mm-457
Subject: Re: What constitutes an implicit use of a function?
(Was: a confusing group theory conundrum)[cu]> It is largely a
matter of notation. Suppose mathematical notation> were defined
in such a way that, when it was clear by context, we> could use
the phrase its' square instead of x^2, as, for instance:> Let x
be a real greater than 1. Then x < its' square> We are not used
to such weird notation. But in more down-toarth> notation, we
say x < x^2, and I believe you will agree that we are> thus
implicitly using the function f(x)=x^2.I don't agree that we
are implicitly using the function f(x) = x^2.The x^2 in x <
x^2 is an element (I assume) in the natural numbers.I am not
sure what of many ways you could be interpreting x^2
asimplictly using the function f(x) = x^2.Consider the
situation where I define the function f to bethe subset of NxN
such that f = { (a,b) in NxN | a in N and b = a*a}.That last
a*a cannot be implicitly using the function fsince I am
defining it (ie f). Otherwise, the definition of f wouldbe
circular.Just because you could replace my use of x^2 by
referring tothe explicit use of the funciton f(x) = x^2 does
not mean thatyou must be doing it implicitly. Sometimes there
is an advantageto putting a particular situation into a more
general context,but that does not mean you were implicitly
invoking that moregeneral context.I am all for casting
mathematical concepts into set theory terms,like functions
being viewed as subset of cartesian product ofdomain and
range. But, it appears that you are trying to castevery
mathematical statement into terms of functions. Aspointed out
by other posters, category theory provides a generalcontext
where ideas viewed as distinct become to be viewed
asparticular cases of a more general catogorical
situation.This unifying viewpoint is more important than just
beingable to say that you have a functor from the class of all
groupsto its identity element.-- Bill Hale
===
Subject: Re: What
constitutes an implicit use of a function? (Was: a confusing
group theory conundrum) Adjunct Assistant Professor at the
University of Montana.>In my previous thread, there seemed to
be a general lack of consensus>with regard to whether
algebraists implicitly invoke a functor when>they say If G is
a group, let G_e be its identityNo, there was no lack of
consensus. Most of us simply did not addressthe claim and
simply went on to say how one could define such a'function'.
But the fact is, there is no need to have a functionwhose
domain is all groups before you can start talking about
theidentity of a given group.There are many ways of describing
a group; it can be a set G with abinary operation * which
satisfies certain axioms, in which case thefact that G
satisfies the axioms automatically gives you anidentity. A
more obvious way is to say it is a set G, with
threeoperations; a binary operation *, a unary operation
^{-1}, and anullary operation e, which satisfies certain
identities. The groupitself already comes equipped with a way
of identitfying its identity:the nullary operation e.In fact,
that is all you need: for every group G, a nullary
operationG->G whose image is the identity. Then you are not
invoking a generalfunction whose arguments are any group, and
whose response is theidentity of the group. You are invoking a
very specific function,associated to the group G itself, which
gives you the identity of G.The existence of such a function
is guaranteed by the very propertiesthat define a group.Or
there are other ways in which one can define a function that
hasthe properties you described. Though such a function is not
NEEDED todo group theory.>More generally, if something is
defined in such a way that it must>possess something else (as,
for instance, a group having to have an>identity, by the very
definition of a group), it seems to me that that>does not mean
we are no longer using a hidden function/functor when we>call
up that something else.Nor does it mean we ->ARE<-. And
certainly, it does ->not<- mean weMUST have a function whose
domain is all such somethings and whosecodomain is all things
that they possess. All you need is for eachsomething G, a
function whose domain is G^{0} = {emptyset}, and whosecodomain
is the singleton consisting of the something else
itpossesses.In the example you mentioned, given a group G, let
e be itsidentity. You do not need to invoke a function defined
for ALL groups:you may simply invoke a function whose domain
is {emptyset} and whosecodomain is {e}. You don't care about
any other groups besides the onethat show up in your
argument.
===
==Arturo
Magidinmagidin@math.berkeley.edu
===
Subject: Re: What
constitutes an implicit use of a function? (Was: a confusing
group theory conundrum) permission for an emailed
response.X-Tom-Swiftie: I just sharpened my pencil, Tom said
pointedly> In my previous thread, there seemed to be a general
lack of consensus> with regard to whether algebraists
implicitly invoke a functor when> they say If G is a group,
let G_e be its identityThere was no lack of consensus.
Everyone agreed that algebraists donot need such a function,
nor do they invoke it.The discussion was about the different
question of whether such afunction is possible, which is a
question of which particular settheory you are using, and
whether your set theory allows functionswith domains or ranges
that are proper classes.> How, then, is it any different to
say,> Let G be a group. Then G's identity is in G> Here the
culprit is the phrase G's identity. The way I see it, this> is
the same as using a functor implicitly, just like its' [x's]>
square.The phrase Let G be a group is really an abbreviation
for:Let G = <g, *, e> be a group. (Where you've decided that
the orderis set, operation, identity.
===
Subject: Re:
Factorization dispute, again> Notice, > (5 a_1(x) + 7)(5
a_2(x) + 7)(5 b_3(x) + 22) = > > 49(300125 x^3 - 18375 x^2 -
360 x + 22)Yeah, yeah. Weren't you supposed to leave this
forum?I guess the troll needs to be fed again.
===
Subject: Re:
Factorization dispute, again>Notice, >(5 a_1(x) + 7)(5 a_2(x)
+ 7)(5 b_3(x) + 22) = > > 49(300125 x^3 - 18375 x^2 - 360 x +
22)>where you see that the constant terms match as now you
have 7(7)(22) =>1078, which is the constant term of the
polynomial>49(300125 x^3 - 18375 x^2 - 360 x + 22).>Various
people have debated me about what happens when you divide
off>49, where for some odd reason, some of them seem to
believe that you>can have>w_1(x), w_2(x), and w_3(x) such that
w_1(x) w_2(x) w_3(x) = 49, and>(5 a_1(x) + 7)/w_1(x) (5 a_2(x)
+ 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = > > 300125 x^3 - 18375
x^2 - 360 x + 22>where the w's vary as x varies, which is a
rather naive notion.Why is it naive?w1(x) = x+7w2(x) =
x^2+7w3(x) = 22/((x+7)(x^2+6))>That's because you can multiply
*everything* out, and simplify to get>(7/w_1(x)) (7/w_2(x))
(22/w_3(x)) = 22As can I (understand that I'm not claiming
that these particularfunctions will work for this problem, I'm
just pointing that thatit isn't naive to assume that they don't
have to be constant). >which should be simple enough for all of
you.>Now those of you who usually work in the field of complex
numbers may>think that it's not a big deal, as you may think
it doesn't matter if>w_3(x) has some factor factor of 7,
despite *seeing* (22/w_3(x)) but>you see, as 22 is coprime to
7 in the ring of algebraic integers, if>w_3(x) isn't coprime
to 7, (22/w_3(x)) does not exist in the ring.Doesn't matter.
What matters is whether or not (5 b_3(x) + 22)/w_3(x)exists in
the ring.Let's try a simpler example: f(x) = x(x+1) over the
integers. Clearlyf(x) is even. So, can we assume that either
x/2 is an integer or(x+1)/2 is an integer? The answer is yes,
but you can't assume itis always the same one. IOW, there
exist a1 and a2 such that x/a1(x)and (x+1)/a2(x) are both
integers and a1(x)a2(x)=2. But you can'tassume that just
because a1(0) = 2 and a2(0) = 1 that a1(x)=2 anda2(x)=1.And
note that 1/2 doesn't exist in the ring, but that doesn't
meanthat a2(x) can't equal 2 every now and then.Alan--
Defendit numerus
===
Subject: Re: Factorization dispute, again Notice, > > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =  49(300125 x^3 - 18375 x^2 - 360 x + 22)> > where you see
that the constant terms match as now you have 7(7)(22) => >
1078, which is the constant term of the polynomial> > > >
49(300125 x^3 - 18375 x^2 - 360 x + 22).> > > > Various people
have debated me about what happens when you divide off> > 49,
where for some odd reason, some of them seem to believe that
you> > can have> > > > w_1(x), w_2(x), and w_3(x) such that
w_1(x) w_2(x) w_3(x) = 49, and> > > > (5 a_1(x) + 7)/w_1(x) (5
a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = > > > > 300125 x^3
- 18375 x^2 - 360 x + 22> > > > where the w's vary as x
varies, which is a rather naive notion.> > > > That's because
you can multiply *everything* out, and simplify to get> >
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22> > which should be
simple enough for all of you.> > > > Now those of you who
usually work in the field of complex numbers may> > think that
it's not a big deal, as you may think it doesn't matter if> >
w_3(x) has some factor factor of 7, despite *seeing*
(22/w_3(x)) but> > you see, as 22 is coprime to 7 in the ring
of algebraic integers, if> > w_3(x) isn't coprime to 7,
(22/w_3(x)) does not exist in the ring.> But 22/w_3(x) *need*
not exist in the ring. It is (5 b3(x)+22)/w3(x)> that must
exist in the ring (and which does exist in the ring. I have>
no idea where you got the idea that 22/w3(x) must be in the
ring.> Consider P(x) = (x + 1)(x + 2), in the integers. It is
always even, so> it can be divided by 2. <deleted>Why? What
I've done is do a basic simplification going from(5 a_1(x) +
7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22to (7/w_1(x)) (7/w_2(x))
(22/w_3(x)) = 22, which is done simply enough by multiplying
everything out and lettingthe terms with x as a factor cancel
each other out.Now you wish to avoid that Dik Winter because
your assertions can't betrue in the ring of algebraic integers
as if w_3(x) isn't coprime to7, there's a contradiction, as
22/w_3(x) can't be in the ring then.You're a crank Dik Winter,
who has been refuted quite simply but youkeep talking as if
convincing *others* changes mathematical truth.It does
not.
===
Subject: Re: Factorization dispute, againwho are these
*others* that you refer to?... perhaps,you have an audience
that is not known to the rest of us,the desingated Peanut
Gallery of would-be critics & helpmeets. or is it just the
entire, 'virtual crowd of allof the folks who are in the
googolplex,that *could* lurk on your bifurcating threads --
ormaybe they should?mea culpa, dood.> keep talking as if
convincing *others* changes mathematical truth.--ils dcues
d'Enron!http://larouchepub.com/
===
Subject: Re: Factorization
dispute, againBut 22/w_3(x) *need* not exist in the ring. It
is (5 b3(x)+22)/w3(x)that must exist in the ring (and which
does exist in the ring. I haveno idea where you got the idea
that 22/w3(x) must be in the ring.> Now you wish to avoid that
Dik Winter because your assertions can't be> true in the ring
of algebraic integers as if w_3(x) isn't coprime to> 7,
there's a contradiction, as 22/w_3(x) can't be in the ring
then.You keep saying that 22/w_3(x) need not be in the
ring.No-one is disputing this. The problem is that you seem to
thinkthat the fact that 22/w_3(x) need not be in the ring is of
greatimportance. Indeed, you seem to assume that it is obvious
why22/w_3(x) must be in the ring. It appears that you are
reasoning thus: The constant term of (b_3(x) + 22) is 22 The
constant term of (b_3(x)/w_3(x) + 22/w_3(x)) cannot be
b_3(x)/w_3(x) so it must be 22/w_3(x) The constant term must
be in the ring, so 22/w_3(x) must be in the ring.The problem
is that the constant term of (b_3(x)/w_3(x) + 22/w_3(x))is
22/w_3(0)= 22/1 = 22. Whether or not 22/w_3(x) is in thering
for values of x other than 0 does not matter. - William
Hughes
===
Subject: Re: Factorization dispute, again> > But
22/w_3(x) *need* not exist in the ring. It is (5
b3(x)+22)/w3(x)> that must exist in the ring (and which does
exist in the ring. I have> no idea where you got the idea that
22/w3(x) must be in the ring.> Now you wish to avoid that Dik
Winter because your assertions can't betrue in the ring of
algebraic integers as if w_3(x) isn't coprime to7, there's a
contradiction, as 22/w_3(x) can't be in the ring then.> You
keep saying that 22/w_3(x) need not be in the ring.> No-one is
disputing this. The problem is that you seem to think> that the
fact that 22/w_3(x) need not be in the ring is of great>
importance. Indeed, you seem to assume that it is obvious why>
22/w_3(x) must be in the ring. It appears that you are
reasoning thus:> The constant term of (b_3(x) + 22) is 22> The
constant term of (b_3(x)/w_3(x) + 22/w_3(x))> cannot be
b_3(x)/w_3(x) so it must be 22/w_3(x)> The constant term must
be in the ring, so 22/w_3(x)> must be in the ring.You're lying
William Hughes, as my exact reasoning was posted. I
justmultiply out the factorization, and simplify.Now readers
should note that posters like William Hughes are cranks,so of
course they have to ignore the actual facts, but consider what
Iposted before:(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5
b_3(x) + 22)/w_3(x) = 300125 x^3 - 18375 x^2 - 360 x + 22
where the w's vary as x varies, which is a rather naive
notion.That's because you can multiply *everything* out, and
simplify to get(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22which
should be simple enough for all of you.> The problem is that
the constant term of (b_3(x)/w_3(x) + 22/w_3(x))> is
22/w_3(0)= 22/1 = 22. Whether or not 22/w_3(x) is in the> ring
for values of x other than 0 does not matter.> - William
HughesDenial of basic algebra is such a sad thing to display
to the world asWilliam Hughes demonstrates a rather odd
irrationality, to all thosereaders who go through my
posts.After all, I'm just talking about multiplying out and
simplifying, andI'll post again because these cranks have a
bad habit of creativedeletion what I said previously:(5 a_1(x)
+ 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22 where the w's vary as x
varies, which is a rather naive notion.That's because you can
multiply *everything* out, and simplify to get(7/w_1(x))
(7/w_2(x)) (22/w_3(x)) = 22which should be simple enough for
all of you.You see, the cranks can't get arouund 22/w_3(x) NOT
being in the ringof algebraic integers if w_3(x) isn't coprime
to 22.And for those of you who wondered, yes, when faced with
a *very* basicrefutation of their positions posters who argue
with me tend to justignore the truth. Later they come back
with the same position. They're irrational and are the real
cranks as you can't reason withthem.They believe false things
but want desperately to believe and
bebelieved.http://mathforprofit.blogspot.com/
===
Subject: Re:
Factorization dispute, again> > But 22/w_3(x) *need* not exist
in the ring. It is (5 b3(x)+22)/w3(x)> that must exist in the
ring (and which does exist in the ring. I have> no idea where
you got the idea that 22/w3(x) must be in the ring. > Now you
wish to avoid that Dik Winter because your assertions can't
be> true in the ring of algebraic integers as if w_3(x) isn't
coprime to> 7, there's a contradiction, as 22/w_3(x) can't be
in the ring then.You keep saying that 22/w_3(x) need not be in
the ring.No-one is disputing this. The problem is that you seem
to thinkthat the fact that 22/w_3(x) need not be in the ring is
of greatimportance. Indeed, you seem to assume that it is
obvious why22/w_3(x) must be in the ring. It appears that you
are reasoning thus: The constant term of (b_3(x) + 22) is 22
The constant term of (b_3(x)/w_3(x) + 22/w_3(x)) cannot be
b_3(x)/w_3(x) so it must be 22/w_3(x) The constant term must
be in the ring, so 22/w_3(x) must be in the ring.> You're
lying William Hughes, as my exact reasoning was posted. I
just> multiply out the factorization, and simplify.Your
reasoning that 22/w_3(x) need not be in the ring has been
postedmany, many times. No-one disagrees with your
conclusion.Your reasoning that 22/w_3(x) must be in the ring
has never been posted.The above is my guess as to what your
reasoning is.Quiz time: What is the constant term of the
following three functions? U(x) = sqrt(x^2 + x)/sqrt(x+7) +
7/sqrt(x+7) V(x) = sqrt(x^2 + x)/7 + 7/7 W(x) = sqrt(x^2
+x)/(x^2 +2x +7) + 7/(x^2 +2x +7)(hint: The constant term of
b(x) is b(0) ) - William Hughes
===
Subject: Re: Factorization
dispute, again> > But 22/w_3(x) *need* not exist in the ring.
It is (5 b3(x)+22)/w3(x)> that must exist in the ring (and
which does exist in the ring. I have> no idea where you got
the idea that 22/w3(x) must be in the ring.> > > Now you wish
to avoid that Dik Winter because your assertions can't be>
true in the ring of algebraic integers as if w_3(x) isn't
coprime to> 7, there's a contradiction, as 22/w_3(x) can't be
in the ring then.> > You keep saying that 22/w_3(x) need not
be in the ring.> No-one is disputing this. The problem is that
you seem to think> that the fact that 22/w_3(x) need not be in
the ring is of great> importance. Indeed, you seem to assume
that it is obvious why> 22/w_3(x) must be in the ring. It
appears that you are reasoning thus:> > The constant term of
(b_3(x) + 22) is 22> > The constant term of (b_3(x)/w_3(x) +
22/w_3(x))> cannot be b_3(x)/w_3(x) so it must be 22/w_3(x)> >
The constant term must be in the ring, so 22/w_3(x)> must be in
the ring.You're lying William Hughes, as my exact reasoning was
posted. I justmultiply out the factorization, and simplify.>
Your reasoning that 22/w_3(x) need not be in the ring has been
posted> many, many times. No-one disagrees with your
conclusion.Actually, it IS in the ring, which is my point
William Hughes. Andyou deleted out the factorization and
simplification which shows thatfact!!!Extraordinary behavior
which is indeed crank.I give you the information, explain it
clearly, and you try to justignore it.> Your reasoning that
22/w_3(x) must be in the ring has never been posted.> The
above is my guess as to what your reasoning is.The assertions
are over the ring of algebraic integers, so operationsare to
be in that ring. I show that you're pushed out of that ring
ina surprising way.> Quiz time:> What is the constant term of
the following three functions?Irrelevant to the issue of
22/w_3(x) having to be in the ring ofalgebraic integers.> U(x)
= sqrt(x^2 + x)/sqrt(x+7) + 7/sqrt(x+7)> V(x) = sqrt(x^2 + x)/7
+ 7/7> W(x) = sqrt(x^2 +x)/(x^2 +2x +7) + 7/(x^2 +2x +7)>
(hint: The constant term of b(x) is b(0) )> - William
HughesWhat I did was multiply out the factorization and
simplify to showthat it's impossible for w_3(x) to not be
coprime to 7 as then22/w_3(x) is NOT in the ring of algebraic
integers.It's simple, direct, and basic, and it refutes
attempts at claimingthat w_3(x) can share non-unit factors
with 7 in the ring of algebraicintegers.
===
Subject: Re:
Factorization dispute, again > > Notice, > > (5 a_1(x) + 7)(5
a_2(x) + 7)(5 b_3(x) + 22) = > > 49(300125 x^3 - 18375 x^2 -
360 x + 22) > > where you see that the constant terms match as
now you have 7(7)(22) = > > 1078, which is the constant term of
the polynomial > > > > 49(300125 x^3 - 18375 x^2 - 360 x + 22).
> > > > Various people have debated me about what happens when
you divide off > > 49, where for some odd reason, some of them
seem to believe that you > > can have > > > > w_1(x), w_2(x),
and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and > > > > (5
a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x)
= > > > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > > > where the
w's vary as x varies, which is a rather naive notion. > > > >
That's because you can multiply *everything* out, and simplify
to get > > (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 > > which
should be simple enough for all of you. > > > > Now those of
you who usually work in the field of complex numbers may > >
think that it's not a big deal, as you may think it doesn't
matter if > > w_3(x) has some factor factor of 7, despite
*seeing* (22/w_3(x)) but > > you see, as 22 is coprime to 7 in
the ring of algebraic integers, if > > w_3(x) isn't coprime to
7, (22/w_3(x)) does not exist in the ring. > > But 22/w_3(x)
*need* not exist in the ring. It is (5 b3(x)+22)/w3(x) > that
must exist in the ring (and which does exist in the ring. I
have > no idea where you got the idea that 22/w3(x) must be in
the ring. > > Consider P(x) = (x + 1)(x + 2), in the integers.
It is always even, so > it can be divided by 2. <deleted> > >
Why?Why? Because your reasoning would lead to the conclusion
that the ringof integers is flawed because it does not contain
numbers that shouldbe in that ring. But you are afraid to
answer the questions I posed inthe part you deleted. > Why?
What I've done is do a basic simplification going from > (5
a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) +
22)/w_3(x) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > to
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22, > > which is done
simply enough by multiplying everything out and letting > the
terms with x as a factor cancel each other out. > > Now you
wish to avoid that Dik Winter because your assertions can't be
> true in the ring of algebraic integers as if w_3(x) isn't
coprime to > 7, there's a contradiction, as 22/w_3(x) can't be
in the ring then.is in the algebraic integers. If you think so
you should answer thequestions in the part you deleted.
