mm-669
===
Subject: Re: Efficiency against frictional resistance
Cut<
> Idiot. You know nothing of friction. Increasing speed does
not
> increase efficiency.
> Paul Cardinale
It's the momentum Paul: The product of mass [m], _and_ its
speed [v];
where [mv = ft] that has to be increased in order to increase
efficiency. Quit playing dumb will ya, and think?
===
===
Subject: Re: Object ring: corrected definition
>>Adjusted it yet again:
The Object Ring is a commutative ring that includes all
numbers such
>>that -1 and 1 are the only members that are both a unit and
an
>>integer, where no non-unit member is a factor of any two
integers 
that
>>are coprime in the ring of integers, and where given a
member x there
>>must exist a nonzero member y and a member z such that xy
is an
>>integer and xz + y is a nonzero integer.
>Oh, that was a bogus addition, the xz + y thing. I've
dropped it.

>>James Harris
> If you drop that condition and keep the others, and
furthermore
> 1. don't want transcendental numbers to be in an object
ring,
> 2. want the object ring to contain the algebraic integers,
> then from the various posts of Winter, Magidin and KRamsey
we
> know how the rings with these properties look like, and that
> there is not only one such ring.
> If you drop 2, which means that you do not require an
object ring
> to be integrally closed, then the situation becomes
complicated.
The object ring does contain algebraic integers.
My thinking was to include *all* numbers that will fit with
the three
conditions:
1. -1 and 1 are the only integers that are units.
2. No member of the ring is a factor of any two integers that
are
coprime in the ring of integers, which I'm thinking might be
that no
member shares non-unit factors with any two integers that are
coprime
in the ring of integers.
Now if you think my second condition is unnecessary you 
can't
just
claim other posters have handled the question. If you wish to
copy an
argument from their posts into your reply and cite that way,
ok, but
I'm not going to read some posts just because *you* claim
they say
something or other.
3. For any member x there is a non-zero member y such that xy
is an
integer.
Now those requirements will leave in algebraic integers. And
I think
they'll keep out certain transcendentals--not all--but ones
like pi
and e, but I'm not certain about how you show that they do.
The idea is to include all numbers such that you don't have 
a
conßict
with coprimeness in the ring of integers. Like 1/2 can't go
in as
then 2(1/2) = 1, and 2 is NOT a factor of 1 in the ring of
integers,
so having 1/2 creates a conßict. Therefore, 1/2 is excluded.
That's the basic concept.
I repeat, the point is to have a ring where you don't 
conßict
with
coprimeness results in the ring of integers.
It's simple, basic, and not a very complicated concept!
James Harris
===
Subject: Re: Object ring: corrected definition
>>>Adjusted it yet again:
>>The Object Ring is a commutative ring that includes all
numbers such
>>>that -1 and 1 are the only members that are both a unit
and an
>>>integer, where no non-unit member is a factor of any two
integers that
>>>are coprime in the ring of integers, and where given a
member x there
>>>must exist a nonzero member y and a member z such that
xy is an
>>>integer and xz + y is a nonzero integer.
>Oh, that was a bogus addition, the xz + y thing. I've
dropped it.
>>James Harris
If you drop that condition and keep the others, and
furthermore
>>1. don't want transcendental numbers to be in an object
ring,
>>2. want the object ring to contain the algebraic integers,
>>then from the various posts of Winter, Magidin and KRamsey
we
>>know how the rings with these properties look like, and that
>>there is not only one such ring.
>>If you drop 2, which means that you do not require an
object ring
>>to be integrally closed, then the situation becomes
complicated.
> The object ring does contain algebraic integers.
> My thinking was to include *all* numbers that will fit with
the three
> conditions:
> 1. -1 and 1 are the only integers that are units.
> 2. No member of the ring is a factor of any two integers
that are
> coprime in the ring of integers, which I'm thinking might
be that no
> member shares non-unit factors with any two integers that
are coprime
> in the ring of integers.
> Now if you think my second condition is unnecessary you
can't just
> claim other posters have handled the question. If you wish
to copy an
> argument from their posts into your reply and cite that
way, ok, but
> I'm not going to read some posts just because *you* claim
they say
> something or other.
> 3. For any member x there is a non-zero member y such that
xy is an
> integer.
> Now those requirements will leave in algebraic integers.
And I think
> they'll keep out certain transcendentals--not all--but 
ones
like pi
> and e, but I'm not certain about how you show that they 
do.
> The idea is to include all numbers such that you don't 
have
a conßict
> with coprimeness in the ring of integers. Like 1/2 can't 
go
in as
> then 2(1/2) = 1, and 2 is NOT a factor of 1 in the ring of
integers,
> so having 1/2 creates a conßict. Therefore, 1/2 is excluded.
> That's the basic concept.
> I repeat, the point is to have a ring where you don't
conßict with
> coprimeness results in the ring of integers.
> It's simple, basic, and not a very complicated concept!
> James Harris
If it were simple, basic, and not very complicated, you would
have
provided an algebraic number that is (1) not an algebraic
integer,
and (2) in the ring you think you've defined.
Apparently, you can't do even this.
Why should anyone pay any attention to a definition, when you
are
unable to specify even ONE element meeting those two
conditions?
Dale.
===
Subject: Re: Object ring: corrected definition
> If it were simple, basic, and not very complicated, you
would have
> provided an algebraic number that is (1) not an algebraic
integer,
> and (2) in the ring you think you've defined.
> Apparently, you can't do even this.
> Why should anyone pay any attention to a definition, when
you are
> unable to specify even ONE element meeting those two
conditions?
Otoh, over the years some people have been interested in the
set of 
positive
integers which are (1) perfect and (2) odd. However, nobody
is currently
able to specify even ONE integer meeting those two
conditions. The simple
fact that no element of a set is known does not mean that the
set should be
ignored.
--
Clive Tooth
http://www.clivetooth.dk
===
Subject: Re: Object ring: corrected definition
>> If it were simple, basic, and not very complicated, you
would have
>> provided an algebraic number that is (1) not an algebraic
integer,
>> and (2) in the ring you think you've 
defined.
>> Apparently, you can't do even this.
>> Why should anyone pay any attention to a definition, when
you are
>> unable to specify even ONE element meeting those two
conditions?
> Otoh, over the years some people have been interested in
the set of
> positive integers which are (1) perfect and (2) odd.
However, nobody
> is currently able to specify even ONE integer meeting those
two
> conditions. The simple fact that no element of a set is
known does
> not mean that the set should be ignored.
I take your meaning: absence of evidence is not evidence of
absence.
However, JSH is maintaining that not only does the class of
numbers
exist, but that it's all manner of simple to establish the
properties
of those numbers. Why, then, shouldn't he exhibit one of
these numbers?
How many people are asserting the existence of odd perfect
numbers?
Besides, the definition of perfect number has lent itself to
some non-
trivial examples, right?
The concept odd perfect number is the combination of two
concepts,
each of which has generated examples, together with a method
for
establishing membership. The question of the intersection of
two
well-defined sets is quite reasonable. I, for one, have never
found
that particular question to be very interesting, but I think
that's a
matter of taste.
Dale
===
Subject: Re: Object ring: corrected definition
> If it were simple, basic, and not very complicated, you
would have
> provided an algebraic number that is (1) not an algebraic
integer,
> and (2) in the ring you think you've defined.
> Apparently, you can't do even this.
> Why should anyone pay any attention to a definition, when
you are
> unable to specify even ONE element meeting those two
conditions?
> Otoh, over the years some people have been interested in
the set of
positive
> integers which are (1) perfect and (2) odd. However, nobody
is currently
> able to specify even ONE integer meeting those two
conditions. The simple
> fact that no element of a set is known does not mean that
the set should
be
> ignored.
Hypothetical: Assume there is an odd perfect number. If some
derivation
from that assumption then shows that there is a contradiction
in the
mathematics of perfect numbers, would it then be fair to say
that the
assumption is proven false?
===
Subject: Re: Object ring: corrected definition
...
> If it were simple, basic, and not very complicated, you
would have
> provided an algebraic number that is (1) not an algebraic
integer,
> and (2) in the ring you think you've defined.
 Apparently, you can't do even this.
....
> Otoh, over the years some people have been interested in
the set of 
positive
> integers which are (1) perfect and (2) odd. However, nobody
is currently
> able to specify even ONE integer meeting those two
conditions. The 
simple
> fact that no element of a set is known does not mean that
the set should 
be
> ignored.
Indeed. However, JSH does not even provide a way to find out
whether a
particular number is in the ring or not.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland, 
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Object ring: corrected definition
>...
> If it were simple, basic, and not very complicated, you
would have
> provided an algebraic number that is (1) not an algebraic
integer,
> and (2) in the ring you think you've defined.
 Apparently, you can't do even this.
>....
> Otoh, over the years some people have been interested in
the set of 
positive
> integers which are (1) perfect and (2) odd. However, nobody
is 
currently
> able to specify even ONE integer meeting those two
conditions. The 
simple
> fact that no element of a set is known does not mean that
the set should 
be
> ignored.
>Indeed. However, JSH does not even provide a way to find out
whether a
>particular number is in the ring or not.
Precisely. The way he states the definition makes no sense at
all -
I'm always puzzled by the fact that people ignore the fact
that
it's incoherent and reply as though he'd said 
something that
means something. In particular:
He says it's the ring of all numbers such that 1 and -1 are
the only units...
All numbers such that [some property that a number might
have]
makes sense, but
The ring is the set of all numbers such that [some property a
ring might have]
simply doesn't make any sense.
I've pointed this out before, to no avail. Oh well.
************************
David C. Ullrich
===
Subject: Re: Object ring: corrected definition
>>Indeed. However, JSH does not even provide a way to find out
>>whether a particular number is in the ring or not.
>> Precisely. The way he states the definition makes no sense
at all -
>> I'm always puzzled by the fact that people ignore the 
fact
that it's
>> incoherent and reply as though he'd said something that
means
>> something. In particular:
>> He says it's the ring of all numbers such that 1 and -1
are the only
>> units...
>> All numbers such that [some property that a number might
have]
>> makes sense,
>That only makes sense when you are defining a subset of an
existing
>set. Or if you are defining in terms of an existing set.
Right. But in all this number has been taken to mean
complex
number, so it _is_ referring to the elements of an existing
set.
>> but
>> The ring is the set of all numbers such that [some
property a ring
>> might have]
>> simply doesn't make any sense.
>Uh, what? That's pretty much how for example the natural
numbers are
>defined. By a set of properties on the _set_.
Yes, they're defined in terms of properties of 
the set.
I didn't say there was anything wrong with that! The
set of natural numbers is _not_ defined to be the set
N = {n : N has a certain property}.
There's a big difference between that and
N = the set with a certain property.
>> I've pointed this out before, to no avail. Oh well.
>The circularity problem is rather that he specifies both
features his
>_set_ should have, as well as features that integers in it
should
>have. But that implies that a subset of his ring is
_identical_ to
>the integers. Which is weird at best. One can reduce this to
>converting this weirdness into an explicit requirement that
other
>conditions imply (which I can't see) that there is exactly
one subset
>to his ring that is isomorphic to the integers, and that the
isomorphs
>of -1 and 1 are the only units in his ring.
************************
David C. Ullrich
===
Subject: Re: Object ring: corrected definition
> Indeed. However, JSH does not even provide a way to find
out
> whether a particular number is in the ring or not.
>> Precisely. The way he states the definition makes no sense
at all -
>> I'm always puzzled by the fact that people ignore the 
fact
that it's
>> incoherent and reply as though he'd said something that
means
>> something. In particular:
>> He says it's the ring of all numbers such that 1 and -1
are the only
>> units...
>> All numbers such that [some property that a number might
have]
>> makes sense,
> That only makes sense when you are defining a subset of an
existing
> set. Or if you are defining in terms of an existing set.
>> but
>> The ring is the set of all numbers such that [some
property a ring
>> might have]
>> simply doesn't make any sense.
> Uh, what? That's pretty much how for example the natural
numbers are
> defined. By a set of properties on the _set_.
No, no. This is a grammatical mistake.  A is a set (or a
subset of B) 
such
that every x in A ... is legal, but A is the set of all x
(in B) such 
that
'P(A)' is not....
>> I've pointed this out before, to no avail. Oh well.
> The circularity problem is rather that he specifies both
features his
> _set_ should have, as well as features that integers in
it should
> have. But that implies that a subset of his ring is
_identical_ to
> the integers. Which is weird at best. One can reduce this to
> converting this weirdness into an explicit requirement that
other
> conditions imply (which I can't see) that there is exactly
one subset
> to his ring that is isomorphic to the integers, and that
the isomorphs
> of -1 and 1 are the only units in his ring.
===
Subject: Re: Object ring: corrected definition
In sci.math, James Harris
<jstevh@msn.comAdjusted it yet again:
The Object Ring is a commutative ring that includes all
numbers such
>that -1 and 1 are the only members that are both a unit
and an
>integer, where no non-unit member is a factor of any two
integers 
that
>are coprime in the ring of integers, and where given a
member x 
there
>must exist a nonzero member y and a member z such that xy
is an
>integer and xz + y is a nonzero integer.
Oh, that was a bogus addition, the xz + y thing. I've
dropped it.
>
James Harris
> If you drop that condition and keep the others, and
furthermore
>> 1. don't want transcendental numbers to be in an object
ring,
>> 2. want the object ring to contain the algebraic integers,
>> then from the various posts of Winter, Magidin and KRamsey
we
>> know how the rings with these properties look like, and
that
>> there is not only one such ring.
>> If you drop 2, which means that you do not require an
object ring
>> to be integrally closed, then the situation becomes
complicated.
>> H
> The object ring does contain algebraic integers.
> My thinking was to include *all* numbers that will fit with
the three
> conditions:
> 1. -1 and 1 are the only integers that are units.
> 2. No member of the ring is a factor of any two integers
that are
> coprime in the ring of integers, which I'm thinking might
be that no
> member shares non-unit factors with any two integers that
are coprime
> in the ring of integers.
> Now if you think my second condition is unnecessary you
can't just
> claim other posters have handled the question. If you wish
to copy an
> argument from their posts into your reply and cite that
way, ok, but
> I'm not going to read some posts just because *you* claim
they say
> something or other.
> 3. For any member x there is a non-zero member y such that
xy is an
> integer.
> Now those requirements will leave in algebraic integers.
And I think
> they'll keep out certain transcendentals--not all--but 
ones
like pi
> and e, but I'm not certain about how you show that they 
do.
If one leaves in algebraic integers one also leaves in numbers
such as 2 - sqrt(3). Now 2 - sqrt(3) is an interesting
number -- for starters, it pairs off with 2 + sqrt(3), such
that the product of the two is a unit -- namely, 1.
Since the number 2 - sqrt(3) fits the definition 
of a unit, I'm
not sure how one can fit definition 1 and the set 
of all
algebraic integers into the same ring.
If one doesn't like 2 - sqrt(3), it turns out 1/2 * (1 +
sqrt(5))
and its cohort, 1/2 * (-1 + sqrt(5)), are units as well.
The second in particular is a solution to the equation
x^2 + x - 1 = 0, as can readily be verified:
(1/2 * (-1 + sqrt(5)))^2 + (1/2 * (-1 + sqrt(5)) - 1
= 1/4 * (1 + 5 - 2 * sqrt(5)) + 1/2 * (-1 + sqrt(5)) - 1
= 3/2 - 1/2 * sqrt(5)) - 1/2 + 1/2*sqrt(5) - 1
= 0.
The first is a solution to x^2 - x - 1; 
verification is left
to the interested reader.
As for being factors of two coprime numbers -- I'm wondering
how one defines factorization in a ring with more than
two units.
> The idea is to include all numbers such that you don't 
have
a conßict
> with coprimeness in the ring of integers. Like 1/2 can't 
go
in as
> then 2(1/2) = 1, and 2 is NOT a factor of 1 in the ring of
integers,
> so having 1/2 creates a conßict. Therefore, 1/2 is excluded.
> That's the basic concept.
> I repeat, the point is to have a ring where you don't
conßict with
> coprimeness results in the ring of integers.
> It's simple, basic, and not a very complicated concept!
It also doesn't appear to work.
> James Harris
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Object ring: corrected definition
Discussion, linux)
> The object ring does contain algebraic integers.
> My thinking was to include *all* numbers that will fit with
the three
> conditions:
> 1. -1 and 1 are the only integers that are units.
> 2. No member of the ring is a factor of any two integers
that are
> coprime in the ring of integers, which I'm thinking might
be that no
> member shares non-unit factors with any two integers that
are coprime
> in the ring of integers.
> Now if you think my second condition is unnecessary you
can't just
> claim other posters have handled the question. If you wish
to copy an
> argument from their posts into your reply and cite that
way, ok, but
> I'm not going to read some posts just because *you* claim
they say
> something or other.
Cute.
> 3. For any member x there is a non-zero member y such that
xy is an
> integer.
> Now those requirements will leave in algebraic integers.
And I think
> they'll keep out certain transcendentals--not all--but 
ones
like pi
> and e, but I'm not certain about how you show that they 
do.
> The idea is to include all numbers such that you don't 
have
a conßict
> with coprimeness in the ring of integers. Like 1/2 can't 
go
in as
> then 2(1/2) = 1, and 2 is NOT a factor of 1 in the ring of
integers,
> so having 1/2 creates a conßict. Therefore, 1/2 is excluded.
> That's the basic concept.
It's not coherent.
You can't define something this way because it 
is circular as
stated.
Your definition amounts to:
Obj = { x in C | Px & Qx & exists y in Obj such that xy in Z }
P and Q are what you numbered 1 and 2. I want to ignore them
for now.
The point is that your definition is circular as stated. You
want to
define Obj, but you do so in terms of Obj. This is 
unacceptable
without some form of justification.
You have to prove that there is some set satisfying this
equation and
that furthermore it is unique in some sense (probably the
largest such
set).
> I repeat, the point is to have a ring where you don't
conßict with
> coprimeness results in the ring of integers.
> It's simple, basic, and not a very complicated concept!
It's simple, basic and evidently incoherent.
--
Quincy, would you rather do epistemology or conceptual
analysis?
You know what? I'd rather fight on an aircraft 
carrier....
And Mama
and Baba (Papa) would fight on an aircraft carrier, too.
-- Quincy P. Hughes, age 3 1/2
===
Subject: Re: Object ring: corrected definition
>>The object ring does contain algebraic integers.
>>My thinking was to include *all* numbers that will fit with
the three
>>conditions:
>>1. -1 and 1 are the only integers that are units.
>>2. No member of the ring is a factor of any two integers
that are
>>coprime in the ring of integers, which I'm thinking might
be that no
>>member shares non-unit factors with any two integers that
are coprime
>>in the ring of integers.
>>3. For any member x there is a non-zero member y such that
xy is an
>>integer.
>>Now those requirements will leave in algebraic integers.
And I think
>>they'll keep out certain transcendentals--not all--but 
ones
like pi
>>and e, but I'm not certain about how you show that they 
do.
>>The idea is to include all numbers such that you don't 
have
a conßict
>>with coprimeness in the ring of integers. Like 1/2 can't 
go
in as
>>then 2(1/2) = 1, and 2 is NOT a factor of 1 in the ring of
integers,
>>so having 1/2 creates a conßict. Therefore, 1/2 is excluded.
>>That's the basic concept.
> It's not coherent.
> You can't define something this way because it 
is circular
as stated.
> Your definition amounts to:
> Obj = { x in C | Px & Qx & exists y in Obj such that xy in
Z }
> P and Q are what you numbered 1 and 2. I want to ignore
them for now.
> The point is that your definition is circular as stated. You
want to
> define Obj, but you do so in terms of Obj. This is
unacceptable
> without some form of justification.
Makes me wonder: how many readers of sci.math
have ever heard of fixpoint definitions?
Herman Jurjus
===
Subject: Re: Object ring: corrected definition
<87brj492bp.fsf@phiwumbda.org> <2kaaqoF19gdetU1@uni-berlin.de>
Discussion, linux)
>The object ring does contain algebraic integers.
>My thinking was to include *all* numbers that will fit with
the three
>conditions:
>1. -1 and 1 are the only integers that are units.
>2. No member of the ring is a factor of any two integers
that are
>coprime in the ring of integers, which I'm thinking might
be that no
>member shares non-unit factors with any two integers that
are coprime
>in the ring of integers.
>3. For any member x there is a non-zero member y such that
xy is an
>integer.
>Now those requirements will leave in algebraic integers.
And I think
>they'll keep out certain transcendentals--not all--but
ones like pi
>and e, but I'm not certain about how you show that they do.
>The idea is to include all numbers such that you don't
have a conßict
>with coprimeness in the ring of integers. Like 1/2 can't
go in as
>then 2(1/2) = 1, and 2 is NOT a factor of 1 in the ring of
integers,
>so having 1/2 creates a conßict. Therefore, 1/2 is
excluded.
>That's the basic concept.
>> It's not coherent.
>> You can't define something this way because 
it is circular
as stated.
>> Your definition amounts to:
>> Obj = { x in C | Px & Qx & exists y in Obj such that xy in
Z }
>> P and Q are what you numbered 1 and 2. I want to ignore
them for now.
>> The point is that your definition is circular as stated.
