"
888o.   ,oo,8oooo88.  888 "Hey, you got your cod liver oil in my chocolate!"
`888oooo88o88o  o888o 888                       -Common occurrence in the
___________________8o888'________________________days before peanut butter__

In article , Ken Fischer  wrote:
>
>        How will you be able to tell if it is a black hole? :-)
>Only one star in 10,000 has greater than two solar masses,
>do I read that right?    Does that mean by the time they
>burn all available fuel, they may be close to the mass of
>the Sun, and then a supernova would leave essentially
>not enough to make a black hole?
>
>Ken Fischer 
Stars don't lose appreciable mass through nuclear fusion; only
the mass equivalent of their energy output.  If a star is 20 sm
before SN, then a signifigant part of that is literally blown
away.  And a BH is inferred from it's gravitational effects
and lack of radiation.   

how big would a black hole be?  i don't know much about astrophysics, so i
don't even know in what units the answer would be.  how big would it be
compared to the earth or to the sun?  when i say big i mean diameter or
circumference.
luke

In article <5b3rn0$u22@news.doit.wisc.edu>,
  hetherwi@math.wisc.edu (Brent Hetherwick) wrote:
> 
> Let F be a field.  Let R be the ring of 3 x 3 matrices over F with (3,1) 
> and (3,2) entries 0.  Thus,
> 
> 	     ( F F F )
> 	R =  ( F F F )
> 	     ( 0 0 F )
> 
> How can I go about calculating the Jacobson radical J(R)?  I assume that 
> this is necessary to go about showing that J(R) is a minimal left ideal, 
> but not a minimal right ideal.
R maps onto both M_2(F) (top left 2x2 square) and F (bottom right entry).
Use the (easy) theorem that if f: R -> S is a ring surjection then f(J(R))
is contained in J(S) to deduce that J(R) is contained in
	I = (0 0 F) .
	    (0 0 F)
	    (0 0 0)
It is easy to see that I is a minimal left ideal, and a non-minimal right
ideal of R. But I is an ideal and if x is an element of I then 1 + x is
plainly invertible in R. Hence I is contained in J(R).
( J(R) = {x \in R: 1 + axb is invertible for all a, b \in R} ).
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

In article <5b1sf0$741@catapult.gatech.edu>, gt4654c@prism.gatech.edu 
(Jeff Cronkhite) wri>:
> 
> 
> 
>    So does the Ricci tensor then actually throw away some of the
> information contained in the Riemann curvature tensor?  
Yes (in dimensions greater than 3).
I guess I need
> to figure out how many truly independent components are in the Riemann
> tensor in four dimensions, and whether there are fewer than that in
> the Ricci (contracted Reimann) tensor.  
20 components for Riemann in 4-dimensions.  10 from Ricci, 10 from the Weyl
tensor which is the completely trace-free part of Riemann.

It's easy to explain why the sum of two negative
numbers is negative, using the example of a barometer.
Does anybody have a similar example to justify that
the product of two negative integers is positive ?
Or else, what are the theoretical reasons ?
Sincerely, Jean-Christophe.

