Newsgroup sci.math 152123

Directory

   Zeros of: 
 a*z^13+b*z^12+c*z^11+d*z10+e*z^9+f*z^8+g*z^7+h*z^6+i*z^5+j*z^4+k*z^3+l*z^
2+m*z+n
where z=x+iy,  a, b, c, d, e, f, g, h, i, j, k, l, m, n,  are real
numbers.
Running the following MATLAB program for particular values of a, b, c, d,
f, g, h, 
i, j, k, l, m, n,  the location of zeros is shown in x,y-plane of an
orthogonal 
coordinate system. An example:
                                                   *** PROGRAM  IS
COPYRIGHTED ***
x=-1.4:.02:1.4;
y=-1.4:.02:1.4;
[X,Y] = meshgrid(x,y);
a=1; b=0; c=0; d=0; e=0; f=0; g=0; h=0; i=0; j=0; k=0; l=0; m=0; n=-1;
R=(a*(X.^13-78*(X.^11).*(Y.^2)+715*(X.^9).*(Y.^4)-1716*(X.^7).*(Y.^6)+1287
*(X.^5).*(Y.^8)-286*(X.^3).*(Y.^10)+13*X.*(Y.^12))+b*(X.^12-66*(X.^10).*(Y.
^2)+495*(X.^8).*(Y.^4)-924*(X.^6).*(Y.^6)+495*(X.^4).*(Y.^8)-66*(X.^2).*(Y.
^10)+Y.^12)+c*(X.^11-55*(X.^9).*(Y.^2)+330*(X.^7).*(Y.^4)-462*(X.^5).*(Y.^6
)+165*(X.^3).*(Y.^8)-11*X.*(Y.^10))+d*((X.^10)-45*(X.^8).*(Y.^2)+210*(X.^6)
.*(Y.^4)-210*(X.^4).*(Y.^6)+45*(X.^2).*(Y.^8)-(Y.^10))+e*((X.^9)-36*(X.^7).
*(Y.^2)+126*(X.^5).*(Y.^4)-84*(X.^3).*(Y.^6)+9*X.*(Y.^8))+f*(X.^8-28*(X.^6)
.*(Y.^2)+70*(X.^4).*(Y.^4)-28*(X.^2).*(Y.^6)+Y.^8)+g*(X.^7-21*(X.^5).*(Y.^2
)+35*(X.^3).*(Y.^4)-7*X.*(Y.^6))+h*(X.^6-15*(X.^4).*(Y.^2)+15*(X.^2).*(Y.^4
)-Y.^6)+i*(X.^5-10*(X.^3).*(Y.^2)+5*X.*(Y.^4))+j*(X.^4-6*(X.^2).*(Y.^2)+Y.^
4)+k*(X.^3-3*X.*(Y.^2))+l*(X.^2-Y.^2)+m*X+n);
I=(a*(13*(X.^12).*Y-286*(X.^10).*(Y.^3)+1287*(X.^8).*(Y.^5)-1716*(X.^6).*(
Y.^7)+715*(X.^4).*(Y.^9)-78*(X.^2).*(Y.^11)+Y.^13)+b*(12*(X.^11).*Y-220*(X.
^9).*(Y.^3)+792*(X.^7).*(Y.^5)-792*(X.^5).*(Y.^7)+220*(X.^3).*(Y.^9)-12*X.*
(Y.^11))+c*(11*(X.^10).*Y-165*(X.^8).*(Y.^3)+462*(X.^6).*(Y.^5)-330*(X.^4).
*(Y.^7)+55*(X.^2).*(Y.^9)-Y.^11)+d*(10*(X.^9).*(Y)-120*(X.^7).*(Y.^3)+252*(
X.^5).*(Y.^5)-120*(X.^3).*(Y.^7)+10*X.*(Y.^9))+e*(9*(X.^8).*Y-84*(X.^6).*(Y
.^3)+126*(X.^4).*(Y.^5)-36*(X.^2).*(Y.^7)+(Y.^9))+f*(8*(X.^7).*Y-56*(X.^5).
*(Y.^3)+56*(X.^3).*(Y.^5)-8*X.*(Y.^7))+g*(7*(X.^6).*Y-35*(X.^4).*(Y.^3)+21*
(X.^2).*(Y.^5)-Y.^7)+h*(6*(X.^5).*Y-20*(X.^3).*(Y.^3)+6*X.*(Y.^5))+i*(5*(X.
^4).*Y-10*(X.^2).*(Y.^3)+Y.^5)+j*(4*(X.^3).*Y-4*X.*(Y.^3))+k*(3*(X.^2).*Y-Y
.^3)+l*(2*X.*Y)+m*Y);
A=sqrt(R.^2+I.^2).*(-sin(atan(I./R))+cos(atan(I./R)));
c=contour(X,Y,A,0:400:800);
clabel(c,'manual');
If there is an interest in the results where MATLAB software is not
available
please request a fax-transmission.
tleko@aol.com