Because if 22/w3(x) must be inthe algebraic integers, for
*exactly the same reasons* 1/2 must bein the integers (see
P(x) = (x + 1)(x + 2), which is divisible by 2in the
integers). > You're a crank Dik Winter, who has been refuted
quite simply but you > keep talking as if convincing *others*
changes mathematical truth.You dodge all my questions, because
you are afraid to answer them. Now,who is the crank?-- dik t.
winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Factorization dispute,
again>There is no fixing stupid because it is not>broken.What
an intruguing comment! I'm not sure I get it, but it had me
giggling while I was tryin to figure it out.-- Let us learn to
dream, gentlemen, then perhaps we shall find the truth... But
let us beware of publishing our dreams before they have been
put to the proof by the waking understanding. -- Friedrich
August Kekul.8e
===
Subject: Re: Factorization dispute, again>
Notice, > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > >
49(300125 x^3 - 18375 x^2 - 360 x + 22)> where you see that
the constant terms match as now you have 7(7)(22) => 1078,
which is the constant term of the polynomial> 49(300125 x^3 -
18375 x^2 - 360 x + 22).> Various people have debated me about
what happens when you divide off> 49, where for some odd
reason, some of them seem to believe that you> can have>
w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) =
49, and> (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x)
+ 22)/w_3(x) = > > 300125 x^3 - 18375 x^2 - 360 x + 22> where
the w's vary as x varies, which is a rather naive notion.>
That's because you can multiply *everything* out, and simplify
to get> (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22> which should be
simple enough for all of you.> Now those of you who usually
work in the field of complex numbers may> think that it's not
a big deal, as you may think it doesn't matter if> w_3(x) has
some factor factor of 7, despite *seeing* (22/w_3(x)) but> you
see, as 22 is coprime to 7 in the ring of algebraic integers,
if> w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in
the ring.> You know, it's like how in integers 1/2 doesn't
exist. It's not an> integer, so it's not in the ring. Yes, you
are right about this. 22/w_3(x) is not an algebraicinteger.
It's obvious, because w_3(x) is a divisor of 7. But it's
irrelevant. What's relevant is that (5*b_3(x) + 22)/w_3(x)is
an algebraic integer. This is easier to see if you
re-writeb_3(x) as it was originally, b_3(x) = a_3(x) - 3. Thus
we have (5*b_3(x) + 22) = (5*a_3(x) - 15 + 22) = (5*a_3(x) + 7)
Thus when we divide by w_3(x), we have 5*a_3(x)/w_3(x) +
7/w_3(x),and both a_3(x)/w_3(x) and 7/w_3(x) *are* algebraic
integers. Well, you are not going to buy this. You are going
to continueto insist (correctly) that 22/w_3(x) cannot be an
algebraic integerand rave about how important that is and how
we are trying to confuse people, etc, etc.. You are totally
hung up on the idea that constant terms are inviolate and
sacred in some way. They are not. No, I am NOT saying that
constant terms are not constant. Of course they are. I am
saying that the important thing here is not divisibility of
the constant term 22 by a factor of 7. Theimportant thing is
divisibility of the *entire expression*5_b3(x) + 22 by a
factor of 7. And that is exactly what occurs. Different topic.
If your method of proof is actually valid,it will apply
similarly to other polynomials than just yourbeloved favorite,
P(x). Thus, consider Q(x) = 7^2*(125*x^3 - 15*x + 22). This has
the essential properties of the JSH polynomial: namely,the
constant term Q(0) = 7*7*22 = 1078, and it can be factoredin
the form (5*b_1 + 7)*(5*b_2 + 7)*(5*c_3 + 22)where b_1, b_2,
and c_3 are functions of x and are algebraicintegers. Note
that when x = 0, b_1 = 0, b_2 = 0, and c_3 = 0. Thus when x =
0, b_1 and b_2 are multiples of 7 (i.e., 7*0in each case).
Thus also when x = 0, the product of theconstant terms equals
the constant term of Q(x): Q(0) = (0 + 7)*(0 + 7)*(0 + 22) =
7*7*22 = 1078. Now according to the JSH argument, b_1 and b_2
should be divisible by 7 when x <> 0 also. So let's see what
happens when x = 1. First, note that Q(1) = 7^2*(125 - 15 +
22) = 7^2*132. This can be factored as Q(1) = r * s * t,where
r = 5*7^{2/3}*w + 7 s = 5*7^{2/3}*w^2 + 7,and t = 5*(7^{2/3} -
3) + 22,where w is a unit in the algebraic integers, w = (-1 +
sqrt(-3))/2. Therefore Q(1) is of the required form, Q(1) =
(5*b_1 + 7)*(5*b_2 + 7)*(5*c_3 + 22),where b_1 = 7^{2/3}*w,
b_2 = 7*{2/3}*w^2,and c_3 = 7^{2/3} - 3,all of which are
algebraic numbers. Note that b_1 and b_2 are *not* divisible
by 7. Neitherare they *coprime* to 7; each has the factor
7^{2/3} incommon with 7. Dividing 7^{2/3} out of each factor
yields: Q(1)/49 = (5*w + 7^{1/3})*(5*w^2 + 7^{1/3})*(5*(1 -
3/7^{2/3}) + 22/7^{2/3}) = (5*w + 7^{1/3})*(5*w^2 +
7^{1/3})*(5 + 7^{1/3}). It is a matter of arithmetic to check
that the right-hand sideequals 132 = Q(1)/49, as it should. It
is worth noting also that the product of the threeconstant
terms in the expression just above is 7^{1/3} * 7^{1/3} *
(22/7^{2/3}) = 22. Note again, this is the factorization when
x = 1. The coefficients b_1, b_2, and c_3 are all functions of
x; thevalues given above are actually b_1(1), b_2(1), and
c_3(1).When x = 0, b_1(0) = 0, b_2(0) = 0, and c_3(0) = 0. For
values of x other than 1 or 0, these functions are difficult to
compute. Note finally that the above factorization of Q(x) is
*not* of the form claimed by JSH, i.e., it is *not* of the
form Q(1) = (5*a_1 + 7)*(5*a_2 + 7)*(5*b_3 + 22),where a_1 and
a_2 are algebraic integers which are divisible by 7. Yet all
the arithmetic checks out; the constant termsas required are
7, 7, and 22. So where does the JSH logicbreak down for this
example, yet succeed for his P(x)?> So you see, my argument is
correct and simple, and mathematicians are> indeed running from
a little gut check in their field. They're> pussies too scared
to handle the truth. Most of your loyal audience has probably
noticed by now that youhave stopped trying to respond to the
substantive parts of my posts. You pounce on some superficial
aspect of them anddelete the rest. Makes you wonder, doesn't
it? Who here is acting like he is too scared to handle the
truth ? Nora B.> But you should also understand, some people
will be able to see that,> which is part of my plan. I can let
mathematicians destroy themselves> proving they can't be
trusted based on what they *see*, while they> forget what they
can't see: the wearing down of the mathematician> mystique.>
James Harris> http://mathforprofit.blogspot.com/
===
Subject:
Re: Factorization dispute, againNotice, (5 a_1(x) + 7)(5
a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x
+ 22)where you see that the constant terms match as now you
have 7(7)(22) =1078, which is the constant term of the
polynomial49(300125 x^3 - 18375 x^2 - 360 x + 22).Various
people have debated me about what happens when you divide
off49, where for some odd reason, some of them seem to believe
that youcan havew_1(x), w_2(x), and w_3(x) such that w_1(x)
w_2(x) w_3(x) = 49, and(5 a_1(x) + 7)/w_1(x) (5 a_2(x) +
7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = 300125 x^3 - 18375 x^2 -
360 x + 22where the w's vary as x varies, which is a rather
naive notion.That's because you can multiply *everything* out,
and simplify to get(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22which
should be simple enough for all of you.Now those of you who
usually work in the field of complex numbers maythink that
it's not a big deal, as you may think it doesn't matter
ifw_3(x) has some factor factor of 7, despite *seeing*
(22/w_3(x)) butyou see, as 22 is coprime to 7 in the ring of
algebraic integers, ifw_3(x) isn't coprime to 7, (22/w_3(x))
does not exist in the ring.You know, it's like how in integers
1/2 doesn't exist. It's not aninteger, so it's not in the
ring.> Yes, you are right about this. 22/w_3(x) is not an
algebraic> integer. It's obvious, because w_3(x) is a divisor
of 7.> But it's irrelevant. What's relevant is that >
(5*b_3(x) + 22)/w_3(x)> is an algebraic integer. This is
easier to see if you re-writeThat ignores the fact
that(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22wouldn't be in the
ring of algebraic integers, if w_3(x) shared anynon-unit
factors with 7.However, that result follows from simplifying(5
a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x)
= 300125 x^3 - 18375 x^2 - 360 x + 22so you have to at least
admit that your claims push you out of thering of algebraic
integers.Can you admit that Nora Baron?
===
Subject: Re:
Factorization dispute, again... > you see, as 22 is coprime to
7 in the ring of algebraic integers, if > w_3(x) isn't coprime
to 7, (22/w_3(x)) does not exist in the ring.... > Yes, you
are right about this. 22/w_3(x) is not an algebraic > integer.
It's obvious, because w_3(x) is a divisor of 7. > > But it's
irrelevant. What's relevant is that > > (5*b_3(x) + 22)/w_3(x)
> > is an algebraic integer. This is easier to see if you
re-write > > That ignores the fact that > > (7/w_1(x))
(7/w_2(x)) (22/w_3(x)) = 22 > > wouldn't be in the ring of
algebraic integers, if w_3(x) shared any > non-unit factors
with 7.Why not? The left hand expression is equal to
49.22/(w1(x).w2(x).w3(x)where I have provided you with w1(x),
w2(x) and w3(x) that multiplytogether to get 49. w3(x) is in
general *not* coprime to 7, andw1(x) and w2(x) are in general
*not* divisible by 7.-- dik t. winter, cwi, kruislaan 413,
1098 sj amsterdam, nederland, +31205924131home: bovenover 215,
1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Factorization dispute,
again[cut]> > That ignores the fact that> > > > (7/w_1(x))
(7/w_2(x)) (22/w_3(x)) = 22> > > > wouldn't be in the ring of
algebraic integers, if w_3(x) shared any> > non-unit factors
with 7.> Why not? The left hand expression is equal to>
49.22/(w1(x).w2(x).w3(x)> where I have provided you with
w1(x), w2(x) and w3(x) that multiply> together to get 49.
w3(x) is in general *not* coprime to 7, and> w1(x) and w2(x)
are in general *not* divisible by 7.I think Harris meant to
say that 22/w_3(x) wouldn't be in thering of algebraic
integers. Therefore, one is not allowedto use it on the left
hand side of that equality.He is not saying that 22 wouldn't
be in the ringof algebraic integers. He is objecting to using
22/w_3(x),which is not an algebraic integer.Others have
explained why this is not a valid objection.-- Bill
Hale
===
Subject: Re: Factorization dispute, again > Notice, >
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3
- 18375 x^2 - 360 x + 22) > where you see that the constant
terms match as now you have 7(7)(22) = > 1078, which is the
constant term of the polynomial > > 49(300125 x^3 - 18375 x^2
- 360 x + 22). > > Various people have debated me about what
happens when you divide off > 49, where for some odd reason,
some of them seem to believe that you > can have > > w_1(x),
w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and  (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) +
22)/w_3(x) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > where
the w's vary as x varies, which is a rather naive notion. > >
That's because you can multiply *everything* out, and simplify
to get > (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 > which should
be simple enough for all of you. > > Now those of you who
usually work in the field of complex numbers may > think that
it's not a big deal, as you may think it doesn't matter if >
w_3(x) has some factor factor of 7, despite *seeing*
(22/w_3(x)) but > you see, as 22 is coprime to 7 in the ring
of algebraic integers, if > w_3(x) isn't coprime to 7,
(22/w_3(x)) does not exist in the ring.But 22/w_3(x) *need*
not exist in the ring. It is (5 b3(x)+22)/w3(x)that must exist
in the ring (and which does exist in the ring. I haveno idea
where you got the idea that 22/w3(x) must be in the
ring.Consider P(x) = (x + 1)(x + 2), in the integers. It is
always even, soit can be divided by 2. Your reasoning about
constant term wouldlead to the result P(x)/2 = (x + 1)(x/2 +
1), because now theconstant terms match. My position is that
there are functions w1(x)and w2(x), defined as follows: w1(x)
= gcd(x + 1, 2) w2(x) = gcd(x + 2, 2)such that P(x)/2 = [(x +
1)/w1(x)] * [(x + 2)/w2(x)]is a valid factorisation. According
to your reasoning above (paraphrase): ...as you may think it
doesn't matter if w1(x) has some factor of 2, despite *seeing*
(1/w1(x)) but you see, as 2 is coprime to 1 in the ring of
integers, if w1(x) isn't coprime to 2, (1/w1(x)) does not
exist in the ring.So, although both factors in the
factorisation are integer, according toyou it is not a valid
factorisation, so the ring of integers is flawed. > So you
see, my argument is correct and simple, and mathematicians are
> indeed running from a little gut check in their field.
They're > pussies too scared to handle the truth.So, it is
your thinking that the ring of integers is flawed?With your
polynomial the situation is similar. I have given
pretty*explicit* definitions of the functions w1(x) to w3(x)
such that (5 a1(x) + 7)/w1(x) , (5 a2(x) + 7)/w2(x) and (5
b3(x) + 22)/w3(x)are all algebraic integer for all integer
x.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland, +31205924131home: bovenover 215, 1025 jn amsterdam,
nederland; http://www.cwi.nl/~dik/
===
Subject: Re: Probability
of a Run>I've appended some code in VBA for the original
algorithm and for the>approximation I mentioned.>I make no
great claims for the code but it should be adequate for
any>investigations you might want to make.> CONGRATULATIONS
IAN SMITH!> With a few lines of Basic code you have cut
through mountains of> mathematical horse  and created a
program that gives quick answers> to what formerly was
considered a difficult problem. This is just> what I hoped
would happen.> I ported your code to BC7 for DOS and ran a few
tests. The results> were exactly the same as what I got using
the recursive program or> through evaluating the coefficients
of the generating function.> May I have your permission to
compile it and put it on the web site of> the Rancocas Valley
Journal of Applied Mathematics, with full credit> to yourself,
of course?> Sam AllenMy standard Conditions of use are No
charges, no conditions and noguarantees. I've no idea if this
is legal in the US but assuming itis, then you are more than
welcome.Ian Smith
===
Subject: Re: Probability of a Run>I've
appended some code in VBA for the original algorithm and for
the>approximation I mentioned.>I make no great claims for the
code but it should be adequate for any>investigations you
might want to make.CONGRATULATIONS IAN SMITH!With a few lines
of Basic code you have cut through mountains ofmathematical
horse  and created a program that gives quick answersto
what formerly was considered a difficult problem. This is
justwhat I hoped would happen.I ported your code to BC7 for
DOS and ran a few tests. The resultswere exactly the same as
what I got using the recursive program orthrough evaluating
the coefficients of the generating function.May I have your
permission to compile it and put it on the web site ofthe
Rancocas Valley Journal of Applied Mathematics, with full
creditto yourself, of course?Sam Allen
===
Subject:
integrability of pfaffian?In the PfaffiandU = A_1.dx_1 +
A_2.dx_2 + ...+A_n.dx_nif A_1, A_2,...are twice continuously
differentiable _functions_ ofthe variables x_1, x_2,...etc.
then a necessary condition for theintegrability of dU is the
equality of the partial derviatives@A_i/@dx_k = @A_k/@x_iand
there is also a much longer expression involving any three of
theA's that will give the sufficient condition for the
existence of anintegrating factor.My question is, what if the
A_i are not simply functions of x_k butrather functionals?
Say, for exmaple, A_1 depends explicitly on (x_1 *x_2), or
(h*x_1), where * denotes the convolution operator and h(t) isa
given function, what can then guarantee the integrability of
dU?
===
Subject: Re: 3 x 3 matrix / eigenvalueKeywords:
eigenvalue real|> If A is a nonsingular 3 x 3 matrix with
nonnegative entries, then why must A|> have a positive real
eigenvalue?|> ...The characteristic equation (setting the
determinant with the eigenvaluesubtracted from the diagonal to
zero) is equivalent to finding the rootsof a cubic polynomial;
so it's clear that there must be a real eigenvalue.Writing
down the definition of the eigenvalue as A*v = lambda *vfor an
eigenvector v and playing around with the signs it is alsoclear
that lambda must stay positive (at least if all threeCartesian
components of v are positive or negative... which is
effectivleythe same as eigenvectors are only defined up to a
const...)
===
Subject: Re: 3 x 3 matrix / eigenvalue<posted &
mailed>> If A is a nonsingular 3 x 3 matrix with nonnegative
entries, then why> must A have a positive real eigenvalue?> >
Mike> This is not true. For example> 1 0 0> 0 0 1> 0 1 0> has
eigenvalues 1,So, 1 is no longer positive ?> 1, and -1. To the
OP: this is (part of) the Perron-Frobenius theorem.-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I
had the last laugh. Alan Partridge, _Bouncing Back_ (14
times)
===
Subject: Re: 3 x 3 matrix / eigenvalue:> If A is a
nonsingular 3 x 3 matrix with nonnegative entries, then why
must:> A have a positive real eigenvalue?:> :> Mike: This is
not true. For example: 1 0 0: 0 0 1: 0 1 0: has eigenvalues 1,
1, and -1. He didn't say *all* the eigenvalues had to be
positive, just one of them.In fact, it is true. One way is to
look at map from R={(x,y,z): x^2+y^2+z^2=1 and x,y,z>=0}to
itself given by v -> Av/||Av||. Since A is nonsingular, Av can
never be 0so this map is defined for all v in R. By the Brouwer
fixed point theorem,this map has a fixed point, which must then
be an eigenvector of A with apositive eigenvalue.This
generalizes directly to n by n matrices.Also, the only fact we
used about A is that for all v in R, Av is nonzero andcontains
all nonnegative entries. So this generalizes to some singular
A's aswell; for example, A's with all positive entries or even
any A with nonnegative entries that does not have an entire
column of 0's. (In fact, since the transpose of A has the same
eigenvalues, this generalizes to any A with nonnegative entries
which does not have an entire row *and* an entire column of
zeroes.)Ted
===
Subject: Re: c-number> Howdy everybody,> I'm
reading Peskin and Schroeder's Intro to QFT and in this> book
the author constantly refers to c-numbers... without> saying
what a c-number is. I couldn't find c-number> on
mathworld.wolfram.com or physicsworld.wolfram.com.> Can
someone please tell me what the hell a c-number is?Adding yet
another interpretation besides commuting and classical: Ithink
they mean a complex number (which obviously is both commuting
andclassical).I think they defined it in this way somewhere in
their book, butscanning the first few pages, I can't find it...
:-(Bye,Bjoern#
===
Subject: Re: c-number>I'm reading Peskin and
Schroeder's Intro to QFT and in this>book the author
constantly refers to c-numbers... without>saying what a
c-number is. I couldn't find c-number>on mathworld.wolfram.com
or physicsworld.wolfram.com.phi(x,y,z,t) with mass m satisfies
the equation: @^2 phi/@t^2 = c^2 (@^2/@x^2 + @^2/@y^2 +
@^2/@z^2)phi - k^2 phiwith k = mc^2 / (h/(2 pi)) c = light
speed, h = Planck's constant. @ = ASCII version of partial
derivative d.The corresponding boundary value problem, with
the boundary surfaceC: {{x,y,z,0): (x,y,z) in R^3} is:
phi''(x,y,z,t) = Wphi (x,y,z,t) phi(x,y,z,0) = Q(x,y,z)
phi'(x,y,z,0) = P(x,y,z) W = c^2 (@^2/@x^2 + @^2/@y^2 +
@^2/@z^2) - k^2where ()' means @/@t.For the classical field,
you have Q(r) Q(r) = Q(r) Q(r) Q(r) P(r') = P(r') Q(r) P(r)
P(r') = P(r') P(r)for any two points r = (x,y,z), r' =
(x',y',z').So, the values (Q(r): r in R^3) and (P(r): r in
R^3) all commutewith one another and so are c-numbers.In
Quantum Physics, you also have: Q(r) Q(r) = Q(r) Q(r) P(r)
P(r') = P(r') P(r)but now: Q(r) P(r') = P(r') Q(r) + i h/(2
pi) delta(r,r')The corresponding quantities don't commute, so
they're q-numbers.You can, in fact, find D(r,t;r',t') =
phi(r,t) phi(r',t') - phi(r',t') phi(r,t)from this by posing
another initial value problem in terms of D: D(r,0;r',t) = 0
D'(r,0;r',t) = -i h/(2 pi) delta(r,r') D''(r,t;r',t') = W
D(r,t;r',t')and solving it (ideally by using Fourier
transforms).Going the other way, the requirement that
phi-phi's all commute oneach equal time surface C_t = {
(x,y,z,t): (x,y,z) in R^3 }that the equations of motion hold
implies that the (@phi/@t-@phi/@t)'swill also commute over
each C_t as well. You can substantially pin downthat the (phi
- @phi/@t)'s have to be by also requiring that
phi-phi'scommute over ANY spacelike surface, in addition to
the C_t's; and thatboth the Q-Q;Q-P;P-P relations and the
equations of motion remaininvariant under Poincare'
transformation.
