mm-1769
===
Subject: for daryl: really cool premium nntp server access -  ugmi - (1/1)
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bozwu hevge
jipko cav lasja zocfe. ...
===
Subject: matrices
does anyone know of a good website that contain the theorems (with
proofs) involving matrix operations, determinants and linear systems?
===
Subject: Re: matrices
<http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/in
-- 
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: matrices
thank you. that was a nice site..
if anyone know of a text website too please tell me.
===
Subject: Re: I need help with a question about maximizing profit
Caskets.
-- 
Darrell 
===
Subject: Re: I need help with a question about maximizing profit
Maybe not. I thought the increased interest in cremation was partly
driven by increased cost of ordinary funerals, including caskets.
Perhaps someone will research this carefully.
bob
===
Subject: Re: I need help with a question about maximizing profit
Or worse, trucks that sit running on the job site and tour buses
running while the riders are in the cafe or tourist attraction.
--Lynn
===
Subject: Re: Exquisite probability
    WHAT DO YOU THINK OF HIS WIZARD ....POWERS??
  MAKES ME THINK ON THE CIG...EHEH
     SO IT STANDS AT 0.5,KNOWING THAT IT HASN'T HAPPENED IN OVER 10 MILLION 

ATTEMPTS..HIHIHI
          It's impossible you say..ok.i guess that that guy in case of 
winning may be the best magical wizard of all times...Perhaps even GW BUSH 
would consult him...
gw that is consulting his c of staff often,trying to learn if bin  L has 
been killed...ok  
  GW HAS THE NATURAL TENDENCE TO GO LOOK FOR TERROR COUNSELING..WONDER 
WHY...AND GW DO NOT MANUFRACTURES  CHEMICAL OR BIO WEPPONS.. HE MAKES 
DESEASES...AND SPREAD THEM ON THE TERRORISTS...THAT DID SO WELL IN 
WTC..WONDER WHERE THEY WILL STRIKE AGAIN... LET ME GUESS?Like texas an nj ny 

is a spartan state...humm and the next attack is in his office right in 
front 
of his noes...Why he's he so concerned about the c of staff..???
===
Subject: Re: Exquisite probability
I don't know how the team mate felt,,,
     BUT TO ME IT JUST FELT EXTRAORDINARY WELL!!A PURE TASTE OF HEAVEN...
 BUT I DON'T ADVISE TO THIS AT HOME.
===
Subject: Re: Exquisite probability
You know I played american football from 11 to 17 years of age?
 They stopped in the 14th pint,I was playing with the 14,and it was drank 28 

pints in total...
  AND I WON THE BET.
Does it satisfies your curiosity ??
===
Subject: Re: Exquisite probability
So what's your name? GE IVEY...ok ,it seems you do enjoy my personal 
computer.
 You're the algebra doc...surething
   It seems you like the stuff that cracks my head..ok you must like it,it 
seems good stuff
 0.05 is  not the right probability.Because,if he scores about 4 by game,it 

means that he can score eight..humm
  I was waiting for a more significant value.let's say... 0.35??
 Did he score the touchdownsor didn't he?
 __________________________
 You need to get a microscope and check what you say it's impossible...
 If the toss of the cig is one single movement isolated from a random xp.. 
and there are only two possible outcomes.... p(x)= 0.5,,,not looking at the 

time..
 But you say it's impossible....why??
IF YOU WANT THE MIRACLE ANSWER TO THIS NAUGHTY PUZZLE..I'LL TELL YOU, BUT 
FIRST...WHY DO YOU SAY IT'S IMPOSSIBLE?Would you bet?
 THE REPUBLIC CHILDREN WILL .... BE THE WIZARDS OF THE NEW RULE!
===
Subject: Software
I hope that someone could help me.
I am looking for some help I guess in the way of a software tutorial
type program.
My math skills are not that great and I need to learn (s) and (z)
domains for DSP and feedback controller applications.
Would anyone know of some software I could use to learn these areas of
math?
Rob
===
Subject: Re: Software
If you are looking for software to simulate systems represented in the
z ans s domains, try Matlab with the Control Systems Toolbox
(expensive, but your university should have it). It will not guide you
step-by-step through the topic, but the help files are quite
comprehensive and you can easily visualize the outputs of the systems
you build, which can be very didactic.
===
Subject: Re: Software
I didnt read the OP's question properly :)  He'd want something far more
advanced than Algebrator...
===
Subject: Re: Software
Yeah theres one called 'Algebrator' that does step-by-step solutions.  Its
abot $70 I think. Google it
===
Subject: implicit function theorem and lagrange multiplier method
I am having difficulties with these theorems. are there anyone here
who (completly) understand this, and can manage to explan the proof of
the lagrange multiplier method? I do not understand the proof in the
book, so I probably need careful explaination of this...
===
Subject: Re: implicit function theorem and lagrange multiplier method
thank you for the clearification.
===
Subject: Re: implicit function theorem and lagrange multiplier method
Not a proof but a way to think about it is this:  The gradient of a 
function, f(x,y,z), always points in the direction of greatest increase.  If 

