mm-1809
I don't know how to proceed to get every negative rational number in the
the image of Q by f.
David
I have a n-dimensional point and m constraints as shown below;
Point P= (p1,p2,...pN)
Constraint 1:  c1*p1+c2*p2+...+C < 0
...
Constraint m:  z1*p1+z2*p2+...+Z < 0
Note: Any Constraint can be viewed as a normalized hyperplane defined
on the n-dimensional space.
It can be argued that, m constraints define a surface (closed or not)
on
the n-dimensional space. Suppose that Point P does not satify the
constraints. I wonder, how can I find the smallest distance between
Point P to the surface determined by the constraints.
Any idea ?
Example: in 2-D
     Point P= (-1,+1)
     Const2:   p2 < 0
It is clear that X does not satify both Const1 and 2. For this case
the distance between Point P to an area (defined by the constraints)
can easly computed such that
   Distance= sqrt(p1*p1+p2*p2)  It is euclidean distance.
However this not always true.
        |
        |
   P    |
        |
---------------- p1 plane
        |
        | An Area where all
        |
        | 
        p2 plane
Consider the 2- dimensional case:
each constraint defines a half-plane. Assuming they intersect at all
(so there is a solution), this will define a region R bordered by a
convex region, which has vertices at some of the points of
intersections of the lines in the constraints. These points can be
easily soved for, and the ones outside R (all points will be either on
the boundry of R or inside) can be discarded (check the equations to
see which ones fail to satisfy them).
Now, get the distance from P to each vertex. For sake of discussion,
assume that the smallest distance is to vertex v2, with adjacent
vertices v1 and v3, defining lines L1 and L3.
Find the projections onto L1 and L3 of P (normal projections, of
course). If this point lies on R (and at most 1 of these points will do
so), then that is the closest point (since R is convex), and you have
the distance. If neither of these projections lie on R, then v1 is the
closest point.
This should translate naturally into more dimensions, checking
distances to the vertices of the bounded region, and using the adjacent
hyperplanes to v1 for finding the distances. If there aren't enough
smallest-dimensional spaces in the intesections. (in the above example,
the z-axis is the intersection of the planes, so you would use the
distance from P to the z-axis, and the distance from z to the xz and yz
planes.)
Consider a set of intersecting halfspaces with vertices
v1=(0,5), v2=(0,-4), v3=(1,0)
with P= (-1,0)
If I understand your proposed algorithm properly, v3 would be 
identified as the nearest vertex, the projections of P to halfspaces 
defined by v1,v3 and v2,v3 would both lie on R, but neither of these 
projections nor v3 is the closest point (which by inspection is (0,0))
A fix to the proposed algorithm is to consider only vertices that come 
from half-spaces that do not contain P.  In this case, only vertices v1 
and v2 would be considered, v2 would be the closest, and the closest point 
to P is found by projecting to the plane v1,v2.
-Rick
such an assumpiton, due to the nature of the problem.
If the half-planes do not intersect, then there is no region (in more
than 2 dimensions no hypersurface) defined, and so you need to rephrase
the problem. For example, what if I choose P to be the origin, and give
What answer would you want to get here? By the way the problem is
phrased, there is no answer, as the restrictions leave you with an
empty set.
Note that since the vertices I look for are the solutions set of your
constraint equations (using equality, rather than inequality), the
assumption that the regions intersect is the same as assumingthat the
system of equations is consistant. If there is no intersection, then
this will be found before any distances are calculated.
Now, let me show you the graph of your question.
R is the surface where all the constrainsts are met.
Now the minimum distance between P to region R is 1.
                |       |
                | R     |
                | Region|
                |       |
-------------------------------------
P(0,0)          |       |
                |       |
                |       |
                |       |
                |       |
                x=1     x=2
Numerically, you can do it this way, with mathematica (3D example) :
pointP = {2,3,4};
pointX = {x,y,z};
distance = Sqrt[(pointP-pointX).(pointP-pointX)];
constraints = x+y+z <= 1 && x+2y+z <= 2;
NMinimize[{distance, constraints}, pointX]
 
OP:
Easier:  minimize the square of the distance.  Astanoff points out that 
this is a nonlineanr programming problem with constraints.  This 
particular example is easy, of course:  You have a quadratic objective 
you know that the closest point lies on at least one boundary.   Thus, 
you have three cases to check:  only one constraint tight (two cases) or 
both tight.  Use the Kuhn-Tucker-Karesh conditions.
There is a vast literature on nonlinear programming.  Luenberger has 
written an elementary introductory text (on both linear and nonlinear 
programming).
-- 
Stephen J. Herschkorn                        sjherschko@netscape.net
I think, you clearly understand what i mean. However, I wonder the
algorithm. Is there a faster way to compute the minimum distance ? I
suppose, it can be solved much simpler and more fast than any other
Optimization problems. Of course, I hope so. Because, the constraints
are normalized.
I've got a question concerning notation. Given a binary vector xi with
i=1,...,n: how can I show with a notation symbol that I just want to
address a part of the vector, for example the elements x3,...,x5   of
this vector?
How can I for example formally express F(x3,...,x5)=1?
Maybe you can sent me an email with an attachment, if it's too
difficult to show the solution with the ASCII symbols.
On 3 Mar 2005 04:21:32 -0800, christoph.stich@gmx.de (Christoph)
etc. so your vector is sum[k = 1 .. n] x_k * e_k
you could let M = { 3, 4, 5} and write y = sum[ k in M] x_k * e_k.
I'm not sure if that is what you mean by addressing those parts of
the vector.
--Lynn
Why is it so? It does not seem obvious to me.
David
|
|Why is it so? It does not seem obvious to me.
assuming that i'm not making some stupid mistake, it's pretty obvious
step-by-step even if not truly obvious all in one first glance.  do
if you like, you can try testing it out on some examples where c and d
are categories each with just a few objects and a few morphisms.
-- 
[e-mail address jdolan@math.ucr.edu]
rlankine@hotmail.com says...
Risto,
Just an observation - what would your reaction be if you 
finally found a way to generate the string from the number 
(with a method that allowed simple quick processing of very 
large numbers)?
If I understand correctly a fair proportion of the 
encryption techniques currently used (and believed to be 
uncrackable in a useful time period) rely on the 
difficulty and time-consumption of factorising very large 
numbers.   
You would therefore be in possession of an algorithm that 
could do one of two things for you personally:
1.  Fame and prestige by publication - we are talking front 
page of Nature and Time etc
2.  Great wealth by selling (secretly) to the highest      
bidder.
These are mutually exclusive - as soon as the world knows 
that large numbers can be factorised easily (even if the 
method is not published) those that need real security will 
stop using them to encrypt and thus nobody would pay that 
highly for your algorithm.
So, do you provide for yourself, your family etc to live in 
luxury and ease and damn your prinicples or do you publish?
Matthew Newell
who would probably give the NSA a single year licence and 
live on the proceeds, pray no-one else discovered in that 
year and then publish.
 
Alas, because the fear of...
3.  Fame and prestige - by getting assassinated trying to sell
secretly
... would be more than just a paranoid delusion, I would settle
for becoming famous.  In fact, I hope I've already managed to
leave some sort of fingerprint into the history of math thru this:
http://tinyurl.com/59u89
Also, with no financial motive, proving that fast factorization
is impossible is an equally worthy goal to me, yet way more
likely as an outcome.  [Furthermore, in case of failing in both,
I think I've already managed to leave enough of margin notes
in the 'net for the wileses of the future to finish the work :-]
Oh well.  Then again, what would you bid for the following
in eBay?
For sale:  Sealed envelope.  The envelope contains _either_
the complete description of an algorithm for factorizing large
integers with such an efficiency, that the time required by the
algorithm would increase proportionally to the logarithm of
the magnitude of the number being factored, _or_ the proof
that no such algorithm can exist.
 ;-)
 - Risto -
Don't pull a James Harris on us. Please.
     --- Christopher Heckman
If by fast you mean polynomial, then proving that fast factorization
is impossible will prove P!=NP.  I believe there are financial rewards
offered for anyone who can prove P?=NP one way or the other.
I personally like your idea.  I have always thought that the factoring
problem will be solved by seeing it as the problem of reversing the process
of multiplication.  But what do I know? :-)
-- 
Daniel Jim.8enez                     djimenez@cs.utexas.edu
I've so much music in my head -- Maurice Ravel, shortly before his 
death.
                              -- John Cage
Are you suggesting that the integer factorization problem is known to
be NP-complete? If so, this is news to me. Reference?
dave
You need NP-completeness to push P=NP home, but why do you need it for
P!=NP? _Anything_ in NPP proves NP!=P.
What matters is that factorisation is in NP. So, if it is not in P,
P!=NP.
Mike.
By the way, a general methos is as follows - suppose that all x,y,z,t
are function of one variable, say s. Then will solve corresponding
system of ODE -
 dx/ds=-y+b*z-t
 dy/ds =(x-z+b*t)
 dz/ds=-b*x+y-t
 dt/ds=x-b*y+z
let 
x(s)=f_1(s,C),
y(s)=f_2(s,C),
z(s)=f_3(s,C),
t(s)=f_4(s,C),
 where C is set of constants C_1,C_2,C_3,C_4.
Its well know - every first integral of the system ODE is a solution
of  yours  PDE. To discover those first integrals you have to
eliminate variable s from last sysytem. The elimination is an art..I
was trying to do it but I have got some complicated expression..but,
in any way , you have now a right directions :)
Here's a Bessel function...
  int(cos(z*sin(x)),x=0..Pi)/Pi = J_0(z);
For the linear case as that you are dealing with the general solution
is straightforward.
Here you have a first order difference equation:
(1) f(n)-f(n-1)=1
with initial condition:
(2) f(0) = 0
You are to find, as with continuous counterpart of differential
equation, the general solution to the associated homogeneous equation
(f(n)-f(n-1)=0) and a particular solution. Then add them toghether for
the general solution.
Finally, the solution comes out applying the initial condition.
- The general solution of the homogeneous:
you have to proceed as with the differential case, that is solving the
characteristic equation.
While for a first-order linear differential equation the general
solution has the following form y = c e^(ax), where a and c are
constants, for a firt-order linear difference equation it has the
following form
(3) f_0(n) = c a^n,
where a and c are constants.
Substituting (3) in (1) you get an algebraic equation known as the
characteristic equation, that you'll solve for a.
(4) a = 2
Putting (4) in (3):
(5) f_0(n) = c 2^n
- The particular solution:
with the forcing term being constant the particular solution is
constant as well:
(6) f_1(n) = k,
with k to be determined substituting (6) in (1)
Putting (7) in (6):
(8) f_1(n) = -1
- The general solution is:
(9) f(n) = f_0(n) + f_1(n)
Putting (5) and (8) in (9):
(10) f(n) = c 2^n - 1
- The solution:
making use of (10) and of the initial condition (2) you can obtain k
(12) c = 1
Finally, putting (12) in (10)
f(n) = 2^n - 1
As you've seen it's very simple, though it's been the simplest case of
linear equation and the simplest case of forcing function.
Sergio de GIOIA
No, for two reasons, both important.
(i) the step 5|x-2| < |x-2| is a little fishy (this
follows from the fact that 5 < 1, right?)
(ii) the step |x-2| = delta is wrong!
There's no reason why we should think that |x-2| = delta.
Because you left out the words Now if 0 < |x-2| = delta then....
The fact that those words appear in all the proofs you read
doesn't mean that they're unimportant so you can leave
them out - just the opposite, without those words the
proof makes no sense.
************************
David C. Ullrich
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God forbid. ;-)
Really! I stand corrected, then, because I couldn't imagine such a proof and 

had never heard of the result. Did he prove that no primitives existed like 

f(x) = a sin (bx + c) for a sufficiently impressive number of functional 
forms or was he somehow able to represent all finite combinations of 
elementary functions? I really can't imagine how a proof would go that could 

handle all such combinations well enough to give a negative result. What 
framework is it that preserves the class of elementary functions and 
generates things like sin(ln(x + cos(x)))? 
  *** Integration in Finite Terms ***
This theory goes back to Liouville, 1835.
Classic text on the subject:
 J. F. Ritt, _Integration in Finite Terms_ (Columbia Univ Pr, 1948)
Introductory papers, aimed at undergraduates:
 A.D. Fitt & G.T.Q. Hoare, The closed-form integration of arbitrary
  functions.  Mathematical Gazette (1993) 227--236.
 E. Marchisotto & G. Zakeri, An invitation to integration in finite
  terms.  College Math. J. 25 (1994) 295--308.
  
