mm-1022

Subject: Re: Mean Value Theorem
> Two places I¹ve seen the equivalent statement: If f is
continuously
> differentiable on [0,1], then |f(x)-f(y)|<=sup|f¹(u)||x-y|
where the
> sup is extended over 0<=u<=1.
> I can¹t see why we need f¹ to be continuous. It seems by
the MVT we
> have
> f(x)-f(y)=f¹(c)(x-y) for some c in [0,1],
> so that
> |f(x)-f(y)| = |f¹(c)||x-y|, for some c in [0,1],
> <= sup_u|f¹(u)||x-y|.
Maybe they intended f have values in R^n, so MVT is not
available.
Assuming f¹ exists everywhere and f¹ is integrable, we can
still
integrate f¹ on [x,y] to get the conclusion. And f¹ bounded
will
imply f¹ is Riemann integrable, so adding the assumption
f¹ is integrable is not needed. So I agree: even in this
case, continuity of f¹ is not requried.
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: How fast is the ConÞnued Fraction factorization
algorithm?
>My impression, based on not very scientiÞc
>(or knowledgeable) experiment
>is that the Quadratic Sieve (with many polynomials)
>is unlikely to factorize number with more than 80 digits
>in a reasonable time on a reasonable computer
>(say 1 day on my 667MHz laptop).
>>What do you mean by unlikely? It makes no sense in this
>>context. QS succeeds with virtual certainty in time that
depends
>>only on the size of n (with some small statistical
variation).
> That¹s just not true in my experience.
> And my comment makes perfect sense.
> The quadratic sieve completes in one or two minutes
> with some 70-digit numbers, and takes hours with others.
> Which is exactly what I would expect.
> What reason do you have to suppose the time has only
> small statistical variation?
> Have you actually tried it?
> What if the number is prime?
> (I gave random 60-digit numbers for factorisation
> as an assignment to a class of 30.
> One was prime, and there was a huge variation
> in the time taken - using quadratic sieve -
> to factorise the others.)
This is probably because your implementation runs trial
division, the
p+-1 and/or a couple of runs of ECM before running the
quadratic seive.
You can probably test this by factoring a few hundred large
numbers
and looking a a histogram of the run times. It will be
extremely
bimodal. The p+-1 and ECM are extremely fast, but
probabilistic
algoritms (they also have expected run times based on the
size of the
factors, not on the size of the original number.)
The
original quadratic seive (if I recall correctly) worked like
this:
To factor n
1) Choose a factor base of small primes.
2) Choose a random number r between 1 and n-1
3) If the least positive residue of r^2 mod n factors
completely using
only the primes in the factor base save r and a list of the
primes
occuring to odd powers.
4) repeat steps 2 and 3 till yo have about as many values of
r as there
are primes in the factor base.
5) You now have almost enough info to Þnd a pair of numbers
such that
x^2=y^2 mod n.
Improvements have increased the probability of Þnding numbers
that
factor completely, but the philosopy is the same. The run
time will be
proportional to the number of trials it takes to Þnd enough
numbers
that completely factor.
===
Subject: Re: covering compact set w/squares
Oh, I think I did misread. How about this:
Given a compact set K in the plane s.t. each pt x is the
center of a square
Q_x, prove that you can Þnd a subsequence Q_x_i of squares
s.t. K is
covered by the Q_x_i and
sum(over i) Char(Q_x_i) <= 4 Char (union(over i) Q_x_i),
where Char(X) is the characteristic (indicator) function of X.
>Given a compact set K in the plane s.t. each pt x is the
center of a
square
>Q_x, prove that you can Þnd a subsequence Q_x_i of squares
s.t. K is
>covered by the Q_x_i and the sum of the areas of each of the
Q_x_i is no
>more than 4 times the area of the union of the Q_x_i.
> Aargh. I think I¹ll just spend the rest of my life posting
replies to
> this message. I have to say that the two counterexamples
I¹ve
> given were wrong (I was reading the statement of the problem
> Note however that the example showing that it¹s not true
that
> in a minimal cover no point is in more than four squares was
> correct. (I¹m still not sure whether the result itself is
true and
> I still wonder whether it¹s really what you wanted to
prove.)
>I think this is a totally geometric thing, and I know it
sufÞces to
show
>that in such a minimal cover no point is contained in more
than 4
squares,
>but how do i show this? I know how to do this with intervals
on the real
>line (there the answer is 2 times the area....), but i get
lost when
going
>up a dimension.
> ************************
> David C. Ullrich
===
Subject: Re: how to understand this statement of Second order
Necessary
Condition?
X-AUTHid: rubin
> I read the statement of SONC for local minimizer as follows:
> Let x* a local minimizer of function f over some constraint
set O,
> and d a feasible direction at this point x*. If
d¹*gradient=0, then
> d¹*H*d>=0.
> where d¹ is the transpose of d column vector...
> I got confused here because this seems to me is a half
statement:
> If d¹*gradient=0, then d¹*H*d>=0.
> but what if d¹*gradient>0 but not =0 at that point x*?
> Is this a possible case in the SONC statment? Is that true
that when
> d¹*gradient>0 but not =0, there is no statement about the
SONC?
> -Walalal
First of all, d having positive inner-product with the
gradient is not a
problem, since x* is a local minimizer, not a maximizer. I
suspect
you¹re really interested in the case d¹*gradient < 0. That
cannot happen
at a local minimizer when d is a feasible direction, since
for some small
t > 0 x*+td would be feasible and f(x*+td) would be less than
f(x*) (by
virtue of the gradient condition), contradicting x* being a
minimizer.
-- Paul
*************************************************************
************
Paul A. Rubin Phone: (517) 432-3509
Department of Management Fax: (517) 432-1111
The Eli Broad Graduate School of Management E-mail:
rubin@msu.edu
Michigan State University http://www.msu.edu/~rubin/
East Lansing, MI 48824-1122 (USA)
*************************************************************
************
Mathematicians are like Frenchmen: whenever you say something
to them,
they translate it into their own language, and at once it is
something
entirely different. J. W. v. GOETHE
===
Subject: strain softening spring
I am a Civil engineer who has been working for the last 20
years in
the Þeld as a result of which my math is somewhat basic. I
have a
good handle of linear systems, but am taking my Þrst baby
steps
with non-lineear systems.
I have read up a few books on non-linear systems looking for a
canned solution for a problem I am trying to solve. But while
I see
a lot of stuff on the DufÞng equation I don¹t see anything
that
resembles the equation I am trying to solve - a
mass-spring-damper
system where the spring softens exponentially.
i.e. the equation I am trying to solve is as follows:
y¹¹ + cy¹ + [ko*e^(-alpha*t)]y = 0
Any help would be most appreciated.
sincerely
Paul Joseph
(pjoseph@excite.com)
===
Subject: Re: Simple numbers
> Hallo;
> Well, p(n) is a n-th prime but if n=123456 how do I know
what is the
> 123456-th prime? I need the correct number (for exampel
1-st prime is ->
2)!
>> Hallo,
>> I know that every simple number (beside 2 and 3) has its
one formula
:
>> S.N =
>> 6*n +/- 1 for some n , but not any n={1,2,3,...+} in
formula 6*n +/-
1
>> gives
>> the simple number (for example: n=6 => 6*6-1=35 , 35 is
not simple ,
>> 5,7|35)
>> By simple number I think you must mean prime.
>> so my question is:
>> If n={1,2,3,...} what is the formula which gives the
simple numbers
>> (beside
>> 2 and 3) for any n?
>> Evidently you don¹t consider p(n) is the n¹th prime to be a
formula.
>> Any
>> particular reason?
>> Computational complexity?
>> Jon Miller
Given any n, there is a simple algorithm that computes p(n).
You could
spend a
lifetime trying to make the algorithm more efÞcient, but the
basic
algorithm
is quite simple.
===
Subject: Re: Mean Value Theorem
>> Two places I¹ve seen the equivalent statement: If f is
continuously
>> differentiable on [0,1], then |f(x)-f(y)|<=sup|f¹(u)||x-y|
where the
>> sup is extended over 0<=u<=1.
>> I can¹t see why we need f¹ to be continuous. It seems by
the MVT we
>> have
>>
>> f(x)-f(y)=f¹(c)(x-y) for some c in [0,1],
>>
>> so that
>>
>> |f(x)-f(y)| = |f¹(c)||x-y|, for some c in [0,1],
>> <= sup_u|f¹(u)||x-y|.
>Maybe they intended f have values in R^n, so MVT is not
available.
>Assuming f¹ exists everywhere and f¹ is integrable, we can
still
>integrate f¹ on [x,y] to get the conclusion. And f¹ bounded
will
>imply f¹ is Riemann integrable, so adding the assumption
>f¹ is integrable is not needed. So I agree: even in this
>case, continuity of f¹ is not requried.
Or you can just say this:
If f : R -> R^n is differentiable then
|f(x)-f(y)| <= sup_u|f¹(u)||x-y|.
Pf: Choose p in R^n with |p| = 1 and
<p, f(x) - f(y)> = |f(x) - f(y)|. Let g = <p, f> and
apply the MVT to g... QED.
Doesn¹t require the fact that f¹ bounded implies that
f¹ is Riemann integrable (the proof of which is not
obvious to me this second...)
************************
David C. Ullrich
===
Subject: Re: Œerf¹ function in C
> BTW I made implementations in Fortran, HP-41C, Pascal,
Modula 2,
> Clipper and recently Visual Basic of the Taylor Series
expansion of
> Phi(x) given in 26.2.11 in A&S since around 1980:
> Phi(x) = 1/2 + phi(x)*(x + x^3/3 + x^5/(3*5) + x^7/(3*5*7)
+ ...)
> Here are my VB results against your algorithm (also
translated to
> Visual Basic). It looks like the VB version loses about 1
digit in
> precision against the C version (I¹m not surprised, but it
is not too
> bad either).
> x Marsaglia (my VB Version)
> Marsaglia (C Version, from your table)
> Klaey (VB)
> Maple 20 places
> 0.123 0.54894645101643700000
> 0.5489464510164369
> 0.54894645101643700000
> 0.54894645101643675909
> 1.2 0.88493032977829200000
> 0.8849303297782918
> 0.88493032977829200000
> 0.88493032977829173198
> 2.4 0.99180246407540400000
> 0.9918024640754040
> 0.99180246407540400000
> 0.99180246407540387055
> 6.1 0.99999999946965800000
> 0.9999999994696578
> 0.99999999946965800000
> 0.99999999946965767370
> -6.1 0.00000000053034350051
> 0.0000000005303427
> 0.00000000053034221459
> 0.00000000053034232638
> -1.1 0.13566606094638300000
> 0.1356660609463828
> 0.13566606094638200000
> 0.13566606094638267517
> 7.2 0.99999999999969900000
> 0.9999999999996990
> 0.99999999999969900000
> 0.99999999999969893721
> I¹m quite pleased how my algorithm still stands up today :-)
> Matthias Kl.8ay
> --
> www.kcc.ch
Excellent suggestion.
While it was not so in the 1960¹s when I
developed the method for Fortran via cPhi(x)/phi(x),
the ordinary Taylor series about zero for the normal
integral,
Phi(x)=.5+integral exp(-t^2/2)/sqrt(2*pi), x=0..t
= .5+(x-x^3/6+x^5/40-x^7/336+...)/sqrt(2*pi)
provides, with current double precision
arithmetic, an excellent method for providing
normal probabilities arithmetic for the range
of x¹s one usually encounters, say to +/- four
or Þve sigmas, letting the CPU do the work
after being given a few simple instructions.
Here is a sample implementation:
-----------------------------------------------
#include <math.h>
#include <stdio.h>
double Phi(double x)
{
long double z,t=0,s=.3989422804014327L;
int i;
s*=x; z=s; t=0;
for(i=3;s!=t;i+=2)
{z*=-x*x*(i-2)/(i*i-i);
t=s; s+=z;}
return .5+s;
}
int main(){
double x;
while(1){
printf(Enter your x value:);
scanf(%lf,&x);
printf(Normal Prob(X<%f)=%20.16fn,x,Phi(x));
}
}
---------------------------------------------
Here are a few results, with the true values
provided by setting Digits:=30 in Maple:
x Phi(x)
.123 .5489464510164368
.5489464510164367590816... true value
2.34 .9903581300546417
.9903581300546416673759...
-2.34 .0096418699453583
.0096418699453583326240...
3.45 .9997197067231839
.9997197067231838225884...
4.56 .9999974423189606
.9999974423189605484746...
6.6 .9999999999809187
.9999999999794421109060...
-6.6 .0000000000190813
.0000000000205578890939...
6.8 .9999999999950689
.9999999999947690424558...
-6.8 .0000000000049310
.0000000000052309575441...
7. 1.0000000000163602
.9999999999987201874561...
-7 -.0000000000163602
.0000000000012798125438...
Again, I hope to access 80-bit
þoating point processors by using
long double¹s. With plain double¹s,
there is little difference for |x|<4;
beyond that accuracy drops to 13 then
12,11,10,... digits until cases such
as Phi(6.5), which returns a value >1.
Those interested are invited to compare
on other platforms the approach using
the Taylor series for cPhi(x)/phi(x)
with that of the Taylor series for plain
Phi(x), using long double¹s then plain double¹s.
Either approach may be preferable to the
sometimes unavailable and/or mysterious erf.
George Marsaglia
===
Subject: Re: covering compact set w/squares
>Oh, I think I did misread. How about this:
>Given a compact set K in the plane s.t. each pt x is the
center of a
square
>Q_x, prove that you can Þnd a subsequence Q_x_i of squares
s.t. K is
>covered by the Q_x_i and
>sum(over i) Char(Q_x_i) <= 4 Char (union(over i) Q_x_i),
>where Char(X) is the characteristic (indicator) function of
X.
Not sure. That inequality is just another way to say that no
point
is contained in more than 4 of the Q_x_i; I¹m not sure whether
one can get that, but we know it doesn¹t follow just from
taking
a minimial cover.
When you say Oh, I think I did misread. How about this: it
sounds like you¹re still not certain what the statement is.
This might be a little simpler if you¹d get the statement
staight to begin with.
(Um: I¹ve been assuming that this is like a homework
problem. If it¹s not, something you¹re trying to do for
some other reason, I can tell you a standard result that¹s
very much like what you¹re trying to prove, and which you
might Þnd equally useful.)
>>Given a compact set K in the plane s.t. each pt x is the
center of a
>square
>>Q_x, prove that you can Þnd a subsequence Q_x_i of squares
s.t. K is
>>covered by the Q_x_i and the sum of the areas of each of
the Q_x_i is
no
>>more than 4 times the area of the union of the Q_x_i.
>> Aargh. I think I¹ll just spend the rest of my life posting
replies to
>> this message. I have to say that the two counterexamples
I¹ve
>> given were wrong (I was reading the statement of the
problem
>> Note however that the example showing that it¹s not true
that
>> in a minimal cover no point is in more than four squares
was
>> correct. (I¹m still not sure whether the result itself is
true and
>> I still wonder whether it¹s really what you wanted to
prove.)
>>I think this is a totally geometric thing, and I know it
sufÞces to
show
>>that in such a minimal cover no point is contained in more
than 4
>squares,
>>but how do i show this? I know how to do this with
intervals on the
real
>>line (there the answer is 2 times the area....), but i get
lost when
>going
>>up a dimension.
>> ************************
>> David C. Ullrich
************************
David C. Ullrich
===
Subject: Re: puzzle: GCDs of InÞnite Set of Integer Pairs
>Playing the physics card isn¹t helpful here. I¹m claiming
that it¹s
>*mathematically* impossible to randomly choose a real from
[0,1]. Not
physically
>impossible. I¹m asking you to convince me otherwise, use a
thought
experiment if
>you want, or not.
>>The canonical example is a spinner which can take on a
random angle.
>>This gets you [0,2pi)
>I think we had this one last time we explored this concept.
At some point
the
>quantum state of the universe takes over. There are a
countable number of
angles
>your spinner can indicate. Not surprising that they all have
non-zero
>probability, and that one actually happened.
Angular momentum is quantized, but angle is not (so far,
anyway). In
any case, bringing quantum physics to a thought experiment is
far
less valid than bringing Lebesgue measure to a question about
probability.
--
Matthew T. Russotto mrussotto@speakeasy.net
Extremism in defense of liberty is no vice, and moderation in
pursuit
of justice is no virtue. But extreme restriction of liberty
in pursuit of