You want to
>> define Obj, but you do so in terms of Obj. This is
unacceptable
>> without some form of justification.
> Makes me wonder: how many readers of sci.math
> have ever heard of fixpoint definitions?
I am certainly familiar with fixed point 
definitions[1], but
that's the
kind of justification I alluded to above. I'm 
not sure if your
comments refer to me, but you snipped the following, which is
explicitly about a fixed point construction.
,----
| You have to prove that there is some set satisfying this
equation and
| that furthermore it is unique in some sense (probably the
largest such
| set).
`----
It's pretty clear that there are solutions for that equation
up
there. Z is one. The algebraic integers are another. Others
have
constructed solutions which properly contain the algebraic
integers.
If I'm not mistaken, there are arguments that there are
maximal
solutions to that equation, but not a largest solution.
If you thought my comments betrayed an ignorance of fixed 
point
definitions, you were mistaken. Of course, the error might be
on my
part. Maybe I didn't express myself well, but I think I
expressed
myself adequately.
(Until recently, my research was exclusively about
coalgebras. Final
coalgebras are greatest fixed points, more or less. 
I've been
introduced to the topic, honest.)
Footnotes:
[1] A matter of taste, but I still dislike the term fixpoint.
--
Not all features that are found on the Security tab are
designed to
help make your documents and files more secure. --Microsoft
on Office
security features (after it was pointed out by a third party
that a
certain password setting is easily bypassed.)
===
Subject: Re: Random number question?
> As others have noted, it's not possible for a computer to
produce 
truly
> random numbers in software.
>>That depends on what you mean by random and software.
>> Indeed. I am curious: Is it possible for a computer to
*distinguish*
>truly
>> random numbers from psuedo random numbers [in software]?
>No.
>However, given a set of numbers, a computer can determine
the degree to
>which the set satisfies or violates any arbitrary measure of
randomness.
This is not correct. Given any test for randomness, a
computer can determine where the set stands with respect
to this test. However, any sequence of numbers has a
positive probability assuming what is usually called
randomness, so neither a computer, nor any other device,
not knowing how the sequence is created, can give a
definite answer to whether or not it is random.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue
University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Random number question?
>> As others have noted, it's not possible for a computer
to produce 
truly
>> random numbers in software.
>That depends on what you mean by random and software.
> Indeed. I am curious: Is it possible for a computer to
*distinguish*
>>truly
> random numbers from psuedo random numbers [in software]?
>>No.
>>However, given a set of numbers, a computer can determine
the degree to
>>which the set satisfies or violates any arbitrary measure of
randomness.
>This is not correct. Given any test for randomness, a
>computer can determine where the set stands with respect
>to this test. However, any sequence of numbers has a
>positive probability assuming what is usually called
>randomness, so neither a computer, nor any other device,
>not knowing how the sequence is created, can give a
>definite answer to whether or not it is random.
So it may be possible to generate sequences in software that
are 
effectively
indistinguishable from truly random sequences. Interesting.
Rich
===
Subject: Re: Random number question?
........................
>However, given a set of numbers, a computer can determine
the degree to
>which the set satisfies or violates any arbitrary measure
of randomness.
>>This is not correct. Given any test for randomness, a
>>computer can determine where the set stands with respect
>>to this test. However, any sequence of numbers has a
>>positive probability assuming what is usually called
>>randomness, so neither a computer, nor any other device,
>>not knowing how the sequence is created, can give a
>>definite answer to whether or not it is random.
>So it may be possible to generate sequences in software that
are 
effectively
>indistinguishable from truly random sequences. Interesting.
ANY sequence cannot be positively identified as not being
truly random. The sequence all of whose elements are 0
is as likely as any particular sequence of 0's and 
1's,
produced by a random process in which, at each place, 0
and 1 are equally likely, and the digits in the various
places are independent.
However, this does not mean that using a sequence like this
will give the type of results which are likely to be obtained
from a random sequence.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue
University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Random number question?
> As others have noted, it's not possible for a computer to
produce
truly
> random numbers in software.
>>That depends on what you mean by random and software.
>> Indeed. I am curious: Is it possible for a computer to 
*distinguish*
>truly
>> random numbers from psuedo random numbers [in software]?
>No.
>However, given a set of numbers, a computer can determine
the degree to
>which the set satisfies or violates any arbitrary measure of
randomness.
Interesting.
First you say:
> This is not correct.
Then you agree with me:
> Given any test for randomness, a
> computer can determine where the set stands with respect
> to this test. However, any sequence of numbers has a
> positive probability assuming what is usually called
> randomness, so neither a computer, nor any other device,
> not knowing how the sequence is created, can give a
> definite answer to whether or not it is random.
===
Subject: Re: generators of SL(2,R)
erudition. But you were wrong...
>>If you know any transfinite math ,
>> you sure don't show it
>>and then, picking fights over it when it isn't 
at issue.
>>So My finite subset is real..
>>Show me a real generator of his type that
>>doesn't go complex?
> That doesn't make much sense to me. What _you_
> need to do is show us an element of SL(2,R) which
> is not in the group generated by those matrices,
> or vice versa.
>>I'm trying to keep it real
>>on more than one level..
>> semantically speaking.
>>Real number domain sets contain many finite sets..
>>We are looking for an overall generator set that remains
real.
>>We know that quadratic number fields are 
infinite but
countable,
>>but they do remain in the real domain:
>>R: a+b*Sqrt[Prime[n]]: real numbers a,b
>>his x goes to--> a*x^2+c*x+b ( as a result of the detm=1
domain
>>definition of SL(2,R))
>>part of which is always complex , real numbers a,b,c.
> Part of _what_ is always complex?
>>If you say that x is defined as real then you have limited
the domain of
>>{a,b,c} so that it is no longer a group under the axioms of
groups.
>>So it is proved that his generators are not generators of
SL(2,R).
> None of this makes much sense. Which probably has something
to do
> with the fact that those matrices _do_ generate SL(2,R).
>>Suppose we use all the quadratic number fields
Q(Sqrt[Prime[n]]) as the
>>basis of the x ( a subset of the reals), but as individual
infinite sets
>>with countable n. We, then, have an infinite set of 
infinite
sets of
>>countable elements that
>>approximates the real domain so well that you can't 
find a
number that
>>isn't covered.
> You can't? Let's see...
> Pi is not covered. (Not that this has anything to do with
> SL(2,R)...)
>>It is still aleph zero, but you can't really tell it from
>>aleph one in reality... not with a Turing machine in a
finite stopping
>>time anyway. As far as I know that is the Chaitin kind of
definition of
>>transfinites with is equivalent to the Cantor one.
>Roger L. Bagula TOP-POSTED:
>>Here's one that might give a real only basis
>>that is based on golden mean number field.
>>gm=(1+Sqrt[5])/2
>>e1={{0,gm},{-1/gm,0}}
>>e2={{gm,x},{0,1/gm}}
>>x={gm,1/gm,1/gm^2}
>>It is always SL(2,R)
>These do not generated SL(2,R). SL(2,R) is
>an uncountable group. Hence it cannot be generated
>by a countable set of generators, let alone a finite
>set such as this.
>>and I think since the
>>f[n]=mod[n*gm,1]=fractionalpart[n*gm]
>>is an infinite set, that the result can cover the
>>whole real domain, but only at an infinite level
>>for n.
>What you think is hardly relevant.
> ************************
> David C. Ullrich
===
Subject: Re: generators of SL(2,R)
>erudition. But you were wrong...
Hmm. Possibly you could be more specific about what I
was wrong about.
Let's see. I said something didn't make much 
sense to me.
That can't be wrong, or if it is you can't 
know that it's
wrong - how could you know that something makes sense _to me_?
Then I asked a few questions about what you'd said - hard
to see how asking a question can be wrong either.
Then I said that Pi is not covered by the elements
of Q(Sqrt[Prime[n]]), for all n. That must be the
part I got wrong.
You should probably explain what's wrong about that.
First you should explain what you _mean_ when you
say that the reals are all covered by Q(Sqrt[Prime[n]]).
(If you mean that every real can be approximated
by elements of Q(Sqrt[Prime[n]]) that would be true.
But that can't be what you mean, because that would
be a strange thing to say, since in fact every real
can be approximated by elements of Q, no need for
that Q(Sqrt[Prime[n]]) stuff. So I assumed you meant
that every real is an element of some Q(Sqrt[Prime[n]]).
That's false - Pi is not an element of any 
Q(Sqrt[Prime[n]]).
So: you should either explain what you mean by
covered or explain how it is that Pi _is_ an
element of some Q(Sqrt[Prime[n]]).)
>If you know any transfinite math ,
> you sure don't show it
>and then, picking fights over it when it isn't 
at issue.
>So My finite subset is real..
>Show me a real generator of his type that
>doesn't go complex?
>> That doesn't make much sense to me. What _you_
>> need to do is show us an element of SL(2,R) which
>> is not in the group generated by those matrices,
>> or vice versa.
>I'm trying to keep it real
>on more than one level..
> semantically speaking.
>Real number domain sets contain many finite sets..
>We are looking for an overall generator set that remains
real.
>We know that quadratic number fields are infinite 
but
countable,
>but they do remain in the real domain:
>R: a+b*Sqrt[Prime[n]]: real numbers a,b
>his x goes to--> a*x^2+c*x+b ( as a result of the detm=1
domain
>definition of SL(2,R))
>part of which is always complex , real numbers a,b,c.
>> Part of _what_ is always complex?
>If you say that x is defined as real then you have limited
the domain of
>{a,b,c} so that it is no longer a group under the axioms
of groups.
>So it is proved that his generators are not generators
of SL(2,R).
>> None of this makes much sense. Which probably has
something to do
>> with the fact that those matrices _do_ generate SL(2,R).
>Suppose we use all the quadratic number fields
Q(Sqrt[Prime[n]]) as the
>basis of the x ( a subset of the reals), but as individual
infinite sets
>with countable n. We, then, have an infinite set of 
infinite
sets of
>countable elements that
>approximates the real domain so well that you can't 
find a
number that
>isn't covered.
>> You can't? Let's see...
>> Pi is not covered. (Not that this has anything to do with
>> SL(2,R)...)
>It is still aleph zero, but you can't really tell it from
>aleph one in reality... not with a Turing machine in a
finite stopping
>time anyway. As far as I know that is the Chaitin kind of
definition of
>transfinites with is equivalent to the Cantor one.
>>Roger L. Bagula TOP-POSTED:

>Here's one that might give a real only basis
>>>that is based on golden mean number field.
>>>gm=(1+Sqrt[5])/2
>>>e1={{0,gm},{-1/gm,0}}
>>>e2={{gm,x},{0,1/gm}}
>>>x={gm,1/gm,1/gm^2}
>>>It is always SL(2,R)

>>These do not generated SL(2,R). SL(2,R) is
>>an uncountable group. Hence it cannot be generated
>>by a countable set of generators, let alone a finite
>>set such as this.

>and I think since the
>>>f[n]=mod[n*gm,1]=fractionalpart[n*gm]
>>>is an infinite set, that the result can cover the
>>>whole real domain, but only at an infinite level
>>>for n.

>>What you think is hardly relevant.

>> ************************
>> David C. Ullrich
************************
David C. Ullrich
===
Subject: Re: generators of SL(2,R)
Roger L. Bagula TOP-POSTED:
> erudition. But you were wrong...
lack of erudition. But you are still wrong.
--
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: generators of SL(2,R)--> answer!
I think you should try for clarification of the question
before making
pronouncements like this.
It is entirely possible that he is talking about something
you don't
know about.
Hyperbolic Poincare disk triangular tesselation of the unit
disk does
have three elements and does form a group, but it is usually
thought of
as a complex group and the metric is different than the one
he quoted.
I remembered where I saw that metric:
 A Walk Along the branches of the Extended Farey Tree by J>
C. Lagaris
and C. P. Tresser. IMB Journal of Research and Development
volume 39,
Number 3, May 1995
Section PSL(2,Z) and the Q tree page290
Called the tiling of H. J. S. Smith
The three generators are on page 291.
T1={{p1,p0},{q1,q0}}
T2={{-p0,p0+p1},{{q0,q0+q1}}
T3={{p0+p1,-p1},{q0+q1,-q1}}
where p1/q1, p0,q0 are rational numbers of the Feray type.
All of you were wrong and way off base...
>>I need to know the generators of Sl(2,R).
>>Someone over here gave me the 3 matrix generators.
>>But i lost it! (damn...)
>>Steve
> If you mean the group of all 2 x 2 matrices with
determinant 1, then 
the
> group doesn't have a set of just 3 generators since the
group is
> uncountable.
===
Subject: Re: generators of SL(2,R)--> answer!
Roger L. Bagula TOP-POSTED:
> I think you should try for clarification of the question
before making
> pronouncements like this.
I think you should learn to read and to understand simple
mathematical
questions before making pronouncements ike this.
> It is entirely possible that he is talking about something
you don't
> know about.
It is completely certain that he is talking about something
that
you don't know about/
> Hyperbolic Poincare disk triangular tesselation of the unit
disk does
> have three elements and does form a group, but it is
usually thought of
> as a complex group and the metric is different than the one
he quoted.
And it's not SL(2,R), the group the OP asked about.
> I remembered where I saw that metric:
>  A Walk Along the branches of the Extended Farey Tree by
J> C. 
Lagaris
> and C. P. Tresser. IMB Journal of Research and Development
volume 39,
> Number 3, May 1995
> Section PSL(2,Z) and the Q tree page290
> Called the tiling of H. J. S. Smith
> The three generators are on page 291.
> T1={{p1,p0},{q1,q0}}
> T2={{-p0,p0+p1},{{q0,q0+q1}}
> T3={{p0+p1,-p1},{q0+q1,-q1}}
> where p1/q1, p0,q0 are rational numbers of the Feray type.
Farey
Note this talks about PSL(2,Z), not SL(2,R) --- a different
group!
> All of you were wrong and way off base...
No, it's you who are wrong and off-base.
--
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: generators of SL(2,R)
I think your kind of math is just lost on Robin Chapman,
David R. Ullrich and Jose Carlos Santos.
As none has been able to give you an adequate honest answer.
I'm familiar with the geometry that you are taking about
which is found in thing like modular forms,
elliptical equations and Farey trees.
As for getting a set of three generators for a surface type
group,
I think none of them seems to have a clue as to what you are
talking
about. It is closer to complex dynamics and noneuclidean
geometry than
Lie algebras. but I' pretty certain you are looking for
SL(2,C) not
SL(2,R) from these equations below. There is an alternative
formulation
SL2(R) that I've seen in some modular form type papers.
These other fellows are so lost in showing off their great
knowledge and
just plain trolling nonsense that
they missed the point that you asked a good question that
they can't
answer adequately.
I would think that you may be looking for something like the
Farey
generators:
DEF bis4(x)
LET x1=1/2
LET y1=1
LET x= mod(x,1)
IF x>=0 and x <= x1 then LET y=x/(1-x)
IF x>x1 and x<=1 then LET y=(1-x)/x
LET bis4=y
END DEF
This is the real numbers that are related to the Feray
rational tree
between [0,1].
It has the inverse function ( x/(x+1),1/(x+1)), I think.
These kinds of groups are closer to Klein groups than matrix
groups
so an Lie algebra approach isn't what you want at all.
I don't know that any three element generator exists.
There is the six element anharmonic group that is in all the
books
and seems closely related to SL(2,2).
These three men aren't going to admit that you stumped them,
but I will. I don't know if such a three element group 
exists
or not.
It is certainly not the one Jose Carlos Santos quoted or
can not be made from it.
>>The question was to write down a set of generators of
SL(2,R),
>>and he did. I took it as read that when he meant those
matrices
>>in SL(2,R) of the form.
>> 1 x
>> 0 1.
>>If you are asked to name generators of a group, then I
think it 
reasonable
>>for people to assume that the elements you name are
intended to be in 
that
>>group.
>>Derek Holt.
> The Mobius transformation:
> Az = (az+b)/(cz+d)
> A in SL(2, R) ---> the group SL.
> z = x + iy
> ==> A(x + iy) = u + iv
> Find u and v...
> Then if you consider the upper half U^2 of R^2.
> With hyperbolic metric (dx^2 + dy^2) / y^2....
> Then to satisfy the action we must have that:
> (du^2 + dv^2) / v^2 = (dx^2 + dy^2) / y^2....
> For the second matrix gaven, this indeed the case!!
> (my mistake)
> A(x + iy) = x + r + iy
> u = x + r
> v = y
> du = dx
> dv = dy
> and (du^2 + dv^2) / v^2 = (dx^2 + dy^2) / y^2
> Steve
===
Subject: Re: generators of SL(2,R)
Roger L. Bagula TOP-POSTED:
> I think your kind of math is just lost on Robin Chapman,
> David R. Ullrich and Jose Carlos Santos.
No, it's lost on you.
> As none has been able to give you an adequate honest answer.
We all did.
> I'm familiar with the geometry that you are taking about
> which is found in thing like modular forms,
> elliptical equations and Farey trees.
ellitpical equations? no doubt these are for
elliptical curves
> As for getting a set of three generators for a surface type
group,
> I think none of them seems to have a clue as to what you
are talking
> about.
No, it's Bagula who has no idea.
> It is closer to complex dynamics and noneuclidean geometry
than
> Lie algebras. but I' pretty certain you are looking for
SL(2,C) not
> SL(2,R) from these equations below.
I expect he was looking at SL(2,R). See his title!
> There is an alternative formulation
> SL2(R) that I've seen in some modular form type papers.
> These other fellows are so lost in showing off their great
knowledge and
> just plain trolling nonsense that
You are the troll
> they missed the point that you asked a good question that
they can't
> answer adequately.
> I would think that you may be looking for something like
the Farey
> generators:
> DEF bis4(x)
> LET x1=1/2
> LET y1=1
> LET x= mod(x,1)
> IF x>=0 and x <= x1 then LET y=x/(1-x)
> IF x>x1 and x<=1 then LET y=(1-x)/x
> LET bis4=y
> END DEF
> This is the real numbers that are related to the Feray
rational tree
> between [0,1].
> It has the inverse function ( x/(x+1),1/(x+1)), I think.
> These kinds of groups are closer to Klein groups than
matrix groups
> so an Lie algebra approach isn't what you want at all.
> I don't know that any three element generator exists.
No three-element set of generators exists. The group
is uncountable. It has no countable generationg set.
> There is the six element anharmonic group that is in all
the books
> and seems closely related to SL(2,2).
> These three men aren't going to admit that you stumped 
them,
> but I will.
No, he stumped you.
> I don't know if such a three element group exists or not.
It does not.
> It is certainly not the one Jose Carlos Santos quoted or
> can not be made from it.
Jose gave a correct, but uncountable, generating set.
--
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: JSH: Posting really so bad?
>> I post what amazes to be absurd to worry about what I
get obsessed
> with I just post to post how I post for YEARS when it's
all exuses
> that ßy about reality trying to control what I post how I
post and
> say that's what they're trying to insult me 
nasty
webpages to any
> journals that dare change my postings to punish me into
not posting
> at all not normal people in an often coordinated one
that's really
> bad I talk about questioning their quo who questions the
quo and
> clearly with a hierarchy in constraining my sick
something wrong
> with you disturbed you disturbed then worrying frenetic
obsessive
> attacks all over the even by email something very wrong
with
> members the very values which hosts it!!!
 James Harris
> I've edited your post down to its main points again,
James.
> Please continue your wonderful posts, but strive to be
concise.
>> LOL!
>> Did you use dissociated press (or a similar tool), or just
take every
>> k-th word for some value of k?
> My conjecture is he did it the old-fashioned way, by hand.
You
> should check Google for his earlier work - the guy's 
simply
a
> genius, I've tried and I can't do anything 
like what he
does.
I second that. For a start, try his FLT proof (the proof
which was actually
*in the margin* all along...)
> ************************
> David C. Ullrich
===
Subject: Re: JSH: Posting really so bad?
 I post what amazes to be absurd to worry about what I
get obsessed
...
> members the very values which hosts it!!!
 James Harris
 I've edited your post down to its main points again,
James.
> Please continue your wonderful posts, but strive to be
concise.
>> LOL!
>> Did you use dissociated press (or a similar tool), or just
take every
>> k-th word for some value of k?
I just edited James like any good editor would: by tightening
up
his prose while retaining his main idea. I'd like to think
that
the result conveys the style of James's inner mental 
workings
better
than his original, turgid ramblings do.
> My conjecture is he did it the old-fashioned way, by hand.
You
> should check Google for his earlier work - the guy's 
simply
a
> genius, I've tried and I can't do anything 
like what he
does.
Aw shucks. I'm blushing. I am but a humble jeweler. James is
the diamond. My only insight is in realizing the futility of
scratching at him with the futile quartz tool known as logic.
> I second that. For a start, try his FLT proof (the proof
which was 
actually
> *in the margin* all along...)
Curiously, I discovered that proof by a similar editing
process.
I just took one of James's proofs and eliminated everything
that
wasn't mathematically valid.
--
Jim Ferry at U of Illinois, Urbana-Champaign, (no, I don't
educate)
2 email me l o o k up 1 row
===
Subject: Re:  f(f(x)) = x , everyone can create such
functions 
>We know functions - f:R->R ,continuous,bijective - like :
> f = 1/x , f = 2 -x , x /(x-1) ... 1)
Well, one of these is continuous from R to R.