A modern text (omitting the algebraic case)
 M. Bronstein, _Symbolic Integration I: Transcendental Functions_
  (Springer-Verlag 1997)
-- 
G. A. Edgar                               
http://www.math.ohio-state.edu/~edgar/
It can be done. After all, you're familiar with series, presumably, and 
could come up with a series for e^(x^2). You could integrate this term by 
term and find a series that would give you the antiderivative. This is 
considerably more messy, however, than int(dx/x) = ln(x) + C.
This isn't the sort of thing we have to do for most functions. We know how 
to write int(dx/x) in a nice, concise way because we have the ln function; 
in fact, some people define ln(x) = int(t = 1 to x, dt/t). You could, if you 

wanted to, write pq(x) = int(t = 0 to x, exp(t^2) dt). So when someone asked 

you what the indefinite integral of e^(x^2) was, you'd say pq(x) + C.
The difference here is that pq, unlike ln, is not an elementary 
function. 
It's not something like a logarithm, sine, cosine, a polynomial, or any 
other beastie that has been used enough to have a name attached to it.
Can't be done? Not at all - don't believe them. Can't be done in closed form 

(no infinite series) in terms of elementary functions? Seems so, but I don't 

think anyone has really proven that.
Remember first that u = e^x = exp(x) and x = ln u, so you can't treat x like 

a constant when you are integrating with respect to u.
Secondly, are you claiming that int[exp(c*ln(u))du] = (1/c) exp[c*ln(u)] + 
C? You'll note that exp[c*ln(u)] = u exp c, so int[exp(c*ln(u))du] = int (u 

exp c du) = (exp c/2) u^2 + C. This is rarely equal to (1/c) exp[c*ln(u)] + 

C, so something must be going wrong.
The formula you are probably half-remembering is int(exp(au)du) = (1/a) 
exp(au) + C. Unfortunately, you have ln(u) instead of u itself in the 
exponent, so you can't use this formula.
Same thing here; u is a function of x and x is a function of u, so in an 
integration with respect to one or the other, you can't treat the other as a 

constant.
I'm not sure what's going on here, but it looks like despite writing du you 

actually took the integral with respect to x.
Justin Davis a .8ecrit :
A guy called Liouville (heard about him?) did that about 150 years ago. 
He *really* proved it. I think in fact this should be in the FAQ somewhere.
What I meant by really proved it is that after a while some things are 
taken as loosely proven by time; that seemed to me to be the case as I 
hadn't heard of the result and couldn't imagine how we might gain a negative 

result working with such complex (and some seemingly arbitrary(*)) objects. 

And yes, I've heard of him; G.A. Edgar was kind enough to give references.
This seems to be the theorem he proved, for those that are interested: If f 

and g are rational functions, g not constant, then int(f(x)^g(x)dx) is 
integrable in finite elementary terms iff there exists a rational function R 

such that f = R' + Rg.
(*) It seems that the class of elementary functions is connected in some way 

that is unclear to me. I will be looking for the proofs, but I wonder if 
anyone is already familiar with the idea. The way I thought, it could have 
been decided a long time ago that pq(x) = int(t = 0 to x, exp(t^2)dt) was an 

elementary function. Were pq deemed elementary, however, the above theorem 
wouldn't hold so neatly. What is so special about the elementary functions? 

Or are we lucky that we picked the elementary functions that we did? 
Well, of course it can be done at all.  e^(x^2) is a
continuous function, so it has an antidervative.
It's just that the antiderivative can't be expressed
as a composition of a fairly arbitrary set of
commonly-used functions.
Is it possible to give a precis of the proof?  I'd guess that
Liouville found a form for an arbitrary combination of elementary
functions, took its derivative, and extracted from that expression
some sort of invariant that differs from that extracted from
functions like e^(x^2) ?  Is that on the right track?
Depends on the value of commonly-used, I suppose.  The
antiderivative can be written as sqrt(pi/4) erfi(x), where
erfi is the imaginary error function.  It could also be 
written in terms of any of an alphabet soup of special functions:
parabolic cylinder functions, the exponential integral function Ei, 
the incomplete Gamma function, Hermite H function, Kummer M and U 
functions, Laguerre L function, Meijer G function, Whittaker M 
and W functions, Dawson's function, hypergeometric 1F1 function, 
etc. 
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada
 
  
This has been posted a number of times.
I really really really ought to polish this up for FAQ inclusion.  I
the general Liouville theory, the second applying this theory to x^x:
We give a fairly complete sketch of the proof that certain functions,
including the asked for one, are not integrable in elementary terms.  The
central theorem is due to Liouville in 1835.  His proof was analytic.  The
sketch below is mostly algebraic and is due to Maxwell Rosenlicht.  See 
his
papers in the _Pacific Journal of Mathematics_, 54, (1968) 153-161 and 65,
(1976), 485-492.
WARNING: Prerequisites for understanding the proof is a first year 
graduate
course in algebra, and a little complex analysis.  No deep results are 
used,
but I cannot take the time to explain standard notions or all the 
deductions.
Notation: a^n is a power n, a_n is a sub n.  C is the complex 
numbers,
for fields F, F[x] is the ring of polynomials in x OR an algebraic 
extension
of F, F(x) is the field of rational functions in a transcendental x, M is
the field of meromorphic functions in one variable.  If f is a complex
function, I(f) will denote an antiderivative of f.
A differential field is a field F of characteristic 0 with a derivation.
Thus, in addition to the field operations + and *, there is a derivative
examples are C(z) and M with the usual derivative map.  Notice a basic
identity (logarithmic differentiation) holds:
   [(a_1 ^ k_1) * ... * (a_n ^ k_n)]'      a_1'             a_n'
   --------------------------------- = k_1 ---  + ... + k_n ---
    (a_1 ^ k_1) * ... * (a_n ^ k_n)        a_1              a_n
The usual rules like the quotient rule also hold.  If a in F satisfies
a'=0, we call a a constant of F.  The set of constants of F is called
Con(F), and forms a subfield of F.
The basic idea in showing something has no elementary integral is to
reduce the problem to a sequence of differential fields F_0, F_1, etc.,
where F_0 = C(z), and F_(i+1) is obtained from F_i by adjoining one
new element t.  t is obtained either algebraically, because t satisfies
some polynomial equation p(t)=0, or exponentially, because t'/t=s' for
some s in F_i, or logarithmically, because t'=s'/s is in F_i.  Notice
that we don't actually take exponentials or logarithms, but only attach
abstract elements that have the appropriate derivatives.  Thus a function
f is integrable in elementary terms iff such a sequence exists starting
with C(z).
Just so there is no confusion, there is no notion of composition 
involved
here.  If you want to take log s, you adjoin a transcendental t with the
relation t'=s'/s.  There is no log function running around, for example,
except as motivation, until we reach actual examples.
We need some easy lemmas.  Throughout the lemmas F is a differential 
field,
and t is transcendental over F.
Lemma 1: If K is an algebraic extension field of F, then there exists a
unique way to extend the derivation map from F to K so as to make K into
a differential field.
Lemma 2: If K=F(t) is a differential field with derivation extending F's,
and t' is in F, then for any polynomial f(t) in F[t], f(t)' is a 
polynomial
in F[t] of the same degree (if the leading coefficient is not in Con(F))
or of degree one less (if the leading coefficient is in Con(F)).
Lemma 3: If K=F(t) is a differential field with derivation extending F's,
and t'/t is in F, then for any a in F, n a positive integer, there exists
h in F such that (a*t^n)'=h*t^n.  More generally, if f(t) is any 
polynomial
in F[t], then f(t)' is of the same degree as f(t), and is a multiple of
f(t) iff f(t) is a monomial.
These are all fairly elementary.  For example, (a*t^n)'=(a'+at'/t)*t^n
in lemma 3.  The final 'iff' in lemma 3 is where transcendence of t comes
in.  Lemma 1 in the usual case of subfields of M can be proven 
analytically
using the implicit function theorem.
--------------------------------------------------------------------------
MAIN THEOREM.  Let F,G be differential fields, let a be in F, let y be in 
G,
and suppose y'=a and G is an elementary differential extension field of F,
and Con(F)=Con(G).  Then there exist c_1,...,c_n in Con(F), u_1,...,u_n, v
in F such that
                        u_1'            u_n'
               a =  c_1 --- + ... + c_n --- + v'.
                        u_1             u_n
In other words, the only functions that have elementary anti-derivatives
are the ones that have this very specific form.
--------------------------------------------------------------------------
This is a very useful theorem for proving non-integrability.  In the usual
case, F,G are subfields of M, so Con(F)=Con(G)=C always holds.
Proof:
By assumption there exists a finite chain of fields connecting F to G
such that the extension from one field to the next is given by performing
an algebraic, logarithmic, or exponential extension.  We show that if the
form (*) can be satisfied with values in F2, and F2 is one of the three
kinds of allowable extensions of F1, then the form (*) can be satisfied
in F1.  The form (*) is obviously satisfied in G: let all the c's be 0, 
the
u's be 1, and let v be the original y for which y'=a.  Thus, if the form
(*) can be pulled down one field, we will be able to pull it down to F,
and the theorem holds.
So we may assume without loss of generality that G=F(t).
Case 1: t is algebraic over F.  Say t is of degree k.  Then there are
polynomials U_i and V such that U_i(t)=u_i and V(t)=v.  So we have
                        U_1(t)'            U_n(t)'
               a =  c_1 ------ + ... + c_n ------ + V(t)'.
                        U_1(t)             U_n(t)
Now, by the uniqueness of extensions of derivatives in the algebraic case,
we may replace t by any of its conjugates t_1,..., t_k, and the same 
equation
holds.  In other words, because a is in F, it is fixed under the Galois
automorphisms.  Summing up over the conjugates, and converting the U'/U
terms into products using logarithmic differentiation, we have
            [U_1(t_1)*...*U_1(t_k)]'
 k a =  c_1 ----------------------- + ...  + [V(t_1)+...+V(t_k)]'.
             U_1(t_1)*...*U_n(t_k)
But the expressions in [...] are symmetric polynomials in t_i, and as
they are polynomials with coefficients in F, the resulting expressions
are in F.  So dividing by k gives us (*) holding in F.
Case 2: t is logarithmic over F.  Because of logarithmic differentiation
we may assume that the u's are monic and irreducible in t and distinct.
Furthermore, we may assume v has been decomposed into partial fractions.
The fractions can only be of the form f/g^j, where deg(f)<def(g) and g
is monic irreducible.  The fact that no terms outside of F appear on the
must be occuring.
Let t'=s'/s, for some s in F.  If f(t) is monic in F[t], then f(t)' is 
also
in F[t], of one less degree.  Thus f(t) does not divide f(t)'.  In 
particular,
all the u'/u terms are in lowest terms already.  In the f/g^j terms in v,
we have a g^(j+1) denominator contribution in v' of the form 
-jfg'/g^(j+1).
a, because a is a member of F.  Thus no f/g^j term occurs at all in v.  
But
(Remember the u's are distinct, monic, and irreducible.)  Thus each of the
u's is in F already, and v is a polynomial.  But v' = a - expression in 
u's,
so v' is in F also.  Thus v = b t + c for some b in con(F), c in F, by 
lemma
2.  Then
                   u_1'            u_n'     s'
          a =  c_1 --- + ... + c_n --- + b --- + c'
                   u_1             u_n      s
is the desired form.  So case 2 holds.
Case 3: t is exponential over F.  So let t'/t=s' for some s in F.  As in
case 2 above, we may assume all the u's are monic, irreducible, and 
distinct
and put v in partial fraction decomposition form.  Indeed the argument is
identical as in case 2 until we try to conclude what form v is.  Here 
lemma
3 tells us that v is a finite sum of terms b*t^j where each coefficient is
in F.  Each of the u's is also in F, with the possible exception that one
of them may be t.  Thus every u'/u term is in F, so again we conclude v'
is in F.  By lemma 3, v is in F.  So if every u is in F, a is in the 
desired
form.  Otherwise, one of the u's, say u_n, is actually t, then 
                        u_1'
               a =  c_1 --- + ... + (c_n s + v)'
                        u_1
is the desired form.  So case 3 holds.
------------------------------------------------------------------QED------
This proof, by the way, is a LOT easier than it looks.  Just work out some
examples, and you'll see what's going on.  (If this were a real expository
paper, such examples would be provided.  Maybe it's better this way.  
Indeed,
if anybody out there takes the time to work some out and post them, I 
would
be much obliged.)
So how to you actually go about using this theorem?  Suppose you want to
integrate f*exp(g) for f,g in C(z), g non zero.  [This isn't yet the asked
for problem.]  Let t=exp(g), so t'/t=g'.  Let F=C(z)(t), G=any 
differential
extension field containing an antiderivative of f*t.  [Note that t is in
fact transcendental over C(z): g is rational and non-zero, so it has a
pole (possibly at infinity) and so t has an essential singularity and 
can't
be algebraic over C(z).]  Is G an elementary extension?  If so, then
                        u_1'            u_n'
             f*t =  c_1 --- + ... + c_n --- + v'
                        u_1             u_n
where the c_i, u_i, and v are in F.  Now the left hand side can be viewed
as a polynomial in C(z)[t] with exactly one term.  We must identify the
coefficient of t in the right hand side and get an equation for f.  But
the first n terms can be factored until the u_i's are linear (using the
logarithmic differentiation identity to preserve the abstract from).  As
for the v' term, long divide and use partial fractions to conclude v is a
sum of monomials: if v had a linear denominator other than t, raised to
some power in its partial fraction decomposition, its derivative would be
terms.  (As in the proof.)  If w is the coefficient of t in v, we have
f=w'+wg' with w in C(z).  Solving this first order ODE, we find that
w=exp(-g)*I(f*exp(g)).  In other words, if an elementary antiderivative
can be found for f*exp(g), where f,g are rational functions, then it is of
the form w*exp(g) for some rational function w.  [Notice that the 
conclusion
would fail for g equal to 0!]
For example, consider f=1 and g=-z^2.  Now exp(z^2)*I(exp(-z^2)) has no
poles (except perhaps at infinity), so if it is a rational function, it
must be a polynomial.  So let (p(z)*exp(-z^2))'=exp(-z^2).  One quickly
verifies that p'-2zp=1.  But the only solution to that ODE is the error
function I(exp(-z^2)) itself (within an additive constant somewhere)!
And the error function is NOT a polynomial!  (Proof?  OK, for one thing,
its Taylor series obtained by termwise integration is infinite.  For
another, its derivative is an exponential.)
As an exercise, prove that I(exp(z)/z) is not elementary.  Conclude that
neither is I(exp(exp(z))) and I(1/log(z)).
For a slightly harder exercise, prove that I(sin(z)/z) is not elementary.
Conclude that neither is I(sin(exp(z))).
Finally, we consider the case I(z^z).
So this time, let F=C(z,l)(t), the field of rational functions in z,l,t,
where l=log z and t=exp(zl)=z^z.  Note that z,l,t are algebraically
independent.  (Choose some appropriate domain of definition.)  Then
t'=(1+l)t, so for a=t in the above situation, the partial fraction
analysis (of the sort done in the previous posts) shows that the only
possibility is for v=wt+... to be the source of the t term on the left,
with w in C(z,l).
So this means, equating t coefficients, 1=w'+(l+1)w.  This is a first
order ODE, whose solution is w=I(z^z)/z^z.  So we must prove that no
such w exists in C(z,l).  So suppose (as in one of Ray Steiner's posts)
w=P/Q, with P,Q in C[z,l] and having no common factors.  Then z^z=
(z^z*P/Q)'=z^z*[(1+l)PQ+P'Q-PQ']/Q^2, or Q^2=(1+l)PQ+P'Q-PQ'.  So Q|Q',
meaning Q is a constant, which we may assume to be one.  So we have
it down to P'+P+lP=1.
Let P=Sum[P_i l^i], with P_i, i=0...n in C[z].  But then in our equation,
there's a dangling P_n l^(n+1) term, a contradiction.
-- 
-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
-- 
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
There's no of course when there's a question about the possibility - the 
original poster said it couldn't be done, period. It ought to be noted that 

