Rudinıs text deŜnes a Borel set as one which can be obtained
by a
countable number of operations, starting from open sets, each
operation consisting in taking unions, intersections, or
complements.
I can understand how the collection B of all Borel sets in
R^k is a
sigma-ring. However, how can one prove that B is the smallest
sigma-ring which contains all open sets? It would appear that
B is a
sigma-algebra as well, since it closed under
complementations. Is
this correct? Lastly, can the above conclusions concerning B
hold for
===
Subject: Re: Sigma-ring proof
> Rudinıs text deŜnes a Borel set as one which can be
obtained by a
> countable number of operations, starting from open sets,
each
> operation consisting in taking unions, intersections, or
> complements. I can understand how the collection B of all
Borel
> sets in R^k is a sigma-ring. However, how can one prove
that B is
> the smallest sigma-ring which contains all open sets? It
would
> appear that B is a sigma-algebra as well, since it closed
under
> complementations. Is this correct? Lastly, can the above
> conclusions concerning B hold for spaces other than R^k?
Many
If a collection S of sets is closed under complementation,
then the
sigma-ring generated by S will also be closed under
complementation;
i.e., the sigma-ring generated by S coincides with the
sigma-algebra
generated by S. In particular, in any topological space, the
sigma-ring generated by the open sets together with the
closed sets is
a sigma-algebra. Now, in a metric space, every closed set is
a G-delta
set (intersection of countably many open sets); in that case,
the
sigma-ring generated by the open sets will also contain the
closed
sets, and it is a sigma-algebra.
===
Subject: Re: Sigma-ring proof
> Rudinıs text deŜnes a Borel set as one which can be
obtained by a
> countable number of operations, starting from open sets,
each
> operation consisting in taking unions, intersections, or
> complements. I can understand how the collection B of all
Borel
> sets in R^k is a sigma-ring. However, how can one prove
that B is
> the smallest sigma-ring which contains all open sets? It
would
> appear that B is a sigma-algebra as well, since it closed
under
> complementations. Is this correct? Lastly, can the above
> conclusions concerning B hold for spaces other than R^k?
Many
> If a collection S of sets is closed under complementation,
then the
> sigma-ring generated by S will also be closed under
complementation;
> i.e., the sigma-ring generated by S coincides with the
sigma-algebra
> generated by S. In particular, in any topological space, the
> sigma-ring generated by the open sets together with the
closed sets is
> a sigma-algebra. Now, in a metric space, every closed set
is a G-delta
> set (intersection of countably many open sets); in that
case, the
> sigma-ring generated by the open sets will also contain the
closed
> sets, and it is a sigma-algebra.
===
Subject: Re: Sigma-ring proof
Your question is not very clear; usually one deŜnes the Borel
set B of R^k
as the smallest sigma-algebra (or sigma-ring, same thing here
since the
space R^k itself is open and whence belongs to B) containing
the open
subsets of R^k. Now your deŜnition deals with constructing
sets using
certain operations. (Btw, you didnıt deŜne that process
precisely and you
should do it (thatıs not thaaaaaat easy to formulate)) What
you can say is
that any sigma-ring containg the open sets of R^k will also
contain (by
deŜnition) all the sets constructed by using the previous
process. The
resulting set (being also a sigma ring) is therefore the
smallest sigma
ring
containing the open sets. This is trivial.
===
Subject: Re: Egyptian topology on Q
> >Oh yes, 1 is isolated.
> For each j in N, pj in S^j is mapped to sum pj, where
> when pj = (u1,..uj) in S^j, sum pj = u1 +..+ uj ?
> Why is 1 isolated? It is in lower limit topology subspace
Q/[0,1],
> but why is 1 isolated by the quotient deŜnition for your
space?
> Theorem of Sierpinski (circa 1920)
> A space S is homeomorphic to the rationals if it is:
> countable, 2nd countable, regular T0, free of isolated
points.
> The more I think about this, the more I think that this
Egyptian
> topology is just the half-open interval topology. In that
case, this
> leads to an interesting counter-example (of a countable
locally
> compact space mod a closed equivalence relation whose
quotient is
> Hausdorff but not locally compact).
> Can anyone give me a reference for Sierpinskiıs theorem?
> It was discussed, not this year, at sci.math and also
> Ask-a-Topologist, http://at.yorku.ca/topology
> in the form for topology. Use their search engine.
> If you understand proof posted there, would you help to
clarify it?
> Iım muddled about the description used to set up the order
upon the
> elements.
I may look at it, but unfortunately my earlier post was wrong
and the
topology is deŜnitely not the half-open interval topology.
Any set
of rationals the form {a_1, a_2, ...} that satisŜes the
following
conditions is closed, while it converges to 0 in the half-open
interval topology:
1. 1/(n+1) < a_n < 1/n
2. length of a_n > n.
The reason (roughly) is that when you look at the inverse
image of
that set in S^k, there are only Ŝnitely many since you get
only the
Ŝnitely many elements whose length is k or less and such a
set is
closed and does not contain (0,0,...,0,k). Hence the inverse
image in
any S^k is closed so it is closed by deŜnition of the quotient
topology. This is not what I wanted, but I have to live with
it. As
a result, Sierpinskiıs theorem is of no help.
===
Subject: the math fails me
I canıt post these equations, they are too long.
However, if any of you can visit where theyıre at,
http://mypeoplepc.com/members/jon8338/polynomial/
and tell me what I did wrong? jongiff2000@yahoo.com
As far as I can tell, the math is perfectly correct, but
the answer just doesnıt make sense.
Jon
===
Subject: Re: the math fails me
> I canıt post these equations, they are too long.
> However, if any of you can visit where theyıre at,
> http://mypeoplepc.com/members/jon8338/polynomial/
> and tell me what I did wrong? jongiff2000@yahoo.com
> As far as I can tell, the math is perfectly correct, but
> the answer just doesnıt make sense.
What standard tells you the math is perfectly correct? If
the answers are wrong, the math is wrong. QED
> Jon Still beating a dead horse.
--
There are two things you must never attempt to prove: the
unprovable -- and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: the math fails me
> As far as I can tell, the math is perfectly correct,
As far as *I* can tell the math is nonsense. You see, real
math has
sentences explaining what is going on. And perhaps a few
equations if
needed.
===
Subject: Re: the math fails me
> I canıt post these equations, they are too long.
> However, if any of you can visit where theyıre at,
> http://mypeoplepc.com/members/jon8338/polynomial/
> and tell me what I did wrong? jongiff2000@yahoo.com
> As far as I can tell, the math is perfectly correct, but
> the answer just doesnıt make sense.
> Jon
Back in grad school, we used to drink Sneaky Petes before
lectures (Sloe
Gin Ŝzzes). After four or Ŝve, everything made sense!
John
===
Subject: Question on SemiDeŜnite Programming
Hi all,
I have a question on a basic semideŜnite programming problem.
Let B=B_0 + lambda_1 B_1+...+lambda_n B_n, where each B_i is
a real
symmetric k by k matrix and lambda_1,...,lambda_n are
parameters.
The problem is: How to Ŝnd the lambda_i such that B is
positive
deŜnite (or semideŜnite (psd)) ?
I know that B is psd iff all eigenvalues are nonegative and
that if
we consider the characteristic polynomial F(y) of B which is
F(y)=y^k+b_{k-1) y^{k-1)+...+b_0 then by Descarteıs rule of
signs F(y)
has only nonnegative rules iff (-1)^{i+k}b_i ge 0 for all
i=0,...,k-1
(b_i are function of lambda_1,...lambda_n). So the goal is to
Ŝnd
lambda_i such that the set
S={(lambda_1,...,lambda_n) in R^n |
(-1)^{i+k}b_i(lambda_1,...,lambda_n) ge 0}
is nonempty and a point is S is a solution to the initial
problem.
Now, my question is how to check that a set like S is
nonempty ? and
how to Ŝnd necessary (and/or sufŜcient) conditions on the B_i
such
that S is nonempty ?
thank you.
===

===

===
Subject: Re: Cosets
> Can anyone suggest a good text book to study about
cosets....I am
> totally confused about them.
> Fraleighıs book is pretty good.
??
An entire book all about cosets? Are we thinking of
the same notions of cosets? Subsets of a group of the
form x.H where H is some subgroup?
Herman Jurjus
===
Subject: Re: Carmichaels theorem
>[...]
>Can you see that Carmichaelıs function is just the LCM of the
>values phi(q), where q ranges over the prime power divisors
of n?
>[...]
> Not quite. Carmichaelıs function is the exponent of the
prime residue
> class group mod n, which in turn is the LCM of the
exponents of the
> prime residue class groups mod q, where q ranges over the
prime power
> divisors of n. Now, those exponents are phi(q), indeed, if
q is odd or
> q=2^e, e<3, since the underlying group is cyclic in this
case. If q=2^e
with
> e>=3, this is no langer the case and the exponent is only
phi(q)/2. This
> accounts for the weird deŜnition of lambda(n).
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: JSH: Math journal review
===
>Subject: Re: JSH: Math journal review
>did they give you any editorial feedback, or
>did they just cave into the assembled masses
>of Usenet?
> do you think Ullrich called him a honky?
>>Argyros already edits several mathematical journals,
including
I didnıt write that.
Idiot.
>--Give Earth a Trickier Dick Cheeny -- out of ofŜce, after
gigayears!
>http://www.benfranklinbooks.com/
>http://members.tripod.com/~american_almanac
>http://www.wlym.com/pdf/iclc/HowTheNation.PDF
>http://www.rand.org/publications/randreview/issues/rr.12.00/
>http://www.rwgrayprojects.com/synergetics/plates/Ŝgs/plate02
.html
--
Mensanator
Ace of Clubs
===
Subject: Re: polya-burnside enumeration
> Can anybody furnish me with some tips/references on
applications of the
> polya-burnside enumeration method to problems involving
symmetries of
> three-dimensional shapes? (This is part of my ongoing
attempts to become
more
> geometrical.) Some broad hints would probably be enough to
get me
started.
Mucho
> appreciado.
> Peace
Try Tuckerıs Applied Combinatorics. Many 2d-applications some
3d. There is
really no difference between the 2d and 3d cases except in
description of
the symmetry groups. For some discussion of 3d symmetry
groups of Ŝgures
try Michal Artinıs Algebra, Ch 5.
Best,
Andy
===
Subject: Re: Name of this group?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i4P52ne02553;
>My group appears to be QxC_3. Here some more info:
>The above group is isomorph to a subset of the
>algebra over the reals generated from the factor group
>F = QxQ/C(QxQ) , C = group center. Now, there is more to
this:
>There are actually two (and very likely only two) such groups
>on this space isomorph to QxC_3 and hence to each other.
>Multiplying elements between these groups amongst themselves
>will, in all likelyhood, produce a single parent group which
>should have more than 60 or so elements at the very least- I
will
>check this soon.
60 or so appears to have been a major understatement.
I now have the order of this (somewhat mysterious)
parent group to be at: 8*32 + 32 + 2 + 2*a
where a very reasonable guess of a is the number of
ways of multiplying the following row vectors pointwise to
create new vectors.
(-1, 1, 1, 1, 1, 1, -1, 1, 1, -1, -1, 1, 1, 1, 1, 1)
(1, -1, 1, 1, 1, 1, 1, -1, 1, 1, 1, -1, -1, 1, 1, 1)
(1, 1, -1, 1, 1, 1, 1, 1, -1, 1, 1, 1, 1, -1, -1, 1)
(-1, -1, -1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, -1)
(1, 1, 1, -1, -1, -1, 1, 1, 1, 1, 1, 1, 1, 1, 1, -1)
(1, 1, 1, -1, 1, 1, -1, 1, 1, 1, 1, -1, 1, -1, 1, 1)
(1, 1, 1, 1, -1, 1, 1, -1, 1, -1, 1, 1, 1, 1, -1, 1)
(1, 1, 1, 1, 1, -1, 1, 1, -1, 1, -1, 1, -1, 1, 1, 1)
That may a lot more than my original guess of 60, but it is
still
a long shot below the Ŝrst upper bound I was able to Ŝnd for
the
parent group, which was 8*32 + 32 + 2 + 2^{16} (!)
Have a nice evening,
C. Dement
>Want to see what this group looks like graphically?
>Check
www.uni-oldenburg.de/~jojo6/Floretions/Floretionspage.html
>Let iı (i with an arrow to the right) = [(1,i)] in F
> Œi (i with an arrow to the left) = [(i,1)] = [(-i,-1)], etc.
> Œijı (a single element written as the
> juxtaposition of an i with an arrow to the right with
> a j with an arrow to the left) = [(i,j)]
>Then the element (i,C_1) in QxC_3 would appear to correspond,
>say, to the element Œi + Œiiı + Œijı + Œikı ; the gray
>shaded area of Floretıs cube.
>Similarly, (j,C_1) to Œj + Œjjı + Œjiı + Œjkı, etc. (could
be slightly
>mistaken in that (j,C_1) actually corresponds to
>ık + Œkkı + Œkiı + Œkjı).
>Sincerly,
>C. Dement
===
Subject: Re: Bijections between uncountable sets
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i4P52qJ02661;
>> ...
>>And thereıs another thing thatıs confusing me. Though no
topology was
>>speciŜed for X and Y, the problem says they are
uncountable. When you
>>say something about the cardinality of a set, arenıt you
automatically
>>assuming a topology is deŜned on it?
>> Good heavens, no. It would be interesting (I mean this
quite
>> sincerely) to know how and where you got that idea, in as
much
>> detail as you can provide.
>This is just a point I have some difŜculty to understand. It
doesnıt
>come from any book. Iıll try to exemplify why it seems to me
you got
>to have some information about a set to say itıs countable or
>uncountable. Again, letıs consider R with the Euclidean
metric. I know
>2 classic proofs that R is uncountable:
You may know only two, but there are many others. In the
statement you are
given two general uncountable sets - not necessarily those
that you know how
to prove are uncountable.
> One of them is based on the
>fact that every real number has a decimal expansion. Then,
we can use
>that well known proof due to Cantor. But to use this proof,
you have
>to know that every real number has a decimal expansion, and
so you
>have to make some assumptions about the objects that form R.
I think
>you donıt actually need a topology, but at least you need
order, you
>have to know R is an ordered Ŝeld.
> You need to know you can add and multilply 2 elements of R
and get another
element of R.
You do not need the order, addition and multiplication. The
proof works just
as well for the set of all functions f:N->{0,1} (which, in
and of itself, has
no inherent notion of order, addition, and multiplication).
>The other proof relies on the fact that R is complete (the
deŜnition
>based on Cauchy sequences), which implies every nested
collection of
>closed non-empty intervals has a common point. And also
relies on the
>fact that R has no isolated point. We can also prove that
every
>compact Hausdorff space with no isolated points is
uncountable.
>But such proofs are based on topolpgies. The Euclidean
metric makes R
>into a complete metric space with no isolated points, wich
implies
>perfect subsets of R are uncountable. If, on the other hand,
you take
>R with the discrete metric, the that proof based on nested
closed
>intervals doesnıt work, because, then, every real number
becomes an
>isolated point of R. And, in addition, as far as I know, itıs
>impossible to talk about closed and open sets, Hausdorff
spaces and
>the like without taking into account some topology.
Here is a simple example of another uncountability proof: for
every
countable set X, P(X) is uncountable (where P(X) is the set
of all subsets of
X). This is a direct result of a theorem known as Cantorıs
theorem. You can
Ŝnd the proof in any elementary text on set theory.
>>Does it make sense to say a set
>>is countable or uncountable if all you know about it is
that itıs just
>>a set of objects? For example, we can prove R, in the
Euclidean
>>metric, is uncountable because we know, at least
axiomatically, that
>>itıs complete, is ordered and every non empty set bounded
above has a
>>supremum.
>> Your because lacks logical force. Consider the entirely
similar
>> claim we can prove that the singleton {0}, in the
Euclidean metric,
>> is uncountable because we know, at least axiomatically,
that itıs
>> complete, is ordered and every non empty set bounded above
has a
>> supremum.
>Yes, thatıs right. But {0} has only one element and itıs and
isolated
>point, because {0} is open.
>>If I only say let A be a set of objects, then thereıs
>>nothing one can say about itıs cardinality.
>> What deŜnition of cardinality are you using (or do you
think you
>> are using)? Are you working by yourself, from a book or
other written
>> source? Are you taking a course? Where do your ideas come
from, here?
>The deŜnition of cardinality Iım aware of is the one
presented in
>most Analysis and Topology books. Cardinality is a rather
relative
>concept, right? We say X and Y have the same cardinality if
thereıs a
>bijection fron X to Y. And if A is Ŝnite, we say A has
cardinality n
>if itıs equivalent to the set {1, 2....n}. Right?
>I study Math on my own, and Iım trying to get into ponts I
couldnıt
>study when I was graduating. I may be wrong, but I really
canıt see
>how someone can say a set is countable or uncountable
without having
>some information about the set, about the relations between
itıs
>elements.
Of course you need some information, but that has nothing to
do with the
statement at hand. When someone tells me x is a number I
canıt say
whether it is even or not unless I have some more
information. But that
doesnıt mean that the proposition x is an even number is
invalid. Your
statement should be true (even though it is not true) for all
uncountable
sets, regardless of whether you can prove them to be
uncountable.
>If someone tells me that A is a set of objects and asks me
>about tıs cardinality, then only honest answer I can give is
I donıt
>know.
If, on the other hand, someone tells you that A is an
uncountable set of
objects and asks you about itıs cardinality, you can say
something: it is
greater than Aleph0.
>Amanda
Cron
===
Subject: @ HOW DESPERATELY DO THE FRIEDMANS NEED INDIANS !!
Comments: This message probably did not originate from the
above address.
It was automatically remailed by one or more anonymous mail
services.
You should NEVER trust ANY address on Usenet ANYWAYS: use PGP
!!!
Get information about complaints from the URL below
X-Remailer-Contact: http://80.65.224.85/POL/ In case my abuse
address is
unreachable: It is because it has been ŝooded by
<mgg@uiuc.edu>, please
contact <abuse@uiuc.edu>
X-Mail2News-Contact: http://80.65.224.85/
It can certainly be very interesting to watch (friends of
Israel)
in their hour of need.
Will Friedmen display this kind of love and humanity for the
poor
in Palestine ??
> Making India Shine
> By THOMAS L. FRIEDMAN
> India just had a stunning election, with incumbents across
the country
> thrown out, largely by rural voters. Clearly rural Indians,
who make
> up the countryıs majority, were telling the cities and the
government
> that they were not happy with the direction of events. I
think I can
> explain what happened, but Ŝrst I have to tell you about
this wild
> typing race I recently had with an 8-year-old Indian girl
at a village
> school.
> The Shanti Bhavan school sits on a once-scorpion-infested
bluff about
> an hourıs drive -- and 10 centuries -- from Bangalore,
Indiaıs Silicon
> Valley. The students are all untouchables, the lowest caste
in
> India, who are not supposed to even get near Indians of a
higher caste
> for fear they will pollute the air others breathe. The
Shanti Bhavan
> school, with 160 students, was started by Abraham George,
one of those
> brainy Indians who made it big in high-tech America. He
came back to
> India with a single mission: to start a privately Ŝnanced
boarding
> school that would take Indiaıs most deprived children and
prove that
> if you gave them access to the same technologies and
education that
> have enabled other Indians to thrive in globalization, they
could,
> too.
> I visited Mr. Georgeıs school in February, and he took me
to a
> classroom where 8-year-old untouchables were learning to
use Microsoft
> Word and Excel. They were having their computer
speed-typing lesson,
> so I challenged the fastest typist to a race. She left me
in the dust
> -- to the cheering delight of her classmates.
> Dust is an appropriate word, because a drought in this area
of
> southern India has left dust everywhere. These kids --
their parents
> are ragpickers, coolies and quarry laborers, said the
schoolıs
> principal, Lalita Law. They come from homes below the
poverty line,
> and from the lowest caste of untouchables, who are supposed
be
> fulŜlling their destiny and left where they are --
according to the
> unwritten laws of Indian society. We get these children at
age 4. They
> donıt know what it is to have a drink of clean water [or
use a
> toilet]. They bathe in Ŝlthy gutter water -- if they are
lucky to
> have a gutter near where they live. They donıt even have
proper scraps
> of clothing. We have to start by socializing them. When we
Ŝrst get
> them, they run out and urinate and defecate wherever they
want. [At
> Ŝrst] we donıt make them sleep on beds because it is a
culture
> shock. Our goal is to give them a world-class education so
they can
> aspire to careers and professions that would have been
totally beyond
> their reach, and have been so for generations.
> After our little typing race, I asked the 8-year-olds what
they wanted
> to be. Their answers were: an astronaut, a doctor, a
> pediatrician, a poetess, physics and chemistry, a scientist
and
> an astronaut, a surgeon, a detective, an author. Looking at
> these kids, Mr. George said, They are the ones who have to
do well
> for India to succeed. (See his Web site, www.tgfworld.org.)
> And that brings us to the lesson of Indiaıs election: the
broad
> globalization strategy that India opted for in the early
1990ıs has
> succeeded in unlocking the countryıs incredible brainpower
and
> stimulating sustained growth, which is the best antipoverty
program. I
> think many Indians understand that retreating from their
globalizing
> strategy now would be a disaster and result in Indiaıs
neighborhood
> rival, China, leaving India in the dust. But the key to
spreading the
> beneŜts of globalization across a big society is not about
more
> Internet. It is about getting your fundamentals right: good
> governance, good education. Indiaıs problem is not too much
> globalization, but too little good governance. Local
government in
> India -- basic democracy -- is so unresponsive and so
corrupted it
> canıt deliver services and education to rural Indians. As
an Indian
> political journalist, Krishna Prasad, told me: The average
Indian
> voter is not saying, `No more reforms,ı as the left wants
to believe,
> but, `More reforms, pleaseı -- genuine reforms, reforms
that do not
> just impact the cities and towns, but ones which percolate
down to the
> grass roots as well.
> India needs a political reform revolution to go with its
economic
> one. With prosperity coming to a few, the great majority
are simply
> spectators to this drama, said Mr. George. The country is
governed
> poorly, with corruption and heavy bureaucracy at all
levels. I am a
> great advocate of technology and globalization, but we must
Ŝnd a way
> to channel their beneŜts to the rural poor. What is
happening today
> will not succeed because we are relying on a corrupt and
socially
> unfair system.
-------------------------------------------------------------
----------
On the one hand the conservative Benjamin Disraeli, a
passionate
advocate of imperial power and glory.
And on the other, his lifelong adversary, the liberal William
Gladstone, who championed the moral vision of Prince Albert
and David
Livingstone.
Gladstone was driven by a sense of high moral purpose, and a
heavy
burden of guilt, in part because his own family had once made
a
fortune from slave labour.
As the leader of the Liberal Party Gladstone campaigned for
the export
of civilised values through commerce, not conquest.
Gladstone feels that the empire is there, thereıs not much
you can do
about it.
He doesnıt want to add to it and he believes that imperialism
is a
creed which can contaminate the British people, make them
warlike,
aggressive.
Whereas he thinks of a world in which there is universal
peace.
When he looks at imperialism he says this, is this Godly and
he
decides it isnıt.
He sees it as might somehow triumphing over right.
And heıs rather frightened, if the British people get
entranced with
empire theyıll go gallivanting off Ŝghting wars here there and
everywhere, spend a lot of money and cease to be a moral
force in the
world.
This view was Ŝercely contested by his great rival, Benjamin
Disraeli. Disraeli Ŝrst moved into the Prime Ministerıs ofŜce
in
1867 and for the next 15 years he and Gladstone would
alternate in
power.
Disraeli believed in the expansion of the British Empire.
He liked to claim that his ancestors had been rich Venetian
merchants
trading with the Orient and this gave him a romantic
enthusiasm for
imperial adventures.
Disraeli viewed the empire as an extraordinary asset. The
empire made
Britain a great power, a global power and also enabled it to
have
plenty of muscle in Europe.
And Disraeli of course likes the glamour of Empire. He sees it
bestowing prestige on the country. He eventually hopes that
the white
colonies will not follow the American course but remain
emotionally
tied to Britain particularly through the person of the Crown.
But Victoria was still in mourning. Since the death of Prince
Albert
she had lost interest in the Empire and all other affairs of
state.
Victoria went into what I call Purdah I think because she felt
incompetent to handle being a Queen. Albert had done the work
for her
so long, Albert had done everything, thought out everything
for her,
arranged everything for her that she did not feel she was up
to it
again.
The Queen found some consolation with the Scotsman John
Brown. She
began writing about him a few months after Albertıs death:
I have an invaluable Highland servant who is my factotum here
and
takes the most wonderful care of me, combining the ofŜces of
groom,
footman, page and maid, I might almost say as he is so handy
about
cloaks and shawls. He always leads my pony and always attends
me out
of doors ...
I think she also enjoyed his picking her up in his arms and
putting
her on her horse. And taking her off her horse again. For the
Ŝrst
time since Albert, she had a strong, brawny man who held her
in his
arms. And I think thatıs as far as the sexuality really went.
But
she enjoyed it
To the dismay of her family and government, the Queen and her
highland
servant became inseparable. A section of press and public
called her
Mrs. Brown and her absence from public duty was widely
condemned.
There were cartoons in the newspapers about this, showing an
empty
throne. There were editorials in the newspapers about it, why
are we
paying so much money to maintain a Royal Family because the
Royal
Family is the symbol of the Empire and of Britain and here we
donıt
have one.
It was Disraeli who would rekindle the queenıs interest in
public
affairs.
His relationship with Victoria had begun badly.
She saw him as an upstart, an opportunist, what the British
call a
chancer, but Disraeli with his considerable charm, set out to
win
her. His ofŜcial dispatches to her were spiced with social
gossip and
witty anecdotes.
Part of Disraeliıs job as Prime Minister was to write an
account of
what was happening in Parliament and what was going on in the
cabinet
to the Queen
And Disraeliıs letters to the Queen were wonderfully detailed
and
rather gossipy, and actually rather indiscreet. He probably
told the
Queen far more than he ought to have done particularly about
divisions
of opinion.
Most people, Prime Ministers, made these letters very brief
and rather
ofŜcial, but Disraeliıs letters to Victoria were full of sort
of
protestations of affection and love and loyalty, really
saying nothing
at all, they were largely sugar, but Queen Victoria lapped it
up.
Vicky: Mr. Disraeliıs reports are just like his novels -
highly
coloured.
She had never had such letters in her life, she declared, and
had
never before known everything. Her attitude to the upstart
underwent
a dramatic change.
Mr Disraeli has achieved his present high position entirely
by his
ability, his wonderful happy disposition ... and I have
nothing but
praise for him.
She sent him primroses that she picked herself.
In return, Disraeli gave her a set of his novels.
Victoria had just published a book of her own, a reminiscence
of her
days with Prince Albert at their palace in Scotland.
Disraeli was awfully good at just saying the tactful remark
that Queen
Victoria would enjoy. For example one of the best was
Disraeli saying
to her, we authors maıam, which was precisely what Victoria
longed to
hear, that they were both part of the same club of writers.
Disraeli bewitched the Queen with his romantic vision of the
British
Empire. It would have horriŜed Prince Albert. In the future,
Victoria and Disraeli would form a powerful alliance for the
imperial
cause - but it would be some time before their partnership
would bear
fruit.
Disraeli Ŝrst term as Prime Minister lasted less than a year.
When
he was voted out of ofŜce, the Queen had to send for the
leader of
the liberals - Gladstone.
Victoria began by liking Gladstone, he seemed to be an
upright man. He
was ambitious but he was also extremely smart.
Prince Albert had warmly approved of Gladstone. When the new
Prime
Minister came to the palace to receive the seals of ofŜce,
the Queen
recorded her approval:
He is very agreeable, so quiet and intellectual, with such a
knowledge of all subjects and is such a good man ...
But her satisfaction did not last. Gladstone embarked on a
whirlwind
of liberal reforms that revived conservative instincts in the
Queen
that had been dormant while Albert was alive.
Mr Gladstone is a very dangerous man. And so very arrogant,
tyrannical and obstinate, with no knowledge of the world or
human
nature ... All this and much want of regard towards my
feelings make
him a very dangerous and unsatisfactory Premier.
She was not amused when he proposed that sailors might be
permitted to
grow beards. And she was horriŜed by moves towards female
equality.
The Queen draws Mr Gladstoneıs attention to the mad and
utterly
demoralising movement of the present-day to place women in
the same
position as men.
But it was Gladstoneıs private life that caused Victoria the
most
concern.
Because of his fanatical religion he felt everyone had to be
converted to his ways of morality and ethics. He would go out
on the
streets at night, even when he was Prime Minister, and solicit
prostitutes, take them back to their rooms, give them bibles.
He
would give them money and he would ask them to tread the
straight and
narrow ways.
Victoria got to know this because her maids in waiting told
her
everything and it repelled her. At one point when Gladstone
was to go
up to visit Victoria at Balmoral, she sent him a letter
telling him
that when he arrived it was to be with a new suit of clothes
that he
had never worn before. It was very clear that she wanted
nothing of
the degrading atmosphere of his involvement with these ladies
of the
evening.
Gladstone was unconcerned by the Queenıs personal disapproval
of him
But he was appalled the imperialist ideas she had picked up
from
Disraeli. His own more liberal views of Britainıs role were
conŜdently being put to the test in Africa.
David Livingstone had returned to his Dark Continent. This
time he had
been sent on an ofŜcial mission to Ŝnd a trading route into
the
interior and to achieve his dream of combining Commerce,
Civilisation
and the Christian religion.
To this end he was provided with generous funds by the British
government and accompanied by six British scientists and his
wife,
Mary, herself a devoted missionary.
Livingstone believed that the Zambezi River could become a
great
highway for British industrial goods.
But as they voyaged along the river, the expedition ran into
dangerous
rapids.
VICTORIA: I saw Mr. Disraeli at quarter to three today. He
knelt down
and kissed hands saying: ŒI plight my troth to the kindest of
Mistressesı. The silver-tongued charmer was back in ofŜce. As
he
had once conŜded to a friend:
DISRAELI: You have heard me called a ŝatterer and it is true.
Everyone likes ŝattery and when you come to royalty you
should lay it
on with a trowel.
Disraeli always loved the company of women and he was very
good at
ŝattering women. And I think with Queen Victoria he was able
to see
that she was lonely. And Disraeli was able to charm her, and
to
ŝatter her. And I think very importantly one of the things
that
Disraeli did was to encourage her to take a far more active
role in
public affairs. The result of this was that basically he had
Queen
Victoria as an ally particularly when he was Prime Minister.
And this was absolutely crucial I think to the success of
Disraeliıs
Ministry. That the Monarchy was behind it. Disraeli set out to
increase Britainıs prestige and expand Victoriaıs Empire, and
within a
year of taking ofŜce fate dealt him a brilliant opportunity.
Just Ŝve years before the Suez Canal had been carved through
the
Egyptian desert. It permitted ocean-going ships to pass
between the
Mediterranean and the Red Sea, linking Europe and the East.
For
Britain it was the lifeline to her greatest imperial
possession.
India with 400 million people was the largest overseas
territory any
empire has ever owned. The Queen called it the most precious
jewel in
her crown and Disraeli feared that a rival power would cut
the new
imperial artery.
The Suez Canal was absolutely crucial to Britainıs Empire.
Suez was
the jugular vein if you like of the British Empire.
It was through the Suez Canal that the route to India, the
short route
to India which was so strategically important happened.
So for Disraeli it was very important and he was quite right
in this I
think, that Britain should have a controlling inŝuence over
the Suez
Canal.
The shares in the Suez Canal company were owned by a number
of French
investors and the ruler of Egypt, the Khedive.
The Khedive had spent Egyptıs wealth on palaces, museums and
railways. Now, he was deep in debt to banks in London and
Paris. The
Canal showed no prospect of paying a dividend for years and he
desperately needed funds. In 1875, he made a secret offer to
the
British.
DISRAELI: Mr. Disraeli with his humble duty to your Majesty:
the
Khedive, on the eve of bankruptcy, appears desirous of
parting with
his shares in the Suez Canal and has communicated,
conŜdentially ...
ŒTis an affair of millions, about four at least but it is
vital to
your Majestyıs authority and power at this critical moment
that the
Canal should belong to England. The Khedive now says that it
is
absolutely necessary that he should have between three and
four
millions sterling by the 30th of this month! Scarcely
breathing time!
But the thing must be done!
The Queen replied by telegram the following day, approving
his course
of action but fearing that it would be difŜcult to arrange.
Normally, Parliament could have granted a government loan. But
Parliament was not in session and a French consortium had
already bid
for the shares.
Disraeli sent his private secretary to seek help from an old
friend.
Baron Rothschild was head of the great banking family and one
of the
richest men in the world. The secretary explained that
Disraeli
needed four million pounds - the price of the Khediveıs
shares in the
Suez Canal.
When? asked Rothschild. By tomorrow, said the secretary.
Rothschild picked up a grape, spat out the pips, and said:
What is
your security? The British Government, was the reply. You
shall
have it, said the Barron
DISRAELI: It is just settled: you have it, Madam. The French
Government has been outgeneraled and the entire interest of
the
Khedive is now yours.
The Queen was delighted. Disraeli treated her not only as his
monarch
but as a woman and a woman of intelligence.
security for India! An immense thing. Mr. Disraeli said that
my
support had been a great help. His mind is so much greater
and his
apprehension of things great and small so much quicker, than
that of
Mr. Gladstone.
Gladstone was strongly opposed the deal because he thought it
would
draw Britain into new imperial commitments. He was right.
The Suez Canal was to drag the British deeper and deeper into
the
murky politics of the Middle East. The overlords of the
entire region
were the Turks but many of their subject peoples were rising
against
them. Faced with rebellion on all sides, the Turks resorted
to mass
slaughter. Russia backed the rebels and Disraeli feared that
the
Turkish Empire would collapse and open the way for the
Russians to
advance on the Suez Canal. However badly they treated their
subjects,
Disraeli thought Britain had to support the Turks. But
Gladstone
thought otherwise.
He was no longer leader of the Liberals. He had retired to his
country estate where he relaxed by chopping down trees and
setting
down his thoughts on God. But he was appalled by stories of
Turkish
atrocities against their Christian subjects. He thought the
corrupt
and crumbling Turkish empire should be brought to an end. He
laid
down his axe and took up his pen.
GLADSTONE: There is not a cannibal in the South Sea Islands
whose
indignation would not arise and overboil at the recital of
what has
been done ... Let the Turks now carry away their abuses in
the only
possible manner, namely by carrying off themselves ... This
thorough
riddance, this most blessed deliverance, is the only
reparation we can
make to the memory of these heaps and heaps of dead; to the
violated
purity alike of matrons, maiden and of child ...
Disraeli called the style of Gladstoneıs protest vulgar,
remarking
that of all the atrocities Gladstoneıs writings were probably
the
worst.
But Gladstone had caught the public mood and in the House of
Commons,
Disraeli was forced to choose his words with more care:
DISRAELI: Our duty at this critical moment is to maintain the
Empire
of England. Nor will we agree to any step, though it may
obtain for a
moment comparative quiet and a false prosperity, that hazards
the
existence of Empire.
Disraeli backed his words with action. As the Russians
advanced on the Turkish capital, he dispatched a British
ŝeet, led by
the most powerful battleship in the world, HMS Devastation.
Public
opinion swung to Disraeliıs side.
SINGERS: We donıt want to Ŝght but by Jingo if we do, Weıve
got the
ships, weıve got the men, weıve got the money, too!
War fever swept through the pubs and music halls of Britain.
The
British may not have liked what the Turks were doing to their
Christian subjects but they shared Disraeliıs determination
to stop
the Russians.
Fearful of war with Britain the Russians agreed to negotiate.
Disraeli
set off to attend peace talks.
He returned in triumph. His diplomacy had forced the Russians
to halt
their advance on the Middle East. The lifeline to India was
secured.
Victoria shared the public rejoicing and decided it was the
right
moment to claim what she considered to be long overdue.
VICTORIA: In common conversation I am sometimes called
Empress of
India. Why have I never ofŜcially assumed this title? I feel I
ought to do so and wish to have preliminary inquiries made.
Disraeli introduced a bill in Parliament to bestow on
Victoria the
title: Queen-Empress of India. Gladstone led the opposition,
calling
the move:
GLADSTONE: Theatrical bombast and folly.
But the title was granted and the Queen was delighted. She
expressed
her gratitude by making Disraeli an Earl.
if you like, embellishing the British Monarchy. And at the
same time
the Queen is given a new sense of responsibility. She is
deeply
interested in India. Immediately she is made Empress she sets
out to
learn Hindustani. Doesnıt make much headway but a lot of good
will
there. And she also hires Indian servants.
Between them, the Queen-Empress and her newly ennobled Prime
Minister
appealed to an imperial spirit that was spreading through
large
sections of the British public.
An aggressive spirit, ŝexing British muscle and lording it
over the
world. Gladstone continued to oppose it. He called it Showy
imperialism.
DERBY: Disraeli believes thoroughly in Œprestigeı and would
think it
in the interests of the country to spend 200 millions on a
war if the
result was to make foreign States think more highly of us.

