Newsgroup sci.math 157664

Ron Winther wrote:
> 
> Clark Cooper wrote:
> >
> > sockeye wrote:
[chop]
> > 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)?
> [Should be]  t = [tan(PI*s)]/PI
> 
> Ron Winther
Ron, 
FWIW you, me(after receiving help), and Mathematica 2.2 agree on
tan(pi*s)/pi. 
Thanks to all who replied

Factor each term: (2^3-1)/(2^3+1)=((2-1)(2^2+2+1))/((2+1)(2^2-2+1)),
etc.  Then notice that a typical term in the numerator x^2+x+1 equals a
term in the denominator one step farther along, namely (x+1)^2-(x+1)+1.
 Then cancel like crazy:
1*(2^2+2+1)*2*(3^2+3+1)*3*(4^2+4+1)*4*(5^2+5+1)      *(n-1)(n^2+n+1)
----------------------------------------------- ...
3*(2^2-2+1)*4*(3^2-3+1)*5*(4^2-4+1)*6*(5^2-5+1)      *(n+1)(n^2-n+1)
After canceling, you are left with
1*2*(n^2+n+1)
-------------
n*(n+1)*(2^2-2+1)
The limit of this as n-->00 is the 2/3 that you found.

macchi@marina.scn.de (Gian Carlo Macchi) wrote:
>I've not followed this thread until now. Anyway, in your example, Roman I 
>always counts 1, independently of its position, and X always counts 10. 
>That is why this numbering system is not positional.
I think it could reasonably be argued that in 'VI' the symbol 'I' has 
a value of 1, but in 'IV' the symbol 'I' has a value of -1.  
Heath

On Mon, 13 Jan 1997, Heath David Hart wrote:
> macchi@marina.scn.de (Gian Carlo Macchi) wrote:
> 
> >I've not followed this thread until now. Anyway, in your example, Roman I 
> >always counts 1, independently of its position, and X always counts 10. 
> >That is why this numbering system is not positional.
> 
> I think it could reasonably be argued that in 'VI' the symbol 'I' has 
> a value of 1, but in 'IV' the symbol 'I' has a value of -1.  
And like the Egyptian and Greek numeration system before it was a form
of cipherization. I'd like to remind a few historians out there that Boyer 
pointed out that cipherization was an important concept that was 
introduced by Middle Kingdom Egyptians.
Milo

Jo E. Chen (jochen@princeton.edu) wrote:
: Odds each time = 6[(1/6)^5] = a, say. 
: 
: From there you can work it out...., e.g.
: 
: Odds this happens at least once = 1- odds never happens 
: 				= 1-(1-a)^3
: 
: Will someone tell me if I got it wrong, please?
 Right.  a=1/6^4=1/1296 is probability of 5 alike.
 (1-a)^3=(1295/1296)^3=0.997687... so
 1-(1-a)^3=0.002313...     where ^ denotes "raise to power"
: 
: ---------
: jean e nichols wrote:
: > 
: > Please forgive me for such a lowly question, but I am not a math type
: > (although I am fascinated by the subject). The question: What are the odds
: > of rolling 5 die of the same out of a total of 5 die, given 3 chances?
: > If this post is inappropiate to this newsgroup please forgive me. I have
: > tried to find an answer to this question all over the net!
: > 
: > jeanne@korrnet.org
One might also try sci.stat.consult
-- 
Michael P. Cohen                       home phone   202-232-4651
1615 Q Street NW #T-1                  office phone 202-219-1917
Washington, DC 20009-6310              office fax   202-219-2061
mcohen@cpcug.org

In article <5bbk9t$5t@news.ox.ac.uk>,
Thomas Womack  wrote:
>Nick Johnson-Hill (nickjh@globalnet.co.uk) wrote:
>: Is it possible to have factorials of non-integers ? I was always taught
>: that the answer is NO but I found the following on a Mathematics web
>: page:
>
>: (1/2)! = Square root of Pi
>
>What you do (I'll over-simplify and be shouted at by one of the Proper
>Mathematicians present) goes something like this.
>
>Let's consider the integral from 0 to infinity of t^x * e^t dt (which
>is a function of x). By integrating by parts, you find that its value for
>x an integer is exactly x!. So we *define* x! by this integral for x not
>an integer. It so happens that (1/2)! is exactly the square root of Pi.
 Fear not, I'll whisper:
 (1) it is t^x * e^(-t) dt to be integrated
 (2) the danger of confusion with Gamma function is always present; 
     in the above notation, x! is Gamma(x+1).
     It is Gamma(1/2) which equals sqrt(pi), and the recursion 
       Gamma(x + 1) = x * Gamma(x) leads to 
          (1/2)! = Gamma(3/2) = (1/2) * Gamma(1/2) = (1/2) * sqrt(pi).
And Gamma(1/2) = integral[0 to infinity] t^(-1/2)*e^(-t) dt (t=u^2)
               = 2 * integral[0 to infinity] e^(-u^2)) du
 can be calculated through its square, using polar coordinates, and comes 
 out as sqrt(pi).
Cheers, Slavek(ZVK).
>The Eternal Union of Soviet Republics lasted seven times longer than
>the Thousand Year Reich
We used to say "With Soviet Union forever, and not a minute longer!"