a modicum of security is a very expensive vice.
===
Subject: Re: Handling the truth
<snip Decker and a guy named Dik Winter say that w_1(x)
w_2(x) = 7, and
> these are *varying* functions that are the factors of
(5a_1(x) + 7)
> and (5a_2(x) + 7), in the ring of algebraic integers.
<snip Can they derive w_1(x) and w_2(x)?
> No.
Yes. Here¹s a repost, in case you missed it.
You¹ve seen what happens when x=1 and x=2. Let¹s try a
different one, where things are simple enough to verify by
hand.
Suppose we take x = -3. Then we have
(5a_1(-3) + 7)(5a_2(-3) + 7) = 7(25(9) + 30(-3) + 2) = 7(137)
where the a¹s satisfy
a^2 + 4a + 7((-3)^2 + (-3)) = a^2 + 4a + 42.
We Þnd that
a_1 = -2 + sqrt(-38) and a_2 = -2 - sqrt(-38)
It¹s easy enough to verify that the a¹s aren¹t divisible by
7 or by sqrt(7).
However, a_1 does share a factor in common with 7,
namely
w_1 = (1 + 3sqrt(-38))^{1/3}
We can verify that
1. w_1^3 divides 7^3, since
(1 + 3sqrt(-38))(1 - 3sqrt(-38)) = 343 = 7^3
So w_1 = (1 + 3sqrt(-38))^{1/3} divides 7.
2. w_1^3 divides a_1^3 = 220 - 26sqrt(-38), since
(1 + sqrt(-38)(-8 - 2sqrt(-38)) = 220 - 26 sqrt(-38)
So w_1 = (1 + 3sqrt(-38))^{1/3} divides a_1
Thus w_1 is a common divisor of a_1 and 7. (In fact,
it¹s a gcd of a_1 and 7.)
In a similar way, we Þnd that
w_2 = (1 - 3sqrt(-38))^{1/3}
is a common divisor of a_2 and 7.
In this case, with
(5a_1 + 7)(5a_2 + 7) = 7(137)
We see that the factor 7 on the right splits as
w_1 * w_2 = 7
and that
f_1 = (5a_1 + 7)/w_1 =
5(-8 - 2sqrt(-38))^{1/3} + (1 - 3sqrt(-38))^{1/3}
and
f_2 = (5a_1 + 7)/w_1 =
5(-8 + 2sqrt(-38))^{1/3} + (1 + 3sqrt(-38))^{1/3}
f_1 * f_2 = 137
exactly as expected. Likewise, it¹s easy to show that
w_1, w_2, f_1, and f_2 are all algebraic integers
and none of them are units.
I hasten to add that this behavior is what happens for
almost all values of x, namely that in
(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
with the a¹s satisfying
a^2 -(x - 1)a + 7(x^2 + x)
we will be able to Þnd algebraic integers w_1(x) and w_2(x)
with w_1(x) * w_2(x) = 7 and w_i(x) dividing a_i(x) so that
with
f_i(x) = (5a_1(x) + 7)/w_i(x)
we will have
f_1(x) * f_2(x) = 25x^2 + 30x + 2
as long as 7 doesn¹t divide 25x^2 + 30x + 2.
With more or less difÞculty, one could do the same
construction for most integers x, just as I did for x=1
and Keith did for x=2. In most cases we¹ll Þnd that 7
splits into two nonunit factors, distributed between
(5a_1+7) and (5a_2+7). In general, as with
w_i(-3) = (1 +/- 3sqrt(-38))^{1/3}
it won¹t be immediately obvious that w_1(x) and w_2(x)
are divisors of 7 and of a_i(x), in the sense that
you can¹t look at them and immediately notice that
they divide 7 or a_i(x).
> Instead you have one of their cohorts Keith Ramsay throwing
out a
> degree 22 polynomial claiming to have found it some kind of
way, yet
> if you look at
> w_1(x) w_2(x) = 7
> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
> and
> a^2 - (x - 1)a + 7(x^2 + x)
> there is no way to derive the w¹s as varying functions of x.
> The only thing that actually makes sense is something like
> w_1(x) = 7, w_2(x) = 1.
No. See above.

> I can point out the obvious, like why 7? Why not 13? Why
can¹t you
> just have something like
> w_1(x) w_2(x) = 13
> (5a_1(x) + 13)(5a_2(x) + 13) = 13(25x^2 + 30x + 2)
> and
> a^2 - (x - 1)a + 13(x^2 + x)?
> Mathematically there has to be a reason, right?
But you can. Just use
(5a_1(x) + 13)(5a_2(x) + 13) = 13(25x^2 + 30x + 3)
where the a¹s satisfy
a^2 - (x - 2)a + 13(x^2 + x)
Then an analogue to what I did above will allow you to
Þnd w_1(x) and w_2(x) for most integers x with the
w¹s being algebraic integer divisors of 13 and the
appropriate a, such that w_1(x) * w_2(x) = 13, just like
we saw above.
Rick
===
Subject: Re: Help needed - primes program
>Hallo,
>I need help with the following :
>I made a program which gives all the primes from selected
area. I made it
in
>Pascal. The only problem is that it hasn¹t a very good
algorithm which I
>need.
>Here is the code:
>------------------------------------------------------------
---------------
-
>---
>Program Primes;
>Var
> x,y:word;
> f:word;
> i,n:word;
> ok:boolean;
>Begin
> Write(ŒEnter X and Y -> Œ);
> Readln(x,y);
> For i:=x to y do
> Begin
> f:=2;
> ok:=true;
> While f<=i-1 do
> Begin
> If i mod f = 0 then
> Begin
> f:=i;
> ok:=false;
> End
> else
> f:=f+1;
> End;
> If ok=true then Write(i,¹ is a prime. Œ);
> End;
> Readln;
>End.
>------------------------------------------------------------
---------------
-
>---
>I wanted to give you attachment with the *.EXE Þle but I
can¹t so if ypu
>need to compile this code download Free Pascal (current
version is 1.0.10)
>form www.freepascal.org .
>Any suggestion and similar is welcome.
Why not start at 3 and ignore all the even numbers in the
loops, it
would double the speed.
http://www.nist.gov/dads/HTML/sieve.html including
implementations in
pascal.
===
Subject: Re: Help needed - primes program
> Hallo,
> I need help with the following :
> I made a program which gives all the primes from selected
area. I made it
in
> Pascal. The only problem is that it hasn¹t a very good
algorithm which I
> need.
Use the Sieve of Eratosthenes.
Google will Þnd several versions and descriptions of it, as
will
the prime links at http://primepages.org/
Phil
--
Unpatched IE vulnerability: XSS in Unparsable XML Files
Description: Cross-Site Scripting on any site hosting Þles
that can be
misrendered in MSXML
Reference: http://sec.greymagic.com/adv/gm013-ie/
Exploit: http://sec.greymagic.com/adv/gm013-ie/
===
Subject: Re: tensors for tots
> I am also familiar with the idea of transforming matrices
themselves
> to a new basis. If B is the transformation matrix (in my
sense),
> then the prescription for transforming a matrix A is:
> A¹ = BAB^(-1)
Observe that this is the Transformation rule for a
(1,1)-Tensor.
> I¹m uncertain how we wish to transform v*
Ok, let there be a Þxed dual vector, say (v*), which is a
linear
mapping (v*) : V -> R. (V a R-vector space)
The value of (v*) at any vector (v) does clearly not depend
on the
basis, let it be x := <(v*), (v)>
(The value of (v*) at (v))
Now we introduce a basis and have coordinate-vectors of (v*)
and (v),
say v* and v. We write v* and v as a column. We get the
value of (v*) at (v) by matrix multiplication of theses two
vectors:
x = <(v*), (v)> = v*^t v.
Now we change the basis. We know the new coordinates of (v),
these are
v¹ = Bv (your notation :-))
Let v*¹ be the new coordinates of (v*).
These are given by
v*¹ = Av* with some Matrix A
What is A? Look at
x = v*^t v = v*¹^t v¹ = (Av*)^t B v = v*^t A^t B v
which implies A^t B = identity matrix,
A = B^(-1)^t.
Here we are.
> I have a hunch this has to do with the metric tensor
A normal matrix, so to say, is a (1,1)-Tensor.
It can be multiplied by a row from the left and by a column
from
the right, that is by a covariant and a contravariant vector.
The metric tensor eats two contravariant vectors, his matrix M
represents a different object. It stands for the mapping
(v,w) -> v^t M w
(here v and w are two columns)
The transposing of v is built in.
This implies a different behaviour under change of
coordinates.
===
Subject: Re: Sets That Resemble Derivatives Somewhat
> 2) [(A-->B)(A¹-->B¹)]¹ = AB¹ U A¹B
> 3) [(A¹-->B)(A-->B¹)]¹ = A¹B¹ U AB
2) is just symmetric difference, assuming the existence of
some universe
===
Subject: Re: e is transcendental (was: classes of
transcendental numbers ?
Why do I even try? Are you even TRYING to learn? Or are you
so closed-minded that you can¹t be bothered?
>><snip>So? The real part of exp(i pi) is cos(pi), and its
imaginary
part
>>is sin(pi), so all you are saying is that cos(pi) = -1 and
>>sin(pi) = 0, and we were already aware of these facts.
There is
>>NO reason to conclude that exp(i pi) = 0.
> I,have a reason , with my due respect.
> Panagiotis Stefanides
>>Yes, but you DON¹T tell us what your reason is. You can¹t
expect us
to
>>accept your claims without giving support for those claims.
So what

