mm-101
===
lim delta->0 [f(x+delta)-f(x)]/(delta)^2 exists.>
wondering if there are functions that the following limit does
exist:> lim delta->0 [f(x+delta)-f(x)] / [(delta)^alpha]
exists. where alpha>1> Any references would be nice
too.f(x+delta) = f(x) + delta fO(x) + delta^2/2
fOO(x) +
o(delta^2)[f(x+delta) - f(x)]/(delta)^2 = fO(x)/delta +
fOO(x)/2 + o(1)Therefore, you must have
fO(x) = 0 ie f
constant as it has been said. Ifthis is not for every x, then
you only have to have a null derivative atthe considered
points.For your second question, you must have the n-th first
derivatives withn = ?or(alpha) equal to zero.-- Nicolas
===
GR56 <spam@me.com> grava .88 la saucisse et au marteau: lim
delta->0 [f(x+delta)-f(x)]/(delta)^2 exists.> wondering if
there are functions that the following limit does exist:>
lim delta->0 [f(x+delta)-f(x)] / [(delta)^alpha] exists.
where alpha>1> Any references would be nice too.f(x+delta)
= f(x) + delta fO(x) + delta^2/2
fOO(x) +
o(delta^2)[f(x+delta) - f(x)]/(delta)^2 = fO(x)/delta +
fOO(x)/2 + o(1)Therefore, you must have
fO(x) = 0 ie f
constant as it has been said. If> this is not for every x,
then you only have to have a null derivative at> the
considered points.For your second question, you must have the
n-th first derivatives with> n = ?or(alpha) equal to
zero.Careful, careful; he didnOt assume the function was
twicedifferentiable, or even once differentiable for that
matter. (Althoughit follows.)If the intent is for the limit
of (f(x+delta)-f(x))/delta^2 to existfor ALL x, then of
course f(x+t) - f(x) f(x+t)-f(x) lim ------------- = lim
----------- * t = 0 t->0 t t->0 t^2for all x, so f is
differentiable and fO(x) = 0 identically: so f isconstant.If
the intent is to ask whether it is possible for this limit to
existfor SOME x, of course it is. Take f(x) = x^3, for
example, at x = 0.--Ron Bruck
===
=About 2 months ago I ran
across a problem on MITOs website to find
theindefinite
integral of a certain function. It was from some
problemsolving seminar, and I think it was rated as the
hardest problem ofall the problems during the semester. I
posted about a while backabout this integral but everybody
just said that since MathematicacanOt do it
itOs probably not
integrable in terms of elementaryfunctions. Well, after about
two months of toiling over this problem,I think I might
finally have a solution. However, thereOs a
fewplaces where,
at least to me, things were a little iffy about whetheror not
I could do what I was doing. So I want to put my solution
andsee if thereOs any ?ws.So, the problem is to
find in
closed formIntegral[f(x) dx], where f(x) =
x/Sqrt[x^4+4x^3-6x^2+4x+1]So, first note two things:1) f(x) =
f(1/x) (assuming of course that x is not 0)2) the polynomial
in the bottom is a reciprocal polynomialReciprocal
polynomials generally can be solved by making asubstitution
of the form t = x+1/x. So I use this to my advantagelater.So
anyway, first we use obvservation 1 to say that if I
=Integral[f(x) dx], then 2I = Integral[f(x) dx] +
Integral[f(1/x) dx]In the second integral, let u = 1/x, and
it transforms toIntegral[f(x) dx] - Integral[f(u)/u^2
du]Since x is just a placeholder, I replace the xOs with
uOs
in the firstintegral and get2I = Integral[(1 - 1/u^2)f(u)
du]Now, we use the second observation. We already have2I =
Integral[(1 - 1/u^2)*(u/Sqrt[u^4+4u^3-6u^2+4u+1]) du]Bringing
that u from the numerator into the denominator, we get2I =
Integral[(1 - 1/u^2)*(1/Sqrt[u^2+4u-6+4/u+1/u^2]) du]Now we
are set up to write the part under the square root as
apoloynomial in x+1/x. Indeed, we get2I = Integral[(1 -
1/u^2)*(1/Sqrt[(u+1/u)^2+4(u+1/u)-8]) du]Let v = u+1/u. Then
dv = (1-1/u^2) dv, which is perfect since thereOsa (1-1/u^2)
sitting right there! So the integral transforms to:2I =
Integral[1/Sqrt[v^2 + 4v - 8] dv]Now this integral is easily
evaluated as 2I = Log[2 + v + Sqrt[v^2 + 4v - 8]]Substituting
v gives 2I = Log[2 + u + 1/u + Sqrt[u^4 + 4u^3 - 6u^2 + 4u +
1]]And substituting x givesI = 0.5*Log[2 + x + 1/x +
(1/x^2)Sqrt[x^4 + 4x^3 - 6x^2 + 4x + 1]]So, is this
right?
===
=Nobody <toolshed37@yahoo.com> escribi.97 en el
mensaje> About 2 months ago I ran across a problem on MITOs
website to find the> indefinite integral of a
certain function.
It was from some problem> solving seminar, and I think it was
rated as the hardest problem of> all the problems during the
semester. I posted about a while back> about this integral
but everybody just said that since Mathematica> canOt do it
itOs probably not integrable in terms of elementary>
functions. Well, after about two months of toiling over this
problem,> I think I might finally have a solution. However,
thereOs a few> places where, at least to me, things were a
little iffy about whether> or not I could do what I was
doing. So I want to put my solution and> see if thereOs any
?ws.>> So, the problem is to find in closed form>>
Integral[f(x) dx], where f(x) = x/Sqrt[x^4+4x^3-6x^2+4x+1]>>
So, first note two things:>> 1) f(x) = f(1/x) (assuming of
course that x is not 0)> 2) the polynomial in the bottom is a
reciprocal polynomial>> Reciprocal polynomials generally can
be solved by making a> substitution of the form t = x+1/x. So
I use this to my advantage> later.>> So anyway, first we use
obvservation 1 to say that if I => Integral[f(x) dx], then 2I
= Integral[f(x) dx] + Integral[f(1/x) dx]>> In the second
integral, let u = 1/x, and it transforms to>> Integral[f(x)
dx] - Integral[f(u)/u^2 du]>> Since x is just a placeholder,
I replace the xOs with uOs in the
first> integral and get>> 2I
= Integral[(1 - 1/u^2)f(u) du]>> Now, we use the second
observation. We already have>> 2I = Integral[(1 -
1/u^2)*(u/Sqrt[u^4+4u^3-6u^2+4u+1]) du]>> Bringing that u
from the numerator into the denominator, we get>> 2I =
Integral[(1 - 1/u^2)*(1/Sqrt[u^2+4u-6+4/u+1/u^2]) du]>> Now
we are set up to write the part under the square root as a>
poloynomial in x+1/x. Indeed, we get>> 2I = Integral[(1 -
1/u^2)*(1/Sqrt[(u+1/u)^2+4(u+1/u)-8]) du]>> Let v = u+1/u.
Then dv = (1-1/u^2) dv, which is perfect since thereOs> a
(1-1/u^2) sitting right there! So the integral transforms
to:>> 2I = Integral[1/Sqrt[v^2 + 4v - 8] dv]>> Now this
integral is easily evaluated as>> 2I = Log[2 + v + Sqrt[v^2 +
4v - 8]]>> Substituting v gives>> 2I = Log[2 + u + 1/u +
Sqrt[u^4 + 4u^3 - 6u^2 + 4u + 1]]Actually is2I = Log[2 + u +
1/u + Sqrt[u^2 + 4u - 6 + 4/u + 1/u^2]]I = (1/2)Log[2 + x +
1/x + Sqrt[x^2 + 4x - 6 + 4/x + 1/x^2]]I = (1/2)Log[2 + x +
1/x + Sqrt[x^4 + 4x^3 - 6x^2 + 4x + 1]/x]ButdI/dx = (x^2 -
1)/(2xSqrt[(x^4 + 4x^3 - 6x^2 + 4x + 1]) = (1/2)(1 -
1/x^2)f(x) ???There are something more wrong ...I think that
it is in> Integral[f(x) dx] - Integral[f(u)/u^2 du] (#1)>>
Since x is just a placeholder, I replace the xOs with
uOs in
the first> integral and get>> 2I = Integral[(1 - 1/u^2)f(u)
du]It would be true if the integrals would be definites. But
not if they areindefinites. The first integral in
#1 is a
function of x, while the secondis a function of u, and x =/=
u, actually is x = 1/u. If you replace x by uin the first
integral, you must to replace dx by -du/u^2, and you get2I =
-2*Integral[f(u)/u^2 du]true, but useless ...-- Ignacio
Larrosa Ca.96estroA Coru.96a
(Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
=[snip]> So
anyway, first we use obvservation 1 to say that if I =>
Integral[f(x) dx], then 2I = Integral[f(x) dx] +
Integral[f(1/x) dx]>> In the second integral, let u = 1/x,
and it transforms to>> Integral[f(x) dx] - Integral[f(u)/u^2
du]>> Since x is just a placeholder, I replace the xOs with
uOs in the first> integral and get>> 2I =
Integral[(1 -
1/u^2)f(u) du][snip]This looks wrong. x is in fact not just a
placeholder, because you aredoing an indefinite integral. Try
writing the integral with explicit limits,say from 1 to
X.However, is this step really necessary? You should be able
to get by withoutit.-Michael.
===
=So, the problem is to find in
closed formIntegral[f(x) dx], where f(x) =
x/Sqrt[x^4+4x^3-6x^2+4x+1]MR1798560 (2002b:11093)van der
Poorten, Alfred J.(5-MCQR-NT); Tran, Xuan
Chuong(5-MCQR-NT)Quasi-elliptic integrals and periodic
continued fractions. (English. English summary)Monatsh. Math.
131 (2000), no. 2, 155--169.11J70 (11Y65)Summary: In this
report we detail the following story. Several centuries ago,
Abel noticed that the well-known elementary integral
$$intfrac{dx}{sqrt{x^2+2bx+c}}=logleft(x+b+sqrt{x^2+2bx+c}
right)$$ is just a presage of more surprising integrals of
the form
$$intfrac{f(x)dx}{sqrt{D(x)}}=logleft(p(x)+q(x)sqrt{D(x)}
right).$$ Here $f$ is a polynomial of degree $g$ and the $D$
are certain polynomials of degree $deg D(x)=2g+2$.
Specifically, $f(x)=pO(x)/q(x)$ (so $q$ divides
$pO$). Note
that, morally, one expects such integrals to produce inverse
elliptic functions and worse, rather than an innocent
logarithm of an algebraic function.Abel went on to study
abelian integrals, and it was Chebyshev who explained---using
continued fractions---what is going on with these
`quasi-ellipticO integrals. Recently, the second author
computed all the polynomials $D$ over the rationals of degree
4 that have an $f$ as above. We explain various contexts in
which the present issues arise. These contexts include
symbolic integration of algebraic functions, the study of
units in function fields and, given a suitable polynomial $g$,
the consideration of the period length of the continued
fraction expansion of the numbers $sqrt{g(n)}$ as $n$ varies
over the integers. But the major content of this survey is an
introduction to period continued fractions in
hyperelliptic---thus quadratic---function fields.Reviewed by
M. Mend?s France-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14 times)
===
=>
About 2 months ago I ran across a problem on MITOs website
to
find the> indefinite integral of a certain
function. It was from
some problem> solving seminar, and I think it was rated as the
hardest problem of> all the problems during the semester. I
posted about a while back> about this integral but everybody
just said that since Mathematica> canOt do it
itOs probably
not integrable in terms of elementary> functions. Well, after
about two months of toiling over this problem,> I think I
might finally have a solution. However, thereOs
a few> places
where, at least to me, things were a little iffy about
whether> or not I could do what I was doing. So I want to put
my solution and> see if thereOs any ?ws.So, the problem
is to
find in closed formIntegral[f(x) dx], where f(x) =
x/Sqrt[x^4+4x^3-6x^2+4x+1]So, first note two things:1) f(x) =
f(1/x) (assuming of course that x is not 0)> 2) the
polynomial in the bottom is a reciprocal polynomialReciprocal
polynomials generally can be solved by making a> substitution
of the form t = x+1/x. So I use this to my advantage>
later.So anyway, first we use obvservation 1 to say that if I
=> Integral[f(x) dx], then 2I = Integral[f(x) dx] +
Integral[f(1/x) dx]In the second integral, let u = 1/x, and
it transforms toIntegral[f(x) dx] - Integral[f(u)/u^2
du]Since x is just a placeholder, I replace the xOs with
uOs
in the first> integral and get2I = Integral[(1 - 1/u^2)f(u)
du]Now, we use the second observation. We already have2I =
Integral[(1 - 1/u^2)*(u/Sqrt[u^4+4u^3-6u^2+4u+1]) du]Bringing
that u from the numerator into the denominator, we get2I =
Integral[(1 - 1/u^2)*(1/Sqrt[u^2+4u-6+4/u+1/u^2]) du]Now we
are set up to write the part under the square root as a>
poloynomial in x+1/x. Indeed, we get2I = Integral[(1 -
1/u^2)*(1/Sqrt[(u+1/u)^2+4(u+1/u)-8]) du]Let v = u+1/u. Then
dv = (1-1/u^2) dv, which is perfect since thereOs> a
(1-1/u^2) sitting right there! So the integral transforms
to:2I = Integral[1/Sqrt[v^2 + 4v - 8] dv]Now this integral is
easily evaluated as 2I = Log[2 + v + Sqrt[v^2 + 4v -
8]]Substituting v gives 2I = Log[2 + u + 1/u + Sqrt[u^4 +
4u^3 - 6u^2 + 4u + 1]]And substituting x givesI = 0.5*Log[2 +
x + 1/x + (1/x^2)Sqrt[x^4 + 4x^3 - 6x^2 + 4x + 1]]So, is this
right?> Well, I definitely made one mistake in that
substituting v should notgive you what I put. I think the
final result should beI = 0.5*Log[(x+1)^2 +
Sqrt[(x+1)^4-12x^2]] - 0.5*Log[x]If someone could
differentiate this using Mathematica and simplify itI would
appreciate it.