In article  <199701122158.NAA19901@alumni.caltech.edu>
ikastan@alumni.caltech.edu (Ilias Kastanas) wrote:
:	Dear Mr. Leko
:
:	1)  I is not imaginary.  Both R and I in  R + iI are real.  Re-
:   member e^z...   I = e^x sin(y), for real x and y.  I is real.
:
:	All my identities are correct by simple math.  Please point out
:   any mistakes.
:
:	2)  atan(tan(y)) = y is true for -pi/2 < y < pi/2 and false for
:   other y.
:
:	3)  I read A as it was.  Yes, "zero" means R = I = 0; but your A
:   is = 0 even when  R = I =/= 0, thanks to  -sin(..) + cos(..).
:
:	That is why your A is wrong, and produces nonexistent "zeros".  I
:   explained this months ago but you ignored me.  So you can hardly
complain
:   for the tone of my post.  And it was _right_: you _still_ refuse to
:   believe elementary trigonometry, and your plots show infinitely many
:   zeros for  z^2... and for  e^z!
:
:	Why not _fix_ it??  A = R^2 + I^2, that's all it takes!
             Thank you for your contribution.
tleko@aol.com

I need some help with this problem dealing with related rates.
A boat is being pulled into a dock with a rope.  The rope passes through a 
pulleyon the dock that is 1.5 metres above the bow of the boat.  If the 
boat is approaching at a rate of 50 cm/s, how fast is the rope being 
pulled in when the boat is 5 m from the dock?
Thanks for your help.
-------------------==== Posted via Deja News ====-----------------------
      http://www.dejanews.com/     Search, Read, Post to Usenet

I understand the fundamentals of binary, but how do you count in Hex?
Ie, how do you translate 8F into base ten, and how do you translate 254
into hex?
-dan

THIS IS A TEST OF THE EMERGENCY POSTING SYSTEM >...

Erwin Morales wrote:
> 
> Does any one know how to express the eigenvalues of (A - BA) in terms
> of the eigenvalues of A and B which are square matrices of the same
> order?
> 
> Thanks a lot.
> 
> Erwin Morales
> Nagoya University
> etm@usa.net
I'm afraid it does'nt possible in simple way.
The only Tr() functional of matrix are linear.
Tr(A+B)=Tr(A)+Tr(B)
Tr(AB)=Tr(A)*Tr(B)
But, if any apropriate solution exist, say for special forms of A B,
please let me know.
                Regards,
                          Dmitry Tikhonov.

Hello,
	recently my sister's math teacher asked a question, and did not accept
her
logical answer (He gave an "F" to the poor girl). I will be meeting with
the teacher this week to discuss this matter, and am seeking support to
show the teacher the error in his ways.
His question for his standard 7th grade math class, in verbatim,  was as
follows:
Q) Take a square piece of paper. Fold it in half. Do it again. Repeat 25
times. How many sheets thick is the final folded piece of paper.
My sister realized that this was a trick question, as she knew a piece
of paper can
not be folded that many times in half, and so far every question had
been based in reality. She came up with this description of her answer :
A) Take a piece of paper and fold it in half. Now, to "Do it again" you
have to unfold the piece of paper and refold it again. (This would be
like closing your hand into a fist and being told to "Do it again." You
first have to open your hand) Repeat this 25 times. The final paper
would have just one fold, being two sheets thick.
The teacher gave this answer an "F", stating that the answer was wrong.
I disagree.
He asked a trick question, and he recieved a very logical answer. In
fact, the only answer that I can concieve that can actually work. She
has been able to answer the questions using first hand examples to
determine other problems. (Such as blocks of wood for area problems).
Further, if he wanted to reach an answer of 2^25 he should have phrased
the question much better, such as "Assuming you can fold a piece of
paper...."
I am seeking opinions on his grade. It appears to me that he does not
consider other answers. I fear that my sister will come to think that
teachers are just out to get her, and that there is only 1 way to do
things. I would like to present this teacher with opinions from other
math teachers, college professors, or degree holders that his refusal of
this answer is professionally wrong.
If you could be so kind as to email me, it would greatly be appreciated.
PS If anyone wants to show other theoretical answers other than 2^25 I'd
also appreciate that. After all, doesn't space have 10 dimensions? That
concept alone would seem to
produce some rather odd answers.
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er.. look, im pretty interested, im 14 in sydney and i can make it
bright and entertaining, im pretty good at getting edutainment going and
i think i do an acceptable job in html, ill be happy to join your group
if at least to make it bold and entertaining.
My homepage is at http://www.fl.net.au/~juzzy/pabs.html but many of the
links are buggered at the moment due to circumstances beyond my control,
im pretty good at grasping ideas and have topped my year in most
subjects.
Well, thanks a lot and remember, dont eat dirty nappies.
Cya
Pabs