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: Moebius Band
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 
1.9 primary) id i5RKUUU23935;
Parameterization of Moebuius Band as per Differential
Geometry  by H.W. Guggenheimer pp205, 1977, Dover is :
{R Cos[th]+u Sin[th/2], R Sin[th]-u Cos[th/2],u},
{u,-.5,.5},{th,0,2 Pi}.It is different from what is
given in the Wolfram MathWorld site :
{(R+u Cos[v/2]) Sin[v],(R+u Cos[v/2]) Cos[v],u
Sin[v/2]}.(th=v);
Is non-orientabilty alone a criterion, is there is no
standard form? The 
plot also looks somewhat different, ßatter than the latter.
===
Subject: Re: Moebius Band
> Parameterization of Moebuius Band as per Differential
> Geometry  by H.W. Guggenheimer pp205, 1977, Dover is :
> {R Cos[th]+u Sin[th/2], R Sin[th]-u Cos[th/2],u},
> {u,-.5,.5},{th,0,2 Pi}.It is different from what is
> given in the Wolfram MathWorld site :
> {(R+u Cos[v/2]) Sin[v],(R+u Cos[v/2]) Cos[v],u
Sin[v/2]}.(th=v);
> Is non-orientabilty alone a criterion, is there is no
standard form? 
The plot also looks somewhat different, ßatter than the
latter.
The standard Moebius band is a square with the edges
identified
in a certain way. Those two you refer to are ways of
embedding it in 3-space; they are merely ways to visualize
it, not the definition.
===
Subject: Re: Solving recurrence relation
> I have an unusual recurrence relation which I am trying to
solve,
> a(n) = ( 1 - sin[pi*(n+1)/2] ) * a(n/2) + 1 ; a(0) = 1
> Given the n/2 index on the r.h.s., I tried solving this via
the
> divide-and-conquer method, by letting n = 2^k. All well
and fine, 
but
> the solution derived only provides correct results for n
odd (since that
> results in a(n) = 1 for n odd) and n a power of 2; it does
not produce 
the
> correct results for n even (except when n is a power of 2).
How is a(n) for odd n defined?
> Now if I let n = 2k (to cover even indices):
> a(2k) = ( 1 - sin[pi*(2k+1)/2] ) * a(2k/2) + 1
> a(2k) = ( 1 - sin[pi*(2k+1)/2] ) * a(k) + 1
> a(2k) = ( 1 - (-1)^k ) * a(k) + 1
For even k, we have a(2k)=(1-1)*a(k) + 1=1. So a(n)=1
whenever n has two
factors of 2.
When n has just one factor of two, we get a(n)=2*a(n/2)+1.
So the sequence is well-defined as soon as you 
define its value
on all odd
numbers, but it is not very interesting.
> Its this last expression that presents a brick wall,
because it appears I
> cannot solve this with conventional methods. I've thumbed
through many
> books on combinatorics and have not seen anything that
discusses (or 
gives
> an example of) solving such r.r.'s which contains a mix of
odd and even
> (2k and k) indices.
> I'd very much like to solve this myself, but I need a
little guidance or
> hints here. Even if someone can point me to some literature
that would
> point me in the right direction, that would be great.
--
hang my head drown my fear
till you all just disappear
reverse my forename for mail! - saibot
===
Subject: Re: Solving recurrence relation
Tobias,
Y'know, I just screwed up on the expression. It should be
a(n) = sin[pi*(n+1)/2] * a(n/2) + 1 ; a(0) = 1
which would make a(n) = 1 for all n odd hold (sorry).
Granted, the index
n/2 is fractional for n odd, but this ends
up being irrelevant when n is odd since sin[pi*(n+1)/2] = 0
is this case.
--john
> I have an unusual recurrence relation which I am trying to
solve,
> a(n) = ( 1 - sin[pi*(n+1)/2] ) * a(n/2) + 1 ; a(0) = 1
> Given the n/2 index on the r.h.s., I tried solving this via
the
> divide-and-conquer method, by letting n = 2^k. All well
and fine,
but
> the solution derived only provides correct results for n
odd (since 
that
> results in a(n) = 1 for n odd) and n a power of 2; it does
not produce
the
> correct results for n even (except when n is a power of 2).
> How is a(n) for odd n defined?
> Now if I let n = 2k (to cover even indices):
> a(2k) = ( 1 - sin[pi*(2k+1)/2] ) * a(2k/2) + 1
> a(2k) = ( 1 - sin[pi*(2k+1)/2] ) * a(k) + 1
> a(2k) = ( 1 - (-1)^k ) * a(k) + 1
> For even k, we have a(2k)=(1-1)*a(k) + 1=1. So a(n)=1
whenever n has two
> factors of 2.
> When n has just one factor of two, we get a(n)=2*a(n/2)+1.
> So the sequence is well-defined as soon as you 
define its
value on all 
odd
> numbers, but it is not very interesting.
> Its this last expression that presents a brick wall,
because it appears
I
> cannot solve this with conventional methods. I've thumbed
through many
> books on combinatorics and have not seen anything that
discusses (or
gives
> an example of) solving such r.r.'s which contains a mix of
odd and even
> (2k and k) indices.
> I'd very much like to solve this myself, but I need a
little guidance 
or
> hints here. Even if someone can point me to some literature
that would
> point me in the right direction, that would be great.
> --
> hang my head drown my fear
> till you all just disappear
> reverse my forename for mail! - saibot
===
Subject: Re: Earliest example of an incomputable real
|>> There exists a TM such that [whatever] is not at
|>> all the same as We can give a procedure for constructing
|>> a TM such that [whatever].)
|But otoh there exists a TM such that [whatever] is
precisely
|enough defined that we can actually prove things about it,
|while the other notion gets a little fuzzy. (For example,
|it's true that there _is_ an algorithm that tells us
|_which_ algorithm solves RH, which shoots down one possible
|precise specification of what the other notion really is...)
Nonconstructive terminology makes expressing some of these
ideas
more awkward. The default constructive meaning of for each
Turing
machine A there exists a Turing machine B such that... is
closer
to this other notion. One small difference is that above
you seem
to be talking about a notion personalized to us, we can
give....
The sense in which one is tempted to misinterpret there
exists a
Turing machine that would indicate the truth or falsity of
the Riemann
hypothesis in this nonconstructive discussion is the default
constructive meaning.
Actually, one would be more likely to leave Turing machines
out of
the picture entirely and just say either the Riemann
hypothesis is
true or it's false, whose constructive meaning is
nontrivial. It
would be quite exciting if someone did prove it, in fact.
Also one
might well leave out the Turing machines from the discussion
of reals
and their decimal expansions.
The crux is that constructively, reals do not necessarily
have decimal
expansions, although if we assume the law of excluded middle
in the
special case of x>=0 or x<0 for reals, then it follows they
do. The
other mentioned facts are all variations on this same theme,
with
ingredients such as Turing machines mixed in. That last
principle is
equivalent to each sequence of natural numbers either having
a nonzero
term or not, and Bishop named this the limited principle of
omniscience (LPO). Allegedly LPO constructively suffices for
the
bulk of classical analysis, and is needed for much of it too.
You refer to precision of concepts. I think one should be
aware of
the bias presumably nearly everyone suffers from, of tending
to
regard concepts that are rigorously expressed in terms of
one's
favorite base theory as being precise.
Constructive terminology does not express itself properly in
terms of
nonconstructive terminology, which makes it harder to make a
persuasive
case for its being precise. Classical terminology can be
expressed in
constructive terms, but it's not clear whether the concepts
once
translated into constructive terms have really the same
meaning as
was originally intended.
It seems to me to be difficult to make a theory-neutral
evaluation
of precision of concepts. The difficulty seems like a
philosophical
difficulty. The important example in this context is the
degree of
precision we attribute to the concept of the existence of a
natural
number N having a certain well-defined property. Both
constructive
and nonconstructive versions of this concept strike their
users as
being sharp. They are well enough developed that each can be
used in
rigorous mathematics.
doesn't have to have anything to do with any ability to
actually find
such an n tends to seem suspect. The double-negative
translation, that
every n failing to have the property leads to a
contradiction, is
adequate up to a point. Some on the nonconstructive side find
it
suspicious that one doesn't feel the need to refer to Turing
machines
for the sake of sharpening the notion of being able to find an
n in
principle.
I don't feel like I have a good way of testing these
perceptions.
Keith Ramsay
===
Subject: Re: Earliest example of an incomputable real
>|>> There exists a TM such that [whatever] is not at
>|>> all the same as We can give a procedure for constructing
>|>> a TM such that [whatever].)
>|But otoh there exists a TM such that [whatever] is
precisely
>|enough defined that we can actually prove things about it,
>|while the other notion gets a little fuzzy. (For example,
>|it's true that there _is_ an algorithm that tells us
>|_which_ algorithm solves RH, which shoots down one possible
>|precise specification of what the other notion really is...)
>Nonconstructive terminology makes expressing some of these
ideas
>more awkward. The default constructive meaning of for each
Turing
>machine A there exists a Turing machine B such that... is
closer
>to this other notion. One small difference is that above
you seem
>to be talking about a notion personalized to us, we can
give....
>The sense in which one is tempted to misinterpret there
exists a
>Turing machine that would indicate the truth or falsity of
the Riemann
>hypothesis in this nonconstructive discussion is the default
>constructive meaning.
>Actually, one would be more likely to leave Turing machines
out of
>the picture entirely and just say either the Riemann
hypothesis is
>true or it's false, whose constructive meaning is
nontrivial. It
>would be quite exciting if someone did prove it, in fact.
Also one
>might well leave out the Turing machines from the discussion
of reals
>and their decimal expansions.
>The crux is that constructively, reals do not necessarily
have decimal
>expansions, although if we assume the law of excluded middle
in the
>special case of x>=0 or x<0 for reals, then it follows they
do. The
>other mentioned facts are all variations on this same theme,
with
>ingredients such as Turing machines mixed in. That last
principle is
>equivalent to each sequence of natural numbers either having
a nonzero
>term or not, and Bishop named this the limited principle of
>omniscience (LPO). Allegedly LPO constructively suffices for
the
>bulk of classical analysis, and is needed for much of it too.
>You refer to precision of concepts. I think one should be
aware of
>the bias presumably nearly everyone suffers from, of tending
to
>regard concepts that are rigorously expressed in terms of
one's
>favorite base theory as being precise.
>Constructive terminology does not express itself properly in
terms of
>nonconstructive terminology, which makes it harder to make a
persuasive
>case for its being precise. Classical terminology can be
expressed in
>constructive terms, but it's not clear whether the concepts
once
>translated into constructive terms have really the same
meaning as
>was originally intended.
>It seems to me to be difficult to make a theory-neutral
evaluation
>of precision of concepts. The difficulty seems like a
philosophical
>difficulty. The important example in this context is the
degree of
>precision we attribute to the concept of the existence of a
natural
>number N having a certain well-defined property. Both
constructive
>and nonconstructive versions of this concept strike their
users as
>being sharp. They are well enough developed that each can be
used in
>rigorous mathematics.
>doesn't have to have anything to do with any ability to
actually find
>such an n tends to seem suspect. The double-negative
translation, that
>every n failing to have the property leads to a
contradiction, is
>adequate up to a point. Some on the nonconstructive side find
it
>suspicious that one doesn't feel the need to refer to 
Turing
machines
>for the sake of sharpening the notion of being able to find
an n in
>principle.
That's interesting - to tell you the truth the idea that the
other notion in question was what constructivists study had
not occured to me.
I'm not sure whether your comments are supposed to be
contradicting
or amplifying what I said. In any case, I'll just clarify 
that
the other notion gets a little fuzzy was supposed to mean
that the notion seems fuzzy to me (which it does - I honestly
have no precise idea what I was talking about, qed), not that
it's necessarily forever imprecise, or imprecise to anyone
but me.
>I don't feel like I have a good way of testing these
perceptions.
>Keith Ramsay
************************
David C. Ullrich
===
Subject: Re: Solving recurrence relation w/ odd and even
indices
>I have an unusual recurrence relation which I am trying to
solve,
> a(n) = ( 1 - sin[pi*(n+1)/2] ) * a(n/2) + 1 ; a(0) = 1
>Given the n/2 index on the r.h.s., I tried solving this via
the
>divide-and-conquer method, by letting n = 2^k. All well
and fine, 
but
>the solution derived only provides correct results for n odd
(since that
>results in a(n) = 1 for n odd) and n a power of 2; it does
not produce the
>correct results for n even (except when n is a power of 2).
Is a(n) supposed to be defined only for integers n?
If not, what does your relation mean when, say, n=1?
But I'll accept your statement that a(n) = 1 for n odd, and
just use the recurrence for n even.
If n is even and n/2 is odd, say n/2 = 2k+1 so n=4k+2,
sin(pi (n+1)/2) = sin(pi (2k+3/2)) = sin(3 pi/2) = -1
so a(n) = 2 a(n/2) = 2.
If n is even and n/2 is even, say n/2 = 2k so n=4k,
sin(pi (n+1)/2) = sin(pi (2k+1/2)) = sin(pi/2) = 1
so a(n) = 1.
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: Re: Solving recurrence relation w/ odd and even
indices
Robert,
a(n) is defined for all positive integers
Y'know, I just screwed up on the expression. It should be
a(n) = sin[pi*(n+1)/2] * a(n/2) + 1 ; a(0) = 1
which would make a(n) = 1 for all n odd hold (sorry).
Granted, the index
n/2 is fractional for n odd, but this ends
up being irrelevant when n is odd since sin[pi*(n+1)/2] = 0
is this case.
--john
>I have an unusual recurrence relation which I am trying to
solve,
> a(n) = ( 1 - sin[pi*(n+1)/2] ) * a(n/2) + 1 ; a(0) = 1
>Given the n/2 index on the r.h.s., I tried solving this via
the
>divide-and-conquer method, by letting n = 2^k. All well
and fine, 
but
>the solution derived only provides correct results for n odd
(since that
>results in a(n) = 1 for n odd) and n a power of 2; it does
not produce
the
>correct results for n even (except when n is a power of 2).
> Is a(n) supposed to be defined only for integers n?
> If not, what does your relation mean when, say, n=1?
> But I'll accept your statement that a(n) = 1 for n odd, 
and
> just use the recurrence for n even.
> If n is even and n/2 is odd, say n/2 = 2k+1 so n=4k+2,
> sin(pi (n+1)/2) = sin(pi (2k+3/2)) = sin(3 pi/2) = -1
> so a(n) = 2 a(n/2) = 2.
> If n is even and n/2 is even, say n/2 = 2k so n=4k,
> sin(pi (n+1)/2) = sin(pi (2k+1/2)) = sin(pi/2) = 1
> so a(n) = 1.
> 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: Re: Solving recurrence relation w/ odd and even
indices
>a(n) is defined for all positive integers
>Y'know, I just screwed up on the expression. It should be
> a(n) = sin[pi*(n+1)/2] * a(n/2) + 1 ; a(0) = 1
>which would make a(n) = 1 for all n odd hold (sorry).
Granted, the index
>n/2 is fractional for n odd, but this ends
>up being irrelevant when n is odd since sin[pi*(n+1)/2] = 0
is this case.
OK, then: if n = 2^j m with j >= 1 and m odd,
a(n) = sin((2^(j-1) m pi + pi/2) a(n/2) + 1
= -a(n/2) + 1 if j = 1, or a(n/2)+1 if j > 1
from which we get a(n) = j-1 for j >= 1.
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: Re: LAPACK FORTRAN code to C?
netlib.org :( I'm going to try use that.
--
Jay Liew
sm AT x DOT jaysern D0T 0RG
===
Subject: Re: LAPACK FORTRAN code to C?
> The original codes for many of these routines were written
in ALGOL and 
can be
> found in the Handbook for Automatic Computation, Linear
Algebra. 
There are
> also a few extensive C libraries which cover similar
ground, including 
A
> Numerical Library in C for Scientists and Engineers, by
Lau, CRC 
Press, and
> the Meschach library available from netlib.
> LAPACK was basically a rewrite of the older libraries
LINPACK and
> EISPACK. It may be that the underlying algorithms were
originally coded
> in ALGOL, but LAPACK was coded in Fortran 77 and was
designed from the
> start to make extensive use of the BLAS library. Neither
f77 nor BLAS
> existed in the heyday of ALGOL.
(hqr) from HACLA which later on became the FORTRAN codes
almost note for 
note.
--
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: A strange series
> Hi Rob,
>I have just sent a mistake! There are no n such that
a(n)<m!
>Replace this argument by : for the n such that a(n)<a(0),
the sequence
>(a(n)^2/A(n)) is bounded (A(n)-> infinity). And for the n
such that
>a(n)>a(0) .... (with a(0) instead of m).
 I am not sure I follow here. I understand that a_n >= m for
all n, so
> you are replacing the condition a_n < m by a_n < a_0. For
the n where
> that condition holds, a_n^2/A_n is trivialy bounded by a_0
(a_n <= a_0
> and A_n >= a_0). However, I missed where you handled the
case in which
> a_n > a_0.
> I made a small mistake again!
> First, you can suppose A(n)->infinity. Thus, for the n such
that
> a(n)<2a(0), the sequence a(n)^2/A(n)->0 so (a(n)^2/A(n)) is
bounded.
> Then, for the n such that a(n)>=2a(0), we consider
> k=ßoor((a(n)-a(0))/M), so kM<= a(n)-a(0)<(k+1)M. Note that
k<=n,
> since a(n)-a(0)<=nM.
> Then A(n)>=a(n)+ a(n)-M + a(n)-2M+ ..+
a(n)-kM=(k+1)a(n)-k(k+1)M/2 (a(n)-a(0))(a(n)+a(0))/(2M)= a(n)^2(1-(a(0)/a(n))^2)/(2M)>=3
> a(n)^2/(8M)
> Finally, a(n)^2/A(n)<=8M/3.
> This is +- your proof.
> Friendly,
> Georges
I've lost track of exactly what you're trying 
to prove.
Perhaps you
could clarify. But maybe I can help get to the heart of the
matter.
I think we're assuming a_n>0. If lim sup a_n is 
finite, then
we're
done: a_n itself is bounded, _a fortiori_ a_n/sqrt(A_n).
So suppose lim sup a_n = infinity. What are we worried about?
We're
worried that a_n grows so fast that it's a 
significant
proportion of
A_n. For example, a_n = 2^n, whence A_n = 2^{n+1}-1, and so
a_n/sqrt(A_n) = O(2^(n/2)) is not bounded. What if a_n has
polynomial
growth in n? Then what about A_n? Don't forget that lim inf
a_n might
be zero, unless you assume otherwise.
To put it in perpective, what I proved a week ago was
boundedness of
a_n/sqrt A_n in the case that a_n had linear growth and
lim sup a_n = lim inf a_n = infinity.
Hope this helps,
Nick
===
Subject: Re: Fermat Algebra
3^2 = (1+1)*(1+2) + 3
4^2 = (1+1)*(1+2+3) + 4
5^2 = (1+1)*(1+2+3+4) + 5
[...]
3^2 = (1+1)*(1+2) +3
3^3 = (1+2+3)*(1+3) + 3
3^5 = (1+2+3)*(1+2+3+4)*(1+3) + 3
3^7 = (1+2+3+4+5+6+7+8+9+10+11+12+13)*(1+2+3)*(1+3) +3
5^2 = 4*5/2 + 5*6/2
13^2 = 12*13/2 + 13*14/2
17^2 = 16*17/2 + 17*18/2
25^2 = 24*25/2 + 25*26/2
5^2 = 10+15
13^2 = 78+91
17^2 = 136+153
25^2 = 300+325
1+2+3+...+N = N*[N+1]/2
X^2 = X*[X-1]/2 + X*[X+1]/2
X^3 = X*[X^2-1]/2 + X*[X^2+1]/2
X^4 = X*[X^3-1]/2 + X*[X^3+1]/2
X^n = X*[X^(n-1) -1]/2 + X*[X^(n-1) + 1]/2
5*2-1 = 3^2
5*3+1 = 4^2
[5*2-1] + [5*3+1] = 5*2 + 5*3 = 5*[2+3] = 5^2
[13*2-1] + [13*11+1] = 5^2 + 12^2 = 13^2
[17*4 - 4] + [17*13 + 4] = 17^2
Z*U + K = X^2
Z*V - K = Y^2
[Z*U + K] + [Z*V - K] = Z*U + Z*V = Z*[U + V] = Z*Z = Z^2
= X^2 + Y^2
Z = U + V
3*(3+4) + 4*|4-3| = 5^2
3*(3+4) + 4*1 = 5^2
3*(3+4) + 4*(1 + 5^2) = 5^3
3*(3+4) + 4*(1 + 5^2 + 5^3) = 5^4
3*(3+4) + 4*(1 + 5^2 + 5^3 + 5^4) = 5^5
etc...
5*(5+12) + 12*|12 - 5| = 13^2
5*(5+12) + 12*(7 + 13^2) = 13^3
5*(5+12) + 12*(7 + 13^2 + 13^3) = 13^4
etc...