I made a point similar to yours in my response, suggesting, for instance, 
that he could use a series. 
If that's what he meant (which isn't so clear), then he was wrong, and this
is an of course.  That's just the Fundamental Theorem of Calculus, 
nothing
deeper.
Sometimes you can use a series, but not always.  You have to check that
the original function has a series expansion that converges to the original
function, and I forget whether that's enough to guarantee that the series
formed by integrating each term converges to the antiderivative of the 
original
function.  Yes, I'm almost sure it *does* work for the specific
case of e^(x^2), but it's just an added complication if the question is
whether e^(x^2) has an antiderivative; *that* follows just by continuity.
The requirement is uniform convergence with continuous terms.
-- 
I'm not interested in mathematics that might have anything
to do with reality. -- Russell Easterly, in sci.math
Pointwise convergence with locally uniformly bounded partial sums 
(or more generally, bounded by some locally L^1 function) is enough.  
That's from the Lebesgue Dominated Convergence Theorem.
In the case of a Taylor series, if the series converges to the original
function on some open interval then the series formed by integrating 
term-by-term
converges to an antiderivative on the same interval.  
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada
Free Rolodex program with documentation.
Go to http://www.mdiecast.com/oldpictures/shambat/Rolodex.htm .
Ilya Shambat.
It's definitely something related to a circle, so I'm guessing you know a 
little about how to solve this problem. You're familiar with the equation 
exp(i * theta) = cos(theta) + i sin (theta)?
You have exp[(i-1)theta] = exp(i * theta) exp(-theta). The exp(i * theta) 
alone would behave as you probably suspect, tracing out counterclockwise a 
circle of radius 1 as theta increases; if this isn't clear to you, it's 
_vital_ that you go back and understand how this function works before you 
go on.
Then consider r exp (i * theta). What is this?
While your exp(i * theta) would be tracing a circle, the term exp(-theta) is 

shrinking as theta increases. This is slightly more complex than having a 
constant r, but the same idea is at play. 
No, it's a spiral.
Jose Carlos Santos
http://share.tickle.com/1a2374
Hi all,
The Cantor set has the same cardinal as the set of real numbers. A way
of proving this fact is this one: a number x belongs to the Cantor set
iff x can be written in base 3 as 0,a_1a_2a_3a_4..., where each a_n is
either 0 or 2. Then, to each such x we can associate the real number
which, in base 2, can be written as 0,(a_1/2)(a_2/2)(a_3/2)... This
defines a surjective function from C onto [0,1].
My question is: does anyone know who is the author of this proof?
Jose Carlos Santos
But it's not a bijection.  What you actually get is a bijection between the
Cantor set and P(N), the power set of the integers.
Cantor certainly knew that |R| = |P(N)|.  One of the most obvious ways of
proving that is to use the Cantor-Bernstein theorem.
-- 
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
I know it's not a bijection! But since there's a surjectiove function
from the Cantor set onto [0,1], the cardinal of the Cantor set is grater
or equal to the cadinal of the set of real numbers. Since, on the other
hand, the Cantor set is a set of real numbers, it follows that both sets
have the same cardinal.
I have explicitly asked who was the author of *that* proof, not who
discovered that theorem.
JOse Carlos Santos
Do you know what the Cantor-Bernstein theorem says?  If you do, then why
did you just repeat what I had already said?
                              ^^^^^^^^^^^^^^^^^^^^^^^^
What proof?  What you described in your original question was not a proof
at all.  I made it into a proof by invoking the Cantor-Bernstein theorem.
If your question is who was the first to describe the bijection between
P(N) and 2^N (one being the set of all subsets of N and the other being
the set of all characteristic functions on N), I think you will find that
it is implicit in Cantor's definition of A^B for cardinal numbers A and
B.
-- 
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
So this is a valid argument, but you might notice that it
can't be formalized without AC or some fragment thereof.
That means it leaves open the possibility that the cardinalities
might not be the same in some otherwise-reasonable model of ZF
that we actually care about (say L(R)).
But in fact there's no such possibility, because you can
get *injections* both ways between the Cantor set
and [0,1], and then apply Schroeder-Bernstein.
My friend Chris K., who used to post on alt.angst, alerted me of a
phenomenon called the externalization of costs. Externalization of
costs consists of economic entitites catering to one set of people
making other people shoulder the ecological, medical or political
burden of the production. So that, for example, the people who leech
out gold in Indonesia with mercury, while they can deliver the gold
to the West cheaply, devastate large tracts of nature and poison the
villagers who live there. The people who log or ranch in the
rainforest kill people who live there; the people who used CFC's
caused holes in ozone layer that sickened many people; etc.
Based on this concept, I, with education both in economics and in
psychology, suggest a different concept. That concept is:
Externalization of psychological costs. The meaning of that concept is
making certain people carry the psychological trash that societies,
communities, countries and individuals put out.
In more familiar terms, externalization of psychological costs is also
known as scapegoating, demonization, character destruction and simple
bullying. It is essentially the case of a society relieving itself on
a person or set of people - a set of a society defecating on
individuals, classes or other communities - and, as they are covered
with - socially manufactured manure - shoving them into the sewers.
This takes place in many societies I have seen, from University of
Virginia to rural South to Midwest to Portland to, well, the Internet.
Any time someone has any idea in their heads that at all differs from
the lie that passes in the society as sanity or righteousness - any
time someone has any propensities or desires that at all differ from
the lie that the society, claiming to derive its validity from
fulfilling the human nature, reflect an aspect of human nature
different from the one the society overbearingly espouses as human
nature - any time someone has any gift, any interest, any capacity for
insight that has the ability to see through the social morass - they
become branded as outcasts or as evil, and the whole weight of society
presses on them.
It has been my experience that, if a man is put into such a position,
the society tries to imprison him or send him away; while if a woman is
put into such a position, the society tries to entomb and destroy her.
And when I remove the woman in that position from her society, the
society as such collapses. There follows a lot of threats, abuse and in
many cases violence. I say in many cases; because for as long as
America remains in any way true to its founding principles, police can
be rightly employed to fight off the overbearing near-totalitarianism
of the entities, wherever they may be, on the Right or on the Left,
that has appropriated for itself the concept of ethics and social
construction at whatever costs and in whatever manner.
I rejoice doing the work of removing scapegoats from cultures. It is my
patriotic duty as an American citizen to ensure that, in a nation whose
government is designed to prevent official oppression, unofficial
oppression does not take form in the name of society or
communitarianism or left-wing Neo-Puritanism or any other pretext for
would-be fascists. I hope more conscientious people perform the same
duty, and in so doing end all dishonesty, oppression and hideousness in
America and make this country what its founders intended for it to be.
Ilya Shambat.
Interesting philosophy you've got there. It sounds like you've been
reading a combination of Rand/Nietzsche/Carson.
////////////////////
www.urine-pimp.com
********************
Feminisim (or sects of)  is hate, it meets every criteria of hate
except recognition.
Nope, just fed up. Too ing obvious I guess. There's every
indication of sever psychological trama while the majority of studies
revolve around hang nails, freezer burn (cuz there's nothing worse in
the world then) and projected denial.
It appears men won't recieve equal application of statistics in
setting policy and influencing judicial decisions. Seems overtly
abusive and a bit too smug and very very provocative.
Anyways, the republican party is out (considering the amount of
infiltration and gender selective policies with far reaching
consequence)
Don't post in sci.math, it looks like bull and no one reads it.
  Post in alt.angst
 
        Really? By algorithmic methods? Boy, you are even more ignorant than 

I
thought. How do you manage to remain so thoroughly ignorant? How come you
are not embarrassed to make a public display of your ignorance, time and
again? Are you ill?
random
I
you
and
That's why he's considered a troll. He's only pretending to be ignorant
to see how many responses he gets.
Note that you do not answer any significant questions in your post.
You still have no idea what a *perfect random number generator* is. Add
that lapse to the growing  list of things you make stupid comments about.
Item 3. was not a possibility. It should have read, 3. I am an ineducable
moron.
--
There are two things you must never attempt to prove: the unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
What is a perfect algorithm? You mean one that always works and terminates 
in poly-time? If that is what you mean, you have never proved this, which 
makes you a liar. If it is not what you mean, then it is crap, and again, 
you are a liar..
And if it is crap that means that you are a liar.
That is nonsense. You have been told many times that any algorithm executing 

on a deterministic computer cannot generate random numbers. You are an 
idiot.
But how fast? There are already plenty of ways to factor anything given 
enough time.
No idiot...
    