In article <32d957c8.0@eclipse.wincom.net>,
Clayton Smith  wrote:
>I have a question regarding the uses of calculus.  Is it possible to
>determine the surface area of a function in two parameters over a
>given range?
Yes.  For a surface z = f(x,y) with x and y in a domain D, the area
is the double integral over D of sqrt(f_x^2 + f_y^2 + 1) dx dy, where
f_x and f_y are the partial derivatives with respect to x and y.
>I have checked through a few calculus textbooks, but all I have
>been able to find so far is a method for determining the arc length of
>a given curve.
Any reasonable text of multi-variable calculus will have this.  For
example, Stewart, Calculus (3rd Edition), section 13.6.  I suspect
you've only been looking at single-variable calculus books.
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             (604) 822-3629
University of British Columbia            fax 822-6074
Vancouver, BC, Canada V6T 1Y4

spectrum wrote:
> 
> I have discovered an EXCELLENT scholarship opportunity while cruising
> around on the WWW.  This scholarship is a contest where 3 students
> team up and design a web page(s) that is entertaining yet
> educationally enlightening...  And if they do a good job they get BIG
> money (there are lots of prizes being given out - the total amount for
> all scholarships is around $1,000,000!).  Unfortunately everyone that
> goes to my school (except myself, of course ;) seems to be a complete
> knucklehead when it comes to internet technology and hard science  =)
> Also, the contest stresses "Collaboration".  You are even scored on
> collaboration.  How do you score high in this area?  Get students from
> all around the country and even the world to work on a project
> together.
> 
> I am a sixteen year old HS junior, and am very science/math/computer
> technology oriented.  If you match this bio (not necessarily 16 or a
> junior, but 18 or under and in high school) PLEASE contact me and I'll
> give you the URLs to more information and we can possibly get a team
> organized.  The prizes are so fantastic that any dynamic, intelligent
> high schooler ought to enter (universities are SO expensive now...).
> 
> We have plenty of time until the completed files are do, but we should
> get the registration filled out within the next couple weeks.  Thanks
> a lot.
> 
>         - sean baxter
> 
>         spectrum@mad.scientist.com
>                 or
>         baxters@eburg.com
> 
> P.S.    That's $25,000 PER student.  Plus that there are two $20,000
> prizes PER student in the group, five $15,000prizes  PER student in
> the group, five $12,000 per student in the group, five $12,000 per
> student, five $9,000 per student... etc.  I doubt it would be possible
> for us to enter and walk away empty handed.
> 
> P.P.S.  The foundation that sponsors this are allowing 3 "coaches" -
> teachers or mentors who are over the age limit and don't want another
> college eduction - to be involved.  So those of you who work with me
> on this could each choose a favorite teacher to help with the
> information and check everything for validity.
> 
> I believe I have some EXCELLENT ideas (plus I am pretty smooth with
> the ol' HTML editor ;) that could possibly put us in the winner's
> circle.
> 
> So email me if you are interested and please include your location of
> residence (City, State, Country will be fine), your interests, your
> computer abilities (i..e programming or HTML or 3D graphics or CGI or
> photoshop or whatever) and ideas with this contest that could help us
> fund an education to CalTech ;)
ah, i sent you an e-mail b4, could u give me the url for this thing too?
thanks :D

I have heard there is a way to fold a square of paper to get a Pentagon or a 5-pointed star.
I am familiar with tying a strip of paper into an overhand knot, but this is not the method
I have in mind.  Does anyone know the method?
I am not a regular on this newsgroup, so I would appreciate e-mail if you have a solution.  I
am at  dschandler@frumble.claremont.edu
Thanks.
--David Chandler