===
Subject: Re: Bible 1, Darwin 0! And we are
talking pure archeology?> Forbidden archeology by Cremo,
Thompson!> Like the name says: Forbidden> I urge you not to
read it!You might try
reading<http://www.talkorigins.org/faqs/mom/groves.html>
instead.Bye,Bjoern
===
Subject: Re: Bible 1, Darwin 0! And we
are talking pure archeology?> part:>>The aim of the book is
not to prove things that the Bible said, Ijust>put a catchy
title.>> I remember looking at it at the library. I thought
its aim was to> prove things the Bhagavad-Gita said.>> So that
we know of the great dispensation offered to us unfortunate>
denizens of the Age of Kali, that we can improve our karma
simply by> chanting a simple hymn of praise to our teachers
and to Lord Krsna.>> John Savard>
http://home.ecn.ab.ca/~jsavard/index.htmlHari Krishna, Hari
Krishna, Ramah, Ramah, Krishna, KrishnaKrishna, Ramah,
Krishna, Ramah, Hari, HariRepeat 100 times while sitting, legs
crossed, ring fingerand thumb touching on a cushioned
floor.....(Note: Only demonstrated effect of this is that you
tend to better accepted in liberal communities at a greater
rate than those that do not......)> but is it at a greater
rate than those eating granola and/or huggingtrees. (i> think
we need a case study on this,... someone write up the
grantproposal.)Involving overseas travel and ....NL
===
Subject:
Re: I NEED HELP BADLY (sorry, maths not psych)> Let's be
serious for once.>> Consider an object being accelerated by a
idealistic jet of water or a> continuous 'stream of elastic
ping pong balls'. What is its subsequent velocity> pattern?>>
accelerated beyond the operating speed of the accelerating
fields, ie at 'c'. I> hope it might also produce a
relationship that is equivalent to mass> 'appearing' to
increase with velocity by gamma.)>>Please define what you mean
by the operating speed of an>>accelerating field. I've been in
physics research for over 20>>years, and I've never heard that
term.>>Do you mean phase velocity, group velocity, how fast
the>>operators turn the knobs, what?>I think you forgot to
answer the question, Henry.>Why is that?> yes, for some
strange reason I 'forget' to answer anything EjP
asks.Considering that he asked you about the meaningof a
meaningless statement, I don't find the reason strange at
all.Paul, not puzzled
===
Subject: Re: I NEED HELP BADLY (sorry,
maths not psych)Expires: 28 days> accelerated beyond the
operating speed of the accelerating fields, ie at 'c'. I> hope
it might also produce a relationship that is equivalent to
mass> 'appearing' to increase with velocity by gamma.)>>Please
define what you mean by the operating speed of an>>accelerating
field. I've been in physics research for over 20>>years, and
I've never heard that term.>>Do you mean phase velocity, group
velocity, how fast the>>operators turn the knobs, what?>>I
think you forgot to answer the question, Henry.>Why is that?>
yes, for some strange reason I 'forget' to answer anything EjP
asks.>Considering that he asked you about the meaning>of a
meaningless statement, I don't find the reason strange at
all.>Paul, not puzzledPaul, is your memory failing?I explained
this to you last year. If fields acted instantaneously, we
would beable to use them for instantaneous comunication.Do you
really want that?Henri Wilson.See why relativity is
wrong:http://www.users.bigpond.com/HeWn/index.htm
===
Subject:
Re: I NEED HELP BADLY (sorry, maths not psych)>I understand
very well why you are desperate to divert>the attention from
the fact that you know very well that>the SR prediction I
showed above is experimentally verified>in accelerators all
over the world all the time, while>the Newtonian prediction is
falsified all the time.> Your idea of the 'Newtonian
prediction' is just a diversion... from the fact that the
prediction of NM fails by ordersof magnitude while the
prediction of SR is correct?> Read what I said to Randy. That
is the real reason why mass appears to increase> and the
required energy shoots up enormously as c is approached.And
Henry knows the REAL reason is: As the electron approaches the
speed at which the field acts, back radiation from the electron
'neutralizes' a volume around itself. The field in its vicinity
is more or less reversed by the electron's own movement. This
requires a great deal of energy; one single charge creating a
'little neutral sphere' in a strong field. This energy also
shows up in bolometer experiments.Paul, dazzled
===
Subject: Re:
I NEED HELP BADLY (sorry, maths not psych)>>I understand very
well why you are desperate to divert>the attention from the
fact that you know very well that>the SR prediction I showed
above is experimentally verified>in accelerators all over the
world all the time, while>the Newtonian prediction is
falsified all the time.Your idea of the 'Newtonian prediction'
is just a diversion.> .. from the fact that the prediction of
NM fails by orders> of magnitude while the prediction of SR is
correct?Read what I said to Randy. That is the real reason why
mass appears to increaseand the required energy shoots up
enormously as c is approached.> And Henry knows the REAL
reason is:> As the electron approaches the speed at which the
field acts,> back radiation from the electron 'neutralizes' a
volume around itself.> The field in its vicinity is more or
less reversed by the electron's own> movement. This requires a
great deal of energy; one single charge creating a> 'little
neutral sphere' in a strong field. This energy also shows up
in> bolometer experiments.I missed this explanation. Having
read it, I can't makeheads or tails of it.Henry: Does whatever
you are trying to describe have anequivalent in the ping-pong
model? If I am trying todrive a vehicle by throwing 100 m/sec
ping-pong ballsat it, so it has a limiting velocity of 100
m/sec,will the kinetic energy continue to rise
beyond0.5*m*100^2 and the momentum continue to rise
beyondm*100 due to some sort of back field?If not, what do
electrons have that a ping-pong-balldriven vehicle does not? -
Randy
===
Subject: Re: I NEED HELP BADLY (sorry, maths not
psych)Expires: 28 days>>I understand very well why you are
desperate to divert>the attention from the fact that you know
very well that>the SR prediction I showed above is
experimentally verified>in accelerators all over the world all
the time, while>the Newtonian prediction is falsified all the
time.> Your idea of the 'Newtonian prediction' is just a
diversion.>.. from the fact that the prediction of NM fails by
orders>of magnitude while the prediction of SR is correct?>
Read what I said to Randy. That is the real reason why mass
appears to increase> and the required energy shoots up
enormously as c is approached.>And Henry knows the REAL reason
is:> As the electron approaches the speed at which the field
acts,> back radiation from the electron 'neutralizes' a volume
around itself.> The field in its vicinity is more or less
reversed by the electron's own> movement. This requires a
great deal of energy; one single charge creating a> 'little
neutral sphere' in a strong field. This energy also shows up
in> bolometer experiments.That's OK Paul. A physical
explanation is worth a thousand meaningless
mathsequations.>Paul, dazzled...by the novelty.....Henri
Wilson.See why relativity is
wrong:http://www.users.bigpond.com/HeWn/index.htm
===
Subject:
Re: I NEED HELP BADLY (sorry, maths not psych)>> And they
claim a simple 'mass increase'.>No they don't. If you actually
read this newsgroup, you'd>know that the term mass is used by
most physicists to>to refer to an invariant quantity.>The term
used is 'relativistic mass increase'.If you actually read this
newsgroup, you'd know that the termrelativistic mass is no
longer favored, and that most physiciststhese days prefer to
reserve the term mass to refer to an invariantquantity. Where
m used to appear in some formulas (such as E =mc^2), gamma*m
is now used. - Randy
===
Subject: Re: I NEED HELP BADLY (sorry,
maths not psych)>It would be nice if you used more text and
fewer programs. I'm not>explanation appears to be contained in
them, your website says nothing>that I am willing to look at.If
the speed of light is a universal constant in all frames of
reference,and in a certain situation light takes a longer path
when seen from oneperspective than it does from another, then
it must take a longer time fromthat perspective than it does
from the other.Or - the longer path takes the longer
time.Usually, the relativist will say something like...The
beam moves vertically in the moving frame, but this is seen as
adiagonal line in the stationary frame.|| | | Therefore time in
the moving frame is less than time in the stationaryframe,
taa-daaaa!, I'm so clever, I understand relativity, mom!Then
he does some elementary algebra which I won't bore you with,
and walksoff, confident he's right.Trouble is, if we bend the
light beam through the angle arctan(v/c) in themoving frame,
this is what happens.t0.........* at1........t1.........*
bt2......t2........t2.........*
ct3..t3.....t3.......t3.........* dThe moving source has moved
left as time passed, but the ray is vertical inthe stationary
frame. The tip of the ray, *, remains the same distance
fromthe edge of the page, at a,b,c,and d, but the source has
moved to the left.So, applying our rule that the longer path
takes the longer time, the longertime is in the moving frame,
not the stationary frame, contrary to therelativists claim
that he understands anything at all.At this point the
relativist will start to splutter about
unsynchronizedclocks.(Did I mention any clocks until now? No,
of course not.)He will claim that in the stationary frame two
clocks are separated|S| | | A Bby a distance vt, and work this
into his argument.(Oh no he won't, you cry!)Oh yes he will,
I've seen it done.Now, of course it is true that the distance
S-A-B is greater than thedistance S-A, so he has found a trick
up his sleeve to make time less in themoving frame once more.
What a clever relativist!But......... that isn't the path the
light takes. It doesn't go from S to A and A to B,bending
through a right-angle. It takes the short-cut, direct from S
to B.The frame with the longer path has the longer time.
Moving clocks speed up,and all because I tilted the
flashlight. Ergo, Einstein's relativity isstupidity, and
relativists are gullible fools.(To Henri: No special math
needed, old chap. Keep it simple :-)Androcles
===
Subject: Re: I
NEED HELP BADLY (sorry, maths not psych)Expires: 28 days>It
would be nice if you used more text and fewer programs. I'm
not>explanation appears to be contained in them, your website
says nothing>that I am willing to look at.>If the speed of
light is a universal constant in all frames of reference,>and
in a certain situation light takes a longer path when seen
from one>perspective than it does from another, then it must
take a longer time from>that perspective than it does from the
other.>Or - the longer path takes the longer time.>Usually, the
relativist will say something like...>The beam moves vertically
in the moving frame, but this is seen as a>diagonal line in the
stationary frame.>|>| >| >| >Therefore time in the moving frame
is less than time in the stationary>frame, taa-daaaa!, I'm so
clever, I understand relativity, mom!>Then he does some
elementary algebra which I won't bore you with, and walks>off,
confident he's right.Yes. All relativists seem to suffer from
the same 'spatial-ability deficiencysyndrome'.As I have
pointed out many times, a vertical beam appears vertical in
allframes. (http://www.users.bigpond.com/hewn/movingframe.exe)
Each infinitesimalelement of the vertical beam follows its own
unique diagonal path in the movingframe. There is no
conceiveable reason to believe that such an astractconstruct
constitutes light, traveling at c. My demo shows that such an
elementcannot be the same as one belonging to a laser beam
deliberately aimeddiagonally in the moving frame. Raindrops
take the same time to reach the ground even though they appear
to bemoving diagonally through the window of your moving
car.>Trouble is, if we bend the light beam through the angle
arctan(v/c) in the>moving frame, this is what
happens.>t0.........* a>t1........>t1.........*
b>t2......>t2........>t2.........*
c>t3..>t3.....>t3.......>t3.........* d>The moving source has
moved left as time passed, but the ray is vertical in>the
stationary frame. The tip of the ray, *, remains the same
distance from>the edge of the page, at a,b,c,and d, but the
source has moved to the left.>So, applying our rule that the
longer path takes the longer time, the longer>time is in the
moving frame, not the stationary frame, contrary to
the>relativists claim that he understands anything at all.>At
this point the relativist will start to splutter about
unsynchronized>clocks.>(Did I mention any clocks until now?
No, of course not.)>He will claim that in the stationary frame
two clocks are separated>|S>| >| >| >A B>by a distance vt, and
work this into his argument.>(Oh no he won't, you cry!)>Oh yes
he will, I've seen it done.>Now, of course it is true that the
distance S-A-B is greater than the>distance S-A, so he has
found a trick up his sleeve to make time less in the>moving
frame once more. What a clever relativist!>But....>.>.>.>.>.>
that isn't the path the light takes. It doesn't go from S to A
and A to B,>bending through a right-angle. It takes the
short-cut, direct from S to B.>The frame with the longer path
has the longer time. Moving clocks speed up,>and all because I
tilted the flashlight. Ergo, Einstein's relativity
is>stupidity, and relativists are gullible fools.Very
true.>(To Henri: No special math needed, old chap. Keep it
simple :-) Nothing can show this more simply than my
demo.However, clocks can appear to run either slow or fast
depending on whether theyare receding or approaching.In
reality, they don't PHYSICALLY change at all with movement.The
term 'gamma' is wrong. It cannot contain 'v^2'.>AndroclesHenri
Wilson.See why relativity is
wrong:http://www.users.bigpond.com/HeWn/index.htm
===
Subject:
Re: I NEED HELP BADLY (sorry, maths not psych)>>It would be
nice if you used more text and fewer programs. I'm
not>>explanation appears to be contained in them, your website
says nothing>>that I am willing to look at.>If the speed of
light is a universal constant in all frames of reference,>and
in a certain situation light takes a longer path when seen
from one>perspective than it does from another, then it must
take a longer timefrom>that perspective than it does from the
other.>Or - the longer path takes the longer time.>Usually,
the relativist will say something like...The beam moves
vertically in the moving frame, but this is seen as a>diagonal
line in the stationary frame.>|>| >| >| >Therefore time in the
moving frame is less than time in the stationary>frame,
taa-daaaa!, I'm so clever, I understand relativity, mom!>Then
he does some elementary algebra which I won't bore you with,
andwalks>off, confident he's right.> Yes. All relativists seem
to suffer from the same 'spatial-abilitydeficiency> syndrome'.>
As I have pointed out many times, a vertical beam appears
vertical in all> frames.
(http://www.users.bigpond.com/hewn/movingframe.exe)
Eachinfinitesimal> element of the vertical beam follows its
own unique diagonal path in themoving> frame. There is no
conceiveable reason to believe that such an astract> construct
constitutes light, traveling at c. My demo shows that such
anelement> cannot be the same as one belonging to a laser beam
deliberately aimed> diagonally in the moving frame.> Raindrops
take the same time to reach the ground even though they
appearto be> moving diagonally through the window of your
moving car.>Trouble is, if we bend the light beam through the
angle arctan(v/c) inthe>moving frame, this is what
happens.>t0.........* a>t1........>t1.........*
b>t2......>t2........>t2.........*
c>t3..>t3.....>t3.......>t3.........* d>The moving source has
moved left as time passed, but the ray is verticalin>the
stationary frame. The tip of the ray, *, remains the same
distancefrom>the edge of the page, at a,b,c,and d, but the
source has moved to theleft.>So, applying our rule that the
longer path takes the longer time, thelonger>time is in the
moving frame, not the stationary frame, contrary to
the>relativists claim that he understands anything at all.>At
this point the relativist will start to splutter about
unsynchronized>clocks.>(Did I mention any clocks until now?
No, of course not.)>He will claim that in the stationary frame
two clocks are separated>|S>| >| >| >A B>by a distance vt, and
work this into his argument.>(Oh no he won't, you cry!)>Oh yes
he will, I've seen it done.>Now, of course it is true that the
distance S-A-B is greater than the>distance S-A, so he has
found a trick up his sleeve to make time less inthe>moving
frame once more. What a clever
relativist!>But....>.>.>.>.>.that isn't the path the light
takes. It doesn't go from S to A and A toB,>bending through a
right-angle. It takes the short-cut, direct from S toB.>The
frame with the longer path has the longer time. Moving clocks
speedup,>and all because I tilted the flashlight. Ergo,
Einstein's relativity is>stupidity, and relativists are
gullible fools.> Very true.>(To Henri: No special math needed,
old chap. Keep it simple :-)> Nothing can show this more simply
than my demo.> However, clocks can appear to run either slow or
fast depending on whetherthey> are receding or approaching.Of
course. Doppler shift.> In reality, they don't PHYSICALLY
change at all with movement.Quite right.> The term 'gamma' is
wrong. It cannot contain 'v^2'.Err...The distance the light
moves from S to A is ct.The distance from A to B is vt.The
distance from S to B is sqrt([ct]^2 +[vt]^2) by Pythagoras.We
can choose an arbitrary unit of time for t, and make t = 1S to
B = hypotenuse = sqrt(c^2 +v^2)Androcles
===
Subject: Re: I NEED
HELP BADLY (sorry, maths not psych)Expires: 28 days> However,
clocks can appear to run either slow or fast depending on
whether>they> are receding or approaching.>Of course. Doppler
shift.> In reality, they don't PHYSICALLY change at all with
movement.>Quite right.> The term 'gamma' is wrong. It cannot
contain 'v^2'.>Err...>The distance the light moves from S to A
is ct.>The distance from A to B is vt.>The distance from S to B
is sqrt([ct]^2 +[vt]^2) by Pythagoras.>We can choose an
arbitrary unit of time for t, and make t = 1>S to B =
hypotenuse = sqrt(c^2 +v^2)>AndroclesWhat moves diagonally is
not light.The diagonal is just a line representing the path of
an infinitesimally smallelement of the light beam. The beam
itself remains vertical but appears to movesideways in the
moving observer's frame. The vertical component of
eachelement's velocity is not affected by observer movement.
How could it be?So Yes, each infinitesimal element moves at
sqrt(c^2+v^2) along it owndiagonal. It is the only element
that moves along that diagonal. Thereforethere is no diagonal
light beam moving at c. Therefore SR is the greatest loadof
codswallop ever dumped on the human race.Henri Wilson.See why
relativity is
wrong:http://www.users.bigpond.com/HeWn/index.htm
===
Subject:
Re: I NEED HELP BADLY (sorry, maths not psych)> So Yes, each
infinitesimal element moves at sqrt(c^2+v^2) along it own>
diagonal. It is the only element that moves along that
diagonal. Therefore> there is no diagonal light beam moving at
c. Therefore SR is the greatestload> of codswallop ever dumped
on the human race.Whoa... ease up, there, H! If I'm flying a
microlite at 100mph across atennis court (along the net) just
as a 100mph ace is served straight downthe middle, the ball
takes a diagonal path in *my frame of reference*. Theonly
silly part would be if I insisted the ball's speed was 100mph
insteadof 141mph.The light certainly does move diagonally. Of
course, if my clock sloweddown, it would appear to move even
faster, and that's the codswallop :)Androcles
===
Subject: Re: I
NEED HELP BADLY (sorry, maths not psych)Expires: 28 days> So
Yes, each infinitesimal element moves at sqrt(c^2+v^2) along
it own> diagonal. It is the only element that moves along that
diagonal. Therefore> there is no diagonal light beam moving at
c. Therefore SR is the greatest>load> of codswallop ever
dumped on the human race.>Whoa... ease up, there, H! If I'm
flying a microlite at 100mph across a>tennis court (along the
net) just as a 100mph ace is served straight down>the middle,
the ball takes a diagonal path in *my frame of reference*.
The>only silly part would be if I insisted the ball's speed
was 100mph instead>of 141mph.But if a stream of balls was
fired from a machine gun along that same line, thestream would
always remain aligned vertically in your frame. Each ball
wouldmove along its own diagonal path at sqrt(c^2+v^2). Take
that to the limit by decreasing ball size and increasing
frequency and youhave the 'light ray equivalent'. And, of
course, each ball takes the same time to reach its target
irrespectiveof your speed. No nonsense about clocks changing
rates.>The light certainly does move diagonally. Of course, if
my clock slowed>down, it would appear to move even faster, and
that's the codswallop :)>AndroclesSure is! Clock rates are not
physically dependent on velocity. My demo 'movingrame.exe' well
illustrates this monumental blunder made byEinstein et al. It
takes only a few seconds to run. See also: 'vertical.exe'It is
quite obvious that a diagonally aimed laser beam in the rest
observerframe is NOT identical to an apparently diagonally
moving element of a verticallight ray that moves sideways
acros that frame.How could SRians be so plainly stoopid?Henri
Wilson.See why relativity is
wrong:http://www.users.bigpond.com/HeWn/index.htm
===
Subject:
Re: I NEED HELP BADLY (sorry, maths not psych)>So Yes, each
infinitesimal element moves at sqrt(c^2+v^2) along it
own>diagonal. It is the only element that moves along that
diagonal.Therefore>there is no diagonal light beam moving at
c. Therefore SR is thegreatest>load>of codswallop ever dumped
on the human race.>Whoa... ease up, there, H! If I'm flying a
microlite at 100mph across a>tennis court (along the net) just
as a 100mph ace is served straight down>the middle, the ball
takes a diagonal path in *my frame of reference*.The>only
silly part would be if I insisted the ball's speed was
100mphinstead>of 141mph.> But if a stream of balls was fired
from a machine gun along that sameline, the> stream would
always remain aligned vertically in your frame. Each
ballwould> move along its own diagonal path at sqrt(c^2+v^2).>
Take that to the limit by decreasing ball size and increasing
frequencyand you> have the 'light ray equivalent'.> And, of
course, each ball takes the same time to reach its
targetirrespective> of your speed. No nonsense about clocks
changing rates.>The light certainly does move diagonally. Of
course, if my clock slowed>down, it would appear to move even
faster, and that's the codswallop :)>Androcles> Sure is! Clock
rates are not physically dependent on velocity.> My demo
'movingrame.exe' well illustrates this monumental blunder made
by> Einstein et al. It takes only a few seconds to run. See
also:'vertical.exe'> It is quite obvious that a diagonally
aimed laser beam in the restobserver> frame is NOT identical
to an apparently diagonally moving element of avertical> light
ray that moves sideways acros that frame.> How could SRians be
so plainly stoopid?> Henri Wilson.Beats me. I think they get
convinced and then there's no going back. But Iwill tell you
something. The debates on this newsgroup used to
beaetherialists against relativists. The aetherialists are
losing ground, andemission theory is gaining. Young people are
now more convinced that everbefore that the speed of light is
source dependent as their intuition tellsthem, and not aether
dependent. As time passes relativity will fall intodisrepute,
and the younger ones will investigate a whole new physics,
rebornout of Newton. They'll build accelerators for EM
radiation that will vastlyimprove interplanetary
communication, and someday FTL probes to investigatedistant
stars and their planets. Other facets, unimaginable to us now,
willcome out of it. This I predict.Glad to have you aboard.