************************
David C. Ullrich
Yes, it does. You've been explaining that mathematicians lie
for eight or ten years now, and so far you haven't convinced
a single person.
Yes, it's a strange situation.
Oh my. I'm on trial? Have been for some time?
Hint: You're sounding totally wacky again.
I wonder why it is that when the rest of us have something to
say about math we're able to convince people we're right
without getting into all this stuff about Fate and the weight
of the future? Just dumb luck, I suppose.
************************
David C. Ullrich
my simple innumeracy aside, I can still see you are a wacho  nitwit. 
clear signs of megalomania.  Delusions that the dribble you say means 
anything to anyone except in your deluded mind.  Your method does not do 
what you say it does   there are many methods for counting primes. Yours 
does not seem to be anything but a misguided attempt.
            josephus
I'm sure the answer to my question is some standard result.
Some definitions:
Let BF be the set of bit sequences---functions from N+ into {0,1}.
For each natural n, and for each f in BF, let C(f,n) be
the the cardinality of the set of all j <= n such that f(j) = 1.
Let r(f,n) be the fraction C(f,n)/2^n
of r(f,n) exists, and is equal to 1/2.
If e is a real number in 0 < e < 1/2, call a function f e-normal
if for all n,
       1/2 - e < r(f,n) < 1/2 + e
The idea is this: e-normality is a falsifiable approximation to normality.
That is, if I observe only a finite sequence of outputs of f, there is no
way to disprove the hypothesis that f is normal. But I *can* disprove the
hypothesis that f is e-normal. If I choose e to be very tiny, then
e-normality is approximately the same as normality.
Now for some measure theory.  Assume the
usual measure on the set BF: for any element f, for any natural
number n, the measure of the set of all functions f' such that
    f(1) = f'(1)
    f(2) = f'(2)
    .
    .
    .
    f(n) = f'(n)
is 2^{-n}.
Question: What is the measure M(e) of the set of e-normal functions?
--
Daryl McCullough
Ithaca, NY
On 3 Mar 2005 08:11:41 -0800, stevendaryl3016@yahoo.com (Daryl
Or rather  C(f,n)/n, right?
So in particular no number is e-normal, since r(f,1) is
1 or 0.
Regardless of f, in fact...
For relatively weak values of approximately the same, yes.
0.
There are probably interesting analogous questions, but this
one is a little too easy.
David C. Ullrich
David C. Ullrich says... 
Right. I should have said something like
        1/2 - (e + 1/2^n) < r(f,n) < 1/2 + (e + 1/2^n)
--
Daryl McCullough
Ithaca, NY
Whoops. I think I made a small mistake in the definition
of e-normal:
That definition needs to be loosened up in the case of small n.
I think this correction works:
If e is a real number in 0 < e < 1/2, call a function f e-normal
if for all n,
       1/2 - e - 2^{-n} < r(f,n) < 1/2 + e + 2^{-n}
The reason for this change is that, in the case n=1, r(f,n) is
either 0 or 1.
--
Daryl McCullough
Ithaca, NY
            days. My association with the Department is that of an alumnus.
What you learned, presumably, is that it is impossible to define an
operations<-.
on the complex numbers such that:
     holds.
you woul dhave -(-1) = 1 < 0, but by (iii) you would also have
-1<0.
HOWEVER, it is quite possible to define an ordering on the complex
numbers. For example, a very easy and natural ordering on the complex
numbers is to define
  (a + bi)  <  (c + di)  if and only if  a<c  or (a=c and b<d).
This is the lexicographic order, and on the complex plane it is the
same as saying that being to the left of, or on the same vertical line
and below, is the same as being smaller than. In this definition, i
is indeed bigger than 0.
This is a perfectly fine ordering that has a lot of good uses. It is
not compatible with the operations, however, so it also fails to be
any good for a number of (algebraic) applications. That does not mean
you cannot define an ordering on the complex numbers.
(Oh, and if you accept the Axiom of Choice, then of course the complex
for this ordering).
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
It's true that the complex numbers aren't well ordered, but that's not
germane because the real numbers aren't either.  A set is said to be
well ordered if every non-empty subset has a minimal element.  For
example the natural numbers are well ordered, the positive rationals
aren't.
The relevant order is that of an ordered field.  So, in addition to
the other axioms for a field, we have
(1) For all x, y, one and only one of
    holds.
(2) For all x, y, z
(3) For all x, y, z
(4) For all x, y, z
It is these rules that the complex numbers don't satisfy.
You did.
You're right.
Hi All,
Is there a way to calculate the volume encased by gaussian function of
a certain value? For example, an asymmetric dumbbell can be described by
Exp[-0.5((x-1.5)^2+y^2+z^2)] + 0.75*Exp[-0.5((x+1.5)^2+y^2+z^2)] = 0.5.
I think that something like the marching cubes algorithm may work, but
don't know if there is any other more elegant way to calculate the 
the area of this isosurface and the volume it surrounds.
Any help would be greatly appreciated.
Minhhuy
This would be a solid of revolution obtained by revolving the curve
 Exp[-0.5((x-1.5)^2+y^2)] + 0.75*Exp[-0.5((x+1.5)^2+y^2)] = 0.5
y = 1/2 sqrt(8 ln(2 exp(3x)+3/2) - (2x+3)^2)
So the volume is pi int_a^b y^2 dx = pi (F(b) - F(a)) where a and b are the 