>>possible reason could you have for expecting us to agree
with your
claim
>>that exp(i pi) = 0?
>The reason is simply that exp(ipi0=-1 should be accompanied
>by the statement that this is the real part solution.
>Is it fair?
>>No, that is not a fair comment. exp(i pi), the complex
number, is
>>equal to -1, the complex number. There is no need to appeal
to the
>>real part. Also, what does exp(i pi) = -1 is the real part
solution
>>mean? You look like you are using terminology in a manner
not
>>recognized in mathematics.
>>David McAnally
>>--------------
>e^[i*pi] ,as accepted ,is a phasor.
>>In most of the relevant Þelds of mathematics, e^[i pi] is a
complex
>>number.
>It is only fair to state that its
>polar representation is :
>e^[i*pi] = MOD 1 , ARG 180 .
>>That is Arg 180 degrees, not just Arg 180. And so what?
That does
>>not lead to your assertion that exp[i pi] = 0, a result for
which
>>you have given absolutely no support. Why don¹t you just
give up?
>>David McAnally
>> At the moment, they (the Time Lords) are far from being
all-powerful.
>> That¹s why it¹s been left up to me and me and me.
>> quote by: Patrick Troughton in The Three Doctors
>>-------
>I, have made myself very explicit.
>My original question of the implication
>of the imaginary component: e^[ipi]=j*0 (to the related
proof)
>Complex Notation, chapter 12 ,Electrical Technology 3RD ED.
>Edward Hughes Longmans ,page 338:
>Stares:
> OA*=OB+jOC=OA(COStheta+jSINtheta)
>* Symbols representing phasors are printed in bold face
italics, while
those representing only magnitudes are printed in ordinary
italics,..
>Here is very clear the difference between PHASOR and
MAGNITUDES
I know the difference between a complex number and its modulus
(or magnitude). You don¹t have to explain the difference to
me.
>[which(MAGNITUDES) I, referred to as REAL PART or IMMAGINAR
PART,
The magnitude of a complex number is generally not equal to
either
its real or imaginary part. How could you claim that it is?
> I should have stated COMPONENTS ].
>>Perhaps you mean that (the imaginary part of e^[i pi]) = 0,
in which
>>case you are correct, but you have had a lot of problem
expressing
>>yourself, especially in view of the way that you initially
made
>>the claim by stating that e^[i pi] = 0, which you described
as the
>>imaginary part solution, using a terminology that nobody
but you
>>knows.
> e[i*theta]=COStheta+iSINtheta
I know that it is true that exp[i theta] = cos(theta)+i
sin(theta).
It follows that exp[i pi] = -1. Incidentally, e[i theta] = i
e theta,
unlike what you have written.
>thetas could be given and calculations could be performed
>for numerical evaluations.
And for other results as well.
>In books is stated that it is FORMULA
>and also terms such as evaluate:(-1+i*sqrt[3])^10
>Are these not solutions to problems?
No. Evaluate (-1+i sqrt(3))^10 is a problem for which you can
get a solution using exp(i theta) = cos(theta)+i sin(theta).
This
does not mean that exp(i theta) = cos(theta)+i sin(theta) is
itself
a solution. You need a problem before you can describe
anything as
a solution.
>>Nobody knows what you mean when you make a statement like
exp[i pi] =
-1
>>is the real part solution of exp[i pi] = -1, or that exp[i
pi] =
0
>>is the imaginary part solution of exp[i pi] = -1. I asked
you to
>>explain your terminology but you haven¹t bothered.
>I, doubt it but ,still I, exlpained it anyway.
I did ask you. And you did not explain your bizarre
terminology. There
was no problem, hence there is no solution, whether real part
or
imaginary part.
>>As I, exlained earlier, but I, am not the kind of not
>being bothered
This was not a quote from me. Why are you pretending that it
was?
>>When I take the imaginary part of the equation exp[i pi] =
-1, I get
>>sin(pi) = 0, a fact which is already known to us.
>>Have you thought also of the fact that if you have exp[i
pi] = -1 (your
>>real part solution) and exp[i pi] = 0 (your imaginary part
solution),
>>then you could conclude that 0 = exp[i pi] = -1, and get a
contradiction?
>Well tthere is always the possibility of an answer such as
>MULTIVALUED , examplum gratias:
It is known that the exponential function is single-valued
(in fact,
it is known that the exponential function is entire).
> e^[2i*pi]=1
> ln[1]=0=2i*pi
The logarithm has a branch point at 0. The exponential
function does
not have a branch point. The analogy is invalid. Because the
exponential function is single-valued, my objection here
still stands.
>I, make use of the comlex notation , but still ,I, have my
>natural questions ,and do not accept everything for granted.
But you shouldn¹t start making up mathematics to suit
yourself.
>I, give an example in the form of question:
>It is required that COS[-i]+i*COS[-i] be evaluated
>so that is its Modulus and Argument
>be evaluated (NUMERICALLY].
cos(-i) = cosh(1), so cos(-i)+i cos(-i) = sqrt(2) cosh(1)
exp(i p/4).
The magnitude is sqrt(2) cosh(1), which is approximately
2.18225,
and its argument is pi/4, which is approximately 0.7854.
Alternatively, the argument is exactly 45 degrees.
David McAnally
--------------
===
Subject: Re: x^2 + y^4 = z^4
> + y^4 = z^4 has no positive-integer solutions. Is the proof
of this
> result short enough for some kind soul to post it, or need
I make a trip
> to the library? (I have citations.)
An elementary proof (no elliptic curves) is based on descent.
That
is, assume a solution exists in positive x, y, z, and show
that
another one can be found with smaller z. The proof is not
particularly difÞcult but too long to type out here (one
single-space
textbook page). It can be found in many basic number theory
books,
e.g. Niven and Zuckerman in the chapter on Diophantine
equations, p.
107 in the 3rd edition.
===
Subject: Re: Mean Value Theorem
> Two places I¹ve seen the equivalent statement: If f is
continuously
> differentiable on [0,1], then |f(x)-f(y)|<=sup|f¹(u)||x-y|
where the
> sup is extended over 0<=u<=1.
> I can¹t see why we need f¹ to be continuous. It seems by
the MVT we
> have
> f(x)-f(y)=f¹(c)(x-y) for some c in [0,1],
> so that
> |f(x)-f(y)| = |f¹(c)||x-y|, for some c in [0,1],
> <= sup_u|f¹(u)||x-y|.
However, it turns out, later in the discussion, they start
talking
about rates of convergence, so that there is in fact a need
for
sup|f¹(u)| to be Þnite--it¹s not just a superþuous nicety.
Does anyone know of a case where a textbook purposely included
superþuous infomation as part of an exercise? Or how about on
a
test question? It seems that it would make the problems
somewhat
more realistic, albeit esthetically unpleasing. It may make
students
think more about the problem. Students seem to develop the
unhealthy
assumption that if it is given, then is must be used.
===
Subject: Re: puzzle: GCDs of InÞnite Set of Integer Pairs
> mmmm... in a sense it does. But then again in a sense, most
real numbers
don¹t
> have a proper existence. In terms of proportions, 100% of
real numbers are
not
> thought of, used, speciÞed, and don¹t occur naturally as a
measure of
> something. I¹m excluding extremely vague usage such as you
specify Œall
reals
> larger than x¹.
The set of real numbers that have a Þnite description is
countable;
this is a superset of the algebraic reals. If your proper
existence
is equivalent to Þnite description, your assertion is trivial.
--
Daniel W. Johnson
panoptes@iquest.net
http://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
===
Subject: Re: Handling the truth
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1IFs8R04646;
>show that he posts from Hamilton College:
You consider mathematicians *incapable* ?
Do you or do you not consider mathematicians *incapable* !
Maurice
===
Subject: intercept length when a random line intersect with
ellipsoid shell
or parallelepiped shell with Þxed thickness
If a random line intersects with a ellipsoid shell or
parellelepiped
shell with a Þxed thickness, how much is the statistical
intercept
length?
===
Subject: Re: covering compact set w/squares
No, it¹s not HW...it was an aside in a measure theory course
that seemed
kind of curious but that I couldn¹t immediately (and at this
point just
plain couldn¹t) verify.
>Oh, I think I did misread. How about this:
>Given a compact set K in the plane s.t. each pt x is the
center of a
square
>Q_x, prove that you can Þnd a subsequence Q_x_i of squares
s.t. K is
>covered by the Q_x_i and
>sum(over i) Char(Q_x_i) <= 4 Char (union(over i) Q_x_i),
>where Char(X) is the characteristic (indicator) function of
X.
> Not sure. That inequality is just another way to say that
no point
> is contained in more than 4 of the Q_x_i; I¹m not sure
whether
> one can get that, but we know it doesn¹t follow just from
taking
> a minimial cover.
> When you say Oh, I think I did misread. How about this: it
> sounds like you¹re still not certain what the statement is.
> This might be a little simpler if you¹d get the statement
> staight to begin with.
> (Um: I¹ve been assuming that this is like a homework
> problem. If it¹s not, something you¹re trying to do for
> some other reason, I can tell you a standard result that¹s
> very much like what you¹re trying to prove, and which you
> might Þnd equally useful.)
>>Given a compact set K in the plane s.t. each pt x is the
center of a
>square
>>Q_x, prove that you can Þnd a subsequence Q_x_i of squares
s.t. K is
>>covered by the Q_x_i and the sum of the areas of each of
the Q_x_i is
no
>>more than 4 times the area of the union of the Q_x_i.
>> Aargh. I think I¹ll just spend the rest of my life posting
replies to
>> this message. I have to say that the two counterexamples
I¹ve
>> given were wrong (I was reading the statement of the
problem
>> Note however that the example showing that it¹s not true
that
>> in a minimal cover no point is in more than four squares
was
>> correct. (I¹m still not sure whether the result itself is
true and
>> I still wonder whether it¹s really what you wanted to
prove.)
>>I think this is a totally geometric thing, and I know it
sufÞces to
show
>>that in such a minimal cover no point is contained in more
than 4
>squares,
>>but how do i show this? I know how to do this with
intervals on the
real
>>line (there the answer is 2 times the area....), but i get
lost when
>going
>>up a dimension.
>> ************************
>> David C. Ullrich
> ************************
> David C. Ullrich
===
> ummm...except it¹s slightly a parody site. it¹s not real.
Why don¹t you
> actually read the whole site before you start condemning
it. Moron.
> Uh..I think he¹s promoting it. You might want to think hard
about that
> last word you used.
> ;^)
Uh...doesn¹t READ like a promotion. I think I¹d rather play
dumb.
- Nick
===
Subject: Re: No Set Contains Every Computable Natural
Russell Easterly says...
>The set of all natural numbers is recursive by Þat.
>Why bother to include computability in the deÞnition?
>A human must decide if x is a member of this set
>before x can be given to the TM.
Right. The notion of recursive set is only nontrivial
for proper subsets of the naturals. For the complete set,
it is recursive by deÞnition.
--
Daryl McCullough
Ithaca, NY
===
Subject: Full Beal Conjecture (revised)
Sorry for the mistake. My proof can be obtained at
===
Subject: Quadratics and transformation
It seems to me that I should be able to reasonably easily get
a general
idea of the type(s) of transformations that have been applied
to x^2
just by looking at the Þnished quadratic.
That is, if I have 3x^2 + 7x + 2, can I easily relate the
coefÞcients
and constant to the dilations, reþections and translations
that this
function represents compared to x^2 ?
I have tried thinking about it as a combined function, that
is,
f(x)=x^2, g[f(x)]=3x^2 + 7x + 2 and trying to Þnd g(x). Well,
I haven¹t
tried with this example but one I tried earlier turned into a
horrible
mess and didn¹t give me what I wanted - I assumed that g(x)
is simply
the inverse of g[f(x)] so maybe that¹s why it got so messy ?
It also seemed

to be a lot of work!
I¹m aware that I could determine the turning point and work
out the
translations and that the roots would indicate what
horizontal dilation
has been applied. I expect that if I calculate two other
points I would
also be able to determine reþection and vertical dilation. Of
course I
could always graph it but I¹m hoping to achieve what I want
from
inspection of the description of the function (is that the
right
terminology?).
Any and all help appreciated.
Ivan.
===
Subject: re:JSH: Why am I so important?
I see what he¹s up to, we¹ll Þll this thread with posts and
he¹ll go
to somewhere else and say that he¹s important and will have
this
thread to back him up, and the chain goes on forever. Don¹t
reply to
him.
http://www.newsfeed.com The #1 Newsgroup Service in the
World! >100,000
Newsgroups
---= 19 East/West-Coast Specialized Servers - Total Privacy
via Encryption
=---
===
Subject: re:Handling the truth
How I hate it when wanna-be mathematicians write out cute
identities
and call it Œresearch¹ and then make wars about the
philosophical
implications of some useless notions. Then they have a go
about their
moral superiority and stuff...
Stick to mainstream math and stop f*cking people¹s brains. A
megalomaniac cat is no way a Œbetter person¹ than the humble
folk.
http://www.newsfeed.com The #1 Newsgroup Service in the
World! >100,000
Newsgroups
---= 19 East/West-Coast Specialized Servers - Total Privacy
via Encryption
=---
===
Subject: Re: HELL :-) (Was: Re: the anticlassicalist }{ ii:
the spectre
continues)
> *MATHEMATICS IS FULL OF TRICKS*
French even allows you the conjugation :
ma th.8ematique
ta th.8ematique
sa th.8ematique
notre th.8fme .88 tics
votre th.8fme attique
leurre, t¹es mat, hic !
> *REDUCING TOPICS TO THE SCIENCE OF NUMBER IS NUMEROLOGY*
I¹d call that accounting.
===
Subject: Re: pentiamond rep-tiles, continued
> That leaves these two:
> ____ __/
> /____ /___/
> I see no reason that these should not be rep-tilable.
Neither do I see
> a rep-tiling of either! Does anybody?
I now see a reason that the pentiamond on the right should
not be
rep-tilable. Suppose you have such a rep-tiling. Look at the
little
pentiamonds along the edge of the big one, starting at the
top vertex
and following the edges on the right and bottom. You will Þnd
that
all the boundaries between the little pentiamonds have the
same orientation
relative to the edge of the big one:
_______________
You cannot escape from this pattern, even when you pass a
120-degree
angle. When you reach the lower left vertex, you¹re stuck.
No solution yet on the I-, or straight, pentiamond. It can¹t
be done
on a scale of 9x9 or smaller.
-:-
To what do I owe the honor of this unexpected visit, Lord
Ruthven . . . alias Lyford Pemberton!
H. C. Artmann, Tom Parker, International Detective
--
Col. G. L. Sicherman
home: colonel@mail.monmouth.com
work: sicherman@att.com
web: <http://www.monmouth.com/~colonel/>
===
Subject: Re: errors in an argument
>Let¹s just look at the probability of a protien made up of
>100 amino acids forming by chance. All the amino acids
>have to line up just right, and the probability of this
>happening in any given try is 1/100! or 10^(-158).
> There are only 20 different amino acids.
> The more subtle error is that nobody says that life required
> exactly this protein to be produced by a random permutation.
> There are probably a huge number of different proteins that
> could do more or less whatever this one does. And its
production
> would not be a matter of random permutations, but rather of
> the gradual accumulation of small improvements. You might
> look up genetic algorithms: this is not just a theoretical
> construct but a method which is actually used to obtain a
> nearly-optimal solution to a difÞcult problem by simulating
> the process of evolution.
> That said, I think it¹s fair to add: my impression, as an
> outsider in this Þeld, is that while evolution itself (for
> organisms that already have the usual genetic machinery)
has pretty
> good mathematical models, we are still far from
understanding
> at a quantitative level how that genetic machinery might
have
> arisen in the Þrst place.
> Robert Israel israel@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
Robert, of course, mentioned exactly the errors I was
thinking of,
with the second being that we don¹t know how many possible
proteins there are with 100 amino acids in them. To say that
for
each one, the probability is essentially 0, and so the
collective
probability for the set of proteins is 0 is quite a mistake.
If I were trying to explain the error to someone, I would
probably
use the analogy of rolling a 20 sided dice 100 times and
recording
the result. By the proof above, one can¹t do this, since the
probability of any particular sequence is (essentially) 0.
I guess what I would really like to know is how poorly I
should
think of the person who made the argument. Here are some
qualiÞers: He boasts of his degree in mathematics. Also, in
his
talk, he indicated that he would try to teach us how to think
critically about what scientists say. On the other hand, he is
now in his mid 60¹s, so he is far removed from his schooling.
Finally, almost certainly (this is a guess of mine) he did not
formulate the proof himself, but it had been presented to him.
He liked the argument and missed the þaws.
Should I not be too critical of his wishful thinking that this
proof actually means something?
By the way, I certainly agree completely with Robert¹s
last paragraph.
John
===
Subject: Re: Big Rip NY Science Times
Is it my imagination, or are this idiot¹s ramblings becoming
even *more* delusional and illucid than before?
-E
===
Subject: Re: How to diagonalize a Hermitian matrix
>Hello group,
> Could somebody please tell me a numerical algorithm to
>diagonalize a Hermitian matrix H, so that I end up not only
with the
>eigenvalues perched along the diagonal, but also with the
unitary
>matrix U that conjugates with H to diagonalize it.
> I should probably say what I really want. I have a
>positive-deÞnite Hermitian matrix H, and I need its matrix
square
>root. The above is simply the approach that occurred to me
for
>obtaining that square-root.
> There are available algorithms to do this diagonalization.
> Any good library should have such.
> I am assuming that by square root you mean the Hermitian
> square root. If you just want a matrix A with H=A*A~, the
> conjugate transpose of A, the Cholesky decomposition is the
> fast way to do it.
> One can use Newton¹s method to obtain the square root of a
> matrix; the rate of convergence is not bad. There are other
> ways as well. I suggest you ask the question in
> sci.math.num-analysis.
think that is what I am looking for. I will try on the group
you
suggest.
Achava
===
Subject: Re: Sets That Resemble Derivatives Somewhat
there¹s a general theme that some people have explored that
says that
the fact that certain sorts of differential operators obey
leibniz-like identities is conceptually subordinate to the
fact that
certain sorts of boundary operators in geometry and topology
obey
leibniz-like identities, roughly (or sometimes precisely,
depending on
the particular concept of boundary being used) boundary(x X
y) =
(boundary(x) X y) + (x X boundary(y)).
--
[e-mail address jdolan@math.ucr.edu]
===
Subject: Big Rip
http://qedcorp.com/destiny/
under construction
DARK MATTERS SURROUND DARK ENERGY
with my comments, additions and physics corrections
ÒTwo big stories from the world of physics may portend the
arrival of
new weapons of mass destruction far more powerful and compact
than
atomic bombs. In recent years it has been discovered that our
universe
is being blown apart by a mysterious anti-gravity effect
called dark
energy. Mainstream physicists are scrambling to explain this
mysterious
acceleration in the expansion of the universe. Some
physicists even
believe that the expansion will lead to The Big Rip when all
of the
matter in the universe is torn asunder - from clusters of
galaxies in
appears to be made of two unknowns - roughly 23% is dark
matter, an
invisible source of gravity, and roughly 73% is dark energy,
an
invisible anti-gravity force. Ordinary matter constitutes
perhaps 4
percent of the universe. Recently the British science news
journal New
Scientist revealed that the American military is pursuing new
types of
exotic bombs - including a new class of isomeric gamma ray
weapons.Ó
That was an original idea of mine in 1963 at Cornell and I
discussed it
with Hans Bethe. That is one of the reasons Ron Bullough
invited me to
Harwell in 1966. No doubt others thought of it but probably
later. I
thought of it while at Tech/Ops in Lexington, Mass working
for George
Parrant Jr.
ÒUnlike conventional atomic and hydrogen bombs, the new
weapons would
trigger the release energy by absorbing radiation, and
respond by
re-emitting a far more powerful radiation. In this new
category of
gamma-ray weapons, a nuclear isomer absorbs x-rays and
re-emits higher
frequency gamma rays. The emitted gamma radiation has been
reported to
release 60 times the energy of the x-rays that trigger the
effect.Ó
Gamma-ray weapons could trigger next arms race 19:00 13
August 03
ÒExclusive from New Scientist Print Edition. Subscribe and
get 4 free
issues.
An exotic kind of nuclear explosive being developed by the US
Department
of Defense could blur the critical distinction between
conventional and
nuclear weapons. The work has also raised fears that weapons
based on
this technology could trigger the next arms race. The
explosive works by
stimulating the release of energy from the nuclei of certain
elements
but does not involve nuclear Þssion or fusion. The energy,
emitted as
gamma radiation, is thousands of times greater than that from
conventional chemical explosives.
The technology has already been included in the Department of
Defense¹s
Militarily Critical Technologies List, which says: Such
extraordinary
energy density has the potential to revolutionise all aspects
of
warfare. Scientists have known for many years that the nuclei
of some
elements, such as hafnium, can exist in a high-energy state,
or nuclear
isomer, that slowly decays to a low-energy state by emitting
gamma rays.
For example, hafnium-178m2, the excited, isomeric form of
hafnium-178,
has a half-life of 31 years. The possibility that this
process could be
explosive was discovered when Carl Collins and colleagues at
the
University of Texas at Dallas demonstrated that they could
artiÞcially
trigger the decay of the hafnium isomer by bombarding it with
low-energy
X-rays (New Scientist print edition, 3 July 1999). The
experiment
released 60 times as much energy as was put in, and in theory
a much
greater energy release could be achieved.Ó
http://www.newscientist.com/news/news.jsp?id=ns99994049
I was thinking in 1963 of a gamma ray laser pumping a nuclear
isomeric
transition. Bethe at the time said it wouldn¹t work and
basically
discouraged me working on it.
Bekkum continued:
ÒIn the summer of 2000 I contacted Nick Cook, the former
aviation editor
and aerospace consultant to Jane¹s Defence Weekly, the
international
military affairs journal. Cook had been investigating black
budget
super-secret research into exotic physics for advanced
propulsion
technologies.Ó
Uh Oh :)
ÒI had been monitoring electronic discussions between various
American
and Russian scientists theorizing about rectifying the
quantum vacuum
for advanced space drive. Several groups of scientists,
partitioned
into various research organizations, were exploring what NASA
calls
Breakthrough Propulsion Physics - exotic technologies for
advanced
space travel to traverse the vast distances between stars.
Partly
inspired by the pulp science Þction stories of their youth,
and partly
by recent reports of multiple radar tracking tapes of
unidentiÞed
objects performing impossible maneuvers in the sky, these
scientists
were on a quest to uncover the most likely new physics for
star travel.
The NASA program was run by Marc Millis, under the Advanced
Space
Transportation Program OfÞce (ASTP). Joe Firmage, a Silicon
Valley
entrepreneur, who at age 28 had found risen to CEO of a three
billion
dollar internet Þrm, began to fund research in parallel with
NASA. He
hired a NASA Ames nano-technology scientist, Creon Levit, to
run the
International Space Sciences Organization,
Joe did that because I suggested it. I introduced Creon to
Joe.
ÒCook was intrigued by the apparent connections between
various private
investors, defense contractors, NASA, INSCOM (American
military
intelligence), and the CIA. While researching exotic
propulsion
technologies Cook had heard rumors of a new kind of weapon, a
sub-quantum atomic bomb, being whispered about in the dark
halls of
defense research.Ó
I think that must have come from me regarding J. P. Vigier¹s
tight
atomic states with experiments in Beograd by Z. Maric and G.
Dragic.
But how did Cook hear about that? We brought Vigier to ISSO
in San
Francisco several times along with physicist Gennady Shipov
from Moscow.
That story with photographs of Vigier and the group is in my
autobiography ÒDestiny Matrix.Ó Dragic A,
Maric Z, Vigier JP; Phys.
Lett. A 265 (2000) 163. New quantum mechanical tight bound
states and
Œcold fusion¹. Creon Levit and Vigier met with Maric in
Budapest in
2000.
Bekkum who is one of my Òon line studentsÓ
continued:
ÒSub-quantum physics is a controversial re-interpretation of
quantum
theory, based on so-called pilot wave theories, where an
information
that the predictions of ordinary quantum mechanics could be
recast into
a pilot wave information theory. Recently Anthony Valentini
of the
Perimeter Institute has suggested that ordinary quantum
theory may be a
special case of pilot wave theories, leaving open the
possibility of new
and exotic non-quantum technologies. Even thought rumors of a
sub-quantum bomb may be purely fantasy ÉÓ
It¹s not fantasy. It might not work, Maric and Dragic in
Beograd, while
not ostensibly trying to make a weapon by any means, were
trying to test
Vigier¹s basic theory of the spatially extended electron,
which I think
is basically a correct idea and Þts my own ideas including
why the
electron appears to shrink to less than 10-16 cm under high
resolution
imaging (i.e. scattering) and how the electric charge
distribution is
contained by the strongly attractive zero point energy exotic
vacuum
dark matter core (Abraham-Becker-Lorentz-Poincare stress
problem of
100 years ago). This only works in the Bohm hidden or extra
variable