The Queen backed Disraeli to the hilt.
VICTORIA: If we are to maintain our position as a Ŝrst rate
power, we
must with our Indian Empire and large colonies be prepared
for attacks
and wars somewhere or other CONTINUALLY
But the strain of
this imperialist policy was beginning to show
British forces in southern Africa had clashed with the most
powerful
warrior nation on the continent - the Zulus.
At the battle of Isandhlwana 600 British soldiers were wiped
out to a
man. It took 17,000 British reinforcements armed with the
latest
artillery to defeat an enemy armed largely with spears.
Back in England a powerful voice was raised in protest.
GLADSTONE: Remember the rights of the savage as we call him.
Remember that the happiness of his humble home.
Gladstone
was no longer in control of Parliament, so he appealed
directly to the
British people.
GLADSTONE: The sanctity of life in the hill villages, is as
inviolable
in the eye of Almighty God as can be your own.
The power of
his oratory drew vast crowds. Ten thousand Zulus had died, he
claimed
.
GLADSTONE: For no other offence than to defend against your
artillery
with their naked bodies their hearths their homes, their
wives, their
families ...
ROY JENKINS: I mean it is one of the great mysteries about
Gladstone,
how his oratory was so effective. Because he wasnıt a
tremendous
phrase maker and he didnıt talk down to his audiences, he
rather
talked up to them. And yet he held them for these very long
periods,
an hour and a half was quite normal in great mass meetings.
I think it was essentially his physical presence, the ŝash of
his
eagleıs eye. The drama of his gestures, the cadence of his
voice.
The Queen was outraged. She complained in her journal: Mr.
Gladstone
is going about like an American stumping orator making most
violent
speeches.
But to her surprise and dismay Gladstone had struck a popular
chord.
Once more, he had appealed to the British sense of justice
and fair
play. They voted the Liberals back into power with a massive
majority.
imperialism he represented.
GLADSTONE: It is like the vanishing of some magniŜcent castle
in an
Italian romance.
Prince Albert would have shared Gladstoneıs
pleasure at the dismissal of Disraeliıs war mongering
government. But
Victoria had turned her back on Albertıs moral vision for the
Empire. She stubbornly refused to accept Gladstone as her new
Prime
Minister.
---------- Cecil Rhodes helped by Jewish Maximes - the
weapons of mass
murder -----------
Africa. The Dark Continent of the early explorers became the
stage for the
Ŝnal act in the story of Queen Victoriaıs Empire. In the
footsteps of
missionaries like David Livingstone, the powers of Europe
conducted a
brutal race for colonies, a race that would become known as
the Scramble
for Africa.
===
Subject: Probability / Combinatorical Problem...
Hello...
I have 3 counters A, B and C that start at 1. For an
experiment X, we
deŜne 3 mutually exclusive events eA, eB and eC that occur
with
probabilities p(eA), p(eB) and p(eC).
Each time an event eA occurs, you gain wA1 if A <=4 and wA2
if A = 5,
for eB itıs wB1 if B <= 4 and wB2 if B=5, and for eC itıs wC1
if C <=
4 and wC2 if C=5.
Then the respective counter is increased by 1 if itıs smaller
than 5.
If it already had reached 5, it is reset to 1 and the other
two
counters are set to 5.
If you choose to rerun X inŜnitely, what are the expected
gains for
eA, eB and eC?
Well if I only had one counter it would be easy, but how do I
deal
with those several interacting counters? My Ŝrst idea was to
consider
a combination of the 3 counters as one state and then solve
the
problem with Markov chains, but that would lead to a
(5^3)*(5^3)-matrix and I donıt think thatıs a clever idea
considering
that I have a similar problem with much larger counters.
Does anyone know which trick is needed for this problem?
===
Subject: Re: Probability / Combinatorical Problem...
>Hello...
>I have 3 counters A, B and C that start at 1. For an
experiment X, we
>deŜne 3 mutually exclusive events eA, eB and eC that occur
with
>probabilities p(eA), p(eB) and p(eC).
>Each time an event eA occurs, you gain wA1 if A <=4 and wA2
if A = 5,
>for eB itıs wB1 if B <= 4 and wB2 if B=5, and for eC itıs
wC1 if C <=
>4 and wC2 if C=5.
>Then the respective counter is increased by 1 if itıs
smaller than 5.
>If it already had reached 5, it is reset to 1 and the other
two
>counters are set to 5.
>If you choose to rerun X inŜnitely, what are the expected
gains for
>eA, eB and eC?
>Well if I only had one counter it would be easy, but how do
I deal
>with those several interacting counters? My Ŝrst idea was to
consider
>a combination of the 3 counters as one state and then solve
the
>problem with Markov chains, but that would lead to a
>(5^3)*(5^3)-matrix
It leads to a 13 by 13 matrix. Since you are going to rerun X
inŜnitely, you do not need to consider the states that are
only
passed once.
> and I donıt think thatıs a clever idea considering
>that I have a similar problem with much larger counters.
>Does anyone know which trick is needed for this problem?
I certainly donıt. :)
===
Subject: Re: Real Number
> There are at least two proofs (both by Cantor) that
> show that a 1-1 and onto mapping f:N -> R is not possible,
> so thereıs at least two inŜnities (and I believe there are
> an inŜnite number of inŜnities, each one constructible
> using a power set construction,
Sure, there are inŜnitely many cardinals. But what is the
cardinality of the set bijectively equivalent inŜnite sets?
;-)
Œcid Œooh
PS - If someone could answer this, that would be really sweet
===
Subject: Re: Real Number
>
> There are at least two proofs (both by Cantor) that
> show that a 1-1 and onto mapping f:N -> R is not possible,
> so thereıs at least two inŜnities (and I believe there are
> an inŜnite number of inŜnities, each one constructible
> using a power set construction,
> Sure, there are inŜnitely many cardinals. But what is the
> cardinality of the set bijectively equivalent inŜnite sets?
;-)
> Œcid Œooh
> PS - If someone could answer this, that would be really
sweet
Depending on your axioms this would not be a set but rather a
proper
class. To see this, weıll show that the collection of all
sets of
cardinality 2 is in 1 - 1 corrrespondence with the collection
of all
sets. Let E be the empty set and S be any set. The ordered
pair
(E, S) or in other symbols, {E, {E, S}}
has cardinality 2 and there is one unique such set for every
set S.
===
Subject: Re: Real Number
>> There are at least two proofs (both by Cantor) that
>> show that a 1-1 and onto mapping f:N -> R is not possible,
>> so thereıs at least two inŜnities (and I believe there are
>> an inŜnite number of inŜnities, each one constructible
>> using a power set construction,
> Sure, there are inŜnitely many cardinals. But what is the
> cardinality of the set bijectively equivalent inŜnite sets?
;-)
There is no such set (as I suspect you are aware), but that
each one
constructibele using a power set construction part is
equivalent to the
generalized continuum hypothesis.
> Œcid Œooh
> PS - If someone could answer this, that would be really
sweet
Not even a little acidic?
--
Dave Seaman
Judge Yohnıs mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: Irreducible polynomials over Q
>> Iıve asked Mathematica to factor (over Q) the polynomial
>> x^{24} - x^{12} + 1 and the answer was that itıs
irreducible.
>> Why? Itıs the 72nd cyclotomic polynomial.
> The cyclotomic polynomials are irreducible.
Of course they are!
> And no, I donıt know how to prove it.
Youıll Ŝnd it in nearly every book that deals with Galois
theory.
See, for instance, Jacobsonıs Basic ALgebra, vol. I.
Jose Carlos Santos
===
Subject: Re: Peanoıs space-Ŝlling curve
message
>><snip>
<more snipping>
>> t = 0.d1d2d3d4...
>>
>> (for example 1/2 = 0.5000..., 1/3 = 0.333..., etc.)
>> If t is as above then deŜne x and y by taking the
>> odd-numbered digits of t and the even-numbered
>> digits of t:
>I had considered a procedure similar to this, but
rejected
>it out-of-hand because it seems to hinge on the axiom of
>choice.
> This has nothing whatever to do with the axiom of choice.
Then one of us has absolutely no idea what the axiom of
choice is. You may choose to identify this person as me,
because I view your function above as a selection function.
>By the way, I donıt recall I actually said
>impossible - very remiss of me if I did and I accept a
slap
>on the wrist. Can you, in your turn point out why? ;-)
> Point out why _what_?
Are you being deliberately obtuse? I accept a slap on the
wrist because I am a scientist and therefore forbidden to
use the word impossible.
< big snip>
> The last three paragraphs are nothing but pompous
> nonsense. (Nonsense because nothing that anyone
> has said in this thread has anything whatever to do
> with the axiom of choice, and pompous because
> youıre trying to sound erudite regarding matters
> you evidently donıt understand one bit.)
Now your not even being obtuse - just downright rude. If you
read more carefully I am being the opposite of erudite,
trying to elicit your aid by letting you know that I have
some slight but passing acquaintanceship with relevant
ideas so that you may post without feeling the need to go
back to square one. I also felt you might respond to a
little humour. It appears I was wrong - tant pis pour toi.
Try to lighten up a little, as you Americans say.
Re: my nonsense. The contention is that you are using a
choice function to map complex numbers into the reals.
Obviously you think otherwise, so be so kind as to address
that speciŜc point in words I can understand, or else let
somebody else do so.
><snip>
John
===
Subject: Re: Peanoıs space-Ŝlling curve
>>
>message
>>
>>><snip>
><more snipping>
> t = 0.d1d2d3d4...
>
> (for example 1/2 = 0.5000..., 1/3 = 0.333..., etc.)
> If t is as above then deŜne x and y by taking the
> odd-numbered digits of t and the even-numbered
> digits of t:
>>
>>I had considered a procedure similar to this, but
>rejected
>>it out-of-hand because it seems to hinge on the axiom of
>>choice.
>> This has nothing whatever to do with the axiom of choice.
>Then one of us has absolutely no idea what the axiom of
>choice is.
Thatıs very clear.
>You may choose to identify this person as me,
>because I view your function above as a selection function.
Itıs not a question of _identifying_ that person, that
person _is_ you.
Can you even _state_ the axiom of choice precisely?
>>By the way, I donıt recall I actually said
>>impossible - very remiss of me if I did and I accept a
>slap
>>on the wrist. Can you, in your turn point out why? ;-)
>> Point out why _what_?
>Are you being deliberately obtuse? I accept a slap on the
>wrist because I am a scientist and therefore forbidden to
>use the word impossible.
>< big snip>
>> The last three paragraphs are nothing but pompous
>> nonsense. (Nonsense because nothing that anyone
>> has said in this thread has anything whatever to do
>> with the axiom of choice, and pompous because
>> youıre trying to sound erudite regarding matters
>> you evidently donıt understand one bit.)
>Now your not even being obtuse - just downright rude. If you
>read more carefully I am being the opposite of erudite,
>trying to elicit your aid by letting you know that I have
>some slight but passing acquaintanceship with relevant
>ideas
But itıs clear that (at least when youıre talking about AC)
you have _no_ acquaintance with the relevant ideas!
You have acquaintance with the _words_.
>so that you may post without feeling the need to go
>back to square one. I also felt you might respond to a
>little humour. It appears I was wrong - tant pis pour toi.
>Try to lighten up a little, as you Americans say.
>Re: my nonsense. The contention is that you are using a
>choice function to map complex numbers into the reals.
You have no idea what the phrase choice function
means. By what seems to be your interpretation of
the axiom of choice, if I deŜne f(x) = 2x I must be
using the axiom of choice, because Iım choosing
to map x to 2x.
(My _guess_ would be that if someone asked you
whether the deŜnition f(x) = 2x requires AC youıd
answer no. If so then Iım at a loss to explain why
you think the deŜnitions above _do_ require AC;
those deŜnitions are no more problematic than
the deŜnition f(x) = 2x, theyıre just a little more
complicated.)
>Obviously you think otherwise, so be so kind as to address
>that speciŜc point in words I can understand, or else let
>somebody else do so.
AC says this: Suppose that A is a set and each element
of A is a nonempty set. Let S be the union of the elements
of A. Then there exists a function f : A -> S such that
f(x) is an element of x for every x in A.
This gets used in proving that various things exist even
though we canıt give explicit constructions of them.
All the functions mentioned above (where above
refers to the entire thread) are deŜned _explicitly_;
no AC involved anywhere.
>><snip>
>John
************************
David C. Ullrich
===
Subject: Re: Peanoıs space-Ŝlling curve
> t = 0.d1d2d3d4...
>
> (for example 1/2 = 0.5000..., 1/3 = 0.333..., etc.)
> If t is as above then deŜne x and y by taking the
> odd-numbered digits of t and the even-numbered
> digits of t:
>>
>>I had considered a procedure similar to this, but
> rejected
>>it out-of-hand because it seems to hinge on the axiom of
>>choice.
>> This has nothing whatever to do with the axiom of choice.
> Then one of us has absolutely no idea what the axiom of
> choice is. You may choose to identify this person as me,
> because I view your function above as a selection function.
I also identify that person as you. David C. Ullrich *deŜned*
a
function explicitly; he didnıt just say that by the axiom of
choice,
thereıs a selection function such that.... If you want to
learn what
the axiom of choice says, read
http://en.wikipedia.org/wiki/Axiom_of_choice
>>By the way, I donıt recall I actually said
>>impossible - very remiss of me if I did and I accept a
> slap
>>on the wrist. Can you, in your turn point out why? ;-)
>> Point out why _what_?
> Are you being deliberately obtuse? I accept a slap on the
> wrist because I am a scientist and therefore forbidden to
> use the word impossible.
Then we mathematicians are not scientists, because we prove
that
the certain things are impossible (such as squaring the
circle with
ruler and compass alone) all the time.
Jose Carlos Santos
===
Subject: Re: Peanoıs space-Ŝlling curve
message
<snip>
> But this is sufŜcient to show that a bijection exists
between the unit
> interval and the unit square. Thatıs because of the
Cantor-Bernstein
> theorem, which says that if you have two sets A and B,
each of which can
> be injected into the other, then a bijection exists. In
the case of the
> interval and the square, the bijection cannot be
continuous, however.
being cast over my darkness, at last!
John
===
Subject: Re: Peanoıs space-Ŝlling curve
>> John Morgan <john.morgan@REMOVECAPSataraxie.fr>
>>
>>
>message
<snippety,snip,snip,snip>
>. Semantically speaking, I was
>showing up my ignorance - only those who are
pig-ignorant
>like to show off their ignorance and I donıt think
either
>of us Ŝt into that category ;-))
> If you donıt _like_ to show off your ignorance then
> you _really_ need to stop making comments about all
> this and stick exclusively to asking questions. Honest -
> I canıt think of anything youıve said in this thread
> thatıs actually been correct.
I have no problem about being ignorant and will ignore the
obvious insult implied above when you use show off instead
of show up immediately after I explained the difference to
you. But question and answer is not much fun. Still, if it
was good enough for Socrates...
You must be using the word Honest in an American rather
than an English sense of the word. Maybe politics is your
calling, that is when you get tired of mathematics or too
old for it like me :-) N.b. In case you donıt understand,
thatıs a winking smiley at the end, which is intended to
show that I am not being too serious in my sarcasm.
<snip>
>> trying to be helpful, which is not the same as trying
to
>be nice.
>Have you ever considered trying to do both at the same
time.
>Iım told itıs possible though I havenıt seen a proof
<grin>
> It can be very difŜcult when youıre dealing with someone
> who clearly understands _nothing_ about the things
> heıs talking about, but who doesnıt let that stop him from
> repeatedly pointing out why various statements must
> be wrong. (Statements the _proofs_ of which are very
> well understood by advanced undergraduates whoıve
> actually studied the Ŝeld.)
Firstly, I have certainly questioned why certain statements
vis-a-vis space-Ŝlling are
considered to be correct. But nowhere have I said that
statements made by you or any other mathy were wrong. I
would never presume to do so. I have however, perhaps
naively, allowed my thoughts to ŝow too quickly from brain
to keyboard.
Secondly, I think I_understand_my religious beliefs as well
as any undergraduate would. Does that make them true for
everybody. If so, why arenıt you all Satanists like me? ;-)
N.b.Hope youıve remembered what the winking smiley means ;-)
Finally, as Iıve said elsewhere, it wouldnıt hurt you to
lighten up a bit. I used to think mathematics was quite a
fun thing, and good exercise for my brain. Then I started
posting to sci.math and am beginning to doubt it :-(
N.b.Thatıs the opposite of a smiley by the way and
signiŜes that I am not laughing. Pity really, just when I
was beginning to get a handle on some of this stuff.
John
===
Subject: Re: Peanoıs space-Ŝlling curve
in
>> whether the curve construct described by Peano that
>> claims to Ŝll a square, does actually do so.
>>
>> There is a simple example of how a discrete process
(if
>inŜnitely applied)can produce a continuous result. Take
the
>inŜnite series 1/2 + 1/4 + 1/8 + ...+ 1/2^n + ...= 1
>> This series represented as points on a straight line
means
>that for
>> reaching the point of abscissa 1 it must run over
>continuous points.
>> The Peano Curve follows this same pattern but in two
>dimensions.
non-mathy.
>Unfortunately I canıt seem to reconcile what you are
telling
>me here with what I thought I knew before!
>SpeciŜcally, the approach to the limiting point of this
>sum would appear denser and denser when represented as a
>graph of the partial sums. Each partial sum can be put
into
>a one-to-one correspondence with an integer so it seems
that
>the graph is a geometrical representation of the Aleph_0
>set. The idea you have offered me suggests that the
partial
>sums form a continuum at the limiting point and so become
>Aleph_1. Now Aleph_1 + a Ŝnite or countably inŜnite
>number of partial sums equals Aleph_1: a contradiction to
>the Ŝrst observation.
> Youıre not making _any_ sense.
> You mentioned the word biology recently. What youıre
> saying here is like if I said that the DNA spiders the
> natural chain polymerase selection, leading to
> mitochondrial dendrites. At last, something
> comprehensible to a non-biologist! Of course
> this contradicts the fact that the gene expresses
> microtubules.
There are grammatical usages that I donıt understand viz.
the verb spider.
There are semantic juxtapositions that I donıt immediately
recognise because of my next point. This is not my area of
biology. Iım a gamekeeper turned poacher - an erstwhile
chemist who latterly became an ecologist. If it were my
Ŝeld perhaps I could point out the semantic difŜculties,
or offer suggestions in plain language as to what I think
it is you want to say. To be fair, this is what you have
tried to do for me, and I would not want you to think for
one minute that I am not grateful - I am. But I need time to
assimilate all of it, because of the contradictions that I
Ŝnd and my inability to dump all my previous knowledge as
you suggest in another post on this subject. Quite simply, I
believe it is ultimately possible to reconcile my previous
understanding with some of what I have already learned here.
For example, you recently told me that aleph_1 was not the
same cardinality as the power set of aleph_0. I have tried
to discard all ideas that I misconceived by believing that
they were. Now Iım trying to understand which of these two
has the higher cardinality and why,if indeed higher has any
meaning. In fact I think I need to read a modern primer. Can
you recommend one?
John
===
Subject: Re: Peanoıs space-Ŝlling curve
message
>>
<snip>
> if he has any sense he pays
> no attention to his queasy feelings and tries to
> understand what theyıre talking about.
Maybe you should become a counsellor for President Dubıyah.
> (Which means we must never question the
> experts? Absolutely not. But we need to
> _understand_ what theyıre saying Ŝrst, before
> deciding itıs all wrong.)
I havenıt decided any such thing. What I am trying to decide
is if what you mathies have discovered about abstract spaces
is usable as a model for real space and time.
>But then scientists can never, ever know if they
>are correct ;-)
> Thatıs a big difference between mathematics
> and the natural sciences. Because on this side
> of the fence we have this thing called _proof_
> (yes, the word exists over there as well, but it
> doesnıt mean the same thing.)
But it does mean the same thing, and we know that we canıt
have it. N.b. I canıt prove that last statement ;-)
>> Especially since weıre talking about
>> abstract mathematics, and youıve stated repeatedly that
>> you donıt _know_ the math.
>Iım trying to know the math and you may be pleased to
hear
>that your posts are helping in this endeavour. But are we
>really talking about abstract mathematics here?
>I sketched part of the Peano curve and it looked pretty
>concrete to me. Perhaps all mathematics is abstract, even
>the applied stuff, so shall we just leave out the word
>abstract ?
> Yes, weıre talking about abstract mathematics. No,
> you have _not_ sketched that curve! You have
> sketched _other_ curves which form _approximations_
> to that curve.
Which gets right back to the nub of my problem. I draw a bit
of the construction and then imagine further recursions. I
note some points exist that will never be traced by any
number of countable recursions. I also Ŝnd points that are
included more than once and topology tells me that I no
longer have a structure that is homeomorphic with a line
(0,1). Any wonder then that I am confused by your assertions
that a piece of abstract algebra can be used to show that
neither of these concrete discoveries are correct.
When you read this you will also need to allow for the fact
that I am not able to put a mathspeak spin on it, but that
perhaps you should try to understand it nevertheless.
Otherwise we can never discuss anything because, as you have
so often pointed out, I donıt have the years and years of
immersion in this stuff that you have. I am assuming you do
want to discuss it as you keep on replying to my posts.
> The fact that those approximations actually converge
> to _something_ has a lot to do with the fact that the
> reals are complete. And the completeness of the
> reals is deŜnitely an abstract thing - there is nothing
> in the physical world that corresponds precisely to
> the real numbers (or if there is thereıs no way we
> could possibly know that.)
Despite this, the reals can be used as adequate models for a
huge range of natural objects. Other objects exist for which
we need two real numbers. For some of which, such as areas,
we can get away with one real, and a unit that is separately
deŜned. Itıs this aspect of the universe that I canıt yet
reconcile with Peanoıs construction, wherein a line and an
area have a correspondence that allows functions to take us
between them. N.b. I am lacking the proper mathspeak here
and hope you can Ŝnd the volition to try to understand what
I am gibbering about.
John
===
Subject: Re: Peanoıs space-Ŝlling curve
>>
>message
>
><snip>
>> if he has any sense he pays
>> no attention to his queasy feelings and tries to
>> understand what theyıre talking about.
>Maybe you should become a counsellor for President Dubıyah.
>> (Which means we must never question the
>> experts? Absolutely not. But we need to
>> _understand_ what theyıre saying Ŝrst, before
>> deciding itıs all wrong.)
>I havenıt decided any such thing. What I am trying to decide
>is if what you mathies have discovered about abstract spaces
>is usable as a model for real space and time.
>>But then scientists can never, ever know if they
>>are correct ;-)
>> Thatıs a big difference between mathematics
>> and the natural sciences. Because on this side
>> of the fence we have this thing called _proof_
>> (yes, the word exists over there as well, but it
>> doesnıt mean the same thing.)
>But it does mean the same thing, and we know that we canıt
>have it. N.b. I canıt prove that last statement ;-)
> Especially since weıre talking about
> abstract mathematics, and youıve stated repeatedly that
> you donıt _know_ the math.
>>
>>Iım trying to know the math and you may be pleased to
>hear
>>that your posts are helping in this endeavour. But are we
>>really talking about abstract mathematics here?
>>I sketched part of the Peano curve and it looked pretty
>>concrete to me. Perhaps all mathematics is abstract, even
>>the applied stuff, so shall we just leave out the word
>>abstract ?
>> Yes, weıre talking about abstract mathematics. No,
>> you have _not_ sketched that curve! You have
>> sketched _other_ curves which form _approximations_
>> to that curve.
>Which gets right back to the nub of my problem. I draw a bit
>of the construction and then imagine further recursions. I
>note some points exist that will never be traced by any
>number of countable recursions. I also Ŝnd points that are
>included more than once and topology tells me that I no
>longer have a structure that is homeomorphic with a line
>(0,1). Any wonder then that I am confused by your assertions
>that a piece of abstract algebra can be used to show that
>neither of these concrete discoveries are correct.
Nobody has denied either of those facts. In fact both
of them were explained to you _many_ times in this
thread. Exactly where did I assert what you say Iıve asserted?
>When you read this you will also need to allow for the fact
>that I am not able to put a mathspeak spin on it, but that
>perhaps you should try to understand it nevertheless.
Understand _what_?
>Otherwise we can never discuss anything because, as you have
>so often pointed out, I donıt have the years and years of
>immersion in this stuff that you have. I am assuming you do
>want to discuss it as you keep on replying to my posts.
>> The fact that those approximations actually converge
>> to _something_ has a lot to do with the fact that the
>> reals are complete. And the completeness of the
>> reals is deŜnitely an abstract thing - there is nothing
>> in the physical world that corresponds precisely to
>> the real numbers (or if there is thereıs no way we
>> could possibly know that.)
>Despite this, the reals can be used as adequate models for a
>huge range of natural objects.
That seems to be true, at least if we donıt look at things at
a Ŝne enough scale. When we start talking about a very
tiny scale itıs not at all clear to physicists what space and
time look like - they _could_ be something like the
reals, they could instead be discrete/quantized.
>Other objects exist for which
>we need two real numbers. For some of which, such as areas,
>we can get away with one real, and a unit that is separately
>deŜned. Itıs this aspect of the universe that I canıt yet
>reconcile with Peanoıs construction, wherein a line and an
>area have a correspondence that allows functions to take us
>between them. N.b. I am lacking the proper mathspeak here
>and hope you can Ŝnd the volition to try to understand what
>I am gibbering about.
>John
************************
David C. Ullrich
===
Subject: Re: 640,000,000,000 TO 1
>> I have seen this Ŝgure calculated out in a previous
>> but basicly if you take ALL the possible DNA options
>> and multiply them out you get a Ŝgure that rounds
>> down to 640 billion.
> Does that mean that there are 640 billion possible human
beings, up to
> differences in nurture? I presume some are more likely to
be generated
> than others so it is difŜcult to calculate exactly when the
human
> genome-space will have been completely Ŝlled, or how long
weıre likely
to
> have to wait before someone is duplicated. But a
back-of-the-envelope
> calculation shows that if the world population climbed to,
say, 12
billion
> (projected within the next half-century) and stayed stable
there, then
> itıs likely that all possible human beings will have been
born after two
> or three thousand years (if the 640 billion Ŝgure is
right). I donıt
like
> this thought at all! (Even though I believe nurture and
environment play
> a big role in determining oneıs personhood.)
Identical twins have identical genomes but are different
people even
with similar nurturing and environment.
> If you factor in possible harmless mutations over such a
period of time
> and such a population, to expand the available DNA options,
how much is
> that likely to affect the 640 billion Ŝgure? Can someone
verify my
> calculation? Iıd be happier knowing we werenıt about to
exhaust our
> genetic possibilities within a (geological) eye-blink.
There are about 30,000 genes in the human genome with any two
reproductively capable people sharing about 94% of their
genes so they
would be able to produce 2^1800 genetically distinct
offspring, not
counting mutations and junk DNA variations. 2^1800 is about
10^541.
There are an estimated 10^80 protons and neutrons in the
universe.
Create a universe of each proton and neutron and a universe
of each of
those protons and neutrons to about 6 iterations and that is
how many
different children could be produced by any two people. Then
consider
how much variation there is in the population of the human
race.
--
Greg G.
A successful acupuncture is a jab well done.
===
Subject: Re: 640,000,000,000 TO 1
> I have seen this Ŝgure calculated out in a previous
> but basicly if you take ALL the possible DNA options
> and multiply them out you get a Ŝgure that rounds
> down to 640 billion.
> Does that mean that there are 640 billion possible human
beings, up
> to
> differences in nurture? I presume some are more likely to be
> generated
> than others so it is difŜcult to calculate exactly when the
human
> genome-space will have been completely Ŝlled, or how long
weıre
> likely to have to wait before someone is duplicated. But a
> back-of-the-envelope calculation shows that if the world
population
> climbed to, say, 12 billion (projected within the next
half-century)
> and stayed stable there, then itıs likely that all possible
human
> beings will have been born after two or three thousand
years (if the
> 640 billion Ŝgure is right). I donıt like this thought at
all!
> (Even
> though I believe nurture and environment play a big role in
> determining oneıs personhood.)
> If you factor in possible harmless mutations over such a
period of
> time and such a population, to expand the available DNA
options, how
> much is that likely to affect the 640 billion Ŝgure? Can
someone
> verify my calculation? Iıd be happier knowing we werenıt
about to
> exhaust our genetic possibilities within a (geological)
eye-blink.
> What is the problem if there is duplication? What do you
expect to
> happen if there is a man identical to a previous man or a
snowŝake
> identical to a previous snowŝake? I canıt imagine the
universe would
> come to an end or the concept of human rights ceases to
have any sway
> over the minds of men. We can cope with twins and triplets
so I see no
> reason why we should be freaked out by identical strangers
of
> different ages.
> --
> Martin Willett
> http://mwillett.org/
Its not like coming up with the 9,000,000 names of god is it?
--
apatriot #23, aa #1779, Grand Pubba, EAC Department of Oxygen
Deprivation
Gary Bohn
Conservatism is not about tradition and morality, hasnıt been
for many
decades...It is about the putative biological and spiritual
superiority of
the wealthy.
Greg Bear
===
Subject: Re: 640,000,000,000 TO 1
> Conservatism is not about tradition and morality, hasnıt
been for many
> decades...It is about the putative biological and spiritual
superiority
of
> the wealthy.
> Greg Bear
And wealth is a heritable trait. More heritable than
intelligence, height,
eye
color, shoe size ...
-- TP
===
Subject: Re: 640,000,000,000 TO 1
>Its not like coming up with the 9,000,000 names of god is it?
Noooooo! Not that! Anything but that!
The peace of God be with you.
Stanley Friesen
===
Subject: Re: How to read partial differentials when reading
aloud...?
so how do you say r (dr) r ?
arrrgghhhh, matey.
> Hello all,
> I canıt Ŝnd this information in an FAQs or textbooks.
> I know that when you are reading out an equation or such,
> you usually read the derivative of x with respect
> to y (dx/dy) in itıs shortened form dee-ex-dee-why.
> But how do you read the partial derivative of x with respect
> to y? Is it still just dee-ex-dee-why as long as the context
> is set that youıre working with partial differentials?
> Yes, and if the context is hostile, you can say
> partial-dee-ex-dee-why
> Also, what about the second partial derivative of x with
respect
> to y?
> I have heard one of my colleague students read it as
> deedee-ex-dee-why-why,
> but I wonder whether he would have said
> deedee-ex-dee-tee-tee
> for derivatives w.r.t. time ;-)
> I always read (or Œthinkı) it as
> dee-second-ex-[over]-dee-why-squared
> or, whem Iım lazy, simply as
> dee-second-ex-dee-why
> and if necessasy I would add the partial preŜx.
> If there is some sort of resource that tells you how to read
> these?
> I think this is it :-)
> hth
> Dirk Vdm
===
Subject: Re: How to read partial differentials when reading
aloud...?
>so how do you say r (dr) r ?
Very amusing.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: How to read partial differentials when reading
aloud...?
> so how do you say r (dr) r ?
> arrrgghhhh, matey.
I have never seen or written an expression even
remotely like
r (dr) r
but it would sound like a bitch indeed ;-)
Give me 2 to Duluth please?
Dirk Vdm
===
Subject: Re: How to read partial differentials when reading
aloud...?
It is a commuted form of d(r^3/3).
> so how do you say r (dr) r ?
> arrrgghhhh, matey.
> I have never seen or written an expression even
> remotely like
> r (dr) r
> but it would sound like a bitch indeed ;-)
> Give me 2 to Duluth please?
> Dirk Vdm
===
Subject: Re: How to read partial differentials when reading
aloud...?
> ....
> I know that when you are reading out an equation or such,
> you usually read the derivative of x with respect
> to y (dx/dy) in itıs shortened form dee-ex-dee-why.
> But how do you read the partial derivative of x with respect
> to y? Is it still just dee-ex-dee-why as long as the context
> is set that youıre working with partial differentials?....
Curiously enough, the same question came up last month on the
little news group alt.math under the title What is it called?
Hereıs what I posted there.
> As youıve probably realized from other replies, thereıs no
> standard name for it. A very careful reading of
y/x is partial d y
> by d x. When I was a student, the local physicists gave
some thought
> to a good short name, and came up with dir (rhyming with
sir, but