solutions of your equation with y = 0, and 
                 3       2
         9 x    x     3 x                              3
F(x) = - --- - ---- - ---- - 2/3 ln(2/3) ln( 1 - --------------)
          4     3      2                         4 exp(3 x) + 3
                           3                        2        2
         + 2/3 dilog(--------------) + 1/3 ln(--------------)
                     4 exp(3 x) + 3           4 exp(3 x) + 3
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada
taken
by
Yes, I thought you had looked at gabriel's stuff so I assumed you knew
what I was talking about.
I will post an entire proof which I think might be correct in the next
few days.
Here is my attempt to prove Gabriel's Theorem:
Have changed some of his wording and am calling it
the AFD rather than ATG as he calls it.
Given an interval [x;x+w] subdivided into n equal parts
and a function f which is continuous on [x;x+w] and
differentiable on [x;x+w), the secant gradient or mean
value of f is equal to the average first derivative of f
[or AFD(f)] defined as follows:
w is the width of the interval
n is the number of partitions
           f(x + w/n)  - f(x)
 AFD(f)  = ------------------             --- L.H.S
                  w/n
                   1   n-1         ws
 AFD(f)  =  Lim    -  SIGMA f'(x+ --- )   --- R.H.S
Proof:
Start with RHS.
                   1
 AFD(f)  =  Lim    - [ f'(x) + f'(x+w/n) + f'(x+2w/n) + ...
                       f'(x+(n-2)w/n) + f'(x+(n-1)w/n)]
Now we simplify Gabriel's proof by keeping the ens fixed and using
t in the definition:
i.e. we let n be fixed for each of f'(x), f'(x+w/n), f'(x+2w/n), etc.
In this case, the following are true:
                 f(x+w/t)-f(x)
  f'(x)  = Lim   -------------
  f'(x+w/n) =
         f(x+w/n+w/t)-f(x+w/n)
  Lim    ---------------------
  f'(x+2w/n) =
        f(x+2w/n+w/t)-f(x+2w/n)
  Lim   -----------------------
  f'(x+ 3w/n) =
        f(x+3w/n+w/t)-f(x+3w/n)
  Lim   -----------------------
  and so on:
 Now,
                   1       f(x+w/t)-f(x)
 AFD(f)  =  Lim    - [ Lim -------------  +
                       f(x+w/n+w/t)-f(x+w/n)
                   Lim ---------------------  +
                       f(x+2w/n+w/t)-f(x+2w/n)
                   Lim -----------------------  +
                       f(x+3w/n+w/t)-f(x+3w/n)
                   Lim ----------------------- +
       + ...
                       f(x+(n-2)w/n+w/t)-f(x+(n-2)w/n)
                   Lim -------------------------------  +
                       f(x+(n-1)w/n+w/t)-f(x+(n-1)w/n)
                   Lim -------------------------------  ]
  Looking at the above we have two limits to infinity.
  i.e. n and t. However, since both n and t tend to infinity,
  the outer limit is taken once over the entire sum.
  With n and t coinciding, we have:
                   1       f(x+w/n)-f(x)
 AFD(f)  =  Lim    - [ Lim -------------  +
                       f(x+2w/n)-f(x+w/n)
                   Lim ------------------  +
                       f(x+3w/n)-f(x+2w/n)
                   Lim -------------------  +
                       f(x+4w/n)-f(x+3w/n)
                   Lim -------------------  +
      + ...
                       f(x+(n-1)w/n)-f(x+(n-2)w/n)
                   Lim ---------------------------  +
                       f(x+w)-f(x+(n-1)w/n)
                   Lim --------------------  ]
  Which leads to (**) :
                   1   f(x+w/n)-f(x)
 AFD(f)  =  Lim    - [ -------------  +
                       f(x+2w/n)-f(x+w/n)
                       ------------------  +
                            w/n
                       f(x+3w/n)-f(x+2w/n)
                       -------------------  +
                            w/n
                       f(x+4w/n)-f(x+3w/n)
                       -------------------  +
                            w/n
      + ...
                       f(x+(n-1)w/n)-f(x+(n-2)w/n)
                       ---------------------------
                            w/n
                       f(x+w)-f(x+(n-1)w/n)
                       --------------------  ]
                            w/n
  Some more simplication:
                   1   f(x+w)-f(x)
  AFD(f) =  Lim    - [ ----------- ]
  Then,
                   1 n   f(x+w)-f(x)
  AFD(f) =  Lim    -.  [ ----------- ]
  So,
                      f(x+w)-f(x)
  AFD(f) =  Lim     [ ----------- ]
  And finally,
             f(x+w)-f(x)
  AFD(f) =   -----------
                   w
  which is the desired result.
  (**) Dropping the inner limit is what I am not sure about.
  Jason Wells
Jason
Still putting lipstick on this pig? :)
Hi Everyone,
of the results and download the program (executable or source code) go 
here:
http://www.craigsarea.com/primes.html
The islands are drawn by taking a list of prime numbers and performing a 
mod
5 on each number in the list.  The result of the mod 5 moves a pen 
around
the screen.  When the pen draws over itself the colour is made lighter,
which creates a height effect.  As all prime numbers (except 2 and 5) end 
in
either 1, 3, 7 or 9 the mod 5 isn't really needed.  The final digit of each
prime number could be used by itself.
The thumbnailed images on the website have been reduced in size and
converted to JPEG.  So some of the detail has been lost.  To download the
original unmodified images just click on any of the thumbnails.  This will
allow you to download a zip file containing a large BMP file.
I can't think of any particular use for the islands other than they look
quite nice.  I'm thinking of plotting them in 3D, which would look even
better.
The website is new so if you have any problems or queries just let me know.
Craig.
Website: www.craigsarea.com
Email: craig@craigsarea.com
Nice pics, Craig. I was surprised that the second could be so
got the same pictures.
One can also draw primes mod 8 (1, 3, 5, 7) and mod 12 (1, 5, 7, 11).
For each modulus there are essentially three ways to map remainders to
directions; other mappings give rotations and reflections of the three.
-- 
Don Reble  djr@nk.ca
constructed up to line n is already contained in line n -13. If the
for the whole diagonal.
If it works in the original version (starting to construct the
antidiagonal with a_{1,1}) it works in my version too (starting with
a_{1,14} and replacing 0 by 1). If you don't believe that, please tell
me why.
tell
It doesn't work in the original version. There's
no sequence, no limit process. There's a number.
We describe the properties of that number. Our
description of those properties demonstrate that
it is not on the original list.
In the original version, the property of the number
is it differs from a_k in position k, for all k. So
it is different from all a_k. So it is not in the list.
If that number is described as it differs from a_k
in position (k+13) for all k, then it is not on the
list. In that sense what works in Cantor's original
version works for a displacement by 13 as well.
But you are attempting to create a version of the proof
that never existed in the first place, then go on
to distort your already distorted and invalid proof.
So let us discuss the single number which you are
trying to describe, the antidiagonal. Does it
differ in at least one position from every number
on the original list?
            - Randy
Impossibility proof means that it is impossible to construct a list
which contains its antidiagonal (= the diagonal constructed by
exchanging the diagonal digits).
Nice that your logic reaches thus far. This means my method starting
at digit a_{1,14}  is equivalent to Cantor's starting at digit
a_{1,1}.
Your ignorance is remarkable. Did you never study Cantor's? His method
leaves only one choice, because it is constructed using the binary
system. But even in the decimal system it is sufficient to take any
fixed rule, saying by what digit d' the digit d has to be replaced.
Give me any such rule, and I will construct a list which shows that
Cantor's method fails.
Thus, there always
Cantor did not pay attention to that problem. However I can use
decimals and any rule of the usual form.
decimal was intended. So we have no problem with sequences 111...
Further I spoke of 1/9 = 0.111...
be
differs
whether
If you produced a number which differed from the n-th
number in the (n+14)th digit, for all n=1,2,..., then the
number would not be in the list. Mueckenheim is correct
as far as that goes.
proof
Mueckenheim's problem seems to be his old nemesis, what
has been so succinctly described as quantor dyslexia.
In this case it goes something like this:
1. Consider the list a1 = 0.(14 1's), a2 = 0.(15 1's), ...
... ak = 0 followed by (k+13) 1's.
2. I will do a diagonal construction to make a number
which is identical to all the a's in the first 14 positions,
but differs from a1 in position 15, a2 in position 16,
... ak in position (k+14), etc for all k. This number, call
it C, is 0.1111....
3. Consider the sequence b1 = 0.(15 1's), b2 = 0.(16 1's),
... bk = 0 followed by (k+14) 1's. I achieve this sequence
by flipping the (k+14)-th bit of each ak.
4. C is at the end of the sequence of b's.
5. All the b's appear on the original list of a's.
6. Therefore C must be on the original list of a's.
The problem is in #4, Mueckenheim's usual inability to
distinguish the limit of a sequence from its elements.
C is not anywhere in the list of b's. Therefore it
does not of course follow that it is in the list of
a's.
             - Randy
Yes. Instead of starting the construction of the antidiagonal at
a_{1,1} as is usual, we can start it at a_{1,14}. The exchange
prescription is: if a_{m,n} = 0 then exchange it by a'_{m,n} = 1. This
is simple. Cantor could have constructed h
is diagonal proof exactly like that.
all finite n we have the sequence as a line contained in the list. Is
there any continuity in infinite calculations. Is 0 the limit of the
sequence 1/n?
If 1/9 is the antidiagonal, then 1/9 must have appeared in a line of
the list. But it does not.
There is always a line with 13 more digits 1 than has the diagonal.
does never happen. Even in the infinit it is wrong.
On the other
Wrong. It shows that Cantor's method does not apply to this special
list. Therefore it does not cover all lists. It fails.
listed above.
Okay, so far, so good...
Yes, but it doesn't follow that there exists an n such that 1/n = 0.
Perhaps I am again misunderstanding you. The point is that the antidiagonal
is produced by exchanging digits appearing in a certain place in a number
with a digit which does not appear in that place in that number.
Once such a number has been constructed, it cannot (not must, as
you have above) have appeared in the list. And, as you have correctly 
pointed 
out, it does not.
See my comment above. The reason that Cantor's method does apply to this
special list is that, as you point out, it can be used to produce
a number not on the list. It seems that Cantor's method succeeds rather
well in this case.
Matt
Or cardinality is an inappliable notion.
For finite numbers n e N we have:
n = Card({1,2,3,...,n}).
As long as the lefthand side remains finite, the righthand side
remains finite too. There is no set of infinitely many natural numbers
Just because you struggle with it doesn't mean there's
a problem with the notion itself.
Yes.
Yes. For every n e N, both sides are finite and equal.
What does this repeated assertion have to do with the
previous discussion?
We can agree with the statement that all sets of the
form {1,2,3,...,n} are finite.
But how does that say anything about whether there are
infinite sets of natural numbers? Not all sets of
natural numbers are of the form {1,2,3,...,n}.
Do you think they are? First you make a statement
about very specific sets (those of the form
{1,2,3,...,n}, then you claim a conclusion about ALL
sets. It doesn't follow.
            - Randy
This looks a bit mixed up. The last sentence does not help.
There are at least two (equivalent) definitions of (right or left)
primitive ideal. I suspect you started with
| An ideal of R is left primitive if it is the largest ideal contained in
| some maximal left ideal of R.
With this definiton, the assumption R is commutative at once
gives the first sentence of your proof attempt, because in this case
left ideal = ideal (in particular left primitive = right 
primitive).
But then you are done: the largest ideal contained in some maximal ideal
has to be that maximal ideal.
If you start with the other definition
| An ideal I of R is left primitive if there exists a faithful irreducible
| R/I - left module.
first observe that such a module is the same as an irreducible R-module
with    I = ann(M) := { r in R | rM = 0 }.
Pick some nonzero m in M and consider  ann(m) := { r in R | rm = 0 }.
Convince yourself that
(1) ann(m) is a left ideal containing I
(2) M = Rm is isomorphic to R/ann(m) as a R-module, hence ann(m) is 
maximal.
(3) any ideal J contained in ann(m) is necessarily contained in ann(M)
This shows that I satisfies the first definition above.
In particular for commutative R, ann(m) is contained in ann(M)=I.
Marc
This can be represented as another free parameter, n0:
  a/x^2 + b/x + c log(x) = (n-n0)*K, n = 0, 1, ..., 8
Alternately,
  n = p/x^2 + q/x + r log(x) + s
where p = a/K, q = b/K, r = c/K, s = n0.
Note that K is completely arbitrary.
Well, it might be statistically questionable as to
whether it's best, but if you were to use least-squares
fitting on n as a function of x, i.e., choose p,q,r,s
to minimize  sum|a/x^2 + b/x + clog(x) + s - n|^2
then you have a simple linear least-squares problem.
Here's how you set it up. Create a matrix A where each
row represents (1/x^2, 1/x, log(x), 1) for an observation
of x. Let z = your parameter vector (p,q,r,s)'.
Then you are trying to find z the least-squares
solution to:
    Az = b
where b = (0,1,2,...)'
That is, you're trying to find z which minimizes
|Az-b|^2 = (Az-b)'(Az-b)
   = z'A'Az + b'b - 2b'Az
The value of z which minimizes this is the solution
to the linear system:
  A'Az = A'b
which you could find as z = (A'A)^(-1) * A'b but it's
more efficient to use a linear solver if you have one.
The solution I get by the way is
   p = 0.2365
   q = -1.211
   r = 1.460
   s = 2.895
I can choose K = anything I like, so I'll choose 1.
The nearest integer to s is 3, so if I round that
off I get:
   m = 0.2365/x^2 - 1.211/x + 1.46*log(x)
where m = n-3 = -3, -2, ..., 5
            - Randy
I look at it, it seems almost trivial to add the extra parameter n0,
but I just didn't see it.
tips were very valuable.
You have a mongrel equation with that log in there. You could do
yourself some good with change of variable  y = 1/x  to get the more
attractive
        ay2 + by -clogy = nK
then    logy = (ay2 + by -nK)/c
so      logy = a'y2 + b'y - n'K
the primes after dividing thru by c to reduce to fewer paramters
Mr. Dual Space
If you have something to say, write an equation.
If you have nothing to say, write an essay
The correct way to ask for help on these groups are to do as much of the 
work as you can, post that, and ask for help with the remainder.
It does not look like you put any thought into the problem.
problem thoroughly.
If you want, I can post the tens of pages I've filled in trying to get
the answer manually. But I don't think you'd read beyond the first
line, so I won't bother the rest of the community.
get
You are trying to do a non-linear fit.  Orthodox least squares fitting
will get you nowhere.  Non-linear least squares fits are no fun.
To do a non-linear fit, you need to guess approximate values for the
parameters, then define new parameters, each being the deviation from
the guessed one.  The problem will then be linearised.
You will do yourself and all your readers a favour if you would change
variables such that x becomes y and n becomes x.
Another method is to decide what the function is which you want to
minimise.  In your case, this will probably be the sum of the square
deviations.  Start with any sensibly guessed set of parameters and use
the numerical method known as the method of steepest decsent to
minimise the square deviation.
Have a look at the following url for a good description of finding a
minimum by themethod of steepest descent
http://trond.hjorteland.com/thesis/node26.html
-- 
Franz
A first-rate laboratory is one in which mediocre scientists can
produce outstanding work
P.M.S. Blackett
In other words... you are too lazy to do your own work.
And if it takes you ten's of pages to fail in solving your problem, 
perhaps you should give up?
values
there
To get started we must irst understand the problem.
Your given data:
points x(n): 0.39, 0.72, 1.00, 1.52, 2.8, 5.2, 9.54, 19.2, 30.1
only makes sense for curve fitting if the following table is used:
 n | x(n)
---------
 1 | 0.39
 2 | 0.72
 3 | 1.00
 4 | 1.52
 5 | 2.8
 6 | 5.2
 7 | 9.54
 8 | 19.2
 9 | 30.1
After all, it is a *curve-fitting* problem, and curves must be at least
two-dimensional, so it only makes sense if our points require at least
two coordinates to specify.
Plot x(n) versus n and you will see a curve. A semilog plot is
recommended.
Now you have 9 data points and 4 parameters, giving 5 degrees of
freedom for minimizing whatever criterion you are using to measure
deviation from best fit.
If this doesn't at least get you started, let us know.
Tom Davidson
Richmond, VA
tadchem a .8ecrit dans le message
You are correct!   If you use the data as above and plot
n on the x axis and x(n) on the log y axis the result is a fairly straight
line.
You might want to look at this site.
http://curvefit.com/
JOL
I took Numerical Analysis back in the '60s, but the basics haven't
changed since then. The algorithms and software have evolved a lot,
though.
In the early '90s I fitted data for the viscosity of about 700 binary
gas mixtures to a one-parameter equation within experimental error.
The parameter was an exponent of 1/3.
Tom Davidson
Richmond, VA
n appears only in the product with the parameter K. So, couldn't you
simply do a fit with a parameter K' = n*K on the right hand side,
and then calculate K=K'/n for different choices of n?
Bye,
Bjoern
Yes, that's a possibility, In  fact, I could divide the whole equation
by K, and get coeffs a', b', c' instead. But the problem still remains
that there seems to be  too many degrees of freedom, because of the
lack of constraints on n. Or perhaps not? This is my concern.
Also, if anyone has some software to do the fitting to get at least ONE
solution, it might throw some light into the problem.
For a function and data set like this I would use maximum likelihood not
least squares.
I would first treat n as a real number to convince myself it is not
something very large.
that done I would simply systematically try all the possibilities in the
neighbourhood of the real value of n.
With such a small data set this should be easy. Graphing the result will
tell you if you are off the mark.
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Summary: Is it itself a geometric lattice?
Content-Length: 480
Originator: rusin@vesuvius
I am looking at geometric lattices to model something, and need to
find an answer to the following question: is the composition of two
geometric lattices itself also a geometric lattice?  I seem to think
it must be, but do not have a proof (lattice theory is not my
subject).  I would appreciate a proof, a counter-example, or a
Shrisha Rao
shrao #AT# acm.org
-- 
http://www.dvaita.org
http://www.dvaita.net
I want to prove the following problem:
Let G be a finite group of order n.
If for every integer d that divides n there is either 0 or 1 subgroups
of G of order d prove that G is cyclic.
What I am trying to do is find an element of order n.
for every pi^mi there exists exactly one sylow pi subgroup of G.
I want to prove that this subgroup has an element of order pi^mi, any
suggestions?
            days. My association with the Department is that of an alumnus.
Ah, you can go ahead with my previous reply, but this should be
easier:
Let H be the p-Sylow subgroup of G. Show that for any x,y in H, you
then use this to show that H must by cyclic.
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
            days. My association with the Department is that of an alumnus.
Since every Sylow subgroup is normal, you have effectively reduced the
problem to the prime power order case; that is, you may assume that G
is of order p^a for some prime p and some positive integer a. If each
Sylow subgroup is cyclic, then the product of their generators will be
a generator for G.
Let us proceed by induction on a (note that the condition on G is
certainly inherited to subgroups). 
If a=1, you're done.
Suppose that you can prove that a group of order p^a must have a
subgroup H of order p^{a-1}. If you can do that, then you know the
(unique) subgroups of orders p, p^2, p^3, ... p^{a-1}, and they are
all generated by powers of x. Let y be any element of G which is not
it would have to be one of the subgroups of order p, p^2, p^3, ...,
p^{a-1}, hence would be generated by a power of x, and thus would be
contained in H; this contradicts the choice of H. So that means that
So, can you prove that every group of order p^a must have a subgroup
of order p^{a-1}?
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
Of the 5 operators you use in that expression, only 1 of them is
traditionally slow -- the ^.  You might try replacing 10^(n-1) with 
an array reference power_of_ten[n-1]. 
--Keith Lewis              klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
The solution is probably an integral or has something to do with an 
integral. I could actually notice that it looks like a Riemann sum: 
Sum(k=0.....(n-1)  f(c_k)(x_k-x_(k-1)
but (x_k-x_(k-1)= x/n (I chose a subdivision into equal parts). In the 
original text I only have 1/n, so I could try:
Sum(k=0.....(n-1)  (1/x)(cos(c_k))(x/n) that would be the integral of 
(Cosx)/x. But my aim is to calculate the limit.
Any help?
Think of the above as a Riemann sum for f(t) = cos(xt) on [0,1].
In that case it converges to the integral of cos(t) dt, wouldn't it?
It lost something in the telling ;-)
After taking some (long) time to read your very long, very boring post I 
already dislike you and your *stupid* ideas.
However, I've got several suggestions to you:
A. Stop posting this nonsense to this newsgroup, it is obvious nobody 
wants you to do so. Even I am tired of you, and it's my first posting to 
this group.
B. If you really think you did something unique why don't you send it to 
(easy to predict they either won't, or laugh at you). But, what are you 
afraid of? you can't lie to all the people all the time - so even if we 
take your theory that all mathematicians are liars as true, there still 
might be a single editor that's not controlled by the evil syndicate of 
C. If you want to prove your claims about the amazing speed of your 
function, why don't you take a crack at the RSA challenge? Shouldn't be 
too hard for an amazing counting function like yours to come up with a 
solution (and it's worth 20,000$). Anyways, such a function is worth 
even more money - I'm surprised you aren't in Maui now, having margarita 
in the sun after selling your amazing function to the highest bidder 
among the US gov., every computer security firm and every spy agency in 
the world.
Please get a life,
G. Diskin
a link to the RSA challenge: 
http://www.rsasecurity.com/rsalabs/node.asp?id=2093
either
post I
to
Your first posting?  Well you already seem to have forgotten where you
are.
You are on Usenet.  Freedom of speech is highly valued, and ordering
other people to talk or not talk is generally frowned upon.
It's arrogant behavior and also rather stupid as on Usenet, people
don't have to listen to you, and it's just rude besides.
I don't tell you whether you should post or not, so what gives you the
nerve to tell me?
Who do you think you are?
The president of the United States or something?
And, hey, he can't tell me whether or not to post either, so who the
hell do you think you are you stupid twerp?
James Harris
He already tried that... several times. It never worked out for him. He now 