We consider  (a+bi)z^3+(c+di)z^2+(e+fi)z+(g+hi)
where  z=x+iy,  i=sqrt(-1)  and  a, b, c, d, e, f, g, h,  are real
numbers.
Running the following MATLAB program for particular values of a, b, c,
d, e, f, g, h,  the location of zeros is shown in a x,y-plane of an
orthogonal
coordinate system.  An example:
                                                *** PROGRAM IS COPYRIGHTED
***
x=-2:.02:2;
y=-2:.02:2;
[X,Y]=meshgrid(x,y);
a=1; b=1; c=1; d=0; e=1; f=0; g=1; h=0;
R=(a*(X.^3-3*X.*(Y.^2))+b*((Y.^3)-3*(X.^2).*Y)+c*(X.^2-Y.^2)-d*(2*X.*Y)+e*
X-f*Y+g);
I=(a*(3*(X.^2).*Y-(Y.^3))+b*(X.^3-3*X.*(Y.^2))+c*(2*X.*Y)+d*(X.^2-Y.^2)+e*
Y+f*X+h);
A=sqrt(R.^2+I.^2).*(-sin(atan(I./R))+cos(atan(I./R)));
c=contour(X,Y,A,0:4:8);
clabel(c,'manual'); 
If there is an interest in the results where MATLAB software is not
available 
please request a fax-transmission.
tleko@aol. com

In article <568mhq$hg9@cc-server9.massey.ac.nz>,
Malcolm Loudon  wrote:
[...]
> Pi/4 = 4atan(1/5) - atan(1/239) 
>
>Can someone give me an easy(ish) argument to demonstrate the veracity of this
>statement ?
[...]
 A standard atan formula (with the right assumptions):
 If x*y<1 then atan(x) + atan(y) = atan((x+y)/(1-x*y))
 Derivable from tan(a+b)=(tan(a) + tan(b))/(1-(tan(a))*tab(b))),
 or by differentiating both sides by x, making sure that we do not step 
over a discontinuity.
So, atan(1/5) + atan(1/5) = ... = atan(5/12)
    atan(5/12) + atan(5/12) = atan(W) (find W)
    atan(W) + atan(-1/239) = ???
(And recall that atan(1)=pi/4.)
Have fun, ZVK (Slavek).

Fredrik Sandstrom (fred@spider.compart.fi) wrote:
: I've got a simple(?) question.  I came to think of it one day, and I
: can't come to a conclusion.  What is
: 
:  lim    0/x         ?
: x -> 0
: 
Just use the definition of the limit and the fact that the function 0/x
is defined on IR-{0}:
lim  0/x = a    iff for each epsilon>0 there is a delta>0 such that 
x->0                |0/x-a|0+) 0/x = 1 and
: lim_(x->0-) 0/x = -1.  Other possibilites are +/- infinity, or perhaps
: 0.  What do you think?
These are not possible answers. lim  0/x = 0 and nothing else.
                                x->0
-- 
Ulrich Lange                       Dept. of Chemical Engineering
                                   University of Alberta
lange@gpu.srv.ualberta.ca          Edmonton, Alberta, T6G 2G6, Canada

Le Compte de Beaudrap wrote:
> 
> On 7 Nov 1996, Archimedes Plutonium wrote:
> 
> > In article <327FD551.4A31@postoffice.worldnet.att.net>
> > kenneth paul collins  writes:
> >
> > > Please, what are "p-adics"?
[Falsely attributed stuff snipped]
Kindly, keep straight what folks post. ken collins
_____________________________________________________
People hate because they fear, and they fear because
they do not understand, and they do not understand 
because hating is less work than understanding.

