mm-430
===
> http://mathforprofit.blogspot.com/Oooh. Well, I'm going to
have to attack this. Sorry James, butthis has the same
problems as your other proofs.1. Let P(x) = 5^3*7^6*x^3 -
3*5^3*7^4*x^2 - 2^3*3^2*5*7^2*x - 2*7^2*11 .2. P(x) =
7^2*(5^3*7^4*x^3 - 3*5^3*7^2*x^2 - 2^3*3^2*5*x - 2*11). P(x) =
(5 * a_1(x) + 7) * (5 * a_2(x) + 7) * (5 * a_3(x) + 7)The
astute reader may notice that 7^3 = 343 cannot be the lastterm
so at least one of the a_i(x) must contain a constant addendof
some sort. I will ignore this bizarre rewrite and instead
notethat P(x) = (5 * b_1(x) + 7) * (5 * b_2(x) + 7) * (5 *
b_3(x) - 22)probably works better, for various reasons. This
is the formyou eventually get to in step 4, by different
methods.We can now attempt to work out the forms for the b_i.
We alreadyknow the following:[1] The product b_1(x) * b_2(x) *
b_3(x) must contain an x^3 term. In fact, this x^3 term must be
7^6 * x^3.[2] If we assume P(x) = 7^6 * 5^3 * (x - x_1) * (x -
x_2) * (x - x_3) for some algebraic numbers x_1, x_2, and x_3,
then we can conclude that, for some permutation of x_i, b_1(x)
= -7*x/x_1 b_2(x) = -7*x/x_2 b_3(x) = 22*x/x_3 is a valid
solution. However, other solutions are possible, such as the
rather trivial b_1(x) = P(x)/5 - 7/5 b_2(x) = -6/5 b_3(x) =
23/5 or something like b_1(x) = -14*x/x_1 - 7/10 b_2(x) =
-7*x/x_1 b_3(x) = -11*x/x_1 + 11/5 In light of this
observation I cannot draw any other conclusions at this time
without additional constraints.[3] If we assume, as you did in
an earlier proof attempt, that P(x) = (5 * c_1 * x + 7) * (5 *
c_2 *x + 7) * (5 * c_3 * x - 22) where the c_i do not depend
on x, then we can indeed state that, for some permutation of
the x_i, c_1 = -7/x_1 c_2 = -7/x_2 c_3 = 22/x_3 However, are
any of these algebraic integers? At this point I can state
that, if the x_i satisfy P(x)/49 = (5^3*7^4*x^3 -
3*5^3*7^2*x^2 - 2^3*3^2*5*x - 2*11) = 0 then it must be the
case that the values y_i = 1/x_i satisfy P(1/y)*y^3/49 =
2*11*y^3 + 2^3*3^2*5*y^2 + 3*5^3*7^2*y - 5^3*7^4 If we further
assume y = z/7, then z = 7 * y = 7/x, and these satisfy the
equation 2*11*z^3 + 2^3*3^2*5*7*z^2 + 3*5^3*7^4*z - 5^3*7^7 =
0 Substitute w = -z and we simply flip signs: 2*11*w^3 -
2^3*3^2*5*7*w^2 + 3*5^3*7^4*w + 5^3*7^7 = 0 This polynomial is
irreducible. If one looks at v = 22/x, then x = 22/v and we get
-v^3 * P(22/v)/49 = 2*11*v^3 + 2^4*3^2*5*11*v^2 +
2^2*3*5*7^2*11^2 + 2^3*5^3*7^4*11^3 We note the common factor
2*11 and divide again, getting: -v^3 * P(22/v)/(2*7^2*11) =
v^3 + 2^3*3^2*5*v^2 + 2*3*5*7^2*11 + 2^2*5^3*7^4*11^2so yes,
c_3 is an algebraic integer -- but that's the only one that
can be.[4] This is not an attack on your person, but on your
proof. However, if you continue to call those in the
mathematical establishment idiots, why are you so surprised at
the shipments of high-explosive vitriol coming on your
doorstep? (This is assuming there *is* a Mathematical
Establishment, admittedly; presumably, there is a
well-established procedure for peer review of mathematical
results, which is about as close to an establishment as I
personally can see at this time -- since I've not attempted to
publish peer-reviewed papers in scientific journals I'm
unlikely to know the details of this process.) Concentrate on
the math. :-)
===
Subject: Logarithm of Stirling numbers of
the second kindI have a problem in which I need to calculate
the logarithm of Stirlingnumbers of the second kind for large
n (in the range 1 to thousands andpreferably millions) and m
in the same range. Due to the large range for nand m I prefer
not to work with the Stirling numbers of the second
kinddirectly, as my working precision is not that great.Are
there any methods that calculate the logarithm of Stirling
numbers ofthe second kind, or good approximations? An
approximation with errors (ofthe natural logarithm of the
Stirling number of the second kind) up to +-10would still be
useful for me, although the more accurate the better
ofcourse.
===
Subject: An urgent question for William Elliot
about a reference.connectivity of a complete graph equals
n-1.I need to write a reference for that proof. Was it yours
so I writeyour name as a reference or there was another
reference like a bookfor example?
===
Subject: [Set Theory] A
set has a disjoint copy of itself ?I am looking for a proof
that any set has a disjoint copy of itself,i.e. that given a
set A, there exists a set B with the samecardinality as A, and
such that A/B=0. I am hoping this is true withZF or
ZFC.Noel.
===
Subject: Re: [Set Theory] A set has a disjoint
copy of itself ?> I am looking for a proof that any set has a
disjoint copy of itself,> i.e. that given a set A, there
exists a set B with the same> cardinality as A, and such that
A/B=0. I am hoping this is true with> ZF or ZFC.If you don't
mind using the Axiom of Foundation (Regularity), youcould just
take B = {{a,A}: a in A}.
===
Subject: Re: [Set Theory] A set has
a disjoint copy of itself ? Adjunct Assistant Professor at the
University of Montana.>I am looking for a proof that any set
has a disjoint copy of itself,>i.e. that given a set A, there
exists a set B with the same>cardinality as A, and such that
A/B=0. I am hoping this is true with>ZF or ZFC.Just paint it a
color. Given any set A, there exists a set B suchthat B is not
an element of A. Take
Ax{B}.
===
=====================================================
===
===============Subject: Re: [Set Theory] A set has a
disjoint copy of itself ? Adjunct Assistant Professor at the
University of Montana.>I am looking for a proof that any set
has a disjoint copy of itself,>i.e. that given a set A, there
exists a set B with the same>cardinality as A, and such that
A/B=0. I am hoping this is true with>ZF or ZFC.>Just paint it
a color. Given any set A, there exists a set B such>that B is
not an element of A. Take Ax{B}.Yuck. This doesn't quite work,
as has been pointed out elsewhere. Butyes, there is a way to do
it in ZF, by choosing an appropriate set Band taking A x {B}. B
can be constructed explicitly, so there is noneed to invoke
Choice.
===
====================================================
===
================Subject: Re: [Set Theory] A set has a
disjoint copy of itself ?>I am looking for a proof that any
set has a disjoint copy of itself,>i.e. that given a set A,
there exists a set B with the same>cardinality as A, and such
that A/B=0. I am hoping this is true with>ZF or ZFC.>Noel.> B
= {0} x A-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
===
Subject: Re: [Set Theory] A set
has a disjoint copy of itself ?> >I am looking for a proof
that any set has a disjoint copy of itself,>i.e. that given a
set A, there exists a set B with the same>cardinality as A,
and such that A/B=0. I am hoping this is true with>ZF or
ZFC.>Noel.> > B = {0} x AThat is B = {(0,a), a in A}.What
ifA = {1, (0,1)} ?-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.hNeedless to say, I had the last
laugh. Partridge, _Bouncing Back_ (14 times)
===
Subject: Re:
[Set Theory] A set has a disjoint copy of itself ?> > >I am
looking for a proof that any set has a disjoint copy of
itself,>i.e. that given a set A, there exists a set B with
the same>cardinality as A, and such that A/B=0. I am hoping
this is true with>ZF or ZFC.>Noel.> > >B = {0} x AThat is B = {(0,a), a in A}.>What if>A = {1, (0,1)} ?> Oops.
You're right. Must be more careful. I trust Herman's
response.-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
===
Subject: Re: [Set Theory] A set
has a disjoint copy of itself ?> > > I am looking for
a proof that any set has a disjoint copy of itself,> i.e.
that given a set A, there exists a set B with the same>
cardinality as A, and such that A/B=0. I am hoping this is
true with> ZF or ZFC.> Noel.> > B = {0} x
A> > That is B = {(0,a), a in A}.> What if> A = {1,
(0,1)} ?> > Oops. You're right. Must be more careful. I trust
Herman's response. Or let y be the least ordinal such that for
all a in A, (y,a) is not in A. Then set B = {y} x A.If you
include the axiom of foundation in ZF, can't we conclude that
{A} x A is disjoint from A, or must we be a bit more careful?
How about {a U {A} : a in A}?-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
===
Subject: Re: [Set Theory] A set
has a disjoint copy of itself ?>I am looking for a proof that
any set has a disjoint copy of itself,>i.e. that given a set
A, there exists a set B with the same>cardinality as A, and
such that A/B=0. I am hoping this is true with>ZF or ZFC.Let
Q_0 = A, Q_{n=1} = U(Q_n), the union of the elements ofQ_n,
and let R be the union of all the Q's. Then if c is any set
not in R, {<c,x>: x in A} will do.-- This address is for
information only. I do not claim that these viewsare those of
the Statistics Department or of Purdue University.Herman
Rubin, Department of Statistics, Purdue
Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX:
(765)494-0558
===
Subject: Re: [Set Theory] A set has a disjoint
copy of itself ? <borab1$33js@odds.stat.purdue.eduI am looking
for a proof that any set has a disjoint copy of itself,>i.e.
that given a set A, there exists a set B with the
same>cardinality as A, and such that A/B=0. I am hoping this
is true with>ZF or ZFC.> Let Q_0 = A, Q_{n=1} = U(Q_n), the
union of the elements of> Q_n, and let R be the union of all
the Q's. Then if c is> any set not in R, {<c,x>: x in A} will
do.Here is a slightly simpler version of your construction,
which doesnot use the Axiom of Replacement.Let U be the set of
all elements of elements of A (your Q_1). Let c beany set not
in U, e.g., let c = {x in U: x is not an element of x).Let B =
{{a,c}: a in A}. Then B is disjoint from A, and you get
abijection f:A --> B by defining f(a) = {a,c}.
===
Subject: Re:
[Set Theory] A set has a disjoint copy of itself ?> I am
looking for a proof that any set has a disjoint copy of
itself,> i.e. that given a set A, there exists a set B with
the same> cardinality as A, and such that A/B=0. I am hoping
this is true with> ZF or ZFC.Sure, if there are no ordinal
numbers in A, then take the the the set{0,1,2,...} that has
the same cardinality. If there is an ordinal number.Do the
same thing, but start counting from the successor of A's
highestordinal.> Noel.
===
Subject: Re: [Set Theory] A set has a
disjoint copy of itself ?> Do the same thing, but start
counting from the successor of A's highest> ordinal.Which is a
problem, if A contains no highest cardinal.... On the
otherhand, if A really is a set, then the ordinals in A are
bounded, so youcan just start off at any upper bound...I'm not
sure if this is better, but here's (IMHO easier) another wayto
see it: If A has cardinality k, take the next cardinal
number,let's call it k+1 (which is in particular a set). Then
k+1A hascardinality k+1, so take any subset of k+1A of
cardinality k. That'sdisjoint to A and has the same
cardinality -> voila :)CheersPhilipp
===
Subject: Re: Riemann's
Zeta Function by H. M. EdwardsIt seems to me that proving the
functional equation for the Riemann Zeta> function is easier
without using complex analysis at all. The proof that> I
referenced earlier basically does it as follows: Let Psi(t) =
sum from n=1 to infinity of exp(-pi n^2 t).> Let R(s) =
Integral from t=0 to infinity of Psi(t) t^{s/2} dt/t> Let Q(s)
= 1/s + Integral from t=0 to 1 of Psi(t) t^{s/2} dt/tThen you
can prove (using Fourier transforms) that R(s) = Q(s) +
Q(1-s)It immediately follows that R(s) = R(1-s). On the other
hand, integrating> the power series for Psi(t) term by term
gives R(s) = sum from n=1 to infinity of pi^{-s/2} n^{-s}
Gamma(s/2)> = pi^{-s/2) Zeta(s) Gamma(s/2)So from R(s) =
R(1-s), you get pi^{-s/2) Zeta(s) Gamma(s/2) = pi^{-(1-s)/2)
Zeta(1-s) Gamma((1-s)/2)Of course, now that I think about it,
this proof is a little> suspicious, because it uses a
definition of the Zeta function> that is only valid for s > 1
(the power series representation),> while the result
necessarily involves Zeta(s) for s < 1.This is Riemann's
second proof. Nothing suspicious about it!> The point is, that
Q(s) is an entire function of s and so> the formula R(s) =
pi^{-s/2) zeta(s) Gamma(s/2)> defines zeta as a meromorphic
function on C.As a matter of fact, I am studying the second
proof. It's on page 15.This one I can follow much easier. The
only line I don't follow is thefollowing(where Theta(x)=
Sum_1^inf exp(-n^2pi x))-Int_inf^1 Theta(1/x)x^(-s/2)dx/x is
made to equalInt_1^inf [x^(1/2)Theta(x)+x^(1/2)/2-1/2]x^(-s/2)
dx/xI get the change of signs, but I don't understand the
expansion in thebrackets[].
===
Subject: Re: Riemann's Zeta
Function by H. M. Edwards> It seems to me that proving the
functional equation for the Riemann Zeta> function is easier
without using complex analysis at all. The proof that> I
referenced earlier basically does it as follows:> Let Psi(t) =
sum from n=1 to infinity of exp(-pi n^2 t).> Let R(s) =
Integral from t=0 to infinity of Psi(t) t^{s/2} dt/t> Let Q(s)
= 1/s + Integral from t=0 to 1 of Psi(t) t^{s/2} dt/t> Then you
can prove (using Fourier transforms) that> R(s) = Q(s) +
Q(1-s)> It immediately follows that R(s) = R(1-s). On the
other hand, integrating> the power series for Psi(t) term by
term gives> R(s) = sum from n=1 to infinity of pi^{-s/2}
n^{-s} Gamma(s/2)> = pi^{-s/2) Zeta(s) Gamma(s/2)> So from
R(s) = R(1-s), you get> pi^{-s/2) Zeta(s) Gamma(s/2) =
pi^{-(1-s)/2) Zeta(1-s) Gamma((1-s)/2)> Of course, now that I
think about it, this proof is a little> suspicious, because it
uses a definition of the Zeta function> that is only valid for
s > 1 (the power series representation),> while the result
necessarily involves Zeta(s) for s < 1.This is Riemann's
second proof. Nothing suspicious about it!> The point is, that
Q(s) is an entire function of s and so> the formula R(s) =
pi^{-s/2) zeta(s) Gamma(s/2)> defines zeta as a meromorphic
function on C.As a matter of fact, I am studying the second
proof. It's on page 15.> This one I can follow much easier.
The only line I don't follow is the> following(where Theta(x)=
Sum_1^inf exp((-n^2)pi x))-Int_inf^1 (Theta(1/x))x^(-s/2)dx/x
is made to equal> Int_1^inf
[x^(1/2)(Theta(x))+(x^(1/2)/2)-(1/2)]x^(-s/2) dx/xI get the
change of signs, but I don't understand the expansion in the>
brackets. > [ ].(I fixed the Ughs)
===
Subject: Re: Riemann's
Zeta Function by H. M. EdwardsHjelmstad says...>(where
Theta(x)= Sum_1^inf exp(-n^2pi x))>-Int_inf^1
Theta(1/x)x^(-s/2)dx/x is made to equal>Int_1^inf
[x^(1/2)Theta(x)+x^(1/2)/2-1/2]x^(-s/2) dx/x>I get the change
of signs, but I don't understand the expansion in
the>brackets>[].This uses the fact 1. Theta(1/x) = [x^{1/2}
Theta(x) + x^{1/2} - 1/2]The proof of this fact is via Fourier
transforms. In general,if f(t) is a suitably well-behaved
function, and we definethe transform F(k) as F(k) = integral
from t=-infinity to +infinity of f(t) exp(-i 2 pi k t) dt then
2. sum from n = -infinity to +infinity f(n) = sum from n =
-infinity to +infinity F(n)So, let f_x(t) = exp(-pi t^2 x)
(treating x as a parameter). Then 3. integral from t=-infinity
to +infinity of exp(-pi t^2 x) exp(-i 2 pi k t) dt = x^{-1/2}
exp(-pi k^2/x)So F_x(k) = x^{-1/2} exp(-pi k^2/x).
Therefore,F_x(n) = x^{-1/2} f_{1/x}(n). Therefore, 4. sum from
n = -infinity to +infinity f_x(n) = x^{-1/2} sum from n =
-infinity to +infinity f_{1/x}(n)Since f_x(n) = f_x(-n), and
f_x(0) = 1, we can write this as 5. 1 + 2 * sum from n=1 to
infinity of f_x(n) = x^{-1/2} [1 + 2 * sum from n=1 to
infinity of f_{1/x}(n)]In terms of Theta, this becomes: 6. 1 +
2 Theta(x) = x^{-1/2} [1 + 2 Theta(1/x) ]which when rearranged
produces equation 1.The proof of equation 2 involves the use
of both Fourier series andFourier integrals.--
===
Subject: Re:
A 'basic' topology question about interiors> Of interest
perhaps are these formulas:> When U subset A subspace S> cl_A
(U) = A / cl_S (U)> int_A (U) = A / int_S (SA / U)Thank you!