Keep the focus on the newbies, and treat them withlogic. It
does no good fighting the diehards, although I'll agree it can
befun.Androcles
===
Subject: Re: I NEED HELP BADLY (sorry, maths
not psych)> Glad to have you aboard. Keep the focus on the
newbies, and treat them with> logic.A fine pair
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/
AndersenLogic.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/
LogicBull.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/
Gibberish.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/
Cornered.htmlDirk Vdm
===
Subject: Re: I NEED HELP BADLY (sorry,
maths not psych)Expires: 28 days> > Henri Wilson.> > See why
relativity is wrong:>
http://www.users.bigpond.com/HeWn/index.htm>It would be nice
if you used more text and fewer programs. I'm not >explanation
appears to be contained in them, your website says nothing
>that I am willing to look at.All my programs have been
discussed at length on the NG. I gather you are newehere.Henri
Wilson.See why relativity is
wrong:http://www.users.bigpond.com/HeWn/index.htm
===
Subject: Q
about random permutations charset=Windows-1252Suppose
x1,x2,...,xn is a random permutation of 1,...,n.Now sweep
through x1,x2,...,xn left-to-right, looking for 1,...,n in
increasing order. Let m be the greatest value found in the
sweep. If m < n, sweep again, looking for m+1,...,n. Repeat
until n is found.Example:Sweep 864513729----- --------- 1 1 2
2 3 3 45 4 6 7 5 8 9What is the distribution of the number of
sweeps? I.e., if S is the number of sweeps, what is
Pr(S=s)(s=1,...,n)?
===
Subject: Re: Q about random
permutations>Suppose x1,x2,...,xn is a random permutation of
1,...,n.>Now sweep through x1,x2,...,xn left-to-right, looking
for >1,...,n in increasing order. Let m be the greatest value
>found in the sweep. If m < n, sweep again, looking for
>m+1,...,n. Repeat until n is found.>Example:>Sweep
864513729>----- ---------> 1 1 2> 2 3> 3 45> 4 6 7> 5 8 9>What
is the distribution of the number of sweeps? I.e., >if S is the
number of sweeps, what is Pr(S=s)(s=1,...,n)?I have not been
able to get a complete solution of theproblem. However, there
are partial results.The distribution is symmetric. That is,
since a permutationand its reverse have their number of sweeps
adding up ton+1, P(S=s) = P(S = n+1-s).For the following,
observe that we do not need that thex's be the integers, but
any n ordered objects will do,and a sweep still consists of
the number of consecutiveobjects, starting from the smallest.
So let G_n(t) bethe generating function of the sweeps from n
objects,i.e., G_n(t) = sum P_n)(S=s)*t^n. If the first
elementof the permutation is k, the contribution to G_n is
P_{n-1}(t)/n, k=1; t*P_{n-1}(t)/n, k=n;
P_{k-1}(t)*P_{n-k}(t)/n, 1 < k < n.So n*P_n(t) =
(1+t)*P_{n-1}(t) + sum P_{k-1}(t)*P_{n-k}(t).If we multiply
this by z^{n-1} and sum, and let Q(t,z) = sum P_n(t)*z^n, we
get Q' - t = (1+t)*Q + Q^2,differentiating being on z. I have
not succeeded in getting the appropriate solution.The first
few generating functions are: 1 t 2 (t + t^2)/2! 3 (t + 4t^2 +
t^3)/3! 4 (t + 11t^2 + 11t^3 + t^4)/4! 5 (t + 26t^2 + 66t^3 +
26t^4 + t^5)/5!-- This address is for information only. I do
not claim that these viewsare those of the Statistics
Department or of Purdue University.Herman Rubin, Department of
Statistics, Purdue Universityhrubin@stat.purdue.edu Phone:
(765)494-6054 FAX: (765)494-0558
===
Subject: Complex numbers
and 2x2 matrixes.How long has the relation between complex
numbers and particular 2x2matrixes been known?Are there any
future plans to 'phase out' complex numbers and replacethem
with 2x2 matrixes which work exactly the same?(...Starblade
Riven Darksquall...)
===
Subject: Complex numbers and 2x2
matrices (was Re: Complex numbers and 2x2 matrixes)> How long
has the relation between complex numbers and particular 2x2>
matrixes been known?As long as matrices have been studied.>
Are there any future plans to 'phase out' complex numbers and
replace> them with 2x2 matrixes which work exactly the
same?No.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
times)
===
Subject: Re: Complex numbers and 2x2 matrices (was
Re: Complex numbers and 2x2 matrixes)Are there any future
plans to 'phase out' complex numbers and replacethem with 2x2
matrixes which work exactly the same?> No.How do you
know?
===
Subject: Re: Complex numbers and 2x2 matrices (was Re:
Complex numbers and 2x2 matrixes)>>Are there any future plans
to 'phase out' complex numbers and replace>>them with 2x2
matrixes which work exactly the same?>No.> How do you know?He
should NOT have answered. The secret committee for
thesupression of matrices will NOT be amused.
===
Subject: Re:
Complex numbers and 2x2 matrices (was Re: Complex numbers and
2x2 matrixes)> Are there any future plans to 'phase out'
complex numbers and replace> them with 2x2 matrixes which work
exactly the same?No.> How do you know?The real question is, are
there any plans to phase out the real numbers and replace them
with 1x1 matrices?
===
Subject: Re: Complex numbers and 2x2
matrices (was Re: Complex numbers and 2x2 matrixes)How long
has the relation between complex numbers and particular
2x2matrixes been known?Are there any future plans to 'phase
out' complex numbers and replacethem with 2x2 matrixes which
work exactly the same?Complex numbers are isomorph
2x2-rotation-matrices are isomorphordinary 2D-vectors (with a
certain multiplication added) -You calculatethem exactly the
same - and you can calculate them in mixed mode too:see the
mixed-mode calculator:http://i-z.eu.tt or
http://i-is-no-longer-imaginary.gmxhome.deThere is the
geometric vector-space (without coordinate-system) too,which
demonstrates, that you don't need this imaginarystuff any
longer,nor the Argand diagramm nor the imaginary axis, just
like CasparWessel started 2oo years ago without these.Have fun
Hero
===
Subject: Re: Equation Object converting?> You'll need to
write a macro using VBA.> Most, if not all of the info you
need, should be in one of the Word VBA> books I have listed in
the list of WordVBA books at my URL below.> I suggest getting
Steve Roman's Writing Word Macros.Thaqnks,MathType 5.2 is a
best solution in my case.
===
Subject: Re: Equation Object
converting?You'll need to write a macro using VBA.Most, if not
all of the info you need, should be in one of the Word VBAbooks
I have listed in the list of WordVBA books at my URL below.I
suggest getting Steve Roman's Writing Word Macros.--
http://www.standards.com/; See Howard Kaikow's web
site.Probably can do so with a quick macro.To what version of
Word are you converting?> Please, give me more details how to
do with a quick macro.
===
Subject: Cardinality of 2^n
numbers?Suppose we had the set of all integers which resembled
an integer 2^nwhere n is any integer. This would start from 1
in the case where n =0 to indefinately high. Let's call this
set Q.There are an infinite amount of such numbers since the
log of infinityis infinity. So the cardinality is aleph-x,
where x is an as yetundetermined variable.Now, by definition,
power sets are larger than the sets associatedwith them. So
the power set of Q would be equal in size to the set ofall
positive integers. Or all non negative integers, if you
includedthe null set.So while Q is not finite, its power set,
which must be larger than it,is supposedly the smallest
infinite set, accordding to a certaintheory which guarantees
that aleph notation is exact and there are no'middle numbers'
in between two consecutive alephs.Isn't this a
contradiction?(...Starblade Riven Darksquall...)
===
Subject:
Re: Cardinality of 2^n numbers?> Suppose we had the set of all
integers which resembled an integer 2^n> where n is any
integer. This would start from 1 in the case where n => 0 to
indefinately high. Let's call this set Q.No, let's call it K.
Q is taken already by the set of rational numbers.However, I
don't get your supposition. It appears that you're taking
allintegers that resemble an integer 2^n. What does it mean
for a numberto resemble an integer? I'll suppose for the
moment that you reallymeant to say all integers of the form
2^n. That set certainly exists.It's just the set of integers
2^n for n a non-negative integer: {1,2,4,8,16,32,...,2^n,
...}> There are an infinite amount of such numbers since the
log of infinity> is infinity. So the cardinality is aleph-x,
where x is an as yet> undetermined variable.Infinity is not a
number, and log(infinity) is undefined. Surely, thelimit
lim_{x -->infinity} log(x) fails to exist, because the
functiongrows without bound, and that fact is typically
written lim_{x -->infinity} log(x) = infinity.That does not
mean that one can substitute infinity for x in theexpression
log(x), and out pops the value infinity. The aboveexpression
for the limit is a *shorthand* expression for what Ibound as x
gets arbitrarily large.> Now, by definition, power sets are
larger than the sets associated> with them. So the power set
of Q would be equal in size to the set of> all positive
integers. Or all non negative integers, if you included> the
null set.No, it isn't *by definition* that power sets are
larger than the setsassociated with them, if you're meaning to
say that, for any set X,one has card(X) < card(2^X).This is a
theorem, proven (if I recall correctly) by Cantor. Theoremsare
hardly ever definitions, and this theorem is surely not
adefinition.> So while Q is not finite, its power set, which
must be larger than it,> is supposedly the smallest infinite
set, accordding to a certain> theory which guarantees that
aleph notation is exact and there are no> 'middle numbers' in
between two consecutive alephs.So, you have the set that I've
renamed K. True, it is not finite: by itsconstruction, K is
clearly of the same cardinality as N_o, the set ofnatural
numbers with zero.Next, you claim that its power set 2^K
(which must be of greatercardinality, according to the theorem
I mentioned relating X and 2^X), is supposedly the smallest
infinite set, accordding to a certain theory which guarantees
that aleph notation is exact and there are no 'middle numbers'
in between two consecutive alephs.Since the alephs are defined
as the successive cardinals, it is *this*that is a definition
(if I understand the definitions correctly).However, 2^K is
surely not the smallest infinite set. The existenceof the
alephs has nothing to do with 2^K, and it is certain that
theydo *not* show that 2^K is the smallest infinite set.>
Isn't this a contradiction?Here's what I think you're doing:
1. Take N_o, the natural numbers with zero, and form the set K
= {2^0, 2^1, 2^2, ..., 2^k ,... } of successive powers of 2. 2.
K is clearly of the same cardinality as N_o. 3. K is clearly
2^N_o. 4. card(N_o) < card(2^N_o), contradiction.Your error is
in step 3. The sequence of powers of 2 is not the same asthe
power set, appearances notwithstanding. The notation 2^X for
thepower set of X is (as with the limit expression I discussed
earlier)simply a shorthand for a particular construction. In
this case, itrefers to the set of functions from the set X to
the twolement set{0, 1}. That is, one can uniquely identify
any subset S of X with amap M_S from X to {0,1} by this
method: if x is in S, let M_S(x) = 1,otherwise let M_S(x) = 0:
M_S(x) = 1 if x is in S, otherwise M_S(x) = 0.It is simple to
verify that distinct subsets are associated to
distinctfunctions, every subset yields a function, and every
function comes froma subset.Simply taking 2^n for every
element n of some set of naturals does notyield the power set.
Note that the method fails already for finite sets.I'll take a
small set X, and form your construction (I'll call it Kagain):
X = {1,2,3} K = {2,4,8}.The power set 2^X is this: 2^X = { {},
{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3} }Note that the power set
2^X doesn't have any elements that areintegers (except {} for
folks who use the von Neumann constructionof the integers, in
which case {} is 0).> (...Starblade Riven Darksquall...)You
might understand some of this if you took an introductory
course in mathematics, or read an elementary set theory text
(I've heard good things about Halmos's Naive Set Theory,
currently in the Springer UTMseries).Dale.CAVEAT: I'm no set
theorist, nor do I play one on TV. There are plentyof regular
contributors to sci.math who are more adept at thisparticular
game than I am, and they may well have more to say on
thistopic.
===
Subject: Re: Cardinality of 2^n numbers?> Suppose
we had the set of all integers which resembled an integer 2^n>
where n is any integer. This would start from 1 in the case
where n => 0 to indefinately high. Let's call this set Q.>
There are an infinite amount of such numbers since the log of
infinity> is infinity. So the cardinality is aleph-x, where x
is an as yet> undetermined variable.> Now, by definition,
power sets are larger than the sets associated> with them. So
the power set of Q would be equal in size to the set of> all
positive integers. Or all non negative integers, if you
included> the null set.No. You are arguing that there is a
function that maps each subset of Q toa natural number. The
obvious mapping would be take the sum of the elementsof the
subset. You are probably thinking that this is the same as
thenatural numbers in base-2, but that is wrong. A natural
number must be*finite*, but a subset of Q can have an infinite
amount of elements. Forexample, the subset {q in Q: lg(q) is
odd} has a sum that is not a naturalnumber (lg x means log
base 2 of x). In fact there are uncountably manysubsets of Q
that are infinitely large. You are thinking of the Set
offinite subsets of Q, which is indeed countable and has a
very clearbijection with the naturals.~Steven> So while Q is
not finite, its power set, which must be larger than it,> is
supposedly the smallest infinite set, accordding to a certain>
theory which guarantees that aleph notation is exact and there
are no> 'middle numbers' in between two consecutive alephs.>
Isn't this a contradiction?> (...Starblade Riven
Darksquall...)
===
Subject: Re: Cardinality of 2^n numbers?
permission for an emailed response. The GNU project will
probably not be Posix conformant, Tom said noncommittally>
Suppose we had the set of all integers which resembled an
integer 2^n> where n is any integer. This would start from 1
in the case where n => 0 to indefinately high. Let's call this
set Q.Do you mean ordinary finite integers? This resembled
relation isunclear to me. I assume you mean:Q = {x in N: x =
2^n for some n in N}> There are an infinite amount of such
numbers since the log of infinity> is infinity. So the
cardinality is aleph-x, where x is an as yet> undetermined
variable.The log of infinity is a meaningless phrase. But Q is
indeedinfinite. x = 0. The construction of Q gives an injection
from N->Q, and theidentity mapping is an injection from Q->N.
So N and Q have the samecardinality, aleph-0.> Now, by
definition, power sets are larger than the sets associated>
with them. So the power set of Q would be equal in size to the
set of> all positive integers. Or all non negative integers, if
you included> the null set.You are confusing compute 2^n with
compute power set. 2^n is the*size of the power set*, not the
size of the *members* of the powerset.The power set of Q has
cardinality of the continuum.> So while Q is not finite, its
power set, which must be larger than it,> is supposedly the
smallest infinite set, accordding to a certain> theory which
guarantees that aleph notation is exact and there are no>
'middle numbers' in between two consecutive alephs.That
certain theory is the definition of the alephs.
===
Subject: Re:
Cardinality of 2^n numbers?> Suppose we had the set of all
integers which resembled an integer 2^n> where n is any
integer.resembled?> This would start from 1 in the case where
n => 0 to indefinately high. Let's call this set Q.Hmmm. Q is
usually reserved to mean the set of rational numbers.Do you
mean that Q = {2^n: n is a nonnegative integer?> There are an
infinite amount of such numbers since the log of infinity> is
infinity.Groan!!! the log of what?! :-(> So the cardinality is
aleph-x, where x is an as yet> undetermined variable.The
cardinaltity of my Q is aleph-zero.> Now, by definition, power
sets are larger than the sets associated> with them. So the
power set of Q would be equal in size to the set of> all
positive integers. Or all non negative integers, if you
included> the null set.size = cardinality? If so the the
power-set of Q (whatever it be)could not have the same
cardinality as N. As no power set has thesame cardinality of
N. (Why did you begin your 2nd sentence in thatparagraph with
so?)> So while Q is not finite, its power set, which must be
larger than it,> is supposedly the smallest infinite set,No.>
accordding to a certain> theory which guarantees that aleph
notation is exact and there are no> 'middle numbers' in
between two consecutive alephs.> Isn't this a
contradiction?No.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
times)
===
Subject: Re: naive geometry questions> Since I'm on
the topic, does anyone know how many dimensions you need> for
a Euclidean space in which you can hyperbolic space?That's a
very good question with a very complicated and, I
think,incomplete, answer. The hyperbolic plane embeds
isometrically intoR^5, but not into R^3. Before you ask
whether R^4 is enough, youhave to decide on things like the
degree of smoothness, embeddingversus immersion, and local
versus global. Here are some threads fromfour years ago:
http://www.math-atlas.org/99/embed_hyperdave
===
Subject: Re:
naive geometry questions> at 03:52 PM, Russell Blackadar
<russell@mdli.com> said:>Indeed, if we're talking only about
embeddings in 3-space, the torus>fails to be an example at
all.>Not an example of what? The OP didn't specify rotational
symmetry, and>it is certainly an example of a surface with a
one parameter family of>isometries.The OP wanted any other
surface S such that a piece of S can be moved around
adlibitum. Some of us interpreted adlibitum to mean that the
group ofself-isometries should act transitively, i.e. that it
contain a 2-parameterfamily of isometries. The torus does not
have that. If you interpret theOP's request merely to be for
the isometry group to be non-discrete, thenyes, clearly a
(regular) torus is an example.dave
===
Subject: How to measure
the linearity of a 3D field?Hi there,I am struggleing with
finding a way to measure the linearity of a 3D field.I have a
set of data with coordinates (x,y,z) and the field value on
each given point f(x,y,z). And we expect the field to be
linear so that the theoretical relationship would
bef(x,y,z)=a*x+b*y+c*z+dWe could do a least square fitting of
the given data to get the a,b,c,d parameters. But the problem
now is how linear the field is?One obvious way is to measure
the sum of residual squared, but suppose the data become
orders of magnitude larger, the residuals are orders of
magnitude larger. But obvously the linearity doesn't change in
this case.I know in curve fitting case, there is a correlation
coefficient, which is given by r^2=a1*a2. where two curve
fittings are done.y=a1*x+b1x=a2*y+b2The a1*a2 is between 0 and
1 and the closer it is to 1, the more linear the data are.This
is a very good measurement of linearity of one-dimensional
data. But I am having some trouble getting the definition of
similiar coefficient for higher dimensions.Could anybody help
me out here?Thank you very much.-- Shi
Jinsj88@cornell.eduhttp://jinshi.dhs.org
===
Subject: An
Approximation Related to a Formula of CusanusFollowing the
introduction, a particularly nice (and perhaps
new)approximation of the inverse sine is presented. That
approximationcan be used to give corresponding approximations
of the sixtrigonometric functions and the five other inverse
functions.------------------------------------
IntroductionNikolaus Cusanus (1401-1464, a.k.a. Nicholas of
Cusa) gave a remarkableformula: In a right triangle with sides
a < b < c, the smallest angle is a/(b + 2*c)*172 degrees.Of
course, this is just an approximation, although it seems that
Cusanusthought it to be exact. His formula is equivalent to
approximatingArcsin(x) by 43/45*Pi*x/(2 + Sqrt(1-x^2)) for 0
<= x <= 1/Sqrt(2).Noting that 43/45*Pi is only slightly larger
than 3, a closely relatedapproximation is 3*x/(2 +
Sqrt(1-x^2)). To approximate Arcsin(x) for1/Sqrt(2) < x <= 1,
using the fact that sin^2(t) + cos^2(t) = 1, we mayreplace x
by Sqrt(1-x^2), and vice versa, and then take the
complement(that is, subtract that result from Pi/2).
Altogether, we obtain ( 3*x/(2 + Sqrt(1-x^2)) for 0 <= x <=
1/Sqrt(2),(1) ( ( Pi/2 - 3*Sqrt(1-x^2)/(2 + x) for 1/Sqrt(2) <
x <= 1as an approximation of Arcsin(x), with |rel. error| <
0.0023 .Of course, due to simple interrelations between the
inverse trigonometricfunctions, one could easily obtain
corresponding approximations forthe other five inverse
functions from (1).Furthermore, due to the algebraic
simplicity of form in (1), it can beinverted easily, giving (
x*(6 + Sqrt(9-3*x^2))/(9 + x^2) for -Pi/4 <= x <= Pi/4(2) ( (
(18 + 3*Sqrt(9-3*(Pi/2-x)^2))/(9 + (Pi/2-x)^2) - 2 for Pi/4 <
x <= 3*Pi/4as an approximation for sin(x), with |rel. error| <
0.0019 . Due to simpleinterrelations between the trigonometric
functions, one could easilyobtain corresponding approximations
for the other five functions from (2).(This ends the
introduction, in which I presume that there is reallynothing
new.)------------------------------------We now primarily
consider approximating Arcsin(x) for 0 <= x <=
1/Sqrt(2),knowing that, as indicated above, corresponding
approximations can beobtained easily for Arcsin(x) for
1/Sqrt(2) < x <= 1, and ultimately forall of the trigonometric
functions and their inverses.The first part of (1) may be
thought of as being in the form(3) x/(1 - h*(1 -
Sqrt(1-x^2)))with h = 1/3. The question then arises: Can the
approximation given by (3)be improved by choosing a different
value of h? Answer: Yes. My favoritechoice is h = (Sqrt(2) +
1)*(Sqrt(2) - 4/Pi) = 0.340341385...Using it, we obtain an
expression giving Arcsin(x) precisely at both 0 and1/Sqrt(2)
and overestimating it for 0 < x < 1/Sqrt(2), withrel. error <
0.00057 . Unfortunately, the improvement in accuracy is
notdramatic. It might also be noted that this approximation is
differentiableat x = 1/Sqrt(2); by contrast, (1) has a jump
discontinuity there.But could the form of (3) be altered
slightly -- say, by introducinganother parameter -- so as to
get much better accuracy? Yes! Choosing theform(4) x/(1 - h*(1
- Sqrt(1-(k*x)^2)))allows relative error to be reduced
substantially, to approximately 1/50of that given by (1). The
parameters h and k were determined so that (4)would give
Arcsin(x) precisely at 1/2 and 1/Sqrt(2). [Slightly
differentvalues of h and k would surely reduce error even a
bit more, but I thinkit's neat to use the two criteria already
mentioned.] Although the exactexpressions for h and k are
indeed messy, namely
(3+2*Sqrt(2))*(2*(3-Sqrt(2))-Pi/2-5/Pi)*(6*Sqrt(2)+Pi*(3-2*
Sqrt(2))-9)h =
--------------------------------------------------------------
-------- (Pi-3)*(6-Pi-2*Sqrt(2))and
Sqrt((Pi-3)*(2*(3*Pi-6*Sqrt(2)+4)-Pi^2))k =
----------------------------------------- ,
(1+Sqrt(2))*(Pi*(3-Sqrt(2))-(Pi/2)^2-5/2)we may of course just
use approximations of these parameters:h = 0.30777349... and k
= 1.0419890...As previously noted, we correspondingly
approximate Arcsin(x) for1/Sqrt(2) < x <= 1 by taking the
complement of (4) once x has been replacedby Sqrt(1-x^2).