interpretation. That is, a classical spatially extended
electric charge
distribution is unstable. It explodes under its own
self-repulsion. This
is why physicists had to postulate a point electron because
they did not
understand that the strong gravity attraction of the positive
zero point
pressure in a possible state of exotic vacuum would hold the
charge
together. As Herbert Frohlich told me at UCSD in La Jolla in
1966
Òthe
basic thing wrong with physics is the idea of the point
electron.Ó The
bad idea of the point electron gives the inÞnite energy in
quantum
electrodynamics. Richard Feynman told me in his ofÞce at Cal
Tech in
1968 that ÒinÞnite renormalization is a shell game, and it
is a scandal
in physics that no one could do betterÓ than what he had
done. They did
not know 100 years ago that 1/3 or so of the universe was
this kind of
exotic vacuum. For example, there is a huge sphere of exotic
vacuum of
w = -1 positive pressure that holds our galaxy together
preventing our
solar system from going off into space on its own. This
sphere looks
like w = 0 Òcold dark matterÓ from our vantage
point. What works on this
large scale also works on the small scale of the single
electron (and
all the charged lepto-quarks). A neutrino has some mass and
is simply a
micro-geon of pure zero point energy with positive pressure.
ÒÉ there is no question that physicists
seriously contemplate a phase
transition in the quantum vacuum as a real possibility. The
quantum
vacuum deÞes common sense, because empty space in quantum
Þeld theory
appear and disappear far too quickly to be detected directly,
but their
existence has been conÞrmed by experiments that demonstrate
their
inþuence on ordinary matter.Ó
A major component of the physical quantum vacuum consists of
virtual
electrons frothing and bubbling at the Fermi surface edge of
the Dirac
negative energy sea. This is because of the Pauli exclusion
principle
that only none or one electron per quantum state. A virtual
electron
pops out of the vacuumÕs Fermi surface leaving a hole behind.
The hole
is the virtual positron. The result is a Òvirtual
electron-positron
pair.Ó However, the virtual electron and the virtual positron
attract
because they have opposite charges and they are exchanging
virtual
photons. Therefore, some of them form a more stable bound
state. An
enormous number of these virtual pairs Bose-Einstein condense
into the
same center of mass quantum wave packet forming the ÒVacuum
Coherence
FieldÓ (AKA ÒInþation
FieldÓ). This is a dynamic steady state of
detailed balance in which there is a continual inþow and
outþow of
virtual pairs into and out of this giant quantum or
Òmacro-quantumÓ
Òsuperþuid.Ó Essentially this is a vacuum
phase transition, similar to
the BCS transition from a normal metal to an electrical
superconductor,
from the globally þat micro-quantum electrodynamic vacuum
without any
gravity at all to the curved macro-quantum electrodynamic
vacuum with
emergent gravity. EinsteinÕs Þeld equation of general
relativity can
be derived from the phase wiggles and ripples in the robust
stable
macroscopically occupied center of mass quantum wave packet
of the bound
state of the virtual electron-positron pair. The exotic
vacuum dark
energy and dark matter are simply the amplitude wiggles and
ripples of
this same virtual pair quantum wave packet. The wave packet
spreads
over the entire 3D space of the post-inþationary bubble on
which our
Hubble-horizoned universe is located along with an inÞnity of
parallel
American. If the world hologram idea is correct, take the
surface area
of the expanding Hubble sphere that is the causal retarded
boundary of
3D space of our past light cone at Earth and divide it by the
quantum of
area. That gives us the number of Bekenstein-Shannon c-bits
and
explains the Òarrow of timeÓ (AKA Second Law
of Thermodynamics) of
increasing thermodynamic entropy in terms of the dynamical
expansion of
the 3D space of the universe. Lenny Susskind calls this
ÒDeSitter Space.Ó
Such research should be forbidden!
Too late. PandoraÕs Box is open. SchrodingerÕs
Cat has jumped out of it.
ÒIn the early 1970¹s Soviet physicists were concerned that
the vacuum of
our universe was in fact only one possible state of empty
space. The
fundamental state of empty space is called the true vacuum.
Our
universe was considered to reside in a false vacuum,
protected from
the true vacuum by the wall of our world. A change from one
vacuum
state to another is known as a phase transition. This is
analogous to
the transition between frozen and liquid water. Lev Okun, a
Russian
physicist and historian recalls Andrei Sakharov, the father
of the
Soviet hydrogen bomb, expressing his concern about research
into the
phase transitions of the vacuum. If the wall between the
vacuum states
was to be breached, calculations showed that an unstoppable
expanding
bubble would continue to grow until it destroyed our entire
universe!
Sakharov declared Such research should be forbidden! since
there was
always the possibility that an experiment might accidentally
trigger a
vacuum phase transition.Ó
British Astronomer Royal, Sir Martin Rees, Master of Trinity
College,
and Director of the Cambridge University Institute of
Theoretical
Astronomy on Madingley Road where Stephen Hawking works
discusses all
this in Chapter 9 of his important book ÒOur Final
Hour.Ó
ÒCould the wall of our universe be breached from within? The
amount of
energy required to punch a hole through the wall appeared to
be
enormous, and no known natural physical phenomena, even the
most
energetic, had punched through either. A recent report
commissioned to
examine potential dangers at the Large Hadron Collider, one
of the next
the best of our existing knowledge. Others are not so
certain, however.
At least one of the Russian physicists I had corresponded
with was said
to have been a former associate of Andrei Sakharov. He
strongly hinted
at new theories the Russians had developed which allow for the
manipulation of the fundamental constants of nature, but he
never
revealed more than a sketch of his ideas. He claimed that a
breakthrough
was within reach, perhaps within Þve years
ÉÓ
Who was that? Not George Ryazanov?
ÒRecent theoretical explorations may suggest another approach
to the
physics of the vacuum. The invisible gravitating dark matter
could be
the other side of the invisible dark energy coin, and that
suggests the
possibility of manipulating the vacuum for energy release.Ó
Now this is my original idea that you got from our
communications over
the past few years. I am the only physicist in the world
today, as far
as I know who has suggested this and has already published it
in my two
books of 2002 so itÕs in the ofÞcial record at the Library
of Congress.
ÒIf a controllable parameter could be found to mediate the
balance
between the invisible dark forces, the result would unleash
the vacuum
energy of creation in all of its awful power and majesty. If
it were
possible to control the dark sides of the force then
spacetime, the
arena where everything we know takes place, could be bent and
twisted
with inÞnitely greater ease than was ever suspected. This
would open
Pandora¹s box to everything from vacuum energy weapons of mass
destruction (capable of destroying the universe!) to
spacetime warp
drives and time machines.Ó
Exactly, the above is the thesis of all my books since 2002
at least.
ÒA quick survey of the international electronic archive of
physics
papers at www.arXiv.org shows that research into the vacuum
of spacetime
for energy production is alive and well.Ó
I do not think that is true. You need to cite speciÞcs here.
There are
lots of þakey new age papers on Òfree energyÓ
on the Web written by
people without any real credentials but they are not on
www.arXiv.org
which is not even allowing competent ÒfringeÓ
papers in controversial
topics like Òcold fusion.Ó So what exactly are
you thinking of here?
Indeed, Paul Ginsparg, who controls the archive, does not
even allow
Carlos Castro to publish conservative competent papers on
ÒClifford
AlgebrasÓ which are not
ÒfringeÓ at all!
ÒMost authors are independent researchers struggling with
limited
funding and resources, yet their theoretical results suggest
that
somewhere in Nick Cook¹s black world, a major breakthrough
has already
taken place. Most likely the United States and Russia are in
the lead,
but China, France, Ukraine, Iran, India and Saudi Arabia all
have
scientists actively pursuing the fundamental physics that
determine the
fabric of our reality, and are seeking the theory and the
means to
access the enormous energies locked inside of the vacuum
since the
creation of the universe. Even if the black budget world has
yet to
unleash the enormous potential of vacuum energy, there are
signs that
those in power may have begun to take notice. Dr. Harold
Puthoff, a
scientist with strong government connections, who has
previously worked
on classiÞed projects for the CIA, is a major proponent of
vacuum
energy physics. Nick Cook¹s book, The Hunt for Zero Point,
and his
recent stories on zero point energy in Jane¹s Defence Weekly
have also
brought attention to the dangers and military potential of
vacuum
research. The American intelligence community Þnanced
so-called psychic
spies for over twenty years and through four presidential
administrations. It is highly unlikely that they would ignore
the
potential of the quantum vacuum. Dr. George Chapline, of the
Lawrence
Livermore National Laboratory, and Dr. Jack Sarfatti in San
Francisco,
knew each other in the sixties in La Jolla, have
independently proposed
that the quantum vacuum may unstable to the formation of
coherent
virtual processes. Sarfatti suggests that gravity is an
emergent
property determined by the physics of the vacuum. His idea is
to Þnd a
means of directly interacting with the physics of the vacuum
that
controls the shape of spacetime. Such a possibility would be
consistent
with the reported success of Evgeny
Podkletnov, the Russian scientist who is experimenting with
spinning
superconducting disks. Podkletnov¹s most recent papers report
the
appearance of a mysterious coherent beam of gravity like
radiation
with a measured force of 1000 G. In an interview on BBC
radio, Nick
Cook pointed out one immediate application of the Podkletnov
beam - the
destruction of missiles and satellites in þight or in orbit
around the
earth. Cook showed the BBC internal documents from Boeing,
the American
aerospace contractor, proving their interest in Podkletnov¹s
research.Ó
This beam stuff I am suspicious of. Of course if the
experiment is good,
I have to think more about it. I am not so sure if
PodkletnovÕs
experiment is any good and has been replicated.
ÒThe connections between Podkletnov¹s results, and the kind
of vacuum
research explored by Sarfatti, beginning in 1999 at the
International
Space Sciences Organization are the latest threads in a trail
that most
likely originates in cold war disinformation, a game played
by East and
West against each other. Glasnost has shifted the balance of
partnerships and the positions of the players, but not the
stakes of an
outcome that would leave the world with even more proliÞc and
powerful
weapons of mass destruction.Ó
That is true, as shown in ÒDestiny Matrix,Ó
however you leave out the
most important evidence -UFOs!
whole business are the connections. Although Nick Cook never
revealed
the identity of his deep throat contact called Dr. Dan
Marckus in
the book The Hunt for Zero Point, there was no question that
the
Podkletnov results had played a major part in Þtting together
the
pieces of the puzzle. The amount of interest was in
Podkletnov¹s
reports by NASA, Boeing, and others in the international
arena of
aerospace and military research communities was evidence that
there was
more here to explore than the latest musings of the
intellectual elite.
The truth is that a fundamental theory of gravity at the
scales of
subatomic nuclear physics does not exist. The fact is that no
one
understands the nature of the gravitational Þeld at very
small scales.
In fact gravity has barely been probed much below one
millimeter.
Every attempt to unify the physical theories of gravity with
the
well-known standard model physics of electromagnetism, and
the strong
and weak nuclear forces, has failed. More importantly there
has been
recent progress in the exotic areas of mainstream research,
such as
superstring theory, which suggest new kinds of physics, which
might
support explanations for Podkletnov¹s impulse gravity effect.
One of
the current fads in theoretical physics involves large extra
dimensions
of space that allow a much stronger version of gravity to
leak off the
membrane world of our ordinary three dimensions. The large
dimensional
picture allows for the well known forces of electromagnetism,
and the
strong and weak nuclear forces, to be conÞned to a three
dimensional
brane-world þoating in a higher dimensional space. Gravitons,
the
are able to slip off of our brane-world, which explains why
the
gravitational force is so much weaker than the other forces
that hold
matter together. Gravitons could be exchanged between our
brane-world
and another brane þoating nearby in the same higher
dimensions.
The Sarfatti picture offers a more direct interaction with
the new
physics than the brane world ideas. Sarfatti¹s vision is to
Þnd a
means of using electromagnetic Þelds in the Josephson effect
to couple
to the virtual electron-positron pair giant coherent
condensate
inþation Þeld inside the vacuum that controls the shape of
spacetime
to the real electron pair giant coherent condensate of a
control high
temperature superconductor. UCBÕs Ray Chiao has a similar
idea using a
superconductor to transduce electromagnetic far Þeld waves to
gravity
waves with high efÞciency conversion. Sarfatti wants to do
the same
thing with non-propagating electromagnetic and gravity near
Þelds. One
wonders if the black budget world may have already produced
some of the
technology needed to explore and test these new realms.Ó
Not a chance. They are clueless about the theory. They are
still stuck
in Hal¹s PV model and Bernie Haisch¹s zero point ideas, which
will
never, in my opinion þy. They are not asking the right
questions and do
not have the right idea in their minds. It is my belief,
until I see
evidence to the contrary, that I am the only physicist on the
planet
today who is doing real theoretical work directly relevant to
the
achievement of practical metric engineering anchored in the
now observed
reality of dark energy. All my work is public. I would love
to be
proved wrong on this especially by Hal Puthoff, but I am not
holding my
breath.;-) Extraordinary claims require extraordinary proof.
Everything
else I have seen is either on the wrong track asking the
wrong questions
like the work of Puthoff, Haisch, Ibison & Rueda for example,
which at
least is real physics that has proved itself wrong in
Ibison¹s PV
cosmology paper, or else the claims are patently obvious
nonsense that
Feynman called Cargo Cult Science. There is also the Russian
torsion
work of Akimov and Shipov and I am not prepared to make a
deÞnitive
statement on that, as the issue is not simple because of
several factors
some political. One must be careful there to separate Shipov¹s
theoretical work from claims made about practical devices
including
weapons applications. I do, however, agree with you that
there is a real
issue here as deÞned in Ch. 9 of Martin Rees¹s Our Final Hour.
Hal Puthoff coined the term metric engineering for þying
saucer
technology. Hal has been working on this problem for many
decades and
has held high USG security clearances and has been privy to
reliable
information that the saucers are real and are alien.
Otherwise he
would not be working on the problem. However, Hal¹s theories,
both of
the zero point energy and of the gravity Þeld will not solve
the
problem because they are too naively based and do not ask the
right
questions. The basic physics required for this task is way
beyond the
depth of PuthoffÕs self-described
ÒengineeringÓ approach and can be
found in Rovelli¹s new book on quantum gravity.
===
Subject: Re: No Set Contains Every Computable Natural
> Russell Easterly says...
>> The usual notion of recursive set is that a set S is
>> recursive as a *subset* of the naturals if there is a
>> Turing machine T such that for any *natural* number n
>> (in unary notation, to be speciÞc)
The set of all natural numbers is not a proper subset of
>the set of all natural numbers.
> But it *is* a subset. A recursive subset.
>> T(n) halts and outputs 1 <-> n is an element of S
>> T(n) halts and outputs 0 <-> n is not an element of S
>> If x is some input that is not a representation of a
natural
>> number, then there is no constraint on what T(n) does.
>
>According to this deÞnition, the set of all natural numbers
>is recursive because there exists a TM that doesn¹t even
>read the input and always outputs a 1.
> Right.
> The set of all natural numbers is recursive by Þat.
No, not by Þat. It is a trivial consequence of a general
deÞnition--
the deÞnition that Daryl McCullough has reproduced for you.
> Why bother to include computability in the deÞnition?
There is no separate deÞnition for the case of all natural
numbers.
There is one deÞnition that applies to all subsets of the set
of natural
numbers. The set of all natural numbers (henceforth N) is a
subset of
N. We apply the deÞntition to N and Þnd out trivially that it
is
recursive.
It is not a trivial deÞnition. It is quite profound. But it is
trivial that N is recursive.
> A human must decide if x is a member of this set
> before x can be given to the TM.
You are seeking to create a new deÞnition and theory of
computability.
Good. You want to consider a larger context of inputs other
than
just natural numbers. For example, you want to include inÞnite
stings as acceptable inputs. Fine. I would agree with you that
if we enlarged the universe of acceptable inputs from just N
to a larger
set containing inÞnite strings, then N would probably not be
recursive in any reasonable new deÞnition of the word. You¹re
right,
a TM would not be able to conclude that an inÞnite string of
1¹s is
not a natural number. I would say that in this enlarged
context, that
N would be recursively enumerable but not recursive.
But your problem is that you are using terms with agreed upon
deÞnitions
by the mathematical community (like recursive) incorrectly.
You have
your own personal deÞnition that you have failed to state.
This is why
mathematicians are arguing with you and pointing out your
obvious
errors. If you would like to re-deÞne recursive by all means
do so.
But make your new deÞnition known. It would also be good to
give
it another name to avoid confusion, like Easterly-Recursive.
Until you do this, you will probably continue to make false
statements
that are easily refuted.
Remember, mathematics is precise.
By the way there are already in existence more generalized
recursion
theories,
which deal with recursion on other domains besides N.
-Leonard Blackburn
> Russell
> - 2 many 2 count
===
Subject: Re: Quadratics and transformation
> It seems to me that I should be able to reasonably easily
get a general
> idea of the type(s) of transformations that have been
applied to x^2
> just by looking at the Þnished quadratic.
> That is, if I have 3x^2 + 7x + 2, can I easily relate the
coefÞcients
> and constant to the dilations, reþections and translations
that this
> function represents compared to x^2 ?
Yes. 3(x^2 + 7/3*x) + 2, complete the square inside the
parentheses. So
you end up with a(x-b)^2 + c. b is translation along the
x-axis, c along
the y-axis, a is the dilation factor. If a is negative,
that¹s a
reþection.
Jon Miller
===
Subject: Re: Understanding taylor expansion for sine
Jonathan Christensen
--
> I¹ve been trying to understand the taylor exapnsion for the
sine
> function,
> sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...
> I¹ve been told that it uses the fact that e^(i*x) = cos(x)
+ i*sin(x),
> and I can easily work out the taylor series for e^x, but
I¹m not sure
> how to use these two pieces to Þnd the expansion for sin(x).
> Any help would be appreciated.
> Jonathan Christensen
---
Outgoing mail is certiÞed Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: e is transcendental (was: classes of
transcendental numbers ?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1IHZOR13942;
>Why do I even try? Are you even TRYING to learn? Or are you
>so closed-minded that you can¹t be bothered?