The equation? :
p is a prime number > 2.
z^p = x*(x+y) + y*( |y-x| +...+ z^(p-1) )
a^3 + b^3 = (a+c)^3
a^3 + b^3 = (b+d)^3
a^3 + b^3 = a^3 + 3ac^2 + 3ca^2 + c^3
a^3 + b^3 = b^3 + 3bd^2 + 3db^2 + d^3
b^3 = 3ac^2 + 3ca^2 + c^3
a^3 = 3db^2 + 3bd^2 + d^3
a^3 - d^3 = 3db^2 + 3bd^2
b^3 - c^3 = 3ca^2 + 3ac^2
b+d = a+c
(a^3+b^3) - (c^3+d^3) = 3ac(a+c) + 3bd(a+c)
(a^3+b^3) - (c^3+d^3) = 3*(a+c)*(ac+bd)
===
===
Subject: Some more derived quantities of mechanics
> Mechanics is the science concerning discrete portions -
objects,
> bodies and masses - of material matter moving relative to
eash other
> during time:
> There are only three fundamental measures essential to
describing
> mechanics; they are: Length; Force, and Duration.
> Length is a linear measure of distension: The distance
something
> extends - as its length width and height; or the distance
something
> moves, or is separated - as the distance between centers of
masses.
> Force is the measure of the pressure exerted between masses
of matter
> when they try to simultaneously occupy and/or pass through
the exact
> same place. This is called impenetrability: Where two or
more masses
> will displace each other; at rates that vary inversely as
the quantity
> of matter each contains. The particular force exerted by
masses
> resting on the terra firma surface of planets like Earth
is called
> 'weight'.
> Duration is the measure of time: A measure of the period of
time
> during which changes occur, and force endures.
> Mass is the measure of matter contained in an object; body,
or mass of
> it, derived as the ratio of the [net] force [f] exerted
on/and/or by
> it, to the rate of displacement [s/t^2] that it causes.
> Space is the place, or room in which matter has a place to
exist, and
> move around in.
Virtually all of the other quantities of mechanics - possibly
with
some exceptions - are derived from the three essential
quantities -
Length; Force, and Duration:
Fer-instance: Speed [v] is derived by dividing the
displacement
distance [l] by the duration [t] during which it occurs, and
velocity
requires specifying a direction for the speed.
Acceleration [a] is a rate of change in speed and/or
direction: Where
the speed and/or direction of a body's motion changes from
one initial
rate (vi), to another rate (vt) during a period of time (t):
So that
the average rate of change is [(vt-vi)/(2t) = v^2/(2r); and is
proportional to the [net] force [f] causing it.
Momentum is the quantity of motion; in that the motion of
[2mv] is
twice as much as [1mv]:
===
Subject: Re: Some more derived quantities of mechanics
> Momentum is the quantity of motion; in that the motion of
[2mv] is
> twice as much as [1mv]:
idiot
Mike
===
Subject: Re: Some more derived quantities of mechanics
> Mechanics is the science concerning discrete portions -
objects,
> bodies and masses - of material matter moving relative to
eash other
> during time:
 There are only three fundamental measures essential to
describing
> mechanics; they are: Length; Force, and Duration.
 Length is a linear measure of distension: The distance
something
> extends - as its length width and height; or the distance
something
> moves, or is separated - as the distance between centers of
masses.
 Force is the measure of the pressure exerted between masses
of matter
> when they try to simultaneously occupy and/or pass through
the exact
> same place.
No, pressure is the force applied uniformly over a surface,
measured
as force per unit of area.
Force is a vector quantity that tends to produce an
acceleration of a
body in the direction of its application.
Double-A
===
Subject: Re: Some more derived quantities of mechanics
> Mechanics is the science concerning discrete portions -
objects,
> bodies and masses - of material matter moving relative to
eash other
> during time:
 There are only three fundamental measures essential to
describing
> mechanics; they are: Length; Force, and Duration.
 Length is a linear measure of distension: The distance
something
> extends - as its length width and height; or the distance
something
> moves, or is separated - as the distance between centers
of masses.
 Force is the measure of the pressure exerted between
masses of matter
> when they try to simultaneously occupy and/or pass
through the exact
> same place.
> No, pressure is the force applied uniformly over a surface,
measured
> as force per unit of area.
> Force is a vector quantity that tends to produce an
acceleration of a
> body in the direction of its application.
> Double-A
Pressure is the _total_ force, divided by the area over which
it is
applied, and varies depending on the area over which it is
spread: But
when the pressure isn't uniformly spread; but is 
concentrated
at
points: Then its imperative to see that force is pressure
exerted over
a specific (unit) area, and must not be allowed to exceed the
allowable bearing strength of the material it bears on, or it
may
crush, pierce or penetrate.
Incidentally, is it alright if I call that double-talk from
Double-A(;^?
===
Subject: Re: Some more derived quantities of mechanics
> Mechanics is the science concerning discrete portions -
objects,
> bodies and masses - of material matter moving relative to
eash other
> during time:
 There are only three fundamental measures essential to
describing
> mechanics; they are: Length; Force, and Duration.
 Length is a linear measure of distension: The distance
something
> extends - as its length width and height; or the distance
something
> moves, or is separated - as the distance between centers of
masses.
 Force is the measure of the pressure exerted between masses
of matter
> when they try to simultaneously occupy and/or pass through
the exact
> same place. This is called impenetrability: Where two or
more masses
> will displace each other; at rates that vary inversely as
the quantity
> of matter each contains. The particular force exerted by
masses
> resting on the terra firma surface of planets like Earth
is called
> 'weight'.
 Duration is the measure of time: A measure of the period of
time
> during which changes occur, and force endures.
 Mass is the measure of matter contained in an object; body,
or mass of
> it, derived as the ratio of the [net] force [f] exerted
on/and/or by
> it, to the rate of displacement [s/t^2] that it causes.
 Space is the place, or room in which matter has a place to
exist, and
> move around in.
> Virtually all of the other quantities of mechanics -
possibly with
> some exceptions - are derived from the three essential
quantities -
> Length; Force, and Duration:
> Fer-instance: Speed [v] is derived by dividing the
displacement
> distance [l] by the duration [t] during which it occurs,
and velocity
> requires specifying a direction for the speed.
> Acceleration [a] is a rate of change in speed and/or
direction: Where
> the speed and/or direction of a body's motion changes from
one initial
> rate (vi), to another rate (vt) during a period of time
(t): So that
> the average rate of change is [(vt-vi)/(2t) = v^2/(2r); and
is
> proportional to the [net] force [f] causing it.
> Momentum is the quantity of motion; in that the motion of
[2mv] is
> twice as much as [1mv]:
That particular spiral curve called an involute changes it's
length
and curvature at a uniform rate; that allows it to act as a
closing
curve between equidistant points along the arc and its
tangent from
the apex, and closing the entire vertex.
The tangent, or slope of this same involute spiral curve on a
distance/time graph changes its slope at a unform rate as it
moves
from one point to the along the curve.
===
Subject: Re: conformal mapping
> I appreciate any help or advise how to solve this problem.
> Assume that sqrt(z) (square root of z) is given in terms of
its principle 
branch, 0<=arg(z)<2*Pi. Find the image of the half-disc
D^+={(x,y) in C | 
x^2+y^2<1, 0<y<+infty} under map w=1/2(sqrt(z)+sqrt(1/z))
I would suggest seeing what the map does to the points 0, 1,
-1, and
i. Since conformal maps map lines (or circles) to lines (or
circles),
and 1, 0, -1 are co-linear, their image will uniquely
determine part
of the boundary. Similarly for the arc through 1, i, and -1.
'cid 'ooh
===
Subject: Re: conformal mapping
> I would suggest seeing what the map does to the points 0,
1, -1, and
> i. Since conformal maps map lines (or circles) to lines (or
circles),
> and 1, 0, -1 are co-linear, their image will uniquely
determine part
> of the boundary. Similarly for the arc through 1, i, and -1.
I don't think this is correct.
A conformal map of the special form f(z) = (az+b)/(cz+d) has
this property,
but not a general conformal map.
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2,
Ireland
===
Subject: Re: conformal mapping
> I would suggest seeing what the map does to the points 0,
1, -1, and
> i. Since conformal maps map lines (or circles) to lines (or
circles),
> and 1, 0, -1 are co-linear, their image will uniquely
determine part
> of the boundary. Similarly for the arc through 1, i, and -1.
> I don't think this is correct.
> A conformal map of the special form f(z) = (az+b)/(cz+d)
has this 
property,
> but not a general conformal map.
D'oh, you're right. This should have been 
obvious to me since
all
complex holomorphic functions are conformal everywhere the
derivative
is non-zero...
'cid 'ooh
===
Subject: Re: conformal mapping
> I appreciate any help or advise how to solve this problem.
> Assume that sqrt(z) (square root of z) is given in terms of
its principle 
> branch, 0<=arg(z)<2*Pi. Find the image of the half-disc
D^+={(x,y) in C | 
> x^2+y^2<1, 0<y<+infty} under map w=1/2(sqrt(z)+sqrt(1/z))
That's not the principal branch. Anyway, I'd 
look at what the
map does to
the unit interval, the upper unit circle, and the negative
unit interval.
===
Subject: Re: conformal mapping
> I appreciate any help or advise how to solve this problem.
> Assume that sqrt(z) (square root of z) is given in terms of
its principle
> branch, 0<=arg(z)<2*Pi. Find the image of the half-disc
D^+={(x,y) in C |
> x^2+y^2<1, 0<y<+infty} under map w=1/2(sqrt(z)+sqrt(1/z))
I think the image is the quarter-plane {(x,y): x<0, y>0}.
With your definition, if z = r exp(it) then 1/z = 1/r
exp(i(2pi - t))
and so
sqrt(z) = sqrt(r) exp(it/2),
sqrt(1/z) = - 1/sqrt(r) exp(-it/2) = - 1/sqrt(z).
Hence
Re(w) = cos(t/2) (sqrt(r) - sqrt(1/r)) < 0
Im(w) = sin(t/2) (sqrt(r) + sqrt(1/r)) > 0.
Conversely suppose Re(w) < 0, Im(w) > 0.
Trying to solve for u = sqrt(z),
w = 1/2 (u - 1/u),
ie
u^2 - 2uw - 1 = 0.
This has 2 solutions u,v satisfying uv = -1, u+v = 2w.
Unless |u| = |v| = 1, just one solution will satisfy |u| < 1.
The other solution is then v = -1/u.
If Re(u) < 0 then Re(-1/u) > 0 and Re(w) > 0. So Re(u) > 0.
If Im(u) < 0 then Im(-1/u) < 0 and Im(w) < 0. So Im(u) > 0.
Thus u lies in the first quadrant and so z = u^2 lies in D.
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2,
Ireland
===
Subject: Re: Fermat Agebra
>You swallowed d^3 and c^3
>a^3 + b^3 = 3ca^2 + 3ac^2 + c^3 + 3db^2 + 3bd^2 + d^3
>Dirk Vdm
> a^3 + b^3 = (a+c)^3
> a^3 + b^3 = (b+d)^3
> a^3 + b^3 = a^3 + b^3
> (a+b) = (b+d)
----- ^ ------------------
don't you mean:
(a+c) = (b+d) ??
---------------------------
> substitution
===
===
Subject: Re: entire function
>>n! is a red herring. Given values v0, v1, ..., and z1, z2,
... with 
|zn|
>>->oo, you can construct an entire function f such that f(zn)
= vn. Hint 
for
>>proof: Assume for simplicity that |z1| < |z2| < ... For
each n > 1 try 
to
>>find a polynomial Pn such that Pn vanishes at z1, ..., zn,
|Pn(z)| is 
>>small
>>for all z with |z| <= |zn| and P(z_(n+1)) = v_(n+1). Such a
polynomial
>>might look like
Cn*(z-z1)*(z-z2)*...*(z-zn)*(z/z_(n+1))^(m(n)) for an 
>>appropriate constant Cn and power m(n). Then add all these
Pn's up to 
get
>>f.
 Can you prove that this sum converges for any z other than
the given 
zn,
>> where the sum is finite?
>Sure, because small means as small as you like, say <
2^(-m). 
Therefore
>P1 + P2 + ... converges uniformly on each disc centered at 0
of radius
>|zn|, hence on compact subsets of C, hence to an entire
function. Details 
>on request.
> I see what you are doing. Make P_n small where |z| <
|z_{n+1}| by using
> a large m_n. This forces C_{n+1} to be big to counteract
the huge
> contribution that (z/z_{n+1})^{m_n} has near z_{n+2}.
Nothing is done
> to control the effect of (z/z_{n+1})^{m_n} where |z| >
|z_{n+1}|, except
> at z_{n+2}, etc, because there are finitely many summands.
So the
> function may behave wildly in between the z_n, but the sum
will converge
> for all z and will have f(z_n) = v_n.
> So, yes; such a function exists. However, actually
constructing this
> function seems a Herculean task using this method.
No actually it's quite natural and simple: We 
find polynomials
Pn so that
|Pn| < 1/2^n on {|z| <= |z_n|}, Pn = 0 on {z_1, ..., z_n},
and Pn is
anything we like at z_(n+1). Then we add them up. It's a
breeze and very
adaptable. (It can also be adapted to handle |z1| <= |z2| <=
...)
===
Subject: Re: Gravity
>> I have what is probably a strange question and I'm not
quite sure how
>> to phrase it. Essentially I'm writing a 
sci-fi short story
and need 
to
>> have an object on earth that is about the size of my fist,
yet so 
dense
>> that there is a noticeable gravitational pull on anyone or
anything 
who
>> gets near it. In fact, by the time someone is within about
half a
>> meter of it, the pull would be so strong, you would not be
able to
>> escape it.
 I have a few questions about this.
 a) Is this possible now? Is there some substance like this?
>Ignoring that, have you considered how heavy such a mass
would be? It's 
>on the order of 200 million tons. How would you support it?
If you
>didn't, what would keep it from sinking to the center of 
the
Earth?
>Answer those and you're ready to go on.
>> b) If it's not possible now, might it ever be possible?
Why? why not?
>> c) Am I totally out of my mind?
>> d) Am I asking this in the wrong group? If so, what group
should I be
>> asking this in.
>This is a good place.
> Nonsense. If he wants good answers to a science question,
sci.physics
> is much better than alt.sci.physics.
Ask _both_ groups, and sci.math; with one post. Nygaard likes
to make
a big deal to get all the attention he can.
===
Subject: Re: Terminal Velocity without drag?
> Why is c the speed limit?
 approaches c?
>
...
 How would you define a tachyon, and what are its properties?
 http://scienceworld.wolfram.com/physics/Tachyon.html
 and tachyon speed are reciprocal.
> Yeah. It's kinda like that. :-)
Why is that so?
travels, in a way, at a speed that is basically from c on out
to
infinity.
The speed of light, c, reminds of the Planck length, scripty
h, and
scripty h slash, the Planck length divided by 2pi. Then
there's the
fine structure constant and stuff.
There is a lot of utility in the application of the
mathematics of
geometric algebras with discussing the space-, time-, and
light-like
vectors, with the 3-1 or 3-2 projections. How many bases
does light
get?
A travelling photon has a frequency. It's generally among
zillions of
other photons ßowing in all directions from the
omnidirectional
source, in a cone from the cone light source, or in a single
line from
the unidirectional point source. The massy, charged, stream of
photons ßoods through the cornea and ßuids of the eyeball to
the
sensitive photorecepting rods and cones of the eye, there
converted to
electrical signals, carried by chemical messengers from
endpoint to
endpoint of the neurons of the thick cable of the optic nerve
directly
into the visual cortex, a clump of a biological network
(neural
network) that is interpreted when visual from the upside-down
image
focussed on the retina to what is then interpreted by the
brain as a
continuously variable image, because natural light is not
coherent.
The photon, with charge measured in electron Volts, eV's,
has about
the same charge as an electron, a difficult kind of subatomic
The standard model of physics has the elements being those
masses of
atoms with equal atomic numbers, the atomic number is the
number of
protons, or comparatively massy to the negative orbiting
electrons composites of, here I got lost, something about
hadrons or
muons or leptons, quarks and quarks of the standard
model, which
follow some proportional constants of multiples of three in
diminishing mass.
At some point in time the mass diminishes so much that the sum
the mass and energy present in that continuous section of
the
coordinate space equate to represent the charged, oscillating,
inertial potential or actuality of the non-empty space.
frequency, the frequency goes to infinity, with 
fixed energy.
As the
What is local frequency adjustment zone.
The neutron is often found within the nucleus of the
atom, in some
respects the neutron is a proton infinitely compacted with an
electron, but the mass and charge of the proton and
electron don't
_sum_ to exactly that of the neutron. In some forms of
radioactive
decay the neutron decomposes to an electron and proton pair,
changing
the atom from being one element to another, in increasing the
atomic
number, but not atomic mass, of the atom.
Within the standard model there are various considerations of
gravity,
strong and weak, and other forces that act upon the
mass/energy that
fill the infinite three-dimensional space that 
characterizes the
observable universe.
I guess I don't yet understand why there are any boundaries
like the
Planck length, or rather, what physical effect occurs at those
boundaries. Can not light vary continuously in frequency from
one to
the next?
Ross F.
===
Subject: Re: limitation to induction on finite bounds
>> Don't try changing the subject. The fact is that every
sequence in the 
list
>> has an infinite tail of 0s, and that 333... does not. Thus
333... is not 
in
>> the list. Do you accept that now?
>the whole is greater than the sum of parts
Not in set theory it isn't. Have you ever heard of
extensionality? I
didn't think so. Maybe if you read some decent introductory
book on set
theory you wouldn't spout so much rubbish (at least on that
subject).
>> Note that I am talking purely about sequences of digits;
interpreting 
them as
>> real decimal fractions is neither necessary nor relevant
to the point I 
am
>> trying to get you to see: that the sequence 333.... is
nowhere in the 
given
>> list of sequences.
>its an in depth analysis of containment of 1/3 in the
sequence
>{0.3, 0.33, 0.333, ...} so its the same subject.
Whatever you are trying to mean by containment, it is
clearly different 
from
being a member of the list. The subject now appears to be
your 
aggressive
ignorance about basic mathematical notions and forms of
reasoning. The good 
news is that it's curable; the bad news is that you 
first have
to recognise 
it
needs curing.
Just in case you feel like exposing yourself to some actual
knowledge, have 
a
look at
<http://planetmath.org/encyclopedia/
AxiomOfExtensionality.html>.
[further Herculean rubbish snipped]
--
===
Subject: Re: limitation to induction on finite bounds
>> Don't try changing the subject. The fact is that every
sequence in the 
list
>> has an infinite tail of 0s, and that 333... does not. Thus
333... is 
not in
>> the list. Do you accept that now?
>the whole is greater than the sum of parts
> Not in set theory it isn't. Have you ever heard of
extensionality? 
I
> didn't think so. Maybe if you read some decent 
introductory
book on set
> theory you wouldn't spout so much rubbish (at least on 
that
subject).
>> Note that I am talking purely about sequences of digits;
interpreting 
them as
>> real decimal fractions is neither necessary nor relevant
to the point I 
am
>> trying to get you to see: that the sequence 333.... is
nowhere in the 
given
>> list of sequences.
>its an in depth analysis of containment of 1/3 in the
sequence
>{0.3, 0.33, 0.333, ...} so its the same subject.
> Whatever you are trying to mean by containment, it is
clearly 
different from
> being a member of the list. The subject now appears to be
your 
aggressive
> ignorance about basic mathematical notions and forms of
reasoning. The 
good
> news is that it's curable; the bad news is that you 
first
have to 
recognise it
> needs curing.
> Just in case you feel like exposing yourself to some actual
knowledge, 
have a
> look at
<http://planetmath.org/encyclopedia/
AxiomOfExtensionality.html>.
You sure get adamant I answer every question when you snip
dozens of mine.
Its impossible to discuss anything with you Barb, I'm lucky
to get a 
question
I want anwered in 6 months from you because you pop in with a
text book 
quote,
degrade all the work, ignore everything then choof off until
next month, 
usually
grumbling 'i'm right he's 
wrong'.
I did get this out of you last episode.
> but is can have all the finite prefixes.
> 0.2
> 0.22
> 0.222
> 0.2222
> ..
> the diagonal of this subset of the computable numbers is
anti_diag 
hence
> there is no finite sequence of digits of anti_diag not
present on the 
list
> of computable numbers.
> So what?
BARB WRITES
>there are an infinite set of digits of anti_diag not present
on the list 
of
>computable numbers. Is that right?
Yes, if by set you actually mean sequence;
So which is it,
all digit strings from anti_diag are on the list of
computable numbers?
OR
there are digit strings from anti_diag which are not on the
list of 
computable numbers?
********* answer goes here ********
Now check this if you can, I'm sure you could shorten my 
defn.
Definition and example {0.3, 0.33, 0.333, ...} setminus 1/3
S setminus r = d
Ax x e S, 0.3, 0.33, 0.333
Ay y e T,
abs(x - r) = y, 0.03.., 0.003.., 0.0003.. = y
Ed d e R, get the 'minimum' of y
Ab b e W,
Az z e T,
Av v e T,
d<=z, d is less than all y
b<=v, b is less than all y
Aa a e W,
d>=a
Because we are dealing with infinite sets, something you have
no grasp on,
we alter the meaning of membership.
