sci.math Google Group

Re: planet earth is the entire universe

zzbun...@netscape.net (zzbunker@netscape.net) — Thu, 20 Nov 2008 11:33:52 UT

But that's what's wrong them then too.
Since the first thing the religous and media cranks did after
people with actual
engineering brains started working on digital processing,
holograms, RISC computers, Optical Computers, HDTV, CD, DVD,
satellites, post Neanderthal robotics, post AT&T phonics,

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Re: Comprehensive Solution Manual for Textbooks

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Re: When do matrices B and C commute?

alainvergh...@gmail.com (alainverghote@gmail.com) — Thu, 20 Nov 2008 11:08:29 UT

On 19 nov, 23:30, Mariano Suárez-Alvarez
Bonjour,
The both given nilpotent matrices might be considered
as cubic 'roots' of the null matrix.
If we want to work with integer powers (all Z ) we must
discard non inversible ones ,that is : singular matrices.
However a null matrix may be equals to pCA) , p a non

Re: #2 preface; new book 2nd edition: Math a subset of Physics, AP-adics

plutonium.archime...@gmail.com — Thu, 20 Nov 2008 10:43:44 UT

I made a mistake here. Since the Reals are the dual opposite of
AP-adics, I need to keep the similarity in multiplication as much as
possible. Perhaps I can end up with total similarity.
In the Reals the dual opposite would be 10,000 x 0.0000....00001 to
the above AP-adics example. And in Reals that product would

Complex analysis with contour integral.

mina_wo...@hanmail.net (mina_world) — Thu, 20 Nov 2008 10:15:45 UT

Hello teacher~
int{0 to oo} 1 / (x^n + 1) dx = pi / {n.sin(pi/n)}
n = 1, 2, 3, ...
------------------------------ ------------------------------ ----------------
Except n = 1,
Choose the circular sector contour with Re^{i.(2.pi)/n}.
With one residue(at z = e^{i.(pi/n)}) and many(?) calculation,
Consequently,

Re: guide to credit derivatives (JP Morgan)

ma...@rdrop.remove.com (William Elliot) — Thu, 20 Nov 2008 09:55:00 UT

Don't use them. That's all the guidance that's
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Re: Factoring question

ma...@rdrop.remove.com (William Elliot) — Thu, 20 Nov 2008 09:49:00 UT

What do you call it when a person can't express himself
clearly and you still divine meaning from his utterance?
But that's not finding phi(n), as OP stated. That's finding p and q.
Once I make the substutions, I get
p + q = n - phi(n) + 1 = p + q
p - q = sqr((p + q)^2 - 4pq) = sqr(p - q)^2 = +-(p - q)

Re: Factoring question

hh...@web.de (Helmut Richter) — Thu, 20 Nov 2008 09:45:49 UT

Let phi(n) = 2^k * r with r odd.
Take any b, and compute modulo n the sequence b^r, b^(2r), b^(4r), ...,
b^phi(n). The last value is 1 (unless b has a common factor with n, in
which case you are done anyway). The last value before the first
occurrence of 1 is either -1 (bad luck) or some other x (good luck). In

Arrange them [ updated ]

azeez...@gmail.com (Nimo) — Thu, 20 Nov 2008 09:23:34 UT

In the previous post
[link]
people didn't understand it clearly or there is a bit of confusing
in understanding the concept.
lets analyze the problem
you just write a random sequence of 1's & 0's

Re: Factoring question

w...@nestle.csail.mit.edu (Bill Dubuque) — Thu, 20 Nov 2008 09:03:37 UT

{p,q} is uniquely determined by its sum s = p+q and product n = pq,
namely: {p,q} is the root set of (X - p)(X - q) = X^2 - s X + n.
The product n = pq is known, and the sum s = p+q occurs in phi(n)
so it too is known: p+q = pq+1 - phi(n) + 1 = n+1 - phi(n).
The reason that this works is because phi(pq) is a symmetric

Re: planet earth is the entire universe

gmearn...@yahoo.com (Mark Earnest) — Thu, 20 Nov 2008 08:59:21 UT

Why not just imagine that you are the only one that exists,
and all of us are your dream? Isn't that easier?