The Vietnam war had to be fought. It was the only way to prevent disruption of
the balance of power, or more bluntly much more bloodshed. It was fought
wrong, that I agree with you. The mistakes were more military than political.
Eddo

In article <32DA6D00.57A6@sunny.iteb.serpukhov.su>,
Dmitry Tikhonov   wrote:
>Erwin Morales wrote:
>> Does any one know how to express the eigenvalues of (A - BA) in terms
>> of the eigenvalues of A and B which are square matrices of the same
>> order?
>I'm afraid it does'nt possible in simple way.
Correct.  Of course 0 is an eigenvalue of A - BA = (I-B)A if and only if
it is an eigenvalue of I-B or A, i.e. 0 is an eigenvalue of A or 1 is an
eigenvalue of B.  Other than that there isn't much that can be said.
For example, for any t<>0 there are two 2 by 2 matrices A and B such that
both A and B have 1 as their only eigenvalue, but AB has t as an eigenvalue.
>The only Tr() functional of matrix are linear.
I don't know what you mean by this.
>Tr(A+B)=Tr(A)+Tr(B)
True.
>Tr(AB)=Tr(A)*Tr(B)
False.  You must be thinking of det(AB) = det(A) det(B).
>But, if any apropriate solution exist, say for special forms of A B,
>please let me know.
If A and B commute, then every eigenvalue of AB is the product of an
eigenvalue of A and an eigenvalue of B.
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             (604) 822-3629
University of British Columbia            fax 822-6074
Vancouver, BC, Canada V6T 1Y4

Kurt Foster writes, "There is no prefix I've heard of for 10^4 or 10^(-4),
so on is probably free to make them up."
Actually, there is a prefix in use for 10^4:  "myria-", like the English
word "myriad", derives from an Old Greek word meaning variously, "10,000"
or "innumerable."
Cited in _The New Webster Encyclopedic Dictionary of The English
Language_, Consolidated Book Publishers, 1952, p. 556:  "myriagram",
"myrialiter" and "myriameter" are French units of measure meaning
respectively 10,000 grams, 10,000 liters and 10 kilograms.
Since the French introduced the metric system, I would accept French
prefixes on metric words.  Note also that "deka-", "hecto-", "kilo-",
"mega-", etc. come to the metric system from Old Greek; it is reasonable
to bring "myria" from the same source.  But is the year a metric unit? 
Unfortunately it is not.
The trouble here is that "-ennium" as a suffix derives from  Old Latin
usage, and all our similar words from "biennium" through "millenium" are
basically Latin words.  Although our metric system uses Greek roots for
positive powers of ten, the Romans used Latin roots for these powers. 
Thus, our English mile is the "milia passum" which is literally the
"thousand paces."  We have already these English words:  "decennial",
"centennial" and "millennial" rather than the "deka-", "hecto-" and
"kilo-" prefixes.
It seems we have four choices:
(1.)  Create the neologism "myriennium" despite its unnatural heritage,
(2.)  Search rare texts for a Latin word for 10,000 and/or 10 thousand
years,
(3.)  Note that the Old Greek word "horos" for "year"  entered Old English
as "gear" and begat both "year" and "yore".  The neologism, "myriayore",
is at least purer than option (1.)
Choice (4.) is to look outside of Latin or Greek.  ( In another 8,000
years, who will know the difference? )  Chinese have a word for "ten
thousand", which, like the Greek, also means "innumerable."  Do they have
a word for "ten thousand years" or is this a phrase?  I recall that the
Hindi and Mayan traditions also had words for large chunks of eternity. 
The Mayan "base 20" system probably rules out that language, but I would
not be surprised to find the right word in Hindi or Sanskrit.
Pedantically yours,
Patrick T. Wahl
( no institutional affiliation )    

Mark Lee wrote:
> 
> Miguel Lerma wrote:
> 
> [snip]
> 
> > Clearly the observation of our world rules out the existence of a being
> > with the following three attributes:
> >
> > 1. All Powerful.
> >
> > 2. All Loving.
> >
> > 3. All Knowing.
> 
An all powerful, all loving and all knowing being. What is the point?? 
In order for this to make any sense at all we must make the assumption 
that the human experience of good and evil is the "absolute truth" and 
above all that the human experience of truth is universal. Otherwise 
why should an all powerful all knowing and all loving being create, 
but to define to itself and to the devil good and evil. But since God 
is all powerful all knowing and all loving then good must triumph over 
evil else evil is as powerful as good and hate as love is and since it 
knew this anyway why bother?? Therefor God is none of the above and I 
suspect not anyway. My question is WHY a God Why do we need or want a 
God.
A faithful athiest
Tim Flaus timf@zip.com.au