These are most helpful formulae.You're welcome. Now prove
them. Hint:> Prove the first and for the second use.> int_A(U)
= A - cl_A (A-U)It appears that these work even if U is not
strictly contained in A. Is thatcorrect?Ben Scott
===
Subject:
Roots of a simple third-degree equationI know there must be a
simple way to calculate the roots of athird-degree equation
where the term in square is missing, like thisone:-r^3 + 3r +
2 = 0How does one do it?Fran.8dois
===
Subject: Re: Roots of a
simple third-degree equationAs several posters have pointed
out, if you are just looking forinteger (or even rational)
roots, it is not too hard.In general, though, it's no easier
to solve a cubic with no quadraticterm than it is to solve the
general case. The reason is as follows:ax^3 + bx^2 + cx + d =
0Substitute x = y - (b/3a). If you multiply it all out, you'll
findthat the new polynomial in y has no quadratic term. So if
you can finda root y=r of this new equation, then x = r-(b/3a)
is a root of theoriginal equation.The method for solvingAy^3 +
By + C = 0was discovered (invented?) in Italy a very long time
ago. It's usuallycalled Caro's formula, although it appears
that Caro actuallystole it from someone else. You can find the
formula (and it'sderivation)
athttp://mathworld.wolfram.com/CubicEquation.hJoeS--------I
know there must be a simple way to calculate the roots of a>
third-degree equation where the term in square is missing,
like this> one:-r^3 + 3r + 2 = 0How does one do it?>
Fran.8dois
===
Subject: Re: Roots of a simple third-degree
equation> I know there must be a simple way to calculate the
roots of a> third-degree equation where the term in square is
missing, like this> one:> -r^3 + 3r + 2 = 0> How does one do
it?> Fran.8doisIf you don't have a square term, jump to the
standard form:
===
===================================x^3 + a
x^2 + b x + c = 0Substitute x = z - a/3giving with p = -a^2 /
3 + b q = 2 a^3 / 27 - a b / 3 + cthe Standard form: z^3 + p z
+ q = 0Calculate discriminant: D = (q/2)^2 + (p/3)^31) if D < 0
calculate t such that cos(t) = q/2 / sqrt( -(p/3)^3 ) calculate
z1 = -2 sqrt( -p/3 ) cos( t/3 ) z2 = -2 sqrt( -p/3 ) cos( 2/3
pi + t/3 ) z3 = -2 sqrt( -p/3 ) cos( 2/3 pi - t/3 )2) if D = 0
(special case of next case) calculate z1 = -2 (q/2)^(1/3) z2 =
z3 = (q/2)^(1/3)3) if D >= 0 calculate u = ( -q/2 + sqrt(D)
)^(1/3) v = ( -q/2 - sqrt(D) )^(1/3) calculate z1 = u + v z2 =
- (u+v)/2 + (u-v)/2 sqrt(3) i z2 = - (u+v)/2 - (u-v)/2 sqrt(3)
iFinally substitute: x1 = z1 - a/3 x2 = z1 - a/3 x3 = z1 -
a/3
===
===================================for your example,
-r^3 + 3 r + 2 = 0directly go to the standard form: z^3 - 3 z
- 2 = 0 so p = -3 and q = -2discriminant D = 1 - 1 = 0D = 0 z1
= -2 (-1)^(1/3) = 2 z2 = z3 = (-1)^(1/3) = -1Two solutions: -1
and 2hthDirk Vdm
===
Subject: Re: Roots of a simple third-degree
equationI know there must be a simple way to calculate the
roots of a> third-degree equation where the term in square is
missing, like this> one:-r^3 + 3r + 2 = 0How does one do it?>
Fran.8doisIf one or more of the roots are rational numbers, as
in the above example, there are a variety of simple methods, as
outlined in other responses to your question.If none of the
roots are rational, things get more difficult. The following
will work in all cases, but is only needed for the tough
ones:First, divide through by the coefficient of the cubic
term to get the standard form x^3 + a*x + b = 0.Let x1, x2 and
x3 represent the 3 roots, and it may be shown that x1 + x2 + x3
= 0, x1*x2 + x2*x3 + x3*x1 = a, and x1*x2*x3 = -bIf one assumes
that x1 = u + v, x2 = u*w + v/w, x3 = u/w + v*w,where w is
either complex root of x^3 = 1, so w^2 + w + 1 = 0,then 3*u*v
= =a and u^3 + v^3 = -b.Let r be any square root of (a/3)^3 +
(b/2)^2, possibly complex,then take u as any cube root of
(-b/2 + r) and v = -a/(3*u), and you have your solutions.For
your example of -x^3 + 3*x + 2 = 0, you have a = -3 and b =
-2, r = 0, u = v = 1, x1 = 1 + 1 = 2 , x2 = x3 = w + 1/w =
-1
===
Subject: Re: Roots of a simple third-degree equationU
find by guess the root r = -1Them -r^3 + 3r + 2 could be
divided for r + 1 to guive-r^2 + r + 2 with roots 2 and -1then
u have a double root -1 and a single root 2-r^3 + 3r + 2 = -
(r-2) (r+1)^2> I know there must be a simple way to calculate
the roots of a> third-degree equation where the term in square
is missing, like this> one:> -r^3 + 3r + 2 = 0> How does one do
it?> Fran.8dois
===
Subject: Re: Roots of a simple third-degree
equationI know there must be a simple way to calculate the
roots of a> third-degree equation where the term in square is
missing, like this> one:-r^3 + 3r + 2 = 0How does one do it?>
Well, there are formulas for the solutions of polynomial
equations of degree <= 5.In this particular case, 2 is easily
seen to be a root. So you can divide the polynomium by (x-2)
and solve the resulting second-degree
equation./David
===
Subject: Re: Roots of a simple third-degree
equationDavid Rasmussen <david.rasmussen@gmx.net> escribi.97
en el mensajeI know there must be a simple way to calculate
the roots of a> third-degree equation where the term in square
is missing, like this> one:-r^3 + 3r + 2 = 0How does one do
it?Well, there are formulas for the solutions of polynomial
equations of> degree <= 5.David meant 'of degree < 5'> In this
particular case, 2 is easily seen to be a root. So you can>
divide the polynomium by (x-2) and solve the resulting
second-degree> equation.To the O.P.:The rational roots of a
polinomial equation with integer cofficients,expressed as a
reduced fraction, have a numerator that divides theindependent
term, and a denominator that divides the leading
coefficient.Then, if the leading coefficient is +/- 1, all the
rational roots areintegers. So, your equation only can has 1,
-1, 2 or -2 as rational roots.If the coefficients of the
equation are rationals no integers, you allwayscan multiply
the equation by the minimum common multiple of
thedenominators, in order to get an equivalent equation with
integercoefficients.--
===
Subject: Sum [n in Set]
(1/n)X-No-Confirm: yesI recall seeing a theorem saying Sum [n
in Set] (1/n) converges iff (some criterion on Set)where Set
is a subset of the positive integers. I seem to recall
itbrought in measure theory. Does anyone know a reference? Or
at least thestatement? (I'm actually only interested in the
statement, not theproof.)Tia,msh210@math.wustl.edu Of a reply,
then, if you have been cheated,http://math.wustl.edu/~msh210/
Likely your mail's by mistake been deleted.
===
Subject: Re: Sum
[n in Set] (1/n)X-No-Confirm: yes>I recall seeing a theorem
saying> Sum [n in Set] (1/n) converges iff (some criterion on
Set)>where Set is a subset of the positive integers. I seem to
recall it>brought in measure theory.> I believe the set of
polynomials { t^n, n in Set } spans a dense> subset of the
vector space of continuous functions on [0,1] iff> Sum [n in
Set] (1/n) diverges. Could that (or something like that)> be
what you meant?Yes!> There is also a conjecture of Erdos which
I think states that> the sum diverges iff Set contains
arbitrarily long arithmetic> progressions.Ah, that I'd never
heard.In the future, you can post or mail; doing both is
unnecessary.msh210@math.wustl.edu Of a reply, then, if you
have been cheated,http://math.wustl.edu/~msh210/ Likely your
mail's by mistake been deleted.
===
Subject: Re: Sum [n in Set]
(1/n)>I recall seeing a theorem saying> Sum [n in Set] (1/n)
converges iff (some criterion on Set)> There is also a
conjecture of Erdos which I think states that> the sum
diverges iff Set contains arbitrarily long arithmetic>
progressions.That should be => , not iff . Obviously we could
have arbitrarilylong arithmetic progressions and still have
convergence, e.g. ifSet = { 10^k + m , 0 < m < k }. I have a
notes saying Erdos offered$3000 for a proof of => . You can
still collect.>In the future, you can post or mail; doing both
is unnecessary.Maybe, but one of these days I'll learn that
what IS necessary is toremember to flag my email so that it
doesn't look like it was (also) anewsgroup post ...
===
Subject:
Re: Sum [n in Set] (1/n)Content-transfer-encoding: 8bit> I
recall seeing a theorem saying> Sum [n in Set] (1/n) converges
iff (some criterion on Set)> where Set is a subset of the
positive integers. I seem to recall it> brought in measure
theory. Does anyone know a reference? Or at least the>
statement? (I'm actually only interested in the statement, not
the> proof.)Perhaps that the span of {x^s : s in S} is dense in
C[0,1] iff sum1/s is infinite? I forget whose theorem it is.
The proof is inRudin's Real and Complex Analysis. (And I may
have some details wrong. Perhaps it's the ALGEBRA generated by
these which is dense iff...)--Ron Bruck
===
Subject: Re: Sum [n
in Set] (1/n)>I recall seeing a theorem saying> Sum [n in Set]
(1/n) converges iff (some criterion on Set)>where Set is a
subset of the positive integers. I seem to recall it>brought
in measure theory. Does anyone know a reference? Or at least
the>statement? (I'm actually only interested in the statement,
not the>proof.)Well, there are infinitely many such statements,
hard to knowwhich one you're referring to. Hmm. Say S is the
set, andd(n) is card(S intersect {1,2,...n}) / n. Then I think
it'sclear that the sum_{n in S} 1/n converges if and only
ifsum(d(2^n)) converges.>Tia,>msh210@math.wustl.edu Of a
reply, then, if you have been
cheated,>http://math.wustl.edu/~msh210/ Likely your mail's by
mistake been deleted.
===
Subject: partial deriviatives!hi!Kinda
basic question...if you have a function F(q(t),t)and you have
the derivatived/dt( d/dq[F(q(t),t) )then it should
bed^2/dq^2*dq/dt[F(q(t),t)] + d^2/dtdq[F(q(t),t)]right?Or am i
mistaken...??/M
===
Subject: Re: partial deriviatives!> Kinda
basic question...if you have a function F(q(t),t)> and you
have the derivatived/dt( d/dq[F(q(t),t) )then it should
bed^2/dq^2*dq/dt[F(q(t),t)] + d^2/dtdq[F(q(t),t)]right?> Or am
i mistaken...??I think mistaken. The notation is a mess. Start
with F(q,r) as a function of two variables. Then G(q,r) =
[dF/dq](q,r) is also function of two variables. You want
dG(q(t),t)/dt, which equals [dG/dq](q(t),t)*dq/dt +
[dG/dr](q(t),t)*1.Recalling what G is, the answer is
[d^2F/dq^2](q(t),t)*dq/dt + [d^2F/drdq](q(t),t).
===
Subject:
Re: partial deriviatives!> Kinda basic question...> if you
have a function F(q(t),t)> and you have the derivative> d/dt(
d/dq[F(q(t),t) )> then it should be> d^2/dq^2*dq/dt[F(q(t),t)]
+ d^2/dtdq[F(q(t),t)]> right?> Or am i mistaken...??>
/MAssuming that you are asking for the total time derivative
(not the partialtime derivative), then the answer
isfrac{partial^2 F}{partial q^2} frac{dq}{dt} +
frac{partial^2F}{partial t partial q}.Kostas.
===
Subject:
Vacuum PropellersJack,I saw this paper while surfing the web
and thought it looked similar to your
work:http://arxiv.org/abs/gr-qc/9612022Yes. There are some
similarities but more important differences. Moese's paper
complements mine as is shown in detail in my paper inProgress
in Quantum Physics Research (Nova Scientific Publishers - a
copy is in http://qedcorp.com/APS/ )1. Moese is dealing with a
rotating real superconducting ground state of a huge number of
real on mass shell electron pair bound states all in dealing
with a virtual superconducting vacuum state of a huge number
of off-mass-shell bound electron-positrons near the -mc^2
Fermi edge inside the 2mc^2 Dirac energy gap. These two
macro-quantum condensates can couple together in a kind of
Josephson virtual-real junction. I have the equation in paper
I cited.2. I derive both Einstein's gravity and the unified
exotic vacuum dark energy/matter field from the vacuum
condensate as an emergent process in the sense of Sakharov and
Anderson. Moese's paper is not fundamental like that. He
assumes Einstein's gravity to start with.PS I corrected an
important typo in a slide in
http://qedcorp.com/APS/StatGate.mov about dark matter with
negative zero point energy density and POSITIVE pressure. The
typo said negative pressure.My Vigier IV paper from Paris in
final form is at http://qedcorp.com/APS/EmergentGravity.doc or
same with .pdf not .docPossible quantum gravity effects in a
charged Bose condensate under variable e.m. fieldAuthors: G.
Moese, J. SchnurerComments: 26 pages, PDF, 7 figures. Final
journal versionReport-no: UTF-391/96Journal-ref: Phys.Essays
14 (2001) 93-105In the weak field approximation to quantum
gravity, a local positive cosmological term mu^2(x)
corresponds to a local negative squared mass term in the
Lagrangian and may thus induce instability and local pinning
of the gravitational field. Such a term can be produced by the
coupling to an external Bose condensate. In the functional
integral, the local pinning acts as a constraint on the field
configurations. We discuss this model in detail and apply it
to a phenomenological analysis of recent experimental
results.Best wishes,Mark
===
Subject: Marketing shift, core
issuesI've been thinking about my problems with getting any
kind ofadmission that my math arguments showing the core error
in mathematicsare correct, so I've gone to marketing books.I
just wanted to warn readers that I may be employing various
tacticsfrom modern research on human psychology to see if I
can't breakthrough the logjam.It seemed to me that it'd be a
good idea to warn you ahead of time, asI'd prefer it that the
math by itself would be enough, but that's notthe way it
works. I guess.At least now I won't have to feel guilty if I
decide to use tactics,as I've given fair warning.James
Harrishttp://mathforprofit.blogspot.com/
===
Subject: Re:
Marketing shift, core issues> I've been thinking about my
problems with getting any kind of> admission that my math
arguments showing the core error in mathematics> are correct,
so I've gone to marketing books.I just wanted to warn readers
that I may be employing various tactics> from modern research
on human psychology to see if I can't break> through the
logjam.You are more liable to break the logjam by playing
Russian roulette solitaire, as the logjam is all in your
imagination.It seemed to me that it'd be a good idea to warn
you ahead of time, as> I'd prefer it that the math by itself
would be enough, but that's not> the way it works. I guess.At
least now I won't have to feel guilty if I decide to use
tactics,> as I've given fair warning.As JSH's past
performances lead one to believe that he would not feel guilty
even for using an Abomb on the mathematical
world.....
===
Subject: Re: Marketing shift, core issues> I've
been thinking about my problems with getting any kind of>
admission that my math arguments showing the core error in
mathematics> are correct, so I've gone to marketing books.>
Just so that we all understand you this once, do you meanthat
you've decided to give up on math and become a bookselleror do
you mean that you've given up on the math and are hopingto use
modern marketing techniques to push your self-describedproof
on a gullible public?Either way, best of luck.p.s. I guess you
missed my earlier post, so I'm reprintingit here.Let's take a
simpler example and see how your argument works.Let Q(x) =
7(25x^2 + 30x + 2) [1] = 7(x^2 + x)(5^2) + 7(x - 1)(5) +
7^2and write Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) [2]Now we
mirror your construction and observe that we may takea_1(x)
and a_2(x) to be roots of r(x) = a^2 - (x - 1)a + 7(x^2 + x)
[3]We see that Q(0) = 14 = (7)(2), and r(0) = a^2 + a has
rootsa_1(0) = 0, a_2(0) = -1. We see that this assignment
isconsistent with your factorization [2], since Q(0) = (5(0) +
7)(5(-1) + 7) = (7)(2) [4]and we see that we can rewrite [4] as
Q(0)/7 = (5(a_1(0)/7) + 1)(5a_2(0) + 7) = (1)(2)or, emphasizing
the constant term in [1] we may write,letting b_2(0) = a_2(0) +
1, Q(0)/7 = (5(a_1(0)/7) + 1)(5b_2(0) + 2) = (1)(2)Now,
however, you would have us write each factorizationover the
algebraic integers as Q(x)/7 = (5(a_1(x)/7) + 1)(5b_2(x) + 2)
[5]regardless of the (rational integer) value of x.This
constraint in [5] does indeed work in some cases. Forexample,
if we take x = -1 we see that Q(-1) = -21 = (7)(-3) r(-1) =
a^2 + 2aand we see that we may take a_1(-1) = 0, a_2(-1) =
-2and that doing so gives us the factorization Q(-1) = (5(0) +
7)(5(-2) + 7)and we see that Q(-1)/7 = (5(a_1(-1)/7) +
1)(5a_2(-1) + 7) = (1)(-3)The problem is that your restriction
in [5] depends stronglyon the polynomial r(x) used to define
the terms a_1(x)and a_2(x). For x = 0, -1 the factorization in
[5] workedbecause of the way r(x) splits into particular linear
factors,but consider the case when x = 1. Now we have Q(1) =
399 = (7)(57) = (7)(3 * 19) r(1) = a^2 + 14which gives us
a_1(1) = sqrt(-14), a_2(1) = -sqrt(-14)and so we have the
factorization Q(1) = (5 sqrt(-14) + 7)(-5 sqrt(-14) + 7)which
is indeed equal to 399, as expected. However, it'simmediately
obvious that sqrt(-14) isn't divisible by 7in the algebraic
integers, so the factorization in [5]does not hold. Instead of
distributing the value 7as 7 in the first factor and 1 in the
second, we havethe distribution sqrt(7) in both factors, so we
havethe factorization in algebraic integers Q(1)/7 = [(5
sqrt(-14) + 7)/sqrt(7)] * [(-5 sqrt(-14) + 7)/sqrt(7)]In fact,
your constraint on the way 7 (or 49, in the exampleyou have
been using) is split among the terms in yournonpolynomial
factorization is valid only in a small numberof cases for
x.R.