===
=> About 2 months ago I ran across a problem
on MITOs website to find the>
indefinite integral of a certain
function. It was from some problem> solving seminar, and I
think it was rated as the hardest problem of> all the
problems during the semester. I posted about a while back>
about this integral but everybody just said that since
Mathematica> canOt do it itOs probably not
integrable in
terms of elementary> functions. Well, after about two months
of toiling over this problem,> I think I might finally have a
solution. However, thereOs a few> places where, at least to
me, things were a little iffy about whether> or not I could
do what I was doing. So I want to put my solution and> see if
thereOs any ?ws.So, the problem is to find
in closed
formIntegral[f(x) dx], where f(x) =
x/Sqrt[x^4+4x^3-6x^2+4x+1]So, first note two things:1) f(x) =
f(1/x) (assuming of course that x is not 0)> 2) the
polynomial in the bottom is a reciprocal polynomialReciprocal
polynomials generally can be solved by making a> substitution
of the form t = x+1/x. So I use this to my advantage>
later.So anyway, first we use obvservation 1 to say that if I
=> Integral[f(x) dx], then 2I = Integral[f(x) dx] +
Integral[f(1/x) dx]In the second integral, let u = 1/x, and
it transforms toIntegral[f(x) dx] - Integral[f(u)/u^2
du]Since x is just a placeholder, I replace the xOs with
uOs
in the first> integral and get2I = Integral[(1 - 1/u^2)f(u)
du]Now, we use the second observation. We already have2I =
Integral[(1 - 1/u^2)*(u/Sqrt[u^4+4u^3-6u^2+4u+1]) du]Bringing
that u from the numerator into the denominator, we get2I =
Integral[(1 - 1/u^2)*(1/Sqrt[u^2+4u-6+4/u+1/u^2]) du]Now we
are set up to write the part under the square root as a>
poloynomial in x+1/x. Indeed, we get2I = Integral[(1 -
1/u^2)*(1/Sqrt[(u+1/u)^2+4(u+1/u)-8]) du]Let v = u+1/u. Then
dv = (1-1/u^2) dv, which is perfect since thereOs> a
(1-1/u^2) sitting right there! So the integral transforms
to:2I = Integral[1/Sqrt[v^2 + 4v - 8] dv]Now this integral is
easily evaluated as 2I = Log[2 + v + Sqrt[v^2 + 4v -
8]]Substituting v gives 2I = Log[2 + u + 1/u + Sqrt[u^4 +
4u^3 - 6u^2 + 4u + 1]]And substituting x givesI = 0.5*Log[2 +
x + 1/x + (1/x^2)Sqrt[x^4 + 4x^3 - 6x^2 + 4x + 1]]So, is this
right?> Well, I definitely made one mistake in that
substituting v should not> give you what I put. I think the
final result should beI = 0.5*Log[(x+1)^2 +
Sqrt[(x+1)^4-12x^2]] - 0.5*Log[x]If someone could
differentiate this using Mathematica and simplify it> I would
appreciate it.The derivative is x^2 - 1
---------------------------- 2 x
sqrt(x^4+4x^3-6x^2+4x+1)--Ron Bruck
===
=Try
this:x^4+4x^3-6x^2+4x+1=(x-a)*(x-b)*(x-c)*(x-d),wherea=-1+1/2
*sqrt(12)+sqrt(3-2*sqrt(3))b=-1+1/2*sqrt(12)-sqrt(3-2*sqrt(3)
)c=-1-1/2*sqrt(12)+sqrt(3+2*sqrt(3))d=-1-1/2*sqrt(12)-sqrt(3+
2*sqrt(3))Thusx/(x^4+4x^3-6x^2+4x+1)=x/((x-a)*(x-b)*(x-c)*(
x-d))Try expressing this as a sum of integrable polynomial
fractions in x, thenintegrate.> About 2 months ago I ran
across a problem on MITOs website to find the>
indefinite
integral of a certain function. It was from some problem>
solving seminar, and I think it was rated as the hardest
problem of> all the problems during the semester. I posted
about a while back> about this integral but everybody just
said that since Mathematica> canOt do it itOs
probably not
integrable in terms of elementary> functions. Well, after
about two months of toiling over this problem,> I think I
might finally have a solution. However, thereOs
a few> places
where, at least to me, things were a little iffy about
whether> or not I could do what I was doing. So I want to put
my solution and> see if thereOs any ?ws.>> So, the
problem is
to find in closed form>> Integral[f(x) dx], where f(x) =
x/Sqrt[x^4+4x^3-6x^2+4x+1]>> So, first note two things:>> 1)
f(x) = f(1/x) (assuming of course that x is not 0)> 2) the
polynomial in the bottom is a reciprocal polynomial>>
Reciprocal polynomials generally can be solved by making a>
substitution of the form t = x+1/x. So I use this to my
advantage> later.>> So anyway, first we use obvservation 1 to
say that if I => Integral[f(x) dx], then 2I = Integral[f(x)
dx] + Integral[f(1/x) dx]>> In the second integral, let u =
1/x, and it transforms to>> Integral[f(x) dx] -
Integral[f(u)/u^2 du]>> Since x is just a placeholder, I
replace the xOs with uOs in the
first> integral and get>> 2I =
Integral[(1 - 1/u^2)f(u) du]>> Now, we use the second
observation. We already have>> 2I = Integral[(1 -
1/u^2)*(u/Sqrt[u^4+4u^3-6u^2+4u+1]) du]>> Bringing that u
from the numerator into the denominator, we get>> 2I =
Integral[(1 - 1/u^2)*(1/Sqrt[u^2+4u-6+4/u+1/u^2]) du]>> Now
we are set up to write the part under the square root as a>
poloynomial in x+1/x. Indeed, we get>> 2I = Integral[(1 -
1/u^2)*(1/Sqrt[(u+1/u)^2+4(u+1/u)-8]) du]>> Let v = u+1/u.
Then dv = (1-1/u^2) dv, which is perfect since thereOs> a
(1-1/u^2) sitting right there! So the integral transforms
to:>> 2I = Integral[1/Sqrt[v^2 + 4v - 8] dv]>> Now this
integral is easily evaluated as>> 2I = Log[2 + v + Sqrt[v^2 +
4v - 8]]>> Substituting v gives>> 2I = Log[2 + u + 1/u +
Sqrt[u^4 + 4u^3 - 6u^2 + 4u + 1]]>> And substituting x
gives>> I = 0.5*Log[2 + x + 1/x + (1/x^2)Sqrt[x^4 + 4x^3 -
6x^2 + 4x + 1]]>> So, is this right?>
===
=> Try this:>
x^4+4x^3-6x^2+4x+1=(x-a)*(x-b)*(x-c)*(x-d),> where>
a=-1+1/2*sqrt(12)+sqrt(3-2*sqrt(3))>
b=-1+1/2*sqrt(12)-sqrt(3-2*sqrt(3))>
c=-1-1/2*sqrt(12)+sqrt(3+2*sqrt(3))>
d=-1-1/2*sqrt(12)-sqrt(3+2*sqrt(3))> Thus>
x/(x^4+4x^3-6x^2+4x+1)=x/((x-a)*(x-b)*(x-c)*(x-d))> Try
expressing this as a sum of integrable polynomial fractions
in x, then> integrate.I wish it were so simple.
Unfortunately, you need to take the squareroot of that
quartic polynomial on the bottom.
===
=>Try
this:>x^4+4x^3-6x^2+4x+1=(x-a)*(x-b)*(x-c)*(x-d),>where>a=-1+
1/2*sqrt(12)+sqrt(3-2*sqrt(3))>b=-1+1/2*sqrt(12)-sqrt(3-2*
sqrt(3))>c=-1-1/2*sqrt(12)+sqrt(3+2*sqrt(3))>d=-1-1/2*sqrt(12
)-sqrt(3+2*sqrt(3))>Thus>x/(x^4+4x^3-6x^2+4x+1)=x/((x-a)*(x-b
)*(x-c)*(x-d))>Try expressing this as a sum of integrable
polynomial fractions in x, then>integrate.... except that
itOs x/sqrt(x^4+4x^3-6x^2+4x+1) that he wants to integrate.
According to Maple,> int(x/sqrt((x-a)*(x-b)*(x-c)*(x-d)),x);
/(a - c) (x - d)1/2 2 /(c - d) (x - b)1/2 2 (d - a)
|---------------| (x - c) |---------------| (a - d) (x - c)/
(b - d) (x - c)/ /(c - d) (x - a)1/2 / |---------------| | (a
- d) (x - c)/ /(a - c) (x - d)1/2 /(c - b) (d - a)1/2 c
EllipticF(|---------------| , |---------------| ) + (a - d)
(x - c)/ (d - b) (c - a)/ (d - c) /(a - c) (x - d)1/2 a - d
/(c - b) (d - a)1/2 EllipticPi(|---------------| , -----,
|---------------| ) (a - d) (x - c)/ a - c (d - b) (c - a)/ /
1/2 | / ((a - c) (c - d) ((x - a) (x - b) (x - c) (x - d)) ) /
/Robert Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
=> About 2 months ago I ran
across a problem on MITOs website to find the>
indefinite
integral of a certain function. It was from some problem>
solving seminar, and I think it was rated as the hardest
problem of> all the problems during the semester. I posted
about a while back> about this integral but everybody just
said that since Mathematica> canOt do it itOs
probably not
integrable in terms of elementary> functions.Mathematica 5
gives an answer -- terribly messy -- in closed form in
termsof elliptic integrals. I have no idea if that answer is
correct or not.> Well, after about two months of toiling over
this problem,> I think I might finally have a solution.
However, thereOs a few> places where, at least to me, things
were a little iffy about whether> or not I could do what I
was doing. So I want to put my solution and> see if thereOs
any ?ws.>> So, the problem is to find in closed form>>
Integral[f(x) dx], where f(x) =
x/Sqrt[x^4+4x^3-6x^2+4x+1][snip]> I = 0.5*Log[2 + x + 1/x +
(1/x^2)Sqrt[x^4 + 4x^3 - 6x^2 + 4x + 1]]>> So, is this
right?Have you forgotten how easy such a thing is to check???
Just differentiatewhat you think to be the antiderivative, and
see if that result is equalto the integrand!Unfortunately, it
seems that you did something wrong...David
===
=> About 2
months ago I ran across a problem on MITOs website to
find
the> indefinite integral of a certain function. It was from
some problem> solving seminar, and I think it was rated as
the hardest problem of> all the problems during the semester.
I posted about a while back> about this integral but everybody
just said that since Mathematica> canOt do it
itOs probably
not integrable in terms of elementary> functions.Mathematica
5 gives an answer -- terribly messy -- in closed form in
terms> of elliptic integrals. I have no idea if that answer
is correct or not.Well, after about two months of toiling
over this problem,> I think I might finally have a solution.
However, thereOs a few> places where, at least to me, things
were a little iffy about whether> or not I could do what I
was doing. So I want to put my solution and> see if thereOs
any ?ws.>> So, the problem is to find in closed form>>
Integral[f(x) dx], where f(x) = x/Sqrt[x^4+4x^3-6x^2+4x+1]>
[snip]> I = 0.5*Log[2 + x + 1/x + (1/x^2)Sqrt[x^4 + 4x^3 -
6x^2 + 4x + 1]]>> So, is this right?Have you forgotten how
easy such a thing is to check??? Just differentiate> what you
think to be the antiderivative, and see if that result is
equal> to the integrand!Unfortunately, it seems that you did
something wrong...DavidI donOt have Mathematica, and
itOs a
painful function to differentiateand simplify. Of course I
could have done it but thereOs alot of roomfor error, and if
I make a mistake differentiating IOm likely to thinkthat my
answer is wrong, when in fact itOs not. However, you said
itis wrong, so I assume you differentiated the result using
Mathematicaand didnOt get the same thing. Unfortunate
indeed.Is it possible someone can point me to the error?
===
=>
Have you forgotten how easy such a thing is to check??? Just
differentiate> what you think to be the antiderivative, and
see if that result is equal> to the integrand!Unfortunately,
it seems that you did something wrong...David> I donOt have
Mathematica, and itOs a painful function to differentiate>
and simplify. Of course I could have done it but thereOs
alot
of room> for error, and if I make a mistake differentiating
IOm likely to think> that my answer is wrong, when in fact
itOs not. However, you said it> is wrong, so I assume you
differentiated the result using Mathematica> and didnOt get
the same thing. Unfortunate indeed.BTW a simple calculator is
a great way to check complicated algebra.Just pick a random
value for X and compute the numericalresult at each stage.
You can quite quickly track down whichstep contained the
error. Numerical differentiation also works quiteeffectively
if you have sufficient precision on the calculatorand use
common sense in selecting delta-X.
===
=theorems. Not being
able to figure out that proof was really buggingme,
IOll sleep
better now.>>If F is given consider the sequence of
functions (defined on |z| = 1)>F[r](z) = F(rz), where r is a
monotone sequence of reals approaching 1, say>r = 1/(1+n)
with n a natural number. Since F is continuous F[r](z)
uniformly>approaches F(z) as r approaches 1. [This is the
point I was missing before ->it holds because the unit circle
is a closed set.] Or more precisely: This works because the
closed _disk_ is a _compact_> set, which implies that f is
uniformly continuous.>Let E/2 be the worst case>error value
of |F(z) - F[r](z)|. Since rz lies within the radius
of>convergence of F, F converges uniformly there, and we can
take a partial sum>P[r](z) that is uniformly within E/2 of
F[r](z). This partial sum is a>polynomial in (z) as well as
in (rz). For all z on the unit circle |F(z) ->P[r](z)| <=
|F(z) - F[r](z)| + |F[r](z) - P[r](z)| < E which goes to zero
as>r approaches 1. So the P[r](z) are the desired polynomials
in z. [My only>nagging doubt here is that this argument would
seem to apply to an F defined>on any circle, not just the unit
circle.]Well you can relax, itOs true for any closed disk.
(In fact itOs true> in much greater generality, although
itOs
not so easy to prove; > RungeOs theorem, included in most
complex books, gives a> generalization, and MergelyanOs
theorem gives the result under> weaker hypotheses yet.)>If
the sequence of polynomials is given they converge and
therefore form a>Cauchy sequence. The difference between any
two of them is an analytic>function and therefore reaches its
maxium modulus on the boundary of the>unit disk. So the
polynomials form a uniform Cauchy sequence on the closed>unit
disk and therefore converge uniformly. [Interestingly none of
the math>books I had included CauchyOs criterion for uniform
convergence, just>CauchyOs criterion for convergence - too
obvious an extension to state>explicitly no doubt. But I did
find explicit mention of it on the web.]>>-The polynomials
converge uniformly on the closed unit disk and
are>continuous, therefore they converge to a continuous
function there.>-The polynomials converge uniformly on the
open unit disk and are analytic,>therefore they converge to
an analytic function there.>>Ian>Been staring at this for
hours and canOt seem to find a proof that>>seems
needlessly
complex or just plain doesnOt work... can someone see>>a
simple argument? I would be much appreciated - I donOt even
need>>this result for anything but somehow I just canOt let
it go.>>Let f be a continuous function on the unit circle T
= {|z| = 1}. Show>>that f can be approximated uniformly on T
by a sequence of polynomials>>in z if and only if f has an
extension F that is continuous on the>>To approximate such an
F, consider dilates F[subscripted r](z) =>>F(rz).>>I canOt
get either direction to work... if F is given then the
hint>>seems to imply that you can obtain the polynomial
sequence from the>>power series expansion of F, but how to
show uniform convergence in a>>simple way?>> The power
series need not converge to F at points of the boundary.>>
But thatOs not what the hint suggests - read the hint
again...>>If the polynomial sequence is given then itOs
easy to use the Cauchy>>integral formula on the polynomials,
take the limit, and get F>>analytic (limit of the integral =
integral of the limit because of the>>uniform convergence,
and analyticity follows from the continuity of>>the limit
funciton f). But how to show that F is continuous at
points>>on the unit circle?>> DonOt use the Cauchy formula.
Use the Maximum Modulus Theorem,>> and look up uniform
convergence and Cauchy sequence in a book>> on adbanced
calculus.>>I found this in Complex Analysis by Gamelin,
section V.4, problem 14.>> David C. Ullrich>> David C.