> without sounding the r). I Ŝnd it quite convenient to say
dir y by

> dir x, but of course itıs not standard.
> Thereıs a similar problem with the Jacobi elliptic functions
> sn(u), cn(u), dn(u), sc(u), etc. (12 of them altogether,
using any
> ordered pair of the letters s, c, d, n). Iıve thought about
a handy
> vowel to pronounce in the middle of those, without
producing any uncouth
> words. :-) My choice is the AW sound, so for example I
pronounce
> dn(u) as dawn u. But of course thatıs not standard either.
> Ken Pledger.
===
Subject: Re: Help Needed Understanding Article
>2. I proved that the NOT rule is valid and made comments
about how the
>other rules can be similarly proven valid.
>>No, you stated the program transformation rule for the NOT
rule. Now you
>>have to prove that it is valid.
> That is not true. I already did that. What is wrong with my
proof?
> Quote what is wrong. Do you want me to copy it here again?
There is something that looks like a proof of the validity of
the NOT
rule. Granted. It is the most trivial rule to prove correct.
Please
provide proof for the other rules as well. (You wonıt get
away with
stating that they go in the same way.)
>>Yes, I know other people have done that
>>already. Refer to their paper, because you are building on
their work.
> Those theorems were developed 10 years before computers
were invented.
So what?
> And, as I said, the fact that I am also able to synthesize
theorems
> from the Theory of Computation does not mean that they did
anything
> that I used.
Yes, you did. You used *their* result that the rules are
valid.
> It means I have a fantastic system with lots of
> functionality that nobody else has ever done before.
No, you have no system at all. You admitted that already. And
you donıt
have a theoretical description of a system either, because
you donıt
prove anything about it (except validity of the NOT rule,
which is
trivial and not your contribution).
>>(And similarly for the other rules: either prove them valid
yourself or
>>refer to the work of others for the proof.)
> What is wrong with the proof that I gave? (I see no point
in copying
> it into a message again, unless you insist.)
You only gave a proof for the NOT rule. You also must prove
it for the
other rules. And you must prove that the programs you gave
for your
axioms are correct.
Note that in order to prove this, you also need some sort of
formal
semantics for your programming language, which you donıt
have. You may
consider to use some well-investigated, theoretical,
Turing-Complete
programming language like WHILE. (But then you have to give
credit to
others of course, which you seem to dislike very much.)
I donıt say these proofs are difŜcult. Only that I donıt see
them in
your paper. And that they are required.
>3. I gave 2 explicit examples, in complete detail, of how
programs are
>synthesized: a program to determine if one number is a
factor of
>another, and a program to list all prime numbers between any
two given
>numbers.
>>Yes, you gave *examples*. Which are not even very
convincing, because
>>for each example you add new axioms to obtain your result.
> Not true. Substantiate.
every time you notice that your 8 *universal* rules of
inference donıt
sufŜce anymore?
[snip question I already answered]
[snip question I already answered]
[snip question I already answered]
(Sorry, I have a job to do. I am not going to repeat myself
every time.)
>Those are just non sequiturs.
>>I donıt see any thus or therefore in my text. Where is the
non
sequitur?
> The sequence of sentences.
There are no sequiturs in the text. There are a number of
points that do
not follow from each other, followed by a *question* (which
doesnıt
follow from something by deŜnition). Do you know what non
sequitur
means, or did you just want to write something in latin in
order to
appear intelligent?
>>I snipped a proof that the program enumerates the natural
numbers, yes.
>>(First you deŜne P to be the natural numbers,
> No I did not.
Yes, you did. You deŜned P to be a set which contains 0, and
of each
element of P its successor. These are the natural numbers.
>>then you prove that the
>>program outputs every element of P.) So there are two
problems with your
>>proof:
>>* It is not a derivation in your logic, but a proof.
> Yes, it is a proof of Induction from my axiom TRUE(x) as
you asked.
You claimed TRUE(x) where equivalent to Peanoıs axioms. So in
particular, it implies induction. You have to give a
derivation of
induction from assumption TRUE(x). You didnıt give that.
> No, it doesnıt use my rules of inference, as they are for
Program
> Synthesis. Those are two separate functions.
Indeed. This was precisely the point I tried to make by
asking you to
provide a derivation. *You* claimed however they were
equivalent.
>>* It does not derive the desired result.
> What is wrong with it? The above is not true. Substantiate.
Enumerating the natural numbers, is something different than
induction.
>>It is relevent because you obviously donıt know what youıre
talking
>>about. In the above, the NOT rule is not a rule but an
axiom. Adding an
>>axiom P->Q to a system is *not* equivalent to adding a rule
P=>Q. You
>>should not confuse the two.
> You can call it anything you want, but it works great, as I
have
> shown.
As I have pointed out, it is very important how you call
things. If you
call it axioms you can prove more things than if you call it
rules.

There are two cases:
* They are really axioms, but you call them rules. But then,
according
to *your own* system, you canıt use transformation rules as
these axioms
are applied, but you have to give programs for these axioms.
Since you
give transformation rules rather than programs, the
conclusion is, in
this case, that you donıt know what youıre talking about.
* They are really rules of inference. But then your claim
that you apply
rules of inference by applying other rules of inference
(substitution
and modes ponens), makes no sense. (Think about it, in order
to apply
the substition rule, you have to use the substitution rule.)
So, also in
this case, the conclusion is that you donıt know what youıre
talking about.
Therefore: you donıt know what youıre talking about. QED.
You have no idea what your own system can do. It is not a
formal system
at all, but an inelegant, informal hack, using ad-hoc axioms.
groente
-- Sander
===
Subject: Improper integral
Let t>-1 . f(x)=log(x) * x^t
Show that the integral of on [0,1] exists.
( Donıt use imtegraion by part)
===
Subject: Re: Improper integral
Select k such that t < k < -1. Compare the given integrand
with (-)t^k.
--OL
===
Subject: Re: Improper integral
>Let t>-1 . f(x)=log(x) * x^t
>Show that the integral of on [0,1] exists.
>( Donıt use imtegraion by part)
Well what _are_ you allowed to use?
Hereıs a hint towards a solution, donıt know whether
itıs legal: Choose r so t > r > -1. Then t = r + s, where
r > -1 and s > 0. Write
log(x) * x^t = (log(x) * x^s) * x^r.
Think about how log(x) * x^s and x^r behave...
************************
David C. Ullrich
===
Subject: Re: Improper integral
> Let t>-1 . f(x)=log(x) * x^t
> Show that the integral of on [0,1] exists.
> ( Donıt use imtegraion by part)
Integral of ??
Let t = 0. Looks very inproper.
===
Subject: Re: Improper integral
William Elliot <marsh@privacy.net> escribi.97:
>> Let t>-1 . f(x)=log(x) * x^t
>> Show that the integral of on [0,1] exists.
>> ( Donıt use imtegraion by part)
> Integral of ??
> Let t = 0. Looks very inproper.
Int(ln(x), x, r, 1) = (1ln(1) - 1) - (r*ln(r) - r) = r -
r*ln(r) - 1
And
Lim( r - r*ln(r) - 1, r, 0+) = -1
Because
Lim(r*ln(r), r, 0+) = Lim(ln(r)/(1/r), r, 0+)
= Lim((1/r)/(-1/r^2), r, 0+) = Lim(-r, r, 0+) = 0
--
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: Improper integral
> Let t>-1 . f(x)=log(x) * x^t
> Show that the integral of on [0,1] exists.
> ( Donıt use imtegraion by part)
There is a h in [0,1], f is integrable on [h,1] and for any x
in ]0,h],
|ln(x)|<=1/x^[(t+1)/2], |f(x)|<=x^[(t-1)/2] which is
integrable on [0,h].
--
Julien Santini
===
Subject: Need help calculating the norm of a linear operator
Let (F_n) be a sequence of linear operators L2 -> R, <x, F_n>
=
int_{t=-1..1}x(t)*cos(pi*n*t)dt. I need to show that (F_n)
does not
converge to 0 operator. There is a hint that it is provable
that
||F_n|| = 1 for all n. Actually proving the original
statement seems
easier for me than proving the hint. For every F_n thereıs an
obvious
x_n from L2 such that ||x_n||=1 and x_n(t)*cos(pi*n*t) =
|cos(pi*n*t)|, so int_{t=-1..1}x(t)*cos(pi*n*t)dt = 2/pi and
sup_{||x||=1}<x, F_n>= 2/pi. It is also easy to show that
||F_n|| is
upper bounded by 1+1/2 using the fact that |cos(pi*n*t)| <=
1. But how
===
Subject: Re: Need help calculating the norm of a linear
operator
>Let (F_n) be a sequence of linear operators L2 -> R, <x,
F_n> =
>int_{t=-1..1}x(t)*cos(pi*n*t)dt. I need to show that (F_n)
does not
>converge to 0 operator. There is a hint that it is provable
that
>||F_n|| = 1 for all n. Actually proving the original
statement seems
>easier for me than proving the hint. For every F_n thereıs
an obvious
>x_n from L2 such that ||x_n||=1 and x_n(t)*cos(pi*n*t) =
>|cos(pi*n*t)|, so int_{t=-1..1}x(t)*cos(pi*n*t)dt = 2/pi and
>sup_{||x||=1}<x, F_n>= 2/pi. It is also easy to show that
||F_n|| is
>upper bounded by 1+1/2 using the fact that |cos(pi*n*t)| <=
1. But how
Somewhere in the book youıll Ŝnd a thing called the Schwarz
inequality, or the Cauchy-Schwarz inequality (together with
a statement of when equality holds.)
************************
David C. Ullrich
===
Subject: Re: Need help calculating the norm of a linear
operator
>>Let (F_n) be a sequence of linear operators L2 -> R, <x,
F_n> =
>>int_{t=-1..1}x(t)*cos(pi*n*t)dt. I need to show that (F_n)
does not
>>converge to 0 operator. There is a hint that it is provable
that
>>||F_n|| = 1 for all n. Actually proving the original
statement seems
>>easier for me than proving the hint. For every F_n thereıs
an obvious
>>x_n from L2 such that ||x_n||=1 and x_n(t)*cos(pi*n*t) =
>>|cos(pi*n*t)|, so int_{t=-1..1}x(t)*cos(pi*n*t)dt = 2/pi and
>>sup_{||x||=1}<x, F_n>= 2/pi. It is also easy to show that
||F_n|| is
>>upper bounded by 1+1/2 using the fact that |cos(pi*n*t)| <=
1. But how
>Somewhere in the book youıll Ŝnd a thing called the Schwarz
>inequality, or the Cauchy-Schwarz inequality (together with
>a statement of when equality holds.)
Given that Alexander Korovyev is posting from .ru, he might
also
want to search for (some version, possibly in Cyrillic) of
Bunyakovski, as well.
Lee Rudolph
===
Subject: Re: Need help calculating the norm of a linear
operator
>Let (F_n) be a sequence of linear operators L2 -> R, <x,
F_n> =
>int_{t=-1..1}x(t)*cos(pi*n*t)dt. I need to show that (F_n)
does not
>converge to 0 operator. There is a hint that it is provable
that
>||F_n|| = 1 for all n. Actually proving the original
statement seems
>easier for me than proving the hint. For every F_n thereıs
an obvious
>x_n from L2 such that ||x_n||=1 and x_n(t)*cos(pi*n*t) =
>|cos(pi*n*t)|, so int_{t=-1..1}x(t)*cos(pi*n*t)dt = 2/pi and
>sup_{||x||=1}<x, F_n>= 2/pi. It is also easy to show that
||F_n|| is
>upper bounded by 1+1/2 using the fact that |cos(pi*n*t)| <=
1. But how
>>Somewhere in the book youıll Ŝnd a thing called the Schwarz
>>inequality, or the Cauchy-Schwarz inequality (together with
>>a statement of when equality holds.)
>Given that Alexander Korovyev is posting from .ru, he might
also
>want to search for (some version, possibly in Cyrillic) of
>Bunyakovski, as well.
(In fact I even realized why it might be relevant here...)
>Lee Rudolph
************************
David C. Ullrich
===
Subject: Multiplication of negative binary numbers.
Gıday Gıday Folks,
Twoıs complement works Ŝne for producing functional positive
and
negative numbers so long as we conŜne ourselves to addition.
Twoıs complement sometimes give the right answer with
subtraction and
sometimes doesnıt.
0011 3
-1001 -(-7)
= 1010 = 10 works,
0111 7
-1001 -(-7)
= 1110 = -2 doesnıt.
Is there a binary code that can deal with
(negative number)*(negative number) = positive number?
Twoıs complement obviously doesnıt. Here is a 4-bit example.
1001
x1110
= 0000
+ 1 0010
+ 10 0100
+ 100 1000
It doesnıt seem to matter if we include the overŝow or not.
In this instance the msb is 1 whatever we choose to do.
Best wishes,
--
Quentin Grady ^ ^ /
New Zealand, >#,#< [
/ /
... and the blind dog was leading.
http://homepages.paradise.net.nz/quentin
===
Subject: Re: Multiplication of negative binary numbers.

!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(
5eZ41to5f%E@ıELIi
$t^
VcLWP@J5p^rst0+(Œ>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
> Gıday Gıday Folks,
> Twoıs complement works Ŝne for producing functional
positive and
> negative numbers so long as we conŜne ourselves to addition.
> Twoıs complement sometimes give the right answer with
subtraction and
> sometimes doesnıt.
It gives the right answer if the right answer is
representable. The
right answer is representable if the carry into the MSB has
the same
sign as the carry out of the MSB.
> 0011 3
> -1001 -(-7)
> = 1010 = 10 works,
> 0111 7
> -1001 -(-7)
> = 1110 = -2 doesnıt.
> Is there a binary code that can deal with
> (negative number)*(negative number) = positive number?
> Twoıs complement obviously doesnıt. Here is a 4-bit example.
> 1001
> x1110
> = 0000
> + 1 0010
> + 10 0100
> + 100 1000
> It doesnıt seem to matter if we include the overŝow or not.
> In this instance the msb is 1 whatever we choose to do.
If one factor is negative, subtract the other factor from the
high
word of the result. And if the other is negative, do the same
the
other way round.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: Length
>
> Force does not always cause motion; as such; but will cause
> deformation and distortions such as simple compression.
> I donıt want you to come to the wrong conclusion, Shead, so
> be aware that if a net force acting on a mass is not zero,
> ^^^^^^^^^
> that mass is going to accelerate.
> Newtonıs Second Law
>
http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html
That says: ŒA force [F] acting on a body gives it an
acceleration [a]
which is in the direction of the force and has magnitude
inversely
proportional to the mass [m] of the bodyı:
According to that Earthıs surface which is acted on all
around by the
force of atmospheric pressure must be accelerating
centripetally
toward its center. No wonder there are so many goofy theories
ŝoating
around.
The fact is that the force of the atmosphere, is due to its
weight,
and along with the weight of the ground itself, as well as
the weight
of the oceans is squeezing Earth from all directions, keeping
it
compressed together.
Any inward motion is so slight as to be unoticable, except
perhaps for
occasional earthquakes; but the pressure increase with depth
is quite
well known.
===
Subject: Re: Length
>
> Force does not always cause motion; as such; but will cause
> deformation and distortions such as simple compression.
> I donıt want you to come to the wrong conclusion, Shead, so
> be aware that if a net force acting on a mass is not zero,
> ^^^^^^^^^
> that mass is going to accelerate.
> Newtonıs Second Law
>
http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html
> That says: ŒA force [F] acting on a body gives it an
acceleration [a]
> which is in the direction of the force and has magnitude
inversely
> proportional to the mass [m] of the bodyı:
> According to that Earthıs surface which is acted on all
around by the
> force of atmospheric pressure must be accelerating
centripetally
> toward its center. No wonder there are so many goofy
theories ŝoating
> around.
The net force on the atmosphere is zero (ignoring the
temperature and
^^^^^^^^^
and pressure variations near the surface driven by the Sun
and known
as weather. You think things are goofy simply because you
donıt
understand
some basic principles of physics. Newtonıs laws are deŜnitely
insightful
in this instance.
===
Subject: det(A+B) once more
If A and B are 2*2, then
det(A+B)=det(A)+det(B) <=> det(A+l*B)=det(A)+det(l*B)
for any (nonzero) scalar l.
Can this be extended somehow to n*n?
(I think yes, except that other certain values than 0
must be excluded.)1
--
Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de
als man ankam wollte man werden, die geschichte schreiben,
die doofen sollen sterben, der plan als man damals nach
hamburg kam
(Kettcar)
===
Subject: Re: det(A+B) once more
>If A and B are 2*2, then
>det(A+B)=det(A)+det(B) <=> det(A+l*B)=det(A)+det(l*B)
>for any (nonzero) scalar l.
>Can this be extended somehow to n*n?
>(I think yes, except that other certain values than 0
>must be excluded.)
Iıll use t instead of l, which looks too much like 1 in the
fonts I use.
P(t) = det(A+tB) - det(A) - det(tB) is a polynomial of degree
at most
n-1 and P(0) = 0. So if it is 0 for n-1 different nonzero
values of t,
you can conclude it is always 0.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Relative Measure & Relative Product of Dynamical
Systems
Does somebody know deŜnitions : Relative Measure and Relative
Product of
Dynamical Systems... or know where they could be on the net ?
M.B.
===
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i4PCYEP09572;
>Suppose that n is positive integer. Suppose that a and b are
positive
>real number.
>How to simplify/compute the following integral.
>Int[0, INFINITE] x^n exp^( - (x-a)^2/ (2b^2)) dx
>In this expression, Int[0 , INFINITE] represents the
integral from
>zero to inŜnite. x is the variable.
>--------------
>ZHANG Yan
Letıs denote your integral by I(n). Then integration by parts
yields
the recursion formula
I(n)= a*I(n-1)+b^2*(n-1)*I(n-2) for n>=2
along with
I(0)= sqrt(pi/2)*b*ERF(a/(sqrt(2)*b)) + sqrt(pi/2)*b
I(1)= a*I(0) + b^2*exp(- a^2/(2*b^2))
where ERF(x)= 2/sqrt(pi)*int(-t^2,t,0,x) is the error
function. (I
have been using the Derive-notation above, but I hope it is
self-
explanatory nevertheless.)
The formula above is already very convenient, if you want to
compute
I(n) for some speciŜc n>=0. If you are very ambitious though,
you
could try to go a step further and consider the generating
function
f(x):= sum(I(n)*x^n,n,0,inf)
which obeys a differential equation of order 2 according to
the
recursion above.
Johann