For a numerical system.
a e S <-> S setminus a = 0
I see no problem with this, set membership now includes the
limits of 
infinite sets
using a simple logical framework to encapsulate infinite sets.
Does this set include a mapping to the digit 1?
9 = (0.9, 0.99, 0.999, 0.9999, ... }
9 setminus 0.9 = 0 0.9 is drawn on the number line
9 setminus 0.99 = 0 0.99 is drawn on the number line
9 setminus 0.98 = 0.01 0.98 is NOT drawn on the number line
9 setminus 1 = 0
Obviously 1 is drawn on the number line along with all the
other members,
just by using a different interpretation of membership.
Is there a number between all the points drawn on the number
line usually 
and 1?
no, it makes sense that the INFINITE set draws the number 1.
Herc
===
Subject: Re: limitation to induction on finite bounds
Herc,
is {0.9, 0.99, 0.999, ...} = 1 ?
If so, where is 1 on the list?
===
Subject: Re: limitation to induction on finite bounds
> the diagonal of this subset of the computable numbers is
anti_diag 
hence
**********************************************************
> there is no finite sequence of digits of anti_diag not
present on the 
list
> of computable numbers.
> So what?
> BARB WRITES
>there are an infinite set of digits of anti_diag not
present on the 
list of
>computable numbers. Is that right?
> Yes, if by set you actually mean sequence;
> So which is it,
> all digit strings from anti_diag are on the list of
computable 
numbers?
> OR
> there are digit strings from anti_diag which are not on the
list of 
computable numbers?
> ********* answer goes here ********
<unsnip question for Barb>
there is no finite sequence of digits of anti_diag not present
on the 
list
of computable numbers.
Will : yes
there are an infinite set of digits of anti_diag not present
on the list 
of
computable numbers.
Barb : Yes, if by set you actually mean sequence
You can see why I want this clarified.
Herc
===
Subject: Re: limitation to induction on finite bounds
In sci.math, |-|erc
<gotcha@beauty.com>
<TmKDc.68080$sj4.55279@news-server.bigpond.net.au>:
[snip for brevity]
> Does this set include a mapping to the digit 1?
> 9 = (0.9, 0.99, 0.999, 0.9999, ... }
> 9 setminus 0.9 = 0 0.9 is drawn on the number line
> 9 setminus 0.99 = 0 0.99 is drawn on the number line
> 9 setminus 0.98 = 0.01 0.98 is NOT drawn on the number line
> 9 setminus 1 = 0
> Obviously 1 is drawn on the number line along with all the
other members,
> just by using a different interpretation of membership.
> Is there a number between all the points drawn on the
number line
> usually and 1? no, it makes sense that the INFINITE set
draws
> the number 1.
> Herc
So are you suggesting that the set
T = {k / 10^n, 0 <= k < 10^n, n > 0, n, k in J}
draws the entire line segment from 0 to 1, inclusive?
After all, T setminus r = 0 for any r, 0 <= r <= 1.
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: limitation to induction on finite bounds
> How much clearer do I need to get on this explanation? :-)
> (ObDumbQuestion: is there a name for the set T? Something
> along the lines of Champernowne's constant
(0.12345678910111213...)?)
That's clear, so T marks 1/3 on the number line!
I didn't need epsilon, but the same effect,
S setminus r = d
Ax x e S, 0.3, 0.33, 0.333
Ay y e T,
abs(x - r) = y, 0.03.., 0.003.., 0.0003.. = y
Ed d e R, get the 'minimum' of y
Ab b e W,
Az z e T,
Av v e T,
d<=z, d is less than all y
b<=v, b is less than all y
Aa a e W,
d>=a
Forall numbers less than all y, d (also less than all y) is
bigger than or 
equal to them,
therefore forall numbers less than (1/3-0.3, 1/3-0.33, ...)
ie. 0, -0.5, -0.001, -10
d is the largest, i.e 0.
therefore
{0.3, 0.33,0.333...} setminus 1/3 = 0
We could call T the 03033 set.
Herc
===
Subject: Re: limitation to induction on finite bounds
In sci.logic, |-|erc
<gotcha@beauty.com>
<3CIDc.67974$sj4.56705@news-server.bigpond.net.au>:
>> How much clearer do I need to get on this explanation? :-)
>> (ObDumbQuestion: is there a name for the set T? Something
>> along the lines of Champernowne's constant
(0.12345678910111213...)?)
> That's clear, so T marks 1/3 on the number line!
And all other numbers in the set [0,1] as well, if one
defines mark in a certain way.
> I didn't need epsilon, but the same effect,
You need epsilon. 1/3 is not in T. 1/3, however, is
arbitrarily close to certain points thereof.
> S setminus r = d
> <- Ax x e S, 0.3, 0.33, 0.333
> Ay y e T,
> abs(x - r) = y, 0.03.., 0.003.., 0.0003.. = y
> Ed d e R, get the 'minimum' of y
> Ab b e W,
> Az z e T,
> Av v e T,
> d<=z, d is less than all y
> b<=v, b is less than all y
> Aa a e W,
> d>=a
> Forall numbers less than all y, d (also less than all y) is
bigger than or 
equal to them,
> therefore forall numbers less than (1/3-0.3, 1/3-0.33, ...)
> ie. 0, -0.5, -0.001, -10
> d is the largest, i.e 0.
> therefore
> {0.3, 0.33,0.333...} setminus 1/3 = 0
> We could call T the 03033 set.
T is what I've been calling {k/10^n, 0 <= k < 10^n, n > 0, 
k,
n in J}.
It's basically the set of all finite decimal 
prefixes. This set
covers the interval [0,1], though only in the sense that
it's dense.
> Herc
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: limitation to induction on finite bounds
> I didn't need epsilon, but the same effect,
> You need epsilon. 1/3 is not in T. 1/3, however, is
> arbitrarily close to certain points thereof.
> S setminus r = d
> <- Ax x e S, 0.3, 0.33, 0.333
> Ay y e T,
> abs(x - r) = y, 0.03.., 0.003.., 0.0003.. = y
> Ed d e R, get the 'minimum' of y
> Ab b e W,
> Az z e T,
> Av v e T,
> d<=z, d is less than all y
> b<=v, b is less than all y
> Aa a e W,
> d>=a
where is epsilon?
S setminus r = d
<->
Ax x e S, 0.3, 0.33, 0.333
Ay y e T,
abs(x - r) = y, {0.03.., 0.003.., 0.0003..} = T
Az e T
Aa, b, a<=d<=z
the only solution is d=0
d is (less than) the smallest difference
and the biggest of those.
say z =0
Aa, b, a<=d<=z
a <= z
so a <= 0
d >= a
so d >= 0
d <= z
so d <= 0
therefore d = 0
ITS LOGIC MAN, nothing to do with approximations arbitrarily
close.
Herc
===
Subject: Re: limitation to induction on finite bounds
In sci.logic, |-|erc
<gotcha@beauty.com>
<e5ODc.68309$sj4.1206@news-server.bigpond.net.au>:
>> I didn't need epsilon, but the same effect,
>> You need epsilon. 1/3 is not in T. 1/3, however, is
>> arbitrarily close to certain points thereof.
 S setminus r = d
>> <-> Ax x e S, 0.3, 0.33, 0.333
>> Ay y e T,
>> abs(x - r) = y, 0.03.., 0.003.., 0.0003.. = y
 Ed d e R, get the 'minimum' of y
>> Ab b e W,
>> Az z e T,
>> Av v e T,
>> d<=z, d is less than all y
>> b<=v, b is less than all y
 Aa a e W,
>> d>=a
> where is epsilon?
Well, if you want, you can take the set T = {k/10^n, 0 <= k <
10^n, n > 0}
and use an epsilon function (which is simply a sequence
generating
a series); the epsilon function's series need merely diverge
to
infinity.
For example, epsilon(j) = 10^-5 / j will result in a complete
cover.
> S setminus r = d
> <- Ax x e S, 0.3, 0.33, 0.333
> Ay y e T,
> abs(x - r) = y, {0.03.., 0.003.., 0.0003..} = T
> Az e T
> Aa, b, a<=d<=z
> the only solution is d=0
> d is (less than) the smallest difference
> and the biggest of those.
> say z =0
> Aa, b, a<=d<=z
> a <= z
> so a <= 0
> d >= a
> so d >= 0
> d <= z
> so d <= 0
> therefore d = 0
> ITS LOGIC MAN, nothing to do with approximations
arbitrarily close.
That is correct. 1/3 is arbitrarily close to elements in S.
1/3 is not *in* S, however.
> Herc
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Can a product of consecutive integers be a
square?
> We know that the product of 3 or 4 consecutive integers
cannot be any 
powers
> of integers. May I know if n > 4, can a product of n
positive integers be 
a
P. Erdos, J. L. Selfridge, The product of consecutive
integers is never
a power, Illinois Journal of Mathematics 19 (1975) 292-301,
MR 51 #12692.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: re:Can a product of consecutive integers be a square?
I haven't thought about the general case, but if n is prime,
the
answer is no. n will appear as a first power and only as a 
first
power.
group Service in the World! >100,000 
Newsgroups
---= 19 East/West-Coast Specialized Servers - Total Privacy
via Encryption 
=---
===
Subject: Re: Can a product of consecutive integers be a
square?
> I haven't thought about the general case, but if n is
prime, the
> answer is no. n will appear as a first power and only as a
first
> power.
2 is a prime, but it appears as a second power in 4 * 5 .
===
Subject: Re: Can a product of consecutive integers be a
square?
> I haven't thought about the general case, but if n is
prime, the
> answer is no. n will appear as a first power and only as a
first
> power.
That is very close to the answer. What if n has p as a prime
factor? n+1
doesn't have p as a prime factor. n*(n+1) could still be a
square, but
only if n has the factor p^2, p^4, p^6 etc.
This is true for _any_ p, and as a consequence n must be a
square. But
for n(n+1) to be square, n+1 must be a square as well. That
is not
possible except for 0*1 = 0, which is a square.
===
Subject: Re: Can a product of consecutive integers be a
square?
>> I haven't thought about the general case, but if n is
prime, the
>> answer is no. n will appear as a first power and only as a
first
>> power.
> That is very close to the answer. What if n has p as a
prime factor?
> n+1 doesn't have p as a prime factor. n*(n+1) could still
be a
> square, but only if n has the factor p^2, p^4, p^6 etc.
> This is true for _any_ p, and as a consequence n must be a
square.
> But for n(n+1) to be square, n+1 must be a square as well.
That is
> not possible except for 0*1 = 0, which is a square.
What is n? In the original question, n was the number of
factors, not
the first factor:
> We know that the product of 3 or 4 consecutive integers
cannot be any
> powers of integers. May I know if n > 4, can a product of n
positive
> to my problem.
===
Subject: Re: Can a product of consecutive integers be a
square?
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(
5eZ41to5f%E@'ELIi
$t^
VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
> I haven't thought about the general case, but if n is
prime, the
> answer is no. n will appear as a first power and only as a
first
> power.
> That is very close to the answer. What if n has p as a
prime factor?
> n+1 doesn't have p as a prime factor. n*(n+1) could still
be a
> square, but only if n has the factor p^2, p^4, p^6 etc.
> This is true for _any_ p, and as a consequence n must be a
> square. But for n(n+1) to be square, n+1 must be a square as
> well. That is not possible except for 0*1 = 0, which is a
square.
Read the original question again. Both of you.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Basic Question about derivatives (sorry I'm new at
this)
A function f that is defined on an open interval (a,b) and
where c is in
(a,b), is said to be differentiable at c whenever the limit
lim f(x) - f(c)
x --> c ----------
x - c
exists, and we call the limit f ' (c).
What I don't understand is the following theorem :
If f is defined on (a,b) and differentiable at a point c in
(a,b), then
there is a function f* (depending on f and on c) which is
continuous at c
and which satisfies the equation f(x) - f(c) = (x-c) f*(x) for
all x in
(a,b), with f*(c) = f ' (c). Conversely, if there is a
function f*,
continuous at c, which satisfies f(x) - f(c) = (x-c) f*(x),
then f is
differentiable at c and f ' (c) = f*(c).
The proof goes as follows in my book : If f '(c), exists, let
f* be defined
on (a,b) as follows :
f*(x) = f(x) - f(c)
----------
x - c
if x does not equal c, f*(c) = f '(c). (ok)
Then f* is continuous at c (why? I am trying to prove using
epsilon delta
definition of continuity but get nowhere) f(x) - f(c) = (x-c)
f*(x) for all
x in (a,b) (why?).
Conversely, if f(x) - f(c) = (x-c) f*(x) holds for some f*
continuous at c,
then by dividing by x - c and letting x ----> c, we see that
f ' (c) exists
(why?) and equals f*(c) (why?)
===
Subject: Re: Basic Question about derivatives (sorry I'm new
at this)
> A function f that is defined on an open interval (a,b) and
where c is in
> (a,b), is said to be differentiable at c whenever the limit
> lim f(x) - f(c)
> x --> c ----------
> x - c
> exists, and we call the limit f ' (c).
> What I don't understand is the following theorem :
> If f is defined on (a,b) and differentiable at a point c in
(a,b), then
> there is a function f* (depending on f and on c) which is
continuous at c
> and which satisfies the equation f(x) - f(c) = (x-c) f*(x)
for all x in
> (a,b), with f*(c) = f ' (c). Conversely, if there is a
function f*,
> continuous at c, which satisfies f(x) - f(c) = (x-c) f*(x),
then f is
> differentiable at c and f ' (c) = f*(c).
> The proof goes as follows in my book : If f '(c), exists,
let f* be 
defined
> on (a,b) as follows :
> f*(x) = f(x) - f(c)
> ----------
> x - c
> if x does not equal c, f*(c) = f '(c). (ok)
> Then f* is continuous at c (why? I am trying to prove using
epsilon 
delta
> definition of continuity but get nowhere)
The definition of f'(c) says that lim(x->c) 
f*(x) = f'(c).
Thus, you
have lim(x->c) f*(x) = f*(c), and that's all you need.
However, if you want to use an epsilon-delta proof, you could
talk about sequences of values of f*(x) and show that they
are Cauchy sequences.
> f(x) - f(c) = (x-c) f*(x) for all
> x in (a,b) (why?).
Well, it's clearly true for x not equal to c by 
definition of
f*(x). For x=c, both sides are equal to 0.
> Conversely, if f(x) - f(c) = (x-c) f*(x) holds for some f*
continuous at 
c,
> then by dividing by x - c and letting x ----> c, we see
that f ' (c) 
exists
> (why?)
Because f'(x) is defined as the limit of f*(x). 
If f* is
continuous
at c, the limit is defined at c.
> and equals f*(c) (why?)
It's equal to the limit of f*(x) as x->c, again by 
definition
of f'(x). That's equal to f*(c) by 
continuity.
- Randy
===
Subject: Re: Basic Question about derivatives (sorry I'm new
at this)
>A function f that is defined on an open interval (a,b) and
where c is in
>(a,b), is said to be differentiable at c whenever the limit
>lim f(x) - f(c)
>x --> c ----------
> x - c
>exists, and we call the limit f ' (c).
>What I don't understand is the following theorem :
>If f is defined on (a,b) and differentiable at a point c in
(a,b), then
>there is a function f* (depending on f and on c) which is
continuous at c
>and which satisfies the equation f(x) - f(c) = (x-c) f*(x)
for all x in
>(a,b), with f*(c) = f ' (c). Conversely, if there is a
function f*,
>continuous at c, which satisfies f(x) - f(c) = (x-c) f*(x),
then f is
>differentiable at c and f ' (c) = f*(c).
>The proof goes as follows in my book : If f '(c), exists,
let f* be 
defined
>on (a,b) as follows :
>f*(x) = f(x) - f(c)
> ----------
> x - c
>if x does not equal c, f*(c) = f '(c). (ok)
>Then f* is continuous at c (why? I am trying to prove using
epsilon delta
>definition of continuity but get nowhere) f(x) - f(c) = (x-c)
f*(x) for 
all
>x in (a,b) (why?).
You don't need epsilon delta if you use the limit 
definition of
continuous:
lim(x->c) f*(x) = lim(x->c)(f(x) - f(c))/(x - c) = f'(c) =
f*(c) where
the first = follows from the definition of f* 
since x != c and
the
last = from the the definition of f* at x = c.
>Conversely, if f(x) - f(c) = (x-c) f*(x) holds for some f*
continuous at 
c,
>then by dividing by x - c and letting x ----> c, we see that
f ' (c) 
exists
>(why?) and equals f*(c) (why?)
Dividing we get (f(x) - f(c))/(x - c) = f*(x) if x != c. Take
the
limit as x -> c on both sides:
lim(x->c)(f(x) - f(c))/(x - c) = lim(x->c)f*(x) = f*(c)
because f* is
continuous. Therefore the limit on the left side exists,
which is
f'(c) and its value is f*(c).
--Lynn
===
Subject: Re: Basic Question about derivatives (sorry I'm new
at this)
> <snipConversely, if f(x) - f(c) = (x-c) f*(x) holds for some f*
continuous at
c,
>then by dividing by x - c and letting x ----> c, we see that
f ' (c)
exists
>(why?) and equals f*(c) (why?)
> Dividing we get (f(x) - f(c))/(x - c) = f*(x) if x != c.
Take the
> limit as x -> c on both sides:
> lim(x->c)(f(x) - f(c))/(x - c) = lim(x->c)f*(x) = f*(c)
because f* is
> continuous. Therefore the limit on the left side exists,
which is
> f'(c) and its value is f*(c).
> --Lynn
A Suggestion, take some time and really understand this one
well, it is key
to future sucess in Calc.
Do all the problems. This is one of the first and main humps
in Calc.
===
Subject: Re: The Definition of 'Analytic 
Function'
> real/complex definition of analytic while learning the
rudiments
> of generating functions. However, intuitively, I've
always 
thought
> of an analytic function as one that can be solved
symbolically.
 Category mistake: functions are not things that can be
solved.
 Anyway, this shows that what you intuitively have
always 
thought
> does not correspond to the general usage of the term
amongst
> mathematicians.
 Is there a name for what I described in the previous post
(assuming 
you
> understand the gist of what I'm saying)?
> It is completely unclear to me what you even might mean
with the quoted
> paragraph.
> First off. I haven't settled on 'what I 
mean' yet. Second:
this,
> unsurprisngly, may not do anything to better explain what
it is that I do
> mean.
> Perhaps I should introduce a notion of computation into
what I'm 
thinking,
> which is unfortunately not something I can rigorously
describe.
> With generating functions: By analytic I simply meant that
an infinite 
power
> series only converges if supplied with specific initial
conditions.
> Informally, if for every point in some given set it
converges to a single
> point in a single (but arbitrary interval) subset of that
set it, is
> considered analytic (in that set). For example, if I'm not
mistaken, the
> series:
> f(s) = 1 + s + s^2 + .... + s^n + ...
> is analytic only if s in (-1,1).
> Computability of ~Analytic Equations
> Essentially when I think of an ~analytic relation, the idea
of 'quickly'
> being able to solve the equation is fundamental. However,
there seem to 
be
> two levels. One is the symbolic level. ie: f(x) = pi*e^x.
This is 
simple
> to solve symbolically (ie: for some constant c, f(c) =
pi*e^c. However, 
the
> problem of calculating the exact decimal number solution,
for say c = O,
> where O is Chaitin's Omega number for a certain computer
language, is
> substantially harder (even though O in this case may be
simple to 
define).
[snip lots of interesting definitions]
Hi Christoff,
I thought it was the method of solution that could be
analytic or
numeric. I can think of a few other contexts in which
'analytic' could
be used as a term in CS or mathematics, but I doubt whether
we should
call a function analytic.
It seems to me, rather, the case that the definition of a
function
could be analytic or numeric. As you have pointed out,
definition of a
function f(x)=pi^x would be convenient for a human, and by
analogy to
method of analytic solution in calculus, one could say that
this
definition, by its symbolic construction, and its connection 
to
symbolic rules of transformation, is already analytic. But I
will
digress, let's take the human out of the equation.
Consider the theory of functions. I could define the same
function
f(x)=pi^x programmatically, on a weird computational device,
such as a
cellular automata, such that, when a representation of x and
a degree
of precision p is placed on an input segment of cells, the
automata
computes f(x) to a certain precision p (according to some
definition
of numeric error). Without being told how this cellular
automata was
devised, with many cells and seemingly complex operation, you
would be
at loss of words to explain the function. It would look
incomprehensible to the human, yet, it is the operational
definition
of the function. I could also print out the automata on a
very large
sheet of paper, and say that this was f(x). However, it is
the same
function as before. The functional properties are the same,
the only
thing that has changed in my opinion the difficulty of program
transformations (ie. Let's say we want to compute the
integral of the
function) It is simply the mode of representation or
definition that
has differed, and not the function itself. [As much as I'd
like to
avoid a platonist approach to the concept]
However, you can still argue that these are remarkably
different kinds
of definitions such that *the entities they define 
are no
longer the
same entity*. Much of the difficulty of our discussion comes, 
I
think, from introducing such numbers like pi or Omega that
can be only
approximated on a finite computer. Before going further, can
you
please tell me whether there is an analytic integer function
by your
definition? Or does analyticity arise, in your opinion, only
when we
are dealing with what cannot be computed precisely?