Ullrich
===
=Two thoughts on why e is good, and a limerick:(1)
If we could choose an irrational number as ourbase (ie
instead of the usual 10 or 2 for binary),then e would be the
most efficient. Why?An n-symbol string in base m can represent
m^n differentnumbers, and to do so we use n*m different
symbols(m letters, but each letter can be n
differentpositions). So if we write the size of
therepresented space (m^n) as a function of the numberof
different symbols k=m*n, we getf(k) = m^(k/m) = (m^(1/m))^k.I
like to think of the number (m^(1/m)) ascapturing the
efficiency of thestandard base-m type encoding. In the
integers,3 maximizes this as 1.4422... But if we usesome
calculus, the global maximum is actually ate, with a value of
1.4446...In other words, our efficiency is maximized at e.This
argument is my own, but IOm sure it has beendone better
elsewhere, because everything I thinkI have thatOs even
slightly original turns out that way...(2) One of the most
beautiful theorems of modernmathematics: the central limit
theorem. Usuallypeople think of it as only applying to
probability, sinceit is easiest to interpret in that realm.
But itview, you could declare the presence of e in
thestatement of CLT to be a consequence of Fourieranalysis
(and sort of from EulerOs formula e^(ix) =cos(x) + i
sin(x)),
but this is a bit of an over-simplification.limit theorem; so
the best elementary idea of it thatI can give is that there
is a single function,something like e^(-x^2/2) (times a
constant), whichrepresents the limiting behavior of virtually
_any_ functionwhich is randomly sampled and then averaged. I
knowthatOs a bad description, but itOs a
start.Finally, the
limerick (written by me :)My favorite number is eit compounds
continuously;if you raise it to pimultiplied by i,negative one
it will be.-Tyler
===
=Part IINote this is all constructive
criticism of HalOs ideas on metric engineering.Excerpts from
Puthoff & IbisonsH.E. PUTHOFF*, S.R. LITTLE AND M.
IBISONInstitute for Advanced Studies at Austin, 4030 West
Braker Lane, Suite 300, Austin, Texas 78759-5329, USA.with my
comments.3. The Quantum Vacuum3.1 Zero-Point Energy (ZPE)
BackgroundQuantum theory tells us that so-called empty space
is not truly empty, but is the seat of myriad energetic
quantum processes. Specifically, quantum field
theory tells us
that, even in empty space, fields (e.g., the electromagnetic
field) continuously ?ctuate about their zero baseline
values.So far, so good. However IMHO it is the QED PV virtual
electron-positron zero point vacuum ?ctuations that dominate
the low energy effective macro-quantum field theory out of
which EinsteinOs gravity together with exotic vacuum
unified
dark energy/matter fields co-emerge like love and marriage all
together as phase and amplitude modulation patterns,
respectively, of the vacuum coherence field that is formed by
the attractive BCS exchange of virtual photons between the
virtual electron-positron pairs all inside the vacuum. Note
that Hal makes no reference at all to the new discovery in
precision cosmology of dark energy as being relevant to his
Quixotic Ahabian quest in search of for the vacuum wind.
:-)The energy associated with these ?ctuations is called
zero point energy (ZPE), re?cting the fact that such
activity remains even at a temperature of absolute zero. Such
a concept is almost certain to have profound implications for
future space travel, as we will now discuss.Agreed.When a
hypothetical ZPE-powered spaceship strains against gravity
and inertia ...ThatOs a strange way of putting it. The local
net random micro-quantu zero point stress-energy density
tensor field isTuv(Exotic Vacuum) = (String
Tension)/ZPEguvusing Ed WittenOs natural units h = c =
k(Boltzmann) = 1Tuv(Exotic Vacuum) = (String
Tension)^2[(String Tension)^3/2|Vacuum Coherence|^2
-1]guvWhat is wrong with PuthoffOs paradigm is that he has
no
idea of Vacuum Coherence in his informal thinking nor
formalized in his mathematics.The zero point energy density
is Too and I use the 3 GR sign conventions in whichToo(Exotic
Vacuum) > 0 means strongly anti-gravity negative pressure dark
energy since the general equation of state(pressure) =
w(energy density)reduces to w = -1 for ALL zero point quantum
fields of all spins.Too(Exotic Vacuum) < 0 means strongly
gravitating positive pressure dark matter.Therefore dark
energy needs(String Tension)^3/2|Vacuum Coherence|^2 > 1Dark
matter needs(String Tension)^3/2|Vacuum Coherence|^2 < 1Where
the critical Vacuum Coherence corresponding to zero Einstein
Cosmological Constant of the non-exotic vacuum in the
large-scale FRW metric is(String Tension)^3/2|Vacuum
Coherence|^2 = 1Or|Vacuum Coherence|*^2 = (String
Tension)^-3/2The Einstein exotic vacuum local geometrodynamic
field equation is thenTuv(Induced Time and Space Warps) +
Tuv(Exotic Vacuum) ~ 0The total covariant 4-divergence
vanishes conserving total local stress-energy Einstein
current density.Unlike 1915 geometrodynamics, the Bianchi
identities break down so that the two pieces of the
divergence are not separately zero!Only the sum is zero. This
is necessary for practical soft metric engineering of the Tech
Gnostic Underground Stream inside the vacuum.That
is,Tuv(Induced Space and Time Warps)^;v + Tuv(Exotic
Vacuum)^;v = 0This is the first small step for Mankind leading
to The Right Stuff to Make Star Trek Real.See my lecture in
Time Travel: The Art of the Possible Disk 2 of Paramount
Pictures DVDSpecial CollectorOs Edition of Star Trek IV: The
Voyage Home indeed! ;-)*Before you read on click on this!
Turn the volume up on your sound.Sarfatti pitches to Puthoff
at bat in the World Series
:-)http://www.niehs.nih.gov/kids/lyrics/ballgame.htmthere are
three elements of the equation that the ZPE technology could
in principle address: (1) a decoupling from gravity,NO! We
want just the opposite! Big error of strategic thinking!The
Question is: What is The Question?Here Hal & Co ask a wrong
question IMHO.STRIKE ONE!(2) a reduction of inertiaNO! Ditto.
ThatOs like preventing your car from being stolen by blowing
it up if it is broken into!STRIKE TWO!or (3) the generation
of energy to overcome both.NO, NO, NO a zillion times NO!
;-)ThatOs the Brute Force approach.That is notThe Tao Chi of
ET!STRIKE THREE!YouOre Out!Take me out to the ball game
....3.2 GravityWith regard to a ZPE basis for gravity, the
Russian physicist Andrei Sakharov was the first to propose
that in a certain sense gravitation is not a fundamental
interaction at all, but rather an induced effect brought
about by changes in the quantum-?ctuation energy of the
vacuum when matter is presentYes, this is essentially correct
and is precisely what my theory is all about. The details are
inhttp://qedcorp.com/APS/EmergentGravity.doc or
http://qedcorp.com/APS/EmergentGravity.doc (smaller file)to be
continued.Engineering the Zero-Point Field and Polarizable
Vacuum for Interstellar Flight1. IntroductionThe concept of
engineering the vacuum found its first expression in the
mainstream physics literature when it was introduced by T. D.
Lee in There he stated:The experimental method to alter the
properties of the vacuum may be called vacuum engineering....
If indeed we are able to alter the vacuum, then we may
encounter some new phenomena, totally unexpected.This
legitimization of the vacuum engineering concept was based on
the recognition that the vacuum is characterized by parameters
and structure that leave no doubt that it constitutes an
energetic medium in its own right. Foremost among these are
its properties that (1) within the and field ?ctuations,
and
(2) within the context of general relativity the vacuum is
the seat of a spacetime structure (metric) that encodes the
distribution of matter and energy. Indeed, on the ?leaf of a
book of essays by Einstein and others on the properties of the
vacuum we find the statement The vacuum is fast emerging as
the
central structure of modern physics [3].Given the known
characteristics of the vacuum, one might reasonably inquire
as to why it is not immediately obvious how to catalyze
robust interactions of the type sought for space-?ght
applications. To begin, in the case of quantum ?ctuations
there are uncertainties that remain to be clarified regarding
global thermodynamic and energy constraints. Furthermore, the
energetic components of potential utility involve very
small-wavelength, high-frequency fields and thus resist facile
engineering solutions.This last remark may not be correct in
general, although it is correct within the framework Puthoff
is pursuing i.e. using the stress-energy density of the
electromagnetic field to attempt to directly warp space-time
geometry with practical utility. Such a brute-force approach
is hopeless and a waste of effort IMHO because Einsteins
equation in this case is simplyInduced Space-Time Warp =
(Applied EM Stress-Energy Density)/(G-String Tension)The
G-String Tension is 10^19Gev per 10^-33 cm, which is simply
too stiff! One would need a way to lower the
G-String-Tension, for example an M Theory that yields
something like:G*-String Tension = e^-(r*/Lp)^?(G-String
Tension)Where r* = compactification scale of extra-dimensions
of hyperspace beyond 4-D space-time andLp^2 = hG/c^3 =
1/(G-String Tension) for h = c = 1Puthoff alludes to this
barrier in his next remark:With regard to perturbation of the
space-time metric, the required energy densities exceed by
many orders of magnitude values achievable with existing
engineering techniques. Nonetheless, we can examine the
constraints, possibilities and implications under the
expectation that as technology matures, felicitous means may
be found that permit the exploitation of the enormous,
as-yet-untapped potential of so-called empty space.This is
the problem that Hal goes into denial and wishing about.
There is no way that will brute-force approach will work in
any practical way. The UFOs do not use high-energy EM field
densities and Hal is primarily interested in the UFOs. That
should tell him right away that he is pursuing the wrong path
to solve the problem.2. Propellantless Propulsion2.1 Global
ConstraintRegardless of the mechanisms that might be
entertained with regard to propellantless or field propulsion
of a spaceship, there exist certain constraints that can be
easily overlooked but must be taken into consideration. A
central one is that, because of the law of conservation of
momentum, the center of mass-energy (CM) of an initially
stationary isolated system cannot change its position if not
acted upon by outside forces. This means that propellantless
or field propulsion, whatever form it takes, is constrained to
involve coupling to the external universe in such a way that
the displacement Engineering the Zero-Point Field and
Polarizable Vacuum for Interstellar FlightIt has been known
since the late 1950s how to circumvent this problem provided
one has exotic stuff with a strong enough negative pressure
to provide what Herman Bondi called negative matter in the
late 1950s and Kip Thorne called exotic matter in the late
1980s in his traversable wormhole time travel papers with
additional work by Igor Novikov and others. Robert Forward
summarized Bondis early work in a paper written in the early
1990s. More details are to be found inWe now know since 1999
from Type 1a supernova data showing the acceleration of the
expansion speed of the Universe that approximately 3/4 of all
the large-scale stuff of the Universe is exotic vacuum
anti-gravitating dark energy with a strong negative zero
point pressure that is equal and opposite to the positive
zero point energy density. This fact is the key to practical
metric engineering of the vacuum that Puthoff et-al
completely misses. The modified vacuum Einstein equation
isTuv(Geometry) + Tuv(Exotic Vacuum) = 0Where Tuv is the
local stress-energy density tensor under the Einstein DIFF(4)
symmetry group.Tuv(Geometry) = (String
Tension)Guv(Geometry)WhereGuv(Geometry) = Ruv [CapitalEth]
(1/2)RguvAndTuv(Exotic Vacuum) = (String Tension)/zpfguv/zpf
= (String Tension)^-1[(String Tension)^3/2|Vacuum
Coherence|^2 [CapitalEth] 1]/zpf > 0 is negative pressure
dark energy exotic vacuum that is ~ 3/4 of large-scale stuff
of the Universe./zpf < 0 is positive pressure dark matter
exotic vacuum that is ~ 1/4 of the large scale stuff of the
Universe.Gravity effect of exotic vacuum stuff ~ c^2/zpfWhere
+ sign on RHS is universally repelling anti-gravity and a
[CapitalEth] sign is universally attracting gravity with
effective short scale coupling strengths much larger than
Newtons.The key idea not found in any of Puthoffs papers on
the subject of metric engineering is the local Einstein
stress-energy tensor current densitys covariant divergence
Tuv^;v where ;v is the Diff(4) covariant derivative relative
to the metric torsion free connection field for parallel
transport of tensors along tangent vector fields in curved
space-time.When there is no exotic vacuum, i.e., /zpf = 0
corresponding to optimal vacuum coherence, the Bianchi
identities work and we have closed current conservation of
the pure geometrodynamic stress-energy tensor local currents,
i.e.Tuv(Geometry)^;v = 0This prevents any practical metric
engineering! In contrast, when there is exotic vacuum
thenTuv(Geometry)^;v =/= 0InsteadTuv(Geometry)^;v +
Tuv(Exotic Vacuum)^;v = 0Where the common string-tension
factor cancels out of this current conservation equation.
This essentially solves the problem for practical metric
engineering bypassing the string tension problem completely!
One uses the Bohm-Aharonov-Josephson effect to tweak
Tuv(Exotic Vacuum) in a soft way.To be continued.The Physical
Principles of Metric Engineeringby Jack Sarfatti5th
draftexcerpt from the book Super CosmosThe term metric
engineering was coined by Dr. Harold Puthoff. Hal, as he
prefers to be called, was a US Naval Officer who then worked
for the National Security Agency before going to the Stanford
Research Institute where he conducted the famous Remote
Viewing experiments with Russell Targ testing Uri Geller,
Ingo Swann, Pat Price and other psychics in a project paid
for by the CIA and Department of Defense Intelligence
Agencies. I first met Hal and Russell at SRI in 1973 and that
story is told in my book Destiny Matrix. Hal has held very
high USG security clearances and it is well known that he is
obsessed with the UFO mystery. Hal, like his co-worker Bernie
Haisch, who is an editor of the Astrophysical Journal, both
strongly believe in the physical reality of ?ing saucers.
They, with Jacques Vallee, have been active in the NASA
Breakthrough Propulsion Program and in UFO groups some
financed by Laurance Rockefeller and the Howard Hughes clone
Las Vegas Hotel Tycoon Robert Bigelow who owns Bigelow
Aerospace Corporation. Since they take the reality of the
?ing saucers seriously so do I. The two lines of theoretical
physics research that Hal has pursued for several decades now,
zero point energy and a polarized vacuum model of gravity, is
primarily motivated by the quest to understand how the
saucers ?. The alleged reality of such advanced alien
technology is clearly of immense importance to US National
Security and beyond. The recent developments in physics shown
in NOVAOs Elegant Universe with Brian Greene, in Stephen
HawkingOs The Universe in a Nutshell, Michio
KakuOs
Hyperspace and Igor NovikovOs The River of Time makes time
traveling alien interference in our history more probable not
less probable. This is all an aspect of metric engineering
defined as the practical control of space and time warps. I
have discussed this in Paramount Pictures DVD Special
CollectorOs Edition of Star Trek IV: The Voyage Home in Time
Travel: The Art of the Possible and in Learning ChannelOs
Ultra-Science: Time Travel. The ultimate form of metric
engineering is seen in Q in Star Trek where a Super Mind is
able to warp space-time. We also need a physics of
consciousness to see if that fiction can be realized in fact.