claims to have a paper submitted to the Annals of Mathematicsat Princeton. 
So sad... 
A sad case indeed :)
That isn't better. It's neither a formula nor an equation but an algorithm.
--
There are two things you must never attempt to prove: the unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
As near as I can tell you believe noone understands mathematics. It is 
clear that you dont.  I think it is a fallacy to assume your thesis is 
more astute than the mathematicians.  It is clear that you separate 
yourself from mathematicians.  Because they do something you dont.  make 
sense.
        josephus
I teach math.  I enjoy it, but I do not recommend it for everyone. 
Teaching depends as much on your communication skills and ability to 
break things down as your understanding of the subject.  See how you 
like tutoring, and that will give you some idea of whether you should 
teach.
Hello
I've got a problem with number theory. Let d(n) is a number of
and I can't do it by counting number of divisors for each i.
I'd be grateful for every hint.
--
Lukasz
            days. My association with the Department is that of an alumnus.
I won't give you every hint, just some hints... 
Imagine listing all the divisors. 1 appears in all the lists.
2 appears exactly floor(n/2) times (the number of multiples of 2
between 1 and n).
3 appears exactly floor(n/3) times. 4 appears exactly floor(n/4)
times... n appears exactly floor(n/n) times.
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
you
  Or, par for the course.
  I can say I think it rocks, Mitch.
  Mitch
My hypothesis would give an average cephalic index of
(2/(1+sqrt(2)))*100 = 82.8. I'm not sure whether I would be measuring in
exactly the same place though.
Neil
-- 
Neil Fernandez
Interesting. I hadn't heard of long hat ovals before. Maybe the
eccentricity of the ellipses for you and your father etc. is relatively
large. I was only suggesting 1/sqrt(2) as an average! :-)
I also suspect that the standard hat oval could usefully be
reengineered, since it doesn't correspond to the shape of most people's
heads, tending to grip - if not pinch - more at the front and back than
at the sides. I guess this might be good for some hats but not all.
Neil
-- 
Neil Fernandez
It was somewhere outside Barstow when Neil Fernandez
Motorbike helmets (which are cheap) are sized on a simple linear
scale. Flying helmets (which are expensive and complicated) have an
eccentricity measure too.
Surely some 19th century text on physiognomy (sp?) is in order ?
A gallon of molten lead? :-)
-- 
Neil Fernandez
Bob Harris   Mar 2, 3:48 pm     show options
rec.crafts.textiles.misc, alt.math.recreational
messages by this author
While we're at it, I've often wondered how much a head weighs.  How can
I
weigh my own head?  Seriously.
Bob H
---
  Pull scale forwards. Lay down on floor. Turn head to side, position
on scale. Obviously you need a second party to the crim--one to read
                                                           Cea
PS. If  former love interest has already informed you that you are a
fathead, or some varient therreof, rely on your source, and skip above
step
Cea said,
on scale. Obviously you need a second party to the crim--one to read
                                                           Cea
PS. If  former love interest has already informed you that you are a
fathead, or some varient therreof, rely on your source, and skip above
step<
Oh my! Coffee everywhere!!!! Cea, I was waiting for you to respond :)
Cindy in WV
  Cea said,
on scale. Obviously you need a second party to the crim--one to read
                                                           Cea
PS. If  former love interest has already informed you that you are a
fathead, or some varient therreof, rely on your source, and skip above
step
---
Oh my! Coffee everywhere!!!! Cea, I was waiting for you to respond :)
Cindy in WVa
---
was done by a 2 Y.O.
It's a crim how some folks don't use spell-chuck or some version
threrrof.
Save 'puter time! Ban the semi-illiterate!!
                                     Cea
Ah, Cea, where was the spew warning here? ROFLOL!!!
Karen
I pondered:
and Carl G replied:
I had considered the Archimedean method but didn't know that heads are as
dense as water.  If I didn't have density information, how might I do it?
For example, suppose my doctor tells me I need to lose 50 pounds, and I'm
trying to decide whether to do it by diet and exercise, or just take the
easy weigh out and have a leg amputated.  Since my leg might not weigh 50
pounds, I'll have to lose some weight by exercise before the amputation.
How will I know how much to do?
Bob H
tangle
The function of sitcoms is to norm responses so that people don't have
to think. Thus Seinfeld and Frazier use the laff track to (1) present
responses outside the norm, to conditions of life which for the sitcom
viewer are often enough intolerable, and (2) resolve the tension by
having the dominant-cool personality of the show (perceived
subconsciously by the viewer as lucky and immortal merely because he
appears, by definition, in all the shows) demonstrate that all
responses except his are ineffectual vis a vis what's happening
offstage.
You are thus welcome to think of the response in these terms, but these
are not the only terms available.
In usenet, flaming and simple bullying form part of this overall
system, in which (in a paradox that should be of interest to sci.math)
Bush's approval ratings, poll results, and election numbers hold at 49%
for no accountable reason...as if individual choices are now taking
place in a political vacuum, like outer space.
Or something.
unique
In gentler times we contented ourselves with kissing, which still
managed to exchange some DNA-containing cells!
dave
OTOH, there's good reason for them to ignore it. An archive with holes
in it is defective. Take the use of X-No-Archive to the extreme and you
have an empty archive: completely useless.
Ned
-- 
Public key:  http://pgp.mit.edu/   http://www.keyserver.net/en/
Fingerprint: D17C FDD5 BBA8 8687 42E3  C8F2 C9FB 0314 E17A 0CD7
 Discussion, linux)
If Google didn't heed that header or have some other opt-out
mechanism, then they surely would have faced legal issues.  In posting
to Usenet, I surely waive some of my rights granted by copyright,
since otherwise posts could not be distributed.  But I don't see any
reason to believe that I've granted the right for Google to present a
permanent public archive of my post.
I don't see that they have that right even with an opt-out header
implemented.  Their archive includes posts that predate Deja, after
all: did those folks implicitly grant the archive rights to Google?
The fact that an archive with holes in it is useless doesn't seem
relevant.  The first question is whether Google has the right to
publish the archive to begin with.
Personally, I'm extremely annoyed at Google's recent decision to edit
posts to obfuscate possible email headers.  I sure as hell didn't give
them the right to edit my material in their public archive, whether
considering using an X-no-archive header to protest this move, but I'm
reluctant to do so because I *like* Google's archive and the service
it provides (even though I don't see how it's consistent with
copyright law).  But I don't like the recent changes.
-- 
All that 'shock and awe' stuff we've just dumped onto the Asian part of 
this
earth - could we have fractured something? Perhaps the earth was just 
reacting
to something that man has done to injure it. The earth is organic, you know. 

It
posting
i'm not sure of the legal aspects, but having posted to a public
facility i don't suppose you can retain exlcusivity on your writings as
far as other people reading it goes.
as with the above, i think it's just there as a public archive, but
probably not stored publicly, the archive itself is probably secreted
Google (well, deja before them) are to be credited with putting it all
together like this.
give
I'm
most folks, at least IT oriented folks, didn't like the changes,
usenet facility because you can search the archive & 'log on' at any
webbrowser.  however, it's worrying to imagine a time in the future
when this great archive might be locked off by making it pay to view.
hope that never happens.
Linux platform
 NewsFleX homepage: http://panteltje.com/panteltje/newsflex/ and ftp 
download ftp://sunsite.unc.edu/pub/linux/system/news/readers/ 
listed above.
On a sunny day (3 Mar 2005 07:06:42 -0800) it happened gswork@mailcity.com
If they make it PPV then they will have to pay me royalties.
ALL text remains copyright.
Whether explicitely stated or not.
Out of context quoting, altered display, and changing of header lines is
explicitely forbidden, as is falsifying the origin / author / date/ 
destination /
and references lines.
I have:
Searched all groups    Results 1 - 10 of 13,400 for panteltje (0.33 seconds) 