><snip
>So? The real part of exp(i pi) is cos(pi), and its imaginary
part
>is sin(pi), so all you are saying is that cos(pi) = -1 and
>sin(pi) = 0, and we were already aware of these facts. There
is

>NO reason to conclude that exp(i pi) = 0.
> I,have a reason , with my due respect.
>> Panagiotis Stefanides
Yes, but you DON¹T tell us what your reason is. You can¹t
expect
us to
>accept your claims without giving support for those claims.
So
what
>possible reason could you have for expecting us to agree
with your
claim
>that exp(i pi) = 0?
>The reason is simply that exp(ipi0=-1 should be accompanied
>>by the statement that this is the real part solution.
>>Is it fair?
No, that is not a fair comment. exp(i pi), the complex
number, is
>equal to -1, the complex number. There is no need to appeal
to the
>real part. Also, what does exp(i pi) = -1 is the real part
solution
>mean? You look like you are using terminology in a manner not
>recognized in mathematics.
David McAnally
--------------

>>e^[i*pi] ,as accepted ,is a phasor.
In most of the relevant Þelds of mathematics, e^[i pi] is a
complex
>number.
>It is only fair to state that its
>>polar representation is :
>e^[i*pi] = MOD 1 , ARG 180 .
That is Arg 180 degrees, not just Arg 180. And so what? That
does
>not lead to your assertion that exp[i pi] = 0, a result for
which
>you have given absolutely no support. Why don¹t you just
give up?
David McAnally
At the moment, they (the Time Lords) are far from being
all-powerful.
> That¹s why it¹s been left up to me and me and me.
> quote by: Patrick Troughton in The Three Doctors
-------
>>I, have made myself very explicit.
>>My original question of the implication
>>of the imaginary component: e^[ipi]=j*0 (to the related
proof)
>>Complex Notation, chapter 12 ,Electrical Technology 3RD ED.
>>Edward Hughes Longmans ,page 338:
>>Stares:
>> OA*=OB+jOC=OA(COStheta+jSINtheta)
>>* Symbols representing phasors are printed in bold face
italics, while
those representing only magnitudes are printed in ordinary
italics,..
>>Here is very clear the difference between PHASOR and
MAGNITUDES
>I know the difference between a complex number and its
modulus
>(or magnitude). You don¹t have to explain the difference to
>me.
>>[which(MAGNITUDES) I, referred to as REAL PART or IMMAGINAR
PART,
>The magnitude of a complex number is generally not equal to
either
>its real or imaginary part. How could you claim that it is?
I, gave a reference:
Complex Notation, chapter 12 ,Electrical Technology 3RD ED.
Edward Hughes Longmans ,page 338:
States:
OA*=OB+jOC=OA(COStheta+jSINtheta)
* Symbols representing phasors are printed in bold face
italics, while those
representing only magnitudes are printed in ordinary
italics,..
What are these magnitudes then ?
>> I should have stated COMPONENTS ].
>Perhaps you mean that (the imaginary part of e^[i pi]) = 0,
in which
>case you are correct, but you have had a lot of problem
expressing
>yourself, especially in view of the way that you initially
made
>the claim by stating that e^[i pi] = 0, which you described
as the
>imaginary part solution, using a terminology that nobody but
you
>knows.
>> e[i*theta]=COStheta+iSINtheta
>I know that it is true that exp[i theta] = cos(theta)+i
sin(theta).
>It follows that exp[i pi] = -1. Incidentally, e[i theta] = i
e theta,
>unlike what you have written.
>>thetas could be given and calculations could be performed
>>for numerical evaluations.
>And for other results as well.
>>In books is stated that it is FORMULA
>>and also terms such as evaluate:(-1+i*sqrt[3])^10
>>Are these not solutions to problems?
>No. Evaluate (-1+i sqrt(3))^10 is a problem for which you can
>get a solution using exp(i theta) = cos(theta)+i sin(theta).
This
>does not mean that exp(i theta) = cos(theta)+i sin(theta) is
itself
>a solution. You need a problem before you can describe
anything as
>a solution.
Of course ,this crops up when theta is substituted by a given
angle.
>Nobody knows what you mean when you make a statement like
exp[i pi] =
-1
>is the real part solution of exp[i pi] = -1, or that exp[i
pi] =
0
>is the imaginary part solution of exp[i pi] = -1. I asked
you to
>explain your terminology but you haven¹t bothered.
>>I, doubt it but ,still I, exlpained it anyway.
>I did ask you. And you did not explain your bizarre
terminology.
There
>was no problem, hence there is no solution, whether real
part or
>imaginary part.
How else would do You call them,I, said:
I should have stated COMPONENTS
>As I, exlained earlier, but I, am not the kind of not
>>being bothered
>This was not a quote from me. Why are you pretending that it
was?
You are correct, this was my quote.
>When I take the imaginary part of the equation exp[i pi] =
-1, I get
>sin(pi) = 0, a fact which is already known to us.
>Have you thought also of the fact that if you have exp[i pi]
= -1 (your
>real part solution) and exp[i pi] = 0 (your imaginary part
solution),
>then you could conclude that 0 = exp[i pi] = -1, and get a
contradiction?
>>Well tthere is always the possibility of an answer such as
>>MULTIVALUED , examplum gratias:
>It is known that the exponential function is single-valued
(in fact,
>it is known that the exponential function is entire).
>> e^[2i*pi]=1
>> ln[1]=0=2i*pi
>The logarithm has a branch point at 0. The exponential
function does
>not have a branch point. The analogy is invalid. Because the
>exponential function is single-valued, my objection here
still stands.
>>I, make use of the comlex notation , but still ,I, have my
>>natural questions ,and do not accept everything for granted.
>But you shouldn¹t start making up mathematics to suit
yourself.
This is not my intention,believe me.
>>I, give an example in the form of question:
>>It is required that COS[-i]+i*COS[-i] be evaluated
>>so that is its Modulus and Argument
>>be evaluated (NUMERICALLY].
>cos(-i) = cosh(1), so cos(-i)+i cos(-i) = sqrt(2) cosh(1)
exp(i p/4).
>The magnitude is sqrt(2) cosh(1), which is approximately
2.18225,
>and its argument is pi/4, which is approximately 0.7854.
>Alternatively, the argument is exactly 45 degrees.
I, thank You.
Panagiotis Stefanides
>David McAnally
>--------------
===
Subject: Re: help with solutions for three questions from the
past contest
> 3. A boat with an ill passenger is 7.5 mi north of a
straight
> coastline which runs east and west. A hospital on the coast
is 60
> miles from the point on shore south of the boat. If the
boat starts
> toward shore at 15 mph at the same time an ambulance leaves
the
> hospital at 60 mph and meets the ambulance, what is the
total
distance
> (to the nearest 0.5 mile) traveled by the boatand the
ambulance? ans:
> 62.5
Choogu
> Boat .
> |
> |
> 7.5| z
> |
> Shore |___________________________________.Hospital
> x 60-x
> Hours along distance z at 15 mph must equal hours along
distance
> 60-x at 60 mph.
> And at constant speeds, distance = speed * time.
I¹ll be curious if that¹s the best solution. (It may be the
one expected.)
A better solution would be to minimize the time to the
hospital. That¹s a
standard calculus question (usually phrased in terms of a
swimming pool),
with the additional proviso that if the ambulance wouldn¹t be
to the
meeting
point at the time the boat gets there, then this solution
should be used.
Of course, I wouldn¹t be a proper kibbitzer if I actually
worked out the
details before posting this criticism.
Dare we point out that, if it¹s a true emergency, the coast
guard sends a
helicopter?
Jon Miller
===
Subject: Re: How to choose a matrix P
No D!=E
so P can¹t be an identity matrix
>Now I am thinking of this question.
>Suppose
>Y1=D - P¹EP
>Y2=D-E
>where D,E,P : nxn matrix
>||P||=1
>How to choose P, so that
>rank Y1 >= rank Y2
>and
>||Y1||>= ||Y2||
> The identity matrix would seem to work...
> Maybe you have some more requirements you¹re not telling us.
> Robert Israel israel@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
===
Subject: Teaching philosophy
I need to write a description of my teaching philosophy. In
order to
do so accurately, I think I need to write at greater length
and in
greater detail on this topic than any hiring committee will
want to read.
Moreover, I need to present my views accurately but, somehow,
in such
a manner as not to vitiate the consideration of my
application.
That being the case, can someone please tell me what my
teaching
philosophy is?
Ignorantly,
Allan Adler
ara@zurich.ai.mit.edu
*************************************************************
***************