===
Subject: Re: Marketing shift, core issues>I've been
thinking about my problems with getting any kind of>admission
that my math arguments showing the core error in
mathematics>are correct, so I've gone to marketing books.You
know, in case you're curious, this sounds really really
stupid.If your results were correct you'd be able to convince
people ofthem by explaining the proofs carefully. But in fact
they're wrong,people continually explain what the errors are,
and tactics frommarketing books are not going to change
that.>I just wanted to warn readers that I may be employing
various tactics>from modern research on human psychology to
see if I can't break>through the logjam.>It seemed to me that
it'd be a good idea to warn you ahead of time, as>I'd prefer
it that the math by itself would be enough, but that's not>the
way it works. I guess.>At least now I won't have to feel guilty
if I decide to use tactics,>as I've given fair warning.>James
Harris>http://mathforprofit.blogspot.com/
===
Subject: Re:
Marketing shift, core issuesI've been thinking about my
problems with getting any kind of> admission that my math
arguments showing the core error in mathematics> are correct,
so I've gone to marketing books.You are thus not merely
incorrigbly stooopid, you are also an out andout fraud and
admit as much (little).http://www.crank.net/harris.h It's not
every braying jackass that gets a whole page at crank.net--
Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe
for children and most mammals)Quis custodiet ipsos custodes?
The Net!
===
Subject: Re: Marketing shift, core issuesI've been
thinking about my problems with getting any kind of> admission
that my math arguments showing the core error in mathematics>
are correct, so I've gone to marketing books.> Many people
have pointed out the the errors in mathematicsare yours. Have
you any idea why crank dot net devotes awhole set of web pages
on you? http://www.crank.net/harris.hExample: For the past
seven years, James Harris has been a regular fixture on the
Usenet newsgroup, sci.math. James, an amateur mathematician,
has produced scores, if not hundreds of self-proclaimed
elementary 'proofs' of Fermat's Last Theorem, none of which,
to date, have withstood serious scrutiny. Due to James's
unfamiliarity with mathematical exposition, the work available
on his main and 'area one' web sites and his contributions to
the newsgroup have often been difficult to understand and have
been augmented and clarified by many others over the years.
This page is my attempt to gather together what has been said
by and to Mr. Harris about FLT and present it in a style which
meets at least minimal standards for clarity.
===
Subject: Re:
Marketing shift, core issues> I've been thinking about my
problems with getting any kind of> admission that my math
arguments showing the core error in mathematics> are correct,
so I've gone to marketing books.> I just wanted to warn
readers that I may be employing various tactics> from modern
research on human psychology to see if I can't break> through
the logjam.> It seemed to me that it'd be a good idea to warn
you ahead of time, as> I'd prefer it that the math by itself
would be enough, but that's not> the way it works. I guess.>
At least now I won't have to feel guilty if I decide to use
tactics,> as I've given fair warning.Let me guess what this
means: your algebra hasn't convinced anyone, so youare going
to try to t people into believing you.
===
Subject: Basic
factorization ideasIf you saw(c_1 x + 7)(c_2 x + 7)( c_3 x +
1) = 49(x^3 + 5x^2 + 3x + 1)with the c's algebraic integers, I
think few of you would have aproblem realizing that only two of
the c's have 7 as a factor.But, of course, you're looking at
*functions* of x, as you have f_1(x) = c_1 x, f_2(x) = c_2 x,
and f_3(x) = c_3 x, so I could also write it as(f_1(x) +
7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1).Notice
that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 +
1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1as long as you're in a
ring where 7 is not a factor of 22.I want to emphasize that
point as notice there's only *one* way todivide through by 49
if 7 is not a factor of 22.Usually you can *see* the other
factors of 7, but I want you toabstract, and generalize.Please
pay careful attention to that example.You may see people who
reply claiming that the word polynomial hassome significance,
as if it's a mystical thing which refutes basiclogic, so if
something isn't polynomial it no longer behaveslogically.Now
then, in my advanced factorization work, I just use functions
of xthat are a lot more complicated than f_1(x) = c_1 x, and
unfortunatelythere are people who can use an unfamiliar leap
in complexity toconfuse others.Some of you have learned
various advanced math topics, now imagine ifin your classrooms
there were some hecklers who continually holleredout at your
teacher, or otherwise disrupted the class?What if when there
were difficult concepts those hecklers would try toconfuse
everyone as they sought to discredit the mathematics?If you
find that hard to imagine, imagine me in your class with
youquestioning the professor and calling him names.How much
would you have learned?I need those of you interested in
mathematics to focus on the basics,so that you can understand
the advanced.James
Harrishttp://mathforprofit.blogspot.com/
===
Subject: [JSH]
Crank Net!: Re: Basic factorization ideas> I need those of you
interested in mathematics to focus on the basics,> so that you
can understand the
advanced.http://www.crank.net/harris.h
===
Subject: Re: Basic
factorization ideas > If you saw > > (c_1 x + 7)(c_2 x + 7)(
c_3 x + 1) = > > 49(x^3 + 5x^2 + 3x + 1) > > with the c's
algebraic integers, I think few of you would have a > problem
realizing that only two of the c's have 7 as a factor.A bit of
ting? Your marketing? Let r1, r2 and r3 be the roots ofthe
cubic in some order, than we have: c1 = 7/r1, c2 = 7/r2, c3 =
1/r3all algebraic integers. So indeed, c1 and c2 are divisible
by 7 inthe algebraic integers. > But, of course, you're looking
at *functions* of x, as you have > > f_1(x) = c_1 x, f_2(x) =
c_2 x, and f_3(x) = c_3 x, > > so I could also write it as >  (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x +
1). > > Notice that dividing both sides by 49 gives > >
(f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x +
1 > > as long as you're in a ring where 7 is not a factor of
22.What has 7 being a factor or not of 22 to do with it? This
is alsotrue when 7 is a factor of 22. > I want to emphasize
that point as notice there's only *one* way to > divide
through by 49 if 7 is not a factor of 22.Oh, you mean that
when 7 is a factor of 22 than there are other waysto do the
division, as in that case 7 is a unit in the ring.Moreover,
there may be values of x such that (f3(x) + 1) is divisible
by7, I think. If this is the case there are other ways to
distribute the7's.... > Now then, in my advanced factorization
work, I just use functions of x > that are a lot more
complicated than f_1(x) = c_1 x, and unfortunately > there are
people who can use an unfamiliar leap in complexity to >
confuse others.Unfortunately, the functions here are trivially
divisible by 7 in thering of algebraic integers. In your
advance factorisation work theyare *not*. But you think that
is a flaw... But indeed, that yourfunctions are much more
complicated *makes* a difference. c1 x *is*divisible by 7
because c1 is divisible by 7, regardless of the valueof x (c1
is a constant), so (c1 x + 7) is trivially divisible by 7.(c3
x + 1) is *not* divisible by 7, unless x has a specific value
as Imentioned above, which might or might not be possible, but
that isirrelevant, because the first two factors are trivially
divisible by7. In your (more complicated case) you *know* that
a1(x) and a2(x)are *not* divisible by 7 for all x (you say that
is a flaw). Ithas been shown that (b3(x) + 22) is divisible by
factors of 7 for almostall x. And those factors compensate for
the lack of factors in a1(x)and a2(x).-- dik t. winter, cwi,
kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home:
bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Basic factorization
ideas> If you saw(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > >
49(x^3 + 5x^2 + 3x + 1)with the c's algebraic integers, I
think few of you would have a> problem realizing that only two
of the c's have 7 as a factor....> I want to emphasize that
point as notice there's only *one* way to> divide through by
49 if 7 is not a factor of 22. > > Oh, you mean that when 7 is
a factor of 22 than there are other ways > to do the division,
as in that case 7 is a unit in the ring. > Moreover, there may
be values of x such that (f3(x) + 1) is divisible by > 7, I
think. If this is the case there are other ways to distribute
the > 7's.Lo and behold, there are indeed (not 7, but factors
of 7)! When x = 7y + 2for some integer y, (c3 x + 1) is *not*
coprime to 7. This example is tooeasy to crack. Define: w3(x)
= gcd(c3 x + 1, 7)(this one is not always 1 (*)), w2(x) =
7/w3(x).We now could divide the first factor by 7, the second
by w2(x) andthe third by w3(x) and remain in the algebraic
integers. So, again,the possible ways to divide 49 through may
very well depend on thevalue of x.----(*) Proof:We have (c3 x +
1) where c3 is 1/r, with r the negative of one of theroots of
x^3 + 5 x^2 + 3 x + 1Now (x/r + 1) = (r + x)/r, so the first
is not coprime to 7 when(r + x) is not coprime to 7. Set x =
7y + 2. (r + x) is notcoprime to 7 when (r + 7y + 2) is not
coprime to 7, and that iswhen r + 2 is not coprime to 7. You
may verify that (r + 2) isa root of y^3 - 11 y^2 + 35 y - 35.
So it is an algebraic integerand not coprime to 7.-- dik t.
winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Basic factorization
ideas>If you saw>(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > >
49(x^3 + 5x^2 + 3x + 1)>with the c's algebraic integers, I
think few of you would have a>problem realizing that only two
of the c's have 7 as a factor.Personally I have a problem with
the fact that I can think of novalues for c_1, c_2 and c_3 such
that c_1, c_2 and c_3 are algebraicintegers and the left and
the right side of the equation are equal.Could you help me out
and give me the values of c_1, c_2 and c_3?
===
Subject: Re:
Basic factorization ideas > >If you saw > >(c_1 x + 7)(c_2 x +
7)( c_3 x + 1) = > > 49(x^3 + 5x^2 + 3x + 1) > >with the c's
algebraic integers, I think few of you would have a >problem
realizing that only two of the c's have 7 as a factor. > >
Personally I have a problem with the fact that I can think of
no > values for c_1, c_2 and c_3 such that c_1, c_2 and c_3
are algebraic > integers and the left and the right side of
the equation are equal.James has ted you. In this version the
roots of the cubic areunits. Part of his marketing strategy, I
think.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland, +31205924131home: bovenover 215, 1025 jn amsterdam,
nederland; http://www.cwi.nl/~dik/
===
Subject: Re: Basic
factorization ideasCrank Information
http://www.crank.net/harris.h http://www.crank.net/usenet.h
http://www.google.com/search?q=harris+site%3Awww.crank.net
http://www.google.com/search?q=%22james+harris%22+site%
3Ausers.pandora.be
===
Subject: Re: Vedic Mathematics --- Myth
and Reality> For instance, 2000 years down the road, the
people of Punjabi > descent may claim that Punjab made great
strides in physics > and chemistry during 1950-2050 and they
can show numerous > physics and chemistry books by Punjabi
authors to support > their claim. But as we know, nearly all
the physics and > chemistry contained in these so-called
Punjabi books > has come from the West.While one does not know
will happen after 2000 years, one thinks one> knows what
happened 2000 years ago, when one is guided by one's trust> in
the people he considers worthy of respect, So claims don't need
to be verified by evidence?Statements made under oath are taken
as evidence in law courts. It ison such visual/oral/aural
(non-literary, that is) evidence that peopleare convicted for
murder. Heard you heard of the term eye witness? By no means
is literary evidence the only sort of evidence. Also,
thegenuineness of the literary evidence is a major issue. So,
just asone takes as genuine evidence the word of a witness who
cannot beproved false, one takes as genuine evidence the
statements of peoplein positions of respect. Of course, there
could be genuine mistakes -a witness can think he is speaking
the truth, when actually he is not. Like, most physicists
genuinely believe in Einstein's Theory ofRelativity. They are
not liars, only deluded, according to the onewho can prove
that theory baseless.> and the various objects> around one
that are of relevance. Such as, a book I picked up at a>
reputable shop, that has been the object of vilification.Why
is it vilification to ask if the math algorithms that > are
being marketed as Vedic mathematics were really > developed by
the Vedic people in the Vedic era?Nobody is just asking that.
They are saying that the writer of thebook is a fraud. This
was the tone of the first post, and somefollowups. To show
that the writer is a fraud, they have to provethat those
algorithms were developed elsewhere, and then they have toshow
where the algorithms came from, who outside India developed
them,where they were used, etc. That they have not done -
those people whoare attacking the writer of the book, I
mean.When one cannot show that, one has to consider it very
plausible thatthey were developed in ancient India, and
followed down the ages. Thebook describes the multiplication
algorithm in a single descriptiveSanskrit word, which goes
into the core of the actual process. As Ihave shown, the
ancient Indians really understood the place valueunderlying
the decimal system, and used it very efficiently.Look, I have
explained the one-line multiplication. But they havealso done
one-line division, one-line square root, etc. If there
areexisting methods elsewhere relating to one-line division or
one-linesquare-root evaluation, then give the references. Or
else, acceptthat the whole thing is really new for the world
outside India. Certainly, I have never come across this method
earlier, and I havewritten multiprecision arithmetic software
routines (consulting Knuth)myself using bases of 10 and higher
(for faster computation). If Iknew about all this Vedic
arithmetic, I could have done a better job,and more easily
too.Arindam Banerjee.
===
Subject: Re: Vedic Mathematics ---
Myth and Reality> While one does not know will happen after
2000 years, one thinks one> knows what happened 2000 years
ago, when one is guided by one's trust> in the people he
considers worthy of respect, So claims don't need to be
verified by evidence?So, just as one takes as genuine evidence
the word of a witness > who cannot be proved false, one takes
as genuine evidence the > statements of people in positions of
respect. Values that are based on respect/reverance for
personalities fall within the boundaries of
temples/mosques/churches.Historical research doesn't depend on
such beliefs.Arindam Banerjee.
===
Subject: Re: Vedic Mathematics
--- Myth and Reality> My father taught me in my childhood days,
12 *> 34> --------> 48> 36> --------> 408Australian primary
schools, and in most Indian schools as well. Asyou can see,
you need three lines to get this result (48, 36 and 408). In
Vedic arithmetic, you get the representation of the result in
justone line, no matter what the length of the factors. Now,
in Indiathere probably still exist a lot of people who never
did this long andrather silly-looking Western method, they
always did it the ancientVedic way, which was so much better.
(I suspect that while they gotthe zero and other Indian
arithmetic stuff into the West, they neverlearnt to use them
the way the ancient Indians did.) Now, they areonly telling us
such methods by publishing them in a book. I havestudied only
the multiplication, but they claim they can do
one-linedivision, square-root, etc. as well. Basically, I
don't see why weshould call them frauds (as the TIFR person
indicated) simply becausethese brilliant methods work out
simpler and faster, and arealtogether better.> cALCULATION IS
> 12 *> 34> --------> 4*1 | 4*2> 3*1 | 3*2 |>
------------------> 3 | 10 | 8 = 408Perfectly correct,
however, computers convert them into binary numbers(1s and 0s)
and then add and multiply using shifting techniques thatare
very efficient. However, for really long numbers, they have
toget a number of integers for the representation,as they use
the samebasic algorithm that you have demonstrated. And there
has to be a lotof linking between these numbers. With Vedic
arithmetic, you candefine a number as an arrays of integers,
and you do the job simplyand also quickly, mainly because the
internal computation does notinvolve extensive array
processing.> So 12*34=508Oh really? Again, never mind!To give
a mathematical (and suitably, one-line!) definition of
Vedicmultiplication: The multiplicative product of two numbers
A and Brepresented as a sequence of characters of length m and
nrespectively, of base X, and of value A(m-1)*X^(m-1)
+A(m-2)*X^(m-2)+...+A(0)*X^0 and B(n-1)*X^(n-1)
+B(n-2)*X^(n-2)...+B(0)*X^0 will be C(m+n-1)*X^(m+n-1)
+C(m+n-2)*X^(m+n-2) +...+ C(0)*X^0, where any C(k) is found by
sum ofthe combinations all the A(i) and B(j) satisfying i+j=k,
where i, jand k range from 0 to m, n and m+n-1 respectively.In
lay language, this means gathering all the units, tens,
hundreds,etc. in just one go, by finding out all the possible
combinations forsame. For instance, to find the units, you can
only multiply the lastdigits of the two numbers.Let anyone try
to multiply say 12345 by 67809 using this method, inone line,
as an exercise! (If you give up, I'll do it!). At present,I am
a bit busy, and have wasted too much time already! :) :)With
best wishes to all,Arindam Banerjee.
===
Subject: Re: Vedic
Mathematics --- Myth and Reality> My father taught me in my
childhood days, 12 *> 34> --------> 48> 36> -------->
408Australian primary schools, and in most Indian schools as
well. As> you can see, you need three lines to get this result
(48, 36 and 408).> In Vedic arithmetic, you get the
representation of the result in just> one line, no matter what
the length of the factors. Now, in IndiaI can also do it in one
line: 12 * 34 -------- 48 36 408But seriously, this vedic
method is no different as our method.Our method is
basically:2*4 + 10*4 + 2*30 + 10*30 = 48 + 360 = 408See,
exactly the same as your method.Wilbert
===
Subject: Re: Vedic
Mathematics --- Myth and Reality> My father taught me in my
childhood days,> 12 *> 34> --------> 48> 36> -------->
408Australian primary schools, and in most Indian schools as
well. As> you can see, you need three lines to get this result
(48, 36 and 408).> In Vedic arithmetic, you get the
representation of the result in just> one line, no matter what
the length of the factors. Now, in IndiaI can also do it in one
line: 12 *> 34> --------> 48 36 408But seriously, this vedic
method is no different as our method.> Our method is
basically:2*4 + 10*4 + 2*30 + 10*30 = 48 + 360 = 408See,
exactly the same as your method.Not at all! > WilbertNo, you
don't see the difference. You have written 48 as 12*4,whereas
in the Vedic method you take only one coefficient at one time.