By publishing this, Google (Deja news) agrees all postings made by
any of its devisions, companies, or representatives asks money to access 
these
access to this work.
Now what.
I can put my own archive on line too in a minute or 2.
Let's see what happens now (nothing).
lack 
Really?  I didn't know that.  That would be a facinating study.
/BAH
Subtract a hundred and four for e-mail.
Wrong.  That's the first exchange of information that occurs,
even before verbal exchange.  Gals check out eye color, manly
physique, ass configuration, hair color and curliness, height,
weight, and smell.  That's a lot of DNA info interpretation.
/BAH
Subtract a hundred and four for e-mail.
You might try to find out if there is a meaning attached to
positions within the field and if there are meanings attached
to each number within that position.  I can see how assigning
such a number would enable a librarian (e.g., at the Library
of Congress) to find the physical book given one simple piece
of information.
That is one of the peskiest problems.  How do you assign identifiers
that cannot be redefined by anybody else?  It just scratches
the surface of some security issues.
The first test would be watching what happens to the identifiers
when two companies merge.  Think about all the number problems
in banking when two very large banks merge.  People make big
bucks resolving these inevitable conflicts.
Another good exercise is to correct an error.
/BAH
Subtract a hundred and four for e-mail.
How did she get from
         there were nine planets before there were any people
to
         there was the number nine, before we had any people.?
Nine planets and the number nine are very different things.
with a
science.
there were
people.
Yes, that's right.
--
Eray
Yes,any 2 points of X can be joined by an injective C^infinity map
(arc) from from R (real numbers) into X (and with non zero tangent
vectors ,i.e. an immersion) .The proof is in section 16.26 of
J.Dieudonne ,Treatisie on Analysis Volume 3 (academic press ,1972) on
Doesn't ASP-completeness (shown in the following paper)
imply NP-completeness?
www-imai.is.s.u-tokyo.ac.jp/~yato/data2/MasterThesis.pdf
--r.e.s.
font,
  1  5 16 13  8 12  2 10 11  7  6 15 14  4  3  9
 11  7 12  6 15 13 14 16  4  9  2  3  5  8  1 10
  3  9 10  8  4 11  1  7 13  5 14 12  6 16 15  2
 15  4 14  2  6  9  3  5 16  8  1 10 12 13  7 11
 10 15  9  1  5  4 13  2  6 16  7 11  3 12 14  8
 12  6  2  3 16 14 15 11  8 13 10  4  7  1  9  5
  5 14 11  4  7 10  8  1  9 12  3  2 13 15  6 16
  7 13  8 16  9  3 12  6  5  1 15 14 11 10  2  4
  4  3 13 14 11  7 16  8 10  2  5  6  1  9 12 15
  2 11  7 15  1  6 10  9 12  3  4 16  8 14  5 13
 16  8  1 12 14  2  5 13  7 15 11  9  4  3 10  6
  6 10  5  9  3 15  4 12  1 14 13  8  2 11 16  7
  9  2  3 11 13 16  7 14 15  6  8  1 10  5  4 12
 13  1  4 10 12  5  6 15  3 11 16  7  9  2  8 14
  8 16  6  5  2  1  9  4 14 10 12 13 15  7 11  3
 14 12 15  7 10  8 11  3  2  4  9  5 16  6 13  1
of course, you are quite right.  blinded by the so called boundary 
conditions of what was a physical system.
cheers
moth 
The main issue for me in showing that surrogate factoring is practical
is in finding a perfect algorithm which will factor a target M for any
non-zero j coprime to M.
I think the answer is a simple one.
Starting as usual with
yx^2 + Ax - M^2 = 0
and
yz^2 + Az - j^2 = 0
where M is the target to be factored, j is some non-zero natural
usually chosen to be less than M, and
T = M^2 - j^2.
Then
y = (M^2 - Ax)/x^2 = (j^2 - Az)/z^2
and you can substitute out y to get
x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
and
z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
where the square roots relate Ax to the factorization of T and j, and
Az is related to the factorization of T and M.
The problem has been that if you find the integer Ax values that follow
from that second expression, you may find that all the solutions for Ax
that are not coprime to M, have M itself as a factor.
So it helps to split A and x up, which is already done for you, as the
expression is actually a relationship between z and x, but z and x are
asymmetrical with regard to factors in common with M.
For instance, if x has p_1^2 as a factor where p_1 is a factor of M,
then z must have p_1 as a factor, but not p_1^2.
You can kind of see that with
y = (M^2 - Ax)/x^2 = (j^2 - Az)/z^2
but it's also easy to rigorously prove it.
So the full algorithm, which splits up A and x requires that you solve
for Ax, and if it has M as a factor, you go ahead and finish out
z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
by solving
(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
denominator.
And then for at least one case, it must be true that the denominator
has a single prime factor of M left to divide out from x so that z has
the proper factor of M.
The proof that Ax must for one case involve A having prime factors of M
and x having the others is all that remains.
Going back to
yx^2 + Ax - M^2 = 0
it is trivial to replace Ax using
Ax = M^2 - yx^2
in
z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
to get
z = x(-Ax +/- sqrt((M^2 - yx^2 - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
so if Ax has M as a factor then yx^2 must as well.
But y = (M^2 - Ax)/x^2, so if x has a single prime factor of M, then A
must as well for y to have that factor.
So, can x have a single prime factor of M?
Assume it can and call that factor g, so x = g^2.
Then
yg^4 + Ag^2 - M^2 = 0, so y = (M^2 - Ag^2)/g^4
and now substituting for y in
yz^2 + Az - j^2 = 0, after solving for z, gives
z = g^4(-A +/- sqrt(A^2 + 4j^2(M^2 - Ag^2)/g^4))/2(M^2 - Ag^2)
and the real issue is whether or not z can be rational in this case so
concentrating on the square root, I have
sqrt(A^2 g^4 + 4j^2 M^2 - 4j^2 g^2 A)
where collecting and completing the square gives
sqrt(g^4 A^2 - 4j^2 g^2 A + 4j^2 + 4j^2 M^2 - 4j^4 )
which is
sqrt((g^2 A - 2j)^2 + 4j^2 T)
so any A that makes that rational will work.
Notice that A may be a fraction with g^2 in the denominator, so let's
look back at
yg^4 + Ag^2 - M^2 = 0
and notice that all that matters is that Ag^2 be an integer, and then
yg^4 must be an integer as well, meaning it is captured in the integer
cases for
z = g^4(-A +/- sqrt(A^2 + 4j^2(M^2 - Ag^2)/g^4))/2(M^2 - Ag^2)
so it must be true that by solving out the full expression you can for
at least one case split up the factor of M, for the case where Ax has M
as a factor, and factor M, for any non-zero j coprime to M.
So the full perfect algorithm can be easily derived now from
z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
as you solve for Ax using
Ax = +/-(f_1 - f_2) + 2j^2
where f_1 f_2 = Tj^2, and you iterate through all possible integer f_1
and f_2, and take the gcd of Ax with M.
If Ax has M as a factor, then you calculate that ratio of z to x.
The numerator is just
(+/-(f_1 - f_2) + 2j^2 +/- (f_1 + f_2))
while the denominator is
2M^2 - 2(+/-(f_1 - f_2) + 2j^2
and you divide out any factors in common between them, and then take
the gcd of the denominator with M, and for at least one case for any
non-zero j coprime to M, you will have a prime factor of M.
The basic algorithm is now perfect.  I think there are far faster more
elegant algorithms out there, but at this point, it's proof of concept
time.
James Harris
Mr. Harris,
after ten years of your moronic postings, I finally understand you!
In the Harrisonian psycho-logic object oriented mathematics (have you
blossomed from the Harrisonian cocoon into the worlds top number theorist
yet?), you define perfect as being not fast, not elegant, not
useful, not understood, not practical, not proven and not 
worth cow
dung!
SIMPLY BRILLIANT!
Why didn't you just state this explicitly ten years ago?
Listen to yourself MAN!!
I don't believe you bothered to define perfect in this context.
So much for the obvious definition of perfect. According to your
standard, perfect apparently means far slower and less elegant than 
other
methods.
--
There are two things you must never attempt to prove: the unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
They don't work.  You need to read between the lines.
I suggest you try it, and just to make it easy, I suggest you
demonstrate with the calulation of pi(10).
What you will have to do is not tell the method that 2 and 3 are prime.
Show how much you know.
I like asking for people to actually put up math with such a simple
test, especially when it's an obnoxious poster.
If I'm wrong, then I'm just wrong, and I'll stand corrected.
But you need to *show* the math this time, and not think you can just
get by with quoting from a book.
James Harris
That was one of the things that are beyond you. 
For example, with k = 5, P = 2*3*5*7*11 = 2310, Q = 1*2*4*6*10 = 480, we 
find Phi (x + P, 5) = Phi (x, 5) + Q; it follows that Phi (x, 5) = Q * 
floor (x / P) + Phi (x mod P, 5); the latter can be found using a 
lookup-table with 2310 entries as 0 <= x mod P < P = 2310. 
This finds Phi (x, 5) with a very simple calculation and a tiny table. 
Now write down your formula so that we can compare. Could have made your 
algorithm a bit faster, but you are just too stupid to listen to people 
who know more than you do and think things through better.
we
*
And that's how math people lie so easily, they throw up a lot of smoke
when you ask them to do something simple.
Most people get intimidated by all the math-ese, and just assume they
must be telling the truth.
Also notice he *still* talked of a lookup table, so it's still some
kind of sieve kind of thing.
Math people are used to getting away with this kind of behavior.
They just don't tell the straight truth if it doesn't suit them.
James Harris
Actually, the code described there which finds primes by using a sieve 
is not very fast; Legendre's method (which is more or less what Harris 
found) will run faster probably by a factor ten or more if it is 
implemented reasonably cleverly. 
Other methods run faster than in linear time; about 1000 times faster 
around N = 10^19.
Probably both.
Actually the key point is that the recursion in question is
not something that nobody else could possibly have thought
of, all it is is a way to make the thing execute incredibly
slowly.
Possibly true - if so there's a _reason_ for that. And the
reason is not that you're the first person to realize that
you can test whether k is prime by checking whether pi(k+1)
equals pi(k).
************************
David C. Ullrich
As a matter of fact, no, the equality cannot prove anything.
We have pi(7) = pi(8) but we also have p(8) = pi(9). To check
But concretly, one should use an other method. This one is
... I don't find the word :-)
mm
I have here an incredible recursive program that adds unsigned integers:
unsigned succ(unsigned a) {
  return a + 1;
}
unsigned pred(unsigned a) {
  unsigned prev, next;
  if(a == 0) {
    printf(0 does not have a predecessor!n);
    exit(1);
  }
  prev = 0;
  while((next = succ(prev)) != a) {
    prev = next;
  }
  return prev;
}
unsigned add(unsigned a, unsigned b) {
  while(b != 0) {
    a = succ(a);
    b = pred(b);
  }
  return a;
}
main() {
  printf(%dn, add(10000, 10000));
}
Can I claim some fame?
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
No need, it will come by itself. But I am not sure that JSH 
can be as inefficient when adding numbers. The way pred
makes use of succ is far too neat.
mm
Recursive? Where do you have a function calling itself?
Sure... why not?
Well, in one post you've gone from nothing changes to everything changes.
Got meds?
--
There are two things you must never attempt to prove: the unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
be
then
just
certain
and
I
and
but
accepted
to
with
it
when
of
all
loss
path
James, please visit this URL and make a phone call:
or
http://tinyurl.com/65cry
You think I'm just trying to make fun of you, but I'm not.  Your
condition is getting worse.  Please make a phone call.
Oh? You are *destined* to succeed?  Could it be...delusions of grandeur?
--
There are two things you must never attempt to prove: the unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
: And, aren't transcendentals reals that are not rationals or true 
: irrationals? 
Transcendentals are irrationals, however not all irrationals are
transcendental.  I have no idea what you mean by true irrationals.
The reals consist of the rationals and the irrationals.
: Seems like it might be more than the general idea, but good 
: enough! I wouldn;t mind the whole thing about transcendentals not lying 
: on the same line as rationals or irrationals, but I already have a lot 
: to learn as it is.....sigh!
The definitions for these terms are all freely available on the web.
        http://mathworld.wolfram.com/TranscendentalNumber.html
        http://mathworld.wolfram.com/IrrationalNumber.html
Lester is using his own private definitions that do not 
correspond to the standard usage and who knows if his
definitions are even consistent.
Stephen
They aren't. Lester speaks to the world in a private language. No wonder 
he is misunderstood.
Bob Kolker
If not, then you haven't yet shown where they are not.
As do mathematikers.  At least Lester has made an effort to 
understand your private language.
And no wonder mathematikers are misunderstood.
-- 
Mercifully free of the ravages of intelligence
        -- Time Bandits
: If not, then you haven't yet shown where they are not.
: As do mathematikers.  At least Lester has made an effort to 
: understand your private language.
: And no wonder mathematikers are misunderstood.
Albert, definitions for every mathematical term are
readily available on the web.  You can also just ask
people and they will give you the definitions.  Do
you honestly think that Lester's definitions are 
equally well defined?
Can you explain tautoglogical regression?  Can you
explain how it falsifies special relativity, general
relativity, quantum mechanics and at the same time
determines the height of Mount Everest?  If you honestly
believe Lester makes sense, can you explain in it some
way that would make sense to somebody else?
Stephen
Mathematical terminology is open and published in thousands of sources. 
It is used as the communication in the mathematical and scientific 
community among tens of thousand of people. Mathematical terminology is 
the very antithesis of private language. Your terminlogy is nowhere used 
in the literature. It is a conglomeration of neologisms used only by 
you. I have asked you repeatedly for references in the peer reviewed 
literature for examples of your terminology and you have not produced 
the references. And for good reason. They do not exist.
Bob Kolker
traveller,
all He
God
Um, actually, the point is that you WILL not do other than what you will 
do,
which is a rather trivial proposition.
The idea would be that the observer sees the outcome of your free 
choice,
and thus you would have to make a free choice by definition (if it was ever
free in the first place).
as
if
it is
must be
you
If the Traveler really sees the future, he cannot be wrong.
It works to preserve ethics -- meaning that we can blame people for their
choices, which without free will is not possible -- while insisting that
various sorts of psychology and behavioural science can still give us
practical information and generalizations.
finite
able
Um, remember that omniscience is absolute knowledge.  The definition of
knowledge certainly plays a role in that, right?
to
about
choice
be a
by
claim
It would be to accept that there is something that it is logically
impossible to know -- since a free choice would indeed, by your logic, be
logically impossible to know -- but note that there is no attempt at proof
or any assumption of free will EXCEPT for the sake of argument, to show 
that
IF free will existed, it would not necessarily clash with omniscience.
There is still no proof that free will exists in any of my comments.
Precisely.  But I never claimed that it did.
cause
God doesn't DETERMINE what's going to happen, but merely KNOWS it.  The
implicit argument you make here is that it is not possible to know the
result of a free choice WITHOUT it meaning that that outcome IS
predetermined.
Depends on what we mean by knowledge.
is
though
If God can know the logically impossible, then since it is only logically
impossible to know the outcome of a free choice then God could indeed know
the outcome of a free choice if omniscience grants the ability to know that
which it is logically impossible to know.  If omniscience only grants the
ability to know that which logically can be known, then God does not know
the outcome of our free choices -- if we have any -- but that does not 
limit
God since to claim that it did would be to insist that God know something
that omniscience cannot grant knowledge of.
Have you talked to God lately to find out what He knows or does not 
know? Are you a prophet of the Lord?
Bob Kolker
Couldn't be mine, since all I was doing was repeating and asking for
clarification of statements other people made ...
Which one would best fit relative number of elements?
What does that mean?
Bob Kolker
But that direct experience of thought somehow misses you completely ...
Just as a hot teapot bubbles and rumbles, a live brain thinks. It is all 
physical.
Bob Kolker
left
Take the set of integers {1, 2, 3} and the set of even integers {2, 4, 6,
8}.  There is a mapping function that I can apply to that set of integers
with is 2*x, just as in your example.  Yet I can clearly show that if I map
every element in the set of integers to the set of even integers that I've
posited here that there is an extra element in the second set, and so they
do not have the same number of elements.  But the mapping worked perfectly
fine, and it was just that the first set ended earlier that caused us 
to
determine the outcome.  Since we can never finish mapping the elements 
in
the infinite sets of integers and even integers, we can never know if such 
a
case could possibly be the case, precisely because there is no last
integer.  But then how can I be certain that my mapping would work?  It's
only because I can never finish that I can never -- and would never -- 
conclude that they didn't map one onto another, since if I stopped mapping
at any N IN the set, it would be clear that the mapping would not succeed.
: left
: Take the set of integers {1, 2, 3} and the set of even integers {2, 4, 6,
: 8}.  There is a mapping function that I can apply to that set of integers
: with is 2*x, just as in your example.  Yet I can clearly show that if I 
map
: every element in the set of integers to the set of even integers that 
I've
: posited here that there is an extra element in the second set, and so 
they
: do not have the same number of elements.  But the mapping worked 
perfectly
: fine, and it was just that the first set ended earlier that caused us 
to
: determine the outcome.  Since we can never finish mapping the elements 
in
: the infinite sets of integers and even integers, we can never know if such 