In the end, I think I will argue that there is no such thing
as an
analytic function, much like there is no analytic number, but
first
let me understand your position more clearly. [Or are you
going to
tell me that Omega is an analytic number?]
--
Eray Ozkural
===
Subject: Re: This should be a no brainer but..
> I've asked this questions numerous times: What is 
earth's
curvature
> per mile?
> I've gotten numerous equations, and none of the 
respondents
end up
> agreeing with each other, and their answers are as varied
as their
> mailing addresses!
> Another issue is answers in metric - I am mathematically
challenged
> and cannot convert answers from KM to Miles to save my life!
Google to the rescue! Just go to Google and type in 534 km in
mi, or 23
trillion mi in light years (or parsecs), or 195 lbs in
stones, or 283
leagues in miles. It's terrific!
===
Subject: Re: This should be a no brainer but..
>> I've asked this questions numerous times: What is 
earth's
curvature
>> per mile?
>> I've gotten numerous equations, and none of the
respondents end up
>> agreeing with each other, and their answers are as varied
as their
>> mailing addresses!
>> Another issue is answers in metric - I am mathematically
challenged
>> and cannot convert answers from KM to Miles to save my
life!
>I don't normally take this tack, but if you _really_ cannot
figure out
>how to do this, it's doubtful the answer will mean much to
you? But
>because you can at least construct the chord of a circle, I
infer you
>are merely my brother in laziness.
>> Now, shouldn't it be as simple as taking a slice through 
a
7,920 mile
>> diameter circle(creating something called a chord I
think), and
>> measuring the highest point of that chord?? In our case,
the straight
>> line slices through the cirle between two points a mile
apart on that
>> circle. What the HECK is the maximum height of the segment
of that
>> circle between the ends of that line?
>Your construction would yield the fall in 1/2 mile, not 1
mile.
>Here's my formula for the drop at 1 mile:
> 7,920 x {1 - cos[arcsin(1/7,920)]} = 7,920 x 7.97e-9
> = 6.31e-5 mile = .33 ft
Why did you use 2R = 7,920? Use the angle not the arcsine.
>Actually, it works out to be a third of a foot to about 8
significant
>digits. Coincidence, or something about the definition of 
mile
>related to the dimensions of the Earth?
>> Please, no more formulae. All my eyes see is Cyrillic.
>Russian is Greek to me.
Oh, for cryin' out loud! We're looking for the 
hump over 1
mile, also
known as the sagitta.
The whole angle A = L/R, diagram will have half angle A2 =
L/2R.
The hump is given by
H = R(1 - cos(A2)) = R versine(A2)
2d term of versine(x) = x^2/2)
= ~R x (L/2R)^2/2 = L^2/8R = 1.98 inches using R = 4000.
===
Subject: Re: This should be a no brainer but..
>> I've asked this questions numerous times: What is 
earth's
curvature
>> per mile?
>> I've gotten numerous equations, and none of the
respondents end up
>> agreeing with each other, and their answers are as varied
as their
>> mailing addresses!
 Another issue is answers in metric - I am mathematically
challenged
>> and cannot convert answers from KM to Miles to save my
life!
 Now, shouldn't it be as simple as taking a slice through 
a
7,920 mile
>> diameter circle(creating something called a chord I
think), and
>> measuring the highest point of that chord?? In our case,
the straight
>> line slices through the cirle between two points a mile
apart on that
>> circle. What the HECK is the maximum height of the segment
of that
>> circle between the ends of that line?
 Please, no more formulae. All my eyes see is Cyrillic.

>> -CC
>If CC didn't want equations, he ought not have submitted 
his
>question to sci.physics. CC's definition of 
curvature is non-
>linear, and therefore, non-viable. Curvature is usually given
>as the inverse radius:
>Curvature = C = 1 / R = arc_angle (radians) / arc_length
(meters)
>In degrees and miles,
>C = 2 / 7920 miles = [2*pi / 360][arc(degrees) / arc_length
(miles)
>Thus,
>C = 0.000253 radian / mile = 0.0145 degree / mile
>The height between cord and arc for d = 1 mile is (d << R):
>h = R [ d / 2R ]^2 = d^2 / 4R = 0 .000631 miles = 0.333 feet
>Note that this relationship is not linear with d. If d = 2
miles,
>then h = 1.333 feet.
>[Old Man]
> With one more OOPS, you can fix your answer.
> Mr. Dual Space
Old Man doesn't think that he did an OOPS. 
It's noted
that, in this thread, above, John had a couple.
[Old Man]
===
Subject: Re: This should be a no brainer but..
> I've asked this questions numerous times: What is
earth's curvature
>> per mile?
>> I've gotten numerous equations, and none of the
respondents end up
>> agreeing with each other, and their answers are as
varied as their
>> mailing addresses!
 Another issue is answers in metric - I am mathematically
challenged
>> and cannot convert answers from KM to Miles to save my
life!
 Now, shouldn't it be as simple as taking a slice through
a 7,920 
mile
>> diameter circle(creating something called a chord I
think), and
>> measuring the highest point of that chord?? In our case,
the 
straight
>> line slices through the cirle between two points a mile
apart on 
that
>> circle. What the HECK is the maximum height of the
segment of that
>> circle between the ends of that line?
 Please, no more formulae. All my eyes see is Cyrillic.

>> -CC
If CC didn't want equations, he ought not have submitted
his
>question to sci.physics. CC's definition of 
curvature is
non-
>linear, and therefore, non-viable. Curvature is usually
given
>as the inverse radius:
Curvature = C = 1 / R = arc_angle (radians) / arc_length
(meters)
In degrees and miles,
C = 2 / 7920 miles = [2*pi / 360][arc(degrees) /
arc_length (miles)
Thus,
C = 0.000253 radian / mile = 0.0145 degree / mile
The height between cord and arc for d = 1 mile is (d << R):
h = R [ d / 2R ]^2 = d^2 / 4R = 0 .000631 miles = 0.333
feet
Note that this relationship is not linear with d. If d = 2
miles,
>then h = 1.333 feet.
[Old Man]
 With one more OOPS, you can fix your answer.
> Mr. Dual Space
> Old Man doesn't think that he did an OOPS. 
It's noted
> that, in this thread, above, John had a couple.
> [Old Man]
nightbat
, you got to admit that's mighty close of deduced arc to
TJ's 8 inches cited one. He must have one heck of a pocket
calculator or
every salty sailor just knows about the eight inches. What
happen John
and Old Man, your sails are sagging in the wind matties.
the nightbat
===
Subject: Re: This should be a no brainer but..
In sci.physics, MorituriMax
<newage@sendarico.net>
<V6nDc.14653$w3.12989@fe2.texas.rr.com>:
>> Another issue is answers in metric - I am mathematically
challenged
>> and cannot convert answers from KM to Miles to save my
life!
 1.6 kilometers per mile.
>> And .6 miles per kilometer.
> Amen!
Ye gods, what imprecision. :-) Then again, as a rule of thumb
these aren't too bad.
I prefer doing it the hard way: 1 mile = 5,280 ft
= 1 mi * 5280 ft/mi * 12 in/ft * 0.0254 m/in = 1609.344 m
exactly.
And of course 1 km = 1000 m / 1609.344 m/mi =
0.621371192237... mi,
though most chop it off well before that point. If one
measures a road 1 km long to 6 digits of precision, one gets
a millimeter of error; that's about the thickness of a
screwdriver blade or thereabouts.
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: This should be a no brainer but..
> In sci.physics, MorituriMax
> <newage@sendarico.net <V6nDc.14653$w3.12989@fe2.texas.rr.com>:
>> Another issue is answers in metric - I am mathematically
challenged
>> and cannot convert answers from KM to Miles to save my
life!
 1.6 kilometers per mile.
 And .6 miles per kilometer.
> Amen!
> Ye gods, what imprecision. :-) Then again, as a rule of
thumb
> these aren't too bad.
> I prefer doing it the hard way: 1 mile = 5,280 ft
> = 1 mi * 5280 ft/mi * 12 in/ft * 0.0254 m/in = 1609.344 m
exactly.
> And of course 1 km = 1000 m / 1609.344 m/mi =
0.621371192237... mi,
> though most chop it off well before that point. If one
> measures a road 1 km long to 6 digits of precision, one gets
> a millimeter of error; that's about the thickness of a
> screwdriver blade or thereabouts.
If you are driving 60 miles per hour (sorry about the
non-metric units) 
then
the degree of inaccuarcy between 0.6 miles and 0.621371192237
miles is 
about
1.3 seconds.
===
Subject: Re: This should be a no brainer but..
charset=iso-8859-1
> I prefer doing it the hard way: 1 mile = 5,280 ft
> = 1 mi * 5280 ft/mi * 12 in/ft * 0.0254 m/in = 1609.344 m
exactly.
> And of course 1 km = 1000 m / 1609.344 m/mi =
0.621371192237... mi,
> though most chop it off well before that point. If one
> measures a road 1 km long to 6 digits of precision, one gets
> a millimeter of error; that's about the thickness of a
> screwdriver blade or thereabouts.
Sorry, I had my newsreader, erm, set to display on 1 digit
after the 
decimal..
yeah thats it.
===
Subject: More straight than geodesic
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 
1.9 primary) id i5S1jCu15366;
Asymptotic lines are also geodesics for generators of a
hyperbolic 
paraboloid of one sheet. The geodesic and normal curvatures
are zero,so is 
the total curvature. [kg,kn,k -> 0 ]. It seems to be a
trivial question, it 
is like a laser beam grazing past a surface but ...
1) Can there be more examples of such surfaces in R^3 ? (
Gauss curvature 
should be < 0 ); If so, which are ?
2) And such examples in R^n , n > 3 ?
===
Subject: Re: Cranks
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 
1.9 primary) id i5S1jCG15374;
(snip)
>Mmmm... Your ignorance of the details seems parallel to that
you had on 
the
>details of proofs of impossibility of trisection... Of
course, your 
global
>feeling can still be right, but usually, the details are
what matters...
>While we are at it, you could *read* Socrates' apology,
Crito, etc. This
>site makes for good reading : <a 
href=http://socrates.clarke.edu/>http://socrates.clarke.edu
/</a>
Well, I did a while ago, and I was summarizing it, in which
(according to 
Plato's account) Socrates was convicted by the authorities 
of
his city-state 
for corrupting the youth with his ideas, and he was required
to kill himself 
by drinking hemlock.
He turned down the offer to escape, but instead voluntarily
drank the 
hemlock to prove a point - that you have to accept the laws
of your 
city-state, and accept the punishments. Otherwise, by
escaping, you 
undermine the laws and the fabric of society.
I left out all of these details to keep my posting short. Do
you like long 
posts?
Now, where was my ignorance of Socrates?
Anthony Natoli
===
Subject: Re: Cranks
> (snip)
>> Mmmm... Your ignorance of the details seems parallel to
that you
>> had on the details of proofs of impossibility of
trisection... Of
>> course, your global feeling can still be right, but
usually, the
>> details are what matters...
>> While we are at it, you could *read* Socrates' apology,
Crito, etc.
>> This site makes for good reading : <a
>>
href=http://socrates.clarke.edu/>http://socrates.clarke.edu
/</a Well, I did a while ago, and I was summarizing it, in which
> (according to Plato's account) Socrates was convicted by 
the
> authorities of his city-state for corrupting the youth with
his
> ideas, and he was required to kill himself by drinking
hemlock.
> He turned down the offer to escape, but instead voluntarily
drank the
> hemlock to prove a point - that you have to accept the
laws of your
> city-state, and accept the punishments. Otherwise, by
escaping, you
> undermine the laws and the fabric of society.
> I left out all of these details to keep my posting short.
Do you
> like long posts?
> Now, where was my ignorance of Socrates?
All right, I apologize (not in the Plato's meaning, of
course). But then it
is simply a matter of judgment, and Plato, or later
philosophers (not to
mention the Apollo oracle), usually describe Socrates as the
wisest of men 
,
and his attitude in drinking the hemlock a noble and proper
one , not at 
all
an act of folly :-) (and, of course, all this began in jest :
the point was
that sqrt(2) *is* constructible, and so crankish attemptsto
construct it
could well succeed...)
> Anthony Natoli
===
Subject: The Floretions- need help in proofreading
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 
1.9 primary) id i5S1jDJ15398;
I previously posted a very rough draft of this work a few
weeks
ago (and has since been upgraded to rough draft- for
whatever
that means). It has changed in part since then and is now (as
it was
before) in dire need of a sound proofreading or two (I am also
content in recieving little tips or to know your opinion of
various
statements made in it.).
Notes:
1. My term left-sided/right-sided division group is
apparently
very similar to the term quasi-group which has already
established itself in literature. Perhaps I should say:
left-sided/right-sided associative quasi-group?
2. The proof of the last proposition, Chung-shu's Division
Group,
is unfortunately incomplete. If I complete it (or disprove
it),
any such proof will be reposted to my website.
3. The proof of the proposition The Big Innercube Group is
a group of order 256 is still giving me headaches to
establish.
4. All claims made in the chapter Floretion Subgroups:
A First Encounter with Floret's Star need to be proofread.
sincerly,
C. Dement
===
Subject: Brick wall: find the distribution
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 
1.9 primary) id i5S1jD515412;
Imagine a 2-D brick wall of arbitrary extent in which each
brick
rests on the 2 bricks immediately beneath. A weight W
rests on the center brick of the top course. If each brick
transmits
forces equally to the 2 bricks immediately beneath, how is
the weight
distributed at the bottom of the wall, which is n courses
high?
Ignore the bricks' own weight.
phil
===
Subject: Re: Brick wall: find the distribution
> Imagine a 2-D brick wall of arbitrary extent in which each
brick
> rests on the 2 bricks immediately beneath. A weight W
> rests on the center brick of the top course. If each brick
transmits
> forces equally to the 2 bricks immediately beneath, how is
the weight
> distributed at the bottom of the wall, which is n courses
high?
> Ignore the bricks' own weight.
Sounds like Pascal's triangle.
-Michael.
===
Subject: Re: Brick wall: find the distribution
> Imagine a 2-D brick wall of arbitrary extent in which each
brick
> rests on the 2 bricks immediately beneath. A weight W
> rests on the center brick of the top course. If each brick
transmits
> forces equally to the 2 bricks immediately beneath, how is
the weight
> distributed at the bottom of the wall, which is n courses
high?
> Ignore the bricks' own weight.
Pascal's Triangle is directly analogous, unless you
consider the turning moments, which I think would
concentrate more force to the center. But transmits
forces equally to the 2 bricks immediately beneath
perhaps rules out that consideration.
-jiw
===
Subject: Re: Brick wall: find the distribution
> Imagine a 2-D brick wall of arbitrary extent in which each
brick
> rests on the 2 bricks immediately beneath. A weight W
> rests on the center brick of the top course. If each brick
transmits
> forces equally to the 2 bricks immediately beneath, how is
the weight
> distributed at the bottom of the wall, which is n courses
high?
> Ignore the bricks' own weight.
> phil
Cute.
The answer may at first surprise you, but then the reason
becomes obvious. Hint: Each brick is also the sum of
the two bricks above it (divided by 2).
How far have you gotten?
- Randy
===
Subject: Re: Sobolev spaces
>Let U be open, bounded in R^n with boudary of U as smooth as
you want.
>Let H^1-denote the Sobolev space on U with k=1,p=2
>Let H^1[o]-denote the subspace of H^1 where functions are
zero on 
boudary
> of U .
>Now i have noticed a lot of PDE books when looking at
>(H^1[o])* don't use fact that its dual is itself under its
inner product
>but instead use a method so we can still use the L^2 inner
product.
>Does anyone know why they do this??
> Took me a minute to recall what you're getting at here...
> You'd really have to ask them why they do this, but it
> seems to me that probably one reason is that studying the
> connections between Sobolev spaces and other spaces works
> better if we use the standard pairing between elements
> of the space and its dual. These things are distributions,
> after all, so it seems natural that when we talk about
> the dual we should use (an extension of) the pairing
> used when we talk about distributions as the dual of
> smooth functions.
>I have heard if you don't want to use Riesz to represent
(H^1)*
>then it becomes fairly complicated.
>Why couldn't we use same method as we did with (H^1[o])* ??
> I don't quite understand the question. The dual of
> W(1,2) _is_ W(-1,2)...
The only time I have seen a defn. for k=negative Sobolev
spaces is when U
is R^n and we are using Fourier transform to define. ((now
this stuff is all 
new
to me)).
Anyhow what would be a defn. for W(-1,2) without just saying
it is the dual of W(1,2)??
I recall reading some PDE book that talked about the
dual of W(1,2)[o](U) using the standard pairing. It then
went on to say
something like  If you want to view the dual of W(1,2)(U) in
the same 
manner
it becomes fairly involved since something is not a space
of 
distrobutions
any more.
>don
> ************************
> David C. Ullrich
===
Subject: Re: Sobolev spaces
>>Let U be open, bounded in R^n with boudary of U as smooth
as you want.
>>Let H^1-denote the Sobolev space on U with k=1,p=2
>>Let H^1[o]-denote the subspace of H^1 where functions are
zero on 
boudary
>> of U .
Now i have noticed a lot of PDE books when looking at
>>(H^1[o])* don't use fact that its dual is itself under its
inner 
product
>>but instead use a method so we can still use the L^2 inner
product.
Does anyone know why they do this??
>> Took me a minute to recall what you're getting at here...
>> You'd really have to ask them why they do this, but it
>> seems to me that probably one reason is that studying the
>> connections between Sobolev spaces and other spaces works
>> better if we use the standard pairing between elements
>> of the space and its dual. These things are distributions,
>> after all, so it seems natural that when we talk about
>> the dual we should use (an extension of) the pairing
>> used when we talk about distributions as the dual of
>> smooth functions.
>>I have heard if you don't want to use Riesz to represent
(H^1)*
>>then it becomes fairly complicated.
Why couldn't we use same method as we did with (H^1[o])* 
??
>> I don't quite understand the question. The dual of
>> W(1,2) _is_ W(-1,2)...
>The only time I have seen a defn. for k=negative Sobolev
spaces is when U
>is R^n and we are using Fourier transform to define. ((now
this stuff is 
all new
>to me)).
> Anyhow what would be a defn. for W(-1,2) without just saying
>it is the dual of W(1,2)??
Hmm. I'd just assumed that... never mind, not sure. Sorry.
>I recall reading some PDE book that talked about the
>dual of W(1,2)[o](U) using the standard pairing. It then
went on to 
say
>something like  If you want to view the dual of W(1,2)(U)
in the same 
manner
>it becomes fairly involved since something is not a space
of 
distrobutions
>any more.
Ok, now I think I understand your question better. Also it's
clear
that I know less about what you're talking about than I
thought:
Exactly how do they define (H^1[o])* using (an extension of)
the standard pairing?
don
>> ************************
>> David C. Ullrich
************************
David C. Ullrich
===
Subject: Re: Maximal Subspaces in a Hilbert space
> Let H denote a real Hilbert space with inner product ( , ).
> Let F:H->(-inf. , inf.]
> fix u in H with F(u)< inf.
> Given a subspace X in H , let G(F,X,u) denote the set of
f's in H
> such that
> (F(u+th)-F(u))/t -> (f,h) as t->0 for all h in X.
> Lets assume there exist some Y ((non-trivial subspace of H))
> such that G(F,Y,u) nonempty.
> Let S denote the set of X with G(F,X,u) nonempty.
> Can we say anything about S having a maximal element?
(w.r.t. inclusion)
> Can we say in general S won't have a maximal element?
> One thing to note is G(F,X,u) is closed and convex in H.
> Also if X dense then G(F,X,u) is empty or a singleton.
> don
Repost
===
Subject: normalizer of Q8 in SL(2,q)
A friend is interested in the following question.
Any help would be appreciated.
Let q be an odd prime power and let N be the normalizer of Q8
in SL(2,q)
where Q8 is the subgroup generated by the two matrices
[ 0 1] [a b]
[-1 0] , [b -a]
where a^2 + b^2 = -1.
Is N/Q8 always either cyclic of order 3 or isomorphic to S_3?
Edwin Clark
===
Subject: Re: normalizer of Q8 in SL(2,q)
> A friend is interested in the following question.
> Any help would be appreciated.
> Let q be an odd prime power and let N be the normalizer of
Q8 in SL(2,q)
> where Q8 is the subgroup generated by the two matrices
> [ 0 1] [a b]
> [-1 0] , [b -a]
> where a^2 + b^2 = -1.
> Is N/Q8 always either cyclic of order 3 or isomorphic to
S_3?
I think so.
Can it be larger?
Let's consider the normalizer of Q8/(+-I) in PSL(2,q).
This is a Klein 4-group. The subgroups of PSL(2,q) of order
prime to q are cyclic, dihedral, A_4, S_4 or A_5.
The only ones normalizing a Klein 4-group are V_4 (dihedral
order 4), dihedral order 8, A_4 and S_4 and so N/Q8 is either
trivial or Z_2 or Z_3 or S_3.
Now what if |N| was divisible by p (q = p^m)? That's not 
going
to happen unless p = 3 --- I suppose one needs to look
carefully
at this case.
Can it be smaller?
Then N/Q8 = 1 or Z_2. To get a contradiction one needs to sit
down and grind out an element of order 3 in the normalizer.