electronic monk wrote:
> 
> Paul Schlyter wrote:
>  >
>  > In article <32D6D8D7.F3C@sqruhs.ruhs.uwm.edu>,
>  > electronic monk   wrote:
>  >
>  > > what i mean is that:
>  > >
>  > > lim   (x * 1/x) = 1
>  > > x->oo
>  >
>  > This is trivial:
>  >
>  >  x * 1/x  =  1
>  >
>  > and
>  >
>  >  lim 1
>  >  x->oo
>  >
>  > remains 1.
>  >
>  > > lim   (x * 1/x) = 0 * oo
>  > > x->oo
>  > >
>  > > so 0*oo can equal one.
>  >
>  > This is different!   Since
>  >
>  >  lim   (x * 1/x) = 1
>  >  x->oo
>  >
>  > then
>  >
>  >  lim   (x * 1/x) = 0 * oo
>  >  x->oo
>  >
>  > implies that 0 * oo not only "can be" one but ALWAYS IS one!  And
> this
>  > is erroneous.
> 
> what i mean is that 0*oo can be anything, and one is just one of it's
> answers.  in the case i stated, the indeterminite form 0*oo is equal to
> 1.  it could equal anything else also, depending on the case at hand.  i
> was just pointing out that it *can* be equal to one, and since oo is the
> only thing that can be multiplied by zero to yield one, oo must be
> zero's inverse.  that is the point i've been trying to make.
> 
> (A) = lim  (x/x) = 1  right?
>       x->oo
Yes.
> (B) = lim  (x/x) = 0*oo  also right?
>       x->oo
No, like I said in another post:
lim(a*b) = lim(a)*lim(b) if both limits exist (thus |lim(a)| < oo)
> but, (A) does not always equal (B) because 0*oo is indeterminante.  only
> in this case does 0*oo equal one.
No, see above.
> > ----------------------------------------------------------------
> > Paul Schlyter,  Swedish Amateur Astronomer's Society (SAAF)
> > Grev Turegatan 40,  S-114 38 Stockholm,  SWEDEN
> > e-mail:  pausch@saaf.se     psr@net.ausys.se    paul@inorbit.com
> > WWW:     http://www.raditex.se/~pausch/    http://spitfire.ausys.se:8003/psr/
> 
> electronic monk
Wilbert

In article <32d9725e.49267882@news.muenchen.org>, bm373592@muenchen.org (Uenal Mutlu) says:
>
>On Fri, 10 Jan 1997 09:22:30 -0700, Karl Schultz  wrote:
>
>>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.
>>> >> >
>...
>>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.
>
>I get a different percent value:
>
>  Since the probability for at least 3 is about 1 in 54 (to be exact
>  p=0.0186375450020), using the usual formula gives:
>
>    E = "at least once at least 3 matching nbrs in 168 consecutive 
>         draws using the same fixed 1 ticket" 
>
>    p(E) = 1 - (1 - 0.018637545)^168 = 0.9576
>
>    meaning 95.76 % of chance of occuring of the event E.
>
>(BUT: I'm still not sure if this is the same as playing 168 different 
>tickets in 1 drawing) 
>
>Which formula did you use?
>
>>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???  
>
>One would need to analyse its guarantee table; I think Adolf 
>mentioned of doing this in another posting:
Here is that winning table:
That the winning table (without regard to the bonus number) of Uenal 168
wheel on 49 numbers that guarantees 3 on 6.
  6-win 5-win  4-win  3-win  blocks      %
                               1        3835601   27.4288
                               2        3058209   21.8696
                               3          821854     5.8772
                               4        3021425   21.6065
                               5+     1228555     8.7855
                      1     0-?      1769036   12.6506
                      2     0-?        139821     0.9999
                      3     0-?          60141     0.4300
                      4     0-?            4226     0.0302
                      5+   0-?            1701     0.0122
              1    0 - 7   0-?         42848     0.3064
              2    0 - 7   0-?             199      0.0014
              3    0 - 7   0-?               32      0.0002
       1   0 - 1  0 - 9   0-?           168      0.0012
Chances to win more than one 3 win are about 72 %, 

Does anyone know if the following is true?
If A is a connected subset of a real vector
space then for any elements x,y in A, there
exists a continuous function f:[0,1]->A so
that f(x)=0 and f(1)=y (using the usual
topology). 
I know that the converse is true, and it
seems intuitively correct, but I cant prove it
or find a counter example. Can anyone help?