you were unconstrained you could find a maximum by moving in the 
direction of the gradient (or a minimum by moving in the opposite 
direction). 
 Continue doing that until you CAN'T: until the gradient is 0.
   If you are constrained- must stay on some curve or surface, then you do 
the best you can: move on the curve or surface in the direction that is 
CLOSEST to the direction of the gradient of f.  (Technically, you move in 
the 
direction of the projection of the gradient on the curve or surface.)  Keep 

doing that until you can't: until there is no projection on the curve or 
surface which means the gradient is perpendicular to the curve or surface.  

Of course, if the curve or surface is given by G(x,y,z)= constant, then you 

can think of that as a level curve/surface for G- and the gradient of G 
is perpendicular to that.  That is, the gradient of f is perpendicular to 
curve and so is the gradient of G: one must be a multiple of the other.
===
Subject: Re: implicit function theorem and lagrange multiplier method
It would have been helpful if you had indicated what the proof in your
book was.
I'm not sure that I _completely_ understand _anything_. Anyway...
Suppose f(x,y) is a function which is continuously differentiable on
some open set and r(t) = (x(t), y(t)) is a differentiable curve with
r(t) in the open set for t in an interval I and r'(t) =/=0. We want the
extrema of f(r(t)). That is, the extrema for f when f is evaluated at
points on the curve.
Suppose f(x_0, y_0) is an extremum of f on the curve and let 
r(t_0) = (x_0, y_0).
Then f(r(t)) is a function of a single variable and its derivative is
zero at t_0. The derivative, by the chain rule, is 
inner or dot product.]
r'(t_0) are perpendicular. But r'(t_0) is tangent to the curve at 
(x_0, y_0) so grad(f(x_0, y_0)) is perpendicular to the curve at 
(x_0, y_0).
Suppose that, instead of r(t), we have a non-parameterized equation for
the curve g(x,y) = 0. You know, I hope, that grad(g(x_0, y_0)) is
perpendicular to the curve g(x, y) = 0. [If not, proceed as above after
observing that the derivative of a constant is 0.]
We now have that both grad(f(x_0, y_0)) and grad(g(x_0, y_0)) are
perpendicular to g(x, y) = 0 and thus the two gradients are parallel.
That is grad(f(x_0, y_0) = Lgrad(g(x_0, y_0). Voila! The method of
Lagrange.
The Implicit Function Theorem gets into it, as far as I know, to assure
that g(x, y) = 0 _is_ a curve - locally y is a function of x.
-- 
Paul Sperry
Columbia, SC (USA)
===
Subject: permutations help!
--------
Let G be the symmetric group S3 (S subscript 3)
Show that G contains just 1 subgroup of order 3, and 3 subgroups of order 
2.
---------
i know that the order of the group G is n! = 3! = 6
The elements of the group together with their orders are:
(1)(2)(3) has order 1
(1)(2,3) has order 2
(1,2)(3) has order 2
(1,2,3) has order 3
(1,3,2) has order 3
(1,3)(2) has order 2
===
Subject: Re: permutations help!
Do you know the Sylow Theorems? If so, that's one way to go without
listing the elements.
If not ...
Label your permutations above e, a, b, c, d, f.
contains an element of order 2.
-- 
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: permutations help!
order 2.
===
Subject: Re: Integer solutions for a circle, conjecture from surrogate 
factoring
You did.
There are 10 more than you found, to wit:
##    (15, 20, 25)
##    (30, 40, 50)
##    (39, 52, 65)
##    (45, 60, 75)
##    (51, 68, 85)
##    (60, 80, 100)
##    (25, 60, 65)
##    (40, 75, 85)
##    (33, 56, 65)
##    (36, 77, 85)
Looks like you missed all the cases where the same
value of c appears more than once (in your list, the
c's are unique).
For c <= 100, the fraction is 0.547, not 0.42.
Not that it helps Harris any.
Idiot.
===
Subject: Re: Integer solutions for a circle, conjecture from surrogate 
factoring
I guess my idiot comments finally sank in, Harris has
deleted all his posts.
I only did c=100 last night. Here's a bigger run and the program I
used.
##    All primitive Pythagorean triples can be generated from the
3-4-5