British Astronomer Royal, Sir Martin Rees who runs the
laboratory where Stephen Hawking works and who is the new
Master of Trinity College, Cambridge has added the dimension
of Doomsday WMD to the quest for a metric engineering
breakthrough in his Chapter 9 of his pessimistic Our Final
Hour on the Ice Nine rip-in-space propagating at the speed of
light that could literally destroy our universe. We are now at
that turning point in our history where, like Mickey Mouse in
Walt DisneyOs Fantasia we could, in our incompetence,
destroy
the entire Universe just as we are now surely destroying
EarthOs Biosphere. Therefore, any aliens out there with Star
Gate Time Travel and Warp Drive Super Technology are already
here in order to stop us from killing them as well as
ourselves. The American Christian Fundamentalist Right has a
strong belief in Apocalypse Now and there is real danger that
this is a self-fulfilling prophecy in a precognitive remote
viewing of our future by Saint John in The BibleOs
Revelations. All the more reason for me, like Paul Revere, to
alert the public to these new developments in physics because
only real knowledge can possibly save us and even then there
is no guarantee.What does mainstream physics tell us about
the possibility of metric engineering? The relevant parts of
mainstream physics for metric engineering are EinsteinOs
general theory of relativity and micro-quantum mechanics
including the More is Different emergence of macro-quantum
super?id order in ground state and vacuum instabilities
developed by condensed matter Princeton physicist P.W. in the
in?tionary chaotic cosmology of A. Linde consistent with the
issue of Scientific American.LetOs start with
EinsteinOs
gravity field equation of ~ 1915. Einstein viewed pure
space-time geometry like the marble in a statue by
Michelangelo with gross matter and radiation as wood.
EinsteinOs equation, in the 21st Century language,
isSpace-Time Curvature = (Stress-Energy Density)/(String
Tension)Curvature has the physical dimensions of 1/Area,
String Tension has Energy/Length and Stress-Energy Density
has Energy/Volume. Simple algebra confirms that this way of
writing EinsteinOs equation is correct dimensionally. The
String Tension is the basic parameter that Ed Witten of the
Princeton Institute of Advanced Study uses in his discussions
of M theory that unifies the five limiting cases
of super string
theory in which the elementary leptons, quarks and gauge force
bosons are vibrating strings of pure energy pulsating in the
extra space dimensions of Calabi-Yau hyperspace. The ordinary
matter and radiation on the right hand side of EinsteinOs
equation are open strings whose ends are stuck to 3Dim brane
worlds of which our universe is one. You can picture a string
as a tiny wormhole whose two ends need not be on the same
brane world because only then can you not have equal numbers
Another equivalent way to look at EinsteinOs 1915 equation
from our Cosmic Justice peeking through the blindfold. That
is, the stress-energy density tensor of pure marble geometry
isStress-Energy Density Tensor of Pure Geometry = (String
Tension)(Space-Time Curvature) (2)EinsteinOs equation is
then
like the static equilibrium balance of forces in architecture
in which the sum of all the contributions to the
stress-energy density tensor add up to a perfect zero! In
this simplest of casesStress-Energy Density Tensor of
Geometry + Stress-Energy Density of Matter etc. = 0 (3)All
physically real objects are either tensors, spinors or
twistors in the theory of relativity. This is because the
coordinate map is not the physical territory. The form of the
laws of physics must be the same locally no matter how the
coordinate map is morphed. To this we must add the Einstein
Equivalence Principle or EEP which says that even in a curved
space-time, we can use the special theory of relativity
locally for a special class of LIF observers to describe what
is happening to a good approximation. The EEP is only meant to
apply in this approximate way and it will break down if one is
falling into a black hole singularity or if the microscope
magnification is so large that quantum gravity zero point
energy density ?ctuations in the space-time geometry itself
get large. There may also be torsion fields not included in
EinsteinOs 1915 geometrodynamics that may require a
modification of the EEP. The spinor is a square root of a
tensor and the Penrose twistor is a spinor in a complex
space-time. The physics of point matrix space-time is in
reality the physics of extended strings and membranes in real
space-time. HawkingOs imaginary time and its connection to
inverse temperature is part of that same story.The key
concept for metric engineering is the Einstein current!
EinsteinOs field equation is always the
statement that the sum
of all the stress-energy density tensors when put on the same
side of the equation all balance out to exactly zero. We then
take a covariant divergence of the equation to get the
Einstein currents, which are conserved when added together.
The covariant divergence depends on something called a
connection field, which tells us how to parallel transport
tensors and spinors along paths or histories in space-time
and beyond including the extra dimensions of hyperspace.
Every time a gauge force field is added, like a torsion
field,
there is an additional piece added to this connection field.
EinsteinOs 1915 theory is a degenerate case of the bigger
unified field theory just like a circle is a
degenerate case of
an ellipse. An ellipse has two centers or foci. When the two
foci merge together the ellipse degenerates into a circle.
EinsteinOs 1915 theory in the form of the Bianchi identities
forbids practical metric engineering because it has an
impenetrable barrier completely separating the marble
geometric current from the wood matter current. Both kinds of
currents must be able to intermingle to transform into each
other in order to have soft practical metric engineering of
Star Gate Time Travel traversable wormholes and weightless
superluminal warp drives using the anti-gravitating exotic
vacuum dark energy with negative zero point exotic vacuum
pressure that is equal in magnitude, but opposite in sign, to
the zero point energy density. The new Type Ia super novae
data showing that our brane world Universe is accelerating in
its rate of expansion, i.e. speeding up rather than slowing
down. Indeed approximately 3/4 of all the stuff of our
Universe on the large scale is made out of dark energy.
Therefore, the idea of ancient traversable Star Gate
wormholes stabilized by dark energy is not so far fetched.
The brute force approach to metric engineering taken by Hal
Puthoff now for at least two decades can never work because
space-time is too stiff to bend directly with electromagnetic
field energy density to do anything worth doing because the
string tension is too large. In contrast, we can used the
quantum interference of the Einstein stress-energy density
currents via the Bohm-Aharonov-Josephson effect for practical
metric engineering provided that the pure geometric currents
are not separately conserved like they are in EinsteinOs
original theory. One way to do that is to bottle or harness
the dark energy as General Douglas Mac Arthur precognitively
remote viewed in his Duty, Honor, Country Farewell Speech to
the Cadets at West Point in 1962 in a speech made even more
remarkable for its reference to the coming war in space with
extra-terrestrials. Michael Turner, a professor of physics at
the impossible to bottle dark energy and if EinsteinOs 1915
theory is the final theory he is correct. However, the real
UFO evidence that drives Hal Puthoff like Captain Ahab after
The White Wale, Moby Dick, is the evidence that suggests
Professor Turner will be proved wrong in that
prediction.ref.http://qedcorp.com/APS/StarGate1.movhttp://
qedcorp.com/APS/EmergentGravity.pdf
===
=IOm attempting to
connect two 3D squares together. They are of 16points each
and I?ve run them through the Bernstein equation.Now, I?m
quite new to 3 dimensional mathematics and new to
theBernstein formula, but its proving to be very
interesting.IOve been trying for days now to get the squares
to match up smoothly.There is a curvature in the surface of
the one square/patch and I?mattempting to keep the same
curvature through the second square (tomake half a ?t oval).
IOve heard about the tangents method, abouthaving tangents
from the control points on the one square to thecontrol
points on the other square, which should create a
smoothsurface. And here in is the problem, there is a definite
line, aninward slope as the two squares/patches meet. As far I
can tell thetangents line up along the join, I just donOt
have
a clue why the joinisnOt perfectly smooth.Many people must
have run into this problem. If you have any thoughts,ideas or
possible directions to pursue, they would all be very
muchappreciated.T. Overton
===
=IOm attempting to connect two
3D squares together. They are of 16points each and I?ve run
them through the Bernstein equation.Now, I?m quite new to 3
dimensional mathematics and new to theBernstein formula, but
its proving to be very interesting.IOve been trying for days
now to get the squares to match up smoothly.There is a
curvature in the surface of the one square/patch and
I?mattempting to keep the same curvature through the second
square (tomake half a ?t oval). IOve heard about the
tangents method, abouthaving tangents from the control points
on the one square to thecontrol points on the other square,
which should create a smoothsurface. And here in is the
problem, there is a definite line, aninward slope as the two
squares/patches meet. As far I can tell thetangents line up
along the join, I just donOt have a clue why the
joinisnOt
perfectly smooth.Many people must have run into this problem.
If you have any thoughts,ideas or possible directions to
pursue, they would all be very muchappreciated.T.
Overton
===
=I am guessing, if we take an (n_1 by n_2 by n_3 by
...n_m) m-dimensional box (where the nOs are positive
integers),and we let the number of lattice points of positive
integer coordinates(k_1,k_2,k_3, ...k_m), where NO common
primes divide EACH k, be q(n),then:limit{n_1->oo, n_2->oo,
n_3->oo, ...n_m->oo}q(n)/(n_1 * n_2 * n_3 * ...n_m) =
1/zeta(m).IE; the fraction of all finite sequences of m
Ouniformly randomly chosenO positive integers,
which are so
that no common prime divides every integer in these
sequences,is 1/zeta(m).(example of counted 4-tuple: (4,3,2,4)
*is* counted because 3 is not divided by 2.)But...what is
meant here by taking the limits of the nOs??I assume that
each n must approach infinity
OindependentlyO of the
others.(Would the limit be the same if n_k was a
monotonically increasing function of n_{k-1}, for
example?)And, what is meant exactly by Ouniformly randomly
chosenO?I am *not* quoting from a book, but rather have
myself UNRIGOROUSLY derived the above well-known result
today, and wish to know if the result has a more rigorous and
technically-correct statement.thanks,Leroy Quet
===
=> Because
KhinchtineOs constant exists and is finite, we
know that the>
distribution of the integers among the terms of the simple
continued> fraction of a sufficently random (whatever is meant
by this) positive> real is not normal.In other words, 1 is
much more likely to occur than, say, 1000 in the> continued
fraction expansion of most positive reals.> So, let us say we
have a real x where the geometric average of xOs> continued
fraction terms approaches KhinchtineOs constant, where
these>
terms are {a(k)}.> So, what can be said about the number
theoretical aspects of {a(k)},> such as:1) What is the
likelyhood that any a(k) will be prime, as k -> oo.ie. If
pcf(m) is the number of primes among a(k) for 1 <= k <= m,
then> what is pcf(m) asymptotical towards?2) Same question as
(1), but replace OprimesO with
Osquarefree> integersO.3) What
is likelihood a(k) and a(k+1) will be coprime?4) What are the
expected number of divisors for all a(k) where 1 <= k> <=
m?Etc etc etc ....> Of course, the number of primes among the
first m CF terms of mostpositive reals would exceed that, on
average, among the first mpositive integers in general, on
average.Right? thanks,> Leroy> Quet
===
=> For some even
positive integer m,> we have a m-by-m grid.In this 2 player
game, each player has (m^2/2) counters, > each counter
numbered with a distinct integer from 1 to (m^2/2).The
players take turns placing the counters into the gridOs
squares in> any order the players wish.(We do not actually
need the counters, for players can simply write> the numbers
in the gridOs squares. But the counters make it easy to>
know
which numbers each player has already used, for each integer
is> to be used one per player.)> (Or we can simply play this
on a computer.)> Scoring: One player is rows, the other
player is columns.For, say, rows, every set of adjacent
integers, where each immediately> adjacent (to left/right)
pair is coprime, is multiplied, then these> groups of
multiplied integers are all added up to get the>
row-playerOs
score.For columns, we do the same, but we consider immediately
adjacent> pairs which are adjacent above/below for
multiplication if coprime.As to help explain what I mean,
here is an example (of a game played> against myself without
using any strategy):> (Who plays which number is
unimportant.) 8 5 8 3> 6 1 2 6> 7 3 1 5> 2 4 7 4 Rows
gets:8*5*8*3 +> 6*1*2 + 6 > + 7*3*1*5> + 2 + 4*7*4Columns
gets: 8 + 6*7*2 +> 5*1*3*4 +> 8 + 2*1*7 +> 3 + 6*5*4> I would
guess that higher m than 4 would be more interesting.We can
use other criteria other than coprimality when
determiningwhich integers to multiply.(One advantage of
coprimality is that if 2 positive integers arelower, thenthey
are more likely to be coprime than if they had been higher,
which{I feel} improves this gameOs strategy.)Any interesting
variations on this game???thanks,Leroy Quet
===
=Does anyone
have any idea how to solve the
following:x+y+z=axy+yz+xz=bxyz=cI have a hunch that x,y,z
should be the roots of the cubic equations
t^3-at^2+bt-c=0
===
=Brian Troutwine
<goofy_headed_punk@msn.com> grava .88 la saucisse et au
marteau:> Does anyone have any idea how to solve the
following:> x+y+z=a> xy+yz+xz=b> xyz=c> I have a hunch
that x,y,z should be the roots of the cubic equations
t^3-at^2+bt-c=0ThatOs true. Now youOve got
many methods to
solve 3rd degreeOsequations, such as Cardan or Ferrari.--
Nicolas
===
=Could anyone help me solve the following:Find six
different nondegenerate triangles with integer sides forwhich
a=16 and A=60 degrees; where the angles of the triangles
arelabeled A,B,C and the sides opposite them a,b,c.
===
=>Could
anyone help me solve the following:>Find six different
nondegenerate triangles with integer sides for>which a=16 and
A=60 degrees; where the angles of the triangles are>labeled
A,B,C and the sides opposite them a,b,c.The law of cosines
says a^2 = b^2 + c^2 - 2 b c cos(A), i.e.256 = b^2 + c^2 - b
c.Sorry: the only positive integer solution is b=c=16. I
think the least value of a for which there are six different
triangles(if you count (a,b,c) and (a,c,b) as different when
b<>c) is 49, for whichyou have the triangles[49, 16, 55],
[49, 21, 56], [49, 35, 56], [49, 39, 55], [49, 49, 49], [49,
55, 16], [49, 55, 39], [49, 56, 35], [49, 56, 21]Robert
Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
=First my original post
copy/pasted, then the post I am replying to byFred the Wonder
Worm, before my reply below.>We have a regular hexagon whose
sides are lables, clockwise from top,>A through F.>>the
hexagonOs inner walls as if the walls were mirrored, but it
also>affects the direction of *itself*; for whenever it
crosses its own>path (as already drawn), it passes through
the path, but is re?cted>as if a mirror has been placed
perpendicularly to the previous pathat>the point of
intersection.>know>if such a particular path actually exists,
because the pathOs final>direction is heavily
dependent upon
the accuracy in which the path is>drawn and the accuracy of
the angles re?cted.>>But I will give the order of the
hexagonOs surfaces as visited by the>path (as drawn at the
time of each particular crossing) below.>>(I know this, if my
by-hand approximation was not too ?wed.)>>(I have no idea.