Jean-Christophe Janodet  writes:
> 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 ?
If I take from you two times a debt of $3, you actually *gain* $6.
-- 
David Kastrup                                     Phone: +49-234-700-5570
Email: dak@neuroinformatik.ruhr-uni-bochum.de       Fax: +49-234-709-4209
Institut f=FCr Neuroinformatik, Universit=E4tsstr. 150, 44780 Bochum, Germa=
ny

electronic monk  writes:
> what i mean is that:
> 
> lim   (x * 1/x) =3D 1
> x->oo 
> 
> lim   (x * 1/x) =3D 0 * oo
> x->oo
> 
> so 0*oo can equal one.
Which is why oo is *not* a real number.  All expressions involving
real numbers are either undefined, or equal exactly one real number.
There is no such thing as constant expressions which "can equal" some
value.  Either they do or they don't.
Your 0*oo can "equal" any real value, and so cannot be called a
constant expression.  So obviously oo does not count as a real
constant.
-- 
David Kastrup                                     Phone: +49-234-700-5570
Email: dak@neuroinformatik.ruhr-uni-bochum.de       Fax: +49-234-709-4209
Institut f=FCr Neuroinformatik, Universit=E4tsstr. 150, 44780 Bochum, Germa=
ny

Felix Dilke wrote:
> 
> Call a class of ordinals sup-closed if it contains the sup of
> each of its subsets.
> 
> Then is it true for each class of ordinals that either it or its
> complement contains a sup-closed proper class?
> 
> Thanks
> 
> Felix
No: divide the class Ord of ordinals in the class C1 of ordinals
whose cofinality is exactly aleph_0, and the class C2 of all other
ordinals.
Assertion 1: C2 does not contain any sup-closed proper class (the
standard term, by the way, is "closed unbounded class"). Indeed,
if C subset of C2 were a closed unbounded class then it would
contain an ordina alpha0 (since it can't be empty), and also an
ordinal alpha1>alpha0, and also an ordinal alpha2>alpha1, and so
on (that is, it contains an increasing omega-sequence alpha_i of
ordinals). The limit of the alpha_i is a limit ordinal (obviously),
and its cofinality is exactly aleph_0, because it is the limit
of a countable sequence of ordinals. So the limit in question is
in C1, therefore not in C, and therefore C is not closed unbounded.
Assertion 2: C1 does not contain any closed unbounded class. The
proof is the same, but this time index the sequence alpha_i by
i in omega_1 (the first uncountable cardinal). The limit of the
sequence is a limit ordinal of cofinality aleph_1, and so it is
in C2, therefore not in C.
The question you're asking is really "is the closed unbounded
filter over the class Ord an ultrafilter?" The same question could
be asked over any cardinal kappa. The answer is always no, even
without AC, if kappa>aleph_1. If kappa is aleph_1, the answer is
no, but this time it requires AC. Interestingly, the Axiom of
Determinacy has the following consequence:
  the closed unbounded filter over aleph_1 is an ultrafilter
(in particular, aleph_1 is a measurable cardinal).
     David A. Madore
    (david.madore@ens.fr,
     http://www.eleves.ens.fr:8080/home/madore/index.html.en)

In article  <199701122158.NAA19901@alumni.caltech.edu>
ikastan@alumni.caltech.edu (Ilias Kastanas) wrote:
:	Dear Mr. Leko
:
:	1)  I is not imaginary.  Both R and I in  R + iI are real.  Re-
:   member e^z...   I = e^x sin(y), for real x and y.  I is real.
:
:	All my identities are correct by simple math.  Please point out
:   any mistakes.
:
:	2)  atan(tan(y)) = y is true for -pi/2 < y < pi/2 and false for
:   other y.
:
:	3)  I read A as it was.  Yes, "zero" means R = I = 0; but your A
:   is = 0 even when  R = I =/= 0, thanks to  -sin(..) + cos(..).
:
:	That is why your A is wrong, and produces nonexistent "zeros".  I
:   explained this months ago but you ignored me.  So you can hardly
complain
:   for the tone of my post.  And it was _right_: you _still_ refuse to
:   believe elementary trigonometry, and your plots show infinitely many
:   zeros for  z^2... and for  e^z!
:
:	Why not _fix_ it??  A = R^2 + I^2, that's all it takes!
             Thank you for your contribution.
tleko@aol.com