Like, you cannot take more than one coefficient, as you have
done. Try doing what you have done for say 12345*567809! That
is the veryimportant difference, and that which will make it
much simpler forcomputer calculation.I know it is very
difficult to comprehend the simple and thebrilliant,
especially when the mind has been narrowly and
closelyformed.Arindam Banerjee.
===
Subject: Re: Vedic
Mathematics --- Myth and Reality> My father taught me in my
childhood days,> 12 *> 34> --------> 48> 36> --------> 408>
Australian primary schools, and in most Indian schools as
well. As> you can see, you need three lines to get this result
(48, 36 and 408).> In Vedic arithmetic, you get the
representation of the result in just> one line, no matter what
the length of the factors. Now, in IndiaI can also do it in one
line: 12 *> 34> --------> 48 36 408But seriously, this vedic
method is no different as our method.> Our method is
basically:2*4 + 10*4 + 2*30 + 10*30 = 48 + 360 = 408See,
exactly the same as your method.Not at all!WilbertNo, you
don't see the difference. You have written 48 as 12*4,>
whereas in the Vedic method you take only one coefficient at
one time.coefficient at the time.> Like, you cannot take more
than one coefficient, as you have done.> Try doing what you
have done for say 12345*567809! That is the very> important
difference, and that which will make it much simpler for>
computer calculation.Our way (including the steps you make in
your mind):10000*9 + 2000*9 + 300*9 + 40*9 + 5*9 + 10000*800 +
2000*800 + ...Precisely the same as your method, only the
ordering of adding is different. Big deal. > I know it is very
difficult to comprehend the simple and the> brilliant,
especially when the mind has been narrowly and closely>
formed.Adding numbers in a different order, yes very
brilliant.Wilbert
===
Subject: Re: Vedic Mathematics --- Myth
and Reality...> To give a mathematical (and suitably,
one-line!) definition of Vedic> multiplication: The
multiplicative product of two numbers A and B> represented as
a sequence of characters of length m and n> respectively, of
base X, and of value A(m-1)*X^(m-1) +>
A(m-2)*X^(m-2)+...+A(0)*X^0 and B(n-1)*X^(n-1) +>
B(n-2)*X^(n-2)...+B(0)*X^0 will be C(m+n-1)*X^(m+n-1) +>
C(m+n-2)*X^(m+n-2) +...+ C(0)*X^0, where any C(k) is found by
sum of> the combinations all the A(i) and B(j) satisfying
i+j=k, where i, j> and k range from 0 to m, n and m+n-1
respectively.If that's one line I'm a Dutchman. It looks like
8 lines to me.This doesn't look like it differs from the
standard definitionof long multiplication. What's Vedic about
it?Phil-- Unpatched IE vulnerability: protocol control
charsDescription: Circumventing content filtersReference:
http://badwebmasters.net/advisory/012/Exploit:
http://badwebmasters.net/advisory/012/test2.asp
===
Subject: Re:
Vedic Mathematics --- Myth and Reality> ...> To give a
mathematical (and suitably, one-line!) definition of Vedic>
multiplication: The multiplicative product of two numbers A
and B> represented as a sequence of characters of length m and
n> respectively, of base X, and of value A(m-1)*X^(m-1) +>
A(m-2)*X^(m-2)+...+A(0)*X^0 and B(n-1)*X^(n-1) +>
B(n-2)*X^(n-2)...+B(0)*X^0 will be C(m+n-1)*X^(m+n-1) +>
C(m+n-2)*X^(m+n-2) +...+ C(0)*X^0, where any C(k) is found by
sum of> the combinations all the A(i) and B(j) satisfying
i+j=k, where i, j> and k range from 0 to m, n and m+n-1
respectively.If that's one line I'm a Dutchman. It looks like
8 lines to me.Well, it is one long sentence, which could have
been written in onelong line. Try explaining the current
Western method so briefly, ifyou can!> This doesn't look like
it differs from the standard definition> of long
multiplication. Try to be a bit more clever, please. Also,
look at my explanation tothe other chap, that I just posted.>
What's Vedic about it?People who invented the method consider
themselves Vedic.Arindam Banerjee.Phil
===
Subject: Re: Vedic
Mathematics --- Myth and Reality> ...> To give a mathematical
(and suitably, one-line!) definition of Vedic> multiplication:
The multiplicative product of two numbers A and B> represented
as a sequence of characters of length m and n> respectively,
of base X, and of value A(m-1)*X^(m-1) +>
A(m-2)*X^(m-2)+...+A(0)*X^0 and B(n-1)*X^(n-1) +>
B(n-2)*X^(n-2)...+B(0)*X^0 will be C(m+n-1)*X^(m+n-1) +>
C(m+n-2)*X^(m+n-2) +...+ C(0)*X^0, where any C(k) is found by
sum of> the combinations all the A(i) and B(j) satisfying
i+j=k, where i, j> and k range from 0 to m, n and m+n-1
respectively.If that's one line I'm a Dutchman. It looks like
8 lines to me.Well, it is one long sentence, which could have
been written in one> long line. Try explaining the current
Western method so briefly, if> you can!The current western
method? If it's small, do it by inspection, else use a
calculator or computer.Much shorter, and applicable to a wider
range of problems too.> This doesn't look like it differs from
the standard definition> of long multiplication. Try to be a
bit more clever, please. Sheesh, so says someone who thinks
that reeordering _the same freaking number of operations_ is
clever. For the kinds of calculations I do, your technique is
_useless_. Utterly useless.It's not clever, it's a chocolate
teapot.Sorry to burst your bubble, but you, or your vedic
cronies, have taken a technique which requires theta(m*n)
single-digit multiplications, and theta(2*m*n) single-digit
accumulations, and turned it into a method that requires
theta(m*n) single-digit multiplications and theta(m*n)
two-digit accumulations -- with exactly the same
multiplicative constants (where m and n are the lengths of the
numbers. That's _exactly_ the same workload.I'm fully aware of
the 'western', 'russian peasant', 'frenchpeasant' and 'vedic'
methods for multiplication, and on the whole I find the
'western' one the easiest to use if I've got a piece of paper
and a pen handy, and I always have a piece of paper and a pen
handy. The 'vedic' method is basically the 'french peasant'
method with JIT multiplications (and thus not requiring the
paperscratch area for the calculations). And it's _exactly_
the definition of multiplication as used inmultiplication by
convolution, which is the most common method of doing big
number multiplication nowadays. However, the multiplication by
convolution has an asymptotically much lower Big-Oh than the
naive counting-on-fingers technique you describe above.> Also,
look at my explanation to> the other chap, that I just
posted.What's Vedic about it?People who invented the method
consider themselves Vedic.And the people who propagate it I
consider vedic-fanboy drones.You're as bad as Mac-zealots and
linux-weenies.If people are too lazy to find the technique for
problem solutionthat suits them best then that's _their_
problem. Stop proselytising.Phil-- Unpatched IE vulnerability:
Security zone transferDescription: Automatically opening IE +
Executing attachmentsPublished: March 22nd 2002Reference:
http://security.greymagic.com/adv/gm002-ie/
===
Subject: Re:
Vedic Mathematics --- Myth and Reality>Now, in India there
probably still exist a lot of people who never did>this long
and rather silly-looking Western method, they always did it
the>ancient Vedic way, which was so much better.Ironically,
multiplication is far easier -- almost combinatoric in fact
--in Roman numerals. And with a minor adjustment, the system
can bemade positional (even without a zero!), with the
algorithm suitableextended.It's not widely known, but a 0 is
not a requirement for a positionalnumber system -- not even
one (in fact) capable of representing 0(!)or even negative
numbers: something the Hindu-Arabic system can'tdo without
extaneous markers (the - symbol).The Roman system is, in fact,
nearly already that kind of system;making it (in conjunction
with the above-mentioned adjustment) bothcombinatorial and
algebraic.The generalized IV rule is rendered as such: For
sequences of M,D,C,L,X,V,I of the form S = z0 a1 z1 a2 z2 ...
an zn with z0 > z1 > z2 > ... > zn each single digits; ai
(i=1,...,n) each a string of 0,1,2, or more digits all of
value less than zi, the value of S is z0 + z1 + z2 + ... + zn
- the sum of the values of a1,a2,...,anThe abovementioned
adjustments are(1) Everything under a bar counts 1000-fold(2)
Everything after a comma counts 1000-fold(3) Everything after
a dash counts 10-fold.(4) Sequences are interpreted only
between dashes, commas and under bars.That makes the Roman
system a superset of a positional 0-less system,with an
interesting combinatorial addition and multiplication
algorithm.>In lay language, this means gathering all the
units, tens, hundreds,>etc. in just one go, by finding out all
the possible combinations for>same.It's the same thing as the
modern grade-school method done in a differentorder.Western
methods include methods based on the fast Fourier
transform,which can multiply 2 N digit numbers using on the
order of N ln(N)multiplications, instead of N^2. Even with a
simple modification ofthe grade school technique, you can get
away with on the order of I thinksomewhere around N^{8/5}
multiplcations.I generally go 36*49 = 1254 + 90*13 - 660 =
1764,or 57*28 = 1056 + 120*10 - 660 = 1596,or 73*39 = 2127 +
100*12 - 480 = 2847.
===
Subject: Re: Vedic Mathematics --- Myth
and Reality>Now, in India there probably still exist a lot of
people who never did>this long and rather silly-looking
Western method, they always did it the>ancient Vedic way,
which was so much better.Ironically, multiplication is far
easier -- almost combinatoric in fact --> in Roman numerals.
And with a minor adjustment, the system can be> made
positional (even without a zero!), with the algorithm
suitable> extended.Try writing 123987739383883938989874 in
Roman numerals. > It's not widely known, but a 0 is not a
requirement for a positional> number system -- not even one
(in fact) capable of representing 0(!)> or even negative
numbers: something the Hindu-Arabic system can't> do without
extaneous markers (the - symbol).They used an abacus to give a
visual representation of a large number.> The Roman system is,
in fact, nearly already that kind of system;> making it (in
conjunction with the above-mentioned adjustment) both>
combinatorial and algebraic.The generalized IV rule is
rendered as such:> For sequences of M,D,C,L,X,V,I of the form>
S = z0 a1 z1 a2 z2 ... an zn> with z0 > z1 > z2 > ... > zn each
single digits;> ai (i=1,...,n) each a string of 0,1,2, or more
digits> all of value less than zi,> the value of S is> z0 + z1
+ z2 + ... + zn - the sum of the values of a1,a2,...,anDid the
Romans have any knowledge of algebra? > The abovementioned
adjustments are> (1) Everything under a bar counts 1000-fold>
(2) Everything after a comma counts 1000-fold> (3) Everything
after a dash counts 10-fold.> (4) Sequences are interpreted
only between dashes, commas and under bars.That makes the
Roman system a superset of a positional 0-less system,> with
an interesting combinatorial addition and multiplication
algorithm.> >In lay language, this means gathering all the
units, tens, hundreds,>etc. in just one go, by finding out all
the possible combinations for>same.It's the same thing as the
modern grade-school method done in a different> order.Tell us
about the grade-school method. What it is, and how it is
thesame as the method given.Only one coefficient is being
taken at a time, in this method, for thecomputation. This
method may well have been rediscovered lately, andused (after
all the book mentioned was published in 1965, and themethods
described there are far older), but certainly this method
isnot taught in schools, as we find out from other posters in
thisthread.My main point is to show that the writer of that
book is not a fraud. His methods can be very useful, as
opposed to current methods taughtin schools. If the Vedic
method is already used, fine!> Western methods include methods
based on the fast Fourier transform,> which can multiply 2 N
digit numbers using on the order of N ln(N)> multiplications,
instead of N^2. Okay, so multiply 12345 by 67809 using FFT,
right here. Show us howit is done more simply than what I have
done below.You have to do a lot better if you want to debunk
this method,convincingly.> Even with a simple modification of>
the grade school technique, you can get away with on the order
of I think> somewhere around N^{8/5} multiplcations.I
generally go 36*49 = 1254 + 90*13 - 660 = 1764,> or 57*28 =
1056 + 120*10 - 660 = 1596,> or 73*39 = 2127 + 100*12 - 480 =
2847.I don't know what you are trying to prove, here!! You
certainly havenot succeeded in multiplying 12345 by 67809 by
*any* method! *)*)Oh, by the way, looks like I have not done
my one-line multiplicationof 12345 by 67809.Not that anyone
seemed really interested! The one line answer
is(6)(19)(40)(61)(91)(85)(67)(36)(45) or 837102105.With a
little bit of practice, one get this quicker. Following
themethod that I had rigorously defined in the earlier post,
(but whichhas been deleted here), and of course using base X
as 10 (our familiardecimal system) :(45) = 5*9(36) =
4*9+5*0(67) = 3*9+8*5+4*0(85) = 2*9+7*5+3*0+4*8(91) =
1*9+6*5+2*0+7*4+3*8(61) = 4*6+1*0+2*8+3*7(40) =
1*8+6*3+2*7(19) = 1*7+2*6(6) = 1*6As is evident, you always
need only to take only one digit at a time. The whole
algorithm is also very well suited for parallel processing.
Especially, when fixed lengths are involved.Good that someone
posted this to sci.math. Hopefully, some among youwill check
out my mathematical derivation for unlimited energy, in
myhomesite, and see if you can find any
flaws.http://www.users.bigpond.com/adda1234/index.htmArindam
Banerjee.
===
Subject: Re: Vedic Mathematics --- Myth and
Reality> My main point is to show that the writer of that book
is not a fraud. > His methods can be very useful, as opposed to
current methods taught> in schools. If current methods are
indeed useless, how did people build the Golden Gate Bridge or
the Queen Mary?> If the Vedic method is already used, fine!Western methods include methods based on the fast Fourier
transform,>which can multiply 2 N digit numbers using on the
order of N ln(N)>multiplications, instead of N^2.
===
Subject:
Re: Vedic Mathematics --- Myth and Reality> > My main point is
to show that the writer of that book is not a fraud. > His
methods can be very useful, as opposed to current methods
taught> in schools. > If current methods are indeed useless,
how did people build the Golden > Gate Bridge or the Queen
Mary?Don't expect a sensible answer, Arindam is a nutjob ...--
---------------------------------------------------------------
---Got to get behind the mulein the morning and
plow
===
Subject: Re: Vedic Mathematics --- Myth and
Reality
===
>Subject: Re: Vedic Mathematics --- Myth and
Reality>Message-id: <bommql$m0m$1@uwm.edu>Now, in India there
probably still exist a lot of people who never did>this long
and rather silly-looking Western method, they always did it
the>ancient Vedic way, which was so much better.>Ironically,
multiplication is far easier -- almost combinatoric in fact
-->in Roman numerals. And with a minor adjustment, the system
can be>made positional (even without a zero!), with the
algorithm suitable>extended.>It's not widely known, Ok.>but a
0 is not a requirement for a positional number systemAre you
sure about that? You're not using 0 and pretending it's not
there areyou?>-- not even one (in fact) capable of
representing 0(!)>or even negative numbers: something the
Hindu-Arabic system can't>do without extaneous markers (the -
symbol).>The Roman system is, in fact, nearly already that
kind of system;>making it (in conjunction with the
above-mentioned adjustment) both>combinatorial and
algebraic.>The generalized IV rule is rendered as such:> For
sequences of M,D,C,L,X,V,I of the form> S = z0 a1 z1 a2 z2 ...
an zn> with z0 > z1 > z2 > ... > zn each single digits;> ai
(i=1,...,n) each a string of 0,1,2, or more digits> all of
value less than zi,> the value of S is> z0 + z1 + z2 + ... +
zn - the sum of the values of a1,a2,...,an>The abovementioned
adjustments are>(1) Everything under a bar counts
1000-fold>(2) Everything after a comma counts 1000-foldWas
that supposed to be 100-fold?>(3) Everything after a dash
counts 10-fold.>(4) Sequences are interpreted only between
dashes, commas and under bars.>That makes the Roman system a
superset of a positional 0-less system,>with an interesting
combinatorial addition and multiplication algorithm.I'm not
following this. Can you actually spell out a few examples of
thissystem?It seems to me that M, D, C, L and X are not
needed, but maybe I just don'tunderstand your system. Why is M
needed if _I represents 1000?>In lay language, this means
gathering all the units, tens, hundreds,>etc. in just one go,
by finding out all the possible combinations for>same.>It's
the same thing as the modern grade-school method done in a
different>order.>Western methods include methods based on the
fast Fourier transform,>which can multiply 2 N digit numbers
using on the order of N ln(N)>multiplications, instead of N^2.
Even with a simple modification of>the grade school technique,
you can get away with on the order of I think>somewhere around
N^{8/5} multiplcations.>I generally go 36*49 = 1254 + 90*13 -
660 = 1764,>or 57*28 = 1056 + 120*10 - 660 = 1596,>or 73*39 =
2127 + 100*12 - 480 = 2847.--MensanatorAce of Clubs
===
Subject:
Re: Vedic Mathematics --- Myth and Reality> >My father taught
me in my childhood days,> 12 *> 34> --------> 48> 36>
--------> 408> >Australian primary schools, and in most
Indian schools as well. As>you can see, you need three lines
to get this result (48, 36 and 408).> In Vedic arithmetic, you
get the representation of the result in just>one line, no
matter what the length of the factors.[...]As I recall from my
teens, there is a book by Trachtenberg on methods of quick
arithmetic. These include long mulitplication where the
calculations are all done on one line.-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
===
Subject: Re: Vedic Mathematics
--- Myth and Reality> >My father taught me in my childhood
days,> 12 *> 34> --------> 48> 36> --------> 408>
Australian primary schools, and in most Indian schools as
well. As>you can see, you need three lines to get this result
(48, 36 and 408).> In Vedic arithmetic, you get the
representation of the result in just>one line, no matter what
the length of the factors.[...]As I recall from my teens,
there is a book by Trachtenberg on methods of > quick
arithmetic. These include long mulitplication where the >
calculations are all done on one line.Very likely. After all,
this is an old method. In the book on VedicMathematics, they
also do one-line division, square-root etc.Arindam
Banerjee.