You best use exact-rational arithetic to get this,if>it is
possible to figure out at all.)>>E, B , D, 2 crossings, F,
E, 1 crossing, F, 3 crossing, D, B, 3>crossings, F, 7
crossings, and back to its staring point.>>(If no crossings
are listed between letters, than no crossings occur>between
them.)>>(If someone solves this, they are going to have to
post some link toa>webpage with the answer, I am
afraid.)>>thanks,>Leroy Quet> If anyone wishes to check my
angle chase, hereOs one description of it.>
IOve tried it
three times now, so either itOs right or IOm
hitting a> blind
spot. Start by drawing in a representative diagram up to that>
point. (Make it big -- this is important!) Let the vertices
in> clockwise order be labelled L, M, N, O, P, Q, so that LM
is side A, MN> is side B, etc. Let the photon start at point
R, and successive points> where it changes direction be S,
T,..., Z. (Z is the point on the last> D-F segment, and I aim
to show that Z does not actually exist.)That is, R is on side
A (LM), S is on side E (PQ), T is on B (MN),> U is on D (OP),
V is on ST, W is on RS, X is on F (QL), Y is on E (QS),> and Z
should be on WX.Let angle LRS = theta. Then we can chase
angles (all degree signs> omitted) as follows: QSR = 120 -
theta> PST = 120 - theta> RST = 2*theta - 60> MTS = 120 -
theta> NTU = 120 - theta> TUO = theta> PUV = theta> UVS =
120> SVW = 120> VWS = 120 - 2*theta> SWX = 120 - 2*theta> RWX
= 60 + 2*theta> LXW = 180 - 3*theta> QXY = 180 - 3*theta> WXY
= 6*theta - 180> XYQ = 3*theta - 120> SYZ = 3*theta - 120>
XYZ = 420 - 6*thetaBut then in triangle XYZ, angles XYZ + ZXY
sum to 240 degrees, which is> impossible. We conclude that the
ray is actually heading at 60 degrees> _away_ from the last
segment from D-F. So that ray must actually cross> segment WS
rather than WX. I believe it then crosses VS and hits E>
again, possibly then going via UV, TV and maybe back to the
starting> point. (I think, but am not sure, that there is
enough leeway for this.)So a modified version of the problem
might have the pattern as A, E, B, D, 2, F, E, 2, E, 2,
startGeoff.I wonder if the ray, after crossing VS does *not*
then head to E,but rather heads more upwards towards UV, then
(perhaps) spiralsinwardly and infinitely towards point V.But a
number of other things could perhaps happen, for my
diagram(hand-drawn) is ambiguous.(Around points V and Z,
there are a few segments that could eachthanks,Leroy
Quet
===
=sum[k^2/k!] from k=1 to k=infSurely there are some
cute algebraic tricks to get 2e out of this, Ijust donOt
know
them - lil help please?thx,cdj
===
=>>sum[k^2/k!] from k=1 to
k=inf>>Surely there are some cute algebraic tricks to get 2e
out of this, I>just donOt know them - lil help
please?>>thx,>>cdjWrite e^x = SUM(n=0 t =oo)
(x^n/n!)Differentiate both sides and then multiply both by x
Again, differentiate both sides and multiply both bu x. Put x
= 1. BINGO!
===
=> sum[k^2/k!] from k=1 to k=infThat equals
sum_(k=1,oo) k/(k-1)! = sum_(k=0,oo) (k+1)/k! = d(xe^x)/dx at
x = 1.
===
=sum[k^2/k!] from k=1 to k=infThat equals
sum_(k=1,oo) k/(k-1)! = sum_(k=0,oo) (k+1)/k! = d(xe^x)/dx at
x > = 1.well arenOt I just a genius now? lolthanks
all,cdj
===
=>>sum[k^2/k!] from k=1 to k=inf> That equals
sum_(k=1,oo) k/(k-1)! = sum_(k=0,oo) (k+1)/k! = d(xe^x)/dx at
x > = 1.Or, in the middle,sum_(k=0,oo) (k+1)/k! =
sum(k>0,k/k!) + sum(k>=0,1/k!)= sum(k>0, 1/(k-1)!) + e= 2eAs
I recall, sum(k>=0, k^n/k!) is an integer multiple of e.The
Bell numbers come to mind.
===
=cdj>> sum[k^2/k!] from k=1 to
k=inf>> Surely there are some cute algebraic tricks to get 2e
out of this, I> just donOt know them - lil help please?Write
k^2 = k(k-1) + kand get two simpler series.LH
===
=>
sum[k^2/k!] from k=1 to k=inf>> Surely there are some cute
algebraic tricks to get 2e out of this, I> just donOt know
them - lil help please?e^(e^x) = 1 + e^x + e^2x/2! + e^3x/3!
+ ...Differentiate twice, then set x = 0.
===
=Here is 2 player
game. (Inspired by math and other preexistingcommonly-played
games.) Each player has an identical set of 15 square tiles
colored asfollowed:Y = yellowR = redB = blueP = purple*Y **
*Y **** R* R* *B*Y ** *Y P* *B RB RB **PY P* PY P*** R* R*
*BPY P* PY *B RB RB Players take turns making a row of tiles,
one tile added at each turn.A player can ONLY add a tile if
the immediately previous tile she/heis to add shares at least
one color with the new tile.Each player get one point for each
color-combination not among his/heropponentOs PREVIOUSLY
placed pieces.(In other words, if the lines of tiles are
expanding to the right andare both aligned, then a player
gets a point for every piece ofhis/hers where the
corresponding piece of her/his opponent is NOT tothe LEFT of
the playerOs own tile in question.)And if any player cannot
use all of his/her pieces (because ofcolor-matching rule),
the player with the most played pieces gets 2points for each
additionally played piece beyond the number played byhis/her
opponent.(We could also just give each player 2 points for
each played piece,since the score-difference would be the
same anyway.)Sample game beginning (played without strategy
here)----------------------------------------------------
Player 1 Player 2-------- ---------*Y **RB RB*Y point **
point** R*PY point *Y point** R*PY point *Y point*B *B**
point P* point*B *BPY point P* pointRB **** P* pointR* RB**
**RB *Betc... etc...(And the points become less frequent as
play progresses.)I would suggest a greater # of colors, such
as 5, 6, 7, or 8 (8 forOadvancedO play). If we
have n colors,
we would have (2^n -1) piecesto get all combinations of colors
(except no colors).thanks,Leroy Quet
===
=I was looking through
my files the other day and found a simple Fermatme, but
thatOs
not saying much since I am not an expert mathematicianso I
thought I would post it here for comment by the
researchcommunity.Note: I am posting this out of mere
curiosity, so please be gentle :-)FermatOs Proof
Below:--------------------------The is a simple proof that
for the equation x^n+y^n=z^n there are nointeger solutions
for n>2.BackgroundFermat must have been considering the
Pythagorean Theorem x^2+y^2=z^2and observed the integer
solution sets x=3I, y=4I, z=5I and mused, dointeger solutions
exist for x^n+y^n=z^n (and had decided no).Restate the
equation: x^n (x^n+y^n)^(n-1) + y^n (x^n+y^n)^(n-1)
=(x^n+y^n)^nNotice the factor (x^n+y^n) has the exponent
(n-1) on the left side ofthe equation and the exponent (n) on
the right side.To solve this equation we must introduce some
exponent factor so thatf(scaler, n-1) and f(scaler, n) have a
common denominator (n).FermatOs proof simply observes (n-1)
and (n) are sequential. Theexponent operators are
A^xA^y=A^(x+y) (addition) or A^x/A^y=A^(x-y)(subtraction).
Sequential numbers do not have common denominatorsgreater
than 1. There is no number that can be added to, orsubtracted
from, sequential numbers that will change the fact thatthey
are sequential.Proof-->Sequential numbers do not have common
denominators greater than 1-->If two numbers are sequential,
one is ODD and one is EVEN and theirdifference is 1-->Factor
the numbers-ODD=ODDa*ODDbEVEN=EVENa*EVENb=EVENa*ODDb-->There
are 2 equations:1=ODD-EVEN 1=EVEN-ODDReplace with factors and
solve.-->1=(ODDa*ODDb)-(EVENa*ODDb)
1=(EVENa*ODDb)-(ODDa*ODDb)1=ODDb(ODDa-EVENa)
1=ODDb(EVENa-ODDa)SolveODDb=1 ODDb=1ODDa=EVENa+1
EVENa=ODDa+1Sequential numbers do not have common
denominators greater than 1.
===
=The Fermat crackpot position
is filled by James Harris and has beenfor a decade or more.
YouOll have to stand in line. Thousands of professional
mathematicians worked on the problem for acentury or so. And
your friend waltzed in with a trivial solution??Truly
amazing!>I was looking through my files the other day and
found a simple Fermat>me, but thatOs not saying much since I
am not an expert mathematician>so I thought I would post it
here for comment by the research>community.>>Note: I am
posting this out of mere curiosity, so please be gentle
:-)>>FermatOs Proof Below:>-------------------------->The is
a simple proof that for the equation x^n+y^n=z^n there are
no>integer solutions for n>2.>>Background>Fermat must have
been considering the Pythagorean Theorem x^2+y^2=z^2>and
observed the integer solution sets x=3I, y=4I, z=5I and
mused, do>integer solutions exist for x^n+y^n=z^n (and had
decided no).>>Restate the equation: x^n (x^n+y^n)^(n-1) + y^n
(x^n+y^n)^(n-1) =>(x^n+y^n)^n>>Notice the factor (x^n+y^n) has
the exponent (n-1) on the left side of>the equation and the
exponent (n) on the right side.>>To solve this equation we
must introduce some exponent factor so that>f(scaler, n-1)
and f(scaler, n) have a common denominator (n).>>FermatOs
proof simply observes (n-1) and (n) are sequential.
The>exponent operators are A^xA^y=A^(x+y) (addition) or
A^x/A^y=A^(x-y)>(subtraction). Sequential numbers do not have
common denominators>greater than 1. There is no number that
can be added to, or>subtracted from, sequential numbers that
will change the fact that>they are
sequential.>>Proof>>-->Sequential numbers do not have common
denominators greater than 1>-->If two numbers are sequential,
one is ODD and one is EVEN and their>difference is
1>>-->Factor the
numbers->ODD=ODDa*ODDb>EVEN=EVENa*EVENb>=EVENa*ODDb>>-->There
are 2 equations:>1=ODD-EVEN 1=EVEN-ODD>Replace with factors
and solve.>>-->1=(ODDa*ODDb)-(EVENa*ODDb)
1=(EVENa*ODDb)-(ODDa*ODDb)>1=ODDb(ODDa-EVENa)
1=ODDb(EVENa-ODDa)>Solve>ODDb=1 ODDb=1>ODDa=EVENa+1
EVENa=ODDa+1>>Sequential numbers do not have common
denominators greater than 1.
===
=(some snips)> Fermat must
have been considering the Pythagorean Theorem x^2+y^2=z^2>
and observed the integer solution sets x=3I, y=4I, z=5I and
mused, do> integer solutions exist for x^n+y^n=z^n (and had
decided no).Restate the equation: x^n (x^n+y^n)^(n-1) + y^n
(x^n+y^n)^(n-1) => (x^n+y^n)^nYou might notice that the
restated equation is always true for any value of x, y, and
n. Eliminating z in that way removes the dependency of the
truth of the equation on the existence of a counterexample to
FLT. Thus, any further proof attempt that focuses on the
restated equation canOt possibly prove FLT.-- Mark
Thornquist
===
=(some snips)Fermat must have been considering
the Pythagorean Theorem x^2+y^2=z^2> and observed the integer
solution sets x=3I, y=4I, z=5I and mused, do> integer
solutions exist for x^n+y^n=z^n (and had decided no).Restate
the equation: x^n (x^n+y^n)^(n-1) + y^n (x^n+y^n)^(n-1) =>
(x^n+y^n)^nYou might notice that the restated equation is
always true for > any value of x, y, and n. Wait a second...
He then proves that this equation has no solution! This is
absolutely brilliant. > Eliminating z in that way removes the
> dependency of the truth of the equation on the existence of
a > counterexample to FLT. Thus, any further proof attempt
that > focuses on the restated equation canOt possibly prove
FLT.
===
=> To solve this equation we must introduce some
exponent factor so that> f(scaler, n-1) and f(scaler, n) have
a common denominator (n).Depending on what this is supposed to
mean, it is either irrelevant orsatisfied by scaler[sic]^(n-1)
.-- Daniel W.
Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes
/039 53 36 N / 086 11 55 W
===
=Unless you are a crank (which I
will assume you arenot (for now)), you should avoid at all
costs posting proofsof FLT. Even if you got it right, people
will alljust assume youOre crazy. Statistically,
itOspretty
much true, and hardly any seasoned mathiewill take the time
to find the ?w in your proof.If you want to learn math,
and
interact withmore experienced mathies, I suggest finding
otherquestions to approach and try to solve - howmany
polyominoes are there? Does the 3n+1sequence always
terminate? If you _insist_ onstudying FLT, then it is most
instructive tofind the mistakes in your proof for
yourself.Believe me, unless you really suck at math(or are a
crank),you truly are capable of finding your
ownmistakes.Finally, and a bit hypocritically, I tried tofind
the ?w in your proof but right away IcanOt tell what
you
mean. What is your overallform of argument? I assume it is to
arriveat a contradiction -- that is, I assume that youare
assuming that x^n+y^n = z^n for some fixedpositive integer
triple x,y,z and fixed n>2; andthat you want to derive a
contradiction from this.So why did you stop using z in your
equation? Youwill have to use that z is a positive
integer_somewhere_. Also, what is this functionf you
introduce? What is an exponent factorand how do you plan to
use it to solve theequation? Why must f(--) and f(---) havea
common denominator of n? Finally, scalar isspelled scalar not
scaler.Tyler> I was looking through my files the other day and
found a simple Fermat> me, but thatOs not saying much since
I
am not an expert mathematician> so I thought I would post it
here for comment by the research> community.Note: I am
posting this out of mere curiosity, so please be gentle
:-)FermatOs Proof Below:> --------------------------> The is
a simple proof that for the equation x^n+y^n=z^n there are
no> integer solutions for n>2.Background> Fermat must have
been considering the Pythagorean Theorem x^2+y^2=z^2> and
observed the integer solution sets x=3I, y=4I, z=5I and
mused, do> integer solutions exist for x^n+y^n=z^n (and had
decided no).Restate the equation: x^n (x^n+y^n)^(n-1) + y^n
(x^n+y^n)^(n-1) => (x^n+y^n)^nNotice the factor (x^n+y^n) has
the exponent (n-1) on the left side of> the equation and the
exponent (n) on the right side.To solve this equation we must
introduce some exponent factor so that> f(scaler, n-1) and
f(scaler, n) have a common denominator (n).FermatOs proof
simply observes (n-1) and (n) are sequential. The> exponent
operators are A^xA^y=A^(x+y) (addition) or A^x/A^y=A^(x-y)>
(subtraction). Sequential numbers do not have common
denominators> greater than 1. There is no number that can be
added to, or> subtracted from, sequential numbers that will
change the fact that> they are sequential.Proof-->Sequential
numbers do not have common denominators greater than 1> -->If
two numbers are sequential, one is ODD and one is EVEN and
their> difference is 1-->Factor the numbers-> ODD=ODDa*ODDb>
EVEN=EVENa*EVENb> =EVENa*ODDb-->There are 2 equations:>
1=ODD-EVEN 1=EVEN-ODD> Replace with factors and
solve.-->1=(ODDa*ODDb)-(EVENa*ODDb)
1=(EVENa*ODDb)-(ODDa*ODDb)> 1=ODDb(ODDa-EVENa)
1=ODDb(EVENa-ODDa)> Solve> ODDb=1 ODDb=1> ODDa=EVENa+1
EVENa=ODDa+1Sequential numbers do not have common
denominators greater than 1.