Taken together then, we get an approximation of Arcsin(x)for 0
<= x <= 1 with |rel. error| < 0.000046 . Using exact values for
h andk, the approximation gives the exact values of Arcsin(x)
for x = 0, 1/2,1/Sqrt(2), Sqrt(3)/2, and 1, and is
differentiable on [0,1).Similar comments could be made
concerning the corresponding approximationsfor the
trigonometric functions and for the five other inverse
trigonometricfunctions.-------------------------------------
How does this approximation of Arcsin(x) compare with other
approximations?Of course there are far more accurate
approximations. Some of them arealso related to Cusanus's
formula; see Lou Talman's Some (Almost) RationalThoughts
<http://clem.mscd.edu/~talmanl/Almost/Rational.html>,
forexample. However, AFAIK, they are of more complicated
forms, thereby makingtheir algebraic inversion more difficult
-- in fact, impossible in mostcases.Abramowitz and Stegun
mention a four-parameter approximation, their item4.4.45 (see
<http://jove.prohosting.com/~skripty/page_81.htm>)
whichprovides accuracy almost identical to our two-parameter
approximation.That polynomial approximation has the form Pi/2
- Sqrt(1-x)*(a_0 + a_1*x + a_2*x^2 + a_3*x^3)in which the
values of the coefficients seem to have been
determinednumerically.------------------------------------Your
thoughtful comments are welcome.David W. Cantrell
===
Subject:
How do you prove that the circumference of a circle is
proportional to it's radius?How do you prove that the
circumference of a circle is proportional to it's
radius?What's the modern version and how did the greeks prove
it?/david
===
Subject: Re: How do you prove that the
circumference of a circle is proportional to it's radius?> How
do you prove that the circumference of a circle is proportional
to > it's radius?> What's the modern version and how did the
greeks prove it?> /davidThe approximations to that ratio are
based on inscribing/circumscribing regular polygons, each of
which may be partitioned into congruent isosceles
triangles.For any such polygon, the odd(circuferential) side
of any of those isosceles triangles is proportional to the
other (radial) sides by similar triangles.Thus each
approximate circumference is proportional to its
radius.
===
Subject: Re: How do you prove that the circumference
of a circle is proportional to it's radius?How do you prove
that the circumference of a circle is proportional to it's
radius?What's the modern version and how did the greeks prove
it?> The approximations to that ratio are based on >
inscribing/circumscribing regular polygons, each of which may
be > partitioned into congruent isosceles triangles.> For any
such polygon, the odd(circuferential) side of any of those >
isosceles triangles is proportional to the other (radial)
sides by > similar triangles.> Thus each approximate
circumference is proportional to its radius.Now if you
know/accept that perimeter of inscribed polygon <
circumference of circle < perimeter of circumscribed polygon,
then it's easy to complete the proof. Now the first
inequality, perimeter of inscribed polygon < circumference of
circle, is clear; a straight line is the shortest distance
between two points. The other inequality, circumference of
circle < perimeter of circumscribed polygon, is certainly
plausible, but is it actually possible to prove it using only
tools available to the ancients?-- Gerry Myerson
(gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Find all
primes of the form 4^n + n^4A problem:* Find all primes of the
form 4^n + n^I just can't do it!4^1 + 1^4 = 5, is prime* If 2
divides n, 2 divides 4^n + n^4 so it can't be prime.* If 2
does noes not divide n and 5 doesn't either 5 divides 4^n +
n^4 so it isn't prime. (4^n always ends in 4 for odd n and n^4
always ends in 1 for odd n that aren't divisible by 5. Thus 4^n
+ n^4 always ends in 5 and is divisible by 5. If anyone has a
more elegant way of proving this, please let me know.)* This
leaves us odd n that are multipiles of 5. I suspect these are
all compsite too. 4^5 + 5^4 = 1649 = 17*97, at least. But how
do you prove it?/david
===
Subject: Re: Find all primes of the
form 4^n + n^4> A problem:> * Find all primes of the form 4^n
+ n^Borrowing one of Einstein's ideas, we have4^N+N^4 =
(2^N+N^2)^2 - 2^(N+1)N^2,and the rest is relatively
easy.
===
Subject: Re: Find all primes of the form 4^n +
n^4david <sad@asf.com> escribi.97 en el mensaje> A problem:> *
Find all primes of the form 4^n + n^> I just can't do it!> 4^1
+ 1^4 = 5, is prime> * If 2 divides n, 2 divides 4^n + n^4 so
it can't be prime.> * If 2 does noes not divide n and 5
doesn't either 5 divides 4^n + n^4> so it isn't prime. (4^n
always ends in 4 for odd n and n^4 always ends> in 1 for odd n
that aren't divisible by 5. Thus 4^n + n^4 always ends in> 5
and is divisible by 5. If anyone has a more elegant way of
proving> this, please let me know.)> * This leaves us odd n
that are multipiles of 5. I suspect these are all> compsite
too. 4^5 + 5^4 = 1649 = 17*97, at least. But how do you prove
it?Fon even n is obvious. For odd n, replace n = 2k+1.--

===
Subject: Re: Find all primes of the form 4^n + n^4> Fon
even n is obvious. For odd n, replace n = 2k+1.For odd n such
that 5 does not divide n, this is obvious too ... but
whatabout the case n = 5*i, where i is odd?
===
Subject: Re:
Find all primes of the form 4^n + n^4Julien Santini
<santini.julien@wanadoo.fr> escribi.97 en el mensajeFon even n
is obvious. For odd n, replace n = 2k+1.> For odd n such that 5
does not divide n, this is obvious too ... but what> about the
case n = 5*i, where i is odd?For n = 2k + 1,n^4 + 4^n = (n^2 +
2^(k + 1)n + 2^n)(n^2 - 2^(k + 1)n + 2^n)-- 
===
Subject: Re:
Find all primes of the form 4^n + n^4> For n = 2k + 1,> n^4 +
4^n = (n^2 + 2^(k + 1)n + 2^n)(n^2 - 2^(k + 1)n + 2^n)Magical
!!Julien Santini
===
Subject: Re: Find all primes of the form
4^n + n^4>Fon even n is obvious. For odd n, replace n = 2k+1.>
For odd n such that 5 does not divide n, this is obvious too
... but what> about the case n = 5*i, where i is odd?Forget
the divisibility by 5, treat all the odd n the same. (you
might benefit from writing one of the 4s
differently)
===
Subject: Re: Find all primes of the form 4^n +
n^4> A problem:> * Find all primes of the form 4^n + n^> I
just can't do it!> 4^1 + 1^4 = 5, is prime> * If 2 divides n,
2 divides 4^n + n^4 so it can't be prime.> * If 2 does noes
not divide n and 5 doesn't either 5 divides 4^n + n^4> so it
isn't prime. (4^n always ends in 4 for odd n and n^4 always
ends> in 1 for odd n that aren't divisible by 5. Thus 4^n +
n^4 always ends in> 5 and is divisible by 5. If anyone has a
more elegant way of proving> this, please let me know.)> *
This leaves us odd n that are multipiles of 5. I suspect these
are all> compsite too. 4^5 + 5^4 = 1649 = 17*97, at least. But
how do you prove it?I set this as a challenge problem in my
number theory
course:http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html
.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to
say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14
times)
===
Subject: Re: Find all primes of the form 4^n + n^4> I
set this as a challenge problem in my number theory course:>
http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html .But you
didn't provide a proof in the answer sheet! =(
===
Subject: Re:
Find all primes of the form 4^n + n^4> I set this as a
challenge problem in my number theory course:>
http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html .> But
you didn't provide a proof in the answer sheet! =(A short
hint. Complete something.
===
Subject: Re: Find all primes of
the form 4^n + n^4>> I set this as a challenge problem in my
number theory course:>>
http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html .> But
you didn't provide a proof in the answer sheet! =(> A short
hint. Complete something.The proof?
===
Subject: Re: Find all
primes of the form 4^n + n^4>> I set this as a challenge
problem in my number theory course:>
http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html . But
you didn't provide a proof in the answer sheet! =(> A short
hint. Complete something.> The proof?Something else
first.
===
Subject: Re: Find all primes of the form 4^n + n^4> >
I set this as a challenge problem in my number theory course:>
http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html .> > But
you didn't provide a proof in the answer sheet! =(Heh, heh,
heh!-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless
to say, I had the last laugh. Alan Partridge, _Bouncing Back_
(14 times)
===
Subject: Re: Help teaching stat> 11/4/03> Dear
teaching colleague,> I could sure use your help. I was just
asked to speak next January at> the National Institute for the
Teaching of Psychology and would like> to ask you for a few
minutes of your time. My presentation at NITOP is> titled,
What's difficult to teach in introductory statistics and how>
to do it. Can you please take ten minutes and help me prepare
for> this presentation by answering the following questions?>
1. In your intro stat class, what three topics do you find
most> difficult to teach your students?> 2. How do you teach
these topics? What techniques, strategies, ideas,> and tools
help you?> 3. What's the one coolest, most fun thing you do in
your intro stat> class to help students learn?Let me partially
answer your questions with an answer to the last one,
Neil.When I teach (HS) introductory statistics, I want the
students to realizethe power of taking a sample to find a
population. I pass out to the class abunch of mini-bags of
M&Ms, and ask them to predict how many red M&Ms are inthe
room. I write their prediction on the board.Then, I have ONE
student open their bag. If they have, say 5 red M&Ms, andthe
class has 20 kids, all the predictions range close to 100
total, with abit of variance (which we can calculate) because
some students astutelypredict that there may be bags out there
loaded with red ones. Then anotherstudent opens their bag, and
the predictions become less variant.In a class of 20 students,
they all start feeling quite certain of theirpredictions after
only 4 or 5 bags being opened. It gives them an excellentsense
of how the number of samples is related to the degree of
certainty ofthe predictions. We graph the range of predictions
vs. the number of openbags, and they see how their variance and
predictions became very stableafter very few samples. Also,
they realize how little an outlier bag affectsthe sum total.
And they get a good understanding of how unusual it would
befor that outlier bag to be one of the first bags sampled,
but how it ispossible.Overall, an excellent intro to sampling.
And then we eat the M&Ms.--riverman
===
Subject: Help teaching
stat11/4/03Dear teaching colleague,I could sure use your help.
I was just asked to speak next January atthe National Institute
for the Teaching of Psychology and would liketo ask you for a
few minutes of your time. My presentation at NITOP istitled,
What's difficult to teach in introductory statistics and howto
do it. Can you please take ten minutes and help me prepare
forthis presentation by answering the following questions?1.
In your intro stat class, what three topics do you find
mostdifficult to teach your students?2. How do you teach these
topics? What techniques, strategies, ideas,and tools help
you?3. What's the one coolest, most fun thing you do in your
intro statclass to help students learn?Your help is
tremendously appreciated and I will send you a copy ofthe
final presentation which should be ready sometime in January.
Canyou please send your responses via email to njs@ku.edu and
pleaseplace the words intro stat in the subject line? Please
be sure toinclude your email address so I may send you a copy
of the finalpresentation.Also, can you please forward this
same letter to two other colleaguesof yours that teach intro
stat?Sincerely,Neil Salkind, Ph.D.University of
KansasLawrence, KS 66045
===
Subject: Trying again!!! PDE book
recommendation,Could someone recommend a good Introduction to
P.D.E.'s book ? I have heardthatBasic Partial Differential
equations by David Bleecker, George Csordas, andDarko Grundler
book is nice, but I haven't had the chance to take a look atit.
Any suggestions would be appreciated!TIALurch
===
Subject: Re:
Trying again!!! PDE book recommendation> ,> Could someone
recommend a good Introduction to P.D.E.'s book ? I have heard>
that> Basic Partial Differential equations by David Bleecker,
George Csordas, and> Darko Grundler book is nice, but I
haven't had the chance to take a look at> it. Any suggestions
would be appreciated!> TIA> LurchThis is not a direct answer
but can't you search onPartial differential equations lecture
notesand then make some kind of judgment about the book,
derived from whatthe professor does in his course. Just a
thought, of course.David Ames
===
Subject: Re: Trying again!!!
PDE book recommendation> ,> Could someone recommend a good
Introduction to P.D.E.'s book ? I have heard> that> Basic
Partial Differential equations by David Bleecker, George
Csordas, and> Darko Grundler book is nice, but I haven't had
the chance to take a look at> it. Any suggestions would be
appreciated!> TIA> Lurch1. Partial Differential Equations: An
Introduction by Walter A. Strauss2. Partial Differential
Equations 4e by Fritz John
===
Subject: Re: Trying again!!! PDE
book recommendation,Could someone recommend a good
Introduction to P.D.E.'s book ? I haveheardthatBasic Partial
Differential equations by David Bleecker, George
Csordas,andDarko Grundler book is nice, but I haven't had the
chance to take a lookatit. Any suggestions would be
appreciated!TIALurch> 1. Partial Differential Equations: An
Introduction by Walter A. Strauss> 2. Partial Differential
Equations 4e by Fritz John
===
Subject: Re: My research,
publication announcement> There's more to my work than just
arguing on Usenet, so I'd like to> point out that my paper
Advanced Polynomial Factorization is slated> to be published:>
See http://www.megasociety.net/NoesisHighlights.html> The Mega
Foundation is an organization of high IQ people, and I'm glad>
to be associated with them. To learn further about the
organization> you can use Google, or see:Actually, it's an
orgainization of people dumb enough to paythe $25 subscription
fee. As far as I can tell, once you'vedone that, their rigorous
application process involves selectingwhether your IQ is 150 or
150-180+ (and sending the recommendeddonation, of
course)http://www.megafoundation.org/Ultranet/
UltranetApp.htmlStrangely, they seem much more concerned about
verifying yourpayment than your IQ. Imagine that!
===
Subject:
Re: My research, publication announcementThere's more to my
work than just arguing on Usenet, so I'd like topoint out that
my paper Advanced Polynomial Factorization is slatedto be
published:See
http://www.megasociety.net/NoesisHighlights.htmlThe Mega
Foundation is an organization of high IQ people, and I'm
gladto be associated with them. To learn further about the
organizationyou can use Google, or see:Why a group like the
Mega Foundation? http://www.ultrahiq.org/Mega/WhyMega.htmI
hope at least some of you will appreciate that often the
mostimportant ideas in history have to get past people limited
by theirlack of imagination and their prejudices, who act
against scientificprogress.What I want you to see is that
there's more to me than Usenet, so thatyou can begin to
understand that the revolution I'm giving you achance to be a
part of is bigger than the small-minded people whocontinually
throughout history work to halt progress.Thank you for your
time and
attention.http://mathforprofit.blogspot.com/
===
Subject: Re: My
research, publication announcementThere's more to my work than
just arguing on Usenet, so I'd like topoint out that my paper
Advanced Polynomial Factorization is slatedto be published:See
http://www.megasociety.net/NoesisHighlights.htmlThe Mega
Foundation is an organization of high IQ people, and I'm
gladto be associated with them. To learn further about the
organizationyou can use Google, or see:Why a group like the
Mega Foundation? http://www.ultrahiq.org/Mega/WhyMega.htmI
hope at least some of you will appreciate that often the
mostimportant ideas in history have to get past people limited
by theirlack of imagination and their prejudices, who act
against scientificprogress.What I want you to see is that
there's more to me than Usenet, so thatyou can begin to
understand that the revolution I'm giving you achance to be a
part of is bigger than the small-minded people whocontinually
throughout history work to halt progress.Thank you for your
time and
attention.http://mathforprofit.blogspot.com/
===
Subject: Re: My
research, publication announcement> There's more to my work
than just arguing on Usenet, so I'd like to> point out that my
paper Advanced Polynomial Factorization is slated> to be
published:> See
http://www.megasociety.net/NoesisHighlights.html> The Mega
Foundation is an organization of high IQ people, and I'm glad>
to be associated with them. To learn further about the
organization> you can use Google, or see:> Why a group like
the Mega Foundation? >
http://www.ultrahiq.org/Mega/WhyMega.htm> I hope at least some
of you will appreciate that often the most> important ideas in
history have to get past people limited by their> lack of
imagination and their prejudices, who act against scientific>
progress.> What I want you to see is that there's more to me
than Usenet, so that> you can begin to understand that the
revolution I'm giving you a> chance to be a part of is bigger
than the small-minded people who> continually throughout
history work to halt progress.> Thank you for your time and
attention.> James Harris>
http://mathforprofit.blogspot.com/Dammit, he's back on top
again. Somoeone will have to reply to pushhim back down
again.
===
Subject: Re: My research, publication
announcement>There's more to my work than just arguing on
Usenet, so I'd like to>point out that my paper Advanced
Polynomial Factorization is slated>to be published:Does the
Usenet Posting Guide you're working on say anything abouthow
to keep your browser from posting multiple copies of the same
post?Doug
===
Subject: Re: My research, publication
announcement>There's more to my work than just arguing on
Usenet, so I'd like to>point out that my paper Advanced
Polynomial Factorization is slated>to be published:>Does the
Usenet Posting Guide you're working on say anything about>how
to keep your browser from posting multiple copies of the same
post?These reposts are intentional. Cf his comments on the
guide to postingon usenet that he was considering
writing<giggle> and things he'ssaid in the past: He thinks
that the best way to read usenet isthrough Google. But the
problem with Google is it displays the most recent post in the
hit list, so he has to keep reposting toprevent his message
from being hidden by replies...Honest.>Doug
===
Subject: Re: My
research, publication announcementThere's more to my work than
just arguing on Usenet> Yes, but that's your crowning
achievement.> , so I'd like topoint out that my paper Advanced
Polynomial Factorization is slatedto be published:See
http://www.megasociety.net/NoesisHighlights.html> How nice
that the cranks have gotten together and put up a website.
actually, it appears that the range of contributors include
non-cranks as well.The Mega Foundation is an organization of
high IQ people, and I'm gladto be associated with them. To
learn further about the organizationyou can use Google, or
see:Why a group like the Mega Foundation?
http://www.ultrahiq.org/Mega/WhyMega.htmI hope at least some
of you will appreciate that often the mostimportant ideas in
history have to get past people limited by theirlack of
imagination and their prejudices, who act against
scientificprogress.> Not any more, now that we have the
Internet . . .What I want you to see is that there's more to
me than Usenet, so thatyou can begin to understand that the
revolution I'm giving you achance to be a part of is bigger
than the small-minded people whocontinually throughout history
work to halt progress.> Is there a publication party? Free
drinks? Are we all invited?
===
Subject: Re: My research,
publication announcementThere's more to my work than just
arguing on Usenet, so I'd like topoint out that my paper
Advanced Polynomial Factorization is slatedto be published:See
http://www.megasociety.net/NoesisHighlights.htmlThe Mega
Foundation is an organization of high IQ people, and I'm
gladto be associated with them. To learn further about the
organizationyou can use Google, or see:Why a group like the
Mega Foundation? http://www.ultrahiq.org/Mega/WhyMega.htmI
hope at least some of you will appreciate that often the
mostimportant ideas in history have to get past people limited
by theirlack of imagination and their prejudices, who act
against scientificprogress.What I want you to see is that
there's more to me than Usenet, so thatyou can begin to
understand that the revolution I'm giving you achance to be a
part of is bigger than the small-minded people whocontinually
throughout history work to halt progress.Thank you for your
time and
attention.http://mathforprofit.blogspot.com/
===
Subject: Re: My
research, publication announcement>There's more to my work than
just arguing on Usenet, so I'd like to>point out that my paper
Advanced Polynomial Factorization is slated>to be
published:>See
http://www.megasociety.net/NoesisHighlights.html>The Mega
Foundation is an organization of high IQ people, and I'm
glad>to be associated with them. They are not, however, the
Mega Society, though the URL above includesthat name. Here is
the Mega Society's
page:http://www.polymath-systems.com/intel/hiqsocs/megasoc/
megasoc.htmlThere is a long and bloody battle about Chris
Langan's attempts totake over usage of the terms Mega Society
and Noesis Journal. One sidecan be found in such places
ashttp://www.polymath-systems.com/intel/hiqsocs/megasoc/
noes152/publish.htmlhttp://www.polymath-systems.com/intel/
hiqsocs/megasoc/noes151/cease&de.html - Randy
===
Subject: Re:
My research, publication announcementIn sci.physics, James
Harris<jstevh@msn.com><3c65f87.0311090900.61517471@
posting.google.com>:> There's more to my work than just
arguing on Usenet, so I'd like to> point out that my paper
Advanced Polynomial Factorization is slated> to be published:>
See http://www.megasociety.net/NoesisHighlights.htmlYour work
is interesting, but highly flawed. I for onecan pick out some
of the flaws without too much trouble;others have done more
detailed analysis. At best, it appearsthat some minor
corrections are needed; at worst, major surgeryis the only
option.I am not sure how your magic polynomial fits into
yourlarger proof, either.[rest snipped]-- #191,
ewill3@earthlink.netIt's still legal to go
.sigless.