===
Subject: Re: Questions about groups, inverses, and continuity.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i4PCYEm09580;
>(Hobby mathematician asks...)
>Let x be an element of the algebra G(A) over the reals
generated from
>some Ŝnite group G = {e, g_2, ..., g_n}.
>Then
>inf {| 1 - |t_1| + |t_2| +...+ |t_n| | : y in G(A),
> x*y = (t_1)e + (t_2)g_2 + ... (t_n)g_n } = 0
>is a necessary condition for the existence of x^{-1}.
>Questions
>1. Is the condition also sufŜcient?
>2. Norm G(A) by setting, for example,
> ||x|| = || (x_1)e + ... (x_n)g_n || = |x_1| + ... + |x_n|.
>
> DeŜne the function f:G(A) -> R^{+},
> f(x) = inf {| 1 - |t_1| + |t_2| +...+ |t_n| | : y in G(A),
> x*y = (t_1)e + (t_2)g_2 + ... (t_n)g_n }
> Under what possible conditions on G(A) is f continuous at...
> (a) points xı such that f(xı) = 0
> (b) every point ?
>C. Dement
I should have checked
http://mathworld.wolfram.com/GroupAlgebra.html
Ŝrst. It gives some helpful hints on notation, but
does not appear to mention that the basis vectors
{e, g_2, ..., g_n} must be linear independent- which
I had implicitely assumed without even mentioning it.
I know this to be true for Ŝnite abelian groups
(see Mackey, for ex.: Unitary Group Representations),
as the one dim. character table of such a group is
nothing more than a collection of orthogonal row vectors.
I am not sure how to prove the general case, however.
C. Dement

===
Subject: Real analyses- measurable sets.... Help urgently
needed
No matter how I start, I always end up with the same and Iıd
really
appreciate some help on this......
(I hope u can understand the problem, as I translate it to
English and
I donıt learn Math in English....)
Prove that for given E>0, there exists a subset A of R (set
of real
numbers) with these characteristics:
1. for every x from R, there exists a surroundings of x such
that no
element from A is in that surroundings
2. A is measurable
3. Lebesgueıs measure of A is <E.
===
Subject: Re: Real analyses- measurable sets.... Help urgently
needed
> Prove that for given E>0, there exists a subset A of R (set
of real
> numbers) with these characteristics:
> 1. for every x from R, there exists a surroundings of x
such that no
> element from A is in that surroundings
What do you call a surrounding ? If itıs a neighbourhood,
then only the
empty set would do.
> 2. A is measurable
> 3. Lebesgueıs measure of A is <E.
===
Subject: Re: Real analyses- measurable sets.... Help urgently
needed
solution says itıs not the empty set, but I guess itıs wrong
(done by
lot.....
Btw, yes, I meant neighbourhood..:)
===
Subject: Re: Real analyses- measurable sets.... Help urgently
needed
> No matter how I start, I always end up with the same and
Iıd really
> appreciate some help on this......
> (I hope u can understand the problem, as I translate it to
English and
> I donıt learn Math in English....)
> Prove that for given E>0, there exists a subset A of R (set
of real
> numbers) with these characteristics:
> 1. for every x from R, there exists a surroundings of x
such that no
> element from A is in that surroundings
> 2. A is measurable
> 3. Lebesgueıs measure of A is <E.
The empty set.
Jose Carlos Santos
===
Subject: Series to Sum
Any help in summing the following series would be welcomed!
4 pi (-1)^k
sum_{k = 1}^{infty } = ------------------------
-1 + exp(pi*(2k+1))
If youıre wondering it is the the integral over R of the
error in the
Sinc Cardinal Series for Sech.
===
Subject: Re: Series to Sum
>Any help in summing the following series would be welcomed!
> 4 pi (-1)^k
>sum_{k = 1}^{infty } = ------------------------
> -1 + exp(pi*(2k+1))
>If youıre wondering it is the the integral over R of the
error in the
>Sinc Cardinal Series for Sech.
-0.00101229164605
using matlab an easy exercise, since this is a rapidly
converging
alternating series, no problems
hth
peter
===
Subject: Entropy of samples on real line segment.
Is there a precise measure/deŜnition of the Œentropyı of N
points on
the real line segment [0,1]? I am guessing that N points
positioned at
(i+.5)/N (where i=0...N-1) gives the conŜguration with
Œmaximumı
entropy.
Sabbir.
===
Subject: False proofs of FLT
Hi!
Could you point me to a good source of interesting and short
false proofs
of
FLT to check my
personal-probably-wrong-too-short-and-simple-proof of FLT?
TIA.
N.
===
Subject: Re: False proofs of FLT
> Hi!
> Could you point me to a good source of interesting and
short false proofs
of
> FLT to check my
personal-probably-wrong-too-short-and-simple-proof of
FLT?
> TIA.
> N.
Just post it and somebody will tell you whatıs wrong with it.
If you
post it while being extremely arrogant, condescending, and
insulting,
then 20 people will tell you whatıs wrong with it, in great
detail.
That probably wonıt help you any more than the other, but
itıs funny
to watch.
Nathan
===
Subject: Re: False proofs of FLT
> Hi!
> Could you point me to a good source of interesting and
short false
proofs of
> FLT to check my
personal-probably-wrong-too-short-and-simple-proof of
FLT?
> TIA.
> N.
> Just post it and somebody will tell you whatıs wrong with
it. If you
> post it while being extremely arrogant, condescending, and
insulting,
> then 20 people will tell you whatıs wrong with it, in great
detail.
> That probably wonıt help you any more than the other, but
itıs funny
> to watch.
:-))
Dirk Vdm
===
Subject: Re: False proofs of FLT
> Could you point me to a good source of interesting and
short false proofs
of
> FLT to check my
personal-probably-wrong-too-short-and-simple-proof of
FLT?
Isnıt it easier to just post your attempt? Short, false
proofs often
apply to the case p = 2 also, so you may want to check that
Ŝrst.
--
J K Haugland
http://www.neutreeko.com
===
Subject: Re: False proofs of FLT
Yes, but I dont want to look like a crank. I prefer to check
if I made some
common mistake Ŝrst ;)
> Could you point me to a good source of interesting and
short false
proofs of
> FLT to check my
personal-probably-wrong-too-short-and-simple-proof of
FLT?
> Isnıt it easier to just post your attempt? Short, false
proofs often
> apply to the case p = 2 also, so you may want to check that
Ŝrst.
> --
> J K Haugland
> http://www.neutreeko.com
===
Subject: Re: False proofs of FLT
days. My association with the Department is that of an
alumnus.
>Yes, but I dont want to look like a crank.
You might want to stop top-posting as a Ŝrst step... (that
refers to
the practice of replying to a post by writing all new
material Ŝrst,
on top, and then quoting the message you are replying to).
Many people
Ŝnd it annoying, and it is hard to read; particularly if and
when you
post an argument and want to ask questions, it makes more
sense to
intersperse your comments with what you are addressing.
> I prefer to check if I made some common mistake Ŝrst ;)
So go through it. If you cannot Ŝnd an error, then you may
post it
saying that you cannot Ŝnd and error and would welcome
comments;
simply saying I have this argument and cannot see anything
wrong with
it... doesnıt make you look like a crank. Posting it and
claiming it
is the best thing since sliced bread and that nobody can
possibly Ŝnd
anything wrong with it would; your reactions to criticisms or
questions may also label you as a crank, but the simple act
of posting
and asking questions will not.
But the suggestion you got already is a good one: run through
your
argument with n=2 (n being the exponent). If you can see no
problem
with the argument when n is 2, then there is certainly
something
wrong, with your perception and your argument (since there
are many
integer solutions to a^2+b^2=c^2). That would give you a clue
as to
whether your ability to spot errors is good enough, and
whether your
argument passes the most standard of crank-detecting tests.
--
Itıs not denial. Iım just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: False proofs of FLT
> Hi!
> Could you point me to a good source of interesting and
short false proofs
of
> FLT to check my
personal-probably-wrong-too-short-and-simple-proof of
FLT?
Underwood Dudley has a book on cranks. I believe one chapter
is on FLT.
===
Subject: Re: False proofs of FLT
Unfortunately Im from spain I cant have
access to such books. I was
looking for a web site.
> Hi!
> Could you point me to a good source of interesting and
short false
proofs of
> FLT to check my
personal-probably-wrong-too-short-and-simple-proof of
FLT?
> Underwood Dudley has a book on cranks. I believe one
chapter is on FLT.
===
Subject: Re: False proofs of FLT
> Unfortunately Im from spain I cant have
access to such books. I was
> looking for a web site.
Ah-so! :-)
There are only simple false proofs on web sites. Choose any
of them.
There are no simple right proofs.
You are welcome
Tapio
> Hi!
>
> Could you point me to a good source of interesting and
short false
> proofs of
> FLT to check my
personal-probably-wrong-too-short-and-simple-proof of
> FLT?
>
> Underwood Dudley has a book on cranks. I believe one
chapter is on
FLT.
===
Subject: Re: False proofs of FLT
> Hi!
> Could you point me to a good source of interesting and
short false proofs
of
> FLT to check my
personal-probably-wrong-too-short-and-simple-proof of
FLT?
> TIA.
> N.
named James Harris.
David
===
Subject: Re: False proofs of FLT
> Hi!
> Could you point me to a good source of interesting and
short false
proofs
> of
> FLT to check my
personal-probably-wrong-too-short-and-simple-proof of
FLT?
> TIA.
> N.
> named James Harris.
> David
Ive heard about him, but I was looking for Œsimplerı proofs.
i.e. Proofs
using elementary math.
===
Subject: Re: False proofs of FLT
One false proof can be this one.
If you take a geometric aproach and start to make a Proof of
Pithagoream
Theorem, then you do the same for grade 3 and then generalize
to higher
degrees you obtain the wrong proof (this false proof can
represent an
insight of what is a 3D representation that Fermat developed
although
Descartes is considered father of 2D Analitical Geometry)
The Œproofı
-----------
Consider three squares A,B and C with sides a>b>c. Then you
draw the
smaller
ones into the greater in the same diagonal but in opposite
corners so you
have a new square which is the intersection of B and C and
also you obtain
two rectangles belonging to A that dont intersect with
nothing. The
resulting equation is
(b+c-a)^2=2(a-b)(a-c)
The same you can do but for cubes and you obtain an
intersecting cube (from
cube B and C) and some prisms that belongs to cube A and
dont intersect
with B or C. You do the same that before and obtain this
(b+c-a)^3=3(a-b)(a-c)(b+c)
and to make the post shorter, the proof is explained here
http://mathpages.com/home/kmath367.htm
So the wrong proof can be obtained due to an ilumination if
you visualize
tessellations of cubes for degrees higher than 3 (x^4=x*x^3
and you have a
prism formed by x cubes and so on)
Also you can say something about Fermatıs Little Theorem
taking this
geometric approach and you obtain equations like
y(y^(n-1)-1)=Prime*F(y)
and it is obvious that if y is not divided by a Prime then
you have the
famous Fermatıs Little Theorem.
My conclusion is that Fermat didnt have formal proofs, only
geometric ones
like his beloved greeks, so he challenged some mathematicians
to obtain the
real formal proofs.
===
Subject: roots of unity in Koblitz
Hi people,
Iım currently trying to understand some stuff from Algebraic
Aspects of
Cryptography and stumbled over the following :
(We have a Field F[p^f] where p is prime)
12. Let d = gcd(k,(p^f)-1). Since d | (p^f)-1, the cyclic
group F*[p^f]
clearly has d d-th roots of unity. Each of them is also a
k-th root.
Problem is, that this is absolutely not clear to me ;)
I have tried an example :
p = 2
f = 4
k = 5
so gcd(5,15) = 5, so I should be able to Ŝnd 5 5th roots of
unity, but I
can only Ŝnd one (Its 1).
Probably Iım missing something here, can anyone else Ŝnd
other 5th roots
of
unity here ?
greets
Dennis
===
Subject: Re: roots of unity in Koblitz
>Hi people,
>Iım currently trying to understand some stuff from Algebraic
Aspects of
>Cryptography and stumbled over the following :
>(We have a Field F[p^f] where p is prime)
>12. Let d = gcd(k,(p^f)-1). Since d | (p^f)-1, the cyclic
group F*[p^f]
>clearly has d d-th roots of unity. Each of them is also a
k-th root.
>Problem is, that this is absolutely not clear to me ;)
>I have tried an example :
>p = 2
>f = 4
>k = 5
>so gcd(5,15) = 5, so I should be able to Ŝnd 5 5th roots of
unity, but I
>can only Ŝnd one (Its 1).
>Probably Iım missing something here, can anyone else Ŝnd
other 5th roots
of
>unity here ?
The polynomial x^5 - 1 factors as
(x - 1)(x^4 + x^3 + x^2 + x + 1)
over GF(2). The roots of the irreducible quartic belong to
GF(2^4).
--
John Adams served two terms as Vice President and one as
President, but
lost
reelection. Later his son became President despite losing the
popular
vote.
That son lost his reelection attempt badly. Now history is
repeating
itself.
pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael,
California
===
Subject: Re: roots of unity in Koblitz
> Iım currently trying to understand some stuff from
Algebraic Aspects of
> Cryptography and stumbled over the following :
> (We have a Field F[p^f] where p is prime)
> 12. Let d = gcd(k,(p^f)-1). Since d | (p^f)-1, the cyclic
group F*[p^f]
> clearly has d d-th roots of unity. Each of them is also a
k-th root.
> Problem is, that this is absolutely not clear to me ;)
> I have tried an example :
> p = 2
> f = 4
> k = 5
> so gcd(5,15) = 5, so I should be able to Ŝnd 5 5th roots of
unity, but I
> can only Ŝnd one (Its 1).
> Probably Iım missing something here, can anyone else Ŝnd
other 5th roots
of
> unity here ?
Sure: exp(2pi i/5), exp(4pi i/5), exp(6pi i/5), and exp(8pi
i/5).
Jose Carlos Santos
===
Subject: Re: roots of unity in Koblitz
>> Iım currently trying to understand some stuff from
Algebraic Aspects
of
>> Cryptography and stumbled over the following :
>> (We have a Field F[p^f] where p is prime)
>> 12. Let d = gcd(k,(p^f)-1). Since d | (p^f)-1, the cyclic
group F*[p^f]
>> clearly has d d-th roots of unity. Each of them is also a
k-th root.
>> Problem is, that this is absolutely not clear to me ;)
>> I have tried an example :
>> p = 2
>> f = 4
>> k = 5
>> so gcd(5,15) = 5, so I should be able to Ŝnd 5 5th roots
of unity, but
I
>> can only Ŝnd one (Its 1).
>> Probably Iım missing something here, can anyone else Ŝnd
other 5th
roots
>> of unity here ?
> Sure: exp(2pi i/5), exp(4pi i/5), exp(6pi i/5), and exp(8pi
i/5).
Hmmm. Not really characteristic 2 entities ... :-(
To answer the OPıs question one needs to know how he is
representing
elements of F_16. Of course the cube of each nonzero element
of F_16
is a 5-th root of unity.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: roots of unity in Koblitz
> so gcd(5,15) = 5, so I should be able to Ŝnd 5 5th roots of
unity, but
I
> can only Ŝnd one (Its 1).
> Probably Iım missing something here, can anyone else Ŝnd
other 5th
roots
> of unity here ?
>> Sure: exp(2pi i/5), exp(4pi i/5), exp(6pi i/5), and
exp(8pi i/5).
> Hmmm. Not really characteristic 2 entities ... :-(
I *really* have to read the questions more carefully before
trying
to answer them. :-(
Jose Carlos Santos
===
Subject: re:a new way round diagonalisation-one inŜnity?meta
set theory
One more example with meta set theory:
We deŜne the power set P(m) as:
x e(t+1) P(m) :<-> ( y e(t) x -> y e(t) m )
( 0 e(t+1) P(m), m e(t+1) P(m) )
The famous result of Cantor was, that there exists no
bijection beween
m and P(m).
In meta set theory this is not so:
If we have for m the set A of all sets, P(m) = P(A) is deŜned
as
x e(t+1) P(A) :<-> ( y e(t) x -> y e(t) A) , this is true for
all sets x,
therefore P(A) = A, with identity as bijection.
Or with Cantors proof:
m be a set, f a bijection m <-> P(m).
x e(t+1) h(f) :<-> x e(t) m : x -e(t) f(x)
(In classical set theory h(f) -= f(x) for all x e m,
as if h(f) = f(x) then : if x e f(x) then x -e h(f)=F(x),
therefore x
-e f(x) therefore x e h(f) and x -e h(f)=f(x).#
In meta set theory if h(f) =(t+1) f(x) (or if h(f) =(...)
f(x)):
if x e(t+1) h(f) then x -e(t) f(x) =(...) h(f).
This is no contradiction, as x may be element an non-element
of h(f)
in different levels.
There is no need for different kinds of inŜnity for sets in
meta set
theory.
Oskar Trestone
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===
Subject: re:Simple Grid-Game #1
oops... from the looks of it it appears as though I proved
the obvious
in a roundabout manner. Actually, in an incorrect manner, I
showed
that a contradiction implies a true statement.
Actually, I left out the actual question. Is a game which
permits a
draw called a futile game because it is gauranteed to end in
a draw
if played properly? If the neither categorical nor futile
above is
changed to not categorical, hence futile (permitting a draw)
does
the above constitute a correct proof that the end result is a
draw?
Does a similar induction work to show that a categorical game
is
unfair.
1. player one is the last to play and wins.
2. player two could not have prevented this win on his last
move,
otherwise, he did not play properly.
3. So player one actually played a win on his move before the
last.
4. player two could not have prevented...
etc. to the Ŝrst move of the game (which may have been made
by either
player).
What if you do not assume as I did that there are only two
players?
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===
Subject: re:Simple Grid-Game #1
I saw a thing in math world that says that there is a
strategy for one
of the two to win since there is no possibility of a draw.
http://mathworld.wolfram.com/CategoricalGame.html
I also saw something else, the deŜnition of a futile game is
one
which permits a draw (tie) when played properly by both
players.
Now given a game, neither futile or Categorical, if player
one plays
properly, he is not gauranteed to win. Then player two plays
case 1: he plays and is guaranteed to win. this is a
contradiction.
case 2: he plays and player one is gauranteed to win. Then he
didnıt
play properly. contradiction.
case 3: since the game is not categorical there is a case 3.
He plays
and neither player is now guaranteed to win.
back to player one.
case 1: he plays and is guaranteed to win. this is a
contradiction
(case 2 above).
case 2: he plays and player two is gauranteed to win. Then he
didnıt
play properly. contradiction.
case 3: He plays and neither player is now guaranteed to win.
etc.
therefore, the winning move is never possible.
Is there a ŝaw in this reasoning, granted that I donıt really
know
what a proper play is, but it seems as though it should be a
play
that doesnıt guarantee the other a win.
If this is right, then any game is either futile or
categorical and
unfair. Even chess.
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===
Subject: normed space
In a proof, they use a fact that I canıt understand. It says:
suppose X is a normed vector space and M is a proper closed
subspace of
X.
We can deŜne norm on X/M as ||x+M||=inf {||x+y||: y in M}.
The book mentions that for any e>0, there is x in X such that
||x||=1 &
||x+M||>= 1-e.
Can anyone see why?
===
Subject: Re: normed space
> In a proof, they use a fact that I canıt understand. It
says:
> suppose X is a normed vector space and M is a proper closed
subspace of
> X.
> We can deŜne norm on X/M as ||x+M||=inf {||x+y||: y in M}.
> The book mentions that for any e>0, there is x in X such
that ||x||=1 &
> ||x+M||>= 1-e.
> Can anyone see why?
Choose a coset with nonzero norm. Say the coset of z has norm
a>0.
Then by deŜnition of the norm, we may choose y in M so that
||z+y|| is as close as we like to a. Your vector will be
x=(z+y)/||z+y|| where y is such that ||z+y||<a/(1-e).
===
Subject: Re: Alexander Grothedieck biography?
> I remember a short biography written by P. Cartier on the
occasion of the
speculative remarks on Grothendieckıs current mental state.
--
Allan Adler <ara@zurich.ai.mit.edu>
* Disclaimer: I am a guest and *not* a member of the MIT
CSAIL. My actions
and
* comments do not reŝect in any way on MIT. Also, I am
nowhere near
Boston.
===
Subject: Re: Alexander Grothedieck biography?
>Has anyone written, or is currently writing, a biography of
Alexander
>timelines. Perhaps there are some biographies written in
other languages

>that my searches arenıt turning up.
> (in french) that unfortunately is not in print. Only
(copies) of
> notes of it are circulating ...
It is being translated (in short takes) by Roy Lisker who
needs
Ŝnancial help to continue it. Google him or fermentpress to
get more
details.
===
Subject: Four-Color Theorem
Am I right in saying that there are no short proofs of the
four-color theorem to date? In 1976 Appel and Haken proved it
using
computers, and in 1996 Robertson, Seymour and Thomas improved
and
shortened the proof. Has there been any news since that time?
And what of Hadwigerıs conjecture?
I ask because I want to prove them, hehe.
-Greg
===
Subject: Re: contravariant tensor Ŝelds, differential
operators and
quantisation
In the meantime, I worked out the transformation rule for the
components of
a second order differential operator in coordinates. By this,
I mean
something of the form
a^ij d^2/(dx^i dx^j)
where the a^ij are the component functions and d is the curly
d partial
differential operator. The transformation will get pretty
complicated for
higher orders, and it does not seem to have to do anything
with tensors -
what kind of structure is it then?
--
hang my head drown my fear
till you all just disappear
reverse my forename for mail! - saibot
===
Subject: Re: 111,111,111 x 111,111,111 =
12,345,678,987,654,321
> Until you mentioned 666 I had no idea what signiŜcance it
had to some
> people.
Actually, the signiŜcance exists only in the mind of the
beholder.
If I showed you a three-leaf clover I found in my yard, you
would
probably say So what? Thereıs a lot of them about. The whole
trick
is to take something thatıs fundamentally easy and give it an
esoteric
appearance.
For example, take the factors of 111,111,111:
3 * 3 * 37 * 333667
Looks like we just missed the boat here. We got 37 instead of
36 and the
last three digits of the last factor are 667, not 666. But a
good
numerologist doesnıt let wrong numbers get in the way.
37 is 36 + 1 and the 36th triangular number (666) added to
the 1st
triangular
number (1) is 667, matching the last three digits of the last
factor. And
note
that the Ŝrst three digits of the last factor (333) are
exactly half of
666.
Not only that, but 3 * 3 * 37 equals 333!
There are so many math operations (addition, subtraction,
mutiplcation,
etc.)
and character string operations (concatenation, splitting,
reversing, etc.)
available that you can relate any two numbers as easily as
you can Ŝnd a
three-leaf clover. All you need is some lame rationalization
(I reversed the

digits because the Beast is the Anti-Christ) for why you used
what you did
and
glossed over the parts that didnıt Ŝt. And the best ally of a
lame
rationalization is repetition. Look, 37 is 36+1. And 667 is
666+1!
Spooky!

> I am not a mathematician but these things are interesting
to play with
and
> learn about..
Yep. Itıs a lot of fun. But is just that: fun.