calvitti@kevin.ces.cwru.edu wrote:
> 
> given that the solution of a linear system dxdt = A.x in R^n is
> 
> (*)     x(t) = exp(A.t).x(0)
> 
> is the time "t" to reach x(t) from x(0) well defined for all A? seems
> that t would be given by the matrix natural log: nl(A). at least this
> works for scalar systems and i imagine that if A can be diagonalized,
> then it should be just as easy in higher dimensions.
> 
> intuitively, the time between two points in the state space - provided
> they are connected by a flowline - is well defined; it could be found
> by numerical integration for instance. however, i don't know how this
> is related to the matrix logarithm.
> 
> for example if A = {{-1,0},{0,-2}} then the (uncoupled) solutions are:
> 
>         x(t) = exp(-t).x(0)
>         y(t) = exp(-2t).y(0)
> 
> where {x(0), y(0)}, {x(t), y(t)} corresponding to the initial and
> final points in R^2 are given. the eigenvalues of A are {-1,-2}.
> however, doesn't the matrix log of a matrix with negative eigenvalues
> have complex entries? how is this matrix log computed to begin with?
> (matlab has a function to compute it numerically)
> 
> can someone email/post some insight or point to the literature?
> thanks for the info,
You have to be careful: Even though tA might have
negative eigenvalues, exp(tA) will not.  Provided
you take the right branch of the logarithm
log(exp(tA)) = tA.
The easiest way to see that the theoretical
solution of dx/dt= Ax with A constant is x(t) =
exp(tA)x(0) is to expand exp(tA) as a MacLaurin
series: exp(tA) = I + tA + (1/2)(tA)^2 + ...
Differentiate this w.r.t. t, and you'll see
immediately that dx/dt= Ax.
Finally, the solution of functions of matrices is
actually quite interesting - and not at all
trivial.  You might, for example be tempted to
compute the matrix exponential by the Taylor's
series.  This turns out to be a bad idea. For
some of the standard techniques, refer to Golub
and Van Loan's classic "Matrix Computations" (now
in third edition!)  
In my opinion, though, the best technique to
compute functions of matrices is the
Schur-Frechet algorithm by C. Kenney and
A.J. Laub.  In fact, the first paper on this
subject was concerned with the computation of the
Matrix Logarithm (Technical Report number
CCEC-95-0714 "A Schur-Fréchet Algorithm for
Computing the Logarithm of a Matrix" C. Kenney
and A.J. Laub, July 14th, 1995.) I'm not sure if
it has been published yet, but you can get a
reprint by snail mail from UCSB.  See the web
page at URL:
http://www-ccec.ece.ucsb.edu/pubs/techrpts/index.html
for further information.
I hope this helps, Cheers, John
-------------------------------------------------
 Dr. J.J. Hench  
 Dept. of Mathematics, Univ. of Reading, England   
 Institute of Informatics and Automation, Prague
-------------------------------------------------

electronic monk wrote:
> 
> ibokor wrote:
>  >
>  > Anonymous (abcd@efgh.net) wrote:
>  > : The limit of 1/x as x --> 0 is infinity.
>  > :
>  >
>  > If you're talking about the real numbers, then
>  > what you have claimed is not true.
>  >
>  > Even if you were to be sloppy about the use
>  > of the word "limit", there would still not be
>  > a limit, since if x is negative, so is 1/x
>  > and if x is positive, so is 1/x. Hence the only
>  > possible limit would be 0, the limit would
>  > have to be non-positive and non-negative.
>  >
>  > d.A.
> 
> lim   1/x = oo
> x->0+
> 
> lim   1/x = -oo
> x->0-
> 
> hence
> 
> lim   1/x = DNE
> x->0
> 
> it does not exist because it is different from the left than from the
> right.  but both limits ARE infinite, meaning that 1/0 is an infinite
> number, either +oo of -oo.  limits were made to evaluate functions at
> infinite and indeterminante values, that is their point in existance.
> 
> electronic monk
Existence is indeed the issue for 1/0. This expression has nothig to do
with limits. 1/0 cannot be defined because contradictions would result
in our reasoning. 0^0, ln(0), also do not exist and therefore can not
equal any object finite or infinite. 0/0 cannot equal anything
(certainly not 1) without admitting inconsistency into our logic.
Owen Holden  (My first try at news groups...I could not resist your
topic)

Let 'A' denote the universal quantifier, 'E' the existential, and '<' the 
"element of" relation.  Then,
	Ey Ax (x < y iff x < z & x !< z)
is an instance of the Comprehension schema.  The y so picked out is the 
empty set.  Its uniqueness follows from Extensionality.
$$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666
		       hetherwi@math.wisc.edu
$$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666

blackj@toadflax.cs.ucdavis.edu (John R. Black) writes:
>What is your favorite "cute" proof?  The irrationality of sqrt(2)?  The
>fact that there are an infinite number of primes?  The proof that all
>numbers are interesting? (This one's more of a joke of course)
1/4 = 16/64 = 166/664 = 1666/6664 = ...
Proof:  Cancel the 6's.  HAW!
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Is God famous?  Or is He just very, very, very, very well-known?