===
Subject: Re: My research, publication
announcement> There's more to my work than just arguing on
Usenet, so I'd like to> point out that my paper Advanced
Polynomial Factorization is slated> to be published:> See
http://www.megasociety.net/NoesisHighlights.html> The Mega
Foundation is an organization of high IQ people, and I'm glad>
to be associated with them. To learn further about the
organization> you can use Google, or see:as UltraHIQ
athttp://www.crank.net/megalomaniacs.html> Why a group like
the Mega Foundation?>
http://www.ultrahiq.org/Mega/WhyMega.htm> I hope at least some
of you will appreciate that often the most> important ideas in
history have to get past people limited by their> lack of
imagination and their prejudices, who act against scientific>
progress.> What I want you to see is that there's more to me
than Usenet, so that> you can begin to understand that the
revolution I'm giving you a> chance to be a part of is bigger
than the small-minded people who> continually throughout
history work to halt progress.> Thank you for your time and
attention.> James Harris>
http://mathforprofit.blogspot.com/
===
Subject: Re: My research,
publication announcement>There's more to my work than just
arguing on Usenet, so I'd like to>point out that my paper
Advanced Polynomial Factorization is slated>to be
published:>See
http://www.megasociety.net/NoesisHighlights.htmlDarn. Looking
at the title of the thread I assumed that your work
wasactually going to be _published_. In math published
meanspublished in a peer-reviewed journal, not published on a
vanityweb site dedicated to 'severely gifted' individuals, who
are allowedto publish all they want, for the proper fee.>The
Mega Foundation is an organization of high IQ people, and I'm
glad>to be associated with them. To learn further about the
organization>you can use Google, or see:>Why a group like the
Mega Foundation? >http://www.ultrahiq.org/Mega/WhyMega.htm>I
hope at least some of you will appreciate that often the
most>important ideas in history have to get past people
limited by their>lack of imagination and their prejudices, who
act against scientific>progress.>What I want you to see is that
there's more to me than Usenet, so that>you can begin to
understand that the revolution I'm giving you a>chance to be a
part of is bigger than the small-minded people who>continually
throughout history work to halt progress.>Thank you for your
time and attention.>James
Harris>http://mathforprofit.blogspot.com/
===
Subject: Re: My
research, publication announcement>There's more to my work
than just arguing on Usenet, so I'd like to>point out that my
paper Advanced Polynomial Factorization is slated>to be
published:>See
http://www.megasociety.net/NoesisHighlights.html> Darn.
Looking at the title of the thread I assumed that your work
was> actually going to be _published_. In math published
means> published in a peer-reviewed journal, not published on
a vanity> web site dedicated to 'severely gifted' individuals,
who are allowed> to publish all they want, for the proper
fee.made all the mistakes which can be made, in very narrow
fields. (NeilsBohr)>The Mega Foundation is an organization of
high IQ people, and I'm glad>to be associated with them. To
learn further about the organization>you can use Google, or
see:>Why a group like the Mega Foundation?
>http://www.ultrahiq.org/Mega/WhyMega.htm>I hope at least some
of you will appreciate that often the most>important ideas in
history have to get past people limited by their>lack of
imagination and their prejudices, who act against
scientific>progress.>What I want you to see is that there's
more to me than Usenet, so that>you can begin to understand
that the revolution I'm giving you a>chance to be a part of is
bigger than the small-minded people who>continually throughout
history work to halt progress.>Thank you for your time and
attention.>James Harris>http://mathforprofit.blogspot.com/>

===
Subject: Re: My research, publication announcementit
appears to be a vanity press, which is just fine,if you
actually want to have copies of your publicationfor sale or
dystribution. if it just goesto the members, that's another
matter ... so,it's not really a vanity press. [NB:that he used
Vantage Press to publish, asit was during the
Depression.]published in a peer-reviewed journal, not
published on a vanityweb site dedicated to 'severely gifted'
individuals, who are allowedto publish all they want, for the
proper fee.> made all the mistakes which can be made, in very
narrow fields. (Neils>http://mathforprofit.blogspot.com/--ils
dcues d'Enron!http://larouchepub.com/
===
Subject: Re: My
research, publication announcementIn sci.physics, James
Harris<jstevh@msn.com><3c65f87.0311081331.1059e80d@
posting.google.com>:> There's more to my work than just
arguing on Usenet, so I'd like to> point out that my paper
Advanced Polynomial Factorization is slated> to be published:>
See http://www.megasociety.net/NoesisHighlights.htmlYe gods,
what horrid typesetting. You'd think that someoneover there
might have investigated CSS1 or CSS2 style sheets?
:-Phttp://www.w3.org> The Mega Foundation is an organization
of high IQ people, and I'm glad> to be associated with them.
To learn further about the organization> you can use Google,
or see:> Why a group like the Mega Foundation? >
http://www.ultrahiq.org/Mega/WhyMega.htm> I hope at least some
of you will appreciate that often the most> important ideas in
history have to get past people limited by their> lack of
imagination and their prejudices, who act against scientific>
progress.> What I want you to see is that there's more to me
than Usenet, so that> you can begin to understand that the
revolution I'm giving you a> chance to be a part of is bigger
than the small-minded people who> continually throughout
history work to halt progress.> Thank you for your time and
attention.> James Harris>
http://mathforprofit.blogspot.com/Oooh. Well, I'm going to
have to attack this. Sorry James, butthis has the same
problems as your other proofs.1. Let P(x) = 5^3*7^6*x^3 -
3*5^3*7^4*x^2 - 2^3*3^2*5*7^2*x - 2*7^2*11 .2. P(x) =
7^2*(5^3*7^4*x^3 - 3*5^3*7^2*x^2 - 2^3*3^2*5*x - 2*11). P(x) =
(5 * a_1(x) + 7) * (5 * a_2(x) + 7) * (5 * a_3(x) + 7)The
astute reader may notice that 7^3 = 343 cannot be the lastterm
so at least one of the a_i(x) must contain a constant addendof
some sort. I will ignore this bizarre rewrite and instead
notethat P(x) = (5 * b_1(x) + 7) * (5 * b_2(x) + 7) * (5 *
b_3(x) - 22)probably works better, for various reasons. This
is the formyou eventually get to in step 4, by different
methods.We can now attempt to work out the forms for the b_i.
We alreadyknow the following:[1] The product b_1(x) * b_2(x) *
b_3(x) must contain an x^3 term. In fact, this x^3 term must be
7^6 * x^3.[2] If we assume P(x) = 7^6 * 5^3 * (x - x_1) * (x -
x_2) * (x - x_3) for some algebraic numbers x_1, x_2, and x_3,
then we can conclude that, for some permutation of x_i, b_1(x)
= -7*x/x_1 b_2(x) = -7*x/x_2 b_3(x) = 22*x/x_3 is a valid
solution. However, other solutions are possible, such as the
rather trivial b_1(x) = P(x)/5 - 7/5 b_2(x) = -6/5 b_3(x) =
23/5 or something like b_1(x) = -14*x/x_1 - 7/10 b_2(x) =
-7*x/x_1 b_3(x) = -11*x/x_1 + 11/5 In light of this
observation I cannot draw any other conclusions at this time
without additional constraints.[3] If we assume, as you did in
an earlier proof attempt, that P(x) = (5 * c_1 * x + 7) * (5 *
c_2 *x + 7) * (5 * c_3 * x - 22) where the c_i do not depend
on x, then we can indeed state that, for some permutation of
the x_i, c_1 = -7/x_1 c_2 = -7/x_2 c_3 = 22/x_3 However, are
any of these algebraic integers? At this point I can state
that, if the x_i satisfy P(x)/49 = (5^3*7^4*x^3 -
3*5^3*7^2*x^2 - 2^3*3^2*5*x - 2*11) = 0 then it must be the
case that the values y_i = 1/x_i satisfy P(1/y)*y^3/49 =
2*11*y^3 + 2^3*3^2*5*y^2 + 3*5^3*7^2*y - 5^3*7^4 If we further
assume y = z/7, then z = 7 * y = 7/x, and these satisfy the
equation 2*11*z^3 + 2^3*3^2*5*7*z^2 + 3*5^3*7^4*z - 5^3*7^7 =
0 Substitute w = -z and we simply flip signs: 2*11*w^3 -
2^3*3^2*5*7*w^2 + 3*5^3*7^4*w + 5^3*7^7 = 0 This polynomial is
irreducible. If one looks at v = 22/x, then x = 22/v and we get
-v^3 * P(22/v)/49 = 2*11*v^3 + 2^4*3^2*5*11*v^2 +
2^2*3*5*7^2*11^2 + 2^3*5^3*7^4*11^3 We note the common factor
2*11 and divide again, getting: -v^3 * P(22/v)/(2*7^2*11) =
v^3 + 2^3*3^2*5*v^2 + 2*3*5*7^2*11 + 2^2*5^3*7^4*11^2so yes,
c_3 is an algebraic integer -- but that's the only one that
can be.[4] This is not an attack on your person, but on your
proof. However, if you continue to call those in the
mathematical establishment idiots, why are you so surprised at
the shipments of highxplosive vitriol coming on your doorstep?
(This is assuming there *is* a Mathematical Establishment,
admittedly; presumably, there is a wellstablished procedure
for peer review of mathematical results, which is about as
close to an establishment as I personally can see at this time
-- since I've not attempted to publish peer-reviewed papers in
scientific journals I'm unlikely to know the details of this
process.) Concentrate on the math. :-)-- #191,
ewill3@earthlink.netIt's still legal to go
.sigless.
===
Subject: Re: My research, publication
announcementSee
http://www.megasociety.net/NoesisHighlights.html> Ye gods,
what horrid typesetting. Not only that; I note that they carry
a book review by William Dembski,one of the leading lights of
the ID movement; people who think thatevolution can not
account for life on earth and that there has to havebeen an
Intelligent Designer.You're in good company, James.V.-- email:
lastname at cs utk eduhomepage: cs utk edu tilde
lastname
===
Subject: Re: My research, publication
announcementIn sci.physics, Victor
Eijkhout<see.sig@for.addy><1g45l67.14bvddq1b6pkjbN%see.sig@
for.addy>:> See
http://www.megasociety.net/NoesisHighlights.html> > Ye gods,
what horrid typesetting. > Not only that; I note that they
carry a book review by William Dembski,> one of the leading
lights of the ID movement; people who think that> evolution
can not account for life on earth and that there has to have>
been an Intelligent Designer.> You're in good company, James.>
V.FSWVO good. :-)-- #191, ewill3@earthlink.net -- insert random
weird value hereIt's still legal to go .sigless.
===
Subject: Re:
My research, publication announcement > There's more to my work
than just arguing on Usenet, so I'd like to > point out that my
paper Advanced Polynomial Factorization is slated > to be
published: > > See
http://www.megasociety.net/NoesisHighlights.html > > The Mega
Foundation is an organization of high IQ people, and I'm glad
> to be associated with them. To learn further about the
organization > you can use Google, or see: > > Why a group
like the Mega Foundation? >
http://www.ultrahiq.org/Mega/WhyMega.htmWhat does it cost to
become a member? What does it cost to get a paperpublished?--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland, +31205924131home: bovenover 215, 1025 jn amsterdam,
nederland; http://www.cwi.nl/~dik/
===
Subject: Re: My research,
publication announcement> There's more to my work than just
arguing on Usenet, so I'd like to> point out that my paper
Advanced Polynomial Factorization is slated> to be published:>
See http://www.megasociety.net/NoesisHighlights.html> The Mega
Foundationis a bad joke, Harris. The sow looks at you and,
when you saySvensk? it snorts and says
Norsk!http://www.crank.net/harris.html It's not every braying
jackass that gets a whole page at crank.netMaybe you can
publish (ha ha ha) in Vanity Fair or My WeeklyReader.-- Uncle
Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for
children and most mammals)Quis custodiet ipsos custodes? The
Net!
===
Subject: Integral of sin(x) / xSomeone asked a way to
calculate integral from 0 to infinity of sin(x) / xwithout
using residues.Probably the most elegant way (imho) is to use
so called Feynman's trickConsider an integral (containing a
parameter p) from 0 to infinity ofe^(-px)*sin(x) wich we
denote by I(p). It is easy to evaluate integrating
byparts:I(p)=1/(p^2+1)Now the trick is to integrate I(p) from
0 to infinity with respect to punder the integral sign
:int(0->oo) 1/(p^2+1) =pi/2
===
Subject: Re: Integral of sin(x)
/ x> Someone asked a way to calculate integral from 0 to
infinity of sin(x) / x> without using residues.> Probably the
most elegant way (imho) is to use so called Feynman'strick>
Consider an integral (containing a parameter p) from 0 to
infinity of> e^(-px)*sin(x) wich we denote by I(p). It is easy
to evaluate integratingby> parts:> I(p)=1/(p^2+1)> Now the
trick is to integrate I(p) from 0 to infinity with respect to
p> under the integral sign :> int(0->oo) 1/(p^2+1)
=pi/2Wonderful! That was just the kind of evaluation I was
thinking of. Quite astandard trick, actually: Introduce an
additional unknown parameter andsolve the more general
problem.It's like one of them factoring algorithm where -
starting from some largenumber n - the first step involves
multiplying the number by some smallfixed constant. I think it
is the Number Field Sieve, but please correct meif I'm
wrong.-Michael.
===
Subject: Everything that has a
beginning...alternative ending to matrix revolutions (no
spoilers) oracle: everything that has a beginning has an end
neo: (thinks long and hard) ...erm, what about the positive
integers 1,2,3,4,5... ?oracle: er...(thinks long and hard,
self destructs, causes system failure) neo: whoa! i did
it.
===
Subject: Re: Everything that has a beginning...>
alternative ending to matrix revolutions (no spoilers) >
oracle: everything that has a beginning has an end > neo:
(thinks long and hard) ...erm, what about the positive
integers> 1,2,3,4,5... ? > oracle: er...(thinks long and hard,
self destructs, causes system> failure) > neo: whoa! i did
it.ted: excellent adventure, dude.
===
Subject: Re: How to
define a function to be smooth?> When we say a function f(t)
is smooth, does this mean that> f has infinite differentials
with respect to t?> Or any other formal definition on this?>
FredReading all these responses is very interesting, as I had
no idea thatsmooth could mean different things. When I was in
grad school inReal Analysis, I recall dealing with smooth
functions which weredefined to be those which had a continuous
first derivative. Eachtime I ran into smooth thereafter, I just
assumed it meant the samething.Jo HoyleGene Codes
Corporation
===
Subject: Re: How to define a function to be
smooth?> > >> >...> I have seen smooth used for once
continuously differentiable, twice>>continuously
differentiable, C^infty and presumably anything in>>between,
...>> >Indeed, I have seen it used for something like the
following function:> f(x) = sqrt(x) when x >= 0> = -sqrt(-x)
when x <= 0.>Has certainly a pretty smooth appearance. There
is no standard>definition in analysis.> >> The graph of this f
is certainly a smooth curve. Hmm, can we come up >>with an
analytic map g: (0,1) -> R^2 whose image is this curve, >Don't think so, haven't thought about it. Probably more
important>is to point out that the smoothness of a curve is
typically not >measured by how smooth a map has the curve as
its _image_.>...>> >True, but a reasonable definition of
smoothness for an implicitly>defined curve is the maximal
degree of smoothness of a *non-singular*>parametrization of
the curve [non-singular means the first derivative>is nowhere
zero).So here is another interesting example: The function f:
R -> R, x |-> cubrt(x) is not even C1, but the curve g:
(-pi/2, pi/2) -> R^2, t |-> (tan^3 t, tan t) is analytic and
non-singular.-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
===
Subject: Re: How to define a
function to be smooth?> ...> I have seen smooth used for once
continuously differentiable, twice> continuously
differentiable, C^infty and presumably anything in> between,
...>> Indeed, I have seen it used for something like the
following function:> f(x) = sqrt(x) when x >= 0> = -sqrt(-x)
when x <= 0.> Has certainly a pretty smooth appearance. There
is no standard> definition in analysis.> The graph of this f
is certainly a smooth curve. Hmm, can we come up >>with an
analytic map g: (0,1) -> R^2 whose image is this curve, >Don't
think so, haven't thought about it. Probably more important>is
to point out that the smoothness of a curve is typically not
>measured by how smooth a map has the curve as its _image_.>
...>>True, but a reasonable definition of smoothness for an
implicitly>defined curve is the maximal degree of smoothness
of a *non-singular*>parametrization of the curve [non-singular
means the first derivative>is nowhere zero).As I pointed out in
the paragraph you omitted. >John Mitchell
===
Subject: Re: How to
define a function to be smooth?>> ...>> I have seen smooth used
for once continuously differentiable, twice>> continuously
differentiable, C^infty and presumably anything in>> between,
...>> Indeed, I have seen it used for something like the
following function:>> f(x) = sqrt(x) when x >= 0>> = -sqrt(-x)
when x <= 0.>> Has certainly a pretty smooth appearance. There
is no standard>> definition in analysis.>> >The graph of this
f is certainly a smooth curve. Hmm, can we come up >with an
analytic map g: (0,1) -> R^2 whose image is this curve, >Don't
think so, haven't thought about it. Probably more important>is
to point out that the smoothness of a curve is typically not
>measured by how smooth a map has the curve as its _image_.>
...>True, but a reasonable definition of smoothness for an
implicitlydefined curve is the maximal degree of smoothness of
a *non-singular*parametrization of the curve [non-singular
means the first derivativeis nowhere zero).John
Mitchell
===
Subject: Re: definite integral> In sci.math and
sci.math.num-analysis alex <heldring@voltor.upc.es> asked
aboutIntegrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 +
2*Cos[x]]],{x,0,Pi}]> I responded in the latter newsgroup but
basically quit simplifying when> it became clear the answer
would involve elliptic integrals.> But as it happens, a grad
student (Hi Geoff!) asked me today to help> him understand
modular forms and I thought I could take this opportunity> to
add some details to my prior post. [Follow-ups set to
sci.math. ]< --- very nice explanations snipped --- >Is there
something readable about that 'geometric' view(not restricted
to modular forms) in a book? I liked it.---remove the no to
mail me
===
Subject: Re: definite integralDistribution: inetIn
sci.math and sci.math.num-analysis alex
<heldring@voltor.upc.es> asked about>
Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 +
2*Cos[x]]],{x,0,Pi}]I responded in the latter newsgroup but
basically quit simplifying whenit became clear the answer
would involve elliptic integrals. But as it happens, a grad
student (Hi Geoff!) asked me today to helphim understand
modular forms and I thought I could take this opportunityto
add some details to my prior post. [Follow-ups set to
sci.math. ]This integrand involves arithmetic operations,
algebraic functions (sqrt)trig functions (cos) and the
logarithm -- yuk! I will simplify a bit,and then connect the
rest to Riemann surfaces.integration by parts. We get two
summands, one of which is sin(x) log( something )and which,
when evaluated at the endpoints, vanishes. The othersummand is
another integral, but this integrand involves no
logarithm:After some simplifying, we see that the original
integral equals int_0^pi (1 - 1/v) sin(x)^2/(v+ 1-cos(x))
dxwhere v = sqrt(d^2 + 2 + 2*cos(x)). That is, the integrand
is a nicerational function of sin(x), cos(x) and this
square-root thing v(which is also a function of the trig
functions). There is no otherx not wrapped into a trig
function (except in the dx).So we can think of the integrand
as a nice function of cos(x) and sin(x)not involving x itself.
That's very important information. Sincesine and cosine are
periodic functions, so is the integrand; we can thinkof the
integrand as living on the interval [0, 2pi], or moreprecisely
as living on the circle R / (2 pi Z) (the quotient of
ourdomain, R, by the periodicity group 2pi Z). We really can
think of thisas the unit circle S^1, since we have a
parameterization R --> S^1given by the parameterization
(X,Y)=(cos(x),sin(x)). (This parameterization passes to the
quotient space, and so effectively establishes a homeomorphism
R/2 pi Z --> S^1.)Now, I don't want to be too specific about
what a nice function is.We always mean to include the
coordinate functions X : S^1 --> R andY : S^1 --> R . We also
mean to include all polynomials in these; thisgives not the
polynomial ring R[X,Y] in two independent variables, butrather
its quotient ring R[X,Y]/(X^2+Y^2-1) since there is this
relationbetween these two coordinate functions on the circle.
We also mean toinclude quotients of these functions; these are
only _functions_ if wethink about the codomain being a
completion of the real line, that is, we are thinking about
meromorphic functions on the circle.In some contexts we might
take nice to include a somewhat largercollection of functions
on the circle; for example we can include this vabove, as well
as the quotient field Q of R[X,Y]/(X^2+Y^2-1).So our integrand
is one of these nice functions built up from X and Y.(The
logarithm AND the trig functions are gone; only algebraic
terms remain.)But the integral also includes this extra dx !
We don't usually askcalculus students to figure out what dx
is, but we do remind them thatthey need to change it when they
change variables. In this case, fromthe relation d cos(x) =
sin(x) dx, we conclude that dx = dX/Y, meaning that the
integrand is still a nice function of X and Y,with a dX thrown
in at the end.Now, it IS possible to define what this dX is.
It's a differential1-form, which is, um, well, it's a, um,
well there's this one-dimensionalvector space over the field Q
of which dX is a member, and in factto every element f of Q we
assign an element df of this vector space;the assignment is
linear (over the field of constants) and satisfiesthe Leibniz
rule d(fg)=f dg + g df. The fact that the vector space
isone-dimensional means that all these df's are multiples of
dX; in factthe multiplier is what we call df/dX (yay!
derivatives really are fractions!)Interestingly, even dY is a
multiple of dX; but you knew that, sinceX^2+Y^2 = 1 means X dX
= - Y dY . So the things you integrate are allof the form
g(X,Y) dX for some nice function g .Now, you can do this sort
of thing over the real line, too, instead ofover the circle.