> Dave
===
>Subject: Re: 111,111,111 x 111,111,111 =
12,345,678,987,654,321
>
>
>> A Ŝend showed me this:
>>
>> 111,111,111 x 111,111,111 = 12,345,678,987,654,321
>
>Iıd have thought a Ŝend would be more interested in 666.
>
> Yes, but the Ŝend is subtle. The LHS has 18 ones, and 18 is
6+6+6.
> And on the RHS, the digits sum to 81, which is 18 backwards.
> So if you ŝip the 81 around and add it to the other 18, you
get
> 36, the sum of integers of which is 666!
>
> Spooky, eh?
>
>
> Hereıs one I learned recently:
> Sum(n = 1 --> 7) [(P_n)^2] = 666
> ( P_n = nth prime )
===
Subject: [tensor algebra] tensorŜeld problem
Hereıs the problem:
DeŜnitions:
M is a real manifold.
T(M) is the space of smooth vectorŜelds on M.
T^1(M) is the space of 1-forms on M.
-------------------------------------------------
An important property of tensor Ŝelds is that they are
multilinear over
the
space of smooth functions.
Given {X_i} smooth vectorŜelds,
and {w^j} 1-forms.
Given a (k,l)-tensorŜeld F.
Then the function F(X_1, ..., X_k, w^1, ..., w^l) is smooth,
and thus F induces a map:
F : T^1(M) x .... x T^1(M) x T(M) x .... x T(M) --->
C^infty(M)
The map F is multilinear over C^infty(M), meaning for
functions f, g, in
C^infty(M) and any smooth vectorŜelds or covectorŜelds alpha
and beta,
F(..., f * alpha + g * beta, ...) = f F(..., alpha , ...) + g
F(..., beta,
...).
-------------------------
The converse is also true.
Any such multilinear map deŜnes a tensorŜeld.
A map t: T^1(M) x .... x T^1(M) x T(M) x .... x T(M) --->
C^infty(M)
is induced by a (k,l) tensorŜeld as above if and only if it
is multilinear
over C^infty(M).
And a map t: T^1(M) x .... x T^1(M) x T(M) x .... x T(M) --->
T(M)
is induced by a (k,l+1) tensorŜeld if and only if it is
multilinear over
C^infty(M).
-------------------
How do i have to prove the converse?
===
Subject: Old Mac-based graphing calculator from Harvard,
circa 1998?
Does anyone have a copy of a graphing calculator application
from
Harvard University, circa 1989? It was standard issue for
Math 21a,
Multivariable Calculus.
It was amazing; if you gave it a multivariable function it
would plot
vector Ŝelds, compute line integrals, curl, ŝux, etc.
I didnıt have a Mac at the time, so I had to schlepp my ŝoppy
downstairs to my dorm-mates room and kick him off his Tetris
game so I
could use his Mac. After graduating I promptly misplaced the
ŝoppy.
No doubt Mathematica would do this also, but this app was
free (with
tuition :), and it Ŝt on a ŝoppy.
Has anyone else even seen this? I donıt even remember the
name of it.
===
Subject: Re: Old Mac-based graphing calculator from Harvard,
circa 1998?
> Does anyone have a copy of a graphing calculator
application from
> Harvard University, circa 1989? It was standard issue for
Math 21a,
> Multivariable Calculus.
> It was amazing; if you gave it a multivariable function it
would plot
> vector Ŝelds, compute line integrals, curl, ŝux, etc.
> I didnıt have a Mac at the time, so I had to schlepp my
ŝoppy
> downstairs to my dorm-mates room and kick him off his
Tetris game so I
> could use his Mac. After graduating I promptly misplaced
the ŝoppy.
> No doubt Mathematica would do this also, but this app was
free (with
> tuition :), and it Ŝt on a ŝoppy.
> Has anyone else even seen this? I donıt even remember the
name of it.
It may be the one named simply Graphing Calculator,
which has since gone commercial,
in which case see
http://www.paciŜct.com/
===
Subject: Re: Old Mac-based graphing calculator from Harvard,
circa 1998?
>> Does anyone have a copy of a graphing calculator
application from
>> Harvard University, circa 1989? It was standard issue for
Math 21a,
>> Multivariable Calculus.
>> It was amazing; if you gave it a multivariable function it
would plot
>> vector Ŝelds, compute line integrals, curl, ŝux, etc.
>> I didnıt have a Mac at the time, so I had to schlepp my
ŝoppy
>> downstairs to my dorm-mates room and kick him off his
Tetris game so I
>> could use his Mac. After graduating I promptly misplaced
the ŝoppy.
>> No doubt Mathematica would do this also, but this app was
free (with
>> tuition :), and it Ŝt on a ŝoppy.
>> Has anyone else even seen this? I donıt even remember the
name of it.
> It may be the one named simply Graphing Calculator,
> which has since gone commercial,
> in which case see
> http://www.paciŜct.com/
It canıt be the same. Appleıs Graphing Calculator was a
PowerPC-only
app, and all Macs in 1989 were 68K.
--
Dave Seaman
Judge Yohnıs mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Bourbaki Volumes (was Re: Cosets)
Originator: jgamble@ripco.com (John M. Gamble)
>> lol. I have a better suggestion : Bourbaki. Any advance ?
>> Pascal
>Unfortunately, Bourbakiıs ones are already lost forever ...
I tried to get
>the Ŝrst topology volumes (in French of course) but I
couldnt -my only
>chance would be to scan every single page of the volumes Iıd
borrow from
my
>uni. library-.
Could you expand on this a little? Why are they lost forever?
Published during the self-destructing-paper era?
Limited print run?
Other?
--
-john
February 28 1997: Last day libraries could order catalogue
cards
from the Library of Congress.
===
Subject: Re: Bourbaki Volumes (was Re: Cosets)
> Could you expand on this a little? Why are they lost
forever?
> Published during the self-destructing-paper era?
> Limited print run?
> Other?
For instance the Ŝrst volumes of Topology (I think Vol. I, II
and III or
so) are not printed anymore (in French, I mean; getting them
in English
would be ok too, but iıd lose much of their semantic value).
The best I could do was to borrow them from my universityıs
library: I got
two old books from the 60ıs, hardly bound with the pages
almost vanishing
into ashes when you touch them (the woman was looking at me
as if I was
some
sorcerer or I dunno which freak, all the more than I had to
bother her for
almost half an hour because the books were in a remote dark
room buried in
a
secret place of the uni. lol); I didnıt have the courage to
scan each of
the
thounsands of pages, and I had to give them back -you can
borrow books only
for one month, even though, hey, those books hadnıt been
borrowed for some
9
years-. Theyıre now lost forever.
===
Subject: Re: Bourbaki Volumes (was Re: Cosets)
>>Could you expand on this a little? Why are they lost
forever?
>>Published during the self-destructing-paper era?
>>Limited print run?
>>Other?
> For instance the Ŝrst volumes of Topology (I think Vol. I,
II and III or
> so) are not printed anymore (in French, I mean; getting
them in English
> would be ok too, but iıd lose much of their semantic value).
> The best I could do was to borrow them from my universityıs
library: I
got
> two old books from the 60ıs, hardly bound with the pages
almost vanishing
> into ashes when you touch them (the woman was looking at me
as if I was
some
> sorcerer or I dunno which freak, all the more than I had to
bother her
for
> almost half an hour because the books were in a remote dark
room buried in
a
> secret place of the uni. lol); I didnıt have the courage to
scan each of
the
> thounsands of pages, and I had to give them back -you can
borrow books
only
> for one month, even though, hey, those books hadnıt been
borrowed for some
9
> years-. Theyıre now lost forever.
Bourbakiıs General Topology in English is available from
amazon.com.
They also have a used copy of Chap. 5 - 10 in French for sale.
Many used copies of Bourbaki books are available from
Abebooks, real cheap
http://www.abebooks.com
===
Subject: Re: Bourbaki Volumes (was Re: Cosets)
> Could you expand on this a little? Why are they lost
forever?
> Published during the self-destructing-paper era?
> Limited print run?
> Other?
> For instance the Ŝrst volumes of Topology (I think Vol. I,
II and III or
> so) are not printed anymore (in French, I mean; getting
them in English
> would be ok too, but iıd lose much of their semantic value).
> The best I could do was to borrow them from my universityıs
library: I
got
> two old books from the 60ıs, hardly bound with the pages
almost vanishing
> into ashes when you touch them (the woman was looking at me
as if I was
some
> sorcerer or I dunno which freak, all the more than I had to
bother her
for
> almost half an hour because the books were in a remote dark
room buried in
a
> secret place of the uni. lol); I didnıt have the courage to
scan each of
the
> thounsands of pages, and I had to give them back -you can
borrow books
only
> for one month, even though, hey, those books hadnıt been
borrowed for some
9
> years-. Theyıre now lost forever.
You might try alt.binaries.e-book. It is surprising what
turns up there
- you might get lucky. Try REQ:Bourbaki as your subject.
--
Paul Sperry
Columbia, SC (USA)
===
Subject: test
test
===
Subject: Question for Harris
have found many different posts of yours with valid
counter-arguments
by other posters.
Since you are still making such contentions, can you please
deŜne a
short and unassailable mathematical proof supporting your
claims?
===
Subject: Re: Question for Harris
> have found many different posts of yours with valid
counter-arguments
> by other posters.
> Since you are still making such contentions, can you please
deŜne a
> short and unassailable mathematical proof supporting your
claims?
Why do you care, crackpot? Are you trying to get some
pointers in being a
kook from Harris?
===
Subject: Re: Question for Harris
> have found many different posts of yours with valid
counter-arguments
> by other posters.
> Since you are still making such contentions, can you please
deŜne a
> short and unassailable mathematical proof supporting your
claims?
JSH _never_ responds to questions.
Gib
===
Subject: Re: Great-circle radius of ellipsoid
> pi r^2 = pi a b
>> or
>> = Pi * [a^3 * b]^.5
> Wrong. The above lacks a logical derivation.
>> ** OR **
>> = Pi * .5 * [a^2 + .5*(a^2 + b^2)]
>> = Pi * .25 * [3a^2 + b^2]
> r = sqrt(ab).
>> or
>> = [a^3 * b]^.25
>> ** OR **
>> = [.5 * (a^2 + .5*<a^2 + b^2>)]^.5 = [.25 * (3a^2 +
b^2)]^.5
>> Look at it this way:
>> Semi-axes . Arcradii
>> | /|
>> | a^2/b |
>> | |
>> | |
>> | = b | = Rr = a *
E2ı{0<Lat<90}
>> | / | / ~=~ [a*b]^.5
>> | | ~=~ [.5*(a^2+b^2)]^.5
>> | b^2/a |
>> |_______________ `.________________
>> = a = a
>> Mean Mean
>> Radius ~=~ [a*b]^.5 Arcradius ~=~ [a*Rr]^.5
>> ~=~ [.5*(a^2+b^2)]^.5 ~=~ [a*(a*b)^.5]^.5
>> = [a^3*b]^.25
~=~ .5*[a+Rr]
>> ~=~ [.5*(a+.5*<a^2+b^2>)]^.5
>> = [.25*(3a^2+b^2)]^.5
>> Thus, *EVEN* if you believe the mean arcradius is the
geometric
>> mean, the actual gm should be [a^3*b]^.25, not [a*b]^.5!
P=)
> The solution you have given is arbitrary and does not
follow a
> consistent derivation, and with that sort of reasoning,
there would
> be inŜnite solutions.
> The answer is simply:
> Area of circle = Area of ellipse
^^^^^^^
> r = sqrt(ab).
Right, for an *ELLIPSE* (the left, semi-axes diagram)!
What is being discussed here is a 3-D ellipsoid (the right,
arcradii diagram).

> If this is wrong as you contend, explain why.
I think the problem is everyone is viewing the ellipsoid as
being
composed of equi-sized meridional ellipses and/or meridionally
parallel semi-ellipses.
While that may be *A* valid interpretation, it is not *THE*
proper
interpretation for this discussion--the Ŝnding of the great
circle arcradius of best Ŝt for the spheroid/ellipsoid.
As such, the relevant approach to analyzing the form of an
ellipsoid
is to view it as being made up of unique great ellipses (GEs),
where a meridian is the maximally elliptic GE and the
equuator is
the minimally elliptic GE, a great circle.
So, for a *3-D ELLIPSOID*, one is averaging the arcradii of
all of
the different GEs, hence the simplest 2-point average would be
between a meridian (~=~ [a*b]^.5 or .5*[a+b] or
[.5*(a^2+b^2)]^.5,
to name the most common)--the maximum GE--and the equator,
which,
being a great circle and having an equal radius throughout,
always
equals a.
GEs, as all geodetic lines travel along them, can also be
viewed as
ARC PATHs (APs).
Let a = 6378.135 and b = 6356.75.
The mean meridional arcradius (Rr) = 6367.4469888342:
[a*b]^.5 = 6367.43352233; 2/[1/a + 1/b] = 6367.42454467;
.5 * [a + b] = 6367.4425; [.5 * (a^2 + b^2)]^.5 =
6367.45147766;
[.5 * (a^1.5 + b^1.5)]^(1/1.5) = 6367.4469888334;
The mid-GE possibilities work out to the following:
Lat1 = 0, Long2 - Long1 = 90,
Lat2 = 45, AP = 45.09601377, mean
arcradius = 6372.80780848;
Lat2 = 45.09601444, AP = 45, mean
arcradius = 6372.78989792;
(.5 * [6372.80780848 + 6372.78989792] = 6372.79885320)
Lat2 = AP = 45.04800712, mean arcradius =
6372.79885325
(the isopathic mid-GE (the concept and process discussed in a
separate
)
Keeping in mind that--given the minimal ellipticity
involved--the
trapezoidal 3-point average provides an integration to about
.0000001, the global great circle arcradius of best Ŝt using
the
three different differentiating methods produce:
.25 * (Rr + 2*6372.80780848 + a) = 6372.7994014;
.25 * (Rr + 2*6372.78989792 + a) = 6372.7904462;
.25 * (Rr + 2*6372.79885325 + a) = 6372.7949238;
= .5 * [6372.7994014 + 6372.7904462]
Now compare the aforementioned approximations:
[a * Rr]^.5 = 6372.78875377;
[a^3 * b]^.5 = 6372.78201486;
.5 * [a + Rr] = 6372.79099442;
[.25 * (3a^2 + b^2)]^.5 = 6372.79547760;
Wouldnıt you agree that [.25 * (3a^2 + b^2)]^.5 is by far the
best
approximation--if not another ACTUAL global mean, via some
other
differentiation?
Letting a = 10000, down to a b value of about 8385, the
maximum
difference from the isopathic is -.74186, whereupon the
deviation
reverses until they equate at a b value of about 7624.5 and
[.25 * (3a^2 + b^2)]^.5 remains greater than the isopathic for
lesser b values (at 7071.067 the difference is +1.94, and at
6000,
+11.03)--thus even if the isopathic IS the only true great
circle
arcradius of best Ŝt, for general purposes [.25 * (3a^2 +
b^2)]^.5
appears adequate.
As I noted in my original reply, given the relationship of
.5 * [a^2 + b^2] to the integrand^2 of the elliptic integral
of the second kind, there is still a good possibility that
[.25 * (3a^2 + b^2)]^.5 may be a closed-form reduction to
another
legitimate great ellipse differentiation.
Or, rather than arcradius, perhaps [.25 * (3a^2 + b^2)]^.5 is
the
global average of the *semi-axes* along each GE--i.e., change
b to
Radius{Lat} for each GE and calculate Rr for each GE (thus
Radius{0}
equals a--with Rr equaling a--and Radius{90}
equals b--with
Rr equaling the meridional Rr)--but I havenıt really explored
that
approach, so it is just a random theory at this point.
~Kaimbridge~
-----
WantedKaimbridge (w/mugshot!):
http://www.angelŜre.com/ma2/digitology/Wanted_KMGC.html
----------
DigitologyThe Grand Theory Of The Universe:
http://www.angelŜre.com/ma2/digitology/index.html
***** Void Where Permitted; Limit 0 Per Customer. *****
===
Subject: Re: Integer Occurs Same # Of Times In These Sequences
[My other reply to this thread is not a duplicate of this
reply.]
We can use the result below to get an integral identity.
(Most likely there is an easier way to prove this.)
Let f1(x) and f2(x) be real->real continuous monotonically
increasing
functions, where f1(0) = f2(0) = 0.
Let g1(x) and g2(x) be such that g1(f1(x)) = x and g2(f2(x))
= x for
all x.
Let a(x) be any real -> real integrable function (where
integral below
exists and converges), where the derivative of a(x) exists.
Let A(x) be the derivative of a(x) at x.
Then:
integral{0 to oo} (a(f1(x)+x) + a(f2(x)+x)
- a(x) - a(f1(x)+f2(x)+x)) dx
=
integral{0 to oo} integral{0 to oo} A(max(g1(x),g2(y))+x+y)
dy dx
(Maybe.)
Using the result in my other reply to this thread, we can get
the
simpler integral identity:
integral{0 to oo} a(f(x)+g(x)+x) dx
=
integral{0 to oo} integral{0 to oo} -A(max(f(x),g(y))+x+y) dy
dx
(Maybe.)
Leroy Quet
> I wonder if anyone has a clever proof of the result below.
> --
> Let f1()and f2() be (almost) any real-real continuous
monotonically
> increasing functions.
> Let g1() and g2() be the inverses of f1() and f2(),
> f1(g1(x)) = x and f2(g2(x)) = x for all x.
> And let each f and g be such that
> f1(1), f2(1), g1(1), and g2(1) all exist and are each >= 0.
> (I am a little unsure if these conditions are necessary or
> sufŜcient.)
> Let m be any positive integer.
> Then:
> (the number of times m occurs in the sequence
> {ŝoor(f1(k)) + k})
> (the number of times m occurs in the sequence
> {ŝoor(f2(k)) + k})
> (the number of times m occurs in the sequence
> {ceiling(max(g1(k),g2(j))) + k + j - 2}),
> is equal to
> (the number of times m occurs in the sequence
> {k})
> (the number of times m occurs in the sequence
> {ŝoor(f1(k)) + ŝoor(f2(k)) + k})
> (the number of times m occurs in the sequence
> {ceiling(max(g1(k),g2(j))) + k + j - 1}),
> for j and k = positive integers (independent between
different
> sequences).
> (max(x,y) means maximum of x or y.)
> Example:
> For m = 12,
> f1(k) = k^2,
> f2(k) = 2^k,
> g1(k) = sqrt(k),
> g2(k) =ln(k)/ln(2),
> we have:
> ŝoor(f1(k)) + k: k = 3
> ŝoor(f2(k)) + k: no kıs
> ceiling(max(g1(k),g2(j))) + k + j - 2:
> k = 9, j = 2;
> k = 8, j = 3;
> k = 7, j = 4;
> k = 6, j = 5;
> k = 5, j = 6;
> k = 4, j = 7;
> k = 3, j = 8;
> k = 1, j = 9
> So we have 9 representations of 12 total among the Ŝrst 3
sequences.
> Counting representations among the last 3 sequences, we
have:
> k: k = 12 (obviously)
> ŝoor(f1(k)) + ŝoor(f2(k)) + k: no kıs
> ceiling(max(g1(k),g2(j))) + k + j - 1:
> k = 9, j = 1;
> k = 8, j = 2;
> k = 7, j = 3;
> k = 6, j = 4;
> k = 5, j = 5;
> k = 4, j = 6;
> k = 3, j = 7;
> k = 2, j = 8
> So, again, we have 9 representations.
> (Amazing)
> As a (possibly fun) exercise, try to prove yourself this
result is
> true for all positive integers m and for all fıs and gıs
(with the
> mentioned restrictions).
> Leroy Quet
===
Subject: Re: Integer Occurs Same # Of Times In These Sequences
[My other reply to this thread {which I still have to post as
of me
writing this} is not a duplicate of this reply.]
Letting f(x) = f1(x), and having the inverse of f(x), g(x),
replace
f2(x),
we can use, in conjunction with the result in this threadıs
original
post, the related but simpler result at
to get the following:
If f(x) is any real -> real continuous monotonically
increasing
function, where
f(1) and g(1) are deŜned and are both >= 0,
g(f(x)) =x for all x,
then:
any particular positive integer m occurs the same number of
times in
{k + j + max(ŝoor(f(k))+1,ceiling(g(j))) - 2}
as in both
{k + j + max(ŝoor(f(k))+1,ceiling(g(j))) - 1}
and
{ŝoor(f(k)) + ceiling(g(k)) + k - 1}
together.
And if f(k) and g(k) never equal an integer for all positive
integers
k, the result above can be simpliŜed slightly to:
the positive integer m occurs the same number of times in
{k + j + ŝoor(max(f(k),g(j))) - 1}
as in both
{k + j + ŝoor(max(f(k),g(j)))}
and
{ŝoor(f(k)) + ŝoor(g(k)) + k}
together.
Leroy Quet
> I wonder if anyone has a clever proof of the result below.
> --
> Let f1()and f2() be (almost) any real-real continuous
monotonically
> increasing functions.
> Let g1() and g2() be the inverses of f1() and f2(),
> f1(g1(x)) = x and f2(g2(x)) = x for all x.
> And let each f and g be such that
> f1(1), f2(1), g1(1), and g2(1) all exist and are each >= 0.
> (I am a little unsure if these conditions are necessary or
> sufŜcient.)
> Let m be any positive integer.
> Then:
> (the number of times m occurs in the sequence
> {ŝoor(f1(k)) + k})
> (the number of times m occurs in the sequence
> {ŝoor(f2(k)) + k})
> (the number of times m occurs in the sequence
> {ceiling(max(g1(k),g2(j))) + k + j - 2}),
> is equal to
> (the number of times m occurs in the sequence
> {k})
> (the number of times m occurs in the sequence
> {ŝoor(f1(k)) + ŝoor(f2(k)) + k})
> (the number of times m occurs in the sequence
> {ceiling(max(g1(k),g2(j))) + k + j - 1}),
> for j and k = positive integers (independent between
different
> sequences).
> (max(x,y) means maximum of x or y.)
> Example:
> For m = 12,
> f1(k) = k^2,
> f2(k) = 2^k,
> g1(k) = sqrt(k),
> g2(k) =ln(k)/ln(2),
> we have:
> ŝoor(f1(k)) + k: k = 3
> ŝoor(f2(k)) + k: no kıs
> ceiling(max(g1(k),g2(j))) + k + j - 2:
> k = 9, j = 2;
> k = 8, j = 3;
> k = 7, j = 4;
> k = 6, j = 5;
> k = 5, j = 6;
> k = 4, j = 7;
> k = 3, j = 8;
> k = 1, j = 9
> So we have 9 representations of 12 total among the Ŝrst 3
sequences.
> Counting representations among the last 3 sequences, we
have:
> k: k = 12 (obviously)
> ŝoor(f1(k)) + ŝoor(f2(k)) + k: no kıs
> ceiling(max(g1(k),g2(j))) + k + j - 1:
> k = 9, j = 1;
> k = 8, j = 2;
> k = 7, j = 3;
> k = 6, j = 4;
> k = 5, j = 5;
> k = 4, j = 6;
> k = 3, j = 7;
> k = 2, j = 8
> So, again, we have 9 representations.
> (Amazing)
> As a (possibly fun) exercise, try to prove yourself this
result is
> true for all positive integers m and for all fıs and gıs
(with the
> mentioned restrictions).
> Leroy Quet
===
Subject: velocities in opposite directions
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i4PIm7b17257;
Question:
If one car has a radar gun and is heading south at appx. 10
mph and another
car is heading north (exact opposite direction) at 40 mph,
will that radar
gun read 50 mph. why or why not?
Deanna
===
Subject: Re: velocities in opposite directions
> Question:
> If one car has a radar gun and is heading south at appx. 10
mph and
another
> car is heading north (exact opposite direction) at 40 mph,
will that radar