===
=Hash: SHA1examine the
statement before posting it, and deemed it worthy
ofdiscussion. With regard to whether or not it qualifies as a
proof,we must first agree on a definition of the
word.My
understanding is that a mathematical proof is a
demonstrationthat, given certain *axioms* (widely accepted
truths), a presentedtheory is necessarily true. Since widely
accepted is a subjectiveassessment, it often becomes
necessary for a mathematician (or otherproblem solver, for
that matter) to break proposed problems down intosmaller
parts that are more easily reconciled with reality.If the
above understanding is correct, then proving or disprovingFLT
involves a demonstration of how existing axioms
eithernecessitate its truth or untruth.Thus it appears that
the submitted work stands upon following axioms:1) The Axiom
of Variable Substitution - used here to restate theequation
to:x^n(x^n+y^n)^(n-1) + y^n(x^n+y^n)^(n-1) = (x^n+y^n)^n2) In
order for this equation to be solved, its writer states
that(n-1) and (n)--sequential numbers--must be shown to have
a commondenominator greater than 1. As a layperson, I can not
confirm theaxiomatic nature of this statement, perhaps someone
else can confirmor deny it.3) Sequential numbers can not share
a common denominator greater than1.In short, the attempted
proof reveals that the root of the problem isthat sequential
numbers can not be shown to have common denominatorsgreater
than 1, explaining why the equation breaks at n > 2, butworks
for 1 and 2.In light of the above, there are only 3 ways to
disprove this theory:1) The definition of proof as proposed is
incorrect.2) One of the three (3) axioms can be shown to be
false (2 lookslike a candidate from my seat, 1 & 3 appear
rather obvious).3) An error in deductive reasoning was made
at one of the three steps(e.g., the variable substitution was
not performed correctly).With all of the brainpower in this
forum, I am confident that we caneither confirm or
rapidly
dispense with (I agree this is more likely)this theory and
perhaps emerge with a greater level of understandingas
well.As iron sharpens iron,so one man sharpens another. -
ProverbsiQA/
AwUBP8O79KDLMcpPQeVjEQLBpQCfXmBOKPS7Z5gwAogY332Z2P8jmkAAoMYOlI
bArqOy/03cGqLxG2ljzlel=IpQk> Unless you are a crank (which I
will assume you are> not (for now)), you should avoid at all
costs posting proofs> of FLT. Even if you got it right,
people will all> just assume youOre crazy. Statistically,
itOs> pretty much true, and hardly any seasoned mathie> will
take the time to find the ?w in your proof.If you want to
learn math, and interact with> more experienced mathies, I
suggest finding other> questions to approach and try to solve
- how> many polyominoes are there? Does the 3n+1> sequence
always terminate? If you _insist_ on> studying FLT, then it
is most instructive to> find the mistakes in your proof for
yourself.> Believe me, unless you really suck at math> (or
are a crank),> you truly are capable of finding your own>
mistakes.Finally, and a bit hypocritically, I tried to> find
the ?w in your proof but right away I> canOt tell what
you
mean. What is your overall> form of argument? I assume it is
to arrive> at a contradiction -- that is, I assume that you>
are assuming that x^n+y^n = z^n for some fixed> positive
integer triple x,y,z and fixed n>2; and> that you want to
derive a contradiction from this.> So why did you stop using
z in your equation? You> will have to use that z is a
positive integer> _somewhere_. Also, what is this function> f
you introduce? What is an exponent factor> and how do you plan
to use it to solve the> equation? Why must f(--) and f(---)
have> a common denominator of n? Finally, scalar is> spelled
scalar not scaler.TylerI was looking through my files the
other day and found a simple Fermat> me, but thatOs not
saying much since I am not an expert mathematician> so I
thought I would post it here for comment by the research>
community.Note: I am posting this out of mere curiosity, so
please be gentle :-)FermatOs Proof Below:>
--------------------------> The is a simple proof that for
the equation x^n+y^n=z^n there are no> integer solutions for
n>2.Background> Fermat must have been considering the
Pythagorean Theorem x^2+y^2=z^2> and observed the integer
solution sets x=3I, y=4I, z=5I and mused, do> integer
solutions exist for x^n+y^n=z^n (and had decided no).Restate
the equation: x^n (x^n+y^n)^(n-1) + y^n (x^n+y^n)^(n-1) =>
(x^n+y^n)^nNotice the factor (x^n+y^n) has the exponent (n-1)
on the left side of> the equation and the exponent (n) on the
right side.To solve this equation we must introduce some
exponent factor so that> f(scaler, n-1) and f(scaler, n) have
a common denominator (n).FermatOs proof simply observes
(n-1)
and (n) are sequential. The> exponent operators are
A^xA^y=A^(x+y) (addition) or A^x/A^y=A^(x-y)> (subtraction).
Sequential numbers do not have common denominators> greater
than 1. There is no number that can be added to, or>
subtracted from, sequential numbers that will change the fact
that> they are sequential.Proof-->Sequential numbers do not
have common denominators greater than 1> -->If two numbers
are sequential, one is ODD and one is EVEN and their>
difference is 1-->Factor the numbers-> ODD=ODDa*ODDb>
EVEN=EVENa*EVENb> =EVENa*ODDb-->There are 2 equations:>
1=ODD-EVEN 1=EVEN-ODD> Replace with factors and
solve.-->1=(ODDa*ODDb)-(EVENa*ODDb)
1=(EVENa*ODDb)-(ODDa*ODDb)> 1=ODDb(ODDa-EVENa)
1=ODDb(EVENa-ODDa)> Solve> ODDb=1 ODDb=1> ODDa=EVENa+1
EVENa=ODDa+1Sequential numbers do not have common
denominators greater than 1.
===
=> examine the statement
before posting it, and deemed it worthy of> discussion. With
regard to whether or not it qualifies as a proof,> we must
first agree on a definition of the word.No, we
already have a
definition. If you wish to use the wordproof, you are well
advised to use its current meaning and not makeup some new
meaning. > If the above understanding is correct, then
proving or disproving> FLT involves a demonstration of how
existing axioms either> necessitate its truth or untruth.Yes,
and the axioms in question are PeanoOs axioms, which are the
verydefinition of the natural numbers. Thomas
===
=> My
understanding is that a mathematical proof is a
demonstration> that, given certain *axioms* (widely accepted
truths), a presented> theory is necessarily true. Since
widely accepted is a subjective> assessment, it often becomes
necessary for a mathematician (or other> problem solver, for
that matter) to break proposed problems down into> smaller
parts that are more easily reconciled with reality.Oh, and
axioms are statements which are too *simple* to be
proved.Example: If a = b and b = c, then a = c. Your axioms
have nobusiness being claimed as axioms, although axiom 1 has
the virtue ofbeing close to the relevant axioms. (You might as
well add the FLTitself as axiom 4 and really shorten your
proof.)You donOt get to pick out the holes in your proof and
call themaxioms.-- Daniel W.
Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes
/039 53 36 N / 086 11 55 W
===
=> Oh, and axioms are statements
which are too *simple* to be proved.> Example: If a = b and b
= c, then a = c. What makes that too simple to be
proved?
===
=> 1) The Axiom of Variable Substitution - used
here to restate the> equation to:x^n(x^n+y^n)^(n-1) +
y^n(x^n+y^n)^(n-1) = (x^n+y^n)^nYouOve restated the equation
as a tautology. This is true for allconceivable values of x,
y, z, and n. > 2) In order for this equation to be solved,
its writer states that> (n-1) and (n)--sequential
numbers--must be shown to have a common> denominator greater
than 1. As a layperson, I can not confirm the> axiomatic
nature of this statement, perhaps someone else can confirm> or
deny it.The equation as stated above certainly does not
require any particularcommon DIVISOR for (n-1) and (n). And
IOm still not sure what thisalleged axiom is trying to
say.Would this axiom assert that a solution for a * k^2 + b *
k^2 = k^3requires a common divisor for 2 and 3?-- Daniel W.
Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes
/039 53 36 N / 086 11 55 W
===
= [.snip.]>My understanding is
that a mathematical proof is a demonstration>that, given
certain *axioms* (widely accepted truths), a presented>theory
is necessarily true. No. Mathematical axioms are NOT widely
accepted truths; they aremerely formal statements. Truth
donOt enter into it.>2) In order for this equation to be
solved, its writer states that>(n-1) and (n)--sequential
numbers--must be shown to have a common>denominator greater
than 1. This is nonsense as written. You mean a common
DIVISOR.>As a layperson, I can not confirm the>axiomatic
nature of this statement, perhaps someone else can confirm>or
deny it.The statement is unsupported. The fact that n-1 and n
are relativelyprime (have no common divisors other than 1 and
-1) was not shown toimply FermatOs Last Theorem.
DonOt know
why he thinks this is the case.>3) Sequential numbers can not
share a common denominator greater than>1.DIVISOR, not
denominator. The OdenominatorO of an integer
isusually taken
to be 1, period.>In short, the attempted proof reveals that
the root of the problem is>that sequential numbers can not be
shown to have common denominators>greater than 1, explaining
why the equation breaks at n > 2, but>works for 1 and 2.No:
the attempted proof ->ASSERTS<- that FermatOs Last Theorem
has ton and n-1 being relatively prime, but that assertion is
notjustified. As for 1 and 2, 1 and 2 have no common divisors
other than 1 and -1, so why would the assertions about n not
work for n=2?>In light of the above, there are only 3 ways to
disprove this theory:>>1) The definition of proof as proposed
is incorrect.Your understanding of what an axiom is is
certainly incorrect.>2) One of the three (3) axioms can be
shown to be false (2 looks>like a candidate from my seat, 1 &
3 appear rather obvious).2 is not and was not asserted to be,
an axiom. Neither is 3, for thatmatter. 3 is a conclusion
derived from the basic properties of theintegers, while 2 was
asserted and not proven:>> Notice the factor (x^n+y^n) has the
exponent (n-1) on the left side of>> the equation and the
exponent (n) on the right side.> To solve this equation
we must introduce some exponent factor so that>> f(scaler,
n-1) and f(scaler, n) have a common denominator (n).The last
paragraph is (a) nonsense as written; what is scaler? Whatis
f(scaler,n-1)? What is f(scaler, n)? What common
denominator?Does he mean common factors? In short:
imprenetable
nonsense.
===
=================================================

===
===============ItOs not denial. IOm just very
selective
about what I accept as reality. --- Calvin (Calvin and
Hobbes)
===
===================================================

===
=============Arturo
Magidinmagidin@math.berkeley.edu
===
=Hash: SHA1Geez...I tried
to preface my post with the fact that I am NOT
amathematician, I was just looking for a little peer review
of an textI received from an associate. In any event:>My
understanding is that a mathematical proof is a
demonstration>that, given certain *axioms* (widely accepted
truths), a presented>theory is necessarily true. No.
Mathematical axioms are NOT widely accepted truths; they are>
merely formal statements. Truth donOt enter into it.- -
-snip-> Your understanding of what an axiom is is certainly
incorrect.I appreciate you attempting to polish up my
definition, but fail tosee how any formal statement is
axiomatic? You may also want toclearthis up with both Merriam
Webster (www.m-w.com) and the MITmathematics lab since they
share the same definition:http://tinyurl.com/wl0m (Section 3:
Axioms)All progress and greater knowledge originate from this
process ofseparating the wheat from the chaff, so I do greatly
appreciate theconstructive comments and illumination that
others have contributedand hope that in some small way the
post may have stimulated
theiQA/AwUBP8QWvaDLMcpPQeVjEQI9sQCgvlW+
v5MIEIMXmn8Xk0mnz6rEiL0AoMYqU53LXg0i2RFQ+cICbmaO/2h8=Nc/3
===
=
> Finally, and a bit hypocritically, I tried to> find the
?w
in your proof but right away I> canOt tell what you mean.He
seems to be tacitly assuming that if m and n are coprime, so
are k^mand k^n. (And the second half of the proof simply
establishes thatn-1 and n are coprime.)-- Daniel W.
Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes
/039 53 36 N / 086 11 55 W
===
=> Unless you are a crank (which
I will assume you are> not (for now)), you should avoid at all
costs posting proofs> of FLT. Even if you got it right, people
will all> just assume youOre crazy.I have never assumed
anyone
is crazy for posting a FLT proof;just optimistic. And if it
even is right, all the better.> Statistically, itOs> pretty
much true, and hardly any seasoned mathie> will take the time
to find the ?w in your proof.ThatOs strange
- every single FLT
proof I have seen here overthe years has been shot down by
someone fairly quickly.(...)---J K
Hauglandhttp://www.neutreeko.com
===
=> I was looking through
my files the other day and found a simple Fermat> me, but
thatOs not saying much since I am not an expert
mathematician> so I thought I would post it here for comment
by the research> community.With many simple proofs of hard
theorems, there is a very simple way to see that is something
fishy, and it works with this proof as well. It is well known
that a^n + b^n = c^n has plenty of solutions in positive
integers a, b, c if n = 2 and none have ever been found if n
> 2. So whatever proof you have that a^n + b^n = c^n _must_
fail if n = 2. If it doesnOt fail for n = 2 then it proves
something wrong, so even if you donOt understand the proof
you _know_ it cannot be correct. The proof you posted never
requires that n > 2.
===
=> The proof you posted never requires
that n > 2.It never even requires that n > 1.-- Daniel W.
Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes
/039 53 36 N / 086 11 55 W
===
=> With many simple proofs of
hard theorems, there is a very simple way > to see that is
something fishy, and it works with this proof as well. >
(SNIP)The proof you posted never requires that n > 2.Care
should be taken, though. The proof of the irrationality
ortranscendance of pi requires that pi be the ratio of a
circleOscircumference to itOs diameter in a
rather subtle,
not-obvious way.
===
=fuffy> I was looking through my files the
other day and found a simple Fermat> me, but thatOs not
saying much since I am not an expert mathematician> so I
thought I would post it here for comment by the research>
community.>> Note: I am posting this out of mere curiosity,
so please be gentle :-)>> FermatOs Proof Below:>
--------------------------> The is a simple proof that for
the equation x^n+y^n=z^n there are no> integer solutions for
n>2.>> Background> Fermat must have been considering the
Pythagorean Theorem x^2+y^2=z^2> and observed the integer
solution sets x=3I, y=4I, z=5I and mused, do> integer
solutions exist for x^n+y^n=z^n (and had decided no).>>
Restate the equation: x^n (x^n+y^n)^(n-1) + y^n
(x^n+y^n)^(n-1) => (x^n+y^n)^n>> Notice the factor (x^n+y^n)
has the exponent (n-1) on the left side of> the equation and
the exponent (n) on the right side.>> To solve this equation
we must introduce some exponent factor so that> f(scaler,
n-1) and f(scaler, n) have a common denominator (n).>>
FermatOs proof simply observes (n-1) and (n) are sequential.
The> exponent operators are A^xA^y=A^(x+y) (addition) or
A^x/A^y=A^(x-y)> (subtraction). Sequential numbers do not
have common denominators> greater than 1. There is no number
that can be added to, or> subtracted from, sequential numbers
that will change the fact that> they are sequential.>> Proof>>
-->Sequential numbers do not have common denominators greater
than 1> -->If two numbers are sequential, one is ODD and one
is EVEN and their> difference is 1>> -->Factor the numbers->
ODD=ODDa*ODDb> EVEN=EVENa*EVENb> =EVENa*ODDb>> -->There are 2
equations:> 1=ODD-EVEN 1=EVEN-ODD> Replace with factors and
solve.>> -->1=(ODDa*ODDb)-(EVENa*ODDb)
1=(EVENa*ODDb)-(ODDa*ODDb)> 1=ODDb(ODDa-EVENa)
1=ODDb(EVENa-ODDa)> Solve> ODDb=1 ODDb=1> ODDa=EVENa+1
EVENa=ODDa+1>> Sequential numbers do not have common
denominators greater than 1.
===
=>> fuffy> fuffy>>
>> Fuffy has refined its tactics. First, all that Fuffy did
was>> to keep the name of the most current poster in a
thread>> from appearing on Google.> Free clue for the
clueless: If you want to see the>> threads have lots of
sub-threads, and the default>> order makes it difficult to
find
recent comments>> in more than one.> What is to be done
about Fuffy?> Duh-h-h. DonOt ask me.> Well, I suspect
virtually every Google user but you>> already knew about sort
by date and how to find >> recent posters.> IOm
a
Mathie.> <snort>People want to know who the last poster
in a thread is,>without having to do a Google Advanced Search.
