We will offer you the latest DVD information about various discount
and wholesale sources You will not only save time and effort but you
can also save anything up to 90%% off retail price, thus increasing
your profit margins by buying at extremely low wholesale prices
Welcome to visit our website !
One order more than 1000 discs,you will get !FREE SHIPPING!
Pls be free to contact with us by email msn livehelp or Skype
Looking forward to hearing from you
Thanks and have a great day!

WWW.DVDSSELL.COM

+1

To all,

I have now completed a paper at the link below, which summarizes the
work I have been doing for the past several months to lay a foundation
for understanding and calculating particle masses:

http://jayryablon.files.wordpress.com/2008/12/finite-amplitudes-without-i-epsilon...

I have also taken the plunge and submitted this for peer review. ;-)?

The abstract is as follows:

By carefully reviewing how the invariant amplitude M is arrived at in
the simplest Yang-Mills gauge group SU(2), we show how to arrive at a
finite, pole-free amplitudes without having to resort to the "+i\epsilon
prescription." We first review how gauge boson mass is generated in the
SU(2) action via spontaneous symmetry breaking in the standard model,
and then carefully consider the formation of finite, on-shell
amplitudes, without "+i\epsilon.

Comments are welcome, and I wish everyone a happy holiday and New Year!

On Dec 9, 4:04 pm, cusanic...@gmail.com wrote:

Unfortunately, that plus everything else will happen in the future.
Too bad Earth doesn't seem to have a viable future, especially if our
hypersphere allows us to get sucked into the GA at 6.111e3 km/s.

Even though we can in fact see into the cosmic future, at least on
behalf of what other galactic realms have coming, or rather of belated
knowledge due to the unusually slow velocity of light, as for most
likely having already encountered their demise, and therefore it’s
another so what's the difference for us?

TAODotology Is About Knowing The TAODot,
Which Is The Entire Uni-Le-Verse And
Every Its Consisting Part:
http://dedanoe.googlepages.com

I am glad I interjected this Calculus preview before the chapter
itself because it clarifies the meaning of
the Second Decimal Point on New Reals which thence replaces the limit
in Calculus.

Working with the Identity Function of y = x

Now let me say that the second decimal point does not come in two
types but only one type. I said the
second decimal point is either a c0 or c1 depending on the remainder
in division out to the
10^(-)9999...99998 place value. So that 1/3 = 0d3333....3333c1 where
the c1 indicates a remainder
of a 1 carryover and that 1/2 = 0d50000...

I said the 1/2 has a second decimal point of c0 so that 1/2 =
0d5000...0000c0 meaning there was
no remainder carryover.

Well, the Calculus limit teaches me that the above conception of the
second decimal point is not
going to work properly.

Please distribute this message as you see fit.

Announcing Maxima 5.17

Maxima is a GPL'd branch of Macsyma, the venerable symbolic
computation system. Maxima 5.17 is a bug fix and feature
enhancement release. The current version is 5.17.1.

Maxima has functions to work with polynomials, matrices, finite sets,
integrals, derivatives and differential equations, linear algebra,
plotting, floating point and arbitrary-precision arithmetic, etc.

Maxima can run on MS Windows and various flavors of Unix,
including MacOS X.
There is a precompiled executable installer for Windows,
source and binary RPM's for Linux, and tar.gz containing
the source distribution.

Maxima is implemented in Common Lisp; several Lisps can compile
and run Maxima, including CMUCL, SBCL, Clisp, GCL, and ECL.

David R Tribble tribble.com> writes:

It would also be nice if he didn't confuse entertaining intellectual
exercises with freedom fighting.

On Thu, 25 Dec 2008 Goo wrote:

Let f be multiplicative and g completely multiplicative. If for every
p and every n f(p^(n+1))=f(p^n)f(p)-g(p)f(p^(n-1)) show that for every
n and m f(n)f(m)=Sum{d|gcd(n,m)}{g(d)f(mn/d^2)}.

I concluded that is if enough to check for n=p^a and m=p^b and tried
to prove that case by induction but failed because I had to prove that
f(p^a)f(p^2)=f(p^(a+1))f(p) as a part of induction base and that
equation does not need to hold.

Thanks in advance.