> gun read 50 mph. why or why not?
> Deanna
At relative vealocities so much less that the speed of light,
the radar
gun should read close enough to 50 mph for all practical
purposes.
===
Subject: Re: velocities in opposite directions
: > Question:
: >
: > If one car has a radar gun and is heading south at appx.
10 mph and
another
: > car is heading north (exact opposite direction) at 40
mph, will that
radar
: > gun read 50 mph. why or why not?
: >
: > Deanna
: >
: At relative vealocities so much less that the speed of
light, the radar
: gun should read close enough to 50 mph for all practical
purposes.
Velocity is a vector (whose maginitude is speed), vectors
add/subtract,
so the net velocity = 50 mph heading north. The velocities
involved are
far less than c = speed of radar beam, so their magnitudes
dominate.
===
Subject: Re: resolving Willıs misunderstanding
>>>>Letıs try a different approach, starting with deŜnitions
of
everything
>>>>under consideration:
>>>>
>>>>We are talking about functions, but the real concern is
what
algorithms
>>>>are used to compute those functions. For simplicityıs
sake, we
can
>>>>restrict our discussion to functions mapping from the
whole
numbers to
>>>>the whole numbers f:W->W. For an algorithm to correctly
represent
>>>>f(x)=y, it should start with input x and produce output y
after
Ŝnitely
>>>>many steps, where each step is computed from the input
and the
previous
>>>>steps. For a given function f, f may or may not be deŜned
for
all/any
>>>>numbers in the whole numbers. As a result, the algorithm
describing f
>>>>is only required to be Ŝnite for those values of x for
which f
is
>>>>deŜned. A function f with such an algorithm is in general
called
a
>>>>partial function. A partial function f which is deŜned on
all
elements
>>>>of the whole numbers is called a total function and is
considered
to be
>>>>computable (or recursive).
>>>>
>>>>At this point, I believe we can start making some clear
observations.
>>>>First: the halting problem is only available for
discussion if we
a
>>>>dealing with partial functions (which have algorithms
that do not
halt
>>>>in Ŝnite time for some input values).
>>>>Second: it appears to be your desire to discuss only total
functions.
>>>>Third: if we restrict ourselves to the total functions,
discussing
the
>>>>halting problem is meaningless.
>>>>
>>>>A question that is relevent to the above: if we are
restricting
>>>>ourselves to total functions, is there a way to specify
the
algorithms
>>>>of all total functions in such a way that there is no
possibility
of
>>>>dealing with potentially partial functions?
>>>>
>>>>Before moving forward, do you have any comments on the
above?
Right now
>>>>Iım just trying to set a framework for a detailed
analysis, but if
the
>>>>basic framework is in dispute, I may need to adjust for
that.
>>>
>>>The term to be deŜned is valid construction. Facts & FALSE
->
anything.
>>>>
>>>>Agreed. I just wanted to make sure there were no
objections before
>>>>diving into the details. If the basic concepts to be
deŜned in
detail
>>>>canıt be agreed on, then there isnıt much point in going
further.
>>>>
>>>>For the moment, letıs allow functions to be n-ary for
ease of
notation.
>>>>(It can be shown that we can restrict ourselves to unary
functions,
but
>>>>they get messier.)
>>>>
>>>>Letıs start by deŜning primitive recursive functions These
functions
>>>>are guaranteed to be total functions and can be
constructed by a
Ŝnite
>>>>number of applications of the following rules:
>>>>
>>>>1) f(x) = x+1 is primitive recursive
>>>>2) f(x1,x2,...,xn) = m is primitive recursive where n and
m >= 0
>>>>3) f(x1,x2,...,xn) = xi is primitive recursive where 0 <
i <= n
>>>>4) If g1, g2, .. gm, are primitive recursive and n-ary,
and h is
>>>>primitive recursive and m-ary then so is n-ary f deŜned as
>>>> f(x1,x2,...,xn) =
h(g1(x1,x2,...,xn),...,gm(x1,x2,...,xn))
>>>>5) For n>=1, if f is n-ary, g is (n-1)-ary, h is
(n+1)-ary, and g
and h
>>>>are primitive recursive, then so is f deŜned by the
following two
rules:
>>>>a) f(0,x2,x3,...,xn) = g(x2,x3,...,xn)
>>>>b) f(x1+1,x2,x3,...,xn) = h(x1,f(x1,x2,...,xn),x2,...,xn)
>>>>
>>>>These rules appear (to my eye) to correspond with the
deŜnition of
a
>>>>function guaranteed to halt you offered in another thread.
>>>>
>>>>Using these it is possible (though messy) to construct
virtually
all
>>>>normal mathematical functions on the whole numbers. In
particular,
>>>>prime decomposition that will allow a number to represent
a sequence
of
>>>>numbers by its prime decomposition. This enables us to
formally
get
>>>>back to unary functions, if need be.
>>>>
>>>>Any comments at this point? The next question will be, do
primitive
>>>>recursive functions describe ALL total computable
functions?
>>>
>>>nope
>>
>>Thereıs two questions, which one are you saying nope to?
>>>
>>>the 1st one, youıre supposed to be establishing the *next*
question.
>>
>>Since there are only 5 rules for deriving primitive
recursive
functions,
>> each one can be indexed based on what rules/functions are
used to
>>construct it. Consider one such indexing f0, f1, f2, f3,
... and let
us
>>construct a new function g0(x) = fx(x) + 1. Since fx can be
formally
>>considered unary, regarless of what n is for n-ary fx, and
x determines
>>the construction of f, g0(x) is deŜned on all x in the
Whole Numbers.
>>Moreover, we have included all partial recursive functions
on our list
>>f0, f1, f2, .... Therefore, g0(x) is a total function, it is
>>computable, but it is not a primitive recursive function. By
prepending
>>g0(x) to the listing of primitive recursive functions, we
can construct
>>g1(x), g2(x), etc. Since there are multiple possible
indexing schemes
>>available for the primitive recursive functions, there are
also
multiple
>>sequences of gi(x) available.
>>
>>Any issues with the above?
>>
>I got no idea what rule 5 does. It looks like a step -1 for
loop that
pushes the
>parameters into the 2nd parameter of the 2nd parameter
recursively,
making
>a longer parameter list every iteration.
>>Itıs a construction similar to f(0) = c, f(x+1) = g(f(x)).
The idea is
>>to deŜne f(0) as something known, and deŜne f(x+1) in terms
of f(x).
>>The parameter list doesnıt actually get longer.
>>construct a new function g0(x) = fx(x) + 1
>this is just diagonalisation all over again, f with godel x
applied to
its own godel number
>with a modiŜcation.
>>You are quite right. Now, does that invalidate the
conclusion?
>whatıs to stop me deŜning consistent p.r.f. where variable
references
are contextually free.
>fx(x) = fx(y) for any x, y
>>Those functions are described in construction 2, they are
the constant
>>functions. This has nothing to do with the validity or
invalidity of
>>the argument.
>this stops dummy functions like the above that double usage
the x in
g0(x)
>Iıve shown that getting a parameter x and using it for 2
arguments is
>enough to change a total system to non halting system.
>>We have not gotten to anything that does not halt. All that
is being
>>said is: a sufŜcient restriction to prevent functions that
donıt halt
>>also excludes some that do always halt.
> ok, but how hard is it to exclude functions of type fx(x),
theyıre the
ones going
> into inŜnite loops. what computation are they performing
that canıt be
> done some other way?
The functions fx do not contain inŜnite loops. Nor do the
functions gi
built from them. The point was that the primitive recursive
functions,
while guaranteed to halt, are not *all* of the functions that
are
guaranteed to halt. The various functions gi are also
guaranteed to
halt, but they were found using a diagonal-style argument,
and are
therefore not primitive recursive.
Having said all that, MOST functions and operations that we
would want
to perform would be primitive recursive functions, including
things like
the addition function, subtraction function (which produces 0
for
negative results), multiplication, integer division, modular
division,
exponentiation, etc. As long as the computations can be
broken down
into the 5 rules applied to various primitive recursive
functions, we
are good to go.
So far, there are no inŜnite loops, and everything is
guaranteed to
halt. Itıs just that not everything that halts is primitive
recursive.
The trick is to see if we can Ŝnd something that will include
ALL
functions that halt, and still exclude those that donıt.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: resolving Willıs misunderstanding
>>>>Letıs try a different approach, starting with deŜnitions
of
everything
>>>>under consideration:
>>>>
>>>>We are talking about functions, but the real concern is
what
algorithms
>>>>are used to compute those functions. For simplicityıs
sake, we
can
>>>>restrict our discussion to functions mapping from the
whole
numbers to
>>>>the whole numbers f:W->W. For an algorithm to correctly
represent
>>>>f(x)=y, it should start with input x and produce output y
after
Ŝnitely
>>>>many steps, where each step is computed from the input
and the
previous
>>>>steps. For a given function f, f may or may not be deŜned
for
all/any
>>>>numbers in the whole numbers. As a result, the algorithm
describing f
>>>>is only required to be Ŝnite for those values of x for
which f
is
>>>>deŜned. A function f with such an algorithm is in general
called a
>>>>partial function. A partial function f which is deŜned on
all
elements
>>>>of the whole numbers is called a total function and is
considered to be
>>>>computable (or recursive).
>>>>
>>>>At this point, I believe we can start making some clear
observations.
>>>>First: the halting problem is only available for
discussion if
we a
>>>>dealing with partial functions (which have algorithms
that do
not halt
>>>>in Ŝnite time for some input values).
>>>>Second: it appears to be your desire to discuss only total
functions.
>>>>Third: if we restrict ourselves to the total functions,
discussing the
>>>>halting problem is meaningless.
>>>>
>>>>A question that is relevent to the above: if we are
restricting
>>>>ourselves to total functions, is there a way to specify
the
algorithms
>>>>of all total functions in such a way that there is no
possibility of
>>>>dealing with potentially partial functions?
>>>>
>>>>Before moving forward, do you have any comments on the
above?
Right now
>>>>Iım just trying to set a framework for a detailed
analysis, but
if the
>>>>basic framework is in dispute, I may need to adjust for
that.
>>>
>>>The term to be deŜned is valid construction. Facts & FALSE
-> anything.
>>>>
>>>>Agreed. I just wanted to make sure there were no
objections
before
>>>>diving into the details. If the basic concepts to be
deŜned in
detail
>>>>canıt be agreed on, then there isnıt much point in going
further.
>>>>
>>>>For the moment, letıs allow functions to be n-ary for
ease of
notation.
>>>>(It can be shown that we can restrict ourselves to unary
functions, but
>>>>they get messier.)
>>>>
>>>>Letıs start by deŜning primitive recursive functions These
functions
>>>>are guaranteed to be total functions and can be
constructed by a
Ŝnite
>>>>number of applications of the following rules:
>>>>
>>>>1) f(x) = x+1 is primitive recursive
>>>>2) f(x1,x2,...,xn) = m is primitive recursive where n and
m >= 0
>>>>3) f(x1,x2,...,xn) = xi is primitive recursive where 0 <
i <= n
>>>>4) If g1, g2, .. gm, are primitive recursive and n-ary,
and h is
>>>>primitive recursive and m-ary then so is n-ary f deŜned as
>>>> f(x1,x2,...,xn) =
h(g1(x1,x2,...,xn),...,gm(x1,x2,...,xn))
>>>>5) For n>=1, if f is n-ary, g is (n-1)-ary, h is
(n+1)-ary, and g
and h
>>>>are primitive recursive, then so is f deŜned by the
following two
rules:
>>>>a) f(0,x2,x3,...,xn) = g(x2,x3,...,xn)
>>>>b) f(x1+1,x2,x3,...,xn) = h(x1,f(x1,x2,...,xn),x2,...,xn)
>>>>
>>>>These rules appear (to my eye) to correspond with the
deŜnition
of a
>>>>function guaranteed to halt you offered in another thread.
>>>>
>>>>Using these it is possible (though messy) to construct
virtually
all
>>>>normal mathematical functions on the whole numbers. In
particular,
>>>>prime decomposition that will allow a number to represent
a
sequence of
>>>>numbers by its prime decomposition. This enables us to
formally
get
>>>>back to unary functions, if need be.
>>>>
>>>>Any comments at this point? The next question will be, do
primitive
>>>>recursive functions describe ALL total computable
functions?
>>>
>>>nope
>>
>>Thereıs two questions, which one are you saying nope to?
>>>
>>>the 1st one, youıre supposed to be establishing the *next*
question.
>>
>>Since there are only 5 rules for deriving primitive
recursive
functions,
>> each one can be indexed based on what rules/functions are
used to
>>construct it. Consider one such indexing f0, f1, f2, f3,
... and let
us
>>construct a new function g0(x) = fx(x) + 1. Since fx can be
formally
>>considered unary, regarless of what n is for n-ary fx, and x
determines
>>the construction of f, g0(x) is deŜned on all x in the Whole
Numbers.
>>Moreover, we have included all partial recursive functions
on our
list
>>f0, f1, f2, .... Therefore, g0(x) is a total function, it is
>>computable, but it is not a primitive recursive function. By
prepending
>>g0(x) to the listing of primitive recursive functions, we
can
construct
>>g1(x), g2(x), etc. Since there are multiple possible
indexing
schemes
>>available for the primitive recursive functions, there are
also
multiple
>>sequences of gi(x) available.
>>
>>Any issues with the above?
>>
>
>I got no idea what rule 5 does. It looks like a step -1 for
loop that
pushes the
>parameters into the 2nd parameter of the 2nd parameter
recursively,
making
>a longer parameter list every iteration.
>>
>>Itıs a construction similar to f(0) = c, f(x+1) = g(f(x)).
The idea is
>>to deŜne f(0) as something known, and deŜne f(x+1) in terms
of f(x).
>>The parameter list doesnıt actually get longer.
>>
>>
>>construct a new function g0(x) = fx(x) + 1
>
>this is just diagonalisation all over again, f with godel x
applied to
its own godel number
>with a modiŜcation.
>>
>>You are quite right. Now, does that invalidate the
conclusion?
>>
>>
>whatıs to stop me deŜning consistent p.r.f. where variable
references
are contextually free.
>fx(x) = fx(y) for any x, y
>>
>>Those functions are described in construction 2, they are
the constant
>>functions. This has nothing to do with the validity or
invalidity of
>>the argument.
>>
>>
>this stops dummy functions like the above that double usage
the x
in g0(x)
>Iıve shown that getting a parameter x and using it for 2
arguments is
>enough to change a total system to non halting system.
>>
>>We have not gotten to anything that does not halt. All that
is being
>>said is: a sufŜcient restriction to prevent functions that
donıt halt
>>also excludes some that do always halt.
> ok, but how hard is it to exclude functions of type fx(x),
theyıre the
ones going
> into inŜnite loops. what computation are they performing
that canıt
be
> done some other way?
> The functions fx do not contain inŜnite loops. Nor do the
functions gi
> built from them. The point was that the primitive recursive
functions,
> while guaranteed to halt, are not *all* of the functions
that are
> guaranteed to halt. The various functions gi are also
guaranteed to
> halt, but they were found using a diagonal-style argument,
and are
> therefore not primitive recursive.
> Having said all that, MOST functions and operations that we
would want
> to perform would be primitive recursive functions,
including things like
> the addition function, subtraction function (which produces
0 for
> negative results), multiplication, integer division,
modular division,
> exponentiation, etc. As long as the computations can be
broken down
> into the 5 rules applied to various primitive recursive
functions, we
> are good to go.
> So far, there are no inŜnite loops, and everything is
guaranteed to
> halt. Itıs just that not everything that halts is primitive
recursive.
> The trick is to see if we can Ŝnd something that will
include ALL
> functions that halt, and still exclude those that donıt.
If fx(x) is a construct that halts, then UTMs must be
impossible. Because
mUTM(mUTM) doesnıt halt, as weıve been through.
If UTMs are not part of primitive recursive functions, then
theyıre not
complete
anyway.
Even so, what does g0(x) do? I still say you are entrenched
in *names* of
functions and not *what* they actually compute. Two
algorithms can perform
the same computation canıt they?
Herc
===
Subject: Polygons With Angles Of Different k-gons
I am wondering generally about irregular polygons which have
angles
each equal to that of different regular polygons.
In other words, each interior angle is equal to
(1 - 2/k)*180 degrees
for a set of positive integers k.
For example, we can have a 5-gon with the angle of an
equilateral
triangle, of a square, of a regular pentagon, of a regular
hexagon,
and of a regular 20-gon.
Or, allowing dupicate angles, we can have a 4-gon with 2
angles of an
equlilateral triangle, one angle of a square, and one of a
dodecagon(12-gon).
If we are allowed to adjust our polygonsı side-lengths as
necessary,
then which such irregular polygons can tile the inŜnite plane
with
congruent copies of themselves?
Leroy Quet
===
Subject: Re: Polygons With Angles Of Different k-gons
3QLpj-NoP*NzsIC,boYU]bQ]Hı<P%JD/,.J>y<#4ga3$21:
> I am wondering generally about irregular polygons which
have angles
> each equal to that of different regular polygons.
> In other words, each interior angle is equal to
> (1 - 2/k)*180 degrees
> for a set of positive integers k.
> For example, we can have a 5-gon with the angle of an
equilateral
> triangle, of a square, of a regular pentagon, of a regular
hexagon,
> and of a regular 20-gon.
> Or, allowing dupicate angles, we can have a 4-gon with 2
angles of an
> equlilateral triangle, one angle of a square, and one of a
> dodecagon(12-gon).
> If we are allowed to adjust our polygonsı side-lengths as
necessary,
> then which such irregular polygons can tile the inŜnite
plane with
> congruent copies of themselves?
A convex polygon that tiles the plane must have at most six
sides, so
the only possible choices come from the Egyptian fraction
representations of 1 with at most six terms, all <= 1/3.
There are only
Ŝnitely many such representations.
In addition, it has to be possible to Ŝt the largest angle
together
with some combination of other angles around a single vertex;
a
combination of four angles is impossible (the four-angle
combination
maximizing the largest angle is 1/3+1/3+1/4+1/12, but all the
Egyptian
fractions I found had a term smaller than 1/12) so we must
have three of
these fractions 1/a + 1/b + 1/c = 1/2 where c is the largest
term in the
Egyptian fraction.
So, it seems, this eliminates all but the following
possibilities
(in the distinct angle case; of course there are more if you
allow
repeated angles):
1/3 + 1/4 + 1/5 + 1/6 + 1/20
1/3 + 1/4 + 1/5 + 1/7 + 1/20 + 1/42
1/3 + 1/4 + 1/5 + 1/8 + 1/20 + 1/24
1/3 + 1/4 + 1/5 + 1/9 + 1/18 + 1/20
1/3 + 1/4 + 1/5 + 1/10 + 1/15 + 1/20
1/3 + 1/4 + 1/6 + 1/7 + 1/12 + 1/42
1/3 + 1/4 + 1/6 + 1/9 + 1/12 + 1/18
1/3 + 1/4 + 1/6 + 1/10 + 1/12 + 1/15
In each of these cases, the internal angles of the hexagon
can be
partitioned into two triples that each add to 360 (in the Ŝrst
solution, consider the pentagon to be a degenerate hexagon
with a 180
degree angle).
If one can form a hexagon in which no two angles from the
same triple
are adjacent, and opposite pairs of edges have equal length,
I think
that would be enough to guarantee a tiling. If we Ŝx one
edgeıs length
and orientation, there are two remaining degrees of freedom
in the other
two edge lengths and two degrees of freedom needed to get the
edge
sequence to form a closed hexagon, so it looks like it may be
possible
in each case, but I havenıt worked out the details.
--
David Eppstein http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer
Science
===
Subject: Re: (Almost)Consecutive-Integer Egyptian-Fraction
Sum =1
> I was led to this question by considering (m+1)-sided
polygons where
> all but one of the sides are that of a k-gon, a (k+1)-gon, a
> (k+2)-gon,..., an (k+m)-gon.
Whooops. Should be (m+2)-sided polygons.
Leroy
> But I wanted the Ŝnal angle to be that of an n-gon, for
some positive
> integer n.
> For example,
> 1/3 + 1/4 + 1/5 + 1/6 + 1/20 = 1.
> So, my question is,
> which integers n are such that:
> 1/k + 1/(k+1) + 1/(k+2) +....+1/(k+m) + 1/n = 1 ?
> What is the sequence of such nıs? Is it in the EIS?
> Leroy Quet
===
Subject: Question about Combinatorics
iım searching the following solution
there are T positions t1,t2, ... ,tT
and L different colors c1,c2, ... ,cL
there exists L^T different combinations.
now i like to calculate the number of
possibilities, where A colors are different.
for example, how many combinations exists,
where 3 entries are different?
is there any closed solution aviable (maybe
very simple, but i didnıt got the solution)?
christoph / www.rune.ch
===
Subject: Re: math bibles
euclid is Ŝne & rather dryly encyclopedic, although
taht in a form that is not in the Britannica sense, but
a better synthetic geometry is _Modern Pure Solid Geometry_
by N.Altshiller-Court -- you donıt acutally need the ŝat stuff
as a prerequisite!
as for calculus, Apostol uses the Leibnizian method
of beginning with the integral, or the Whole (Enchilada .-)

Knuth has never given any idea of the problem
of the ŝoating-point algorithms & hardware (IEEE-755, -855,
I think); they are inherently chaotic ...
Chaos is the mother of Chronos!
> Is there a classic bible of differential/integral calculus?
Iım
> looking for the book that is to calculus what Euclidıs
Elements is to
> geometry or, even better, Knuthıs TAOCP is to algorithm
analysis --
--ils duces dıEnron!
http://tarpley.net
===
Subject: Re: math bibles
> Is there a classic bible of differential/integral calculus?
Landau, Edmund (1877-1938). _Differential and Integral
Calculus_
===
Subject: Re: math bibles
> Is there a classic bible of differential/integral calculus?
Iım
> looking for the book that is to calculus what Euclidıs
Elements is to
> geometry or, even better, Knuthıs TAOCP is to algorithm
analysis --
> encyclopedic in scope, but with enough depth to be suitable
for
> learning/refresher in addition to academic reference. Iıve
spent the
> last few days searching Amazon, et al, but all the books I
Ŝnd seem
> to be either too application-focused, or copies of the same
awful
> modern textbooks. Anyone have any suggestions?
> I also welcome suggestions of other biblical tomes in
mathematics
> and related Ŝelds. (I love books, and Iım always looking to
add
> another great one to my collection.) For example, Iım often
surprised
> how few people have sat down and spent a few evenings with
George
> Booleıs The Laws of Thought...
For number theory, Disquisitiones Arithmeticae by C.F. Gauss.
If your
Latin is rusty, an English translation was published by Yale
U Press.
===
Subject: Re: math bibles
> Is there a classic bible of differential/integral calculus?
Iım
> looking for the book that is to calculus what Euclidıs
Elements is to
> geometry or, even better, Knuthıs TAOCP is to algorithm
analysis --
> encyclopedic in scope, but with enough depth to be suitable
for
> learning/refresher in addition to academic reference. Iıve
spent the
> last few days searching Amazon, et al, but all the books I
Ŝnd seem
> to be either too application-focused, or copies of the same
awful
> modern textbooks. Anyone have any suggestions?
> I also welcome suggestions of other biblical tomes in
mathematics
> and related Ŝelds. (I love books, and Iım always looking to
add
> another great one to my collection.) For example, Iım often
surprised
> how few people have sat down and spent a few evenings with
George
> Booleıs The Laws of Thought...
The Bourbaki volumes were an impressive achievement when they
were
published, and still worth reading today.
Any chance that someone will try to do what Bourbaki has
done, updated
for the 21st century?
===
Subject: Re: math bibles
> Is there a classic bible of differential/integral calculus?
Iım
> looking for the book that is to calculus what Euclidıs
Elements is to
> geometry or, even better, Knuthıs TAOCP is to algorithm
analysis --
> encyclopedic in scope, but with enough depth to be suitable
for
> learning/refresher in addition to academic reference.
> I also welcome suggestions of other biblical tomes in
mathematics
> and related Ŝelds.
Apostol for calculus. Herstein for undergrad abstract algebra.
Royden for Real Analysis. Rudin has two analysis books - we
used
to call them Rudin and baby Rudin, but I guess you could call
them the old & new testaments. Hardy & Wright for number
theory.
OıNan or Strang or Hoffman & Kunze for linear algebra. Marsden
and Tromba for vector calculus. Ahlfors for complex analysis.
Feller for probability.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: math bibles
> Is there a classic bible of differential/integral calculus?
Iım
> looking for the book that is to calculus what Euclidıs
Elements is to
> geometry or, even better, Knuthıs TAOCP is to algorithm
analysis --
> encyclopedic in scope, but with enough depth to be suitable
for
> learning/refresher in addition to academic reference.
>
> I also welcome suggestions of other biblical tomes in
mathematics
> and related Ŝelds.
> Apostol for calculus. Herstein for undergrad abstract
algebra.
> Royden for Real Analysis. Rudin has two analysis books - we
used
> to call them Rudin and baby Rudin, but I guess you could
call
> them the old & new testaments. Hardy & Wright for number
theory.
> OıNan or Strang or Hoffman & Kunze for linear algebra.
Marsden
> and Tromba for vector calculus. Ahlfors for complex
analysis.
Donıt forget Kelley for point-set topology, the
Dunford-Schwartz
trilogy for functional analysis, Stanley for enumerative
combinatorics, and Spivak for differential geometry.
> Feller for probability.
===
Subject: Re: math bibles
> Apostol for calculus. Herstein for undergrad abstract
algebra.
> Royden for Real Analysis. Rudin has two analysis books - we
used
> to call them Rudin and baby Rudin, but I guess you could
call
> them the old & new testaments. Hardy & Wright for number
theory.
> OıNan or Strang or Hoffman & Kunze for linear algebra.
Marsden
> and Tromba for vector calculus. Ahlfors for complex
analysis.
> Feller for probability.
You forgot logic, combinatorics, and topology
--
Mitch Harris
(remove q to reply)
===
Subject: Re: Euclidıs algorithm for quadratic sequence
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i4PJUdT22001;
>Here is a problem that I have:
>Given real constants a, b and eps > 0, I can quite easily
Ŝnd the
>smallest k >= 0 such that (a + b*k) mod 1 < eps, using
Euclidıs
>algorithm or a slight variation of it. Instead of requiring
about 1/eps
>steps if I just tried k=0, k=1 etc. sequentially, this takes
only about
>log (1/eps) steps, quite a difference (I need to solve this
usually for
>eps around 10^-15).
>The problem is: I have real constants a, b and c, and I need
to Ŝnd the
>smallest k >= 0 such that (a + b*k + c*k^2) mod 1 < eps. c
is very
>small, about the same size as eps, and both are around
10^-15. I found a
>method that takes about O ((1/eps)^(2/3) * log (1/eps))
steps - this
>makes the problem solvable for me in a reasonable amount of
time for eps
>= 10^-15, but not for much smaller values like eps = 10^-18.
Is there
>any better algorithm for this problem known?
Look at some of the work of Noam Elkies, speciŜcally
http://arxiv.org/abs/math/0005139
===
Subject: Re: Euclidıs algorithm for quadratic sequence
>Here is a problem that I have:
>Given real constants a, b and eps > 0, I can quite easily
Ŝnd the
>smallest k >= 0 such that (a + b*k) mod 1 < eps, using
Euclidıs
>algorithm or a slight variation of it. Instead of requiring
about 1/eps
>steps if I just tried k=0, k=1 etc. sequentially, this takes
only about
>log (1/eps) steps, quite a difference (I need to solve this
usually for
>eps around 10^-15).
>The problem is: I have real constants a, b and c, and I need
to Ŝnd the

>smallest k >= 0 such that (a + b*k + c*k^2) mod 1 < eps. c
is very
>small, about the same size as eps, and both are around
10^-15. I found a

>method that takes about O ((1/eps)^(2/3) * log (1/eps))
steps - this
>makes the problem solvable for me in a reasonable amount of
time for eps

>= 10^-15, but not for much smaller values like eps = 10^-18.
Is there
>any better algorithm for this problem known?
> Look at some of the work of Noam Elkies, speciŜcally
> http://arxiv.org/abs/math/0005139
That will keep me busy for a while...
===
Subject: Re: Journal editors and reviewers, speak up
you rely completely upon *argumentarium ad hominem*, as if
there really were no mainstream mathematicians in the
googolplex,
or as if all of us amateurs (except for you) are just
replaying our roles dıecole haute-grammaire (or
what ever).
it is sad that that group of your peers at the journal,
would not give you a reason for their late refusal, although
it is obvious that they got outside advice
of your former illegible proofs of Fermatıs Last.

were you one of the high-school abusers?

> My assessment is that most mainstream mathematicians simply
donıt
> bother with talking in such a public forum where *anyone*
can reply.

> American high school type social scene because you failed
so utterly

> people who refuse to grow up, who relive high school
nightmares out in
> public by taking the role of the socially popular abusers,
and joy in
--ils duces dıEnron!
http://tarpley.net
===
Subject: Re: Journal editors and reviewers, speak up
> You know as well as I, Dr. Harris
> Dr. Harris? Dr. Harris? Letıs not get carried away.
> Doug
I wouldnıt want to get carried away. No, Iıve gone that route
before,
and it led me astray. James probably doesnıt even read my
posts anymore
because of my past behavior. He has no reason to trust me.
But that doesnıt mean I canıt point out to the rest of you
some of the,
umm, interesting patterns in your behavior. My previous post
in this
thread made the mild suggestion that there is, perhaps, a
bias in the
mathematical community against those who are not ofŜcial
members of its
club -- i.e., against those who do not hold doctorates in
mathematics.
I called James, Dr. Harris, employing the utmost irony: if he
really
were Dr. Harris, he wouldnıt be having these difŜculties, of
course.
I expected that some might disagree with me (of course weıre
not an
exclusive club, wink wink), but I found it extremely
*interesting* that
what raised a hackle was simply the string
Dr. Harris
It reminds me of a situation at my nephewıs Catholic
kindergarten. Kids
are allowed to bring in cake mix on their birthday, and the
teacher helps
them make cake. Any ŝavor you want . . . except devilıs food.
I
wonder
if she crossed herself when she said devil. Or if she would
cower and
quail should passing rufŜans say Ni! to her.
Mathematicians may think themselves too rational to tremble
at mere words,
so I say unto you all:
Prof. James S. Harris, Ph.D.
Director, Institute of Advanced Study
Von Neumann Fine Professor of Mathematics
Princeton University
Did you make a plus sign? Oh, the horror of imagining James
Harris to be
one of us! Itıs not enough to actually do mathematics, oh no,
it has to
be published in a peer-reviewed journal. But wait! Itıs not
enough to
publish in a peer-reviewed journal -- the journal has to be
reputable.
Now what reputable journal is going to publish anything by
someone without
a Ph.D.? Not true you say? Oh, a graduate student at a
prestigious
university with a powerful advisor is also allowed to publish?
Yes, yes, I know. Itıs not a matter of allowed. I know the
party
line.
To be publishable in a reputable journal, work must be of a
caliber that is
rarely achieved by those without a doctorate. I was expecting
this sort of
reply, and I agree that it is plausible, but that is not the
response I
got.
Instead:
Dr. Harris?!?!? Aaaaaah! Horrors!!!
But in reality, Dr. Harris would not be enough. The doctorate
would
have
to come from a reputable university, right? And even if it
did, it would
have
to be in the right Ŝeld . . . I mean, if James were to
actually write a
dissertation on something *original* (say Object Oriented
Mathematics)
which
didnıt neatly Ŝt into one of our pigeonholes, well, that
wouldnıt count,
would it? Or, okay, suppose that James actually did stiŝe his
creativity
long enough to write a dissertation on something traditional
and boring --
the
cohomology of Banach algebras, say. Would that be enough?
No! Weıd just make up some other criterion!!! Ha ha ha ha ha!
And why would we do this? I know that you know the answer
James. Youıre
probably not reading this because you donıt want to hear it,
but the reason
is simply this: there is a conspiracy against you. Youıve
rufŝed the
feathers of too many eminent mathematicians, and theyıve
decided to shut
you
out.
You may wonder: is their control complete? No! Youıve proven
that!
There are some journals out there, apparently, that didnıt
get the word.
That eluded the gaze of the overseers. But all math journals
are networked
together in such a way that you donıt really have a hope of
making it all
the way to publication.
But keep on trying. Doomed efforts are so noble. Better than
clogging up
these newsgroups, I suppose. Better than giving a conference
talk no one
will attend. Better than just giving up and starting a new
life.
--
Jim Ferry at U of Illinois, Urbana-Champaign, (no, I donıt
educate)
2 email me l o o k up 1 row
===
Subject: Re: Journal editors and reviewers, speak up
actually, I multiŝected.
as to why he didnıt wait til they actually published
-- or is it an e-journal,
from which the paper was edited-out? --
he wonıt answer that, either, I guess.

> Prof. James S. Harris, Ph.D.
> Director, Institute of Advanced Study
> Von Neumann Fine Professor of Mathematics
> Princeton University
> Did you make a plus sign? Oh, the horror of imagining James
Harris to be

> You may wonder: is their control complete? No! Youıve
proven that!
--ils duces dıEnron!
http://tarpley.net
===
Subject: Re: Journal editors and reviewers, speak up
> You know as well as I, Dr. Harris
> Dr. Harris? Dr. Harris? Letıs not get carried away.
> Doug
> I wouldnıt want to get carried away. No, Iıve gone that
route before,
> and it led me astray. James probably doesnıt even read my
posts anymore
> because of my past behavior. He has no reason to trust me.
> But that doesnıt mean I canıt point out to the rest of you
some of the,
> umm, interesting patterns in your behavior. My previous
post in this
> thread made the mild suggestion that there is, perhaps, a
bias in the
> mathematical community against those who are not ofŜcial
members of its
> club -- i.e., against those who do not hold doctorates in
mathematics.
> I called James, Dr. Harris, employing the utmost irony: if
he really
> were Dr. Harris, he wouldnıt be having these difŜculties,
of course.
> I expected that some might disagree with me (of course
weıre not an
> exclusive club, wink wink), but I found it extremely
*interesting* that
> what raised a hackle was simply the string
> Dr. Harris
> It reminds me of a situation at my nephewıs Catholic
kindergarten. Kids
> are allowed to bring in cake mix on their birthday, and the
teacher helps
> them make cake. Any ŝavor you want . . . except devilıs
food. I
wonder
> if she crossed herself when she said devil. Or if she would
cower
and
> quail should passing rufŜans say Ni! to her.
> Mathematicians may think themselves too rational to tremble
at mere
words,
> so I say unto you all:
> Prof. James S. Harris, Ph.D.
> Director, Institute of Advanced Study
> Von Neumann Fine Professor of Mathematics
> Princeton University
> Did you make a plus sign? Oh, the horror of imagining James
Harris to be
> one of us! Itıs not enough to actually do mathematics, oh
no, it has to
> be published in a peer-reviewed journal. But wait! Itıs not
enough to
> publish in a peer-reviewed journal -- the journal has to be
reputable.
> Now what reputable journal is going to publish anything by
someone
without
> a Ph.D.? Not true you say? Oh, a graduate student at a
prestigious
> university with a powerful advisor is also allowed to
publish?
> Yes, yes, I know. Itıs not a matter of allowed. I know the
party
line.
> To be publishable in a reputable journal, work must be of a
caliber that
is
> rarely achieved by those without a doctorate. I was
expecting this sort
of
> reply, and I agree that it is plausible, but that is not
the response I
got.
> Instead:
> Dr. Harris?!?!? Aaaaaah! Horrors!!!
> But in reality, Dr. Harris would not be enough. The
doctorate would
have
> to come from a reputable university, right? And even if it
did, it would
have
> to be in the right Ŝeld . . . I mean, if James were to
actually write
a
> dissertation on something *original* (say Object Oriented
Mathematics)
which
> didnıt neatly Ŝt into one of our pigeonholes, well, that
wouldnıt count,
> would it? Or, okay, suppose that James actually did stiŝe
his
creativity
> long enough to write a dissertation on something
traditional and boring --

the
> cohomology of Banach algebras, say. Would that be enough?
> No! Weıd just make up some other criterion!!! Ha ha ha ha
ha!
> And why would we do this? I know that you know the answer
James. Youıre
> probably not reading this because you donıt want to hear
it, but the
reason
> is simply this: there is a conspiracy against you. Youıve
rufŝed the
> feathers of too many eminent mathematicians, and theyıve
decided to shut
you
> out.
> You may wonder: is their control complete? No! Youıve
proven that!
> There are some journals out there, apparently, that didnıt
get the word.
> That eluded the gaze of the overseers. But all math
journals are
networked
> together in such a way that you donıt really have a hope of
making it all
> the way to publication.
> But keep on trying. Doomed efforts are so noble. Better
than clogging
up
> these newsgroups, I suppose. Better than giving a
conference talk no one
> will attend. Better than just giving up and starting a new
life.
That was rather long winded. What was your point?
===
Subject: Re: Journal editors and reviewers, speak up
> All in all my analysis is that Usenet here is like, as I
mentioned
> before, an American high school, and maybe some of you
recreated an
> American high school type social scene because you failed
so utterly
> when you were kids.
> Here you live the fantasy that youıre on the top of the
heap socially,
> and engage in insults and attacks on others as you may have
received
> attacks in the past.
> For those not aware of the American high school social
scene the movie
> currently in wide release here called Mean Girls will give
you
> plenty of information.
> Sadly, while it had much potential, Usenet has been taken
over by
> people who refuse to grow up, who relive high school
nightmares out in
> public by taking the role of the socially popular abusers,
and joy in
> trying to inŝict their emotional pain upon others: adults
in body but
> adolescents in reality as they never got past that stage,
and still
> canıt escape their scars.
> James Often in error, but never in doubt. Harris
Having failed miserably at mathematics, you now appear to be
embarking on a
career as a psychologist. However, you are dangerously
close to practicing that profession without a license. See
your own
therapist!
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: Journal editors and reviewers, speak up
>> You donıt seem to have noticed that the mainstream
mathematicians
>> youıve been attacking here (like Professors Magidin and
Ullrich)
>> also make use of those established and reputable outlets
for their
>> mathematical work. Itıs not difŜcult to Ŝnd published
papers for
>> the professionals who post here. Have you forgotten that
*they*
>> were the ones who encouraged you to try to get your work
>> published? USENET is a hobby and a diversion for them, not
the
>> professional outlet *youıve* tried to make of it.
> I donıt really recall either Arturo or David encouraging
James to
> submit his work to journals and waste the time of people
doing real
> work (which us folk posting here are typically *not* doing,
at least
> not while weıre posting).
> I may be mistaken in my recollections.
Perhaps I wasnıt clear enough in my writing. Arturo and David
were
mentioned just as examples of mainstream mathematicians in
general.
When I said *they* were the ones who encouraged you I meant
some of
the
mathematicians who post here, not those two in particular.
Actually,
I donıt remember who in particular made those suggestions to
James,
but I do recall that several of the regulars here did so.
--
Wayne Brown (HPCC #1104) | When your tailıs in a crack, you
improvise
fwbrown@bellsouth.net | if youıre good enough. Otherwise you
give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
Subject: Re: My paper passed peer review
I have never submitted a paper for review.
the question is, is it supposed
to be a sort of hermetically-sealed effort
of the reviewers, in a room-for-review, or
do they take copies home in a pile & go through them,
with liesurely reference to their other peers and
non-human references?

did Harris recieve *any* polite refutation/refusal,
at all -- did he call the editor, a honky, or
did the editor call him, a fermatiste?
(although the advance factorization paper may not
have acutally mentioned Pierre de,
did it even have a recognizable hypothesis or abstract?)