*
*
* Disclaimer: I am a guest and *not* a member of the MIT
ArtiÞcial
*
* Intelligence Lab. My actions and comments do not reþect
*
* in any way on MIT. Moreover, I am nowhere near the Boston
*
* metropolitan area.
*
*
*
*************************************************************
***************

===
Subject: Re: proving integrability of binomial numbers
Achava Nakhash, the Loving Snake <achava@hotmail.com>
escribi.97:
> Ignacio Larrosa Ca.96estro
<ilarrosaQUITARMAYUSCULAS@mundo-r.com PS - I have never heard
of Pascal¹s triangle referred to as
> Tartaglia¹s triangle. I only know of Tartaglia as a
contributor to
> solving the general cubic. Is this because I mostly have
access to
> American and other English language sources?
Google [Tartaglia triangle] or [Tartaglia triangulo], or of
course,
[Tartaglia triangolo] ...
At secondary school, I always saw Tarataglia¹s triangle
(well, actually
ŒTri.87ngulo de Tartaglia¹).
--
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: question about periodic function
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1IHh7M14607;
>If you have two periodic functions f(t) and g(t) with
periods a and b,
>then the function f(t) + g(t) is periodic with period c
where c is the
>least-common-multiple of a and b. Here the Þrst period is
1/.018=500/9 nd
>the second period is 1/.02=50. So the least-common-multiple
is 500, i.e.
>1/.002. That should be the period of the combined function.
How to prove the theorem? Any hint is welcome. TIA.
===
Subject: Re: strain softening spring
>i.e. the equation I am trying to solve is as follows:
>y¹¹ + cy¹ + [ko*e^(-alpha*t)]y = 0
>Any help would be most appreciated.
>sincerely
>Paul Joseph
>(pjoseph@excite.com)
Maple gives a solution in terms of Bessel functions. I posted
a Þle
called DE.pdf showing the solution at:
http://math.asu.edu/~kurtz/de/
--Lynn
===
Subject: Re: intercept length when a random line intersect
with ellipsoid
shell or parallelepiped shell with Þxed thickness
ChenHS <chenhuisu@hotmail.com> escribi.97:
> If a random line intersects with a ellipsoid shell or
parellelepiped
> shell with a Þxed thickness, how much is the statistical
intercept
> length?
You must deÞne in a precisse way how the random line is
choosed. The
answer
depend on it and it can be very different.
--
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: CAN ANYONE HELP ME?????
> CAN ANYONE HELP ME?????
> The Œcaps lock¹ key is on the left-hand side of the
keyboard, above
> Œshift¹ and below Œtab¹. It should look like this:
> +--------+
> | |<--- | <-- tab key
> | --->| |
> +--------+-+
> | Caps | <-- caps lock key
> | Lock |
> +--------+-+
> | / |
> | || | <-- shift key
> +--------+
> Over on the right-hand side of the keyboard, above the
numeric keypad,
> should be three LEDs, labelled (from left to right) ŒNum
Lock¹, ŒCaps
> Lock¹ and ŒScroll Lock¹
> Press your Œcaps lock¹ key until the LED marked ŒCaps Lock¹
is no longer
> lit.
> Now you are ready to post on usenet.
> 1.DETERMINE DOMAIN OF THE FUNCTION
> f(X)=X3/X2+2X+1
> What happens when x^2 + 2x + 1 = 0?
Excellent! I (along with Baghdad Bob and most readers)
assumed he meant
(x^3/x^2) + 2x + 1 (following the standard rules for order of
operations),
but you ferreted out the probable real meaning, which for OP¹s
clariÞcation
should have been written x^3/(x^2 + 2x + 1).
> FIND EXTREME FUNCTIONS f AND DRAW ITS GRAPH
extreme points? extreme points of the function?
> 2.BY NEWTONS METHOD SOLVE EQUATION:
> X3-X-2=0
Your book or your notes have Newton¹s method (or
Newton-Raphson iteration).
Google also gets about 37,000 hits (only 4,780 for
Newton-Raphson), but
some
of these are advanced treatments. You can still get good
info, but you
have
to look for elementary treatments.
Jon Miller
===
Subject: Re: the anticlassicalist }{ iv: from maps to logic
> -=-=-=-=-= from maps to logic =-=-=-=-=-=-
> Now let us look at interpretting this arrow symbol, such as
in
process,
> categorial inclusion, rule reÞnement, or progressions of
state. We
want
> to be able to evaluate logical relationships between
objects. So we look
to
> deÞne on a recognition map new objects deÞned by any two
objects which
> allows us to collectively compare two objects logically.
The Þrst one
of
> these, which can be called the least upper bound or
disjunction of
two
> states Œa¹ and Œb¹, written (a / b) which obeys the rules
that
> a -> (a / b)
> b -> (a / b)
> forall x in the lattice with a -> x and b -> x, we have (a
/ b) -> x
> where it is important to differentiate when these are
derivationally
found
> on a lattice (objects already deÞned through previous
axioms with the
> appropriate universal property derivable) and when the
existence of new
> objects is being deÞned (for each new deÞnition requires
new consistency
/
> completeness / etc. checks).
You mean it provides a natural framework for conjectures ?
> duality
Doesn¹t tell much about that. Duality is negation in boolean
algebras, isn¹t
it ?
===
Subject: Re: question about periodic function
>>If you have two periodic functions f(t) and g(t) with
periods a and b,
>>then the function f(t) + g(t) is periodic with period c
where c is the
>>least-common-multiple of a and b. Here the Þrst period is
1/.018=500/9
nd
>>the second period is 1/.02=50. So the least-common-multiple
is 500, i.e.
>>1/.002. That should be the period of the combined function.
>How to prove the theorem? Any hint is welcome. TIA.
If f(t) has period a then f(t + na) = f(t) for any integer n,
right?
Same idea for g: g(t + mb) = g(t) for any integer m.
Hmmmm....how could I Þnd an integer that works for both.....?
--Lynn
===
Subject: Re: Teaching philosophy
> I need to write a description of my teaching philosophy. In
order to
> do so accurately, I think I need to write at greater length
and in
> greater detail on this topic than any hiring committee will
want to read.
> Moreover, I need to present my views accurately but,
somehow, in such
> a manner as not to vitiate the consideration of my
application.
> That being the case, can someone please tell me what my
teaching
> philosophy is?
What kind of philosophy? Metaphysics? Epistemology? Ethics?
Aesthetics?
Kantian? Pragmatic? Platonic? Aristotelean?
Bob Kolker
===
Subject: Re: Teaching philosophy
>I need to write a description of my teaching philosophy. In
order to
>do so accurately, I think I need to write at greater length
and in
>greater detail on this topic than any hiring committee will
want to read.
>Moreover, I need to present my views accurately but,
somehow, in such
>a manner as not to vitiate the consideration of my
application.
>That being the case, can someone please tell me what my
teaching
>philosophy is?
>Ignorantly,
>Allan Adler
>ara@zurich.ai.mit.edu
I would be happy to do so, but I only know how to take
multiple choice
tests. What are the choices?
:-)
--Lynn
===
Subject: Re: How many different resistances with n resistors?
Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow)
>> The Puzzle Corner in MIT¹s magazine Technology Review
raised the
>> question of the number of distinct resistances achievable
by connecting
>> ten unit resistors.
>Tony Bartoletti (seq. A048211) considers only cases that can
be reduced by
>applying the formula for parallel/series. If you have 6
resistors there is
1
>conÞguration where this is not possible. I suppose we get an
additional
>resistance value.
What¹s the 6-resistor conÞguration you have in mind?
And has anyone calculated the answer to the problem if we
don¹t assume
series/parallel circuits?
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the
artillery---however great, will
never exceed four of those miles of which as many thousand
separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two
New Sciences
===
Subject: Re: errors in an argument
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1IIGEO17931;
>Should I not be too critical of his wishful thinking that
this
>proof actually means something?
His argument is an overly sophisticated version of one of the
most basic
arguments
against evolution, one that I can¹t rule out. How does super
evolved,
complex, Þne tuned
to the millimeter biological structures appear when their
evolutionary
beneÞt seems to be
absolutely nothing even with 90% or 95% of them installed.
People used to point out that eyes are one of these
mechanisms from the very
beginning,
but that, I think, is a bad example because any ability to
sense this way,
even on a very
primitive level, must be a huge advantage.
My favorite is birds : How can one evolve into þying ? So
many things have
to happen in
order to create functional wings it seems impossible any
awkward animal
developing in
this direction would survive it¹s cost, or any failing þying
attempts. The
evolutionist
arguments sound totally absurd : they ran fast or jumped from
tree to tree
and got an
advantage slowly developing wings. Sounds like absolute
voodoo to me.
===
Subject: Re: the anticlassicalist }{ v: universal truths
> And on these lattices we can make more explicit our
theories of negation.
> On sci.logic, mitch recently gave a very interesting post
on Stone
algebras.
> When we want to model uses of various deÞnitions, we need
to show them
as
> an axiom set. Stone algebras study the axiom set:
> ~(a) / ~(~(a)) <--> T
Btw, can complex numbers model it ? Let¹s see
~ : r exp(i s) --> r/2 exp(i s/2)
a = 1
~(a) = -1/2
~(~(a)) = i/4
let¹s say T is 0...
Well, then I am sure a Lorentz-Moebius transform can save the
situation, I
mean, provide a natural interpretation to the same formula
graphics. The
Moebius group is strictly 3-transitive, and
z / w <---> u
can pass as denoting -the- moebius transform that swaps z and
w while
keeping
u invariant. Now idea if the thing can mimick a lattice any
further,
though.
Isn¹t the symmetric group attached to the boolean algebra in
such a manner
that extremely transitive (but non-symmetric) groups shine as
interesting
analogues ?
===
Subject: Re: . 53ab2750 is an anonymous user .
> AHahahahaha....AHAHAhahahaha...ahahaha....
> ..... you too get three attaboys for your reply !
> Life¹s a bowl of cherries, pits and nuts and all....
ahahahahaha.......ahahahahanson
> Yeah. My niece used to laugh like you ... when she was 2
years old
ehhhhh, hmmmmmm.........ahahahaha......
Your retort has a pH 12+ for some reason. Is it due to an
after effect
from ~ ....after chemo, I, Peter Brown was never the same?
Cheer up, Peter, *laugh*, that¹s all that¹s there, that¹s all
you got left!
Life¹s a bowl of cherries, pits and nuts & all....Do you git
it? git it git
it?
hanson, bonvivant and commentator on all walks of
life...AHAhhahaha..
Ahahahahahaha..........AHAHAHHAHHAAHAHAHHAHa........
ahahahahaha..
===
Subject: Re: Teaching philosophy
>I need to write a description of my teaching philosophy. In
order to
>do so accurately, I think I need to write at greater length
and in
>greater detail on this topic than any hiring committee will
want to read.
>Moreover, I need to present my views accurately but,
somehow, in such
>a manner as not to vitiate the consideration of my
application.
>That being the case, can someone please tell me what my
teaching
>philosophy is?
>Ignorantly,
>Allan Adler
A more serious answer:
I promise to cheerfully accept students who don¹t have the
prerequisites and can¹t add fractions without a calculator or
solve
linear equations. I will have them work in groups so,
hopefully,
someone in each group can do the work. I will be their best
friend so
they don¹t think of me as a teacher. We will have nice
pleasant
classes where we all discover things together so nobody feels
left out
or has their self-esteem bruised. Before each exam I will
give out a
practice exam ****very**** similar to the real exam so they
will know
what to expect. I will be very generous with part credit for
those
that still don¹t get it, even giving re-takes if necessary. I
**guarantee** a very high success rate in my classes. There
will be so
many A¹s and B¹s that I will get awesome student evaluations.
Students
will þock to my classes because I am such a great teacher.
The Dean
will notice the improvement in my department¹s success rate
and will
probably increase it¹s funding. Over time, as my methods
catch on,
even the University administration and maybe even the Regents
will
hold our department up as a shining example of what is being
done to
improve higher education.
--Lynn
===
Subject: Re: HELL :-) (Was: Re: the anticlassicalist }{ ii:
the spectre
continues)
> *MATHEMATICS IS FULL OF TRICKS*
> French even allows you the conjugation :
> ma th.8ematique
> ta th.8ematique
> sa th.8ematique
> notre th.8fme .88 tics
> votre th.8fme attique
> leurre, t¹es mat, hic !
> *REDUCING TOPICS TO THE SCIENCE OF NUMBER IS NUMEROLOGY*
> I¹d call that accounting.
lol
:-)
mitch
===
Subject: Re: Simple numbers
Well what is that algorithm?
> Hallo;
> Well, p(n) is a n-th prime but if n=123456 how do I know
what is the
> 123456-th prime? I need the correct number (for exampel
1-st prime is
->
2)!
>> Hallo,
>> I know that every simple number (beside 2 and 3) has its
one
formula :
>> S.N =
>> 6*n +/- 1 for some n , but not any n={1,2,3,...+} in
formula 6*n
+/- 1
>> gives
>> the simple number (for example: n=6 => 6*6-1=35 , 35 is
not simple
,
>> 5,7|35)
>> By simple number I think you must mean prime.
>> so my question is:
>> If n={1,2,3,...} what is the formula which gives the simple
numbers
>> (beside
>> 2 and 3) for any n?
>> Evidently you don¹t consider p(n) is the n¹th prime to be a
formula.
>> Any
>> particular reason?
>> Computational complexity?
>> Jon Miller
> Given any n, there is a simple algorithm that computes
p(n). You could
spend a
> lifetime trying to make the algorithm more efÞcient, but
the basic
algorithm
> is quite simple.
===
Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational
>
> I will use this to show that
> if cos x = (1 + sqrt(5))/2 then x is transcendental.
(Hence, it is

> irrational.)
>
> But that wasn¹t the original question. The original question
> was whether x / pi is irrational. A whole nother kettle of
Þsh.
>
> To make this question actually look interesting, note that
if
> cos(x)=( (1+sqrt(5)) / 4 ), then x/pi is equal to 1/5.
>
> (Also, the original question has sqrt(5)-1 instead of
sqrt(5)+1 .
> Again, cos(x) = ( (sqrt(5)-1)/4 ) implies x/pi = 2/5 . The
question is