Arild Kvalbein wrote:
> On 12-Nov-96 09:31:22 Maurizio Paolini wrote:
> >One widely used definition says that a function is strictly
> >increasing (in the whole domain of definition) if for any two
> >distinct points x, y it holds  [f(x) - f(y)](x - y) > 0.
>                                 ^^^^^^^^^^^^^^^^^^^^
> I don't quite follow you here. What is meant by this expression? In words,
> please.
Not speaking for Mr. Paolini, but I believe he is indicating the product
of the difference of two points in the domain and the difference of the
function evaluated at those respective points.  For exaple, let's consider
the function f(x) = x^3.  Now let's choose two points in the domain, say
x=-2 and y=0.  Thus f(x) = f(-2) = -8 and f(y) = f(0) = 0.  From the
above definition, we see that [f(x) - f(y)](x - y) = [-8 - 0](-2 - 0) = 16.
Since this product is greater than zero, and would be greater than zero
*for any two distinct points in the domain*, the function f(x) = x^3 is
strictly increasing.
Regards,
Scott Hewitt

Bill Dubuque wrote:
> 
> Bill Dubuque  writes:
> >
> > More generally, for any squarefree integer m, and any integer x,
> >
> >       phi(m) + 1
> >     x            = x  (mod m)                                [1]
> > ...
> 
> I should have pointed out the special case m=2*p, p an odd prime:
> 
>         p
>       x   = x  (mod 2*p)
> 
> which is a slight generalization of Fermat's Little Theorem, and
> for p=3 yields the result in the original post.
> 
> -Bill Dubuque
This is a lot of firepower.  x^3-x = (x-1) * x * (x+1), which is the
product of three consecutive integers.  One of these must be divisible
by 2 and one of them must be divisible by 3.
-- 
John Wilkinson

 e^(i pi) = - 1
e^(2i pi) = 1
    2i pi = ln(1)
    2i pi = 0
  -4 pi^2 = 0
     pi^2 = 0
       pi = 0
therefore,
e^(i pi) + 1 = 0   is false
Where did I go wrong?

In article <560d4c$gsv@news-central.tiac.net>, numtheor@tiac.net (Bob
Silverman) wrote:
> 
> lv54308@ltec.net (Lyle VonSpreckelsen) wrote:
> 
> >Solve this
> 
> >Three Pipes supply an oil storage tank.  The tank can be filled by
> >pipes  A and B running for 10 hours, by pipes B and C running for 15
> >hours, or by pipes A and C running for 20 hours.  How long does it
> >take to fill the tank if all three pipes run?
> 
> >Got my andvanced math teacher (MR. V.) stumped 
If he's stumped he's not very advanced.
The problem is ill-posed as we don't know where the pipes are
fed from. If they are from the same source, they may not act
independently.
Assume they are independent. Let a, b, and c be the flow rates of
A, B and C respectively, and let the volume of the tank be V
(which is not really needed). Then find the time to fill the
tank under the above scenarios in terms of a, b and c. This
gives 2 equations to solve for these variables. As Bob hinted,
the equations are linear in 1/a etc. Then it's trivial to
aggregate the results.
Terry Moore, Statistics Department, Massey University, New Zealand.
Imagine a person with a gift of ridicule [He might say] First that a
negative quantity has no logarithm; secondly that a negative quantity has
no square root; thirdly that the first non-existent is to the second as the
circumference of a circle is to the diameter. Augustus de Morgan