For every function f : R --> R you can define a differential
df which is a multiple of dx, namely df = f'(x) dx
Significantly, every (nice) differential is of this form: if
you giveme something in this vector space, i.e. some
expression g(x) dx, I canthink of it as a differential of a
function G, that is, g(x) dx = dG;all I need to do is let G be
any antiderivative of g, and I know sucha thing exists because
I can define it as the integral G(u)=int_a^u g(x)dx.Then the
fundamental theorem of calculus tells us that int_a^b g(x) dx
,which is now int_a^b dG, can be evaluated as G(b) -
G(a).Well, that sort of thing works on the circle, too: if you
have a 1-form g(X,Y) dX which happens to be the differential dG
of somefunction G(X,Y), then the integral of g(X,Y)dX from one
point (X0,Y0)of the circle to another point (X1,Y1) is just
G(X1,Y1)-G(X0,Y0).A funny thing happens on the circle, though:
not every 1-form is thedifferential of a function. Try applying
all this theory to computeint_0^{2pi} 1 dx (which you know
comes out to 2pi). (If you like,you may write the integrand as
cos(x)^2+sin(x)^2 !) Make all thesubstitutions as above and you
find the integral becomes an integralof dX/Y, computed
(counterclockwise) around the circle. If you like,you may
write this as the integral of dX/sqrt(1-X^2), but you have
toremember which branch of the sqrt you mean as you move
around thecircle (and you have to remember that you start
integrating from X=1and finish for a moment at X=0, not the
other way around). Well, you'llstill get 2pi if you work it
all out, of course, but the point isthat dX/Y _cannot_ be dG
for any function G since if it were, the integral would become
G(1,0) - G(1,0) = 0.On the other hand, this is about as bad as
you can get on the circle:If the integral of a 1-form w around
the whole circle is zero, thenthe integral from any point a to
any other point b is independentof the path chosen to get from
a to b. Hence (after fixing a , say)we can define a function
G(b) = integral of w from a to b andthen we find dG = w . If
the integral of w around the whole circleis some other number
k, then w - (k/2pi)(dX/Y) has an integral of 0and so from the
previous sentence, is equal to dG for some G.In short: the set
of differentials is a complex subspace of the set of all1-forms
-- in fact, is a subspace of codimension 1. This codimension,
1, is a topological invariant: precisely the problem is that
the circle has 1 hole.The upshot of all this is that (a) it's
more natural to consider theintegral as living on a quotient
of R rather than R itself; (b) we cancontinue to do many
calculations formally; (c) a key computation is tosee how far
an integral differs from being an actual differential.OK, now
all this would go very smoothly if we took nice to
meanrational (function), but our integrand is not a rational
functionon the circle, because of the v defined by square
roots. That willmove us to a different space instead of
S^1.Before we do that, I want to reduce the number of
variables. Working onthe circle is natural, all right, but it
forces us to use X and Y both.It turns out that the circle
(minus one point) can be reduced back to the line using
stereographic projection. This mapping is a bijection; you
canwrite out formulas either way. As I noted in a prior post,
we can writeit this way: there is a bijection R --> S^1 - {
(-1,0) } given bysending t to the point ( (1-t^2)/(1+t^2),
2t/(1+t^2) ). So X nowmeans this function (1-t^2)/(1+t^2) on
the real line, and similarly for Y.In particular, the ring Q
of rational functions in X and Y is simply the ring C(t) of
rational functions in t. (All this is a purelyalgebraic
result, if you prefer: Q was defined as a field extension of
Cof transcendence degree 1, and now we are observing that it
happens tobe a purely transcendental extension of C, with t as
a generator.)Also note that dX has a differential dX =
-4t/(1+t^2)^2 dt.So with this change of variables, our
integral becomes int_0^pi (1 - 1/v) sin(x)^2/(v+ 1-cos(x)) dx=
int_{half-circle} (1 - 1/v) Y^2/(v + 1 - X) dX/Y= int_0^infty
(1 - 1/v) (2t/(1+t^2))^2/(v + 1 - (1-t^2)/(1+t^2))
(-4t/(1+t^2)^2)/(2t/(1+t^2)) dt= int_0^infty (1 - 1/v)
(-8t^2)/( (1+t^2)^2(v + vt^2 + 2t^2) ) dtwhere v = sqrt(d^2 +
2 + 2*cos(x)) = sqrt(d^2 + 4/(1+t^2)).Now, here's the clever
idea: when we were integrating around the circle,we allowed
ourselves to use _two_ coordinates X and Y, with
theunderstanding that they are related (X^2+Y^2 = 1) so that,
in a pinch,we could reduce those integrals to integrals of one
variable. Butof course to do so we would need to keep track of
the signs of thesquare roots, so we preferred to keep both
variables handy. Now in thisnew integral, we again have two
variables, v and t, which arerelated (v^2 = d^2+4/(1+t^2) )
and we instinctively think of this asan integral in one
variable t with v as a mere shorthand. But -- whynot use the
previous perspective?Thus, we view this as an integral of a
rational (!) function of t and v,the coordinate functions on a
curve defined in the t,v-plane. (The curveappears to have two
components, which makes it easy to keep the sign ofv straight,
but as I will explain below is actually an artifact of
therepresentation of the curve; pretend it is just as
connected as S^1 !)At the expense of moving to an integrand
with two variables, we havedispensed with the square roots as
well as the trig and log functions. Phew!Now, as in the case
of an integral on the circle, we have the polynomialfunctions
on this curve, and the rational functions, and the 1-forms,
someof which are differentials of rational functions, others
not. The rational functions form the quadratic extension
C(t)[v] of the rational functionsin t, given by the relation
v^2 = d^2 + 4/(1+t^2). (This is a field of transcendence
degree 1, but it's not a purely-transcendental extension ofC
as we had with the circle. That's not obvious! But it does
mean we can'tfind an algebraic substitution which will get rid
of the square rootsfor us.) Some familiarity with algebra
immediately suggests that this meansevery rational function in
this ring may simply be written as f1(t) + f2(t) v for some
(unique) pair of rational functions of one variable, f1 and
f2. It's not hard to find the decomposition explicitly. For
example, for our integrand (1-1/v) (-8t^2)/( (1+t^2)^2(v +
vt^2 + 2t^2) )we can rationalize the denominators to get (1 -
v/v^2) (-8t^2)(-v(1 + t^2) + 2t^2)/( (1+t^2)^2(-v^2(1 + t^2)^2
+ 4t^4) )and expand a portion of the numerator, and then
replace every instance ofv^2 with d^2+4/(1+t^2). I get the two
f's to be f1 =
8*t^2*(3*t^2+1)/(d^2*t^4-4*t^4+2*d^2*t^2+4*t^2+d^2+4)/(1+t^2)^
2 f2 = -8*(d^2*t^2+2*t^2+d^2+4)*t^2/
((d^2*t^4-4*t^4+2*d^2*t^2+4*t^2+d^2+4)*(1+t^2)*(d^2+d^2*t^2+4)
)So now our integral is of ( f1(t) + f2(t) v ) dt. See? It
gets betterall the time!The first part is easily
antidifferentiated, as it's a rationalfunction. That
antiderivative involves some arctangents and so on.The
definite integral becomes improper if |d|<2 but otherwise
converges andcan evaluated numerically very easily. The answer
is an algebraic multipleof Pi (the algebraic number lying in
some quartic extension of therational numbers; the extension
varies with d, so the answer looksquite messy when expressed
as a function of d. It gets a little betterlooking if you
write d in the form d=(u-5/u)/2 for some number u.)Then we are
reduced to trying to integrate f2(t) v dt along a path on
ourcurve. This problem can be reduced by expressing f2 in
terms ofpartial fractions. This is a very general technique,
and even coverscases in which f2 has repeated factors in the
denominator (ours does not). For example if a
partial-fractions summand is of the form 1/L^2 with Llinear,
then we can use the expansion int (1/L^2)(v dt) = int( d(v/L)
+ 4t/(L (t^2+1)(4+d^2(1+t^2)) ) v dt )to reduce the integrand
to one with no square factors in the denominator.That is, we
take advantage of the opportunity to subtract off
differentialsof known functions. (This is a
homological-algebra sort of thing: all ourdifferentials w have
dw=0 and with a full accounting of our rationalfunctions G we
can figure out which w's are of the form w = dG.Kernel-of-this
modulo image-of-that, and we are, as it turns out, computingthe
homology of our underlying space!)For our particular f2, the
partial fractions decomposition leaves uswith just a few
integrands: two of the form v/(t^2+constant) dt, andthen a
more complicated one which, as above, can be split for any
particulard, or can be split symbolically into two more
integrands of the same type,using the substitution d=(u-5/u)/2
.At the end, then, we have rewritten the integral to involve
knownelementary functions, finally including integrals of
functions of the formsv dt/(t^2 + K). You can ask your
favorite symbolic-algebra program toevaluate these integrals
and get some complicated answers involvingelliptic functions.
If you read up a bit, you'll see that these ellipticfunctions
are defined precisely as antiderivatives of very
similarfunctions; that is, there's probably no advantage to
rewriting our lastintegrals in terms of those functions,
except to conform to tradition.In retrospect, we decide that
there had to have been an answer of thissort. That is, most of
the 1-forms w which we need to integrate aregoing to be
differentials dG of functions G define on our curve.We can
construct this G exactly as we did on the circle (and as we
doin the Fundamental Theorem of Calculus) -- by integrating
the1-form along a curve. All we need to do is to ensure that
the integralis path-independent, i.e. that the integral around
any closed loop is zero.We can accomplish this by subtracting
off suitable multiples of just afew special 1-forms. Thus, the
whole integral ought to be expressed interms of some known
functions on our curve in the t,v-plane, plus afew special
summands, each relating to one of the holes in the curves.This
construct is just a little tricky, because our curve in
t,v-space is notconnected; one can't define G to be the
integral of w from a to bif there are NO PATHS from a to b !
It turns out that a much moresatisfying answer works all the
calculations in C, not R. The equationv^2 = d^2 + 4/(1+t^2)
defines a relation satisfied by certain pairs (t,v);these
pairs form a complex 1-dimensional manifold (i.e. a complex
curve) in complex 2-space. Viewing each complex dimension as a
pair of real ones,this solution set is now not a curve in the
plane but a real surface in R^4.It happens to be true that
this surface is a torus for our example.We can do all the
integrals we want along this surface. You will findthat
integrals are path-independent precisely when the integrals
around thetwo disjoint loops of the torus both vanish.So the
general integral of a function on the torus will be the sum
ofsome elementary functions defined on the torus, plus a
couple ofsummands which relate to the extent that our 1-form
f2(t) v dt doesnot integrate to zero around the two
independent loops in the torus(the generators of the first
homology group). That's were the ellipticfunctions come from:
they are the usual choices of functions used topick off the
nonzero contributions of integrals around the loops.I'd like
to close by making the connection to covering spaces once
again.Just as we started with a covering R --> S^1, it is true
that there isa covering of a Riemann surface in t,v-space,
usually by the upper half-planeof C. And it's just as true
that we can pull back 1-forms fromthe Riemann surface to get
1-forms on the upper half-plane. Now, the kicker is that even
though the integrand itself livesdownstairs on the Riemann
surface, and is thus invariant under the fundamental group
when we pull it back to the universal covering space,when we
pull back a 1-form, we must also pull back the dx (or
whatever). This thing is NOT invariant, unlike the case of the
real line coveringthe circle. In that other case, the
fundamental group consisted oftranslations f(x) = x + 2 pi n,
and df(x) is equal to dx (i.e. f'(x)=1).In this Riemann
surface case, the fundamental group consists of
certainfractional-linear transformations on the upper-half
plane, i.e.f(z) = (az+b)/(cz+d), so that df(z) = f'(z) dz is
NOT the same as dz; it's off by a factor of (ad-bc)/(cz+d)^2
(which isn't even constant).There are several settings in
which one can make this seem natural, butone way or the other,
you just have to get used to seeing the integrandas being a
function G(z) defined on the covering space and subjectto a
transformation law G(f(z)) = G(z)*(ad-bc)/(cz+d)^2.By the way,
one can change the exponent 2 to other integers; functionsG
which transform this way are modular forms of other weights.So
I didn't really provide the brief form of the integral which,
perhaps,the OP was waiting for, but I hope I provided some
information aboutwhy the answer he gets will have the form it
does.dave
===
Subject: eigen vectors of the matrix productGiven
the product of two matrices A and B, denoted as C=AB;suppose A
is on site 1, and B is on site 2; if it is not possible to move
A and B together to compute C, is it possible to compute
theeigenvector of C from A and B, respectively?I remember in
Quantum Computation, if C is the tensor product (ordensity
product, sorry, I am not sure) of two matrices A and B,
theeigenvectors of Ccan be computed by the eigenvectors of A
and eigenvectors of B.But how about the regular matrix
product?Any ideas? thanks a lotroy
===
Subject: Re: eigen
vectors of the matrix product>Given the product of two
matrices A and B, denoted as C=AB;>suppose A is on site 1, and
B is on site 2; if it is not possible to >move A and B together
to compute C, is it possible to compute the>eigenvector of C
from A and B, respectively?Well, if you give me the (complex)
eigenvectors and the eigenvalues of bothA and B, I can sort of
rebuild A and B, compute C, and then computeeigenvalues and
eigenvectors. But without full information aboutthe
eigenstructure of A and B, I don't see how you can get
muchinformation about C. Think geometrically. Suppose you have
a rubber sheet nailed at one pointto the table top. One person
comes in and stretches the sheet, first in one direction
(pulling both sides away from the nail) and then stretchesit
in the perpendicular direction by a different amount. A second
personthen comes in and does likewise, but with a whole new
pair of directionsand a new pair of stretching factors. With
complete informationabout the four factors and two directions,
you can express how thecombination of their moves can be
accomplished with just a singlepair of stretches. With
incomplete information, there's not much youcan say in
general. (Try a simple case: each person simply doubleslengths
in a single direction and then stops. That's not enough
informationto know whether the sheet is stretched by a factor
of 4 somewhere, or by a factor of 2 in every direction, or by
some intermediate amounts.)BTW, there is not really any
mathematical meaning to moving two
matricestogether.dave
===
Subject: Re: Sets before logic : > : >
Random thoughts on creating a theory of sets prior to a theory
of : > propositions and quantifiers: : > : > Let's start with
the empty set, 0, and logical identity, =, then we can : >
define T, for true, by : > : > T =def 0 = 0 : > : > Let's
define ordered pair a la Kuratowski, then we can define : >
conjunction by : > Kuratowksy defines <a,b> as {a,{a,b}}. But
how do you make sense of : > that latter notation at this
stage of the presentation? Note that in : > ZF, you can only
make sense of it because of the Axiom of Pairing; you : > can
only verify that it satisfies the ordered pair axiom because
of : > the Axiom of Extensionality. Those axioms both seem to
require the : > apparatus of first order logic to be
formulated. I already said that. Why don't *I* rate a reply? :
Well, we cannot define un-ordered pair in context by : u in
{x,y) iff u = x or u = y : since we don't have or. So let's
take {x,y} as primitive and define : {x} =def {x,x} : <x,y>
=def {{x},{x,y}}Nobody is impressed. You'd be much better off
just taking<x,y> as primitive. Your mission, now, which will
be much harder,is, given that you've taken {x,y} as primitive,
how on earth isanybody supposed to parse {x,y,z} ?More to the
point, you said before that we don't have 'or',but you have a
much bigger problem: you don't have *in*, EITHER.What good
does it do you to take {x,y} as primitive, and to claimthat
you've presumed some set theory, if you STILL have NO wayof
deciding whether z is or isn't *in* {x,y} ?
===
Subject: Re:
Sets before logicRandom thoughts on creating a theory of sets
prior to a theory ofpropositions and quantifiers:Let's start
with the empty set, 0, and logical identity, =, then we
candefine T, for true, by T =def 0 = 0Let's define ordered
pair a la Kuratowski, then we can defineconjunction by>
Kuratowksy defines <a,b> as {a,{a,b}}. But how do you make
sense of> that latter notation at this stage of the
presentation? Note that in> ZF, you can only make sense of it
because of the Axiom of Pairing; you> can only verify that it
satisfies the ordered pair axiom because of> the Axiom of
Extensionality. Those axioms both seem to require the>
apparatus of first order logic to be formulated. Well, we
cannot define un-ordered pair in context by u in {x,y) iff u =
x or u = ysince we don't have or. So let's take {x,y} as
primitive and define {x} =def {x,x} <x,y> =def {{x},{x,y}} >
So how do you> propose to make sense of {a,{a,b}} without
quantifiers?By using free variable formulae.> It's not that
this seems like a waste of time; maybe it's interesting.> But
what I'd want to see is a formal system (in the way that we
can> write formal systems for first order logic), in which you
show me how> derivations in your set theory are going to
proceed.> -- G.C.
===
Subject: Ordering a Power SetThis question
arises from economics, but I'll give you the questionfirst, the
context later:Suppose we have a set of n objects, and a binary
relation S that is alinear ordering, ie irreflexive,
asymmetric, complete, transitive. Isthere any way to use this
relation to induce some soft of ordering onthe power set of
our set of objects?This arises from an examination of
equilibrium in the jungle: we haveN agents, and S is
strength(1 is stronger than 2, etc.) To see ifthis equilibirum
is coalition-proof we need some way to ordercoalitions of
agents. Which I don't know how to do.
===
Subject: Re: Ordering
a Power Set> This question arises from economics, but I'll
give you the question> first, the context later:> Suppose we
have a set of n objects, and a binary relation S that is a>
linear ordering, ie irreflexive, asymmetric, complete,
transitive. Is> there any way to use this relation to induce
some soft of ordering on> the power set of our set of
objects?> This arises from an examination of equilibrium in
the jungle: we have> N agents, and S is strength(1 is stronger
than 2, etc.) To see if> this equilibirum is coalition-proof we
need some way to order> coalitions of agents. Which I don't
know how to do.Subsets of an nlement set correspond to strings
of length n made upof the symbols 0 and 1... 1 means this
object is in the set 0 meansthis object is not in the set. But
strings of 0s and 1s of length ncorrespond to binary
representations of the integers from 0 to 2^n-1. There is a
natural way to order THIS set, right?
===
Subject: Re: Ordering
a Power Set Adjunct Assistant Professor at the University of
Montana.>This question arises from economics, but I'll give
you the question>first, the context later:>Suppose we have a
set of n objects, and a binary relation S that is a>linear
ordering, ie irreflexive, asymmetric, complete, transitive.
Is>there any way to use this relation to induce some soft of
ordering on>the power set of our set of objects?There are many
ways to do it.One way would be to use a variant of the
lexicographic ordering:(a) The empty set is the smallest
set.(b) Let A and B be nonempty. Let a be the smallest element
of A, and bthe smallest element of B. Then we say that A is
smaller than B if andonly if (i) a<b; or (ii) a=b and
A-{a}<B-{b}.This sets up a recursion process.This works
because any total ordering on a finite set is awell-ordering,
so it makes sense to talk about smallest element.However, this
ordering has a number of unnatural features. Aslightly more
interesting and natural way is to extend the naturalpartial
ordering that exists in P(S) under inclusion.Order P(S) as
follows:(1) Two subsets of different size are ordered by their
size; that is, if A and B are subsets, and A has fewer elements
than B, then A<B.(2) Order the singletons by letting {a} < {b}
if and only if a<b.(3) Assuming all subsets of fewer than k
elements have been ordered, k>1, order the subsets of k
elements as follows: if A and B both have k elements, let a be
the smallest element of A and b the smallest element of B. Then
A<B if and only if a<b or (a=b and A-{a}<B-{b}).But of the many
ways to order P(S) in some way induced by theordering of S, you
need to consider your problem.>This arises from an examination
of equilibrium in the jungle: we have>N agents, and S is
strength(1 is stronger than 2, etc.) To see if>this
equilibirum is coalition-proof we need some way to
order>coalitions of agents. Which I don't know how to do.How
you order the coalitions will depend on the properties of
thestrength. Does a coalition involving strength 2 and 3 beat
oneinvolving 1 and 4? You seem to be putting the horse before
the cart:you want to understand how the strength of a
coalition works, so youcan then order the coalitions; you
don't first order the coalitions insome way and then use that
order to figure out their relativestrengths.
===
==Arturo
Magidinmagidin@math.berkeley.edu
===
Subject: metabelian group
questionHi I am doing some algebra problems for fun.( I am not
math major and haven't taken this course, but I am learning it
by myself, I am trying to solve allthe problems of Issaac's
book)A group is called metabelian if there exists some abelian
normal subgroup A of G such that G/A is also abelian.a) show
that G is metabelian iff G''=1 (G' is the commutator
subgroup)b) show that homomorphic images of metabelian groups
are metabelianc) show that subgroups of metabelian groups are
metabelianI don't have clue for itcan any one give me some
hints?thanks a lot!
===
Subject: Re: metabelian group question
Adjunct Assistant Professor at the University of Montana.>Hi I
am doing some algebra problems for fun.( I am not math major
and haven't >taken this course, but I am learning it by
myself, I am trying to solve all>the problems of Issaac's
book)>A group is called metabelian if there exists some
abelian normal subgroup >A of G such that G/A is also
abelian.>a) show that G is metabelian iff G''=1 (G' is the
commutator subgroup)>b) show that homomorphic images of
metabelian groups are metabelian>c) show that subgroups of
metabelian groups are metabelian>I don't have clue for it>can
any one give me some hints?The commutator subgroup of G is
defined to be the subgroup generatedby all elements of the
form [x,y] = x^{-1}y^{-1}xy, with x and y in G.(a) Prove that
G/G' is abelian, and if N is any normal subgroup of G such
that G/N is abelian, then G' is contained in N. Deduce that if
G is metabelian, with A the witness, then G'<A. Then prove that
if H<G, then H'<G'. So G''<A'. What can you say about A'?(b)
Use the isomorphism theorems. Say G is metabelian, with A
abelian and G/A abelian. Let N be any normal subgroup. The
obvious candidates to show that G/N is metabelian would be
AN/N. Use the isomorphism theorems to show that: (i) AN/N is
abelian; and (ii) (G/N)/(A/N) is abelian.(c) Say G is
metabelian, with A abelian and normal, G/A abelian, H a
subgroup. Use the isomorphism theorems to show that: (i) A
intersect H is normal in H; (ii) H/(A intersect H) is abelian.