> to show that this 2/5 value turns transcendental when that
Œ4¹ on the
> denominator changes to a Œ2.¹)
>
> J
> So sorry for the bleeps in my non-solution.
> Yes, the original question did involve (sqrt(5)-1)/(2 *p)i
not
> sqrt(5)+1 and Lindemann¹s lemma and
> my proof are only true if x ne 0. Of course e^0 = 1.
> I will go back to the drawing board and see if my original
proof can
> be Þxed.
> Sorry for the whole nother kettle of Þsh!
> Ray Steiner
Well, I Þnally got the solution(hopefully).
I don¹t have time to post all the details right now, but here
is a
brief outline of the steps:
1). Use cos(2*pi/5)= (-1 + sqrt(5))/4 to Þnd all the roots of
unity
in Q(exp(2*pi*i/5)).
(There are 10 of them.)
2). Now consider cos(pi*r)= (-1 + sqrt(5))/2.
Then, by the usual formulas of trigonometry, we get
cos(2*pi*r)= 2-sqrt(5)
and
sin(2*pi*r)= 2*sqrt( -2+sqrt(5)).
If r were rational, cos(2*pi*r) + i*sin(2*pi*r) (*)
would be a root of unity in Q(exp(2*pi*i/5)).
3). Since (*) doesn¹t match any of the answers of step 1, r
cannot be
rational.
Hence, r is irrational.
Hope this does the job!
Ray Steiner
===
Subject: Re: Full Beal Conjecture (revised)
> Sorry for the mistake. My proof can be obtained at
Did you win the dosh?
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: strain softening spring
> I am a Civil engineer who has been working for the last 20
years in
> the Þeld as a result of which my math is somewhat basic. I
have a
> good handle of linear systems, but am taking my Þrst baby
steps
> with non-lineear systems.
> I have read up a few books on non-linear systems looking
for a
> canned solution for a problem I am trying to solve. But
while I see
> a lot of stuff on the DufÞng equation I don¹t see anything
that
> resembles the equation I am trying to solve - a
mass-spring-damper
> system where the spring softens exponentially.
> i.e. the equation I am trying to solve is as follows:
> y¹¹ + cy¹ + [ko*e^(-alpha*t)]y = 0
> Any help would be most appreciated.
Well, if you want an intuitive (albeit approximate) solution,
you should
start by considering the relative size of the parameters.
SpeciÞcally, you
have ordinary damping due to the Œc¹ parameter, and then you
have the
exponential softening, i.e. the Œalpha¹ parameter. These both
have units of
inverse time and you should start by determining which range
of values you
are interested in, e.g. c << alpha and/or c >> alpha. In
either of these
cases, the time scales are very different, and you can
therefore make some
assumptions. For instance, if alpha is very small, then the
softening
happens on a very slow time scale and therefore the spring
constant is
almost, well, constant. You can therefore assume that the
solution is an
ordinary damped oscillation and simply replace k0 in the
solution with k0 *
e^(-alpha*t).
Another line of approach is to try to eliminate the damping
by substituting
y(t) = z(t) * e^(-beta*t) for some suitable chosen value of
beta.
A third idea is to make a vector plot. Introduce z=y¹ and for
each point
(y,z) draw a vector whose direction is (y¹,z¹). Try it Þrst
with alpha =
0,
because then the vectors are independent of time. You will
see spirals that
lead to the point (0,0). When alpha is nonzero, the vector
plot is
time-dependent, but you might be able to say something
qualitative about
them anyway, e.g. whether they spiral faster or slower
towards (0,0).
-Michael.
===
Subject: A newbie¹s question -- about real number
Two real numbers r1 = 0.89, r2 = 0.889999999..... (with
inÞnite 9s),
===
Subject: Re: A newbie¹s question -- about real number
lisong@iastate.edu asks this about real numbers:
>Two real numbers r1 = 0.89, r2 = 0.889999999..... (with
inÞnite 9s),
Theoretically NO.
In real life measurement or result calculations in which only
two
signiÞcant
Þgures are reasonable, then r1=r2. The context must be
appropriate.
G C
===
Subject: DeÞnition of Separable Space (basic topology
question)
Let A be a subset of (X,T). Then A is dense in X iff for
every non-empty
open subset U of X, A / U != {}.
I have seen two deÞnitions of Œseparable topological space¹:
a) (X,T) is separable if there exists A <subset> X, where A
is countable
and
dense in X
b) (X,T) is separable if there exists A <proper subset> X,
where A is
countable and dense in X
Given def (a), its simple to prove that all countable spaces
X are
automatically separable since the subset X is countable and
non-trivially
intersects every non-empty open subset. However, this won¹t
work given def
(b). I Þrst became suspicious when I saw a proof that all
countable
spaces
were separable that seemed ridiculously complex in comparison
to the
seemingly obvious 1 step proof above. I later found
deÞnitions of
separable like def (b).
l8r, Mike N. Christoff
===
Subject: Re: A newbie¹s question -- about real number
Adjunct Assistant Professor at the University of Montana.
>Two real numbers r1 = 0.89, r2 = 0.889999999..... (with
inÞnite 9s),
Both expressions are taken to represent the same real number.
The
Þrst is the real number 8/10 + 9/100.
The second is taken to represent the limit of the inÞnity
series
8/10 + 8/100 + (9/10^3 + 9/10^4 + 9/10^5 + ....)
Since Sum_{i=3}^{inÞnity} 9/10^i =
9/10^3*(Sum_{i=0}^{inÞnity} 1/10^i),
and the series (1/1 + 1/10 + 1/100 + 1/1000 + ...) converges
to
1/(1-(1/10)) = 1/(9/10) = 10/9, we have that the original
inÞnite
series
8/10 + 8/100 + (9/10^3 + 9/10^4 + 9/10^5 + ....)
converges to
8/10 + 8/100 + 9/10^3*(10/9) = 8/10 + 8/100 + 1/100 = 8/10 +
9/100,
which is equal to r1.
--
It¹s not denial. I¹m just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Cauchy¹s Interlacing Theorem
Hi folks,
I¹m looking for a proof of the following...
Let A = [a0 a^h ; a A2] >= 0 be a NxN hermitian matrix
where a0 is a scalar > 0 and A2 is the lower-right N-1xN-1
(principal) submatrix of A.
According to Cauchy¹s interlacing theorem (1829 I guess),
the N Eigenvalues gamma_i of A and the N-1 Eigenvalues alpha_i
of A2 interlace, i.e., gamma_1 >= alpha_1 >= gamma_2 >= ...
Interestingly, having played a little bit with MATLAB,
also the Eigenvalues gamma_i of A and the Eigenvalues beta_i
of the N-1xN-1 matrix A2-a*a^h/a0 seem to interlace. However,
this seems to be true only for nonnegative-deÞnite A.
Does anyone have a proof, or reference to a proof for that??
Initially, I wanted to show that if A >= 0 with rank A >= 2,
then also A2-a*a^h/a0 >= 0; but, then I discovered Cauchy¹s
theorem... ;)
(If rank A = 1, then A2-a*a^h/a0 yields the N-1xN-1 zero
matrix,
and if rank A >= 2, then rank A2-a*a^h/a0 >= 1, at least this
is easily seen).
Once the above assumption had been proven, A2-a*a^h/a0 >= 0
would directly follow as beta_N-1 >= gamma_N >= 0
(comparing the smallest Eigenvalues)
Christian.
===
Subject: Re: Collatz Conjecture : Symmetry question.
> Every time I post on this problem I get beat up so please
be kind.
I hope that wasn¹t directed at me. I never beat anyone. It
may seem
that way to Ernst Berg, but I was just trying to help him by
offering
constructive criticism. Apparently some people can¹t get past
the
criticism part and get all bent out of shape to the point
where they
ignore the constructive part. But that¹s Ernst¹s problem.
> I have found many properties in the landscape of this
problem and the
> one that has most of my interest the gross amount of
non-regularity
> tightly coupled to large a domain of symmetry. I have
noticed 2 forms
> of symmetry in this problem which have me thoroughly
intrigued, but do
> not know where to go with these notions at this point.
> Symmetry 1;
> I have noticed that given any tree pattern of any size, it
is exactly
> duplicated inÞnitely many times. For example, take the
entire tree
> starting at A and extend out all branches to a depth of any
N. Then
> that pattern taken generically is repeated at an offset of
A +
> n*(constant value) with (n = 1, 2, 3, ... inÞnity), with
the constant
> value is a function of N.
That seems reasonable since any sequence vector evaluates to
a simple
formula
a¹ = (X*a - Z)/Y
which means there are always an inÞnte set of integer
solutions to every
sequence vector.
> Symmetry 2;
> I have also noticed that all values at N (here I include all
> rationales and integers by taking all possible paths from A
to N,
> keeping NON_INTEGER branches in this process) produces N¹
sets of a
> collective 2^(N-1) number of values where the number of
items in each
> set follows the Binomial CoefÞcients or Pascal¹s Triangle
(which ever
> you normally call it). If you take each of these N¹ sets of
values and
> calculate the ordered set difference (N¹¹) that the
difference sets
> (N¹¹) are constant and equal for all N independent of A.
I¹m not sure I follow this, but since there are only two
operations (3x+1
and x/2) every number has two possible ancestors, so carried
to N levels
(and ignoring whether the ancestors are integers) there would
certainly
be 2^N numbers at distance N.
> This means that using only the N¹¹ sets, all sets at N in
groups of
> N¹, thus all values at N, for any A, can be calculated
without
> extending the tree from A to N through all the intermediate
N¹s.
> In other words, all values of absolute stopping distance of
N, for any
> A, can be calculated directly with out traversing the tree
to Þnd
> each of those values.
But there are vastly more total numbers at a given level than
there are
integers, so although this is true, I¹m not sure it leads
anywhere.
For example, I recently found all the integers that are at
level 84
(from 1):
179,441,377
If I included all the potential non-integer ancestors, I
would have had to
sift through 2^84 or
19,342,813,113,834,066,795,298,816
numbers to Þnd which ones are integers. In this case, it is
much quicker to

traverse the tree from level 1 up to level 84. I have a
program that can
build each level from the previous one. Since it only
generates integer
results, it is much more efÞcient than testing each binary
permutaion
of the two operations (provided you record the integers you
collected from
the previous level).
It would be nice to be able to just start at an arbitrary
level and Þnd
all the integers. So although it can be done in theory, it
can¹t be done
in practice for high levels.
Unless you have some means of determining which of the 2^84
numbers at
level
84 are integers without having to iterate through them all.
> Is this interesting to anyone?
Yes.
> Or better yet, Is this anything of interest for this
problem.
I¹m not an expert, so I can¹t say.
===
Subject: Re: A newbie¹s question -- about real number
>lisong@iastate.edu asks this about real numbers:
>>Two real numbers r1 = 0.89, r2 = 0.889999999..... (with
inÞnite 9s),
>Theoretically NO.
Hence the rationals are uncountable!
Rich
===
Subject: Graph Theory Textbook
I¹m an undergraduate, and this summer I will be participating
in an
REU for discrete math and combinatorics. I am looking for a
good
graph theory textbook to learn the basics from. I have seen
the books
that Dover (which of course are very cheap) has to offer on
the
subject, but unfortunately they seem too elementary. Is the
Springer
GTM Graph Theory by Reinhard Diestel any good? Any
suggestions are
highly welcomed.
Jack
===
Subject: Re: errors in an argument
> My favorite is birds : How can one evolve into þying ? So
many things
> have to happen in order to create functional wings it seems
impossible
any
> awkward animal developing in this direction would survive
it¹s cost, or
> any failing þying attempts. The evolutionist arguments
sound totally
> absurd : they ran fast or jumped from tree to tree and got
an advantage
> slowly developing wings. Sounds like absolute voodoo to me.
That depends on the size of the animal. If the animal is
small enough,
the terminal velocity can result in a non-lethal impact with
the
ground. (I believe the threshold is around the size of a
mouse or a
cat.) Incremental levels of control over the landing point
then have an
obvious beneÞt.
--
Daniel W. Johnson
panoptes@iquest.net
http://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
===
Subject: Re: 3 Squares Covering 1 Circle
> For b, what does a C3v (invariant under 120 deg rotation)
> symmetric arrangement yield? (Place the outermost edge of
> the squares tangent to the circle.)
>
> Here¹s a similar situation that does better than the 80.4%
> with part (a). Use this 120 deg symmetry, but place the
diagonals
> of the square along the radii. The optimum placement of the
> squares (they can be shifted along the radii) covers 83.0%
of
> the circle¹s area. This is the best that I¹ve found, but I
don¹t
> know if it¹s the best possible.
> To be a little more precise, the area is 2.60808705... I
have an exact
> form, but it¹s pretty messy. This happens when the diagonal
of each
> square extends .13980931 past the center of the circle
(along the
> diameter).
Oh, that¹s a nice conÞguration to consider!
I checked your result, and I agree with your area Ac =
2.60808705,
although the extension of the square beyond the center of the
circle is
a little off: it should be rc = 0.13977317. (I would guess
that you
obtained the value of Ac to single precision, which yields
only half the
precision for rc).
For general r (distance diameter extends beyond the circle¹s
center),
the area A of the resulting Þgure can be found as follows:
Let u = sqrt(2) - r, v = sqrt(2 - u^2), d = (u - v)/2, and
p = (1 + u v)/2, then
A = 3 (p - sqrt(p) d - (3 + sqrt(3)) r^2/2 + Arcsin(d)).
This area is maximized when r equals the critical value rc:
rc = (19 - 5 sqrt(3) - sqrt(176 - 86 sqrt(3))) / (26 sqrt(2)).
The formula for Ac = A(rc) simpliÞes greatly (as one would
expect).
Let dc = rc (3 + sqrt(3))/2. Then
Ac = 3 (1 - sqrt(2) dc + Arcsin(dc)).
To summarize the results so far, the best result posted for
case (a)
(no overlap) is
(a) pi/6 - 1/2 + sqrt(3)/4 + sqrt(55)/8 + Arccos(3/8) =
2.57003584 = 81.8067816% of the circle,
and the best result posted for case (b) (overlap allowed) is
(b) Ac (see above) =
2.60808705 = 83.0179893% of the circle.
-Jim Ferry
===
Subject: Re: A newbie¹s question -- about real number
> Hence the rationals are uncountable!
Nice try, but no cigar. The set of decimal sequences that
terminate in all
0¹s or all 9¹s is countable.
===
Subject: Re: A newbie¹s question -- about real number
> Theoretically NO.
> In real life measurement or result calculations in which
only two
signiÞcant
> Þgures are reasonable, then r1=r2. The context must be
appropriate.
Theoretically YES, and given that we discuss mathematics
here, that is the
only meaningful answer.
===
Subject: Re: DeÞnition of Separable Space (basic topology
question)
> Let A be a subset of (X,T). Then A is dense in X iff for
every non-empty
> open subset U of X, A / U != {}.
> I have seen two deÞnitions of Œseparable topological space¹:
> a) (X,T) is separable if there exists A <subset> X, where A
is countable
and
> dense in X
> b) (X,T) is separable if there exists A <proper subset> X,
where A is
> countable and dense in X
> Given def (a), its simple to prove that all countable
spaces X are
> automatically separable since the subset X is countable and
non-trivially
> intersects every non-empty open subset. However, this won¹t
work given
def
> (b). I Þrst became suspicious when I saw a proof that all
countable
spaces
> were separable that seemed ridiculously complex in
comparison to the
> seemingly obvious 1 step proof above. I later found
deÞnitions of
> separable like def (b).
> l8r, Mike N. Christoff
I have never seen deÞnition (b). I have seen the symbol
$subset$
used to mean subset, not proper subset, so maybe that is what
you saw.
Certainly a one-point space with the discrete topology is to
be
considered separable, even though it has no dense proper
subsets.
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: How many different resistances with n resistors?
>> The Puzzle Corner in MIT¹s magazine Technology Review
raised
the
>> question of the number of distinct resistances achievable
by
connecting
>> ten unit resistors.
>Tony Bartoletti (seq. A048211) considers only cases that can
be reduced
by
>applying the formula for parallel/series. If you have 6
resistors there
is 1
>conÞguration where this is not possible. I suppose we get an
additional
>resistance value.
> What¹s the 6-resistor conÞguration you have in mind?
An H with asymmetric legs. I thought that 6 is the smallest
number with
this
property, but obviously a H with 5 resistors gives us r=1
which seems
impossible
with other combinations of 5 resistors. I did not yet check,
whether the 6-H
has
a resistance not occuring among the (other) 53 possibilities.
I will play
with
this a little tomorrow. I think we would Þnd a reference in
encyclopedia
if
someone had solved the general case.
Klaus
> And has anyone calculated the answer to the problem if we
don¹t assume
> series/parallel circuits?
> --
> Tim Chow tchow-at-alum-dot-mit-dot-edu
> The range of our projectiles---even ... the
artillery---however great,
will
> never exceed four of those miles of which as many thousand
separate us
from
> the center of the earth. ---Galileo, Dialogues Concerning
Two New
Sciences
===
Subject: Re: Help needed - primes program
Well how about this:
-------------------------------------------------------------
---------------