In article <3288ABC2.27F6@ix.netcom.com> Judson McClendon  
 writes:
=Le Compte de Beaudrap wrote:
=>         But back on the original topic, if christianity preaches "Thou  
shalt
=> not think for thyself", how does one account for people like Netwon,  
Descartes,
=> Euler? Each of them was very religious, very christian, and each one  
was a
=> mathematician, philospher, and physicist (even by today's definition,  
as
=> opposed to the definition in their time, when one group almost  
inevitably
=> implied the other two). None of them was condemned by any church at any  
time
=> (to my knowledge). So, ~(For all cases): (Religion Rots Reason).
=> Unproved, and with three counterexamples anyhow.
=
=Christianity consists of this: A personal relationship with Jesus Christ
=as your Savior and Lord.  Many millions of people will testify that they
=are far better at all aspects of living as Christians than before,
=including mentally.
=
I think many millions who use various illicit drugs testify that they feel  
they are in a better state when under the influence of drugs.  But I  
wouldn't advocate that as an excuse to use drugs.
EDEW

In a previous article, Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) says:
no, man; it is just a syllogism (when combined
with your "1st", it is a nice, little tautology .-)  you assume that
any "infinite" regression is self-similar or just scaled, but
it could be (or plainly *is*) qualitatively differnt at different scales ...
not that I believe in many-worlds etc.ad vomitorium.
	As Riemann emphasizes [4],
the incorporation of any valid new principle into mathematical physics,
requires us to depart the domain of math.fromalism for the realm
of experimental physics, and, thereafter, to redesgin math.
to accomodate what the old math.could never develop, or represent.
That quality of invention, is the only source
of "not-entropy" upon which all economy depends."
  --by [deletive expleted], Copyr.9august
that's all that Archie was trying to present,
about the quantum hall effect, but I have no idea
if he has any experimentss to "prove" his plutonium crappola --
not that he actually made them, of course.
>connects with the first. Second is: If atoms are the last cut, the last
>division and that all things subatomic have no independent existence,
>this implies that there exists no infinite regression downwards nor
>infinite progression upwards.
-- 
You *don't* have to be a rocket scientist.  (College Career Counselor
					     to me, again )
There is no dimension without time.  --RBF (Synergetics, 527.01)

Makes sense to me.
Irving H. Anellis
Managing Editor/Publisher
Modern Logic Publishing
2408 1/2 W. Lincoln Way (Upper Level)
Ames, IA 50014-7217, USA
email: IrvAnellis@aol.com; ModerLog@aol.com;
f1.mlp@isumvs.iastate.edu (preferred)
tel. 515 / 292-1819

In article <563g2p$4e6@q.seanet.com>, ksbrown@seanet.com (Kevin Brown) writes:
[...]
|> 
|> In general I think these expansions can be generated from partitions
|> of any odd abundant number. 
In the nomenclature of UPiNT this conjecture is "every odd abundant number
is pseudoperfect", or even more charmingly "there are no odd weird numbers".
[A weird number is a primitive abundant number that is not pseudoperfect.]
Sadly, this seems too good to be true... Counterexample, anyone?
Chris Thompson
Email: cet1@cam.ac.uk

In article <562646$ca@niamh.indigo.ie>, gerryq@indigo.ie (Gerry Quinn)
wrote:
> Many pupils of elementary teachers will become shopkeepers.  They will not 
> want to confuse their customers by posting prices such as 3 5/4 instead of 
> 4 1/4.  The concept of proper fractions is useful to their business.
But neither of these is a proper fraction. They are both mixed numbers.
Of course the 1/4 is proper while the 5/4 isn't.
But this is a minor quibble. Most countries use decimal currency.
And you can't make a case using weights and measures - all
civilised countries use the metric system :-)
Terry Moore, Statistics Department, Massey University, New Zealand.
Imagine a person with a gift of ridicule [He might say] First that a
negative quantity has no logarithm; secondly that a negative quantity has
no square root; thirdly that the first non-existent is to the second as the
circumference of a circle is to the diameter. Augustus de Morgan