##    triangle by using the 3 linear transformations below, where
a ,b, c
##    are sides of a triple:
##
##       new side a        new side b      new side c
##      a - 2b + 2c       2a - b + 2c     2a - 2b + 3c
##      a + 2b + 2c       2a + b + 2c     2a + 2b + 3c
##     -a + 2b + 2c      -2a + b + 2c    -2a + 2b + 3c
##
##    If one begins with 3, 4, 5 then all other primitive triples will
##    eventually be produced. In other words, every primitive triple
will
##    be a parent to 3 additional primitive triples. example:
##
##        Let a = 3, b = 4, c = 5.
##
##      new  side a                new side b                new side
c
##     3 - (2[Times]4) + (2[Times]5) = 5     (2[Times]3) - 4 + 
(2[Times]5) = 12   (2[Times]3) -
(2[Times]4) + (3[Times]5) = 13
##     3 + (2[Times]4) + (2[Times]5) = 21    (2[Times]3) + 4 + 
(2[Times]5) = 20   (2[Times]3) +
(2[Times]4) + (3[Times]5) = 29
##    -3 + (2[Times]4) + (2[Times]5) = 15   -(2[Times]3) + 4 + 
(2[Times]5) = 8   -(2[Times]3) +
(2[Times]4) + (3[Times]5) = 17
def create ppt(n):
    ppt = [(3,4,5)]
    new ppt = []
    done = 0
    start = 0
    while done==0:
        for p in ppt[start:]:
            a1 =    p[0] - 2*p[1] + 2*p[2]
            a2 =    p[0] + 2*p[1] + 2*p[2]
            a3 =   -p[0] + 2*p[1] + 2*p[2]
            b1 =  2*p[0] -   p[1] + 2*p[2]
            b2 =  2*p[0] +   p[1] + 2*p[2]
            b3 = -2*p[0] +   p[1] + 2*p[2]
            c1 =  2*p[0] - 2*p[1] + 3*p[2]
            c2 =  2*p[0] + 2*p[1] + 3*p[2]
            c3 = -2*p[0] + 2*p[1] + 3*p[2]
            if c1 <= n:
                new ppt.append((a1,b1,c1))
            if c2 <= n:
                new ppt.append((a2,b2,c2))
            if c3 <= n:
                new ppt.append((a3,b3,c3))
            start = len(ppt)
            ppt.extend(new ppt)
            new ppt = []
        else:
            done = 1
    return ppt
for e in xrange(2,7):
    n = 10**e
    ppt = create ppt(n)
    print 'For c <= ',n
    print 'primitive Pythagorean Triples: %8d' % len(ppt)
    pt = []
    for p in ppt:
        pt.append(p)
        d = n/p[2]
        for q in xrange(2,d+1):
            pt.append((p[0]*q,p[1]*q,p[2]*q))
    print '      all Pythagorean Triples: %8d' % len(pt)
    print
##    For c <=  100
##    primitive Pythagorean Triples:       16
##          all Pythagorean Triples:       52
##
##    For c <=  1000
##    primitive Pythagorean Triples:      158
##          all Pythagorean Triples:      881
##
##    For c <=  10000
##    primitive Pythagorean Triples:     1593
##          all Pythagorean Triples:    12471
##
##    For c <=  100000
##    primitive Pythagorean Triples:    15919
##          all Pythagorean Triples:   161436
##
##    For c <=  1000000
##    primitive Pythagorean Triples:   159139
##          all Pythagorean Triples:  1980642
I guess Surrogate Factoring doesn't explain these results after all.
===
Subject: Re: Integer solutions for a circle, conjecture from surrogate 
factoring
No, that was intentional (assuming that your c is James' r). James
My program finds the number of c's for which at least one {a,b}
exists, not the total number of triples {a,b,c}.
Who, me? Or Harris?
-Rotwang
===
Subject: Re: Integer solutions for a circle, conjecture from surrogate 
factoring
Ok, I assumed he meant total since he used it in his example,
he just overlooked one of the multiples.
Who deleted their posts?
===
Subject: Re: Integer solutions for a circle, conjecture from surrogate 
factoring
Just checking. Incidentally I'm surprised that James abandonned this
he said he still thought he was correct despite the figures I posted;
I wonder what made him change his mind.
By the way, if any brainy individual knows how to calculate the actual
limiting value (if it exists) of the proportion of natural numbers
that are the radius of a pythagorean triple, I would be interested to
hear about it.
-Rotwang
===
Subject: Prime Numbers
Theorem: If p is prime, and a + b = p - 1 for a, b positive integers, then 
a!*b! is congruent to (-1)^(b + 1) (mod p)
a.) Illustrate two examples (w/ primes that are greater than 5) showing the 