Second clue for the clueless:Sort by date is on every thread
with more than one message. Lookaround. Nothing to do with
Advanced Search.> For>example, if Google shows Justin Van
Twinkie as the last>poster, I wonOt bother to read the last
postingThen youOll have no idea whether or not 15 people
have
posted sincethe last time you read the thread.
===
=>>
fuffy> fuffy> Fuffy has refined its tactics.
First, all that Fuffy did was>> to keep the name of the most
current poster in a thread>> from appearing on Google.>
Free clue for the clueless: If you want to see the>> threads
have lots of sub-threads, and the default>> order makes it
difficult to find recent comments>> in more than
one.> What
is to be done about Fuffy?> Duh-h-h. DonOt ask me.>
Well, I suspect virtually every Google user but you>> already
knew about sort by date and how to find >> recent posters.>
IOm a Mathie.> <snort>People want to know who the last
poster in a thread is,>without having to do a Google Advanced
Search. Second clue for the clueless:Sort by date is on every
thread with more than one message. Look> around. Nothing to
do with Advanced Search.For>example, if Google shows Justin
Van Twinkie as the last>poster, I wonOt bother to read the
last postingThen youOll have no idea whether or not 15
people
have posted since> the last time you read the thread.All
right, now I see what you mean. Now I know how to keep
McCallum from wasting my time. On the otherhand, IOll bet
there are Google Users other than myself who donOtyou seem
to
want to believe, and for people who donOt know how itworks,
McCallum fucks things up in a way that he has no right
to.
===
=Suppose you have some random variables X_j that
satisfythe conditions of the central limit theorem
(identialmeans and variances). We all know that the SUM of
suchrandom variables tends toward a Gaussian distribution.Now
suppose that you have some function f(X) that maps each of
these random variables X_j. Suppose for simplicitythat the
function f(X) is invertible, at least in some
restricteddomain, but perhaps nonlinear.Is there any
statement that can be made about the distributionof the sum
of f(X_j)?References would be greatly appreciated, also.Much
thanks,JB
===
=>Suppose you have some random variables X_j that
satisfy>the conditions of the central limit theorem
(idential>means and variances). We all know that the SUM of
such>random variables tends toward a Gaussian
distribution.>Now suppose that you have some function f(X)
that maps >each of these random variables X_j. Suppose for
simplicity>that the function f(X) is invertible, at least in
some restricted>domain, but perhaps nonlinear.>Is there any
statement that can be made about the distribution>of the sum
of f(X_j)?Well, obviously you want the f(X_j) to satisfy
conditionson means and variances. They donOt have to be
identical, though. LindebergOs Theorem says if Y_j are
independent with means mu_jand variances sigma_j^2, M_n =
sum_{j=1}^n mu_j and S_n^2 = sum_{j=1}^n sigma_j^2, with
1/S_n^2 sum_{j=1}^n E[(Y_j - mu_j)^2 I(|Y_j - mu_j| >= S_n
epsilon) -> 0as n -> infinity for all epsilon > 0 (I(A) being
the indicator function of A, i.e. 1 if A is true and 0
otherwise), then 1/S_n sum_{j=1}^n (Y_j - mu_j) converges in
distribution to a standard normal random variable.A condition
that may be easier to apply is that 1/S_n^3 sum_{j=1}^n E
|(Y_j - mu_j)^3| -> 0 as n -> infinity.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
=>Suppose you have some random
variables X_j that satisfy>the conditions of the central limit
theorem (idential>means and variances). We all know that the
SUM of such>random variables tends toward a Gaussian
distribution.Actually we donOt know that: assuming your X_j
are independent, but not identically distributed, identical
means and variancesare not enough for a Central Limit
Theorem: you need the Lindebergcondition. An amusing example
isX_j = 0 with probability 1-1/j^2 +/- j with probability
1/(2 j^2) eachso all X_j have mean 0 and variance 1. But
since sum_j 1/j^2 < infinity,with probability 1 all but
finitely many X_j are 0, and thus 1/n sum_{j=1}^n X_j -> 0
almost surely. >Now suppose that you have some function f(X)
that maps >each of these random variables X_j. Suppose for
simplicity>that the function f(X) is invertible, at least in
some restricted>domain, but perhaps nonlinear.>Is there any
statement that can be made about the distribution>of the sum
of f(X_j)?Well, obviously you want the f(X_j) to satisfy
conditions> on means and variances. They donOt have to be
identical, though. > LindebergOs Theorem says if Y_j are
independent with means mu_j> and variances sigma_j^2, M_n =
sum_{j=1}^n mu_j and > S_n^2 = sum_{j=1}^n sigma_j^2, with >
1/S_n^2 sum_{j=1}^n E[(Y_j - mu_j)^2 I(|Y_j - mu_j| >= S_n
epsilon) -> 0I left out a bracket: that should be 1/S_n^2
sum_{j=1}^n E[(Y_j - mu_j)^2 I(|Y_j - mu_j| >= S_n epsilon)]
-> 0> as n -> infinity for all epsilon > 0 (I(A) being the
indicator function of > A, i.e. 1 if A is true and 0
otherwise), then > 1/S_n sum_{j=1}^n (Y_j - mu_j) converges
in distribution to a standard > normal random variable.E.g.
suppose for simplicity all X_j and f(X_j) have mean 0, f(0) =
0, and there are positive constants a and b such that a(y-x) <
f(y) - f(x) < b(y-x) whenever x < y. Thena^2 Var(X_j) <=
Var(f(X_j)) <= b^2 Var(X_j). IOll write S_n^2 = sum_{j=1}^n
Var(X_j) and SO_n^2 = sum_{j=1}^n Var(f(X_j)),so a S_n <=
SO_n <= b S_n.In LindebergOs conditionf(X_j)^2
I(|f(X_j)| >
SO_n epsilon) <= b^2 X_j^2 I(|X_j| > S_n epsilon a/b)so if
X_j satisfy the condition, so do f(X_j). Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
=OK I see where I went wrong
now, I think. Let me try again.Consider the Category, (I use
capital C to indicate it may have the cardinality of all
classes, thus have to be a super-category or whatever);whose
objects are all ordinary categories, and whose arrows are
functors.This Category has natural transformations defined
within it, which are(I think?) mappings from functors to
functors. So they are a sort of2nd-order Arrow in this
Category.So, given their properties, what do Natural
Transformations (in this Category) correspond to in the way
of mappings-of-arrows in an ordinary category?Does this make
sense now? What special named sort of arrow-transformation do
they correspond
to?----------------------------------------------------------
-------------------- Bill Taylor
W.Taylor@math.canterbury.ac.nz-------------------------------
----------------------------------------------- I shot an
arrow in the air And down it came, I know not where. I looked
for it up in a tree Which was, though, just a
categOry!----------------------------------------------------

--------------------------
===
=>A and B have different
dimensions,>but AB and BA are both square.>Can anyone provide
a proof showing the eigenvalues are the same for AB and BA
?Side question: WhatOs your favorite linear algebra text for
answeringthis sort of question? I pulled out my Horn and
Johnson but realized Ihad no idea where to look to find a
proof, if there is one.Eigenvalue was too broad an index
entry.Related question: Do you have a favorite book
coveringspecially-structured matrices? That would include,
for example, fastToeplitz inversion methods? Or could easily
answer the question whensomebody asks, is there a name for
this kind of matrix? - Randy
===
=0072>> IOve got a list of
points (in x,y form) that are the corners of a> polygon, in
order. ItOs not necessarily a convex polygon. Is there a>
standard algorithm for determining whether a given point is
inside> or outside the polygon? All this takes place in the
plane... For> practical purposes, an algorithm is ok if it
can handle up to 7 points> decently.>> Many thanks
-Adrian> This is probably in the faq at
comp.programming.games.algorithms, or> something like that.
Google will probably also help.>> I know itOs in the
comp.graphics.algorithms FAQ.>>
http://www.faqs.org/faqs/graphics/algorithms-faq/>> Check out
question 2.03.>> - Randy
===
=>> I have looking at a few web
pages dealing with the largest known> calculated primes and
a great deal of computational time is taking> into searching
for these numbers and verifying they are primes. I have>
seen the EuclidOs proof of infinitude primes and
it occurs
that me> that super-large prime numbers can be calculated
using the following:>> <snip> Others have already given you
an example of why your idea wonOt ?, butIthere> is no
largest
prime (and therefore an infinite number of primes) is as>
follows:>> If there is a largest prime, P, then the number of
primes is finite. Wecan> therefore form a new number, N, which
is one greater than the product of> all primes. Therefore, N
> P (and, indeed, it would be very much greater> than P).>>
When divided by any prime in our list, it leaves remainder 1.
Therefore,> /either/ it is prime /or/ it is the product of two
or more primes, atleast> one of which is greater than P.>> In
/either/ case, P has been shown not to be the largest prime
number.>> Nope. Nowhere have you assumed that the list of
primes known is> complete up to P. YouOve simply proved that
the finite list of> known primes is not the list of all
primes.>> If you use EuclidOs proof as constructive, then
the
sets you generate0079> {2,3,7}> {2,3,7,43}> {2,3,7,13,43,139}>
{2,3,7,13,43,139,3263443}> and thence you find the new primes
{547,607,1033,31051}, all of which> are less than the
currently largest known prime 3263443.>> DonOt forget the
second case! :-)>> DonOt forget that if youOre
pretending
that you donOt know what> the primes are, you
canOt suddenly
pull otherwise well-known> features of the prime numbers out
of a hat.>> Phil> --> Unpatched IE vulnerability:
ADODB.Stream local file writing> Description: Planting
arbitrary files on the local file system>
Exploit:
http://ip3e83566f.speed.planet.nl/eeye.html> (but unrelated
to the EEye exploit)
===
=>Well duh!>All of _what_,
though.Primes.John
===
=> Nope. That is prior art. However
taxing folks on their exusions and > bodily productions is
not far in the future. The jism tax. > ItOs the coming
thing.HOWWWWWWWWWWLL !!!!!! Pun of the week!But seriously, we
already have, here in NZ, proposed taxes on emissions,not of
people but cows. In order to comply with the perceived
obligationsunder the Kyoto protocol, (hey thereOs a song
line
there... Kyo-to pro-to..),our PC government proposed a new
cattle emission tax on farmers for the allegedharm to the
ozone layer due to emitted methane!Naturally, this was
immediately dubbed the fart tax by the predatory media,and
attracted so much derision, joking, and protest from the
farming communitythat itOs unlikely to see the light of day;
which is in some ways a pitybecause a fart tax would have
been a wondrous thing! (Even if smellinga little
fishy!)-------------------------------------------------------
----------------------- Bill Taylor
W.Taylor@math.canterbury.ac.nz-------------------------------
----------------------------------------------- Camou?ge
condoms: So they wonOt see you
coming.------------------------------------------------------
------------------------
===
=>I should note that the above
carries no attribution. It is>>copyrighted material
originally published in The Onion,>>but I couldnOt locate
any
of the original Onion material>>online anywhere. >>IOve
already posted the Onion material (actually, IOve posted the
link>to the material fron the OnionOs own website) elsewhere
in this thread.Huh. I didnOt realize their archive went back
that far. I once triedand eventually gave up. ItOs the one
with several eminent physicistsdiscussing Big Bang theory.
Like man, what if the whole universe is,like, a big Atom in
another Universe? Wow, heavy, man. That sort ofthing. -
Randy
===
=When a Riemann surface is constructed by joining
copies of the cutplane the arbitrariness in the choice of
cuts carries over into anarbitrariness in the Riemann
surface. And where there is no uniquedomain there is no
unique function so it is incorrect to speak ofthe function
w(z) defined by an initially given relation f(w,z)=0.It is
rigor not pedantry to remark that the derivable functions
aremany and that the initially given relation is just what
they have incommon. When Riemann introduced his Riemann
surfaces it may be that he didso to eliminate the
multiplicity that comes with branches. But themultiplicity
reappears in this altered form. To understand it one hasto
stay with analysis and not move on to a more abstract
geometricaltreatment. For other notes on this topic visit my
websitewww.riemannsurfaces.info.
===
=>When a Riemann surface
is constructed by joining copies of the cut>plane the
arbitrariness in the choice of cuts carries over into
an>arbitrariness in the Riemann surface.Lucky that thatOs
not
the only way to construct a Riemann surfacethen...> And where
there is no unique>domain there is no unique function so it
is incorrect to speak of>the function w(z) defined by an
initially given relation f(w,z)=0.Look up analytic
continuation in an advanced book on complexanalysis: the
standard construction of _the_ Riemann surface heredoes not
have anything to do with cuts, and thereOs nothingarbitrary
about it. The surface is just the set of all functionelements
which can be reached by continuation along a path,
startingwith a given function element.>It is rigor not
pedantry to remark that the derivable functions are>many and
that the initially given relation is just what they have
in>common.> When Riemann introduced his Riemann surfaces it
may be that he did>so to eliminate the multiplicity that
comes with branches. But the>multiplicity reappears in this
altered form. To understand it one has>to stay with analysis
and not move on to a more abstract geometrical>treatment.>
For other notes on this topic visit my
website>www.riemannsurfaces.info.> David C.
Ullrich
===
=>Message-id:
<2410d7e.0311241320.5b02c68e@posting.google.com>Can someone
provide with a mathematical algorithm which explains
the>horrendous pixel-crime perpetred by the infamous Mr.
Mensanator?>>http://www.thequantummachine.com/images.phpHey,
the new image is a big improvement over the original. A
littleconstructive criticism helps doesnOt it?>>I thought
that my TFT screen had been degraded by some air
pollutant,>but then I realized that it was only his image
which had some fungus>on it... :-)--MensanatorAce of
Clubs
===
=>> Given an integral domain R, can we not
construct a field in the same >> way we construct Q from Z?
I.e., equivalence classes on R x >> (R{0}), where (a, b) ~
(c, d) iff a d = b c . (Integral domains >> are by definition
commutative, no? So says Herstein.)>>If R has a 1, then the
mapping f(r) = (r,1) gives the desired>embedding; but without
the 1, how do we embed R?You are not really mapping to (r,1),
you are mapping to the class of> (r,1). So map r to the class
of (r^2,r) if r is nonzero, and map 0 to> the class of (0,r)
for arbitrary r different from 0. > Doh! Of course; because
cancellation holds in R (follows fromintegrality), if we map
r,s to (r^2,r) and (s^2,s), resp., then r^2.s= s^2.r iff r =
s...>

===
==========================================================

===
======> ItOs not denial. IOm just very
selective about>
what I accept as reality.> --- Calvin (Calvin and Hobbes)>

===
==========================================================

===
======Arturo Magidin> magidin@math.berkeley.edu====Please
help with equations/sketches for a circular torus
intersectionwhen the cutting plane is not parallel to its
axis of symmetry.Equations may be expressed in terms of torus
tube radius, tubedisplacement from axis,inclination/length of
perpendicular from torusaxis/center to the intersecting
plane. References requested preferablyfrom the internet.