In article <567rt0$on7@zrt7a.gfz-potsdam.de>,
Patrick Denny  wrote:
)In article ,
)Alexander Anderson   wrote:
)
)>>Hello! I think that the limit
)>>                        1
)>>                 lim -------
)>>                     n sin n
)>>doesn't exist, but I can't prove it: help!
)
)   1        1   1                                           1
)-------  =  - -----  which greater in size than or equal to - for all n.
)n sin n     n sin n                                         n
)
)So, the limit of the latter, if it exists, is greater in size than the
)limit of the former. But the latter is divergent, i.e. it becomes infinitely
)large. So, the former is divergent.
)
)Patrick Denny
Sorry, Pat, no cigar. 1/n -> 0. Note that the question is about limits,
not limits of sums.
Mike
-- 
----
char *p="char *p=%c%s%c;main(){printf(p,34,p,34);}";main(){printf(p,34,p,34);}
I don't speak for DSC.         <- They make me say that.

tleko@aol.com wrote:
[Unreadable MATLAB-code, which is supposed to find the roots of a 13th
 degree polynomial]
Please stop posting these awful MATLAB-programs and read your MATLAB
manual. Zeros of arbitrary degree polynomials 
a(n)*z^n + a(n-1)*z^(n-1) + .... + a1*z + a0
are found automatically by MATLAB using the command "roots". Example:
z^5 + 2 z^4 + 3 z^3 + 4 z^2 + 5 z + 6
>> roots([1 2 3 4 5 6])             
ans =
   0.5517 + 1.2533i
   0.5517 - 1.2533i
  -1.4918          
  -0.8058 + 1.2229i
  -0.8058 - 1.2229i
-- 
Ulrich Lange                       Dept. of Chemical Engineering
                                   University of Alberta
lange@gpu.srv.ualberta.ca          Edmonton, Alberta, T6G 2G6, Canada

Hitech (Hitech@cris.com) wrote:
:  e^(i pi) = - 1
: e^(2i pi) = 1
:     2i pi = ln(1)
      ^^^^^^^^^^^^^
This is the mistake. The implication (e^a = b ==> a = ln(b)) does not hold
for arbitrary complex numbers. The standard definition of the logarithm of
a complex number is
ln(z) = ln(|z|) + i arg(z)  
where arg(z) is the argument of z (i.e the angle t in the polar
representation z=re^(i t)) _between -pi/2 and pi/2_ ! 
Thus ln(e^(2 i pi)) = ln(|e^(2 i pi)|) + i arg(2 e^(i pi)) 
                    = ln(1) + i*0 = 0
Your mistake is analogous to writing  (-b)^2 = b^2 ==> -b = b. This does
not hold for real numbers since the standard definition of the squareroot
of a is the _positive_ number b, such that b*b=a.
:     2i pi = 0
:   -4 pi^2 = 0
:      pi^2 = 0
:        pi = 0
: 
: therefore,
: 
: e^(i pi) + 1 = 0   is false
: 
: 
: Where did I go wrong?
-- 
Ulrich Lange                       Dept. of Chemical Engineering
                                   University of Alberta
lange@gpu.srv.ualberta.ca          Edmonton, Alberta, T6G 2G6, Canada

"Brian Christopher Sapp"  wrote:
>I would be very pleased if someone could assist me with the following
>questions:
>1.  Given that a number is a three digit palindrome, what is the
>probability that it is divisible by 11?
Consider 11N.  For this to be 3 digits, N must be 2 digits, say  10a+b
which I write as ab.  Now   write 11ab as digits,  say  a'zb,   where z = a+b
mod 10  and a' = a  or  a+1 depending whether a+b > 10.   Now you
want a' = b.  Count the cases that make this possible.
>2.  Some of the factors of a locker number are 2, 5, and 9.  If it has
>exactly nine additional factors, what is the locker number?
If  N =  p1^a1   * p2^a2 * .....
then the number of factors is (a1+1)(a2+1) .....
So  p1 = 2,  a1 = 1,    p2 = 3  a2 = 2,   p3 = 5, a3 = 1....
The number of factors is therefore   2 * 3 * 2 * (a4+1)(a5+1) ....  = 12K  say.
But we are already given 3 factors and are told there are 9 more.
Therefore K = 1. and a4 = a5 = a6 = ... = 0.  
'You can lead a horse's ass to knowledge, but you can't make him think.'