I presume you've done this :-)
--
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: maximal subspace in L^2
>> Let L^2 denote L^2(R). R is reals
 Given a subspace X of L^2 we say X in S
>> if there exists a f[X] in L^2 such that
>> int(f[X]*h)=int(h) for all h in X.
 all integrals are over R.
>> clearly any X is in L^1 intersect L^2.
 now my question is , Is there a maximal X in S using
inclusion ??
 We aren't ((at least i couldn't)) use 
Zorn's lemma
>> since you can find a chain which doesn't have 
an upper
bound.
 don
> Don't see that I helped much, everything I said was wrong
> (except for the parts where I explained that everything
> I said was wrong.)
>This question was just a feable attempt of mine to find a
counterexample
>to something.What i am really interested is in the question
>posted  maximal subspaces in a Hilbert space.
> ??? What is it you want to know about them?
The question is posted in thread called Maximal Subspaces in
a
Hilbert space.
This was originally posted a few days ago.I just posted a
reply to it
so i think it will rise to top of list.
>don
> ************************
> David C. Ullrich
don
===
Subject: Re: maximal subspace in L^2
> Let L^2 denote L^2(R). R is reals
 Given a subspace X of L^2 we say X in S
> if there exists a f[X] in L^2 such that
> int(f[X]*h)=int(h) for all h in X.
 all integrals are over R.
> clearly any X is in L^1 intersect L^2.
 now my question is , Is there a maximal X in S using
inclusion ??
 We aren't ((at least i couldn't)) use 
Zorn's lemma
> since you can find a chain which doesn't have 
an upper
bound.
 don
 Don't see that I helped much, everything I said was wrong
>> (except for the parts where I explained that everything
>> I said was wrong.)
>>This question was just a feable attempt of mine to find a 
counterexample
>>to something.What i am really interested is in the question
>>posted  maximal subspaces in a Hilbert space.
>> ??? What is it you want to know about them?
>The question is posted in thread called Maximal Subspaces
in a
>Hilbert space.
Ah, right. Since there have been no replies to that post
I would conjecture that nobody has anything useful to say
about it.
>This was originally posted a few days ago.I just posted a
reply to it
>so i think it will rise to top of list.
Ok, there's been _one_ reply.
>>don
>> ************************
>> David C. Ullrich
>don
************************
David C. Ullrich
===
Subject: Re: Cranks
 I got a question for you. Would you think a noncrank
person, who
>> is not insane at all, dares to try to work on angle
trisectors?
>> His/her project is completely nonsense, because as we
know, it is
>> already proven that angle trisection is impossible. It
seems to me
>> that no sane person tries to do so, but rather accepts the
theorem.
 If Yes, then I think I have to modify my definition a bit.
>I don't think trying to prove angle trisection (of any
angle) makes one
>a crank. I guess your reasoning is that if something is
known to be
>impossible to the general mathematical community, to attempt
to prove
>it is just insane or something.
> No no. After all, that guy is trying to do something
impossible,
> 100 %! What that guy is doing is trying to make 1 + 1 = 2 to
> 1 + 1 = 3.
>
appear on my newsserver until now.
What I'm saying is that the person may not know 
it's
impossible. I
think we both agree that the person must first know 
confidently
that
mathematicians say it's impossible in order to be on the 
first
step of
the journey to crankhood. There's enough misinformation out
there that
I think a lot of times people don't know this.
>But I don't buy that. First, why would anybody believe a
proof they've
>never seen the proof of? If someone came up to me with a
gun, and said
>I'll shoot you if you say this theorem is right this but it
is actually
>wrong, and I've never even seen even a part of the proof of
it, I might
>very well, depending on other details, abdicate and say
maybe the
>theorem isn't right. My point is that there are reasons to
believe a
>proof I've never seen, but these are reasons more along the
lines of
>I'm pretty sure rather than I am absolutely utterly sure,
like 
when
>I prove a theorem myself.
>What does this have to do with the topic? Well, consider an
average
>person that knows some geometry (is this average?). S/he
doesn't know
>enough to understand a proof of the impossibility of angle
trisection.
> I think any average person can understand the proof, if
s/he tries
> (or wants to) I am assuming that anyone can understand any
available
> proofs. UNDER THAT ASSUMPTION, I am talking.
> But ignoring the proof is, I think, definitely an action of
c.
> Ignoring the proof and still continuing to work on the
problem is also.
That's a big assumption. See below (I apologize for not
writing it
here, but I think it fits well below).
I think it entirely depends on why someone ignores the proof.
Just
because someone says he has a proof doesn't mean 
I'm going to
go look
at it, especially if I have reasons for thinking his
conclusions are
ßawed.
I think the crucial thing here is that somebody may ignore a
proof, but
if he does so even knowing that every living mathematician on
the
planet believes the proof (and it's even taught to
undergraduates!),
then that is a sign of crankhood.
I believe it's a sign because it shows a kind of mentality
that doesn't
allow for any mistakes by the person. This kind of person
isn't
willing to have some doubt that his conclusions may be wrong,
not even
the least bit.
>Why would such a person be 100% convinced or even close to
that, that
>angle trisection is impossible? I think it's really quite
reasonable
>to remain skeptical in this circumstance.
> I think that a sane person buys the statement, Angle
trisection is
> impossible; it doesn't matter if that person knows about
geometry,
> or about any bit of the impossibility proof.
> An insane person seems to reject the statement.
I think your assumption that anyone can understand the
impossibility
proof is far-fetched. Most people can't understand how you
can even
show something is impossible. Most people don't understand
proof by
contradiction. There's a funny story about Erdos trying to
explain the
proof that square root of two is irrational to a colleague's
wife. He
starts with assuming it is rational, as usual. Then he
derives the
standard contradiction and says he has proven it. She
responds, If
square root of two is irrational, why did you assume it was
rational at
the beginning?
So I think if you were to tell someone angle trisection is
impossible, he would be very confused.
>...
>I fear you are thinking that somehow ignorance of something
makes one a
>crank.
> If that person knows that there is a proof, then that person
> can no longer be considered as ignorant.
>That's really not what I would say at all. I would say that
>when ignorance is no longer an issue,
>if someone persists in
>maintaining angle trisection is possible (in general), then
that is the
>seed of crankdom.
> What would you mean by when ignorance is no longer an
issue?
> Are you saying that if that person just starts and keeps
believing
> that angle trisection is possible, then the guy is a c? We
do have
> to think if that guy is completely ignorant or not about
the subject.
> Cantor worked on the continuum hypothesis. But now, we know
that
> we cannot do anything about it. Is he a c? I think he was
awesome.
Well, first, he wasn't working in the same 
framework so it's
entirely
possible he could have proved CH, I think. The independence
results
about CH deal with ZF. My understanding of Cantor's set
theory is that
it was never formalized.
What I meant is that say someone sits down with the
trisector, and
explains the proof of impossibility. Then there are two
possibilities
that would indicate crankhood: 1) the trisector nods and nods
and
seemingly follows it all. But then says the proof is correct,
but
doesn't apply to his situation!
2) the trisector can't follow a part of the argument and
refuses to try
and understand, even knowing that this is a standard, famous
proof
known to all mathematicians
The other possibilities don't necessarily imply crankhood
(unless I
missed one).
Possibility 1) shows there's something fundamentally wrong 
in
the
trisector's thinking. Possibility 2) is more along the lines
of what I
think of crank most of the time. This is someone who sees the
mathematical community as a mafia, controlling what 
he's
supposed to
believe or do.
I can perfectly understand why mathematicians may look like
the mafia.
In fact, there are plenty of cultural studies folk that think
we are.
I can understand that from the outside it looks like more
prestigious
folk force the little people to go one way or another on
whether
something is true or not. But I think all this overlooks the
remarkable level of agreement among mathematicians and the
great
success in resolving disputes about correctness. I said
before that I
don't have a definition of crank, but I do have 
a kind of
operational
definition. Get two cranks in the same room and 
they'll never
agree.
Get two mathematicians in the same room and they'll agree on
most
things.
Unfortunately, the operational definition 
doesn't help when
you get a
mathematician and a crank in the same room :-)
> By the way, I got another question for you. I read a math
book and
> it says that a proof acceptable in one generation might not
be
> correct in another generation. Now, considering this
statement,
> is a person c if s/he tries to find a ßaw in an accepted
proof?
Sure, that's right, standards of proof change. This is
something
instructors are very careful to gloss over in introduction to
proof
classes :-)
It's ambiguous what you mean by accepted proof. If you mean
it was
previously accepted, when standards were clearly different, I
doubt
people would call you a crank if you said you were going to
go through
the proof and see if it holds up to today's standards.
Depending on
whether other people have done this and think it does hold
up, people
might think you were wasting your time, but that's a
different issue.
Now, if you mean accepted proof as in it's a known proof
but not
necessarily with sufficiently different standards, then things
get
tricky. For example, if you were to go through Andrew Wiles,
et al's
proof of Fermat's Last Theorem, I think people would look at
you a
little funny. Just because a lot of smart people have already
gone
through it. And there's been several books written on it.
They'd look
at you a lot less funny if you were an expert on the material.
On the other hand, if you wanted to go through Hales and
company's
proof of the Kepler Conjecture, I don't think the look you
get would be
as funny.
So it depends on the complexity of the proof and how well
documented it
is, and how trusted it is. Of course, I haven't explained
what
trusted means. But I said it was tricky :-)
Even a theorem that is not documented *at all*, say a
well-known
folklore theorem can be very trusted and you may well look
like a
crank for trying to disprove it. Even if you can't 
find anyone
that
knows how to prove it precisely! That's because some 
theorems
like
that fit so well with many trusted theorems that for many
mathematicians, that's as good as a proof.
And then of course, comparing levels of standards of proof is
often not
black and white either.
There's even been recent (within the last 25 years) proofs,
where very
few people could understand it. The author(s) say that it's
because
not enough people have the background. But this just looks
suspicious
to a lot of people. So for years people say something's a
conjecture
(instead of a theorem), and then, whaddya know, a new
generation of
mathematicians with the right background appears, and they
say it's
clearly a proof. So now it's a theorem. But then people 
argue
about
whether the original author deserves sole credit or whether
all the
people that clarified the proof actually helped prove the
theorem also.
So some people credit one guy with a theorem and some people
credit 6
guys. And some people don't even credit the original guy. 
And
some
sticklers don't even want to admit there's a 
proof!
Actually, this kind of thing is going on all the time, but
the general
public is ignorant of this. I've even witnessed some
interesting
occasions when one mathematician who thinks something is
proven meets
another mathematician who doesn't think it is! And 
it's the
most
amusing when one of them is giving a talk with the other in
the
audience :-)
===
Subject: Re: Cranks
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 
1.9 primary) id i5S1jCk15379;
...
>That is, you're no Socrates.
Neither are you, but you are a gadßy.
Anthony Natoli
===
Subject: Re: Why are Elliptic Curves So Important?
> Elliptic curves ........
> I hate to just be a cheerleader, but that was an
exceptionally
> fine post.
Yes, it was. I don't think he's a regular, but 
sci.math could
always
use more of what he's got.
===
Subject: Re: Why are Elliptic Curves So Important?
> Elliptic curves ........
> I hate to just be a cheerleader, but that was an
exceptionally
> fine post.
> Bart
I second that. Here's another question: I've 
done some
reading on
elliptic curves and regularly come into contact with what is
called
the Tate-Shafarevich group. What is its significance? It seems
important. Could someone give an example?
Craig
===
Subject: Re: Why are Elliptic Curves So Important?
>> Elliptic curves ........
>> I hate to just be a cheerleader, but that was an
exceptionally
>> fine post.
>> Bart
> I second that. Here's another question: I've 
done some
reading on
> elliptic curves and regularly come into contact with what
is called
> the Tate-Shafarevich group. What is its significance? It
seems
> important. Could someone give an example?
> Craig
the importance of the Tate-Shafarevich group.
Let's back up a bit and look at ring theory. Unique
factorization is 
useful
and some rings have it and some don't. In those that 
don't it
turns out
that the class group is non-trivial. So we say that a
non-trivial class
group is an *obstruction* to unique factorization.
If we look at a genus zero curve (conic) over the rationals
and ask if it
has a rational point we may think of looking for points
modulo a prime p.
Clearly, if it has no points modulo p it cannot have a
rational point.
There is a theorem by Hasse that says if a conic has a
rational point
modulo every prime then it has a point over the rational
numbers. This is
not as difficult to check as it sounds since in practice one
only needs to
check primes dividing the discriminant of the conic (and the
prime at
infinity which is another way of saying that the conic must
have a
solution over the real numbers).
We move up now to curves of genus one (elliptic curves) and
hope that the
Hasse theorem applies. Sometimes it does and sometimes it
doesn't. Some
elliptic curves seem to have an *obstruction* to the Hasse
principle. They
have solutions modulo all primes together with a real
solution but no
solution over the rationals. It turns out that these curves
have
non-trivial Tate-Shafarevich groups and this is an
obstruction in the same
way a non-trivial class group is an obstruction to unique
factorization.
The analogy between the two groups is even deeper. The
Dirichlet class
number formula contains the order of the class group as one
of its factors.
The Birch and Swinnerton-Dyer conjecture is a formula which
has the order
of the Tate-Shafarevich group as one of its factors. Both
involve special
values of L-functions.
You asked for an example. The first elliptic curve in
Cremona's tables
which has a non-trivial Tate-Shafarevich group is 571A with
formula
y^2 + y = x^3 - x^2 - 929 - 10595
It has a Tate-Shafarevich group of order 4.
Jack Fearnley
===
Subject: Re: Why are Elliptic Curves So Important?
> at explaining the importance of the Tate-Shafarevich group.
I felt like sitting at the fireplace and listening to
fascinating fairy tales. Someone should take a guitar now
and we should sing a bit.
all the other friendly, helpful and learned people, too.
Rainer Rosenthal
r.rosenthal@web.de
===
Subject: Re: Why are Elliptic Curves So Important?
 Elliptic curves ........
 I hate to just be a cheerleader, but that was an
exceptionally
>> fine post.

>> Bart
 I second that. Here's another question: I've 
done some
reading on
> elliptic curves and regularly come into contact with what
is called
> the Tate-Shafarevich group. What is its significance? It
seems
> important. Could someone give an example?
 Craig
> the importance of the Tate-Shafarevich group.
> Let's back up a bit and look at ring theory. Unique
factorization is 
useful
> and some rings have it and some don't. In those that 
don't
it turns out
> that the class group is non-trivial. So we say that a
non-trivial class
> group is an *obstruction* to unique factorization.
> If we look at a genus zero curve (conic) over the rationals
and ask if it
> has a rational point we may think of looking for points
modulo a prime p. 
> Clearly, if it has no points modulo p it cannot have a
rational point.
> There is a theorem by Hasse that says if a conic has a
rational point
> modulo every prime then it has a point over the rational
numbers. This 
is
> not as difficult to check as it sounds since in practice one
only needs 
to
> check primes dividing the discriminant of the conic (and
the prime at
> infinity which is another way of saying that the conic must
have a
> solution over the real numbers).
> We move up now to curves of genus one (elliptic curves) and
hope that the
> Hasse theorem applies. Sometimes it does and sometimes it
doesn't. Some
> elliptic curves seem to have an *obstruction* to the Hasse
principle. 
They
> have solutions modulo all primes together with a real
solution but no
> solution over the rationals. It turns out that these curves
have
> non-trivial Tate-Shafarevich groups and this is an
obstruction in the 
same
> way a non-trivial class group is an obstruction to unique
factorization.
> The analogy between the two groups is even deeper. The
Dirichlet class
> number formula contains the order of the class group as one
of its 
factors.
> The Birch and Swinnerton-Dyer conjecture is a formula which
has the order
> of the Tate-Shafarevich group as one of its factors. Both
involve 
special
> values of L-functions.
> You asked for an example. The first elliptic curve in
Cremona's tables
> which has a non-trivial Tate-Shafarevich group is 571A with
formula
> y^2 + y = x^3 - x^2 - 929 - 10595
> It has a Tate-Shafarevich group of order 4.
> Jack Fearnley
First, let me echo the earlier posts and say that Mr.
Fearnley's posts
are really excellent. However, let me add one point of
clarification
to his explanation of the Tate-Shafarevich group. He says
that:
Some elliptic curves seem to have an *obstruction* to the
Hasse
principle. They have solutions modulo all primes together
with a
real solution but no solution over the rationals. It turns
out that
these curves have non-trivial Tate-Shafarevich groups.
This is not quite correct. The elliptic curve itself always
has a
rational point, namely the point at infinity that acts as the
identity
element for the group law. Instead, associated to an elliptic
curve E
is a collection of other curves C_1,C_2,... called
homogeneous spaces
for E. These homogeneous spaces all have genus 1, but they
may or may
not have rational points. (More precisely, they are
isomorphic to E
over the algebraic closure of Q, but may not be isomorphic
over Q.)
Anyway, the elements of the Tate-Shafarevich group consist of
the
homogeneous spaces having the property that they have
solutions
modulo all primes together with a real solution, and such a
homogeneous space C gives a nontrivial element of the
Tate-Shafarevich
group if and only if it has no rational rational points. So
Mr.
Fearnley's explanation is essentially correct, except that 
it
applies
to the homogeneous spaces for E, not to E itself.
Examples of genus 1 curves with solutions modulo every prime
power and
in the reals, but no rational solutions, are y^2 = 68x^4 - 1
and
3x^3 + 4y^3 + 5z^3 = 0.
ax^3+by^3+cz^3=0 has this property.
One final quibble. It's not enough to have 
solutions modulo
all
primes, one actually needs solutions modulo all prime
powers, which
by the Chinese remainder theorem, is that same as having
solutions
modulo m for all integers m. A somewhat fancier way to say
this is
that there should be p-adic solutions for all p and also real
solutions, which is the same as saying there should be
solutions over
every completion of the rational numbers.
Joe Silverman
===
Subject: Re: Why are Elliptic Curves So Important?
> First, let me echo the earlier posts and say that Mr.
Fearnley's posts
> are really excellent. However, let me add one point of
clarification
> to his explanation of the Tate-Shafarevich group.
As a matter of interest, is this group known to be finite?
Does the fact that all elliptic curves over Q are modular
have any bearing on that?
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2,
Ireland
===
Subject: Re: Why are Elliptic Curves So Important?
> First, let me echo the earlier posts and say that Mr.
Fearnley's posts
> are really excellent. However, let me add one point of
clarification
> to his explanation of the Tate-Shafarevich group.
> As a matter of interest, is this group known to be finite?
> Does the fact that all elliptic curves over Q are modular
> have any bearing on that?
It is conjectured to be finite (in which case, its order is a
perfect
square), but this is only known in certain cases. My
recollection is
that the modularity combined with rank E(Q) being 0 or 1 is
enough to
prove that SHA(E/Q) is finite. If E has CM, this is due to
Rubin, and
in general, for elliptic curves that are modular (so, after
Wiles
et.al., that's all of them) it is due to Kolyvagin. This is
just for
elliptic curves over Q, not over general number fields. 
I'm
not sure
if there are any curve E/Q with rank E(Q) two or larger for
which it
is known that SHA(E/Q) is finite.
JS
===
Subject: Re: Sobolev spaces 2
>can anyone make sense of following
>let U be open,bounded in R^n.
>let H^1 denoted the Sobolev space on U with k=1,p=2
>I am reading a paper where it says
>let H denote the zero-average subspace of the dual of H^1
>If i view the dual of H^1 as H^1 then this makes sense to me.
>But if i view the dual in the way most PDE books do it ((see
previous 
post
>Sobolev Spaces )) then i have no clue.
> Well, _since_ we're defining the dual the way 
they do, the
dual
> of H^1 is another Sobolev space, hence it's a space of
distributions
> with compact support, and such things do have mean values.
> To put it another way: Since U is bounded the function 1 is
in
> H^1. If L is an element of the dual of H^1 then the mean of
> L would be <1, L>, and the space in question is the space
> of all L with <1, L> = 0.
yes that makes sense
> (Here I'm assuming that we've 
defined the dual of W(1,2)
> as W(-1,2) with the standard pairing, not as W(1,2).
> If we defined the dual to be W(1,2) then we could also
> talk about <1, L>_W, where <,>_W is the inner product
> in W(1,2), but since that inner product is not just
> (the extension of) an integral I don't think that
> <1, L> would deserve to be called the mean of L
> in that case.)
>don
> ************************
> David C. Ullrich
don
===
Subject: Re: Sobolev spaces 2
>>can anyone make sense of following
let U be open,bounded in R^n.
>>let H^1 denoted the Sobolev space on U with k=1,p=2
I am reading a paper where it says
>>let H denote the zero-average subspace of the dual of H^1
If i view the dual of H^1 as H^1 then this makes sense to
me.
>>But if i view the dual in the way most PDE books do it
((see previous 
post
>>Sobolev Spaces )) then i have no clue.
[1]
>> Well, _since_ we're defining the dual the way 
they do, the
dual
>> of H^1 is another Sobolev space, hence it's a space of
distributions
>> with compact support, and such things do have mean values.
[2]
>> To put it another way: Since U is bounded the function 1
is in
>> H^1. If L is an element of the dual of H^1 then the mean of
>> L would be <1, L>, and the space in question is the space
>> of all L with <1, L> = 0.