In article <01bbfc36$cc5b7540$85160ccb@michaelj>, Michael Joyce
 writes
>>      One purpose of Goals 2000 is to have K-12 students 
>> achieve worldclass standards (by the year 2000) in math and 
>> science.
>
>Err... does that mean to make American students as good as the other
>countries by that WORLD CLASS bit? hmmmm...
Exactly. Hmmmmm...
-- 
Goddess
The girl who cried "MONSTER!" and got her brother....
E-mail : goddess@segl.demon.co.uk
Homepage: http:/www.segl.demon.co.uk/frances

In article ,
jhallen@world.std.com, aka Joseph H Allen wrote:
> Now the statement "all non-black things aren't ravens" is logically
> equivalent to "all ravens are black".  Thus all of the non-black things you
> find which aren't ravens (your red coat, the white ceiling, etc.) also
> support your generalization that "all ravens are black".
Nevertheless an aristotelian view would suggest that the statement
'all ravens are black' actually means 'all ravens are black ravens'
where these ravens obviously haven't yet got a colour attribute.
Constructing quasi-paradoxons by using weak language doesn't prove
much but entertains simply. The equivalence of the suggested correction
would be boring: 'the statement 'there is one raven which is not a
black raven' is false'.
-- 
Michail Brzitwa                        +49-511-343215

The book  
                 LES R\'ESEAUX PARFAITS DES ESPACES EUCLIDIENS
                     (PERFECT LATTICES IN EUCLIDEAN SPACES) 
                                      by 
                                Jacques MARTINET
                        Professor, University of Bordeaux 
recently appeared. It is sold at the price of 385 FF, 
and can be ordered at 
	                Editions MASSON, 
	                5, rue Laromigui\`ere, 
	                F-75241 Paris cedex 05. 
                        *********************** 
Here is a translation into English of the back cover of the book, followed 
by the table of contents: 
                        *********************** 
This book is dedicated to beginning graduate (or to advanced undergraduate)  
students in mathematics or computer science, as well as to researchers. 
The reader is expected to be somewhat familiar with the basic techniques 
of algebra and Euclidean geometry that one can usually learn 
in graduate courses. 
To a given lattice in a Euclidean space in naturally attached a 
``regular'' sphere packing. To exhibit dense sphere packings is one 
of the main problems in the geometry of numbers. The property of 
perfection, the central topic of this book, is a property of 
a linear nature, which is fullfilled by all extreme lattices, 
those on which the density attains a local maximum. 
The book contains many (164) exercises. Two introductory chapters and 
four appendices will help the reader to master the language of the theory 
as well as the techniques from algebra which are needed. 
                        *********************** 
                    TABLE DES MATI\`ERES  --  CONTENT 
Introduction (Introduction)
Chapter I 
   G\'en\'eralit\'es sur les r\'eseaux 
   Generalities on lattices 
Chapter II 
   In\'egalit\'es g\'eom\'etriques 
   Geometrical inequalities 
Chapter III 
   Perfection et eutaxie 
   Perfection and eutaxy 
Chapter IV 
   Les r\'eseaux de racines  
   Root lattices 
Chapter V 
   R\'eseaux li\'es aux r\'eseaux de racines 
   Lattices related to root lattices 
Chapter VI 
   R\'eseaux parfaits de petitie dimension 
   Low-dimensional perfect lattices 
Chapter VII 
   L'algorithme de Vorono{\"\i} 
   The Voronoi algortithm 
Chapter VIII
   R\'eseaux hermitiens 
   Hermitian lattices 
Chapter IX 
   Les configurations de vecteurs minimaux 
   Configurations of minimal vectors 
Chapter X 
   Extr\'emalit\'e dans des familles de r\'eseaux 
   Extremality in families of lattices 
Chapter XI 
   Op\'erations de groupes 
   Group actions 
Chapter XII 
   Sections des r\'eseaux 
   Sections of lattices 
Chapter XIII 
   Extensions de l'algorithme de Vorono{\"\i} 
   Enlargements of the Voronoi algorithm 
Chapter XIV 
   Donn\'ees num\'eriques 
   Numerical data 
Appendix 1 
   Formes quadratiques et anneaux de Dedekind 
   Quadratic forms and Dedekind domains 
Appendix 2 
   Les groupes de quaternions 
   Quaternionic groups 
Appendix  3 
   Alg\`ebres semi-simples 
   Semi-simple algebras 
Appendix 4 
   Arithm\'etique dans les alg\`ebres semi-simples 
   Arithmetic in semi-simple algebras 
Bibliographie 
References 

kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) wrote:
>
> it's a definite plus if you find a buddy to study with: compare notes, 
> ask questions, etc. Often I got closer to the answer just by asking the 
> question out loud (i.e. realizing what to ask about);
        IMO, this might be the  function of the newsgroups to the
        people that can't find a person that enjoys math to  talk
        with.
        Even  here  in  my  University,  there is only a selected
        group of 6 (maybe)  persons  with  whom  I can talk about
        Abstract Algebra, for instance.
        The  newsgroup  is a way (at least to myself) to bring me
        closer with people that really enjoys what they're doing.
        It is often a place  where  I find motivation to subjects
        that I thought were not interesting or that I didn't even
        imagine they existed.
        Just some  thoughts  that  I  guess  someone  might  find
        interesting, Roger...
--
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
  Rogerio Brito - rbrito@ime.usp.br - http://www.ime.usp.br/~rbrito 
  "Master, Master, where's the dreams that I've been after?
   Master, Master, you promised only lies!
   Laughter, laughter, all I hear or see is laughter.
   Laughter, laughter, laughing at my cries."
             James Hetfield (Metallica), Master of Puppets
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

Christian Wieschebrink  wrote in article
<5a8t4g$8t1@news00.btx.dtag.de>...
> 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?
> 
> 
Your conjecture is (lucky you) true, and the proof is relatively simple.
Write the equation in the form q^2=p^3-1. Thus it becomes equivalent with
	q^2=(p-1)(p^2+p+1).
Now it would be nice if p-1 and p^2+p+1 were mutually prime. Let
d=(p-1,p^2+p+1) so that d|(p-1) and d|(p^2+p+1).
It is easy to see that d|3. Now, either d=1 or d=3.
Suppose d=1, then both p-1 and p^2+p+1 must be squares, but
p^2<(p^2+p+1)<(p^2+2p+1) and the latter can't be a square. Contradiction!
Next, suppose d=3. Then, write p-1=3m, p^2+p+1=3n, and subsequently q^2=mn,
where m and n are mutually prime. Hence m=u^2 and n=v^2.
Now we have p^2+p+1=3v^2, and, multiplying by 4 we get (2p+1)^2+3=12v^2,
(6u^2+3)^2+3=12v^2, and, finally,
	(2v)^2-3(2u^2+1)^2=1.
Hence, (2v) and (2u^2+1) are solutions of the Pell's equation x^2-3y^2=1.
First nontrivial solution of this equation is (x_1,y_1)=(2,1), and
by the known theory, all of its solutions are generated by
x_n+y_n=(2+\sqrt{3})^n. Writing x_n-y_n=(2-\sqrt{3})^n, we obtain a simple
formula
y_n=((2+\sqrt{3})^n-(2-\sqrt{3})^n)/(2*\sqrt{3}). Now the easy induction
shows that y_n satisfies the recurrence formula y_n=4*y_{n-1}-y_{n-2},
and hence the parity of y_n alternates (y_n is 1,4,15,56,209,...). Since we
need y_n=2u^2+1, it suffices to prove that among the members of
an array z_n=(y_{2n-1}-1)/2 there are no perfect squares.
By substituting the formula z_n=(y_{2n-1}-1)/2 into
y_n=((2+\sqrt{3})^n-(2-\sqrt{3})^n)/(2*\sqrt{3}) and using a trivial
induction, we can see that
z_n satisfies the reccurence formula z_n=14*z_{n-1}-7*z_{n-2}+6. Clearly,
z_2=7 is not a square (while z_1=0 corresponds to the trivial solution
(p,q)=(1,0)), and, for n>=3, we have that z_n is congruent 6 modulo 7. But
it is easy to see that any square is congruent 0,1,2,4 modulo 7, and
it follows that z_n cannot be a perfect square, qed.
Djordje Milicevic, Belgrade (student). Email: djolem@eunet.yu,
mm96133@alas.matf.bg.ac.yu.
PS. There is a similar problem p^3-q^2=5, which is efficiently solved using
the quadratic residues theory.

There is an error in the previous solution I posted here!
Djordje Milicevic  wrote in article
<01bbfa68$3492cf80$LocalHost@djolemlite>...
> 
> 
> > 
> >     There are no positive integers p,q, such that 
> >     p^3 - q^2 = 1 
> >     is true.
> Your conjecture is (lucky you) true, and the proof is relatively simple.
blah, blah
> By substituting the formula z_n=(y_{2n-1}-1)/2 into
> y_n=((2+\sqrt{3})^n-(2-\sqrt{3})^n)/(2*\sqrt{3}) and using a trivial
> induction, we can see that
*** Now comes the error! ***
> z_n satisfies the reccurence formula z_n=14*z_{n-1}-7*z_{n-2}+6. Clearly,
*** hence the rest of the message is nonsense.
> z_2=7 is not a square (while z_1=0 corresponds to the trivial solution
The recurrence equation is z_n=14*z_{n-1}-z_{n-2}+6.
At the moment I don't know a nice way to complete the solution.
Sorry for inconvinience.