===
Subject: Re: Vedic Mathematics --- Myth and
Reality> As I recall from my teens, there is a book by
Trachtenberg on methods of> quick arithmetic. These include
long mulitplication where the> calculations are all done on
one line.> Very likely. After all, this is an old method. In
the book on Vedic> Mathematics, they also do one-line
division, square-root etc.Do you know how old exactly?And...
do you have a reference for that book? (Online somewhere?)The
'one-register' method of multiplication is indeed very
nice.But... it is arithmetic, rather than mathematics. That
may also explainwhy the math teaching world is not very fond
of ts like this.Mathematics teachers are desperately fighting
the general prejudicethat mathematics is 'just'
arithmetic.Cheers,Herman Jurjus> Arindam Banerjee.
===
Subject:
Re: Vedic Mathematics --- Myth and Reality> > > >As I
recall from my teens, there is a book by Trachtenberg on
methods of>quick arithmetic. These include long
mulitplication where the>calculations are all done on one
line.> >Very likely. After all, this is an old method. In
the book on Vedic>Mathematics, they also do one-line
division, square-root etc.> >Do you know how old
exactly?>And... do you have a reference for that book? (Online
somewhere?)>The 'one-register' method of multiplication is
indeed very nice.>But... it is arithmetic, rather than
mathematics. That may also explain>why the math teaching world
is not very fond of ts like this.>Mathematics teachers are
desperately fighting the general prejudice>that mathematics is
'just' arithmetic.> I doubt this had anything to do with Vedic
mathematics (though I may just be unaware of the sources). It
was a system partly invented (or perhaps merely collected) by
Jacow Trachtenberg in the twentieth century. The standard
reference in English is The Trachtenberg Speed System of Basic
Mathematics by Cutler
<http://www.amazon.com/exec/obidos/search-handle-url/index=
books&field-author=Cutler%2C%20Ann/104-9836485-8643947> and
McShane, though this was translated and adapted from
Trachtenberg's original German (I think). You can find it at
Amazon.The descriptions available on the web (search for
Trachtenberg and arithmetic) mainly talk about ts for
multiplying by single digits, 11, and 12, which is the first
part of the book. Later chapters discuss techniques of long
addition, multiplication, and division.-- Stephen J.
Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Turner in NY
Times Science 11/11/03I think we are so confused that we should
keep an open mind to tinkering with gravity, said Dr. Michael
Turner, a cosmologist at the University of Chicago.previouslyQ
to Ed Witten: How can the cosmological constant be so close to
zero but not zero?Ed Witten's answer: I really don't know.
It's very perplexing that astronomical observations seem to
show that there is a cosmological constant. It's definitely
the most troublesome, for my interests, definitely the most
troublesome, observation in physics in my lifetime. In my
career that is.My answer to the same question is at
http://qedcorp.com/APS/StarGate1.movand
http://qedcorp.com/APS/EmergentGravity.doc or same with .pdf
extension for AcrobatGeneral relativity predicted the bending
of light, the expansion of the universe and black holes, and
has served as the foundation for modern cosmology, but
theorists have never presumed that it would be the last word
on gravity.For one thing, it is mathematically incompatible
with the quantum laws very small distances or very high
energies corresponding to the first moments after the Big
Bang, where space and time become discontinuous, general
relativity has to be merged with quantum theory, a project
that has engrossed the present generation of physicists.But
recently some experts have been wondering out loud if it is
time to rewrite Einstein's version of the law as it applies to
the other end of the length scale, to very long distances. The
motivation comes from the predominance of what is sometimes
called the dark sector in the universe.According to what has
recently become a highly celebrated standard model, ordinary
atoms make up only 5 percent of the stuff of the cosmos. Some
kind of mysterious dark matter, perhaps consisting of while
the rest  a whopping 70 percent  consists of something even
more mysterious, known as dark energy.Obviously a theory that
leaves 95 percent of the universe unexplained is less than a
complete triumph.Neither dark energy nor dark matter has been
observed or detected directly. Each has been inferred from its
gravitational effects on the tiny fraction of stuff we can see.
As a result, some scientists have suggested that what
astronomers have discovered in the last 20 years is their own
ignorance of gravity.In particular, the discovery, five years
ago, that the expansion of the universe is apparently
accelerating, under the influence of that dark energy, has
occasioned a re-evaluation of the old certainties.The simplest
explanation for dark energy is something called the
cosmological constant, first invented by Einstein, a cosmic
repulsion caused by the energy residing in empty space. But
attempts to calculate this energy have resulted in numbers
1060 bigger than what astronomers have measured  so large that
the universe would have blown apart before atoms or galaxies
could have formed  causing theorists to throw up their hands.I
think we are so confused that we should keep an open mind to
tinkering with gravity, said Dr. Michael Turner, a cosmologist
at the University of Chicago.As a result of all this, physics
literature has become peppered with suggestions of ways to
change gravity. This fall, given a choice of explanations for
dark energy during cosmology workshop at the Kavli Institute
for Theoretical Physics in Santa Barbara, Calif., 20 of the 44
participants voted for some variation of Einstein was
wrong.Some of these proposals take their cue from the
science-fiction-sounding string theory, the putative theory of
everything, which holds out the possibility that our universe
might be a 4-dimensional membrane (or brane) in an
11-dimensional space.nature in string theory would be stuck to
the brane, like the nap on a rug. But the strings responsible
for transmitting gravity would be able to drift away or leak
into the meta-space surrounding the brane as they traveled
along it from distant objects, according to a theory set forth
in 2000 by Dr. Gia Dvali, Dr. Gregory Gabadadze and Dr. Massimo
Porrati of New York University. The effect, they say, would be
to make distant galaxies appear as if they were accelerating
as they moved away from us.Also in a stringy vein is
Cardassian expansion, named after the villainous race on Star
Trek, and dreamed up by Dr. Katherine Freese and Dr. Matthew
Lewis of the University of Michigan. According to their
theory, the universe accelerates as a result of other branes
tugging on our own. One can get an accelerating universe
without having any dark energy, Dr. Freese said.Other
theorists are going back and modifying general relativity
possible equations that would carry out his ideas. But more
complicated equations might be necessary. That was the
approach taken by Dr. Turner and his colleagues, Dr. Sean
Carroll and Dr. Vikram Duvvuri of Chicago, and Dr. Mark
Trodden of Syracuse. The result was a universe that would
speed up as it got bigger and emptier.That might sound crazy,
Dr. Turner said, but not any crazier than the idea 80 years
ago that the universe would be expanding.The model raises as
many questions as it answers, but it and others like it are
still worth pursuing, Dr. Carroll said.Something funny is
going on when the universe gets to be 10 ion years old, he
said, and none of our current ideas is standing up and
declaring itself to be the right answer, so we have to be
bold.
===
Subject: Re: Turner in NY Times Science 11/11/03> I
think we are so confused that we should keep an open mind to>
tinkering with gravity, said Dr. Michael Turner, a cosmologist
at the> University of Chicago.> previously> Q to Ed Witten: How
can the cosmological constant be so close to zero> but not
zero?Close to zero? WTF does close mean? Is there a maximum
value we cancompare it to to see how close it is?--Denis
Loubetdloubet@io.comhttp://www.io.com/~dloubet
===
Subject: Re:
JSH: $100,000 US offer, Abel Prize> One thing I've been
fascinated by as I've considered replies to my> posts is a
loose group coordination between posters, as some try to> post
with math, and others just post various jibes, but all keep>
focused on pushing the false notion that my rather basic
argument> showing a problem with the definition of algebraic
integers is wrong.Now I've tried to use money before, but
failed as I think people in> the math community understand
that NONE of you can overcome that kind> of coordinated
effort, and even if you post your own version of my own>
argument, the rest of math society will shred you.However, the
shortness and simplicity of the proof of the problem in> core
offers another solution--machine proof checking.That means I
can make the offer of sharing at least $100,000 US with> ONE
person or one group that collaborates to produce the check by>
machine from the Abel Prize, assuming, of course that I win
it.This offer is rescinded. That is, I have changed my mind
and am nolonger offering to give $100,000 US or promise of any
sum of money toanyone for helping me with my current research,
including doing acomputer check, or any other help
whatsoever.I repeat there is no monetary offer on the table
for helping me withrecgnition for my work.Thank you for your
time and attention.James
Harrishttp://mathforprofit.blogspot.com/
===
Subject: Re:
Classification error, algebraic integer issue> Despite the
history of science and physics in particular, it seems to> me
that many of you believe that mathematics as a field is>
invulnerable to a simple error. However I *have* found a>
classification error that is over a hundred years old, as>
mathematicians failed to realize there were numbers that had
slipped> between the cracks of their classification
scheme.Basically, the ring of algebraic integers is too small
to include all> the numbers they believe it includes, and it's
relatively easy to> show.[Very bad math DELETED ...]> > which
proves that f must have 2 itself as a factor for g to be an>
algebraic integer.> OOPS! I've concluded that nothing in the
post proves that statement.It turns out that based on what's
given, maybe it's true, but it's not proven. OOPS indeed! Wait
just a minute on this. We need to remind you of some things you
said about this previously ---> > which proves that f must
have 2 itself as a factor for g to be an> algebraic
integer.>So simply beautiful that it handles the question of
whether or not>people arguing with me are correct, in such a
short space.> OOPS! Looks like the people arguing with you
WERE correct, right?Just for future reference: Were they
lying? Were they cranks? Werethey conspiring to suppress your
startling discovery, even though they secretly knew your proof
was right ? Is that what happened ? And will you remember this
the next time a disagreement with your b.s. comes up, and you
accuse us of lying ???>Possibly I moved too quickly here, as
notice that the coefficient>prove that two of the roots share
factors with 2, while just *one*>does not, as the last two are
even, while the second is not. OOPS! Possibly you *did* move
too quickly here - but you have often been quick to conclude
what you *want* rather than what is correct.>Easy mathematics,
with such a clear result is quite satisfying. OOPS! Easy
mathematics, yes, but also *wrong* mathematics. *No*clear
result, and *not* quite satisfying. Looks like you gloated too
soon!> So c_1 and c_2 while not algebraic integers cannot be
written as a> ratio of non-unit coprime algebraic integers
either. OOPS! This whopper contradicts a famous well-known
theorem!>Which is fascinating in and of itself, but remember,
it's the same>thing for algebraic integers themselves, which
should give you some>perspective on why these numbers belong
in the same category. OOPS! Not so fascinating after all! Has
it ever occurred to you that you are over-using that word? You
are not, after all, thebrainy second-in-command on the original
Starship Enterprise, eh?>It was a simple classification error
made before any of you were born>as it's over a hundred years
old, where I'm the one man who stepped>forward to point it out
and fix it. OOPS! Apparently the one man who stepped forward
here ought to step back and think a little harder ...>That
puts me in a singular place in human history. Hard to disagree
with this one! You continue to build your reputation as a
world-class boob!> These numbers represent an unexplored
frontier of mathematics, and an> opportunity for many of you
to make an impact like you never could> have dreamed of
before in the field.> OOOPS! These numbers represent the
unexplored frontier of, uh, ... theempty set! Might be kind of
hard to get a PhD thesis out of that!> It's fresh ground.
OOOPS again! It's fresh ... what ? The point here: try, try,
try very hard to remember this little fiasco, which you
reproduced over the last few days several times inseveral
different threads, the next time you claim your critics
areliars, incompetents, cranks, or whatever. WE'RE NOT!!! Nora
B.>James Harris><a
href=http://mathforprofit.blogspot.com/>http://
mathforprofit.blogspot.com/</a>
===
Subject: Re: Classification
error, algebraic integer issue> It's fresh ground.> OOOPS
again! It's fresh ... what ?Fresh ground coffee. Hey,
sometimes people just forget a when they type.After all,
didn't someone just say JSH is doing marketing research?He's
selling _coffee_ fergoodnesssake. No wonder we didn't
recognizeit as _mathematics_!
===
Subject: Re: Newsgroup survey:
Math and personality assessment Discussion, linux)> It seems
to me that there have been debates over math concepts I>
thought basic, so here's a quick survey:> James is the person
who's always whining about social issues in> mathematics and
accusing mathematicians of caring more about each other's>
opinions than about the Truth. So why is he also the only
person who> conducts these ridiculous opinion surveys? Does he
*really* think that> *this time* he's going to find that the
masses are on his side? And why> does he care *what* everyone
else thinks, if he's so sure he's right?Duh. He's doing
marketing research. -- Well *supposedly* a correct and
profound math paper can get publishedin a 'reputable journal'
which means that the journals I've faced sofar may lose a lot
of their luster once the full story comes out.s--- James
Harris, on the quality of math journals rejecting his
paper
===
Subject: Semdirect product of categories?Semidirect
product is a common construction in group theory. But it
isobvious that if one relaxes the required assumptions, it can
begeneralized to monoids.Now it occurs to me that another
simple generalization step can bringthe concept to a
categorical formulation: let A,B be categories andlet (A,A)
(the monoid of functors A->A) be regarded as a categoryitself.
Then choose a functor sigma:B->(A,A) and
define(a2,b2)(a1,b1):=(a2sigma(b2)a1,b2b1)for all a2,b2,a1,b1
s.t. Dom(a2)=sigma(b2)Cod(a1), Dom(b2)=Cod(b1).It is
straightforward to check that the definition is well-posed.
Butnow the question is: is it of any utility? Is it taken into
accountsomewhere?Michele-- > Comments should say _why_
something is being done.Oh? My comments always say what
_really_ should have happened. :)- Tore Aursand on
comp.lang.perl.misc
===
Subject: Re: Semdirect product of
categories?|Semidirect product is a common construction in
group theory. But it is|obvious that if one relaxes the
required assumptions, it can be|generalized to monoids.||Now
it occurs to me that another simple generalization step can
bring|the concept to a categorical formulation: let A,B be
categories and|let (A,A) (the monoid of functors A->A) be
regarded as a category|itself. Then choose a functor
sigma:B->(A,A) and
define||(a2,b2)(a1,b1):=(a2sigma(b2)a1,b2b1)||for all
a2,b2,a1,b1 s.t. Dom(a2)=sigma(b2)Cod(a1),
Dom(b2)=Cod(b1).||It is straightforward to check that the
definition is well-posed. But|now the question is: is it of
any utility? Is it taken into account|somewhere?actually i'm
not sure i understand your notation well enough to seewhether
what you're describing actually works, but i'm a bit
skepticalabout it because you don't seem to be giving the most
straightforwardgeneralization of the semi-direct product
construction to a contextinvolving arbitrary categories.the
most straightforward such generalization is some version of
thehomotopy colimit construction. you start with a category c
and afunctor (or in some versions a pseudo-functor) f from c
to thecategory of categories. the homotopy colimit of f is the
categorywhere an object is a pair (x,y) with x an object in c
and y an objectin f(x), and a morphism from (x,y) to (z,w) is
a pair (m,n) withm:x->z in c and n:f(m)(y)->w in f(z), with
composition of morphismsdefined in a semi-obvious
way.semi-direct product of monoids is then the special case
where c has aunique object x and f(x) has a unique object
y.one version of the homotopy colimit construction is called
thegrothendieck construction and has some pretty important
uses inmathematics. the name homotopy colimit that i used
above is perhapsnot very standard terminology, but considering
the many differentversion of the construction which have been
studied, i'm not surewhether standard terminology really
exists in this case. some namesfor other versions (or in some
cases maybe the same version) of theconstruction that you
might want to look up include 2-colimit,pseudo-colimit, lax
colimit, and so forth.by the way the most general version of
the construction is generalenough to encompass not just
semi-direct products of groups but infact arbitrary group
extensions. the extra twisting data needed todescribe an
arbitrary group extension can be considered as part of thedata
forming a pseudo-functor.-- [e-mail address
jdolan@math.ucr.edu]
===
Subject: Call for ParticipationHello
All,NFSNET is a distributed computing project devoted to
pushing factorization limits. It allows users to run client
software which sends data back to acentral site. The data is
collected and eventually used to produce a
factorization.However, the data is sent via an HTTP
connection, which means that clients must be on-line.I am
working on trying to do some slightly smaller numbers, but are
on the 'MOST-WANTED' list for factorizations. I could use some
help, as I have veryfew machines.I will provide executables
(even source if people want to look at it) and data via
email.People will then run the code OFF-LINE then email the
results back.The current project is to factor 2^653+1, which
is the smallest unfactored number of the form 2^n+1 and is in
fact THE 'Most-Wanted' factorization. Wewill then proceed with
other numbers of the form 2^n+1.I solicit your help. Please
email me at rsilverman@draper.com. I am posting from my AOL
account because I do not have newsgroup access at work.The
method for doing this work is the way NFSNET used to work,
before it was automated. But this older method allows off-line
processing. SilvermanYou can lead a horse's ass to knowledge,
but you can't make him think.