Why does this imply that H is metabelian?The ->very<-
high-powered approach (akin to swatting flies with anuclear
device; my suggestion is to just skip it, but I'm including
itjust because I can...): Show that a group G is metabelian if
and onlyif it satisfies the identity [ [a,b],[x,y] ] = e for
all a,b,x,y in G.Conclude that the class of all metabelian
groups is a variety (in thesense of general algebra), and
therefore is closed under bothsubgroups and quotients. Then
show that the verbal subgroupcorresponding to [[a,b],[x,y]] is
G'', to conclude (a).
===
==Arturo
Magidinmagidin@math.berkeley.edu
===
Subject: Re: metabelian
group question>Hi I am doing some algebra problems for fun.( I
am not math major and haven't >taken this course, but I am
learning it by myself, I am trying to solve all>the problems
of Issaac's book)>A group is called metabelian if there exists
some abelian normal subgroup >A of G such that G/A is also
abelian.>a) show that G is metabelian iff G''=1 (G' is the
commutator subgroup)And G'' = (G')' is the commutator subgroup
of G'. And the commutatorsubgroup of a group is a trivial group
if and only if the originalgroup is an abelian group. So, G'' =
1 if and only if ... ?>b) show that homomorphic images of
metabelian groups are metabelian>c) show that subgroups of
metabelian groups are metabelian>I don't have clue for itName
everything in sight (G is a metabelian group, A is an
abeliannormal subgroup of G, h: G->H is a homomorphism onto a
group H, ...)and then consider that you are trying to find a
certain thing which, after all, if it exists must depend on
the things you havenamed--and if it can be shown to exist
presumably depends on thosethings in a more-or-less natural
way. What could it be?...This sounds like Herman Rubin's
standard advice for word problems(well, except for the part
after --, where natural usually shouldbe replaced by chosen by
the teacher, or maybe chosen by Naturebut by no means
`natural'). Maybe it is.Lee Rudolph
===
Subject: Re: Naive Q:
Set theory, logic - which comes first?>Two questions whose
answers I am seeking are:>1) What exactly might we mean by
representation?>2) How do we define a mathematical object if
we do not want to>identify it with some representation of
it?[...]>A mathematical object is uniquely determined if we
know what counts as>a model of it, so instead of working with
a mathematical object we>could work with the property of being
a model of it, but I think that>would be inellegant. [Analogy:
in geometry, instead of talking about>points, we could talk
about the property a line may have of passing>through that
point, but this is inellegant.]> This is the extensional view.
But the same numbers can> be used as both finite ordinals and
finite cardinals, and> these are not the same intensional
concept. That the same> numbers can be used for both is
useful.I am not sure I have understood the point of your
remark.When I said mathematical object above, I did not mean
something thatincludes intension. Nor did I mean to include
intension when Imentioned the property of being a model of
[the natural numbers].When I wanted to distinguish between a
mathematical object and theproperty of being a model of it,
the distinction I wanted to draw hadnothing to do with the
extension-intension distinction.As you state, finite ordinals
and finite cardinals are the same inextension (I am not sure
if one can say without reservation that theyare different in
intension).To the list above of questions whose answers I am
seeking may beadded:3. What might we mean by intensional
object?However, I do not see any reason why knowing the answer
of thisquestion should be a prerequsite for knowing the answers
of the othertwo.Mattias
===
Subject: Re: Naive Q: Set theory,
logic - which comes first?>>Two questions whose answers I am
seeking are:>>1) What exactly might we mean by
representation?>>2) How do we define a mathematical object if
we do not want to>>identify it with some representation of
it?>> This is done quite often. The Peano Postulates
CHARACTERIZE>> the positive (or non-negative) integers. There
are many other>> ways to characterize them.
.....................>By protesting when I said that a natural
number is a sequence of>symbols in which only symbol the symbol
'|' occurs, you said something>about what a natural number
is.>A mathematical object is uniquely determined if we know
what counts as>a model of it, so instead of working with a
mathematical object we>could work with the property of being a
model of it, but I think that>would be inellegant.This is the
extensional view. But the same numbers canbe used as both
finite ordinals and finite cardinals, andthese are not the
same intensional concept. That the samenumbers can be used for
both is useful.-- This address is for information only. I do
not claim that these viewsare those of the Statistics
Department or of Purdue University.Herman Rubin, Department of
Statistics, Purdue Universityhrubin@stat.purdue.edu Phone:
(765)494-6054 FAX: (765)494-0558
===
Subject: Worthwhile? (Re:
Continuum)> I will take the proof with me to my grave. A
noblest state of the mind. The attainment of the nobility
should not require any 'elitestatus'. Besides, entire theories
should be just as weightless as asingle proof to carry to one's
grave effortlessly. So what is at stake here? BTP? Seriously,
please let me know if there is any interest out thereabout my
giving it a try to settle BTP. (I can assure you that this
isnot something worth my taking to the grave with
me.)
===
Subject: Re: Continuum> Ich spreche Deutsche nicht so
schlecht. Wie angegeben in anderen Post, Z> steht vor Ze Set
which bla bla bla. Haben Sie etwas mehr Fragen ?> Then what is
your set Z? If it is the set of integers, you are wrong> about
its cardinality, and I am not aware of any common>
interpretation as a set other than as the set of integers ( Z
for> Zermelo, IIRC).> More likely, it's Z for Zahlen.Ich habe
nichts gefragt. Was heisst Troll auf Deutsch?-- Dave
SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>
===
Subject: Re: ContinuumLerne dem Killfile zu
nuetzen, als weil versuchen dem Mumia konservieren.Troll, in
some cases, is what some people who'd like to run usenet
butfortunately don't call those whose opinions they cannot
comment on, but makethemselves look really smart by saying I'm
so above it. Troll heisst einZwerg, in diesem Fall - ein
Giftzwerg, ob Sie hatte nichts gewuesst bisheute.Ich spreche
Deutsche nicht so schlecht. Wie angegeben in anderen
Post,Zsteht vor Ze Set which bla bla bla. Haben Sie etwas mehr
Fragen ?>> Then what is your set Z? If it is the set of
integers, you are wrong>> about its cardinality, and I am not
aware of any common>> interpretation as a set other than as
the set of integers ( Z for>> Zermelo, IIRC).>More likely,
it's Z for Zahlen.> Ich habe nichts gefragt. Was heisst Troll
auf Deutsch?> --> Dave Seaman> Judge Yohn's mistakes revealed
in Mumia Abu-Jamal ruling.>
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: Continuum> Lerne dem Killfile zu
nuetzen, als weil versuchen dem Mumia konservieren.> Troll, in
some cases, is what some people who'd like to run usenet but>
fortunately don't call those whose opinions they cannot
comment on, but make> themselves look really smart by saying
I'm so above it. Troll heisst ein> Zwerg, in diesem Fall - ein
Giftzwerg, ob Sie hatte nichts gewuesst bis> heute.The only
opinion I have expressed in this thread is that the usage of
Zfor the integers derives from Zahlen.*Plonk*-- Dave
SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>
===
Subject: Re: ContinuumIch spreche Deutsche
nicht so schlecht. Wie angegeben in anderen Post, Zsteht vor
Ze Set which bla bla bla. Haben Sie etwas mehr Fragen ?>I am a
novice to the redneck notations accepted in the US math
world,and>intend to stay this way. To cater whims of
philistines,card(N):=aleph_0 and>card(Z):=2**aleph0Then what
is your set Z? If it is the set of integers, you are
wrongabout its cardinality, and I am not aware of any
commoninterpretation as a set other than as the set of
integers ( Z forZermelo, IIRC).> More likely, it's Z for
Zahlen.> --> Dave Seaman> Judge Yohn's mistakes revealed in
Mumia Abu-Jamal ruling.>
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Even LatticesHi.One knows that E_8 is
the smallest example of an even unimodularlattice, and the
Leech lattice (dim=24) is the smallest example of aunimodular
lattice with minimum norm = 4.Are there interesting unimodular
lattices with minimum norm = 6 (orlarger)? What about odd
minimum norms (other than 1)?
===
Subject: Divisibility
ProblemSuppose p is an odd prime congruent to 3 (mod 4). If r
is any positive oddnumber, and 2pr+1 is prime, then is2pr+1 |
(p^r) + 1Ever true?Any help or ideas would be greatly
appreciated.GREG
===
Subject: Question about a C^infty functionI
am going through a proof that needs the existance of a
real-valued function F(x)F : R^n --> Rthat is C infinity,
i.e., its n-th derivative exists for every n=0,1,2,...The
condition on F is that it has to satisfyF(x) = 0 for norm(x)
<= rF(x) = 1 for norm(x) => Rwhere 0 < r < RMy problem is that
I remember (probably I'm very wrong), from complex analysis,
that there is no holomorphic function that satisfies these
conditions. However, this is in the Reals, so I can't use
those results.Is it possible to construct this function or
prove its existance?Any help is appreciated,Fernando G. del
Cueto
===
Subject: Re: Question about a C^infty function> I am
going through a proof that needs the existance of a
real-valued > function F(x)> F : R^n --> R> that is C
infinity, i.e., its n-th derivative exists for every
n=0,1,2,...> The condition on F is that it has to satisfy>
F(x) = 0 for norm(x) <= r> F(x) = 1 for norm(x) => R> where 0
< r < R> My problem is that I remember (probably I'm very
wrong), from complex > analysis, that there is no holomorphic
function that satisfies these > conditions. However, this is
in the Reals, so I can't use those results.> Is it possible to
construct this function or prove its existance?> Any help is
appreciated,> Fernando G. del CuetoUmm... sure.Infinitely
differentiable isn't *nearly* as strong as holomorphic - it's
not even as strong as analytic. In particular local behaviour
doesn't tell you a lot about global behaviour, and the zeroes
of your function can be (almost) arbitrarily badly behaved.
The set of zeroes has to be a closed set, but I suspect that
given any closed set you can probably construct an infinitely
differentiable function that is zero only on that set. Or not.
I don't really know, but it looks highly plausible and a brief
sketch proof with some major details in need of filling in
suggests that it probably works.All of which is really just a
long-winded way of saying that you mustn't let your intuition
about complex analysis colour your intuition about real
analysis. It simply doesn't work the same way.In this
particular case you should be able to express your function F
as a function of ||x||. ||x|| is differentiable everywhere
except at 0, and what do you know about F near
0...?David(E-mail address spam-blocked in the obvious
way)
===
Subject: Re: Question about a C^infty function> I am
going through a proof that needs the existance of a
real-valued > function F(x)> F : R^n --> R> that is C
infinity, i.e., its n-th derivative exists for every
n=0,1,2,...> The condition on F is that it has to satisfy>
F(x) = 0 for norm(x) <= r> F(x) = 1 for norm(x) => R> where 0
< r < R> My problem is that I remember (probably I'm very
wrong), from complex > analysis, that there is no holomorphic
function that satisfies these > conditions. However, this is
in the Reals, so I can't use those results.> Is it possible to
construct this function or prove its existance?> Any help is
appreciated,> Fernando G. del CuetoWhat you need is what I've
heard called a bump function, that is,a C-infinity function
B(x) for which B(x) = 0 for x <= a B(x) > 0 for a < x < b B(x)
= 0 for x >= bNote that this function q(x) (the standard
C-infinity functionthat has all zero derivatives at x = 0):
q(x) = exp(-1/x^2) for |x|>0, q(0) = 0can be used to fashion
B(x) as follows:Let q1(x) = 0 for x <= a q(x-a) for x > aand
q2(x) = q(x-b) for x <= b 0 for x >= bThen q1 and q2 are
C-infinity functions with support(q1) = {x | x >= a}
support(q2) = {x | x <= b}Let B(x) = q1(x)q2(x), and notice
that B(x) hassupport in {x | a <= x <= b}.Note that B is
positive on the open interval (a,b).Next, a smooth step s(x)
is formed by integrating B(t)from -infinity to x. Note that
s(x) is identically 0for x <= a, and takes a positive constant
value for x >= b.That constant can be adjusted to your favorite
value bymultiplying s(x) by the appropriate factor.Your
function F can be produced by the appropriate selectionof
constant values for a and b, and then taking F(x) =
B(|x|)where |x| is the Euclidean norm |x| = sqrt(x'x).Pretty
standard stuff, I think.Dale.
===
Subject: Other Gamma-Function
AsymptoticsMost of us know thatx! = Gamma(x+1) ~ x^(x+1/2)
exp(-x) sqrt(2 pi)as x -> oo, by Stirling's asymptotical
formula.But what about the asymptotic behavior of Gamma(x+1)
as x appoachesother limits other than real positive
infinity?For example:In the vicinity of x = 0,Gamma(x+1) =
exp(-c*x +x^2 pi^2/12 - x^3 zeta(3) +O(x^4)),where c is
Euler's constant.-And, as x -> -oo,Gamma(x+1) ~ -(-x)^(x+1/2)
exp(-x) csc(pi x) sqrt(pi/2).But I am not absolutely certain
about this asymptotic formula, andwould not use it to
investigate the behavior of Gamma(x+1) at x verynear negative
integers.-So, what, for example, about the asymptotics if x =
1/2 + i y, where yapproaches infinity?Or how about if x
approaches any other interesting finite/infinitecomplex/real
constants?Leroy Quet
===
Subject: Advice Needed: Bachelor and
Master's Distance, Correspondence, Online.Background: I am a
48-year old adjunct partial-load professor employed in a
localcollege and very happy teaching on contract. My
qualifications arefrom the techy/industry side, i.e.
engineer/technologist associationexams, college diploma, 28
years of technical and teaching experience.I have aways
enjoyed an interest and ability in applied mathematics.Both
kids have moved away and home, finances, and marriage, are
nowsecure, hence, its time for *me*!It has always been my
dream to publish papers, write text books,supervise courses,
and generally influence education in applied mathareas. My
area of expertise is instrumentation and control,
requiringLaplace, differential equations, complex numbers,
matrices, etc.In order to achieve these goals and to enhance
my knowledge,qualifications, and employability, I am looking
for an online,correspondence, or distance Bachelor's degree
which can lead to aMaster's Degree or Ph.D.Q1. Has anybody out
there done this?Q2. Will a math degree of this type be
considered acceptable to teachin university?Q3. Will I be
qualified to teach engineering students?Q4. Will publishers
publish?Thank you in advance for any and all comments and
opinions.George
===
Subject: AN: GuruGram #27..is newly
available for free download
athttp://www.tinaja.com/glib/msquant.pdfSourcecode is provided
as http://www.tinaja.com/glib/msquant.pslIt is on Magic
Sinewave Quantization optimizations.Also includes some DFT
Fourier Transform fundamental routines.Magic sinewaves are a
brand new way to synthesize high power digitalsinewaves with
arbitrarily low distortion at the highest possible
energyefficiency.Other GuruGrams at
http://www.tinaja.com/gurgrm01.asp-- Don LancasterSynergetics
3860 West First Street Box 809 Thatcher, AZ 85552voice:
(928)428-4073 email: don@tinaja.com fax 847-574-1462Please
visit my GURU's LAIR web site at
http://www.tinaja.com
===
Subject: Re: Array Recursion
PuzzleSeems as if no one has any clue...(snicker)So I will
give a couple clues below...> We have a triangular array of
integers {a(m,n)}.> a(1,1) = 0;> And, for 1 <= n <= m, m >= 2,
recursively:> a(m,n) = > > m-1> --- ---> > 1 + > ( > mu(j) )
a(m-1,k)> / /> --- ---> k=1 j|k> j>= kn/m> Ascii-mode:> a(m,n)
=> 1 + sum{k=1 to m-1} (sum{j|k, j >= kn/m} mu(j)) a(m-1,k)>
The inner-sum is over the divisors, j, of k,> where each j is
>= kn/m.> And mu() is the Mobius (Moebius) function.> Now, the
array's elements can be easily described with a closed-form>
(ie. non-recursive) definition.> What is this closed-form for
{a(m,n)}?> I will first give a clue in a few days, then the
answer a few days> after that, if no one posts an answer
before I do this.> Leroy QuetFirst, every a(m,n) is a positive
integer form >=2, 1 <= n <= m.Second, a(m,1) DOES always =
m-1,and a(m,m) does always = 1 for m >= 2.Leroy
Quet
===
Subject: Array Recursion PuzzleWe have a triangular
array of integers {a(m,n)}.a(1,1) = 0;And, for 1 <= n <= m, m
>= 2, recursively:a(m,n) = m-1 --- --- 1 + > ( > mu(j) )
a(m-1,k) / / --- --- k=1 j|k j>= kn/mAscii-mode:a(m,n) =1 +
sum{k=1 to m-1} (sum{j|k, j >= kn/m} mu(j)) a(m-1,k)The
inner-sum is over the divisors, j, of k,where each j is >=
kn/m.And mu() is the Mobius (Moebius) function.Now, the
array's elements can be easily described with a
closed-form(ie. non-recursive) definition.What is this
closed-form for {a(m,n)}?I will first give a clue in a few
days, then the answer a few daysafter that, if no one posts an
answer before I do this.Leroy Quet
===
Subject: Re: Key Core
Error Argumentsorry; I shouldn't make an implication thatan
inductive proof is just the reverseof a deductive proof.>
there is also a proof of the isomorphism> of deductive proofs
with inductive ones,> which may perhaps be amenable to
combining the two forms> into a tautology, or necklace.--ils
duces d'Enron!http://larouchepub.com/
===
Subject: Re: Key Core
Error Argumentsci.physics snipped.>If y is not 0, you can't
use the constant term tricks that depend on y>being 0.>>That's
stupid. The constant term once found is distinguished by
being>>constant.>>All the trick is doing is finding it.>>So
let me give you the example that I've used elsewhere which
is>>a_1(x) + 7, which has a constant term that is 7. Do you
understand>>what it means for it to be constant?>>Here's a
test.>>If I have x=11, then I necessarily have a_1(11) + 7,
right?>>Notice the constant term is STILL there, do you
understand?>>Now then, if I now divide P(11) by 49, what
should the constant term>>be?>>If you answer honestly I'll be
shocked.>Let me try again. (a_1(y) + 7)/x is not the same as
7/x, unless a_1(y) = 0.> Ok, I can go from there Richard
Henry, and note that necessarily> a_1(y) has *some* factor in
common with 7, right?But for some values of y, that factor
might be 1. This is the possibility you have not been dealing
with.> So let's call that factor f, now dividing 49 from P(y)
will divide f> off from a_1(y) + 7, understand?Again, f might
be 1 for some values of y.> Now then, you have a_1(y)/f + 7/f,
and if f does not equal 7, what> does that tell you about the
*constant* term Richard Henry?Nothing, because the constant
term is determined by a_1(0), and that does not deal with
a_1(y).> Necessarily, you have 7/f left as the constant term
for a factor of> P(11)/49, and if 7/f does not equal 1, you
have a contradiction.This is assuming f is not a function of
y. The whole point has been that most of us believe that f
*is* a function of y.> Understand?> The trick is to FIND the
constant term, as you know that it's not> affected by the
value of x, and it sits there like a rock, unaffected> by the
value of x, and none of your protestations against
mathematical> reality will change that fact.This trick is not
as useful as you believe it is.-- 
===
Subject: Re: Key Core
Error Argument>Which line contains the error?Actually, I think
most of JSH's proofs can be characterized by thefollowing
anecdote I think I heard from my old high-school mathteacher
about one professor or another:A lecture involved a somewhat
algebraically heavy calculation thatfilled the entire
blackboard while time was running out for the day.Since the
desired answer did not appear at the end of the
calculation,the professor simply ended his lecture by circling
the entirecalculation, commenting that the error was contained
somewhere inthis region and giving the assignment to find
it.
===
Subject: proving manifolds are metrizableIn Lie Groups,
Lie Algebras, and their Representations, V. S.Varadarajan
defines a C^infinity manifold to be a second
countableHausdorff space M with a C^infinity differentiable
structure, thatbeing an assignment D:U|->D(U), for all open U
contained in M, suchthat(i) for each open U contained in M,
D(U) is an algebra ofcomplex-valued functions on U containing
1(ii) if V, U are open, V is contained in U, and f is in D(U),
then therestriction of f to V is in D(V); if V_i (i in J) are
open, V is theunion of the V_i, and f is a complex-valued
function defined on V suchthat the restriction of f to V_i is
in D(V_i) for all i in J, then fis in D(V)(iii) there exists
an integer m>0 with the following property; for anyx in M, one
can find an open set U containing x, and m real functionsx_1,
..., x_m from D(U) such that (a) the
mapxi:y|->(x_1(y),...,x_m(y)) is a homoeomorphism of U onto an
opensubset of R^m, and (b) for any open set V contained in U
and anycomplex-valued function f defined on V, f is in D(V) if
and only if fcircle xi^-1 is a C^infinity function on xi[V].He
then remarks that M is obviously locally connected
andmetrizable.It sure is obviously locally connected, but
metrizable I can't see.Can anyone shed some light?
===
Subject:
ProbabilityI want to determine the probability that if you
pick B,C randomly from[0,1], and A randomly from (0,1], that
the polynomial Ax^2 + Bx + Chas two distinct roots. Obviously
this boils down to B^2-4AC >= 0. So I try doing a double
integral of the function C=B^2/(4A), but Ican't figure out
what to put as my limits of integration. Anyone havea
suggestion?