>As far as I know, the refutations sent to the journal under
discussion
>havenıt passed peer review.

> Wouldnıt the proper thing to do have been to return the
refutations
> to the senders and suggest they submit the refutation as a
formal
> paper for future publication, subject to peer review?
--ils duces dıEnron!
http://tarpley.net
===
Subject: Re: Four people crossing a bridge
Teabag wibbled:
>
> What % of the bridge can this ŝashlight illuminate?
> It seesm to me that the returner need not cover 100% of the
bridge on

> most of his trips, saving hundreds if not thousands of
seconds.
>
> any properly constructed bridge will have guard rail or
hand rail
> so that you can cross in pitch dark
> You guys are picking apart the puzzle, which is something I
love to
> do, but you are also wanting to CHANGE the puzzle, and
thatıs a Bozo
> No-No.
Iıll tell you what it is: the puzzle posits something
apparently
RealWorld, but it isnıt. This annoys me.
Ask me something to do with JustNumbers, Ŝne, but start
talking about
bridges and ŝashlights and Iıll ask for more details about
them.
> 35 minutes, as Mu-Pi said.
> Denny Crane.
--
Am I in Bolton?
===
Subject: Re: Four people crossing a bridge
35 ... this really is a question for alt.stupidity
===
Subject: Re: Four people crossing a bridge
> 35 ... this really is a question for alt.stupidity
What is even more stupid is that you gave the answer days
after it was
Ŝrst
posted. Did you copy it?
===
Subject: Re: Four people crossing a bridge
So why donıt you smart ers from the Dorm Radio in the
Basement of
the Science Building post the ing solution... to the original
puzzle as posed? Very few posters on this thread have even
tried to
post a solution.
> Iım quite sure now that it was meant as a joke.
> Of course it was meant as a joke. It helps to know from
sci.math that
> Clive Tooth is not an idiot, and it helps to know how often
this puzzle
> (used to) get posted to rec.puzzles... in the usual
version, the
obvious
> answer, which is that the fastest person should accompany
each of the
> others one by one, is wrong. So whatıs particular about
this version is
if
> you donıt look carefully you expect to use the same trick
(to have the
two
> slowest go over together and neither of them come back),
and you probably
> say ha, think you can catch me out with this old chestnut?
but of
course
> youıd be the one caught out.
> And naturally the person who multiplied the times together
was joking
too,
> in the spirit of the stupid puzzle, and someone caught that
mistake
too.
===
Subject: Re: Four people crossing a bridge
> So why donıt you smart ers from the Dorm Radio in the
Basement of
> the Science Building post the ing solution... to the
original
> puzzle as posed? Very few posters on this thread have even
tried to
> post a solution.
Itıs been posted several times: 35.
===
Subject: Re: Four people crossing a bridge
> So why donıt you smart ers from the Dorm Radio in the
Basement of
> the Science Building post the ing solution... to the
original
> puzzle as posed? Very few posters on this thread have even
tried to
> post a solution.
> Itıs been posted several times: 35.
Yes, numbnuts, itıs been posted several times (by my count,
4)... the
second time was by me, conŜrming the original solverıs
solution,
along with a hint as to why that was a solution. Suck me.
Denny Crane.
===
Subject: Re: Four people crossing a bridge
> So why donıt you smart ers from the Dorm Radio in the
Basement of
> the Science Building post the ing solution... to the
original
> puzzle as posed? Very few posters on this thread have even
tried to
> post a solution.
> Itıs been posted several times: 35.
There seems to be some confusion. A ŝashlight is simply
US-English for
the object called a torch in UK-English.
--
Clive Tooth
http://www.clivetooth.dk
===
Subject: Re: Four people crossing a bridge
The Last Danish Pastry wibbled:
> So why donıt you smart ers from the Dorm Radio in the
Basement of
> the Science Building post the ing solution... to the
original
> puzzle as posed? Very few posters on this thread have even
tried to
> post a solution.
> Itıs been posted several times: 35.
> There seems to be some confusion. A ŝashlight is simply
US-English
for
> the object called a torch in UK-English.
Ah, but one mustnıt confuse this kind of torch with a
juggling torch.
That would be a serious mistake.
--
Am I in Bolton?
===
Subject: Re: Four people crossing a bridge
>So why donıt you smart ers from the Dorm Radio in the
Basement of
>the Science Building post the ing solution... to the original
>puzzle as posed? Very few posters on this thread have even
tried to
>post a solution.
>>Itıs been posted several times: 35.
> There seems to be some confusion. A ŝashlight is simply
US-English
for
> the object called a torch in UK-English.
As far as how language differences may cause variations in
the outcome
of the puzzle, we can burn that bridge when we come to it.
--Bill
--
The World Wide Web is the hugest vanity press
in the history of the human race!
http://billwilkinson.home.mindspring.com/index.html
===
Subject: Re: Four people crossing a bridge
>> So why donıt you smart ers from the Dorm Radio in the
Basement of
>> the Science Building post the ing solution... to the
original
>> puzzle as posed? Very few posters on this thread have even
tried to
>> post a solution.
>> Itıs been posted several times: 35.
> There seems to be some confusion. A ŝashlight is simply
US-English
for
> the object called a torch in UK-English.
Hahaha quite good Clive, sounds like you have been studying
the masters
e.g. Kevin S. Wilson who trolled rec.puzzles extremely
successfully in
January 2002... This is a new departure for you though. Good
luck.
===
Subject: Re: Four people crossing a bridge
>So why donıt you smart ers from the Dorm Radio in the
Basement of
>the Science Building post the ing solution... to the original
>puzzle as posed? Very few posters on this thread have even
tried to
>post a solution.
>> Iım quite sure now that it was meant as a joke.
>> Of course it was meant as a joke. It helps to know from
sci.math that
>> Clive Tooth is not an idiot, and it helps to know how
often this puzzle
>> (used to) get posted to rec.puzzles... in the usual
version, the
obvious
>> answer, which is that the fastest person should accompany
each of the
>> others one by one, is wrong. So whatıs particular about
this version is
if
>> you donıt look carefully you expect to use the same trick
(to have the
two
>> slowest go over together and neither of them come back),
and you
probably
>> say ha, think you can catch me out with this old chestnut?
but of
course
>> youıd be the one caught out.
>> And naturally the person who multiplied the times together
was joking
too,
>> in the spirit of the stupid puzzle, and someone caught
that mistake
too.
It looks like these two guys, together posted the solution:
>It takes 43 minutes.
>A and B cross the path ( 10 minutes)
>A comes back. ( 11 minutes)
>C and D cross the path. (23 minutes)
>B comes back (33 minutes)
>A and B cross the path (43 minutes)
>> I think itıs implied theyıre trying to cross as quickly as
possible. So
>> why not 10 + 1 + 11 + 1 + 12 = 35 minutes?
Merged into one:
A and B cross the path ( 10 minutes)
A comes back. ( 11 minutes)
A and C cross the path. (23 minutes)
A comes back (24 minutes)
A and D cross the path (35 minutes)
--
dgates@spamfreelinkline.com
===
Subject: Re: Balls
Ioannis wibbled:
>  Lynn Kurtz
<kurtzDELETE-THIS@asu.edu>
[CapitalEth][EDouble
Dot][Micro]
.b3
.b9[EDo
ubleDot].b9
> Of course, pool players frequently put spin on the cue ball
(top,
> bottom, side) which not only affects how it caroms off the
target ball
> but how it reacts to the cloth and rails. So itıs a bit (a
lot) more
> complicated than just elastic collisions.
> --Lynn
> You are right, of course. My use of in principle is almost
always
> guaranteed to be exaggerated :*)
> Players also often strike the cue ball in such a way that
it completely
> bounces off the table and falls back. In principle all
those actions
can
> be approximated, although it would be extremely difŜcult.
The most recent snooker world championship featured one game
where one
player managed to get the ball off the table. How we laughed.
I saw a roulette ball shoot out of the wheel and across the
casino once.
Donıt know what they did about the bets: I wasnıt playing, I
was waiting
for the poker tourney to start.
--
Am I in Bolton?
===
Subject: Re: Balls
> I saw a roulette ball shoot out of the wheel and across the
casino once.
> Donıt know what they did about the bets: I wasnıt playing,
I was waiting
> for the poker tourney to start.
They should pay off on the bets for 0 and 00, and everybody
else gets to
ride for the next spin.
===
Subject: Re: Balls
> In the game of snooker (see footnote) it is sometimes
necessary to break
> up a pack of touching red balls with the cue ball. Does the
> unsolvability(?) of the three body problem mean that the
player
cannot
> predict just where the cue ball will end up after such a
shot?
> I ask because commentators sometimes criticize a player for
the way he
> plays such a shot if the outcome is poor, especially in
regard to the
> position that the cue ball comes to rest in. My thought is
that the
> player can only trust to luck in such a situation.
gravitating bodies, but wondered if it might also apply to
three bodies
interacting in other ways.
--
G.C.
===
Subject: Re: Balls
The original post, resurrected from 1997, asks -- essentially
-- the
jelly-bean problem: Is there a formula for determining the
maximum
number of jelly beans in a jar (or other shape).
In an effort to win a contest at work, I spent considerable
time and
thought trying to solve this enigma. I failed miserably.
I had ultimately decided that a straight line of jelly beans
in one
plane was a proper measure since any overlapping jelly beans
would be
in a different line, and I decided that I would take two
measurements
of the heigth, and two of the circumfrance, take an average
of both
measurements, and work out all the cross-multiplied
possibilities, and
take the average, or at least what seemed most reasonable, of
the
cubic volume calculations in unit-measures of jelly beans.
I discarded my highest number, but it was closest to correct.
So, apparently, I should have counted a line of jelly beans
and its
adjacent overlapping line AS ONE COMPACT LINE.
Still, volume-wise, the depth of a cubic line is not quite
two jelly
beans, and I am unclear about how to measure or solve for the
unseen
depth of a cubic line of jelly beans in a jar.
Very Respectfully,
Ray Donald Pratt
===
Subject: Help with testing for nonlinearity
Its been awhile since Iıve done stats work and I am very
rusty. Any
help would be greatly appreciated.
Iım testing for linearity of some datasets against a time
trend with a
comparison of the correlation ratio and the correlation
coefŜcient.
Iım running into a few cases where my derived Eta-squared is
less than
my R-squared, which, by deŜnition, cannot be true, right? Iıve
doubled over my Ŝgures and canıt Ŝnd where Iım going astray.
My
correlation coefŜcients look sound, and Iım deriving my
Eta-squared
as the SS(between groups)/SS(total)
Any help/suggestions/observations would be fantastic!
-FX
===
Subject: Re: Open letter to James Harris - Self-Publish
>
> Why donıt you self-publish your paper?
> Excellent idea: he may seek tips from Mr Wolfram who
> self-published his book.
No, Mr. Wolfram was rich enough and inŝuential enough already
to
self-publish.
James Harris would have to self-publish because nobody but an
incompetent person would publish his work.
===
Subject: SimpliŜcation Assitance needed
I havenıt used my math skills much at all for 20 years now,
and I canıt even seem to do the problem below.
Could one of you whizzes help me out on this?
I think Iıve got my equation right,
I just need to get the Theta (T) out on the right
and get the equation in terms of N and P so I can calulate
Theta,
but canıt for the life of me remember the trig simpliŜcations.
Here it is:
/ N / N cos(T-P) -N +1
ArcTan( -------- ) + ArcTan( ---------------- ) = 90 +P
cos(T) / N sin(T-P) /
Fred
p.s. in case the multiple lines get out of whack, here it is
on one line:
ArcTan(N/cos(T)) + ArcTan( (N*cos(T-P)-N+1) / (N*sin(T-P)) )
= 90 +P
===

===
Subject: Re: InŜnite Series: Summation [(-1)^(n-1)][(2+sin
n)/sqrt n] ->
Converge?
> Does this series converge? By direct comparison to
summation 1/sqrt n
this
> series is not absolutely convergent. But could it be
conditionally
> convergent? The absolute values of the terms do not
approach zero
> monotonically so ... ?
Divide and conquer:
sum[n=1 to inf]((-1)^(n-1)*2/sqrt(n)) converges
conditionally (standard alternating series test);
sum[n=1 to inf]((-1)^(n-1)*sin(n))/sqrt(n)
also converges conditionally, for a slightly more
subtle reason (Dirichletıs Test).
[Rest can be skipped]
The second series directly:
Denote x=pi-1, then sin(n*x) = (-1)^(n-1) * sin(n),
so the second series is
sum[n=1 to inf] sin(n*x)/sqrt(n)
and we manipulate the partial sums (N >= 2)
sum[n=1 to N] sin(n*x)/sqrt(n)
= cos(x/2) - cos((N+1/2)*x)/sqrt(N+1)
+ 1/(2*sin(x/2)) * sum[n=1 to N-1] A(n)
where
A(n) = (1/sqrt(n) - 1/sqrt(n+1)) * cos((n+1/2)*x)
(check it out; itıs called summation by parts).
The last sum (extended to inŜnity) is convergent
absolutely, and the initial terms converge as N
goes to inŜnity.
Oh, and sin(x/2) is different from 0.
===
Subject: Re: Can the deduction theorem be used recursively?
>> No, Godelıs Completeness Theorem is the converse:
>> logically valid wffs are theorems.
I meant iff.
>>
> [...] it depends on who you ask
> Only if you ask someone who is clueless [...] (Chris
Mentzel)
> :-)
> Introduction to Mathematical Logic, Mendelson 1964, pg.
> 68, Godelıs Completeness Theorem: In any Ŝrst order
predicate
> calculus, the theorems are precisely the logically valid
wffs.
> Idiot, of course this is a *corollary* of G.9adelıs
Completeness
Theorem.
Maybe so, but thatıs not how Mendelson describes it (as you
can see.)
I looked it up on the internet (the Ŝrst time Iıve actually
read much
about it - Iıve seen it but never got into it) and saw that
it was
different than what Mendelson says (=> vs. <=>) so I opted for
Mendelson (for obvious reasons) and said It depends on who
you ask.
Mendelson is clueless?!?!?
> I knew about the resistance that I quoted earlier
> Right. Never forget: _The resistance_ actually PROVES that
you are
> right, that you MUST BE right!
Well actually yeah. You know why? They try so hard to prove
me wrong
but fail! They foam at the mouth at the possibility that Iıve
developed something new, which might add legitimacy to my
observations
about BS in CS (BoguS Computer Science publications). So they
try
like hell to Ŝnd someone else who has developed a Program
Synthesis
system or axiomatized the Theory of Computation - and thatıs
exactly
what I need. When they come up empty-handed or drag out
nonsense like
the Boyer-Moore paper (not a single character of which is
program
output), it conŜrms that Iım the only one whoıs done it. What
better
reafŜrmation than someone who is DYING to prove me wrong
falling ŝat
on their face?
The funniest part is when I talk about something that I have
spent
very little time with (other than looking it up on the
internet
because someone posted something about it and making a quick
response
to their post) like some part of Set Theory or Proof Theory,
and make
a mistake, and they go crazy! Charlieıs stupid, Charlieıs a
liar,
Charlieıs paperıs no good. And my paper has nothing to do
with Set
Theory or Proof Theory! LOL I guess thatıs easier than
actually
reading it and saying something about its contents. :-p
> from The Mystery of the Aleph:
>
> Conservative thinkers sought to restrict novel ideas, since
these new
> point of view, he should have been shaped in their image,
and had no
> right to challenge his teachersı philosophy. Cantor
believed that
> their arbitrary restrictions impeded the growth of the
Ŝeld. He gave
> many examples of mathematical ideas that never would have
had a chance
> to succeed and evolve if their proponents had encountered
the kind of
> opposition he faced.
>
> Right. First came CANTOR - and now you. I see.
Well, there is a lot of similarity, actually. The resistance
without
addressing the issues, the incompleteness connection. I even
read
that Cantor published some of his work himself. I guess some
things
never change.
Charlie Volkstorf
Cambridge, MA
> F.
===
Subject: Re: Can the deduction theorem be used recursively?
>>
>> No, Godelıs Completeness Theorem is the converse:
>> [all] logically valid wffs are theorems.
>>
> I meant iff.
Itıs still wrong. Look, Charly, Iıve read G.9adelıs paper, I
KNOW what he
has proved in it.
> Introduction to Mathematical Logic, Mendelson 1964, pg.
> 68, Godelıs Completeness Theorem: In any Ŝrst order
predicate
> calculus, the theorems are precisely the logically valid
wffs.
>
> Mendelson is clueless?!?
No. See D. Ullrichıs comment:
That iff is traditionally broken into two parts, one the
soundness
and one the completeness. The soundness theorem is trivial,
the completeness isnıt. The iff is indeed true; if someone
called
that Godelıs Completeness Theorem it would be because
the completeness, which Godel proved, is the non-trivial half.
(David Ullrich)
And in my own words:
FIRST you have to prove that the system is consistent -
thatıs the easy
part. And then you have to prove that it is complete - thatıs
the hard
part. G.9adel was THE FIRST to present a proof for that,
concerning FOPL.
predicate calculus, the theorems are precisely the logically
valid
wffs.
F.
===
Subject: Re: Can the deduction theorem be used recursively?
>
> And formal systems giving sound and complete axiomatizations
> for FOL with MP as the only inference rule also exist.
>
> You really think you do a lot for your (giggle) status as a
genius
> by continuing to dispute something this elementary?
>
> I donıt have to dispute that. Godel proved in 1931 that
there is no
> sound and complete axiomatization for FOL.
>
> You appear to be confused about what Godel has shown. In
fact, Godel
> proved the (semantic) *completeness* of FOL -- in 1930
IIRC. He proved
> the incompleteness of *arithmetic* (that is, of any
consistent,
> recursive axiomatization of it) in 1931. Perhaps you are
also
> confusing the sense in which FOL is complete with the sense
in which
> arithmetic is not. They are not the same.
> I think the question is that of terminology. I thought he
was
> referring to completeness in the sense that is denied by
Godelıs
> Incompleteness Theorem, rather than in the sense that it is
conŜrmed
> by Godelıs Completeness Theorem.
But, see, you really shouldnıt have, because it is impossible
-- simply
*by deŜnition*, not because of Godelıs theorem -- to have a
sound and
complete axiomatization of FOL in that sense of complete. You
didnıt
realize that. Do you see why that might lead one to think you
donıt
understand the subject matter?
> You seem to be saying that calling the former a refutation
of a sound
> and complete axiomatization for FOL is deŜnitely wrong,
Confused might be more appropriate than wrong.
> although I have heard it described with so many different
terms that I
> am a little surprised to hear that. Perhaps you can explain
why there
> is such a lack of standardization of terminology in the
literature,
> but there is a standard that is being violated here.
But that terminology is completely standard. And really,
there is no
problem here. Like many terms with two meanings, occurrences
of
complete are wholly disambiguated by context -- though, of
course, you
have to understand the contexts.
> What are you babbling about here? Name something that
Hilbert
> said was so that turned out to be false.
>
> There are absolutely no unsolvable problems.
>
> Hilbert explicitly takes the solvability axiom to be an
unproved
but
> important methodological conviction that motivates the
mathematical
> researcher. He *hoped* it was true, and perhaps believed it
was (in
> 1900), but he never asserted it with the sort of dogmatic
certainty
your
> paraphrase above suggests.
> Believe me that I am not the source of the translation. I
know of at
> least 6 sources for that quote.
Yes, one bad translation and 5 copies. Weıll no doubt see
more because
of Google.
> Please complete the following by Ŝlling in the blank with
an English
> sentence: When Hilbert became an honorary citizen of
K.9anigsberg, he
> gave a lecture, published in 1931, in which he said, ___.
Instead of
> the foolish ignorabimus, our answer is on the contrary: We
must know,
> We shall know.
F.9fr uns gibt es kein Ignorabimus, und meiner Meinung nach
auch f.9fr
die
Naturwissenschaft .9fberhaupt nicht.
For us there is no Ignorabimus, and in my opinion none at all
in the
natural sciences.
Both for us [mathematicians] and in my opinion pretty clearly
indicate this was a methodological principle, perhaps even a
reasonable
belief, but not something Hilbert thought had been (or even
perhaps
could have been) conclusively demonstrated. The translation
you cited
is obviously very poor.