a
: case could possibly be the case, precisely because there is no last
: integer.  But then how can I be certain that my mapping would work?  It's
: only because I can never finish that I can never -- and would never -- 
: conclude that they didn't map one onto another, since if I stopped 
mapping
: at any N IN the set, it would be clear that the mapping would not 
succeed.
I can finish the mapping by describing it.  f(x)=2x is a mapping
between the integers and even integers.  In what way is it
not complete?  Which integer has not been mapped to an even integer
and which even integer has not been mapped to an integer?
Stephen
the
0.5
element.
but
You should probably stop trying to help yourself here, because you have
pretty much just argued that your mapping view of the number of elements on
infinite sets results in a contradiction with the definition of proper
subset.  Remember why you were insisting that I was simply relying on the
definition of subset, which was to argue that that argument didn't hold for
infinite sets -- as proven by Cantor.  But here you have just posited a
definition of proper subset that pretty clearly seems to insist that a
proper subset will have less elements than its superset.  Thus, you have a
contradiction if you want to maintain both definitions -- which it seems
that you do.
However, this is immaterial to MY claim.  MY claim was not that a proper
subset had to have less elements or that I was appealing to the definition
of proper subset.  Mine was based on the definitions of the sets 
themselves,
and had nothing to do with the definition of proper subset, since I agreed
that in one of your examples IF it was a proper subset THEN they would
indeed have the same number of elements by the definitions of the sets (the
octal-decimal string example), clearly showing that I did not claim that 
any
proper subset MUST have less elements than the superset.
represent
as
The string 11 as an octal string and the string 11 as a decimal 
string
do NOT represent the same thing.  Nor did I ever comment about them
magically changing size, but that the reason why the proper subset 
thing
COULD work is that the represented thing of the strings is inherently
different and so we can then say that each octal string represents,
ultimately, all integers just as all decimal strings represent all integers
not
Um, Stephen?  You are now insisting that proper subsets must have less
elements than the set, which you vigorously opposed earlier.
As for me, it isn't that I didn't understand what subsets are, but that I
felt no need to care about them to make my point.  The fact that a
mathematical definition agrees with me only bolsters the argument against
the mapping view, but does not affect my argument in any way.
: the
: 0.5
: element.
: but
: You should probably stop trying to help yourself here, because you have
: pretty much just argued that your mapping view of the number of elements 
on
: infinite sets results in a contradiction with the definition of proper
: subset.  
How does it possible contradict the definition of proper subset.
A is a proper subset of B is every element of A is in B and
there exists an element of B that is not in A.  There is nothing
in that definition that has anything to do with bijections,
or anything in that definition that says that a bijection cannot
exist between a set and a proper subset.
:  Remember why you were insisting that I was simply relying on the
: definition of subset, which was to argue that that argument didn't hold 
for
: infinite sets -- as proven by Cantor.  But here you have just posited a
: definition of proper subset that pretty clearly seems to insist that a
: proper subset will have less elements than its superset.  Thus, you have 
a
: contradiction if you want to maintain both definitions -- which it seems
: that you do.
No.  You have to define by what you  mean by less elements.
I already gave you the definition for same number of elements that
I am using, which is that there is a bijection between two sets.
By this definition, if there exists a bijection between A and B
than A cannot have less elements than B.
: However, this is immaterial to MY claim.  MY claim was not that a proper
: subset had to have less elements or that I was appealing to the 
definition
: of proper subset.  Mine was based on the definitions of the sets 
themselves,
: and had nothing to do with the definition of proper subset, since I 
agreed
: that in one of your examples IF it was a proper subset THEN they would
: indeed have the same number of elements by the definitions of the sets 
(the
: octal-decimal string example), clearly showing that I did not claim that 
any
: proper subset MUST have less elements than the superset.
Then why do you claim that [0,1] must have less elements than [0,2]?
You apparently have some reason that does not have to do with
the fact with that [0,1] is a proper subset of [0,2] but you
have yet to articulate it.
: represent
: as
: The string 11 as an octal string and the string 11 as a decimal 
string
: do NOT represent the same thing.  Nor did I ever comment about them
: magically changing size, but that the reason why the proper subset 
thing
: COULD work is that the represented thing of the strings is inherently
: different and so we can then say that each octal string represents,
: ultimately, all integers just as all decimal strings represent all 
integers
: not
: Um, Stephen?  You are now insisting that proper subsets must have less
: elements than the set, which you vigorously opposed earlier.
No.  I have never said that.  Show me where I ever said that?
You keep claiming that (0,1) has fewer elements than (0,2).
The only reason  you have provided is that (0,1) is a proper subset
of (0,2).  However you also claim that if A is a proper subset
of B than A does not necessarily have fewer elements than B.
So again, why do you claim that (0,1) has fewer elements than (0,2)?
: As for me, it isn't that I didn't understand what subsets are, but that I
: felt no need to care about them to make my point.  The fact that a
: mathematical definition agrees with me only bolsters the argument against
: the mapping view, but does not affect my argument in any way.
What mathematical definition?  There is no mathematical definition
that relates subsets and sizes of infinite sets.
Stephen
Just as an aside, are we sure that Kant is talking about the same sort of
space that you are when you talk about space being curved?  The reason I 
ask
is that in studying Kant it seems that the notion of flat or curved 
is
not something that he would associate with the concepts of space (he
basically talks about ordering things in space and providing diffentiation
in place to synthesize over).  I suspect that the curved space-time is what
Kant would consider an appearance of a thing-in-itself (but not space as 
the
concept).  Thus, as the appearance of a thing-in-itself, we can 
interact
with it, and find out things about it through science.  And so we discover
that it is a thing, and a thing out there that we can only have the
appearance of.
In short, in order to claim that space-time is curved you'd have to 
consider
it a substance.  But if it's a substance, it isn't the space that Kant 
talks
about, and so we might not have the refutation of space and Euclidean
geometry.
 Where he went
I think you make the same mistake here that Bob did.  Even if it is true
that Euclidean geometry is not an a priori synthetic, it does not mean that
there are NO a priori synthetics, even for Kant.  I doubt that he claimed
that either Euclidean geometry is a synthetic a priori or nothing is ...
I've never understood why space should be a substance, so I don't really 
understand what you (and Kant, presumably) intend. the space I imagine 
intuit is waht I can move through, it's a relationship between things. 
Thus it's not a thing-in-itself. Etc. Of course the problem here is the 
vagueness, the ambiguity, the slipperiness of thing (which is just as 
bad in Germn, Ding can be just about anything - see, there it goes 
again. Catch it quick!)
True, careless of me.
: I've never understood why space should be a substance, so I don't really 
: understand what you (and Kant, presumably) intend. the space I imagine 
: intuit is waht I can move through, it's a relationship between things. 
: Thus it's not a thing-in-itself. Etc. Of course the problem here is the 
: vagueness, the ambiguity, the slipperiness of thing (which is just as 
: bad in Germn, Ding can be just about anything - see, there it goes 
: again. Catch it quick!)
No, Allan is making the common error in thinking that in order
for space to be curved it must be a substance of some sort
that can be physically bent.  However that is not what space
being curved means.
Stephen
What space being 'curved' means is nothing more than mathematical 
abstractions regarding a hypothetical 'curved' space just 
coincidentally happen to match certain interpretations of 
observations.  Theories are falsifiable. The map is not the 
territory.  You and others here, who make the mental leap of 
faith from an as yet unfalsified theory to the belief that space 
really is curved, could avoid a lot of controversy by simply 
using the phrase 'as if' more often. i.e. It is as if space is 
curved.
-- 
Mercifully free of the ravages of intelligence
        -- Time Bandits
And not only that we know from experiment that a flat non-curved model 
cannot be right.
Bob Kolker
method
equation,
the
It depends on the relation between the methods.  I certainly wouldn't
suspect the method that we had used to prove the new methods, at least not
without really good reason.  And even in independent methods I'd return to
the definition of root to see which one best fits that definition,
particularly if it was a definition that was not simply invented, but was
derived from other notions.
same
saying
Um, the methods for determining the number of elements in a set ARE
algorithms.  You're welcome [grin].
I'm not sure if that's the case, but for the most part this agrees with my
view.  We know that infinite sets act differently, and so which
differently should we consider?  My answer is that showing that mapping
doesn't apply to infinite sets is more consistent with both general rules 
of
subsets and intuitions/definitions of more and should be accepted.  You can
disagree, but I doubt anyone can prove I'm wrong [grin].
element
Well, what I meant was that when we match we don't normally do any
mathematical translation on it, but instead select one from set A and one 
of
set B and repeat this until we've dealt with all of the elements in at 
least
one of the sets.  Thus saying that you can map by multiplying and 
dividing
by two does not seem to follow DIRECTLY from matching.
2?
Well, the point here, though, is that I certainly wouldn't say that the
number of elements in the set of tables is equal to the number of people.  
I
WOULD say that the number of chairs is equal to the number of people, but
the division mapping you did does not map the number of chairs onto the
number of people.
Please don't take this as an angry rant, but you have the strangest 
tendency
to make all sorts of great arguments against things that not only did I not
say, but that I agree with [grin].  I never said that set operations 
weren't
common or used in everyday life, but had some issues with this particular
one as applied to infinite sets (and how it is justified).  So it seems 
that
we are vigourously agreeing here ...
You're right, the fact that aruithmetic operations are actually a 
mapping functions was a discovery, and a very useful one.
[...]
Except that I see no reason not to apply mapping to infinite sets, nor 
to using mapping functions to show that the mapping is both 1 to 1 and 
complete. I still don't see why you object to that. Apart from the fact 
that such procedures lead to very odd conclusions. But so what? If 
reason didn't lead to odd conclusions, we wouldn't need it.
the
is
Probably true, but waiting seems to have the same problem.  If I have 
no
concept of before and after, waiting seems to have as little meaning.  
Which
was my entire point, here ...
He also kind of contradicts it in his refutation of idealism ....
I pretty much agree here (although am not certain that newborns don't have
one, even if they are just lacking permanence.  They might just be little
solipsists [grin]).
that
not
You cannot place octal and decimals numbers in the same set and call it a
set of any kind of number.  The operations are defined at least slightly
differently, and have their own meanings.  They are separate number 
systems.
That's more than just the fact that the strings are different.  
Certainly,
they both represent the same thing (an integer number) but what we mean
when we talk about a binary or octal or decimal number what we talk about 
is
a change in the BASE that we are using, that leads to different operations,
that leads to different things we can do.
You cannot say that a set of octal numbers would ever be a proper subset of
a set of decimal numbers, and that is sufficient to prove my case and
alleviate your concerns about my confusion.
I don't think so, and you have yet to provide any evidence that it is 
except
to insist that it should apply to sets that I never claimed it applied to 
by
elevating it to the definition, which is precisely what you have done with
cardinality -- elevated the mapping method to the status of the 
definition
of more.
I'm
this
Nope.  If the mathematicians in this thread aren't insisting that they have
the proper and universal definition of more, they were free to take me up 
on
my offer to allow them to define the mapping relation as per infinite sets
as an axiom of their mathematics and happily ignore everyone else who says
that even for mathematical purposes that axiom is not required to be
accepted.  In short, admit that you cannot justify that claim universally -- 

even in mathematics -- and I, at least, will go away.
Recall that I only entered this thread to oppose the mathematicians who 
were
insisting that it is obvious that more or equinumerous (as Bob puts it) is
expressed by mapping, even on infinite sets when Tony was commenting that
that move was at a minimum counterintuitive and even likely contradictory
with the definition of proper subset.
[...]
What operations are different? Are you suggesting that you can/can't add 
octals, etc? Or are you referring to the mechanics of carrying, etc? 
Can't be the former, as it would be sheer nonsense. Must be the latter.
But just because the number systems have different bases doesn't mean 
that the operations of addition, subtraction, multiplication, etc are 
different. If you have in mind oddities such as the digit sum of a 
number in the decimal system predicting divisibility by three, that's a 
side effect of the base used, but it doesn't chnage the operations you 
can perform. Ie, six plus six (I've purposely not numerals) is twelve, 
and twelve is divisible by three in any number system whatsoever.
[...]
Oh, I see, you're talking about numerals, not numbers. Well, so what? 
The fact that one set of symbols isn't included in another set of 
symbols doesn't prove anything about what the symbols represent.
        iD8DBQFCJ5g2vmGe70vHPUMRAiIpAKCby+8B4Dz2RZHcXl4+uHMUr0qJYQCeP16N
        sh8scHJVgylreGAuvIBbXfA=
        =7Gn2
That's silly.
What is different, are the operations on the string representations.
The operation on the numbers are the same, for they are the same
numbers.
You explicitly deny that you are talking about the string
representations.  But everything distinctive about octal, compared
with decimal, has to do with the string representations.  The number
fifteen is the number fifteen.  Whether I write it as 15 in
decimal, or as 17 in octal, it is still the same number.  There are
not distinct octal numbers and decimal numbers.
The BASE has to do with forming string representations.
The operations on the *numbers* are the same.  We carry out
operations by means of manipulation of their string representations.
The mechanics of how we manipulate the string representations depends
on whether we are using octal representations or decimal
representations.  But, either way, these are a proxy for the intended
operation on the numbers.  And the operations on the numbers are the
same.
: That's silly.
: What is different, are the operations on the string representations.
: The operation on the numbers are the same, for they are the same
: numbers.
: You explicitly deny that you are talking about the string
: representations.  But everything distinctive about octal, compared
: with decimal, has to do with the string representations.  The number
: fifteen is the number fifteen.  Whether I write it as 15 in
: decimal, or as 17 in octal, it is still the same number.  There are
: not distinct octal numbers and decimal numbers.
Don't forget XV. :)  
Stephen