TIA
===
=> Please help with equations/sketches for a circular
torus intersection> when the cutting plane is not parallel to
its axis of symmetry.> Equations may be expressed in terms of
torus tube radius, tube> displacement from
axis,inclination/length of perpendicular from torus>
axis/center to the intersecting plane. References requested
preferably> from the internet. TIAjust take a
three-dimensional parametrisation of the torus like ( cos s *
(cos t + 2) )f(s,t) = ( sin s * (cos t + 2) ) ( sin t )This is
for a torus of displacement 2 and tube radius one. With
vectorcalculus you now can easily deduce the cutting with any
plane.Rene.-- Ren.8e MeyerStudent of Physics &
MathematicsZhejiang University, Hangzhou, China
===
=An
interesting integral just appeared in alt.math, and there
should bea neater solution than I found. The problem is to
evaluate int_0^1 int_0^{1-x} 1/(1-axy) dy dx,where |a| < 4.My
solution was to expand into a geometric series (justified
since
|a|< 4, 0 <= y <= 1-x ==> |4axy| < 1), obtaining int_0^1 a^n
x^n (1-x)^(n+1) dx,and then use the standard identity int_0^1
x^a (1-x)^b = Gamma[a+1] Gamma[b+1]/Gamma[a+b+2].This yields
the integral as a power series, sum_{n=0}^infty a^n
n!^2/(2n+2)!,which is then recognizable as 2
arcsin(sqrt(a)/2)^2 --------------------- . aBut there ought
to be a direct way to see this. Any takers?--Ron Bruck
===
=Two
minor corrections: that should read |a|<4, 0 <= y <= 1-x ==>
|axy| < 1,and we obtain for the value of the integral
(1-x)^(n+1) sum_{n=0}^infty int_0^1 a^n x^n ----------- dx.
n+1But the final answer remains unchanged.--Ron Bruck> An
interesting integral just appeared in alt.math, and there
should be> a neater solution than I found. The problem is to
evaluate int_0^1 int_0^{1-x} 1/(1-axy) dy dx,where |a| < 4.My
solution was to expand into a geometric series (justified
since
|a|> < 4, 0 <= y <= 1-x ==> |4axy| < 1), obtaining int_0^1 a^n
x^n (1-x)^(n+1) dx,and then use the standard identity int_0^1
x^a (1-x)^b = Gamma[a+1] Gamma[b+1]/Gamma[a+b+2].This yields
the integral as a power series, sum_{n=0}^infty a^n
n!^2/(2n+2)!,which is then recognizable as 2
arcsin(sqrt(a)/2)^2> --------------------- .> aBut there
ought to be a direct way to see this. Any takers?--Ron
Bruck
===
=Set R_1 = x+y, R_2 = xyIs there an easy way of
writing the polynomial x^n + y^n in terms ofR_1 and R_2? I
know it can be done by expanding out with R_1^n, butthis is
obviously not practical for large n. Is there some sort
ofidentity that allows you to write the answer down
immediately,similar to the binomial theorem?
===
=> Set R_1 =
x+y, R_2 = xyIs there an easy way of writing the polynomial
x^n + y^n in terms of> R_1 and R_2? I know it can be done by
expanding out with R_1^n, but> this is obviously not
practical for large n. Is there some sort of> identity that
allows you to write the answer down immediately,> similar to
the binomial theorem?The buzzword is Dickson polynomial.Ley u
= x + y and x = xy. Form the generating functionF(t) =
sum_{n=0}^infinity (x^n + y^n)t^n.ThenF(t) = 1/(1-xt) + (1-yt)
= (2 - ut)/(1 - ut + vt^2) = (2 - ut)sum_{m=0}^infinity (u +
vt)^m t^m = (2 - ut)sum_{n=0}^infinity t^n sum_m {m choose
n-m}u^{2m-n} v^{n-m}etc. Pulling out the coefficient of t^n
gives D_n(u,v)the Dickson polynomial of the first kind with
D_n(u,v) = x^n + y^n.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14 times)
===
=>
Set R_1 = x+y, R_2 = xy>> Is there an easy way of writing the
polynomial x^n + y^n in terms of> R_1 and R_2? I know it can
be done by expanding out with R_1^n, but> this is obviously
not practical for large n. Is there some sort of> identity
that allows you to write the answer down immediately,>
similar to the binomial theorem?HereOs my attempt:Write f_n
=
x^n+y^n. Then expand (x+y)*(x^n+y^n) to get the
recurrencerelation:f_(n+1) - R_1 * f_n + R_2 * f_(n-1) =
0with the initial conditions: f_0 = 2 and f_1 = R_1.The
solution to the recurrence relation may be written as:f_n = A
* u^n + B * v^n, whereu and v are roots of the quadratic
equation t^2 - R_1 * t + R_2 = 0.The coefficients A and B may
be found by considering the initial
conditions.-Michael.
===
=Set R_1 = x+y, R_2 = xy> HereOs my
attempt:Write f_n = x^n+y^n. Then expand (x+y)*(x^n+y^n) to
get the recurrence> relation:> f_(n+1) - R_1 * f_n + R_2 *
f_(n-1) = 0> with the initial conditions: f_0 = 2 and f_1 =
R_1.That is a special case of the NewtonOs identities. > The
solution to the recurrence relation may be written as:> f_n =
A * u^n + B * v^n, where> u and v are roots of the quadratic
equation t^2 - R_1 * t + R_2 = 0.DonOt you think that the
solutions to this equation would be x and y.... > The
coefficients A and B may be found by considering the initial
conditions.... with A=1 and B=1? So this approach wonOt give
much information, as thesolution you get will be (drums,
please)f_n=x^n+y^n :)Jyrki
===
=>> Set R_1 = x+y, R_2 =
xy>> HereOs my attempt:>> Write f_n = x^n+y^n. Then expand
(x+y)*(x^n+y^n) to get the recurrence> relation:> f_(n+1) -
R_1 * f_n + R_2 * f_(n-1) = 0> with the initial conditions:
f_0 = 2 and f_1 = R_1.>> That is a special case of the
NewtonOs identities.>> The solution to the recurrence
relation may be written as:> f_n = A * u^n + B * v^n, where>
u and v are roots of the quadratic equation t^2 - R_1 * t +
R_2 = 0.>> DonOt you think that the solutions to this
equation would be x and y....Indeed....>> The coefficients A
and B may be found by considering the initialconditions.>>
... with A=1 and B=1? So this approach wonOt give much
information, as the> solution you get will be (drums,
please)>> f_n=x^n+y^n :)Oh well, it seemed like such a good
idea at the time.... :-)-Michael.
===
=Set R_1 = x+y, R_2 =
xyIs there an easy way of writing the polynomial x^n + y^n in
terms of> R_1 and R_2? I know it can be done by expanding out
with R_1^n, but> this is obviously not practical for large n.
Is there some sort of> identity that allows you to write the
answer down immediately,> similar to the binomial
theorem?Search for NewtonOs identities (a useful recurrence
relation) andWaringOs formula. They are also helpful for a
larger number of variables(x,y,z,etc.). I donOt know what
would be the simplest formula for twovariables. WouldnOt any
formula for DicksonOs polynomials do? Searchfor any of these
key phrases!Jyrki Lahtonen, Turku, Finland
===
=> Ok, I
apologize.I did not say it was fraud, I _asked_ if it was.You
say it was not, that answers the question.Ok. I accept your
apologies. I am little tired of this controversy,and seems
stupid to get angry with people I donOt personally meet.>
ItOs just that extraordinary claims require extraordinary
evidence. What you > present on your web site cannot even be
considered evidence, let alone> extraordinary evidence. IOm
supposed to take your word that if presented> correctly, the
lines would be parallel? ThatOs not how science is done.I
agree with that. As I said, the main investigation was done
byCrater&McDaniel. I didnOt reproduce step by step, but made
my owncalculations and simulations that were consistent. At
that time thecomputer simulation was very cost, so I made
some simulations thatshowed the same level that theirs.Also,
revised the protocol, and the pattern can be fitted with
thatmean deviation inside of the mounds, the problem is
trying to do byhand. But it is easy, just using least squares
(if you insist, asCrater, in a coordinate fit).I will prepare
when I get some time to do it by computer, you mustnotice
that at the time I did the diagram, as it is rotated, it
wouldhave took me a considerable time to locate the
correspondingcoordinates and there was the issue of the
non-rectified image.What I have thought is to use a
superimposed image, with the centersclearly marked, so it is
visually discernible how far is the deviationin each mound.>
And letOs not forget that you solicited comments. Do you
think that I am the> only one who sees problems with what you
presented? You should be happy to > see such comments so that
you can correct the situation. The problem is that the
critics here and other posts were far fromgentle. I am not
accustomized to this. Also notice that I am littletired of
writing some statements, and getting them ignored.For
example, I stated that the simulations (and calculation by
handusing area ratios will do) show that you need about 75 or
more moundsto get reasonable levels of that pattern being by
chance.And a poster in my forum came with 106 dots and showed
some nicealignments. Perhaps I loose patience very fast, but I
had to explainthat not only those nice alignments were
present, but very likely themounds pattern. Other thing was
to extract it, which would likelyrequire a computer.> I will
clarify also in my webpage soon, Good.givin links to your
webpage.DonOt you have your priorities wrong?
ShouldnOt you
focus on what _you_ are> presenting? It is now closed. I
think the best test would be to perfom a completescan of the
martian surface. It could be used to detect crater chainsand
make crater population counts also (they are similar to
mounds,but with lightning inverted). However, in order to
avoid bias, Ishould make it independent from the mounds, and
when I detectautomatically, this may be somehow indicative of
non-randomness.> You will be acussed of dishonesty.
Dishonesty? I just presented what I saw. You admit that the
original image> was not orthorectified, so I was honest in
saying that the lines do not appear> parallel. Now if I made
a mistake, it was an honest one, because I used the> image
that _you_ provided. Rather than explaining in the text why
the image> was incorrect, wouldnOt it have been better to
post a correct image?But that was said in the webpage. I
retire the accusation ofdishonesty here and in my webpage,
accepting that you did not notice.And yes, I will do that. I
will also publish the source code I use andthe link to the
original image in NSSDC website.> You can sue me also, if you
want. I am wishing it.Sue you for what? Calling me dishonest?
Calling me pathetic? Calling me a slow> thinker? Calling me
stupid? These comments wonOt make you look good.I retire
those comments, but you must notice that I have
heardeverything, from being a pseudoscientist, to not knowing
probabilitynotions, fraud, and countless others.The problem is
a mathematical one. For example my claim that this isthe only
pattern (after a square-based one) with the greater number
ofDIFFERENT parallel/perpendicular directions for 5 points
(BAGED, plusrepeating P) goes unchallenged. And this is what
irritates me. Thatclaim is just a very defined one, testable
with analitical geometry.But based on a diagram, so... could
I have missed some configuration?I recommend to read most of
the material I present, at leastsuperficially, before making
comments, and let the childish I am moreintelligent than you
issues out of the conversation.
===
=>Message-id:
<2410d7e.0311220235.6c366ff9@posting.google.com> You are
one of those who confuse the probability of an event
occuring> with the event itself. Duh...It is you who confuse
it. You still believe that the probability ofthat event has
turned to 1. This is a claim I have heard countlesstimes. Yet
a math PhD laughed at it.see how they do that.Also the crater
chains are analysed probabilistic, but still they arethere
since some million years ago... > In the mounds problem, the
probabilistic experiment is having 6 mounds> formed by some
process. The process has a great number of pausible>
outcomes, depending on its particular physical nature. If you
assume> some type of a random formation process, then the
probability of> getting the formation shown is extremely
low.Instead of wasting your time uploading gifs and accusing
people, why> donOt you program a simulation of 6 mounds
forming at random, set your> accuracy range for parallel and
perpendicular lines and see of you can> get that or similar
formations and what the frequency of those are.> That will
convince you the probability of getting that formation or>
similar ones is extremely low. ThatOs all there is to it.I
have done that particular experiment. I used parallel lines
only fora computer simulation of a number of points, not just
6 but greater.Still, for such a number of parallel
relationships I got 1/1000 for 6points. I draw some of them
at hand (I got numbers in the screen onlybecause at that time
I didnOt get any notion of image processing).Most were
unremarkably, and there was a tendency to form what I
calleddegenerated lines (a bug).You can do by yourself and
check. Otherwise you are talking withoutfoundation, just as
counting sheeps, and you know. DO THE SIMULATIONAND WHEN YOU
HAVE A FIGURE, WE TALK AGAIN. If not, you are
makingpseudoscience.
===
=>Message-id:
<1f366ae.0311240955.b79503f@posting.google.com>>Message-id:
<2410d7e.0311220235.6c366ff9@posting.google.com>[snip]>
>> You just canOt _prove_ itOs not random.
Probability has no
meaning>> for something that has already happened. You claim
the odds are>> 200,000,000,000 to 1 and then you say that no
other configuration>> has this property.> Duh.>>
>>[snip]>>You are one of those who confuse the probability of
an event occuring>with the event itself. Duh...>>In the mounds
problem, the probabilistic experiment is having 6
mounds>formed by some process. The process has a great number
of pausible>outcomes, depending on its particular physical
nature. If you assume>some type of a random formation
process, then the probability of>getting the formation shown
is extremely low.The formation? DonOt you mean a
formation?>>Instead of wasting your time uploading gifs and
accusing people, why>donOt you Why doesnOt the
OP have a
mathematically correct drawing on his web site?>program a
simulation of 6 mounds forming at random, What, exactly, is
the geological process I am supposed to simulate? Or should
Ijust sprinkle random pixels onto an image?>set your accuracy
rangeAccuract range? This is math, not physics. There is no
range allowed indetermining parallel and perpendicular.>for
parallel and perpendicular lines and see of you can>get that
or similar formations and what the frequency of those
are.>That will convince you the probability of getting that
formation or>similar ones is extremely low. ThatOs all there
is to it.YouOre wrong. There is a world of difference
between
that formation andsimilar formations.>>By the way, has you
ever been to a casino? Have you ever looked at the night sky
and noticed all the polygons formed bythe stars?Do you think
there is some mysterious reason for this?>The number you see
the>roullette ball sitting on of course just happended but
you canOt bet>on it. Have you thought about _why_ they
wonOt
let you bet on it? Do you know what theprobability is of an
event that has already happened? >If you think talking about
its probability has no meaning>because it just happened,
Whatever the probability was before the event, after the
event the probabilityis simply 1.Possibly you didnOt see
this
example, so IOll repeat it:Put 64 coins in a box and shake.
Note the pattern of heads and tails. Theprobability of that
pattern occuring is1/18446744073709551616and yet it happened!
Shake the box again. The pattern you see now also had
a1/18446744073709551616chance of occuing and it happened
also. The chance of BOTH those patternsoccuring back to back
is1/340282366920938463463374607431768211456What conclusion do
you draw from this?>then if you have just won, why should
they>pay you back 36 times your bet? They shouldnOt,
roullette payouts are 35 to 1. Of course, you would know
thisif youOve ever been to a casino.>Just take your money
back and thatOs>it.:)--MensanatorAce of Clubs
===
=> Whatever
the probability was before the event, after the event the
probability> is simply 1.The probability is related to the
chances of that event happening. Iwould not apply to events
that have already happened. The p value doesnot get altered,
because p is predictive, and you are talking of avalue which
is not predictive.I would think that it is more a semantic
question.I would assign 0 to event which are impossible and 1
to event that aresurely TO HAPPEN (in the
future).
===
=>Message-id:
<2410d7e.0311250420.2e0c3caf@posting.google.com>> Whatever
the probability was before the event, after the event
the>probability>> is simply 1.>>The probability is related to
the chances of that event happening. I>would not apply to
events that have already happened. The p value does>not get
altered, because p is predictive, and you are talking of
a>value which is not predictive.>>I would think that it is