===
Subject: Re: partial
deriviatives!> if you have a function F(q(t),t)> and you
have the derivative> d/dt( d/dq[F(q(t),t) )> then it should
be> d^2/dq^2*dq/dt[F(q(t),t)] + d^2/dtdq[F(q(t),t)]>
right?> Or am i mistaken...??> /M>Assuming that you are
asking for the total time derivative (not the partial>time
derivative), then the answer is[shnip]but the real question
is... is the d/dq a total q derivative?So (with the notation
for partial derivative as pF/px):let q(t) be invertabled/dq[
F(q,t(q)) ] = (pF/pq) + (pF/pt)(dt/dq)let pF/pq ==
G(q(t),t)let pF/pt == H(q(t),t)d/dt[ G(q(t),t) + H(q(t),t)
(dt/dq) ] = pG/pt + (pG/pq)(dq/dt) + pH/pt (dt/dq) +
(pH/pq)(dq/dt)(dt/dq) = pG/pt + pH/pq + (pG/pq)(dq/dt)
+(pH/pt)(dt/dq) == (p^2F/ptpq) + (p^2F/pqpt) +
(p^2F/pq^2)(dq/dt) + (p^F/pt^2)(dt/dq) = ((p^2F/pq^2)(dq/dt) +
(p^2F/pqpt)) + (p^F/pt^2)(dt/dq) + (p^2F/ptpq)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ where this is the answer if
only the OP d/dq means partialadam
===
Subject: Re: Four Color
Graphs> Every complete 4-partite graph is four-colorable. The
Mathworld> descriptions of Complete Graph, Complete k-Partite
Graph,> k-Partite Graph are essentially appropriate to the
following> discussion.The complete graph Kn is of course
n-partite. In Mathworld, the> complete k-partite graph (Ck) is
denoted Kp,q, ... ,r; where p+q+ ...> +r = n. Here, a slightly
different nomenclature is adopted; ie, Ck => (P1,P2,P3, ...
Pk).> To minimize confusion, let Pi represent partition 'i'
and pi represent> the nummer of vertices in partition 'i'. So
we can also write that Ck = {p1, p2, p3, ... , pk}Then, C4 =
{p1, p2, p3, p4}. The number of edges (Ec4) in C4 is, Ec4 =
p1*(p2+p3+p4) + P2*(p3+p4) + p3*p4 > For example, let p1 = p2
= p3 = p4 = n/4; then Ec4 = n/4*(n/4+n/4+n/4) + n/4*(n/4+n/4)
+ n/4*n/4 => [6*(n/4*n/4)}> Ec4 = .375*(n^2)Let n = 16, then
pi = 4; and Ec4 = .375*(16*16) = 96. To check> Ec4 = 4*(4+4+4)
+ 4*(4+4) + 4*4 = 48 + 32 + 16 = 96.If we let Ec4 =
INT(0.375*(n^2)), then the formula works for all> values of n
when p1 ~ p2 ~ p3 ~ p4. For example; for n = 14, C4 = {4,> 4,
3, 3}. Without wrting out the calculations in detail; Ec4 =
4*10 + 4*6 + 9 = 40 + 24 + 9 = 73> Ec4 = INT(.375*(14^14)) =
INT(.375*(196)) = 73At the other end of the spectrum, let p1=
(n-3), p2 = p3 = p4 = 1. > Then Ec4 = (n-3)*(1+1+1) + 1*(1+1)
+ 1*1 = 3*n -6 > > We propose that 3*n-6 <= Ec4 <= 0.375*(n^2)
for all n and all possible> partitionings thereof.Let n =8. The
following partitionings are possible; {2,2,2,2),{3.2.2,1},
{3,3,1,1}. {4,2,1,1} & {5,1,1,1}. Howwever, no planargraph may
have more than INT(n/2) vertices in any one partition. Sothe
partitioning {5,1,1,1} is not applicable. Then Partitioning
max edges max diagonals {2,2,2,2} 24 16 {3,2,2,1} 23 15
{3,3,1,1} 22 14 {4,2,1,1} 21 13Note that any planar graph may
be depicted as an n-sided polygon withup to (2*n-6) diagonals.
If n = 8, then 2n-6 = 10. Therefore, thepartitioning {2,2,2,2)
can be reduced to Binomial(16,10) = 8,008different 4-color
graphs with (3n-6)= 18 edges. But there are only2,772 planar
graphs with8 vertices and 18 edges. (see HYPOTHESIS below) The
total number ofconfigurations due to all four partitionings is
12,298. It isimpossible to determine the number of duplicate
configurations.Further, there are Binomial(20,16) = 4,845
complete 4-partite graphswith n=8 and {2,2,2,2} partitioning.
So it is conservatively safe tosay that there are > 8,008
4-colorable graphs with 8 vertices and 18edges. Of which, only
2,772 may be planar. HYPOTHESIS, The number of maximal planar
graphs is [C_(n)*C_(n-1)]/2;whereC_(n) and C_(n-1) are the Cat
numbers for (n) and (n-1). Specifically; C_(8) = 132, C_(7) =
42. Afterthought. A possible counter example to the FCT would
be acomplete 4-partite graph with less than 3n-6 edges; if one
existed.As n increases, the ratio of all 4-color graphs to
4-color planargraphs increases rapidly.
===
Subject: JSH:
Deprogramming needed?(Marketing ploy warning)What if indeed I
*am* wrong, and I don't have these great mathdiscoveries?After
all, I've been at this since April 1995 having spent a lot
oftime and effort, with literally thousands of posts along
with allkinds of other activities, websites, and email to
mathematicians allover the world.But, what if I'm wrong?What
if all the time and energy I've invested in my work has made
itdifficult for me to see, along with the *harsh* and
unforgivinghostility from several people who seem to have made
it their missionto make me miserable and find pleasure in
mocking or trying tohumiliate me, have made it extremely
difficult for me to see thetruth?Think about the crushing
sense of shame and misery if indeed I findout that the logical
connections I so carefully and impatientlydiscovered over the
years are simply not really there, but are a needinduced
delusion.It seems to me that marketing ploy though these
statements may be,dealing with people who've made it their
business to try and make memiserable, only to at times claim
they're trying to help me is justtoo much.Aren't there any
*other* people who suppose they are rational, who canfollow a
logical argument, who might comment?Why is it always the same
people? Or people imitating them in hostileand mocking
displays of animosity or anger?Aren't there any rational
people who can trace out the steps in thework I've presented,
who might step forward at this time, anddemonstrate an ability
to just be objective?I don't want any replies of people
offering, but if you might considerit, I just want you to
think about it. Sure I've been reading thismarketing book, but
it seems to me that still *someone* out theremight be the right
person, as of course, I don't believe I'm wrong. I've traced
out every step in the arguments I have, and I think
thatirrational people have been dominating the discussion
using groupeffects to hide the truth.Think about it. I'll come
back to the subject later.James
Harrishttp://mathforprofit.blogspot.com/
===
Subject: Re: JSH:
Deprogramming needed?> (Marketing ploy warning)What if indeed
I *am* wrong, and I don't have these great math>
discoveries?We get more random numbers (from your post):
ed6af0c92fa24f0810455fd94f4ccdaaDavid Bernier
===
Subject: Re:
Deprogramming needed?> (Marketing ploy warning)> What if
indeed I *am* wrong, and I don't have these great math>
discoveries?<<<Snipped The tear Jerker>James Harris>
http://mathforprofit.blogspot.com/What If? That question on
its face causes one toconsider the state of mind of the
speaker... Afteryears of people specifically detailing your
mentalconstructs and pointing out your errors in logic
youstill have a doubt of your pass or fail status in life....
R.
Mays----------------------------------------------------------
-------------------Some where within the Quantum
StateHttp://.Mays.Com/story.hhttp://.mays.com/mayday.hhttp://.
mays.com/rainy.hOf all the systems of religion that ever were
invented, there is no more derogatory to theAlmighty, more
unedifying to man, more repugnant to reason, and more
contradictory to itself than this thing called
Christianity.Too absurd for belief, too impossible to
convince, and too inconsistent for practice, renders the heart
torpid or produces only atheists or fanatics. As an engine of
power, it serves the purpose ofdespotism, and as a means of
wealth, the avariceof priests, but so far as respects the good
of man ingeneral it leads to nothing here or hereafter. [
Paine, The Age of Reason]
===
Subject: Re: Deprogramming
needed?> (Marketing ploy warning)> What if indeed I *am*
wrong, and I don't have these great math> discoveries?> After
all, I've been at this since April 1995 having spent a lot of>
time and effort, with literally thousands of posts along with
all> kinds of other activities, websites, and email to
mathematicians all> over the world.> But, what if I'm wrong?>
What if all the time and energy I've invested in my work has
made it> difficult for me to see, along with the *harsh* and
unforgiving> hostility from several people who seem to have
made it their mission> to make me miserable and find pleasure
in mocking or trying to> humiliate me, have made it extremely
difficult for me to see the> truth?> Think about the crushing
sense of shame and misery if indeed I find> out that the
logical connections I so carefully and impatiently> discovered
over the years are simply not really there, but are a need>
induced delusion.So if you don't get your way, you're going to
cry?
===
Subject: Re: JSH: Deprogramming needed?(Marketing ploy
warning)What if indeed I *am* wrong, and I don't have these
great math> math discoveries?Then you would be empirically
sane. No chance (note your insertedif despite an
extraordinarily large body of trivial disproofsubmitted by
more than a score of posters),http://www.crank.net/harris.h
It's not every braying jackass that gets a whole page at
crank.net-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic
URL! Unsafe for children and most mammals)Quis custodiet ipsos
custodes? The Net!
===
Subject: Re: JSH: Deprogramming needed? (Marketing ploy warning)> > What if indeed I *am* wrong,
and I don't have these great math> math discoveries?>Then you
would be empirically sane. No chance (note your inserted>if
despite an extraordinarily large body of trivial
disproof>submitted by more than a score of
posters),>http://www.crank.net/harris.h> It's not every
braying jackass that gets a whole page at crank.netAh, but
wouldn't he be likely to take that as evidence that they
arepaying great attention to his ideas because of the ideas'
great merit(rather than great crankiness)? How does one get
the considered attentionof a megalomaniac anyway?--
---------------------------| BBB b Barbara at LivingHistory
stop co stop uk| B B aa rrr b || BBB a a r bbb | | B B a a r b
b | | BBB aa a r bbb | -----------------------------
===
Subject:
[JSH] Crank Net!: Re: Deprogramming
needed?http://www.crank.net/harris.h> (Marketing ploy
warning)http://www.crank.net/harris.h> What if indeed I *am*
wrong, and I don't have these great math>
discoveries?http://www.crank.net/harris.h> Think about it.
I'll come back to the subject
later.http://www.crank.net/harris.h> James
Harrishttp://www.crank.net/harris.h
===
Subject: Re:
Deprogramming needed?> (Marketing ploy warning)> What if
indeed I *am* wrong, and I don't have these great math>
discoveries?> After all, I've been at this since April 1995
having spent a lot of> time and effort, with literally
thousands of posts along with all> kinds of other activities,
websites, and email to mathematicians all> over the world.>
But, what if I'm wrong?> What if all the time and energy I've
invested in my work has made it> difficult for me to see,
along with the *harsh* and unforgiving> hostility from several
people who seem to have made it their mission> to make me
miserable and find pleasure in mocking or trying to> humiliate
me, have made it extremely difficult for me to see the> truth?>
Think about the crushing sense of shame and misery if indeed I
find> out that the logical connections I so carefully and
impatiently> discovered over the years are simply not really
there, but are a need> induced delusion.> It seems to me that
marketing ploy though these statements may be,> dealing with
people who've made it their business to try and make me>
miserable, only to at times claim they're trying to help me is
just> too much.> Aren't there any *other* people who suppose
they are rational, who can> follow a logical argument, who
might comment?> Why is it always the same people? Or people
imitating them in hostile> and mocking displays of animosity
or anger?> Aren't there any rational people who can trace out
the steps in the> work I've presented, who might step forward
at this time, and> demonstrate an ability to just be
objective?> I don't want any replies of people offering, but
if you might consider> it, I just want you to think about it.
Sure I've been reading this> marketing book, but it seems to
me that still *someone* out there> might be the right person,
as of course, I don't believe I'm wrong.> I've traced out
every step in the arguments I have, and I think that>
irrational people have been dominating the discussion using
group> effects to hide the truth.> Think about it. I'll come
back to the subject later.> James Harris>
http://mathforprofit.blogspot.com/Does your definition of
objective mean: agreeing with you? I think thesepeople are
being objective.Lurch
===
Subject: Re: Deprogramming needed? :
Does your definition of objective mean: agreeing with you?No,
unless it is about matters of objective fact. : I think these
: people are being objective.Well, they aren't. There is no
objective answer to the questionof whether anybody does or
doesn't deserve the sort of abusethat JSH has been getting
from some of these people, and moreto the point, given that
everybody participating in the discussionis doing so from
mindsets that SHARE a certain kind of academic-Americanness,
it IS objectively knowable that that kind of abuse,EVEN it is
deserved, is counter-productive. JSH arrives herein 2 kinds of
error. One involves a series of small errors inthe
factorizations, in the computations, in the alleged
possibilitiesaround the reasoning steps. The other involves
idiotic over-generalizations about the character of
mathematicians generally.Tragically, some of the people in the
discussion are reacting tohim in ways that tend to make them
individual supporting examplesof the insupportable
generalization. If everybody else could emulateDik Winter and
confine their criticisms to the math, this would allgo away
very quickly.
===
Subject: Re: JSH: Deprogramming needed?>
(Marketing ploy warning)How long does it take your Marketing
ploy warning to become a threat?--There are two things you
must never attempt to prove: the unprovable --and the
obvious.----http://www.crbond.com
===
Subject: Re: JSH:
Deprogramming needed?James Harris <jstevh@msn.com> grava .88
la saucisse et au marteau:> Aren't there any *other* people
who suppose they are rational, who can> follow a logical
argument, who might comment?I don't know the french equivalent
for algebraic integers. If someonetells me what set it is (is
it the positive and negative integers?), Imay have the courage
to take a look. But anyhow, your proof is hard toread. You
should at least follow the indications the others have
givenyou (clearly define your functions, explicitly when you
can and so on).> Why is it always the same people? Or people
imitating them in hostile> and mocking displays of animosity
or anger?Because I suppose very few people have had the
courage to read yourproof. And the one who didn't saw the
former(s) commenting it andpointing out some problems you
didn't solve (you keep claiming they'rewrong, but your
explications are more confuse than their). I've knownbetter
motivations.--
===
Subject: Re: JSH: Deprogramming needed?>
James Harris <jstevh@msn.com> grava .88 la saucisse et au
marteau: Aren't there any *other* people who suppose they are
rational, who can> follow a logical argument, who might
comment?I don't know the french equivalent for algebraic
integers. Nombres alg.8ebriques . Ciao, Renaud
Dreyer
===
Subject: Re: JSH: Deprogramming needed?James Harris
<jstevh@msn.com> grava .88 la saucisse et au marteau:> Aren't
there any *other* people who suppose they are rational, who
can> follow a logical argument, who might comment?I don't know
the french equivalent for algebraic integers. Nombres
alg.8ebriques . Ciao,D.8esol.8e, c'est  entiers alg.8ebriques
, of course. Ciao, Renaud Dreyer
===
Subject: Re: JSH:
Deprogramming needed?Renaud Dreyer <rdreyer@math.berkeley.edu>
grava .88 la saucisse et au marteau:> D.8esol.8e, c'est
entiers alg.8ebriques , of course. Ciao,Le probl.8fme, c'est
que j'ai jamais entendu l'expression entieralg.8ebrique en
frn.8dais, c'est bien le probl.8fme. Je connais les
entiersnaturels, les entiers relatifs et les nombres
alg.8ebriqu.8es qui sont lesnombres r.8eels non transcents
(donc pas forc.8ement tr.8fs entiers).Quid?--
===
Subject:
question about canonical commutation relations [P,Q]=i*hbar*II
am a college student who heard this problem from one of my
linearalgebra professors.Question:Find matrices of infinite
size Q and P such that[P,Q]=i*hbar*I, hbar is a constant,I is
the identity matrix, i^2=-1I've tried to prove that there do
not exist nxn complex matrices Q andP such that
[P,Q]=i*hbar*I.Proof. Suppose there exist nxn complex matrices
Q and P such that[P,Q]=i*hbar*I.[P,Q]=i*hbar*I=>
trace([P,Q])=trace(i*hbar*I)trace([P,Q])=trace(P*Q-Q*P)=trace(
P*Q)-trace(Q*P)=0trace(i*hbar*I)=i*n*hbar<>0=>
trace([P,Q])<>trace(i*hbar*I).It's a contradiction.Q.E.D.I've
also tried to find the matrices of infinite size.Let P,Q be
matrices of infinite
size.p[i,1]=1p[i+1,2]=1...p[i+n,n+1]=1...q[1,i]=
-i*hbarq[2,i+1]= -i*hbar...q[n+1,i+n]= -i*hbar...when
i->infinity[P,Q]=P*Q-Q*P=i*hbar*I.Can anybody explain where
the error (if any!) comes into myderivation?Any comments would
be appreciated.
===
Subject: Re: question about canonical
commutation relations [P,Q]=i*hbar*I>Question:>Find matrices
of infinite size Q and P such that>[P,Q]=i*hbar*I, hbar is a
constant,I is the identity matrix, i^2=-1>I've tried to prove
that there do not exist nxn complex matrices Q and>P such that
[P,Q]=i*hbar*I....Your proof is correct.>I've also tried to
find the matrices of infinite size.>Let P,Q be matrices of
infinite size.>p[i,1]=1It's best not to use i here because it
will be confused with sqrt(-1).Note that the product of two
infinite matrices is well defined if they have only finitely
many nonzero entries in each row(there are more general
conditions, but this one is enough foryour case).Hint: I think
you'll want to define P and Q so each has only one nonzeroentry
in each row: one of them only above the main diagonal, the
otheronly below the main diagonal. Department of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: question about
canonical commutation relations [P,Q]=i*hbar*I
<halfsearch@hotmail.com> grava .88 la saucisse et au marteau:>
Find matrices of infinite size Q and P such thatWhat is a
matrice of infinite size? IMO, this is a complete
nonsensesince results on matrices specifically use that we are
in finitedimension.--
===
Subject: Re: question about canonical
commutation relations [P,Q]=i*hbar*I> <halfsearch@hotmail.com>
grava .88 la saucisse et au marteau:> > Find matrices of
infinite size Q and P such thatWhat is a matrice of infinite
size? IMO, this is a complete nonsense> since results on
matrices specifically use that we are in finite>
dimension.Given a finite-dimensional vector space V over a
field F and a basis B, matrices correspond in an easy way to
linear maps V -> V.Now suppose that V is of infinite
dimension. Then linear maps V -> V correspond in the same way
to functions from BxB -> F.Since B is a basis, these functions
have a finiteness propertythat allows us to take inner products
of rows and columns ...