Simon Read  wrote:
>jac@ibms46.scri.fsu.edu (Jim Carr) wrote:
>>davk@netcom.com (David Kaufman) writes:
>>>
>>>The square root of 2 can be written in Basic computer language
>>>as follows: 2^.5 or 2^(1/2) or SQR(2).
>>                     =======
>>
>> If this is valid Basic, no wonder it is commonly said that those 
>> who learn Basic first are often crippled for life as programmers. 
>> That expression is equal to 1 in other high-level languages. 
>rubbish nonsense rubbish nonsense drivel nonsense garbage
>spew bilge tosh tripe rhubarb moonshine nonsense hogwash
>2^(1/2) is the square root of two unless you deliberately do
>something strange, like using integer variables. There are
>probably lots of compiler options to chane the rules about this,
>but that is highly system-specific and not relevant here.
You are quite incorrect. It isn't simply a matter of selecting
compilier options. The result of the expression 2^(1/2) is dependent
on the features of the language used.
For example, this is a legal expression in ANSI C. A compiler that is
compliant with the ANSI C specification will do an interger divide and
compute 1/2 as 0. It will then do a bit wise XOR of this result with
the value 2 with a final result of 2.
For some other language, there might well be an integer divide
followed by an exponentiation with the result of 1.
For still yet another language the result might well be a floating
point divide followed by an exponentiation.
For still others this might not even be a legal statement and will
simply cause a compiler error.
It is rather futile and silly to argue as to which of these
possiblities are more correct. It is critical to know how the language
you are programming in will evaluate this expression as well as any
other expression you might use. If you don't like the way ANSI C
evaluates the expression 2^(1/2) then you should use a different
programming language.

In article <32D03E45.5D76@idirect.com>,
Owen Holden  wrote:
>Existence is indeed the issue for 1/0. This expression has nothig to do
>with limits. 1/0 cannot be defined because contradictions would result
>in our reasoning. 0^0, ln(0), also do not exist and therefore can not
>equal any object finite or infinite. 0/0 cannot equal anything
>(certainly not 1) without admitting inconsistency into our logic.
One nit.  The expression 0^0 does indeed have a value, which has
nothing to do with limits.  This is discusses in the sci.math FAQ,
which gives several reasons why 0^0 = 1, at least in the case where the
exponent is considered to be an integer.
My favorite reason:  0^0 is the cardinality of the class of functions
mapping the empty set to itself, which is one.
Dave Seaman
-- 
Dave Seaman			dseaman@purdue.edu
      ++++ stop the execution of Mumia Abu-Jamal ++++
    ++++ if you agree copy these lines to your sig ++++
++++ see http://www.xs4all.nl/~tank/spg-l/sigaction.htm ++++

David Kastrup  wrote:
>Pretty much unthinkable in most civilized states with a separation of
>religion and state.  Now of course I am aware that the Jewish canon of
>law which is supposedly valid for Christians as well contains *very*
>strict outruling of homosexuality.  But it contains a host of other
>rules with equally strict penalties which nobody cares a bit about any
>more.  We don't lock women away for the time of their period, for
>example.  No gynaecologist and his patients get sentenced to death
>because they uncovered the "blood flow" of the woman.
>
>But Christians have always been very selective in what laws they want
>to be zealots about, and the American are traditionally pretty zealous.
>
    No, the death penalty is only for serious crimes such as homosexual 
perversion. There is some mention of uncleanness regarding menstruating, 
but that is about it. A great nation would outlaw homosexual perversion 
and promote a good culture for quality people to live in. And the Old 
Testament is not Jewish canon. The chosen race are the White people and 
described as White in the Bible. Jews falsely claim to be the chosen, 
when in fact they are the race of Anti-Christs and enemies of God.

On Mon, 06 Jan 1997 22:45:49 -0700, Mark Friesel
 wrote:
>Sorry for asking a question that seems rather trivial, but I was 
>wondering if someone could review, in a reasonable space, why black 
>holes form and the conditions under which they do so. 
Black Holes in the GR sense remain hypothetical.