---
Program PB;
Var
x,y:word;
f:word;
i,n:word;
ok:boolean;
Begin
Write(Enter X and Y -> Œ);
Readln(x,y);
For i:=x to y do
Begin
f:=3;
ok:=true;
If (i=1) or (i mod 2 = 0) then
Begin
f:=i+1;
ok:=false;
End;
While f<=sqrt(i) do
Begin
If i mod f = 0 then
Begin
f:=i;
ok:=false;
End
else
f:=f+2;
End;
If ok=true then Write(i,¹ is a prime. Œ);
End;
Readln;
End.
-------------------------------------------------------------
---------------

---
It ignores all number with form k*2 , k{2,3,4,...} and it
goes till sqrt(i)
insted of i.
Is this faste or slower than sieve of eEatosthenes? Why?
===
Subject: Re: A newbie¹s question -- about real number
> Two real numbers r1 = 0.89, r2 = 0.889999999..... (with
inÞnite 9s),
> can we say r1 = r2 ?
Yes. The problem is that we present inÞnite decimal
expansions in school
before their meaning can be properly understood. Once this
meaning is made
clear, proving r1 = r2 is a very simple exercise.
===
Subject: Re: Teaching philosophy
> I promise to cheerfully accept students who don¹t have the
> prerequisites and can¹t add fractions without a calculator
or solve
> linear equations. I will have them work in groups so,
hopefully,
> someone in each group can do the work. I will be their best
friend so
> they don¹t think of me as a teacher. We will have nice
pleasant
> classes where we all discover things together so nobody
feels left out
> or has their self-esteem bruised. Before each exam I will
give out a
> practice exam ****very**** similar to the real exam so they
will know
> what to expect. I will be very generous with part credit
for those
> that still don¹t get it, even giving re-takes if necessary.
I
> **guarantee** a very high success rate in my classes. There
will be so
> many A¹s and B¹s that I will get awesome student
evaluations. Students
> will þock to my classes because I am such a great teacher.
The Dean
> will notice the improvement in my department¹s success rate
and will
> probably increase it¹s funding. Over time, as my methods
catch on,
> even the University administration and maybe even the
Regents will
> hold our department up as a shining example of what is
being done to
> improve higher education.
Hah, pretty good!
===
Subject: Re: Math Too Advanced For Mainstream Economists
Robert Vienneau <rvien@see.sig.com>
threatening to hold his breath until he dies once again,
> The pitiful child who posts below is teaches economics at a
leading
> second-tier department of economics in the U.S.A. Hard to
believe.
Maybe you can get your big brother to beat him up.
===
Subject: Re: help with solutions for three questions from the
past contest
> 3. A boat with an ill passenger is 7.5 mi north of a
straight
> coastline which runs east and west. A hospital on the coast
is 60
> miles from the point on shore south of the boat. If the
boat starts
> toward shore at 15 mph at the same time an ambulance leaves
the
> hospital at 60 mph and meets the ambulance, what is the
total
distance
> (to the nearest 0.5 mile) traveled by the boatand the
ambulance?
ans:
> 62.5
Choogu
> Boat .
> |
> |
> 7.5| z
> |
> Shore |___________________________________.Hospital
> x 60-x
> Hours along distance z at 15 mph must equal hours along
distance
> 60-x at 60 mph.
> And at constant speeds, distance = speed * time.
> I¹ll be curious if that¹s the best solution. (It may be the
one
expected.)
> A better solution would be to minimize the time to the
hospital.
This does minimize the time to the hospital!
===
Subject: Re: question about periodic function
>If you have two periodic functions f(t) and g(t) with
periods a and b,
>then the function f(t) + g(t) is periodic with period c
where c is the
>least-common-multiple of a and b. Here the Þrst period is
1/.018=500/9
nd
>the second period is 1/.02=50. So the least-common-multiple
is 500, i.e.
>1/.002. That should be the period of the combined function.
> How to prove the theorem? Any hint is welcome. TIA.
The sum will only be periodic if the ratio of periods is
rational.
If that ratio is irrational f(t) + g(t) is not periodic.
===
Subject: Re: Simple numbers
> Well what is that algorithm?
It¹s called the Sieve of Eratosthenes.
--
Dave Seaman
Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Anyone knows a formula for this? (combinatorics)
Hi
Let n be the number of cards drawn from an ordinary deck (52
cards).
For example according to the book, if n=5
i.e if we pick 5 cards randomly from a deck of 52 cards
The probability of getting
a) A full house is:
[C(4,2)*C(4,3)*2*C(13,2)]/C(52,5)
b) Exactly two pairs is:
[C(4,2)*C(4,2)*C(44,1)*C(13,2)]/C(52,5)
So i was wondering if there¹s a general result (a formula)?
for a) and b)
where n is the number of cards drawn.
===
Subject: Re: errors in an argument
> If I were trying to explain the error to someone, I would
probably
> use the analogy of rolling a 20 sided dice 100 times and
recording
> the result. By the proof above, one can¹t do this, since the
> probability of any particular sequence is (essentially) 0.
What about the probability of any particular sperm actually
being the one
to fertilize an egg? That would seem to preclude the
possibility of sexual
reproduction altogether.
--
Joe Bramblett, KD5NRH
===
Subject: Re: I¹d like to join this group!
>> Does the Equation below really prove that 2=1 ?? Y N
>> 1) X=Y ; Given
>> 2) X^2=XY ; Multiply both sides by X
>> 3) X^2-Y^2=XY-Y^2 ; Subtract Y^2 from both sides
>> 4) (X+Y)(X-Y)=Y(X-Y) ; Factor
>> 5) X+Y=Y ; Cancel out (X-Y) term
>> 6) 2Y=Y ; Substitute X for Y, by equation 1
>> 7) 2=1 ; Divide both sides by Y
> Go back to step 4. What it really proves is 2x0=1x0.
> Now try this:
> 1) I am nothing without her. (Shakespeare)
> 2) ME - SHE = 0 (Algebraic representation)
> 3) ME = SHE (Added SHE to both sides)
> 4) M = SH (Divided by E)
> 5) MIT = SHIT (Multiplied by IT)
> So there you have the mathematical proof of what Shakespeare
> thought of the future American techincal education. :-)
You Guys are too sharp.
That was actually printed in OMNI magazine Nov 79 , as proof
that 2=1 . ;-)
===
Subject: Re: Dik Winter¹s claims revisited, dependency issue
> James Harris:
> |> They¹ve gone so far that Keith Ramsay even claimed to
have posted a
> |> solution for the w¹s, but how did he pick w¹s from
inÞnity?
> I¹m not sure what you mean here by pick w¹s from inÞnity.
> | He didn¹t. There is a well-deÞned algorithm for computing
> |the w¹s, starting with the number (1 + sqrt(-167))/2. If
you
> |had replaced 7 with 17, the starting number would have been
> |(-1 + sqrt(-407))/2 and the result would be different.
> Let¹s recall what I posted.
> |It appears that Q(sqrt(-167)) has a class group of order
11.
> |Let r stand for (1+sqrt(-167))/2 (which is an algebraic
integer,
> |in spite of the 2 in the denominator).
> |r^11 = (-592764018-86559857*r)
> | = (44555-222*r) (-12882-2017*r)
> |and
> |7^11 = (44555-222*r)(44555+222*r).
> As an intermediate step, I found the GCD of
((1+sqrt(-167))/2)^11
> and 7^11, which is an 11-th power of the GCD of
(1+sqrt(-167))/2
> and 7.
> Harris to check the result using a calculator or something
like that.
> |Also (-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11,
which is
> |relatively prime to 7^11.
> |The factor 44555-222*r = 44444 - 111 sqrt(-167) and its
conjugate
> |44444 + 111 sqrt(-167) are roots of x^2-88888*x+7^11=0.
One GCD of
> |r with 7 is then (44444-111 sqrt(-167))^(1/11), which is a
root of
> |x^22 - 88888 * x^11 + 7^11 = 0.
All those numbers are hard to parse through, so I just went
back to
considering that
a_1(x) + a_2(x) = (x-1)
so assuming factors f_1(x), and f_2(x), with your
w_1(2) = (44444-111 sqrt(-167))^(1/11), and
w_2(x) = (44444+111 sqrt(-167))^(1/11),
I have w_1(2) f_1(2) + w_2(2) f_2(2) = 1
and assuming the w¹s are symmetric by sign, I have that
f_1(2) must
equal f_2(2).
But f_1(2)f_2(2) = 6
as the problem I see is that you¹re stuck, if the w¹s are
symmetric by
sign, so are you saying they are not?
James Harris
===
Subject: Re: How many ways to put 5 balls into 500 ordered
cups?
> You are given 500 numbered cups and Þve identical balls.
Any cup can
> hold up to Þve balls. How many ways can you put the Þve
balls into
> the 500 cups?
Three; you can use your left hand, your right hand, or both.
Of course,
you could use some other sort of mechanical aid, but that
would be a
question for the engineering groups.
--
Joe Bramblett, KD5NRH
===
Subject: Re: A newbie¹s question -- about real number
>> Hence the rationals are uncountable!
>Nice try, but no cigar. The set of decimal sequences that
terminate in all

>0¹s or all 9¹s is countable.
No cigar? Bummer. But I *can* Þnd an uncountable number of
non-terminating
representations of 89/100 and Þgured that since r1<>r2 the
conclusion
followed. If that is not true then clearly 89/100 is
irrational! Even
better.
Rich
===
Subject: Re: No Set Contains Every Computable Natural
> Russell Easterly says...
> The usual notion of recursive set is that a set S is
>> recursive as a *subset* of the naturals if there is a
>> Turing machine T such that for any *natural* number n
>> (in unary notation, to be speciÞc)
The set of all natural numbers is not a proper subset of
>the set of all natural numbers.
But it *is* a subset. A recursive subset.
> T(n) halts and outputs 1 <-> n is an element of S
>> T(n) halts and outputs 0 <-> n is not an element of S
>> If x is some input that is not a representation of a
natural
>> number, then there is no constraint on what T(n) does.
>
>According to this deÞnition, the set of all natural numbers
>is recursive because there exists a TM that doesn¹t even
>read the input and always outputs a 1.
Right.
> The set of all natural numbers is recursive by Þat.
> No, not by Þat. It is a trivial consequence of a general
deÞnition--
> the deÞnition that Daryl McCullough has reproduced for you.
Using Daryl McCullough¹s deÞnition, I can claim the set of
all natural numbers is (1,2,3). Since the string 1111 does
not represent a natural number in my system, it is too large
to be the input to a Turing machine.
> Why bother to include computability in the deÞnition?
> There is no separate deÞnition for the case of all natural
numbers.
> There is one deÞnition that applies to all subsets of the
set of natural
> numbers. The set of all natural numbers (henceforth N) is a
subset of
> N. We apply the deÞntition to N and Þnd out trivially that
it is
> recursive.
> It is not a trivial deÞnition. It is quite profound. But it
is
> trivial that N is recursive.
I Þnd the idea that N=(1,2,3) quite profound.
> A human must decide if x is a member of this set
> before x can be given to the TM.
> You are seeking to create a new deÞnition and theory of
computability.
> Good. You want to consider a larger context of inputs other
than
> just natural numbers. For example, you want to include
inÞnite
> stings as acceptable inputs.
Who decides the string is Þnite?
A TM will always think the input is Þnite.
I have given proofs of this.
> Fine. I would agree with you that
> if we enlarged the universe of acceptable inputs from just
N to a larger
> set containing inÞnite strings, then N would probably not be
> recursive in any reasonable new deÞnition of the word.
You¹re right,
> a TM would not be able to conclude that an inÞnite string
of 1¹s is
> not a natural number.
A TM will say that any string of 1¹s represents a natural
number.
> I would say that in this enlarged context, that
> N would be recursively enumerable but not recursive.
> But your problem is that you are using terms with agreed
upon deÞnitions
> by the mathematical community (like recursive) incorrectly.
You have
> your own personal deÞnition that you have failed to state.
I used a standard deÞnition of decidable language.
The standard deÞnition states that a human must decide
that a string is Þnite before giving it to the TM.
Ultimately, a human being must decide if a string, x,
represents a natural number.
Why pretend there is an algorithm that can do this?
> This is why
> mathematicians are arguing with you and pointing out your
obvious
> errors. If you would like to re-deÞne recursive by all
means do so.
I am not redeÞning recursive.
I am pointing out that if the deÞnition of decidable languange
says there is a TM that can decide if string, x, is a member
of the language, then the language of all unary
representations
of the natural numbers is NOT a decidable language.
If we include the requirement that a human being decides
whether x is a Þnite string, then I agree N is recursive.
This additional requirement allows me to choose N to be
any set I want it to be.
> But make your new deÞnition known. It would also be good to
give
> it another name to avoid confusion, like Easterly-Recursive.
> Until you do this, you will probably continue to make false
statements
> that are easily refuted.
Are you refuting my claim that N=(1,2,3)?
> Remember, mathematics is precise.
N is precisely equal to (1,2,3).
> By the way there are already in existence more generalized
recursion
theories,
> which deal with recursion on other domains besides N.
Can you give references?
Obviously, I need to learn more about this stuff.
Russell
- 2 many 2 count
===
Subject: Re: Anyone knows a formula for this? (combinatorics)
Adjunct Assistant Professor at the University of Montana.
>Let n be the number of cards drawn from an ordinary deck (52
cards).
>For example according to the book, if n=5
>i.e if we pick 5 cards randomly from a deck of 52 cards
>The probability of getting
>a) A full house is:
>[C(4,2)*C(4,3)*2*C(13,2)]/C(52,5)
>b) Exactly two pairs is:
>[C(4,2)*C(4,2)*C(44,1)*C(13,2)]/C(52,5)
>So i was wondering if there¹s a general result (a formula)?
for a) and b)
>where n is the number of cards drawn.
You should try to understand what the numbers mean.
For example, with full house, you have C(52,5): that¹s the
number of
different ways in which you can randomly pick 5 cards from a
deck of
52 cards. It is the total space. If you are randomly pickin k
cards
from a deck of 52, then that number should obviously be
replaced by
C(52,k).
What do the other numbers mean? I.e., C(4,2)*C(4,3)*2*C(13,2)
should be the number of different 5 card hands which are a
full
house. How is this accomplished? Why does this count that?
Well, a full house is determined by stating the following
things:
(1) The rank of the 3-of-a-kind (i.e., is it 3 aces, 3 jacks,
3
tens, etc); and
(2) The rank of the pair; and
(3) The suits involved in the 3-of-a-kind; and
(4) The suits involved in the pair.
That completely determines the hand, since there are no cards
leftover
in a 5-card hand.
Now, C(4,2) means how many ways