===
Subject: *OT* BBS's Remeber
these??Remember BBS's?? I was looking through some oldBBS
programs last night then decided to check out the internet to
see ifanyone still uses BBS's. I came along http://Synchro.net
(Synchronet) lastnight (some of you may remember it). They went
open source and are nowoffering it free - it uses Telnet... so,
I downloaded and installed it on myCompaq Proliant 2500 Server.
tried it out, woah - the memories... am going to leave it up as
its pretty neat... so come check it out...(don't mind the BBS
name, I was tired). You'll need to have outgoing accessto port
23 (telnet). I even went as far to get it working with
DOVE-Net!!!Maybe we can get a SCI.Math BBS going. lolAnyways
here is the Link:telnet://cyberhood.servehttp.com
Curry
===
Subject: Practical Statistics (Help)Hello
All,Background Info:I'm looking for help on what I precieve as
real world statistics problem.I'm an electrical engineer
currently working on a problem involving theproduction testing
of a wireless network device. We need to come up with atest
that can be performed on every device built. The test must
verify thatthe recieve sensitivity is within specification.
The specification is thatthe device must operate up to 100 m
away from a certain type of transmitterin a certain
environment. Performing this test directly is very
impracticalin a mass production environment. Therefore, I
would like to perform a testsuch that the Device Under Test
(DUT) is only inches from a lower powertransmitter in a
controlled environment (a shielded box). To do this, Imust
determine what power level (P) the lower power tranxmitter
must be setto in order to simulate a normal transmitter 100 m
away. The customer hasrejected all proposals of theoretical
calculations of P and wants us toperform a study to determin
P. My job is to come up with a plan to do this.I would like to
take a sample of known good devices and test them bylowering
transmitter power (P) until they fail. After I collect this
data,I would like to use it to determine what P should be set
to for testing ofthe 1000's of new devices that will be built
every day until the end of theproduct's life.The Math
Questions:What method should I use to determine the sample
size for the study? I havea pretty good educated guess at the
range of values that I will get for P.dev. with high
confidence. I can then use this information to find a good
Pfor my future testing.I know this probably sounds like a
problem from a text book, but it's not.I've been reading
through an old statistics book and I think that I'mstarting to
talk the talk a little, but I still can't walk the walk.
Anysuggestions would be appreaciated.Warm
Wishes,-Andy
===
Subject: Re: Cube (the movie) by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hAA2wVZ22788;>Saw it yesterday.>In
German.>Uhm, did the synchro botch everything up or is it>a
lot of senseless mathbabble already in the original? :-)>--
>Hauke Reddmann <:-EX8 >Private
email:fc3a501@math.uni-hamburg.de>For our chemistry
workgroup,remove math from the address>For spamming, remove
anything elseSaw it too.The math is not completely senseless:
the idea to interpret thenumbers as coordinates is quite
obvious, isn't it?Later on they used the same numbers to
describe the path a subcube follows over time. It's not so
clear how to do that, although it'spossible.Then suddenly the
property of being a prime power played a role inidentifying
those rooms, that are traps. At that point I guess itwas just
nonsense. It seems difficult to me to encode two types
ofinformation into the number triplets shown in the
movie.H
===
Subject: Prove: by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hAA2wUQ22782;If g.c.d (x,y)=1and x divides zand y divides
zthen xy divides z
===
Subject: Re: if x has normally
distributed digits, does 1/x? by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hAA2wmw22889;> >Is there a simple proof, or
counterexample, to the proposition>that if we consider the
expansion of the fractional part of some>(irrational) number
x in some base, and know that it is normal,>then it follows
that the expansion of 1/x is also normal?> > I seriously
doubt it. Here's an indication of why:> > Let's talk about
base 2. Start with x = .01010101... .> Of course x is not
normal, but it does contain the right> number of 0's and
1's; it's weakly normal in base 2, in> a sense. But x = 1/3,
so 1/x = three = 11.000... , which is> far from
normal.>You're in the land of rationals. >Rationals and
normality don't mix.>(Take that out of context, man!)>Well,
that's amusing enough to make it seem possible that you>were
just trying to be funny. In case you were also trying to>make
a serious point:>I certainly didn't mean that this was a
counterexample, or>something that would lead to a
counterexample with no new>ideas required. But what I had in
mind was something>_vaguely_ like this: Suppose we could find
x_n such that>each n-bit sequence in the binary expansion of
x_n occurs>with the right frequency, but 1/x_n has a
terminating>binary expansion. Then let x consist of a long
stretch>of the bits of x_1, followed by a much longer
stretch>of the bits of x_2, etc. For suitable values of
long>and longer x will be normal, but it doesn't seem
so>unlikely that 1/x would turn out to be abnormal.>Phil>--
>Unpatched IE vulnerability: window.open search
injection>Description: cross-domain scripting,
cookie/data/identity > theft, command execution>Reference:
<a
href=http://safecenter.net/liudieyu/WsFakeSrc/
WsFakeSrc-Content.HTM>http://safecenter.net/liudieyu/WsFakeSrc
/WsFakeSrc-Content.HTM</a>Exploit: <a
href=http://safecenter.net/liudieyu/WsFakeSrc/
WsFakeSrc-MyPage.htm>http://safecenter.net/liudieyu/WsFakeSrc/
WsFakeSrc-MyPage.htm</aSince most numbers are normal, I think
that the best approach is tostart with an abnormal irrational
number. Its reciprocal will mostlikely be normal and thus be
the example that you are looking for.My guess is that if you
clear to zero every other digit of Pi andthen take the
reciprocal of the result, you will get a normal
number.
===
Subject: Mathematical Integrity by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hAA2wuo22943;I am a math student
and I love math more than anything else. Istarted to love math
when I found out that I wasn't good at it andstarted studying
it for over a year now. I love math for its purityand
integrity. However, lately, I surfed through several math
forums and I found many unpleasant posts. Sometimes a student
would post an innocent question but he or she would get scold
for posting such a stupid question and would be called a
moron. And sometimes a math fanatic would post something of
interest, but other mathematicians would scold that person for
posting something so ridiculous. It seems like these dese days,
some mathematicians would use profanity against others
mathematicians. Why? Where is the purity and integrity in
mathematics? I know that after posting this, many other great
mathematicians would scold me, but that's what I've seen so
far. Anyhow, I still love math by heart and I would like to
give my respect to many great math fanatics in this
forum.
===
Subject: Re: Algebraic integers issue, quick example
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hAA2wqg22928;>It tutns out there's
a quick and basic argument which conclusively>proves that
there are numbers left out by the definition of
algebraic>integers.>Consider>(c_1 x + 7)(c_2 x + 7)( c_3 x +
2) = > > 49(x^3 + 5x^2 + 3x + 2)>where you might wonder what
the c's are, and that's what is covered in>this post.>Now
then, you have as a zero of the factorization x = -7/c_1, so
let>x= -7/c_1, so you have>49(-7^3/c_1^3 + 5(49)/c_1^2 -
21/c_1 + 2) = 0>which is>2c_1^3 - 21 c_1^2 + 245 c_1 - 343 =
0.>But that is a non-monic primitive irreducible over Q, so
c_1 and by>symmetry c_2 cannot be algebraic integers. However
they must be>algebraic numbers, and it can be shown that any
algebraic number can>be written as the ratio of algebraic
integers.Yep, so far.>So then there must exist f, such that
fc_1 is an algebraic integer,>and letting g = fc_1 and
multiplying both sides by f, I have>(gx + 7f)(c_2 x + 7)( c_3
x + 2) = > > 49f(x^3 + 5x^2 + 3x + 2)>so a zero is now x =
-7f/g, which gives>49f(-7^3f^3/g^3 + 5(49)f^2/g^2 - 21f/g + 2)
= 0>which is>2g^3 - 21f g^2 + 245f^2 g - 343 f^3 = 0>which
proves that f must have 2 itself as a factor for g to be
an>algebraic integer.>For instance, letting f=2, gives>g^3 -
21 g^2 + 245(2) g - 343 (4) = 0.>But looking back again
at>(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x
+ 2)>that would mean that c_3 has a factor that is 2, which can
distribute>to c_1 x + 7, but that leaves c_2 x + 7, which also
needs a factor of>2 by symmetry.>But you see, you only have
just that one 2.>So c_1 and c_2 while not algebraic integers
cannot be written as a>ratio of non-unit coprime algebraic
integers either.Well as you say clearly there is a
factorization of the formyou want where c_1, c_2 and c_3 are
algebraic NUMBERS.And every algebraic number can be written as
the ratioof an algebraic integer and an ordinary integer, ie,
c_1 = A/qfor some a.i. A and integer q. And there is a great
big honker of a theorem which says any two nonzeroalgebraic
integers have a GCD which is an a.i. also.Therefore cancelling
out the GCD of A and q, you can write c_1 = A'/q',where A' and
q' are coprime a.i.'s.So you're wrong.>These numbers represent
an unexplored frontier of mathematics, and an>opportunity for
many of you to make an impact like you never could>have
dreamed of before in the field.>It's fresh ground. Fresh
manure, more like.B.P.>James Harris><a
href=http://mathforprofit.blogspot.com/>http://
mathforprofit.blogspot.com/</a>
===
Subject: Squares that end
with four identical digits by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hAA2woZ22908; Which integers can have squares that end with
four identical digits?
===
Subject: Re: please tell me
primes100-200 as fast as possible by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hAA2wwc22953;><preCan you please tell me the prime numbers
between 100-200 as soon as></pre>
===
Subject: Re: Who
contributed most to mathematics? by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hAA2weo22836;>I was just trying to see who in fact borned the
most mathematicians. >My SWAG list was not intended to be
complete but just a start. Ok, I >blew it after France at
least according to MacTutor link below. >However I was close
with Italy, Germany and former Russia but totally
>underestimated England and the U.S. and way over estimated
India and>Greece. This is a quantative analysis. The next
step is qualitative>analysis which is why I asked you guys
the question.>I doubt that anybody would claim that being in
the MacTutor list >is all that significant as an indication of
mathematical importance: >it's mainly a question of which
people somebody has bothered to write>a biography of. Is this
generally true, ie., of other disciplines -
physics,medicine,literature etc? What other list would do it.
E.T Bell's Men of Mathematics, Morris Kline's Mathematical
Thought from Ancient to Modern Times etc, Biography Top 100
Men of the Millennium? As I recall Guttenburg got top honors
and Newton second. Guttenburg gotit because his printing press
got the word out to the masses. I question Newton as second.
Had Fermat published he may have got thecalculus in his
resume. >There sure have been a lot of important French
mathematicians;>Fourier, Galois, d'Alembert...>Fermat - but
his last theorem was published after his death. >So during his
life time it did not play a major role in
mathematical>developement.>Fermat's importance in mathematics
does not rest on his Last Theorem.>He made many contributions
in various areas. Although he didn't publish>much, he
corresponded with many important mathematicians, so he
was>quite influential in his lifetime.Did these other
Mathematicians publish much? I doubt it. > The same can be
said of Da Vinci and other Greats whose>certain works were
suppressed during their lifetime.>Although da Vinci did study
a lot of mathematics, I'm not aware of>any really significant
contributions he made to mathematics. No because he did not
publish.>but the Greeks started it.>Just about every thing
the Greeks did was rediscovered by others>as it was needed in
their time. Much of Archemedies work was for War >efforts as
they were needed. Sure they have being first name
>recognition but played no more role than India and other
countries >to thought today.>This is just silliness. The main
contribution of the Greeks was>to discover the idea of
mathematical proof. Nearly all Western >mathematicians until
very recent times got their first taste ofYou really believe
the infinity of primes by contradictionnecessitated Euclid? Or
the pythagorean theorem? How many proofsare out there
now.>mathematics from studying Euclid.Come on. Euclid did not
invent deductive reasoning - the syllogism -A pair of
warranties and a consecutant, the warranties statingcertain
things and the consecutant following of necessity from
themEuclid was a compiler,structurer,sequencer,denotator and
doder on that which comes next cannot exist a priori. Because
of this, Euclidgrasped by the masses who passed it on years
later by others throughthe best recording device known at the
time - human memory. Certainly,methods of higher order such as
analysis today were know but made noheadway untill much later
because it was not passed on theough humanmemory. Indeed, when
the great knower died so did the idea. Eg., skiping several
hundred years, the proof that this margin is to narrow to
contain.Rediscovery is not a silly issue. How many
re-discoveries have youyou made? I would guess many. As youths
(16) we played with the 5 cent 25 hole pinball machines in the
pool room. These paid off 5 cents per game that you won. When
you decided to cash in your games which of course was a rare
event,we called the clerk Gillie,who had bad eyesight, to come
over andverify the counter,flip the reset switch and pay us the
money which of course went right back into the machine. When we
won, Gillie would come over, squint, and look at the counter.
Then he would tellus to flip the switch. In doing so we
noticed that the switch couldbe toggled back and forth quickly
thus stoping the won game count down. So if we had 100 games,
we could quickly flip the switch on and off to stop the
counter at 99. Then 98 etc. Here in came the problem. If we
could t Gillie to cashing in 100 game,99 games,98etc., How
much money could we win? That evening I sat with pencil
andpaper and figured it out by examining a smaller set of
numbers andgeneralizied it to 100(100+1)/2*.05 = $247.5. This
is the summationformula for an arithmetic progression that I
rediscovered when the need arised.Around the same time I
discovered if the sum pf the digits of a number is divisible
by 3 the number is divisible by 3. SimilarlyI found the rule
for 9 and 11. 7 and other numbers defied me butI tried.Before
that I was looking at some numbers and wondered how many ways
you can arrange them. Numbers1,2 = (1,2),(2,1) 2 ways1,2,3 =
(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1) = 6
ways1,2,3,4 = blah blah = 24 ways1,2,3,4,5, blah,blah yeah I
took it out to this level = 120 waysLo and behold, the numbers
2,6,24,120 have a pattern! Yes,2 = 1*26 = 2*324 = 4*6120=
5*24so 120 = 5*24= 5*4*6 = 5*4*3*2 = 5*4*3*2*1. That's it! The
number ofarrangements of n things is n*(n-1)*(n-2)*...1) Before
that, fooling around with differences between two squaresI came
up with x^2-y^2 = (x-y)(x+y)Later in the 60's fooling around
with FLT, I pounced upon this.I a number can be expressd as
the difference between 2 squaresin more than 1 way the number
cannot be prime. Indeed, every oddnumber can expressd as the
difference between 2 squares in a trivialway. Eg., 17 =
9^2-8^2, 9 = 5^2-4^2, 2k+1 = (k+1)^2-k^2 are trivialsolutions.
Since x^2-y^2 = ab a>b > 1 it follows that from my
earlierdiscovery (Dog gone it it happened again. I had the
science channel on and the announcer said discovery as I typed
it above)n = ab = x^2-y^2 = (x-y)(x+y)we havex-y = b x+y = aas
factor of n.This is a non-trivial difference between 2
squares. Thus n is not prime. Around the same time I realized
empirically that for prime px^(p-1)/2 +- y^(p-1)/2 is
divisible by p. This is sufficient butnot necessary p be prime
since x=10,y=1, p=9 we have 10^4 - 1^4 = 9999 which is
divisible by p=9. This can be expanded by my diff 2 squares
idea to x^(p-1) - y^(p-1) is div by p if p is prime.Around the
same time I found an infinity of solutions to the righttriangle
with hypotenuse of the form 4n+1There is a lot more but no need
to go on. I will summarize this asfollows.Ask your self or a
student a question and chances are you or thestudent will make
a discovery or rediscovery with the answer.Necessity is the
mother of invention.Find the indespensible man and fire himIn
fact we may have been better off with out Euclid because
analogous to strict church domination, a great deal of
Euclid's rigour stiffledcreativity and forward thinking as the
Pythagoreans did in theirtime. Or another view is WWIII would
have already come and gone hadcreativity not been tamed by
bigshots in power.CinoObservation lends to analysis and
analysis lends to law
===
Subject: Re: Greek Alphebet by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hAA2wgv22865;><pre> > In my
studies on triginometry, and calculus, I am comming across
many> symbols which appear to be the Greek Alphebet, as
there was a chart of> the Alphebet in the front of the book.
I am wondering if someone could>Do the Greeks use Latin and
Greek letters in a similar way to us LatinWithin Greece ,
books , education etc. the Greek Alphabetis used( for
Geometry).To mention also,that there are, some other letters
that are in latin,but are not used in the Greek Alphabet any
more like F (called Digama -double gama),and as far as I,
remember the letter J (calledyiot.It is worth pointing ,that
(according to Britannica under ALPHABET)the Alphabet was a
Greek invention based upon North Semitic (proto-Canaanitish)
writing which indicated only consonants,a procedure suitable
enough for a Semitic language but not for an Indo-European
one.Panagiotis Stefanideshttp://www.stefanides.gr
===
Subject:
Re: ellipse circumference approximation by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hAA3OXr25023;For a proof for an
exact equation for the circumference of an ellipse, check out
my recently published book entitled: Circular Elliptics at:
ww.Trafford.com
===
Subject: Re: ellipse circumference
approximation> For a proof for an exact equation for the
circumference of an ellipse,> check out my recently published
book entitled: Circular Elliptics at:> ww.Trafford.comHmm. The
title of this thread mentions approximation. But it seems
thatyou are claiming to have an _exact_ expression for the
perimeter. Ofcourse, such expressions are well known, but they
involve an ellipticintegral (or something equivalent). The
perimeter cannot be expressed inclosed form in terms of
elementary functions.Are you, perchance, claiming to have done
what I just said can't be done?If not, then briefly, what's
interesting about your work?BTW, for anyone interested,
see<http://www.trafford.com/4dcgi/view-item?item=2270&
315194254-5408aaa>.It says that the book Relates the ellipse
to the circle in ways you neverdreamed of, offers a proof for
an equation for the circumference...Price: US$9.80l
===
Subject:
Re: TI-89: wrong answer for integration! by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hAAI8fq19654;Be carful when you
type the equation in.You have to say a*x in the exponent else
the calculator thinks ax is a single variable.