theorem above.
b.) Prove the above theorem.
WORK:
a.) Showing it:
I'm assuming you pick a prime. So, let's pick 7 and 11.
a + b = 7 - 1 = 6
a!*b! is congruent to (-1)^(b + 1) (mod 7) ... not sure where to go from 
here.
And similarly for 11:
a + b = 11 - 1 = 10
a!*b! is congruent to (-1)^(b + 1) (mod 11)
And no clue how to do part b.
===
Subject: Prime Numbers
Theorem: If p is prime, and a + b = p - 1 for a, b positive integers, then 
a!*b! is congruent to (-1)^(b + 1) (mod p)
a.) Illustrate two examples (w/ primes that are greater than 5) showing the 

theorem above.
b.) Prove the above theorem.
WORK:
a.) Showing it:
I'm assuming you pick a prime. So, let's pick 7 and 11.
a + b = 7 - 1 = 6
a!*b! is congruent to (-1)^(b + 1) (mod 7) ... not sure where to go from 
here.
And similarly for 11:
a + b = 11 - 1 = 10
a!*b! is congruent to (-1)^(b + 1) (mod 11)
And no clue how to do part b.
===
Subject: Re: Prime Numbers
  Part A at least is easy.
  Remember that you are just asked to give examples.  Pick two number whose 

sum is 6.  Calculate a!*b! and see if the result is congruent to (-1)^(b+1) 

mod 7.  For example, 2+4= 6 and 2!4!= 48.  48= -1 mod 7 (since it is 1 less 

than 49).  Since b (did you notice that it doesn't matter which number you 
call b) is even, b+1 is odd and so (-1)^(b+1) is -1.
 
  First, since p must be a prime larger than 5, p is odd and so p-1 is even. 

 a and b are either both even or both odd.  If a and b are both even then 
(-1)^(b+1) is -1, if both odd, (-1)^(b+1) is 1.  You need to show that if a 

and b are both even, then a!*b! is one less than a multiple of p, if both 
odd, then a!*b! is one more than a multiple of p.
===
Subject: Re: Prime Numbers
Sorry for the double post, not sure how that happened.
===
Subject: Re: Possible Solution to Indeterminite Operation 1^ Infinity
As 1^inf,there are,0^0 and inf.^0.
 Logs are used in solving these indeterminations.The trick to this account 
is the way you use the logs.You need to consider the indetermination 
type,and 
the choose the right way to apply the logs.
 What I don't get is your equation,you should be using  analytical functions 

of real variable.
 Try something like this ::
 L,-é0, (tan x)^(sin x)   --é 0^0
=
 ln y = ln (tan x)^(sin x)
 ln y = [ln (tan x)]/(1)/(sin x)
 L,-é0, [ln (tan x)]/(1)/(sin x)
 =
 L,-é0, tan^2(x)/cos x
 =
 L,-é0, tan^2(x).sec (x) = 0
  __ ___ ___ ___ ___ ___ ___ ___ ___ __ ___ ___ ___ Not the vertical line 
x=2.....
===
Subject: Re: Possible Solution to Indeterminite Operation 1^ Infinity
===
Subject: SOLVE IT
find the coefficient of x^5 in (x-1)*(x^2-2)*(x^3-3)..............
(x^100-100)
===
Subject: Re: SOLVE IT
26*IteratedProduct[n, {n, 6, 100}]
I don't suppose you care how I figured it out...
===
Subject: for bendick: truly interesting news server -  edli ayta - (1/1)
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