Le Compte de Beaudrap  wrote:
>On Fri, 8 Nov 1996, Peter Abrahams wrote:
>> Is there a difference between an antilog and an exponent?
>> Thanks, Peter Abrahams, telscope@europa.com
>> 
Only in form.  The Antilog consists of two parts separated by a
decimal.  The first part gives the "order" of the number  (10, 100,
1000... if base 10), the rest the precision.
If the first part is negative (shown by a bar over the digits), the
number equivalent is less than 1.  Thus Bar3..08991 may be  the
antilog (base 10) of .0001213.  As an exponent, Bar3.08991 means
-3+0.08991 = -2.91009, and .00123 would be equal (approx) to 10
exp(-2.91009).

WAPPLER FRANK (fw7984@csc.albany.edu) wrote:
: 
: 
: David K. Davis wrote:
: > Frank Wappler wrote:
: > ...
: > : But now stipulate a procedure which can change individual digits of those
: > : number representations in the list. Change the first digit of the first
: > : number, the second digit of the second number, etc. (that's another
: > : recursive method).
: > ...
: > I think this is basically right but there's no need to speak of recursive
: > procedures here. Consider ANY enumeration of the reals (between say 0 and
: > 1 - to avoid annoying complications). Then every such number can be
: > represented as an infinite string of digits following a '.', (all digits
: > are zero after a certain point to make all strings infinite). We've made
: > no significant assumptions here except that such an enumeration can be
: > done.  But now construct another such real: take as the first digit any
: > digit not equal to the first digit of the first enumerated number, take as
: > the second digit any digit not equal to the the second digit of the second
: 
: Two questions:
: 
: 1) If you want to be sure that the number was not in the list each digit must
: be changed in a systematic way (- let's call this recursive), giving you a
: `choice within a systematic range around that digit sequence'. Is it correct
: that therefore for every proposed >>truely infinite<< sequences of digits
: (i.e. `real number') one can construct an individual list as part of a 
: (hypothetical) `Cantor List' of real numbers, and that (the recursively
: accessible part of) this `Cantor List' could be formed by `starting with' 
: individual real numbers? 
The change need NOT be systematic - it doesn't matter! This is an 
existence proof as opposed to a constructive proof - we are showing that
some number wasn't included - no matter how we enumerate. Sometimes a
writer will prescribe a definite way in which this number that differs in
each digit can be constructed - but it's really not important - all that's
important is that such a number exists.
The rest of 1) I don't understand. Clarify, or maybe someone else sees
what you're getting at. 
: 2) Which set of axioms are the numbers `you are talking about' based on?
I don't know whether Cantor had some axiomatization of the reals in mind
when he gave this proof. Later, others axiomatized set theory to make sure
that everything Cantor did was kosher (meaning at least consistent). I'm
also not sure - now that I think about it - whether this proof doesn't
involve the axiom of choice (which says one can form a new set by picking
one element out of each set of a collection of sets). Help! 
BTW, Ilias K's amendment is right on - and the above may need more
amending.
-Dave D.

binklmj5@wfu.edu (Mathew J. Binkley) wrote:
>Is there a way of defining the derivative 
>
> n
>d L
>---    with n not an integer.
>  n
>dx
Yes, actually more than one. You could look at "Fractional Calculus
and Its Applications" in the Lecture Notes in Mathematics Series (Vol
457) published by Springer-Verlag, 1974, edited by A. Dold and B.
Eckmann. It includes several papers you might find helpful, including
a history of the ideas behind fractional calculus by Bertram Ross, and
others which discuss in more detail the various alternative
definitions of a fractional derivative or integral, e.g.
Riemann-Liouville or Weyl.
The paper "Fractional Derivatives and Special Functions" by Lavoie,
Osler and Tremblay (SIAM Review Vol. 18 No.2 April 1976) also has a
discussion of the various representations of a fractional derivative
or integral and their limitations. Note particularly the approach
taken by Grunwald in the 19th century which defines the fractional
derivative in terms of a limit of (an infinite sum of) ratios of
finite differences, which looks most like the usual definition of an
integer derivative.
Frank Peseckis
frank@5points.com
http://www.5points.com/

Hitech@cris.com (Hitech) wrote:
> e^(i pi) = - 1
>e^(2i pi) = 1
>    2i pi = ln(1)
>    2i pi = 0
>  -4 pi^2 = 0
>     pi^2 = 0
>       pi = 0
>therefore,
>e^(i pi) + 1 = 0   is false
>Where did I go wrong?
Logarithms are not unique in the complex numbers.   exp(2 pi i) = 1,  but so is
exp (2k pi i) for k = 0,1,2,3,4,...  So log(1) = 0 or 2pi i  or 4pi i or .....
'You can lead a horse's ass to knowledge, but you can't make him think.'