===
Subject: Re:
Oneness of a number by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hAAI8W719627;>if(n is even) n = n/2;>if(n is odd) n = 3*n +
1;>Keep doing this until n = 1. The amount of necessary steps
for this is>called the oneness of n.Not sure what you mean
about 5 not going to 1one(5) = 16one(16) = 8one(8) = 4one(4) =
2one(2) = 1
===
Subject: Number Of Squares Within A Square (nxn)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hAAI8T719617;For my homework we had
been set the question of finding the numbers of squares within
a square.I managed to work out an answer for myself but out of
interest found your Website and a really neat solution that had
been posted sometime ago.However, I have some problems with the
proof.Please could you explain how to get the
Sum(n^2)=(n+1)n(2n+1)/6I'd be really grateful for your
help.
===
Subject: Re: Marvellous developments of FLT by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hAALClW01298; Hi to all, To the
moment, once my procedures could becomment with better english
You should bepatient again. I can present now tiny part with
myparameters for n=3: Consider Z and X as odd numbers and
Z=(p+q)(r-s) X=(p-q)(r+s) where p;q;r;s natural numbers then
once Z;X are of gcd=1 so p;q of gcd=1 and r;s of gcd=1but also
we'll take (p+q);(r-s) of gcd=1 and (p-q);(r+s) of gcd=1but
then Z-X = 2(qr-ps) will be of gcd=1 to Z+X = 2(pr-qs) except
of 2from these (qr-ps);(pr-qs) of gcd=1and finally p;q;r;s of
gcd=1 Now ordering similar Z^3=(P+Q)(R-S) X^3=(P-Q)(R+S)where
P;Q;R;S can be expressed withspecial formula by p;q;r;s
parameters we'll eventually develop inconsistency...
Ro
===
Subject: Re: A 1st semester Calc proof that e^(i*t) =
cis(t) by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id hAALChi01289;Ok, the real
log function can be defined as the definite integral of 1/x.
The real exponential can then be defined as itsinverse. sin
and cos can be defined as the parameterization of the unit
circle in cartesian coordinates. e(z) for z complex is can
then be defined as e(z+iy) = e(z)(cos(y) + isin(y).
Admittedly, this takes out all the wonder from Euler's
formula, but its logically sound and I have seen it done in
some texts.I assume, then, that everything required can be
derived from these defintions, which only require the integral
but absolutely no series.
===
Subject: Re: Algorithm for square
root? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hAALRxx02332;><pre>
<1998051218145200.OAA19142@ladder03.news.aol.com>...> Does
anyone know of a site with the algorithm> for finding the
square root of a positive number?><a
href=http://www.worldserver.com/turk/ComputerGraphics/
FixedSqrt.pdf>http://www.worldserver.com/turk/ComputerGraphics
/FixedSqrt.pdf</adescribes a binary square root extraction
mechanism. The decimal version>is similar, but requires a bit
more guesswork, as does long division.></pre>
===
Subject: two
new math fora to announce by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hAANMCA11419;http://www.math.utexas.edu/mailman/listinfo/
math-talk-- started primarily for discussions at UT-Austin,
but anyone can
subscribehttp://forums.austin.craigslist.org/?forumID=1394--
Craigslist easter egg site could be really cool if enough
people get into it. Similar niche to this site, maybe. A
different feel, though. Best wishes,Joe
===
Subject: Godel's
universeI'am looking for literature concerning Godel's
Rotating Universe.Suggestions?KB
===
Subject: two additional
math fora to announce by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hAANMBb11415;Started this primarily for discussions at
UT-Austin, but anyone can
subscribe.http://www.math.utexas.edu/mailman/listinfo/
math-talk/A Craigslist easteregg. This could be really cool...
check it
out.http://forums.austin.craigslist.org/?forumID=1394
===
Subject: Re: Why Derivative is Inverse to Integral; geometric
explanation formath education by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hAB06uH15289;> just this Archimedes Plutonium meant when said
of a slope of one of> rectangle sides. To simplify a trapezium
to a rectangle is simpler> than to resume the lost information
of the slope. So in most cases we> are unable to integrate.
The same, we can easily expand a function> into Fourier
series, but to reconstruct the original by the Fourier>
series - it's practically unsolvable problem. Or, if a
function has>better than picketfence.>Can you think of a term
for a rectangle that has an end-sides of just a>mere point
rather than a true rectangle whose four sides are more
than>just a point?>In the derivative there are no strange
objects but in the integral there>is this>strange object of a
rectangle whose end-sides are mere one point.>So in
differentiation, there appears to be no real strange objects
for a>set of trapezoids is normal geometric objects but in
integration we have>this strange set of objects of rectangles
whose end sides are one point.>And thus, I see geometrically
the inverse relationship between derivative>and integral as
that between trapezoids to one-point-sided rectangles.
When>differentiating one makes trapezoids and when one
integrates they collapse>the trapezoids into these strange
rectangles.>Sergey, can you comment on the issue that
derivative has normal geometric>objects of trapezoids (what I
call picketfences) yet the integral relies>upon these>strange
abnormal geometric objects of a collapsed rectangle whose
end-sides>consist of one mere point.>i think trapezoids are
stupid!!!! >So can we say that the essence of the Inverse
relation between>derivative and>integral is the geometric idea
that one is to expand strange rectangles into>trapezoids and
the other is to collapse trapezoids into one-point end
sided>rectangles.>P.S. I do not know if I should bother with a
geometric explanation of the>2nd derivative that is
acceleration in physics.>Archimedes Plutonium,
a_plutonium@hotmail.com>whole entire Universe is just one big
atom where dots>of the electron-dot-cloud are
galaxies
===
Subject: Re: Squares that end with four identical
digits by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id hAB2Ehe24376;> Which
integers can have squares that end with four identical
digits?I believe that all integers that end in two zeroes have
squares that end in four identical digits.
===
Subject: Re:
Trying again!!! PDE book recommendation by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hAB2EhG24388;Charlie,If you're
coming from the engineering perspective of PDE's, I'd
recommend Partial Differential Equations for Scientists and
Engineers by Stanley Farlow.Eric Ritchson>Hi all,>Could
someone recommend a good Introduction to P.D.E.'s book ? I
have heard>that>Basic Partial Differential equations by David
Bleecker, George Csordas, and>Darko Grundler book is nice, but
I haven't had the chance to take a look at>it. Any suggestions
would be appreciated!>TIA>Lurch
===
Subject: Re: {Group Theory}
Confusing group theory conundrum by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hAB2Ehv24380;>questions.>OK, every group has an identity
element, right? True by definition>It seems to me that, given
a group, that group therefor determines a>unique identity
element. (The uniqueness of a given group's identity>element
is easily proved)>Therefor, it seems to me that there exists a
function (indeed, a>function which algebraists subtly draw upon
without even realizing it)>that associates with each group it's
identity element.That really depends on how you interpret the
word function. This is not a function in the normal set
theoretic sense.>For instance, an algebraist says: We have a
group G. Then it must>have the identity element G_e. But by
saying this, she has>implicitly used such a function.Not
really. For instance, consider this analog: We have a set S
which is nonempty. Then it must have some member S_x. This is
true by definition, but this does not imply the existence of a
function mapping sets to their elements. In fact, even if you
restrict your domain so that it is well defined, you still
can't prove the existence of this function (in the general
case) without invoking the axiom of choice.>But I am told that
you cannot have a set that contains every group... >so this
function's domain is nonsense! In fact the whole function>is
nonsense!Not nonesense, just not a function in the usual
set-theoretic sense.>Now before you reply that the identity
element isn't unique because a>set can be a group under many
diverse different operations, I already>am taking this into
consideration. When the algebraist says Then it>must have the
identity element G_e, the operation is either obvious>from
context or else has already been specified. So really
the>elusive function I am talking about is one which takes as
input a>group-operation couplet and then outputs the identity.
But this is>all a lot more opaque, and though necessary for
rigour, not necessary>for understanding.>So it seems very much
like such a function exists, and yet-- it>cannot. Is it some
hyperfunction that transcends set theory and is>taught at the
postgraduate level? I am very confused and very much>eager for
you to shed some of your very, very highly esteemed wisdom>on
the subject.>Sniz PilborCron
===
Subject: Re: How do you prove
that the circumference of a circle is proportional to it's
radius? by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id hAB76AB12633;> > How do
you prove that the circumference of a circle is proportional
to > it's radius?> What's the modern version and how did
the greeks prove it?> > The approximations to that ratio
are based on > inscribing/circumscribing regular polygons,
each of which may be > partitioned into congruent isosceles
triangles.> > For any such polygon, the
odd(circuferential) side of any of those > isosceles
triangles is proportional to the other (radial) sides by >
similar triangles.> > Thus each approximate circumference
is proportional to its radius.>Now if you know/accept that
>perimeter of inscribed polygon > < circumference of circle
> < perimeter of circumscribed polygon, >then it's easy to
complete the proof. Now the first inequality, > perimeter of
inscribed polygon < circumference of circle, >is clear; a
straight line is the shortest distance between two points.
>The other inequality, > circumference of circle < perimeter
of circumscribed polygon, >is certainly plausible, but is it
actually possible to prove it >using only tools available to
the ancients?>Well, I don't know what Euclid's _definition_ of
the circumference>of a circle was, but if he had one I can't
imagine that it was not>something essentially equivalent to
what we would call the least>upper bound of the perimeter of
an inscribed polygon. Assuming>that, then first we need to
show that >(*) perimeter of inscribed polygon > < perimeter of
circumscribed polygon;>that shows that the least upper bound
mentioned above>_exists_, and also makes the other
inequlalities clear.>(Regardless of how much of this they were
explicit about,>I do tend to suspect that they would have said
the question>was the same as (*), for some reason.)>The
circumcribed polygon is the union of triangles with
height>equal to the radius of the circle and base one of the
sides of>the polygon, so the _area_ of the circumcribed
polygon is>equal to r times its circumference.>Otoh, let r' <
r. Note that adding points to the inscribed>polygon increases
the circumference, by the triangle inequality.>So starting
with an inscribed polygon we can find another>with larger
circumference, and with area > r' * perimeter.>Since the
relation between the areas is clear, it follows>that r' *
inscribed perimeter < r * circumscribed perimeter>for all r' <
r, hence (*).>Euclid did not discuss the length of curved
lines, or the areaof curved surfaces, apparently because he
had no satisfactory definition of such things. I believe
Archimedes was the first.He proposed several postulates, from
which he could concludethat any convex curve is longer than an
inscribed polygon,and shorter than a circumscribing polygon.
Effectively, he had adefinition for the length of convex
curves, and similarly forconvex surfaces. To these he applied
the method of exhaustion. Kleinhans
===
Subject: Re: How do you
prove that the circumference of a circle is proportional to
it's radius? by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id hABKvka05954;> How do you
prove that the circumference of a circle is proportional to >
it's radius?Please refer
to:http://mathforum.org/discuss/sci.math/a/m/107260/
107288andhttp://mathforum.org/discuss/sci.math/a/m/107260/
107289Panagiotis Stefanideshttp://www.stefanides.gr
===
Subject:
Re: Basic factorization ideas by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hABNIYZ16695;>If you saw>(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) =
> > 49(x^3 + 5x^2 + 3x + 1)>with the c's algebraic integers, I
think few of you would have a>problem realizing that only two
of the c's have 7 as a factor. Yep, this polynomial can be
factored as you sayand yep again, if it is factored that way
then c_1 andc_2 are divisible by 7. No prob with that.>But, of
course, you're looking at *functions* of x, as you have >f_1(x)
= c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, >so I could also
write it as>(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5
x^2 + 3x + 1).>Notice that dividing both sides by 49 gives
>(f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x
+ 1>as long as you're in a ring where 7 is not a factor of 22.
What??? Where'd the 22 come from ???>I want to emphasize that
point as notice there's only *one* way to>divide through by 49
if 7 is not a factor of 22. You've flipped out! There is no 22
in sight in what you said above.>Usually you can *see* the
other factors of 7, but I want you to>abstract, and
generalize.>Please pay careful attention to that example. I'm
a-tryin', a-generalizin', and a-abstractin'. I'm
a-payin'careful attention! But I still don't *see* no 22 !>You
may see people who reply claiming that the word polynomial
has>some significance, as if it's a mystical thing which
refutes basic>logic, so if something isn't polynomial it no
longer behaves>logically. Yep, the logic is a problem here.
Like, where do you get22 out of the stuff at the top?>Now
then, in my advanced factorization work, I just use functions
of x>that are a lot more complicated than f_1(x) = c_1 x, and
unfortunately>there are people who can use an unfamiliar leap
in complexity to>confuse others. You mean us, or you mean
you?>Some of you have learned various advanced math topics,
now imagine if>in your classrooms there were some hecklers who
continually hollered>out at your teacher, or otherwise
disrupted the class? If he spouted wrong math for months or
years on end he would deserve it.>What if when there were
difficult concepts those hecklers would try to>confuse
everyone as they sought to discredit the mathematics?>If you
find that hard to imagine, imagine me in your class with
you>questioning the professor and calling him names.>How much
would you have learned?you out of the class to learn anything
at all.B.P.>I need those of you interested in mathematics to
focus on the basics,>so that you can understand the
advanced.>James Harris><a
href=http://mathforprofit.blogspot.com/>http://
mathforprofit.blogspot.com/</a>
===
Subject: Re: out-of-kilter
algorithm by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id hAC2Vns31064;Try
http://citeseer.nj.nec.comand google search with keyword 'out
of kilter' or 'network flow'>Hello!>I was just wondering if
there is someone who could help me in finding some>usefull
site with explanation of out-of-kilter algorithm and examples
of>problems solved with this algorithm.>Thank you in
advance>Mimmy
===
Subject: Re: differentiable...problem... by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hAC2VoM31077;>if f is
differentialbe on (0, infinite)>and lim [f(x) +f'(x)] = L
(x->infinite)>show that lim f(x) = L (x->infinite) and lim
f'(x) = 0 (x->infinite)This seems incorrect as stated. What
about f(x)=e^(-x)? It's differentaible,
and:f(x)+f'(x)=e^(-x)+(-e^(-x)) = 0So:lim f(x)+f'(x) = 0 (x ->
inf)>------------------------------------>it seems to be
trivial. but i no
touch.....oops........>help.....me......please......
===
Subject
: Re: Newsgroup survey: Math and personality assessment by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hAC2VoS31086;Hey James, you forgot
a question.8. Is JSH a f&$#wit that gives eveyone else the
s#its?Your mathematical legacy will be a footnote recognising
you as a prizecrackpot, you complete and utter social
misfit.It'd be great if the first (and last) reply to any of
your futureposts was always the same, namely,JSH is crackpot.
Please do not reply to his posts as it justencourages him.It'd
be fascinating to see what would happen to you in the fullness
oftime if this was the only response you got. I predict
descent fromyour current lofty height of idiocy into complete
insanity.Do everyone a favour and go play in the traffic.>It
seems to me that there have been debates over math concepts
I>thought basic, so here's a quick survey:>1. Before I
mentioned it, had you ever heard of the
distributive>property?>2. In your experience, is math
quirky?>3. Do you think that mathematics is an extremely
difficult discipline>that only experts are really good at
handling?>4. What is the distributive property?>5. Is a math
proof perfect?>6. Do you consider yourself to be a reasonable
person?>7. If a mathematical argument is explained to you in
detail, using>basic algebra, if it's correct, would you admit
that, even to a>hostile crowd, like even if many posters on
sci.math would call you>names and insult you for admitting
it?>Thank you for your time and attention.>James Harris><a
href=http://mathforprofit.blogspot.com/>http://
mathforprofit.blogspot.com/</a>
===
Subject: Re: Stroboscopic
effect by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id hAC3VSW02735;>Is it
possible to have a stroboscopic effect (where wheel
spokes>appear to be rotating backwards) in real life, rather
than in>movies where it is common?> Yes, easily. Try
spinning something under a fluorescent light.> Are you too
young to remember record turntables with stroboscopic> speed
indicators?>Not at all. (But I did always wonder why they even
bothered put the>indicator for 78rpm on them!)>But on an open
road, under a clear sun, is it possible for your>eyes to
create this effect?> You need a varying source of light;
seeing one spoked wheel> through another might do it.>But
under sunlight, no interference, etc? I'm busy convincing my
9th>grade students of their inability to differentiate between
reality>memories and fantasy (TV) memories. Several of them
swear that>they've seen it on cars passing them on highways,
on open roads,>in broad daylight.>--rivermanIf your eyes are
fixed on the road instead of the car, the bottomspokes will be
visible since their velocity relative to the roadis low. Fast
eye movements can produce momentary stoboscopiceffects.The
image compression method that the retina uses is a mystery,but
it is highly efficient. It seems to be very good at
preventingartifacts of a stroboscopic nature. There are color
artifacts though.The color effects of the Benham disk probably
have something to dowith the way color is encoded in the optic
nerve.http://faculty.washington.edu/chudler/benham.h
===