Angelo Vistoli wrote:
> 
> 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.
> 
> ...
> 
> How about proof by hallucination? It is a technique I have used a lot.
Or a similar method: From the holy bible we find that...
Doesn't work for everybody, but you can convince (fool?) a lot of people
with this argument.
Gr., Wilbert
-- 
Was sich uberhaupt sagen lasst, lasst sich klar sagen

Dear Mr. Kastanas,
There are too  many extraneous comments in your writings unrelated
to mathematics. Regrettably your mathematics is wrong.
1. In your first comment 'a' is a real number. 'I' represents an imaginary
    value and all your following identities are false.
2. atan(tan(y)) is correct in the periodic interval.
3. For a complex function having zeros requires R=0 and I=0 at the
    same point which is not the case, and you misread the the value of A.
Best regards,  tleko@aol.com

In article , dik@cwi.nl (Dik T. Winter) writes:
> 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
Yes, there is a sad mathematical fact here.
The standard real numbers are a subset of the standard complex numbers,
but this is NOT TRUE of their compactifications (which is really what we
are talking about here), since R needs a 2-point compactification (+oo and
-oo  whereas C needs a 1-point compactification (projective oo).
To my mind, this sad mathematical fact means that hardware should concentrate
on standard R (which can therefore be extended to standard C in the cartesian
way) and leave the interpretations of infinity to software, since no one
interpretation will suffice for both R and C (unless we also build C and
its 1-point compactification in).
James Davenport

andrew harford (cs4-03@cslan.ud.ie) wrote:
: Does anyone know if the following is true?
: 
: If A is a connected subset of a real vector
: space then for any elements x,y in A, there
: exists a continuous function f:[0,1]->A so
: that f(x)=0 and f(1)=y (using the usual
: topology). 
: 
: I know that the converse is true, and it
: seems intuitively correct, but I cant prove it
: or find a counter example. Can anyone help?
No, this is not true. Your question can be reput as
"is every connected subset of a real vector space path-connected?"
In R^2 there are well-known counterexamples to this, such as the sin(1/x)-
continuum. If your space is locally path-connected (e.g locally convex, like
R^n) then the result would be true for open subsets. 
Hope this helps,
Henno Brandsma.