Dru Morgan (drum@loop.com) wrote:
: Is there a formula for finding the lenght of  a groove that goes around a 
: spiral (such as a record album)?  If you know the diameter of the record and 
: the distance (period?) between grooves, what would be the formula for the 
: length of the groove?  You would also have to subtract the part in the 
: middle where there is no groove.  Please make sure to cc-by-mail your response 
: to me at drum@loop.com
: 
: Thanks.
: Dru Morgan
:
I think all of these posting which assume it is easy to determine, say,
the groove pitch, are sort of silly.  In practical terms, how would you
measure this pitch to any degree of accuracy?  Look at the record under
a microscope with a reticle?  And, it is sometimes the case that the
pitch is not constant--passages with low amplitude can be recorded at
higher density than those with high amplitude, and this was, I believe,
done back when we still made records.  So if you want to know how long
the groove is, time the record.  But if you want to predict how long 
the record will play, read the jacket.

Robert Israel (israel@math.ubc.ca) wrote:
> Well, almost... 
> -1 + x^4 y^4 =0 is an unbounded set, but 1 + x^4 y^4 = 0 is not, 
> despite having the same greatest-degree terms.  
> If your homogeneous polynomial has no nontrivial zeros, then it's bounded.
> If the homogeneous polynomial has changes of sign, rather than just zeros, 
> then it's unbounded. 
> But if there are zeros and no sign change, the question is more delicate.
Right. My idea was to homogenize the original bivariate polynomial 
with a third variable z, and look at the points at infinity z=0. 
If there are no points at infinity, then the polynomial defines 
a bounded set. However the reciprocal does not seem to be true 
in general, since there could be isolated points at infinity, 
as in your example (actually in your example the only points 
are at infinity). 
Miguel A. Lerma

: In article , Anthony Potts  says:
: >
: >No, I could, if I wanted either go to CERN full time and pick up $100k per
: >year tax free, or head out to California, and get a bit less, doing
: >research in HEP there.
: >
: >Neither appeals to me. As I said, physics is not challenging. 
Anthony,
Look, if you're after a challenge, try embracing a theory which goes against
everything most physicists believe and then try to peddle it.
Very challenging, let me assure you.
-- 
John August
You can pick your friends, and you can pick your nose. But you can't
pick your friend's nose.

In a previous article, ba137@lafn.org (Brian Hutchings) says:
4. e.g., 1854 habilitation dissertation.
>	As Riemann emphasizes [4],
>the incorporation of any valid new principle into mathematical physics,
>requires us to depart the domain of math.fromalism for the realm
>of experimental physics, and, thereafter, to redesgin math.
>to accomodate what the old math.could never develop, or represent.
>That quality of invention, is the only source
>of "not-entropy" upon which all economy depends."
>  --by [deletive expleted], Copyr.9august
-- 
You *don't* have to be a rocket scientist.  (College Career Counselor
					     to me, again )
There is no dimension without time.  --RBF (Synergetics, 527.01)

In a previous article, fc3a501@GEO.math.uni-hamburg.de (Hauke Reddmann) says:
I don't get it; what is the equation supposed to "do" ??
>: >What can you get when you deform the surface of a cube into the surface
>: >of a sphere?
>: 
>: >the surface x^4+y^4+z^4=1  @8^)
>: 
>: or more generally the surface x^2n + y^2n + z^2n = 1
-- 
You *don't* have to be a rocket scientist.  (College Career Counselor
					     to me, again )
There is no dimension without time.  --RBF (Synergetics, 527.01)