> I am wondering if anyone has been wise enough to point out
that the
> set of axioms should be Ŝnite - which would preserve the
analogy with
> programming languages
Why on earth should *that* be a constraint on what counts as
an
axiomatic system? Programming languages are a relatively
recent
invention. Clear instances of inŜnite axiomatic systems were
around
some 50 years prior to the development of the Ŝrst programming
languages.
> Actually, ZF is not even a true axiomatic system,...
>
> Youıre quite wrong about that. You donıt get to decide what
counts as
> an axiomatic system (unless of course you arenıt interested
in
> conversing with the mathematical community).
> I thought mathematicians were allowed to make judgments.
This isnıt a matter for judgment. What counts as an axiomatic
system at
this point in time is simply a matter of deŜnition. You can
decide not
to make use of the deŜnition if you donıt like it for some
reason, but
the fact is that it is now pretty much universally accepted
within the
mathematical community, and the fact is that that deŜnition
permits
axiomatic theories with inŜnitely many axioms. The *proper*
response
for you, with your strictly Ŝnitistic leanings, is *not* to
try to say
that axiomatic theory doesnıt in fact mean what it
universally does,
but rather that the only theories that you think are
philosophically
legitimate have only Ŝnitely many axioms.
> Offhand (speculating), I would say that ZF is probably not
Ŝnitely
> axiomatizable because of the Axiom of Foundation.
>
> Wrong again, provably so.
>
> (I have not pursued a proof, and would welcome an on-line
source.)
>
> Google non-well-founded set theory.
> I donıt see a proof off-hand, and would still like to. What
about ZF
> gets in the way, then?
Well, in a nutshell, the fact that (due to the replacement
schema) it
has inŜnitely many independent axioms.
> The rest of ZF looks, prima facie, easily Ŝnitely
axiomatizable.
>
> Er, how would you suggest revising the replacement schema?
> As with any axiom schema that substitutes any particular
formula for a
> variable, we simply need axioms that deŜne the set of
formulas.
I havenıt a clue as to what you think you are suggesting.
> (What does Er mean?)
Same as welllll or ummmmm...
> Thus we have the awkward situation in which a system tells
us both
> what is and what is not.
>
> A confusing and misleading remark; been reading Sein und
Zeit? ;-) The
> only sense I can charitably make of it is that ZF entails
both that
> certain sets exist and also rules out the existence of
certain
purported
> sets (non-well-founded ones, for instance, or ones
containing members
of
> unbounded rank).
> You got it!
> Nothing awkward about that.
> How do you generate legitimate sets when some axioms say
what are sets
> and others say what arenıt?
Err, wellll, ummm, assuming by generate you mean prove the
existence
of, suppose I give you some axioms about what ducks are:
Ducks are birds.
Ducks have oily feathers.
Ducks have a rounded beak.
Ducks have webbed feet.
Ducks quack.
Then you come along, and having never seen a duck, but lots
of dogs, ask
whether ducks have fur, bark, and have teeth. Accordingly, I
add some
axioms about what ducks arenıt:
NO ducks have fur.
NO ducks bark.
NO ducks have teeth.
Now, didnıt that qualiŜcation about what ducks arenıt help us
to better
understand what ducks are? It wasnıt contradictory at all to
say what
is not in duck theory, right?
Same with set theory.
> In ZF, however, we know that there are in fact formulas
that do not
> deŜne legitimate sets. The axiom of foundation prohibits
them and we
> must be careful not to consider such a formula as deŜning a
set (to
> avoid a paradox.) But axiomatic systems have no mechanism
to generate
> both what is and what isnıt (a set), or formulate a
reaction to the
> potential of there being an inconsistency. The
mathematician has to
> be careful not to make that mistake. That is why I say that
ZF is not
> a true axiomatic system, since in practice we do need to
take
> additional steps to avoid that potential inconsistency.
> An axiomatic system, by its nature, tells us only what is.
>
> It just ainıt so. You are making stuff up that contradicts
standard
> deŜnitions. This is not a good way to make headway in the
mathematical
> community.
> An axiomatic system is a way to generate theorems (an r.e.
set.)
Well, thatıs one sort of operational way to look at it.
> And in ZF the theorems are really just statements that
generate
> allowable sets.
You can use them to *prove* that certain sets exist, but they
donıt
generate sets. They describe them.
> Except that some theorems say what is NOT a set.
Just as some axioms of duck theory say what a duck is NOT.
You think
this is some sort of problem. It isnıt. Trust me.
> That goes against the grain of generating an r.e. set (of
sets.)
This thing about axioms generating sets might be the source
of your
confusion, though I confess I donıt know what that confusion
is.
> We canıt say that the result of applying a rule to theorems
generates
> allowable sets.
I donıt know what this means.
> (Iıd rather be right than popular.)
Alas, currently you are neither.
> Typically (at least, in all the texts I can think of) a
formal theory
is
> deŜned in such a way that the question of whether or not a
sequence of
> sentences is a legitimate proof from the set A of axioms of
the theory
> is completely independent of issues of decidability. But,
of course,
if
> you want to be able to *tell* when something is a proof
from A, then A
> in general has to be not merely r.e. but fully recursive.
> I tend to agree,
What do you mean you *tend* to agree? Thatıs like saying you
tend to
agree that, say, the complement of a recursive set is
recursive.
> although typically you use an axiomatic system to generate
proofs, not
> to decide if a particular thing is a proof.
How axiomatic systems are typically used is an empirical
matter, one
whose relevance is far from clear, but, be that as it may,
you canıt
very well *use* an axiomatic system to generate proofs if you
donıt have
a sureŜre way of deciding whether or not a certain sentence
that you
need at some step in the proof is an axiom. Either way, you
need the
axioms to be recursive. R.e. wonıt cut it.
> So having an inŜnite number of axioms is even worse than I
said.
Youıve yet to identify a problem, let alone something even
worse. If
the axioms are recursive, there isnıt one.
> I admittedly have yet to give it a close look, but your
confusion
> over a number of fairly elementary points here raises
suspicions
> about the rather grandiose claims you make for your theory
of
> computation.
> All I see is a potential issue regarding terminology, and
some
> disagreements over what constitutes a reasonable axiomatic
system.
Unfortunately, these are not mere issues or disagreements.
There is a
good deal of confusion on your part.
> In any case, I would caution you to not draw too many
conclusions
> about a paper that you havenıt read yet. (We have enough of
that here
> already.) :)
I have drawn no Ŝrm conclusions about it. Rather, as noted
earlier,
your confusions over numerous elementary points of
mathematical logic
and set theory simply raise suspicions, though at this point
for me,
very strong ones. It is quite inconceivable to me that your
work could
have the virtues and signiŜcance you claim for it when you
are so
unclear about such elementary matters. These doubts are
strengthened
when you expend a lot of energy defending your mistakes
instead of
acknowledging them and moving on.
Chris Menzel
===
Subject: Re: Can the deduction theorem be used recursively?
> How can anybody take mathematical logic seriously? Goedel
proved his
> incompleteness theorem, and he *also* proved his
completeness theorem.
Tarski
> proved the undeŜnability of truth, and he *also* gave a
deŜnition of
truth.
Set theorists seem content with coming up with silly names
for technical
concepts. I mean, come on, pre-mouse, universal weasel,
handgranade, ... Not that I mind, though.
--
Aatu Koskensilta (aatu.koskensilta@xortec.Ŝ)
Wovon man nicht sprechen kann, daruber muss man schweigen
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
===
Subject: Re: Can the deduction theorem be used recursively?
> No itıs not why you said it was impossible! Youıre lying
about
> what you said - when you do that you should probably at
> least not quote what you said.
You know why I said something and Iım a liar because I denied
the real
reason why I said it?? Terrible!
> You said it was impossible because of Godelıs Incompleteness
> Theorem. (Itıs not clear to me which is the more charitable
> assumption here: youıre lying about what you said, or you
> think that there exists P such that neither P nor ~P is a
> theorem of FOL is a proof of Godelsı Incompleteness
Theorem.)
Yeah, right. Sounds like a proof to me.
> Excellent point. Or in any case it _would_ be an excellent
point
> if exhibiting ignorance was the same thing as boldly going
where
> no man has gone before.
Well, what do you expect from a stupid guy like me?
>> No, the axioms of ZF have nothing
>> to do with deŜning what constitutes a wff.
>Youıve got half of it. Now see if you can understand the
second half:
> If they did, then you wouldnıt need axiom schemata.
> Youıre catching on, omitting the bit I was replying to,
where
> you said that the axioms of ZF come in two parts, the ones
> that deŜne what a wff is and the other ones.
I omitted something? Oh, so Iım a liar again. Oh, I see.
Charlieıs
a big liar. Thatıs 2 or 3 times in one message. He lies about
why he
says things and he omits things in a message. Gad! Not only
stupid,
but a big liar, too.
> Again, fascinating technique. You say A includes B, I
> say no it doesnıt, and you say Iıve got half of it.
Well, if you really want to know: ZF has nothing to do with
deŜning
what a wff is since it doesnıt, so thatıs the Ŝrst half. The
second
half is that if it did, then instead of axiom schemas you
would have
axioms for both what constitutes a wff and for the sets
deŜned from
them.
Or am I just trying to talk my way out of lie # 3? (or is it
# 4?)
> The deŜnition of a wff has nothing to do with axioms. The
> deŜnition of a wff is part of the deŜnition of a _language_;
> the axioms of set theory are wffs _in_ that language.
Itıs back to the principle of processes. Iım talking about
whatıs
needed  a process that deŜnes (generates) the wffs, combined
with
the process that takes those wffs and deŜnes sets from them.
All of
that could be done with axioms, rather than axiom schemata.
> Nobodyıs called you stupid for saying things contrary to
whatıs
> been published. People have called you stupid
Oh, itıs plural now? Must be, then.
> for saying things
> that are _false_ (and well-known to be false) - the way you
go
> on about how youıve done things that nobodyıs done, and
> the way you reply to comments has something to do with
> that.
(whatever it says.) No need to waste time thinking about that!
Ok, reality check: You confuse reading and memorizing certain
theorems
with intelligence and creativity. You spend your life
memorizing. I
spend my life creating.
To each his own.
Charlie Volkstorf
Cambridge, MA
> Anyway, Chris has already given a big hint as to _why_
> ZF is not Ŝnitely axiomatizable (which is not quite the
> same as a hint regarding how one actually proves
> that). So Iıll just call you stupid for saying that you
> suspect that the axiom of foundation has something
> to do with the fact that ZF is not Ŝnitely axiomatizable:
> Thatıs _one_ axiom, not an axiom scheme, so itıs
> equisitely hard for me to see how it could have
> anything to do with it. (Every nonempty set x
> has an element y such that x and y are disjoint.)
>Charlie Volkstorf
>Cambridge, MA
>> ************************
>>
>> David C. Ullrich
> ************************
> David C. Ullrich
===
Subject: Re: Can the deduction theorem be used recursively?
> Godel proved in 1931 that there is no sound and
> complete axiomatization for FOL. (Charlie-Boo)
>>
>> You really donıt have any understanding of _any_ of this
stuff,
>> right?
the wrong completeness. Now everything you ever spouted out
is true,
vindicated, authenticated. You Ŝnally have your ironclad
proof that
Charlie is a dumb-ass who doesnıt know anything about
anything. He
misused the word completeness. You put him to shamw. Giggle.
Gaggle. Guggle. Gafŝe.
Can we all say that together now, children? Can you say
Goggle -
Gaggle - Guggle - Gafŝe? Maybe if you studied Mathematics and
Logic
for 10 years you could learn to be say it too! Gaggle Guggle
Giggle
Gafŝe Gufŝe Goggle
# of Posts: 534
# of Ideas Presented: 0
# of Problems Solved: 0
# of Systems Developed: 0
# of Cut & Paste from famous books: 534
LOL You are pitiful. You are so jealous of the smart people.
COMPLETE canıt be ambiguous  the famous guys put it in the
titles of
they Ŝrst published it  it didnıt exist yet! LOL Or maybe
theyıre
using it in the same way. Who cares? What difference does it
make?
Youıre such a fame-worshiper. You canıt even judge if a word
is
ambiguous or not  you ask the famous people.
I come up with more ideas in a day than youıve ever posted in
your
entire life. Of course I make mistakes  Iıve posted hundreds
of
ideas. Naturally some of them are going to be
misinterpretations,
misreadings. That has nothing to do with intelligence or
creativity.
Those are just stupid little slip-ups. I thought he meant
completeness as in Godelıs Incompleteness Theorem, not Godelıs
Completeness Theorem. So what? Itıs just a word - a term.
Does that
have anything to do with intelligence? Creative ability? Or,
for
that matter, honesty or maturity? Of course not. But to you,
famous
words, famous authors, famous titles - are your security
blanket.
Your fetish. Your self-reassurance. Your friends. The people
you
bond with. Your weapons. Your life.
The only way you would ever make a mistake is when you
misspell
someoneıs name as you type in their work.
Frege  itıs not your work. Itıs not you. Those people donıt
really
exist. Theyıre mostly dead. You arenıt really Frege. Itıs all
a
dream. Pretend. Make believe.
Fame. Prestige. Reputation. They donıt rub off, Frege.
Quoting a
famous person doesnıt make you famous. Quoting a famous person
doesnıt make you smart.
Youıre a secretary. A photocopy machine. A carbon copy. A
blurry,
faded carbon copy of a man.
C-B
>>
> I assumed you meant complete in the sense of every sentence
being
> provable or refutable
>
>> Right. In other words, you didnıt know what sound and
complete
>> axiomatiziation for FOL means.
>>
> No, I donıt know what you mean when you use an ambiguous
term like
> complete.
>
> G.9adel, Kurt, Die Vollst.8andigkeit der Axiome des
logischen
> Funktionenkalk.9ŝs, Monatshefte f.9fr Mathematik und Physik,
> 37: 349-360, 1930.
> Henkin, Leon, The Completeness of the First-Order Functional
> Calculus, Journal of Symbolic Logic, 14: 159-166, 1949.
> Right, G.9adel and Henkin used ambiguous terms in the
titles of
their
> papers. _Giggle_ :-)
>
> rather than in the sense of the theorems being logically
valid
> (assured by Godelıs Completeness Theorem.)
>
>> No, Godelıs Completeness Theorem is the converse:
>> logically valid wffs are theorems.
>>
> [...] it depends on who you ask
> Only if you ask someone who is clueless [...] (Chris
Mentzel)
> :-)
> Introduction to Mathematical Logic, Mendelson 1964, pg.
> 68, Godelıs Completeness Theorem: In any Ŝrst order
predicate
> calculus, the theorems are precisely the logically valid
wffs.
> Idiot, of course this is a *corollary* of G.9adelıs
Completeness
Theorem.
> FIRST you have to prove that the system is consistent -
thatıs the easy
> part. And then you have to prove that it is complete -
thatıs the hard
> part. G.9adel was THE FIRST to present a proof for that,
concerning
FOPL.
> In any Ŝrst order predicate calculus, the theorems are
precisely the
> logically valid wffs.
>>
>> You should really drop all this soon - each post is more
>> hilarious than the last.
>>
> I have nothing against being entertained. More importantly,
it is
> serving one of my original objectives: to get into the
heads of the
> people so entrenched in the old boyıs club where reputation
takes
> the place of technical analysis.
> Right, you already got in my head as a typical usenet crank.
> I am seeing a couple of things.
> Sure... Much more than ordinary persons, thatıs for sure.
:-)
> I found out a long time ago that journals like authors who
validate
> existing dogma.
> Ehrr?
> I knew about the resistance that I quoted earlier
> Right. Never forget: _The resistance_ actually PROVES that
you are
> right, that you MUST BE right!
> from The Mystery of the Aleph:
>
> Conservative thinkers sought to restrict novel ideas, since
these new
> point of view, he should have been shaped in their image,
and had no
> right to challenge his teachersı philosophy. Cantor
believed that
> their arbitrary restrictions impeded the growth of the
Ŝeld. He gave
> many examples of mathematical ideas that never would have
had a chance
> to succeed and evolve if their proponents had encountered
the kind of
> opposition he faced.
>
> Right. First came CANTOR - and now you. I see.
> But now I am seeing even more.
> Sure...
> Itıs not just to reafŜrm what they
> teach. Quoting others is the only thing that they know. If
you try
> to make a point that questions the validity of an existing
school of
> thought, it involves actually understanding what the
existing system
> is doing, what it is trying to accomplish, and what is
wrong. Itıs
> just so much easier to quote other people.
> Indeed! :-)
> The problem isnıt completely psychological.
>
> Iım not so sure about t h a t.

===
Subject: New (?) Technique for Drawing a Tetrahedron
Iıd Ŝgured-out that isometric drawings were the same
as taking an extreme telephoto lens to your subject (or
looking at the stars) in my mechanical drawing class
in junior high.
recently, I also realized that you donıt have
to go with making all three XYZ coords equal,
at the same apparent angle of view.
yesterday, while making my poster for a poster-session
at UCLA, a simple diagram of a tetrahedron, used
to represent an airplane for a weight & balance determination,
I realized that itıs simple, isometrically,
to exactly represent any tetrahedron,
just by using a circle to represent the circumsphere
of the tetrahedron; see, How?

actually, this is the perfect place to use the dual
terminology,
tetra-asteron, since we afŜx the 4 points to the
circumsphere, or
to the celestial sphere.
--ils duces dıEnron!
http://tarpley.net
--Give Earth a Trickier Dick Cheeny -- out of ofŜce, after
gigayears!
http://www.benfranklinbooks.com/
http://members.tripod.com/~american_almanac
http://www.wlym.com/pdf/iclc/HowTheNation.PDF
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://www.rwgrayprojects.com/synergetics/plates/Ŝgs/plate02.
html
===
Subject: yıı x^2 + 2yı x - y^3 = 0 ?
Does anybody have any suggestions on how to Ŝnd an analytic
solution to
yıı x^2 + 2yı x - y^3 = 0 ?
===
Subject: Re: smallest disk covering a set of points
<snip>
> A = { (x,0) in Q^2 : x > 0 and x^2 < 2 }.
<snip>
> Itıs not completely self explanatory. You seem to be
missing the fact
> that the center of the disk must be a point belonging to
the space under
> consideration (in this case, Q^2).
This is an unnecessary complication in my view, as Iıve
already stated.
Since Q^2 is a subset of R^2, Iım quite happy to accept an
R^2 solution.
> Look at the set A above, and think of it as a subset of
R^2. That set
> has a minimum covering disk in R^2, namely the disk
centered at
> ( sqrt(2)/2, 0 ) with radius sqrt(2)/2.
Agreed. This would be my solution.
> However, there is no disk in Q^2 that is centered at
sqrt(2)/2, since
> sqrt(2)/2 is not in Q. Therefore, the best you can do in
Q^2 is to
> choose a point with rational coordinates that is close to
> ( sqrt(2)/2 , 0 ) and then choose a radius slightly larger
than sqrt(2)/2
> in order to cover the whole set.
Fair enough.
> The closer the center is to ( sqrt(2)/2 , 0 ), the smaller
the radius can
> be. There is no closest point to the desired center in Q^2,
and
> therefore there is no smallest radius. Hence, no MCD in Q^2.
Do you have proof that there is no Œclosestı point? Bear in
mind that
Iım not familiar with the restrictions of the coordinate
system.
Iım starting to get the picture that youıre arguing about the
word
Œminimumı... speciŜcally that there are two Œclosestı points
in Q^2 to
the point (sqrt(2)/2,0), thus two discs of minimum radius,
therfore no
MINIMUM covering disc.
If thatıs what youıre trying to point out, why didnıt you
just point it
out in the Ŝrst place? Sheesh :>
--
Corey Murtagh
The Electric Monk
Quidquid latine dictum sit, altum viditur!
===
Subject: Re: smallest disk covering a set of points
><snip>
>> A = { (x,0) in Q^2 : x > 0 and x^2 < 2 }.
><snip>
>> Itıs not completely self explanatory. You seem to be
missing the fact
>> that the center of the disk must be a point belonging to
the space under
>> consideration (in this case, Q^2).
> This is an unnecessary complication in my view, as Iıve
already stated.
> Since Q^2 is a subset of R^2, Iım quite happy to accept an
R^2
solution.
Not every metric space can be realized as a subset of R^2.
Besides, the
whole point of this discussion is to realize that R^2 has
special
properties that make it possible to deduce a solution in that
speciŜc
context, even though similar solutions may not exist in
arbitrary metric
spaces.
You are just attempting to sweep the special properties of
R^2 under the
rug by pretending that all metric spaces share the same
properties. They
donıt.
>> Look at the set A above, and think of it as a subset of
R^2. That set
>> has a minimum covering disk in R^2, namely the disk
centered at
>> ( sqrt(2)/2, 0 ) with radius sqrt(2)/2.
> Agreed. This would be my solution.
That is the solution to a different problem. It is not a
solution to the
problem that was posed.
>> However, there is no disk in Q^2 that is centered at
sqrt(2)/2, since
>> sqrt(2)/2 is not in Q. Therefore, the best you can do in
Q^2 is to
>> choose a point with rational coordinates that is close to
>> ( sqrt(2)/2 , 0 ) and then choose a radius slightly larger
than
sqrt(2)/2
>> in order to cover the whole set.
> Fair enough.
>> The closer the center is to ( sqrt(2)/2 , 0 ), the smaller
the radius
can
>> be. There is no closest point to the desired center in
Q^2, and
>> therefore there is no smallest radius. Hence, no MCD in
Q^2.
> Do you have proof that there is no Œclosestı point? Bear in
mind that
> Iım not familiar with the restrictions of the coordinate
system.
Consider a covering disk for A that is centered at the point
(p,q) in
Q^2. If q is nonzero, then we note that the point (p,0) is
also in Q^2
and there is a covering disk centered at (p,0) that is
smaller than any
covering disk centered at (p,q). Therefore we only need to
consider
rational points on the x-axis as candidates for the center of
a covering
disk.
The closer p is to sqrt(2)/2, the smaller the covering disk
centered at
(p,0) can be. But p cannot coincide with sqrt(2)/2, since p
is required
to be rational. By the density of the rationals, there is a
rational pı
that lies between p and sqrt(2)/2, and the covering disk
centered at
(pı,0) can be smaller than any covering disk centered at
(p,0). Hence,
no smallest covering disk is possible, for the same reason
that there is
no rational number that is closest to sqrt(2)/2.
> Iım starting to get the picture that youıre arguing about
the word
> Œminimumı... speciŜcally that there are two Œclosestı
points in Q^2 to
> the point (sqrt(2)/2,0), thus two discs of minimum radius,
therfore no
> MINIMUM covering disc.
Two closest points? Name one of them.
> If thatıs what youıre trying to point out, why didnıt you
just point it
> out in the Ŝrst place? Sheesh :>
Because there most certainly are *not* two closest points.
There are no
closest points whatsoever. Sheesh.
--
Dave Seaman
Judge Yohnıs mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: smallest disk covering a set of points
> ) So my approach was to Ŝnd an MCD in RxR and then pick a
smallest MCD in

> ) QxQ that contains the MCD of RxR - it neednıt have the
same radius or
> ) centre.
> There are discs in RxR that donıt have a smallest covering
disc in QxQ.
> (That is, for any covering disc you supply, I can Ŝnd a
smaller one.)
> Worse, for any number in R thatıs not in Q, the set of
larger numbers in
Q
> doesnıt have a smallest member. (i.e. itıs open-ended.)
Whatıs the smallest real number greater than 1? :>
--
Corey Murtagh
The Electric Monk
Quidquid latine dictum sit, altum viditur!
===
Subject: Re: smallest disk covering a set of points
)> Worse, for any number in R thatıs not in Q, the set of
larger numbers in
Q
)> doesnıt have a smallest member. (i.e. itıs open-ended.)
)
) Whatıs the smallest real number greater than 1? :>
Sorry, I should have said Œthe set of not-smaller numbersı,
which in this
case happens to be exactly the same, but only because the
number R is
speciŜed to not be in Q.
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the
statements
made in the above text. For all I know I might be
drugged or something..
No Iım not paranoid. You all think Iım paranoid, donıt you !
#EOT
===
Subject: Re: smallest disk covering a set of points
> As a small exercise: Is there a smallest rational number
that is not
> smaller than sqrt(2) ?
Given that this is cross posted to comp.programming - yes, if
you are
representing that number in a speciŜc datatype, there is a
smallest
number that is not smaller than sqrt(2):
dave@feathers:~/progs/c> cat smallest_over_root2.c
int main()
{
double value,increment;
value=3;
increment=1;
while (increment)
{
if (value*value >= 2)
{
value-=increment;
}
else
{
value+=increment;
increment/=2;
}
}
printf(Smallest double over root(2)=%.80fn,value);
}
dave@feathers:~/progs/c> ./smallest_over_root2
Smallest double over root(2)=
1.41421356237309492343001693370752036571502685546875000000
Same answer if the test is (value*value > 2).
So two questions that are long overdue, if they hasnıt
already been
answered, are: did the OP want a mathematical answer or a
programmatic
answer to the problem, and, if the latter, does he have
mathematical
software that can deal with surds or is he using bog-standard
datatypes?
Fascinating discussion though. One of the most interesting
Iıve read
for a long time.
Dave.
===
Subject: Diagonal approximation of complex matrices
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i4PLeGk03366;
Hello all,
Given a square complex matrix, I am trying to Ŝnd the optimal
diagonal
approximation of the matrix. Precisely, if G is the given
complex matrix and
H is the set of all diagonal complex matrices, then I want to
Ŝnd Hd in H
such that
Hbd = arg min sigma_max (G - Hbd)
where sigma_max is the maximum singular value.
When G is a 2x2 matrix, using Ŝrst and second order
optimality conditions,
I am able to show that the optimal diagonal approximation is
simply the
diagonal elements of G. However, when the dimensions of G are
higher than
2x2, numerical evidence suggests that this simple result does
not hold.
I did some literature survey, but could Ŝnd any papers
dealing with this
problem. I came across some papers (e.g.
http://db.cwi.nl/rapporten/abstract.php?abstractnr=890) where
the spectral
radius of (G - Hbd) is minimized, but the approach is only
numerical.
I would much appreciate, if someone can point me towards any
papers where
this problem has been discussed. Any helpful suggestions
towards Ŝnding the
optimal diagonal approximation (and parametrizing all
solutions, if
non-unique) will be extremely helpful.
Vinay
===
Subject: Re: Differentiation
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i4PLl8v03873;
>Give an example of a unifomrly continuous function on [0, 1]
that is
differentiable on (0, 1) but whose derivative is not bounded
on (0, 1)
i m not sure about this question, but how about f(x):=
sqrt(X) on
I:=[0, 1] it is unifomrlly continuous since I is close
bounded interval, n
f(x) is differentiable on (0,1) except x=0 by using the
deŜnition of limit,
fı(x) is not bounded on (0, 1) since the limit of fı(x) does
not exists when
x=0;
===
Subject: SMSU Problem Corner
The new problems have been posted. Please visit us at
http://math.smsu.edu/~les/POTW.html
I am (Ŝnally) slogging through the backlog of old problems.
===
Subject: Re: Subset of Irrationals
>Any help with this is greatly appreciated. Let u be a subset
of (0,1)
such
>that (a) every member of u is irrational and (b) if s and t
are any two
>members of u, s-t is irrational. Let K be the set of all
subsets of
(0,1)
>satisfying uıs requirements and L is a set listing the
cardinality of
the
>sets in K. What is the largest cardinal in L?
> DeŜne an equivalence relation on (0,1) by letting r~s if
and only if
> r-s is rational.
> Clearly, r~r, if r~s then s~r, and if r~s and s~t, then
since r-s and
> s-t are rational, so is their sum, r-t, so r~t. Thus this
is an
> equivalence relation.
> A subset of (0,1) will satisfy your conditions (a) and (b)
if and only
> if it intersects each equivalence class in at most one
element, and
> contains no elements in the equivalence class of 1/2 (the
rationals).
> I claim that each equivalence class is at most countable.
For given an
> equivalence class, and a Ŝxed representative r, then the set
> { r - s: s in [r] }
> is a subset of Q, and clearly in one-to-one correspondence
with [r].
> Thus, there are 2^{aleph_0} equivalence classes. If you
accept the
> Axiom of Choice, there is a set that contains exactly one
> representative from each class; remove the rational
representative,
> and obtain a set in K of cardinality 2^{aleph_0}. Since
clearly no
> largest cardinal can be in L, this is the answer.
past two days Iıve been trying to prove that 2^Aleph_0 is the
answer by
direct construction. After some trial and error I discovered
this
approach.
The largest cardinal in L will be at most 2^Aleph_0 since any
subset of the
irrationals can be no bigger than this. Map every irrational
number in
(0,1) to some subset of the irrationals near 1/10 in the
following way.
For
the decimal expansion of any irrational in (0,1), let each
digit correspond
to its numerical value except zero, which corresponds to 10.
Then,
starting
with 0.1, the number of zeros after the 1 corresponds to the
factorial
value
of the Ŝrst digit. All subsequent digits are found by Ŝrst
adding
another
1 digit and then adding additional zeros; with the number of
additional
zeros equaling the factorial of the sum of the previous and
current digit
of
the original number. For instance we can Ŝnd the new
irrational that
0.1219... maps to by starting with 0.1. The Ŝrst digit in
0.1219... is 1
and 1!=1 so we add one zero to 0.1 and add another 1 digit to
get 0.101.
The second digit in 0.1219... is 2 and adding that to the
Ŝrst digit, 1,
gives 3 and 3!=6. So we add 6 zeros followed by another 1
digit to 0.101
to
get 0.1010000001. The third digit in 0.1219... is 1 and
adding it to the
sum of its previous digits gives 1+2+1=4 and 4!=24. So we add
24 zeros to
0.1010000001 and add another 1 digit to get
0.10100000010000000000000000000000001. Continuing in this way
we get:
0.1219... <----> 0.10100000010000000000000000000000001000...
Another example:
0.2212... <----> 0.10010000000000000000000000001000000000...
In fact, this technique should map every real in (0,1) to a
unique
irrational. So we have 2^Aleph_0 irrationals consisting of
1ıs separated
by
ever larger sequences of zeros and the difference between any
two of these
irrationals is another irrational. For instance we can
subtract
0.2212...ıs
counter part from 0.1219...ıs counter part to get:
0.00090000009999999999999999999000000999...
This number is irrational. A rational number must have a
repeating decimal
sequence, like 00090999:
0.0009099900090999000909990009099900090999...
And this cannot be added to something like:
0.1010010000001000000000000000000000000100...
to get a number consisting of only 1ıs and 0ıs. So we have
constructed a
set of 2^Aleph_0 irrational numbers and the difference
between any two of
them is another irrational number. There must be sets in K
with
cardinality
2^Aleph_0 and, since this is the largest cardinal that could
possibly be in
L, this is the answer.
RL
===
Subject: Re: Big Bertha Thing blogs
Big Bertha Thing Chair
Cosmic Ray Series
Possible Real World System Constructs
http://web.onetel.com/~tonylance/chair.html
Access page JPG 18K Image
Astrophysics net ring Access Site
Newsgroup Reviews including uk.railway
Detail from frontspiece painting showing,
so called arms up to chair sedan.
Caption;-
A Chinese Street
The World and Its People
Asia With Special Reference to British Possessions
Published by Thomas Nelson and Sons 1903
Without Author or Editor Name
(C) Copyright Tony Lance 1998
Distribute complete and free of charge to comply.
Big Bertha Thing Serpico
Serpico was an honest cop in New York, that was all he ever
wanted.
As he lay, in his hospital bed, recovering from being shot in
the face,
all he could bear to watch on TV was Sesame Street.
At one time or another, he had upset every cop on the force,
except one.
Now, he even had mayor John Lindsay, dancing in attendance.
He still thought, it was worth it.
Now he lives in Switzerland, on a disability pension.
There is this incredible sense of deja vu,
even down to the walking stick.
Most days he feels Ŝne, even if occasionally he could sleep
for a week.
Bibliography;-
Serpico by Peter Maas
(C) Copyright 1973 by Peter Maas and Tsampa Company, Inc.
Published by William Collins Sons & Co.Ltd. Glasgow.
Serpico Ŝlm by Paramount Pictures
True story covered by
New York Times (3rd February 1971)
Tony Lance
mickalice@big-bertha-thing.com
Big Bertha Thing handcart
1. Gardening section of Daily Telegraph, on Saturday
17th November 2001, shows full front page spread
picture of pin-wheel rickshaw.
2. For another picture see;-
http://web.onetel.com/~tonylance/pinwheel.html
3. Both show load perfectly distributed.
4. Both are idealized pictures of a rickshaw that does
not exist.
5. A 19th century cheque-book journalist requested a
picture of an unnusual rickshaw, which was promised
for tomorrow.
6. A portrait painter, a landscape artist and a
cartoonist submitted quotes, which the cartoonist won.
7. A wheelborrow uses straight arm technology.
8. In common with a handcart, the so called pin-wheel
rickshaw, uses bent arm technology.
9. It is physically impossible and so unviable.
10. This is a scientiŜc cartoon, some work, some donıt.
Tony Lance
mickalice@big-bertha-thing.com
===
Subject: Matlab 6.5
I can send you 3 CDs containing Matlab 6.5 (free of charge)
if you can
provide ONE of these packages:
1) Mathcad 11.
2) Maple v.9.0 or v9.5.
3) Multisim 7 (circuit simulator).
4) Mathematica 5.0.
No money involved in the exchange.
Trucho
===
Subject: Re: Matlab 6.5
That would be the *legitimate* version of Matlab, correct?
Trucho Nemesis <trucho1270@hotmail.com> schrieb im Newsbeitrag
> I can send you 3 CDs containing Matlab 6.5 (free of charge)
if you can
> provide ONE of these packages:
> 1) Mathcad 11.
> 2) Maple v.9.0 or v9.5.
> 3) Multisim 7 (circuit simulator).
> 4) Mathematica 5.0.
> No money involved in the exchange.
> Trucho
===
Subject: 3rd D.P. Roots
Is there any rule (or rule of thumb) for Ŝnding roots (real
or complex) of
polynomials with this form?
f(x) = Ax^3 + Bx + C
(A, B, C not 1 or 0)
For example (or speciŜcally in this case, if there are
different cases,
even with the above restriction):
f(x) = 2x^3 + x - 2
In advance, I thank you for your help, even if this question
is simple.
John
===
Subject: Re: 3rd D.P. Roots
Googling solving cubic equation gives 69,000 hits, the Ŝrst
one
http://www.1728.com/cubic2.htm looks particularly simple.
> Is there any rule (or rule of thumb) for Ŝnding roots (real
or complex)
of
> polynomials with this form?
> f(x) = Ax^3 + Bx + C
> (A, B, C not 1 or 0)
> For example (or speciŜcally in this case, if there are
different cases,
> even with the above restriction):
> f(x) = 2x^3 + x - 2
> In advance, I thank you for your help, even if this
question is simple.
> John
===
Subject: simple limit laws
boundary=----=_NextPart_000_0027_01C44254.9626AE60
-------------------------------------------------------------
--------
charset=utf-8
In one of my books, I have the following
f Œ(x).89x = f(x + .89x) - f(x)
By how did they get there (step by step)
What I know is the following
f Œ(x).89x = [lim(.89x->0)
(f(x+.89x) - f(x))/.89x)] *
.89x
=...?..
= f(x + .89x) - f(x)
(using of course the well-known limit laws)
vielen dank fuerıs nachdenken
===
Subject: Re: simple limit laws
> In one of my books, I have the following
> f Œ(x).89x = f(x + .89x) - f(x)

> By how did they get there (step by step)
They didnıt. Itıs false for most f.
> What I know is the following
> f Œ(x).89x = [lim(.89x->0)
(f(x+.89x) - f(x))/.89x)] *
.89x
> =...?..
> = f(x + .89x) - f(x)
> (using of course the well-known limit laws)
Thatıs false too. If delta x -> 0, both sides give 0 in the
limit.
===
Subject: Re: simple limit laws
X-NFilter: 1.2.0
> In one of my books, I have the following
> f Œ(x)?x = f(x + ?x) - f(x)
> By how did they get there (step by step)
> What I know is the following
> f Œ(x)?x = [lim(?x->0) (f(x+?x) - f(x))/?x)] * ?x
> =...?..
> = f(x + ?x) - f(x)
> (using of course the well-known limit laws)
The deŜnition of differentiation is:
fı(x) = [lim(dx --> 0) (f(x+dx) - f(x))/dx ]
so it shouldnıt be too much of a stretch to see that
fı(x)*dx = f(x+dx) - f(x)
for inŜnitessimal dx.