I recently posted a simplified, two-dimensional version of a question to which
several kind people sent me solutions.  I had hoped that knowing how to solve
the 2D case would let me solve my real problem, a 3D analog of the 2d case,
but it turns out the 3D case is much more difficult than I realized.  So now
I hope that someone can help me with this new problem.
In polar coordinates, I need to sum together elliptical Gaussian "mountains"
placed at regular intervals in a circle such that their peaks add to a sum
that is as flat across the top as possible.  A simpler case is to look at
two partially overlapping Gaussians mountains: then, what I need is to find
how far part they can be spaced before a saddle point appears between them.
To visualize this, in two dimensions, picture the sum of two Gaussians
    s(x) = g(x) + g(-x)
When the function s is graphed with increasing values of x > 0, the two
Gaussians are spaced farther and farther apart, and you get a range of shapes:
          _
         / \                 __                _  _
        /   \               /  \              / \/ \
       /     \             /    \            /      \
      /       \           /      \          /        \
   --'         `--     --'        `--    --'          `--
       Fig. 1              Fig. 2             Fig. 3
        x = 0              x small            x larger
In two dimensions, I wanted to find the largest value that x could take on
without the dip appearing in the top of the sum in Figure 3.  The analog in
three dimensions is that the mountains are arranged as shown in the following
horizontal cross section through their bases (and the separation is
exaggerated):
    _         ^         _ rho      Assume polar coordinates of theta and
   |\         |         /|         rho, and a sum of Gaussians of the form
     @ @      |      @ @
    @ \  @    |    @  / @          s(theta,rho) = g(theta,rho) + g(-theta,rho)
    @  \  @   |   @  /  @
     @  \ @   |   @ /  @           When theta = 0, the two blobs will lie on
  _    @ @    |    @ @             top of each other and add into one peak
  /|      \   |<->/                (a local maximum).  As theta increases, 
one        \  |  / theta           the blobs will spread apart and the sum 
Gaussian    \ | /                  will decrease, and eventually a saddle 
sliced       \|/                   point will appear.
horizontally -+---------
              |
What I need to do is to find the largest value of the angle theta before the
saddle point appears between the two Gaussian "mountains".
If I understand it correctly from my basic calc (and I'm quite rusty at
this), one way of testing for an extreme value in some function f(x,y) of two
variables is also to use a second derivative test, which goes like this:
  1) Find where the first partial derivative in each direction = 0 or
     one or both fails to exist.
  2) If there is a point f(a,b) at which they're both zero, examine the 2nd
     partial derivatives in each direction as well as the combined sum
     fxx fyy - fxy^2 (where fxx is the second partial derivative in the x
     direction, fyy is the 2nd deriv. in the y direction, and fxy is the
     partial in each direction successively).  The conditions are:
        a) if fxx < 0 and fxx fyy - fxy^2 > 0, it's a relative max
        b) if fxx > 0 and fxx fyy - fxy^2 > 0, it's a relative min
        c) if fxx fyy - fxy^2 < 0 it's a saddle point
        d) the test is inconclusive if fxx fyy - fxy^2 = 0.
First, it is not clear to me how to express the transition point when the
saddle point begins to appear.  I thought that maybe it would be correct to
look for where fxx fyy - fxy^2 = 0, but actually that is explicitly a
condition where the test is supposed to be inconclusive.
Second, I've tried to solve the equations with the help of Mathematica, but
the combination of cosine & sine terms and exponentials makes solving for the
various zeros very difficult.  The particular Gaussian I need to use is:
                                           2   2   2   2           2   2
                                   n  -2 Pi rho (sx Cos[theta] + sy Sin[theta])
g(theta,rho)= (2 Pi rho Cos[theta])  E
The exponent n is an integer >= 1.  The sigmas, sx and sy, are positive
reals.  (This equation was converted from one in cartesian coordinates to
polar coordinates, because I think it is easier to express the angular
relationship between the Gaussians.  That's why the sigmas are expressed in
the x and y directions.)  Although n is shown as a parameter, I have only a
finite number of value of n for which solutions are needed (n = 1-7), so I
can in fact examine the cases individually and thus eliminate n from the eq.
I am hoping the solution(s) will have the form of an expression of theta in
terms of sx and sy.
As mentioned above, for my purposes, it's enough to look at the sum of two
such Gaussians,
s(theta,rho) = g(theta,rho) + g(-theta,rho)
Taking partial derivatives with respect to rho and theta of the sum leads to
quite an ugly mess.  I've tried various manipulations and simplifications,
and using the solving facilities in Mathematica 2.2 to try to obtain a
solution to this.  It appears that the first partial derivatives in the rho
and theta direction will go to zero when
                        _
                      \/n
rho = ------------------------------------
              ____________________________
             /  2   2           2   2
      2 Pi \/ sx Cos[theta] + sy Sin[theta]
Next, I tried looking at the second partial derivative of s(theta,rho) with
respect to rho.  It is possible to substitute back the expression above for
rho into this second derivative, and then to try to solve for the value of
theta.  To make it easier, I've set n = 1 for the first case.  Unfortunately,
the result is still either too complex to solve, or I'm doing something badly
wrong, because I get indeterminate results.
Can anyone offer suggestions for alternative (and hopefully easier) solution
paths to this problem?
-- 
Mike Hucka     hucka@umich.edu     
 PhD wanna-be, computational models of visual processing (AI Lab)    University
    UNIX systems administrator & programmer/analyst (EECS DCO)      of Michigan

Dan Larsen (danlarsen@centuryinter.net) wrote:
: I understand the fundamentals of binary, but how do you count in Hex?
: Ie, how do you translate 8F into base ten, and how do you translate 254
: into hex?
If you understand binary, base 16 is very easy; just expand each digit
into four binary digits, and then write then next to one another.
ie 8 = 1000 F = 1111, so 8F = 10001111 = 143.
Or use the definition of hex : &12345 = 5 + 4 * 16 + 3 * 16^2 + 2 * 16^3 +
1 * 16^4
To translate 254 into hex, either write it in binary, group the digits into
fours starting from the end, and translate each group-of-4 to a hex digit,
or use a loop like this one
Given N
1 Work out N's remainder on division by 16 (which I'll write N%16 to
  save typing)
2 Write it at the left-hand end of the number (ie working backwards)
3 Divide N by 16, ignoring any remainder
4 If N isn't zero, go back to step 1
EG N=12345
N%16 = 9, so the hex version ends 9
N/16 = 771. Now, set N=771
N%16 = 3, so the hex version's penultimate digit is 3
N/16 = 48. Now, set N=48
N%16 = 0, so the antepenultimate digit is 0
N/16 = 3. Set N=3
N%16 = 3, so the first digit is 3
N/16 = 0, so we can stop
So 12345 = &3039
: -dan
--
Tom
The Eternal Union of Soviet Republics lasted seven times longer than
the Thousand Year Reich