>yes that makes sense
Or at least [2] makes sense - having seen your post in the
other thread I now realize that [1] makes less sense than
I thought...
>> (Here I'm assuming that we've 
defined the dual of W(1,2)
>> as W(-1,2) with the standard pairing, not as W(1,2).
>> If we defined the dual to be W(1,2) then we could also
>> talk about <1, L>_W, where <,>_W is the inner product
>> in W(1,2), but since that inner product is not just
>> (the extension of) an integral I don't think that
>> <1, L> would deserve to be called the mean of L
>> in that case.)
>>don
>> ************************
>> David C. Ullrich
>don
************************
David C. Ullrich
===
Subject: Re: Guess the Continued Fractions' Value
In both cases, the answer is 1/2.
I am only posting this answer as an attempt to revive
interest in this
topic (and let the author know that the World is not ignoring
him). I am
not spoiling anybody's fun in the process, since the fun 
part
is to prove
that this is indeed so...
Note that the term continued fraction is normally used only
when the
so-called partial quotients [the sequence of quantities
given here] are
integers (all of them should be positive, with the possible
exception of 
the
first one).
Since we are dealing with recreational mathematics here, 
it's
probably OK 
to
use the term to denote identical algebraic manipulations of
unrestricted
quantities, but the lack of a preliminary explanation may
well have hurt 
the
initial appeal of this puzzle for most readers... [Just a
thought.]
For an introduction to continued fractions, I'll shamelessly
plug my own
www.numericana.com/answer/fractions.htm
Gerard P. Michon, Ph.D.
www.numericana.com
> I am posting this as a puzzle, even though it should be
easy to solve
> for many math-types.
> I find the result interesting, even if not earth-shattering.
> Puzzle:
> Find the closed form for these continued fractions:
> [2;-2/3,-2*3,2/3^2,2*3^2,-2/3^3,-2*3^3,2/3^4,2*3^4,....]
> and
> [2;-1/2,-2^2,1/2^2,2^3,-1/2^3,-2^4,1/2^4,2^5,-1/2^5,....]
> their patterns.
> In each, we have 2 -'s alternating with 2 
+'s.)
> Leroy Quet
> Come on, this is easy.
> For the answer to both is
> one...
> of the most simply understood mathematical constants known.
> ;)
> Just to be clear,
> the terms of the first continued fraction follow the 
pattern:
> ...-2/3^(2k-1), -2*3^(2k-1), 2/3^(2k), 2*3^(2k),...
> as k increases by 1.
> And the second CF's terms follow the pattern:
> ...-1/2^(2k-1), -2^(2k), 1/2^(2k), 2^(2k+1),...
> (Actually, I hope I have not made a mistake. So if someone
comes up
> with a different solution than I have, and can prove their
solution is
> correct, I will easily accept their solution as the correct
answer.)
> Leroy Quet
===
Subject: Re: Guess the Continued Fractions' Value
> In both cases, the answer is 1/2.
Are you sure?
I got 1 for both continued fractions' values.
I have been error-prone lately, I know. But I checked the
second CF
numerically up to a few terms on my hand-calculator, and it
seems to
be nearing 1, not 1/2.
Maybe someone who reads sci.math or rec.puzzles should check
the
continued fractions numerically up to, say, their 50th
convergents,
just to see if 1 is a likely value for both CFs.
> I am only posting this answer as an attempt to revive
interest in 
this
> topic (and let the author know that the World is not
ignoring him). I am
> not spoiling anybody's fun in the process, since the fun
part is to prove
> that this is indeed so...
> Note that the term continued fraction is normally used
only when the
> so-called partial quotients [the sequence of quantities
given here] 
are
> integers (all of them should be positive, with the possible
exception of 
the
> first one).
> Since we are dealing with recreational mathematics here,
it's probably OK 
to
> use the term to denote identical algebraic manipulations of
unrestricted
> quantities, but the lack of a preliminary explanation may
well have hurt 
the
> initial appeal of this puzzle for most readers... [Just a
thought.]
> For an introduction to continued fractions, I'll
shamelessly plug my own
> www.numericana.com/answer/fractions.htm
Interesting link.
I should say that, in my opinion, the term continued
fraction should
apply when the terms are not necessarily positive integers as
well.
I think I read that a continued fraction with terms which
specifically
are all (after first term) positive integers is called a
simple
continued fraction.
Is this right?
By the way, I am posting another such puzzle in another
thread (with a
similar title).
Leroy Quet
> Gerard P. Michon, Ph.D.
> www.numericana.com
> I am posting this as a puzzle, even though it should be
easy to solve
> for many math-types.
> I find the result interesting, even if not
earth-shattering.
 Puzzle:
 Find the closed form for these continued fractions:
 [2;-2/3,-2*3,2/3^2,2*3^2,-2/3^3,-2*3^3,2/3^4,2*3^4,....]
 and
 [2;-1/2,-2^2,1/2^2,2^3,-1/2^3,-2^4,1/2^4,2^5,-1/2^5,....]
 their patterns.
> In each, we have 2 -'s alternating with 2 
+'s.)
 Leroy Quet
> Come on, this is easy.
> For the answer to both is
> one...
> of the most simply understood mathematical constants known.
> ;)
> Just to be clear,
> the terms of the first continued fraction follow the 
pattern:
> ...-2/3^(2k-1), -2*3^(2k-1), 2/3^(2k), 2*3^(2k),...
> as k increases by 1.
> And the second CF's terms follow the pattern:
> ...-1/2^(2k-1), -2^(2k), 1/2^(2k), 2^(2k+1),...
> (Actually, I hope I have not made a mistake. So if someone
comes up
> with a different solution than I have, and can prove their
solution is
> correct, I will easily accept their solution as the correct
answer.)
> Leroy Quet
===
Subject: Re: Guess the Continued Fractions' Value
ERRATUM: The answer in both cases is (indeed) 1.
About the issue of vocabulary, the term continued fractions
has been 
used
for several other things in the past (including expressions
where the
successive numerators were not unity). The qualifier simple
and/or
regular was then added to distinguish what has become the
modern 
meaning
of the [unqualified] term (see
www.numericana.com/answer/fractions.htm ).
As the other types of continued fractions are now
deprecated, the
qualifier has been dropped from most discussions of the
subject. What you
are discussing here is STILL a simple and/or regular
continued 
fraction,
except that you allow partial quotients that are not
positive integers.
You are free to do so but should warn your readers ahead of
time [that most
of the literature on continued fraction does not apply to
your use of the
term]. The reason for this is to cherish and preserve the
existing harmony
which allows everyone to use of classic continued fractions
in their
writings without the need for a similar warning [in the other
direction] or
the fear that the accepted meaning of the term is under
attack...
For example, at
www.numericana.com/answer/fractions.htm#functions I do
discuss the expansion of (some) analytical functions in the
spirit 
[sic]
of continued fractions. Mathematics is all about the search
for patterns
and similarities, and it's useful to do such things, as long
as one is
careful not to switch [without warning] the meaning of basic
words...
Gerard P. Michon, Ph.D.
www.numericana.com
> In both cases, the answer is 1/2.
> Are you sure?
> I got 1 for both continued fractions' values.
> I have been error-prone lately, I know. But I checked the
second CF
> numerically up to a few terms on my hand-calculator, and it
seems to
> be nearing 1, not 1/2.
> Maybe someone who reads sci.math or rec.puzzles should
check the
> continued fractions numerically up to, say, their 50th
convergents,
> just to see if 1 is a likely value for both CFs.
> I am only posting this answer as an attempt to revive
interest in 
this
> topic (and let the author know that the World is not
ignoring him). I
am
> not spoiling anybody's fun in the process, since the fun
part is to
prove
> that this is indeed so...
> Note that the term continued fraction is normally used
only when 
the
> so-called partial quotients [the sequence of quantities
given here]
are
> integers (all of them should be positive, with the possible
exception 
of
the
> first one).
> Since we are dealing with recreational mathematics here,
it's probably
OK to
> use the term to denote identical algebraic manipulations of
unrestricted
> quantities, but the lack of a preliminary explanation may
well have 
hurt
the
> initial appeal of this puzzle for most readers... [Just a
thought.]
> For an introduction to continued fractions, I'll
shamelessly plug my 
own
> www.numericana.com/answer/fractions.htm
> Interesting link.
> I should say that, in my opinion, the term continued
fraction should
> apply when the terms are not necessarily positive integers
as well.
> I think I read that a continued fraction with terms which
specifically
> are all (after first term) positive integers is called a
simple
> continued fraction.
> Is this right?
> By the way, I am posting another such puzzle in another
thread (with a
> similar title).
> Leroy Quet
> Gerard P. Michon, Ph.D.
> www.numericana.com
> I am posting this as a puzzle, even though it should be
easy to
solve
> for many math-types.
> I find the result interesting, even if not
earth-shattering.
 Puzzle:
 Find the closed form for these continued fractions:
 [2;-2/3,-2*3,2/3^2,2*3^2,-2/3^3,-2*3^3,2/3^4,2*3^4,....]
 and
 [2;-1/2,-2^2,1/2^2,2^3,-1/2^3,-2^4,1/2^4,2^5,-1/2^5,....]
 their patterns.
> In each, we have 2 -'s alternating with 2 
+'s.)
 Leroy Quet

 Come on, this is easy.
> For the answer to both is
> one...
> of the most simply understood mathematical constants
known.
> ;)
 Just to be clear,
> the terms of the first continued fraction follow the
pattern:
 ...-2/3^(2k-1), -2*3^(2k-1), 2/3^(2k), 2*3^(2k),...
 as k increases by 1.
 And the second CF's terms follow the pattern:
 ...-1/2^(2k-1), -2^(2k), 1/2^(2k), 2^(2k+1),...

> (Actually, I hope I have not made a mistake. So if
someone comes up
> with a different solution than I have, and can prove
their solution 
is
> correct, I will easily accept their solution as the
correct answer.)

> Leroy Quet
===
Subject: Re: Fermat's Last Theorem: A short proof?
> Again, you have an interesting point concerning the
> convergence/divergence of infinite series.
> This point was also raised by Jeroen Boschma concerning the
symmetry
> of the series with respect to a and b (Sorry, Jeroen, for
the lack of
> a separate post; it would look just like this one.)
> If the series is posed as I posed it in my attempted proof,
the series
> converges because a/b < 1, which was an initial assumption
(i.e., b  a). This assumption is valid because a = b guarantees that
c is never
> an integer via c^n = a^n + b^n, so a = b can be further
ignored. It
> follows that one has to be larger than the other, so I
chose b > a.
> Making the opposite choice does not change the result, only
the labels
> within the proof.
> With this choice made, I am free to pose infinite series
within their
> radii of convergence. This is all that I did, and the rest
of the
> proof apparently follows.
You've missed the point. It's not a question 
of whether or not
the series converge, it's a question of what manipulations 
are
permissible with infinite series.
I'll try again.
Infinite series are sensible only as limits.
a_1 + a_2 + ... means limit as n -> infinity of s_n,
where s_n = a_1 + a_2 + ... + a_n.
Now suppose the terms a_1, a_2, ..., all depend on some
parameter p,
so we're really talking about a_1(p) + a_2(p) + ....
The value of this infinite series (if it has one) is
the limit as n -> infinity of s_n(p),
where s_n(p) = a_1(p) + a_2(p) + ... + a_n(p).
Now you also want to take a limit as p -> infinity.
Do you take the limit on p first, then the limit on n?
or do you do it the other way around?
The point is, it makes a difference;
lim n -> infinity (lim p -> infinity s_n(p)) and
lim p -> infinity (lim n -> infinity s_n(p)) are 
not always
equal.
But if you think about your argument you may find that it
depends
precisely on these two quantities being equal.
You can't, in general, find the limit of a sum 
of terms by
finding
the sum of the limit of the terms.
And, anyway you've ignored my main point, which 
I'll repeat
here,
for your benefit:
> That's a small problem. A much bigger problem is that some
of the best
> mathematical minds in history worked on Fermat, and over
the course of
> 300 years got nowhere, so you aren't going to get anywhere
using nothing 
> more advanced than the binomial theorem. You have a better
chance of
> using a chemistry set to find a cure for cancer than you
have of finding 
> a proof of Fermat using this kind of stuff.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: approximating a composite integer number
> I was wondering: suppose I have a composite number N and a
given set of
> primes P={p1, p2, ..., pm}.
> Some elements of P may be equal to eachother (i.e. some
primes may occur
> multiple times in P). I
> want to approximate N by using only primes element of P.
> For a certain N, the factors of N may not all be element of
P. Assuming 
that
> prod_{i=1}^m pi > N,
> then I can (at least) approximate N using the elements of P.
> Question: how do choose elements of P to get the best
approximation of N?
Looks to me like a bin-packing problem.
You've got a bin of capacity log N,
and you are trying to fill it, as nearly as possible,
with parcels of volume log p_1, log p_2, ..., log p_m.
The bad news is that bin-packing is, in general, hard.
The good news is that some efficient approximate algorithms
are known.
Have a look.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Fermat's Last Theorem: A short proof? (rev.)
> You have an interesting idea, but I do not think that it
helps or
> hurts my attempted proof.
> I think that it is enough to show that any combination of
positive
> integers (a,b,n>2) cannot produce an integer value for c^2
via c^n =
> a^n + b^n. In my view, it is much easier to consider
positive-integer
> triplets (a,b,n>2) to learn about values of c^2 than to
consider
> positive-integer triplets (a,b,c) to learn about values of
n. I
> submit that the former requires less searching to achieve
the same
> proof.
You've utterly missed the point.
What distinguishes a mathematician from a crank is that a
crank
tries to prove himself right, whereas a mathematician tries to
prove herself wrong. A mathematician puts all her effort into
trying to show that her own work can't possibly be correct,
and
only when repeated attempts have utterly failed does she
suggest
that just maybe she has found something new and worthwhile.
I've given you the opportunity to prove yourself wrong -
something
which you should have seen yourself, but in any event an
opportunity
that you should run to embrace, if you have any intellectual
integrity
whatsoever.
I'll try again.
Assume that a, b, and n are positive integers with b > a.
Define c by c^n = b^n + a^n.
Raise both sides to the power 3/n;
c^3 = (b^n + a^n)^(3/n).
Expand the right side using the binomial theorem (just as you
do
in your argument with (b^n + a^n)^(2/n)).
If n = 2 the right side will be an infinite series.
Now in your argument you do a bunch of stuff and wind up
concluding
that c^2 can't be an integer (so c can't be an 
integer, so
Fermat).
I say, do the analogous stuff in the new context and see
whether you
don't conclude that c^3 can't an integer. If 
you do conclude
that
c^3 can't be an integer, you have succeeded in the worthy 
goal
of showing that your original argument is bogus, and
congratulations
will be in order.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Surrogate factoring
>Hint: if you haven't worked out the details, in particular
given
>a proof that it _is_ better than chance, then you _are_ just
>speculating.
>Exactly _how_ did you confirm that this is better than just
>trying random factors?
I see this was crossposted to sci.crypt from sci.math... so
you did
not have to use Google to find that sci.crypt was under
attack, and
John Savard
http://home.ecn.ab.ca/~jsavard/index.html
===
Subject: Re: Surrogate factoring
> (jk - Tk + sT)(jk + Tk + sT) = T^4
> so
> k = (-jsT +/- T^2 sqrt(j^2 - T^2 + s^2))/(j^2 - T^2)
> whee you can pick s.
 So now, factoring T is dependent on the factorization of
T^2-s^2
> Would you care to elaborate how, from the factorisation of
T^2-s^2,
> one finds a suitable integer j?
Well, amazingly enough it's as simple as with my quadratic
example.
In my original post if you'll remember I had
x^2 + ax - T = 0, so
x = (-a +/- sqrt(a^2 + 4T))/2
and if you already knew the factors of T then calculating 
'a'
is
trivial.
(In case you don't know how, if T = f_1 f_2, then a = f_1 -
f_2 or a =
f_2 - f_1, so yes, it's trivial.)
The problem, of course, is that if T is your target, you
don't know
what its factors are!
However, using my factorization, you can just pick some s,
and then
factor T^2-s^2 instead, and then calculate j.
I've typically considered j=b/a, assuming it's 
a fraction,
but I just
realized that an integer j might solve things as well.
What I have is a wide open area, with very little theory or
methodology worked out, which came from noticing that there
was this
factorization
(jk - Tk + sT)(jk + Tk + sT) = T^4
where unlike with my x^2 + ax - T example, you get
k = (-jsT +/- T^2 sqrt(j^2 - T^2 + s^2))/(j^2 - T^2)
where the square root gives an integer dependent NOT on the
factors of
T, but on the factors of T^2-s^2, which is rather
extraordinary, and
YOU get to pick s.
I just thought about this a few weeks ago, and have made only
initial
efforts at exploring it.
The reason it might be a big deal is that public key
encryption
depends on finding large numbers with only two large prime
factors
specially chosen so that their product is hard to factor, BUT
maybe
it's possible to shift to another large number that will 
have
many
prime factors.
My hope is that some number theorists will show interest.
> One way to explain would be to illustrate with an example.
Say I take
> T = 178765164017621646307.
> Trying to factor T^2-s^2 for odd s>0, I find that for s=15
> T^2-s^2 =
8*11*13*307*347*373*50153*124471*42027701*326147*8215783
> and indeed, factoring T^2-s^2 is easier than factoring T.
Now how do
> we proceed?
> Feel free to use a smaller example; or give an algorithm,
I'll compute
> the example.
I know it may be hard for some of you to handle this
situation as
cryptology is so well-worked, and there are so many theories
that have
been put together with so many methodologies, but at this
point it's
at the ground stage.
There is no well-worked out theory to lean on, and no
full-scale
methodology that I'm aware of, but just an idea.
If it's too early for you, fine, 
don't worry about it.
My hope is that some number theorists might get interested in
this
development.
Algebraically it seems to be a way to factor by using a
surrogate
factorization which leads to a factoring of your target.
If it can be made to work, then that would probably be
significant.
James Harris
===
Subject: Re: Surrogate factoring
> Would you care to elaborate how, from the factorisation of
T^2-s^2,
> one finds a suitable integer j?
> Well, amazingly enough it's as simple as with my quadratic
example.
Would you please explain this technique? Could not extract it
from
earlier posts.
> I just thought about this a few weeks ago, and have made
only initial
> efforts at exploring it. (..)
> My hope is that some number theorists will show interest.
Interest will come if you explicit a method to find j. Until
you get
one, any enthusiam is way premature
Claude Leroy
===
Subject: Re: Surrogate factoring
> Would you care to elaborate how, from the factorisation
of T^2-s^2,
> one finds a suitable integer j?
 Well, amazingly enough it's as simple as with my quadratic
example.
> Would you please explain this technique? Could not extract
it from
> earlier posts.
> I just thought about this a few weeks ago, and have made
only initial
> efforts at exploring it. (..)
> My hope is that some number theorists will show interest.
> Interest will come if you explicit a method to find j. Until
you get
> one, any enthusiam is way premature
> Claude Leroy
I don't disagree. Like I said at this point 
it's just an idea
that I
came up with a few weeks ago. For most of you I'd think 
it'd
be of no
interest as you're more into well-established methods with a
lot of
theory worked out.
I'm more posting for those who might be interested in
something that
looks like it is new. Then again, maybe someone can connect
it fully
to established approaches.
So why do I post about it?
It's like with open source. I could sit on this idea, try to
develop
it and maybe go for the RSA Challenge, but with something
this new,
one of you might notice something quick that shoots it down
completely. Or someone out there might have some insights I
don't and
advance the theory rapidly...more rapidly than I can alone.
So I've put it out there. I don't mind talking 
about the
idea, but
mostly I think I'll be saying that it's a new 
idea, and
there's not a
lot of theory worked out yet.
It may impact public key encryption as if it can be made
practical,
it'd mean that given a number T to factor you could shift to
factoring
T^2-s^2 instead, where you pick s, which should nullify
picking large
primes p_1 and p_2 carefully so that their product is hard to
factor.
That's the pull of the idea. The question is, can this
approach be
made practical?
James Harris
===
Subject: Re: Surrogate factoring
>> (jk - Tk + sT)(jk + Tk + sT) = T^4
>> so
>> k = (-jsT +/- T^2 sqrt(j^2 - T^2 + s^2))/(j^2 - T^2) whee
you can
>> pick s.
 So now, factoring T is dependent on the factorization of
T^2-s^2
>> Would you care to elaborate how, from the factorisation of
T^2-s^2, one
>> finds a suitable integer j?
> Well, amazingly enough it's as simple as with my quadratic
example.
> In my original post if you'll remember I had
> x^2 + ax - T = 0, so
> x = (-a +/- sqrt(a^2 + 4T))/2
> and if you already knew the factors of T then calculating
'a' is
> trivial.
> (In case you don't know how, if T = f_1 f_2, then a = f_1 
-
f_2 or a =
> f_2 - f_1, so yes, it's trivial.)
> The problem, of course, is that if T is your target, you
don't know what
> its factors are!
> However, using my factorization, you can just pick some s,
and then
> factor T^2-s^2 instead, and then calculate j.
> I've typically considered j=b/a, assuming 
it's a fraction,
but I just
> realized that an integer j might solve things as well.
> What I have is a wide open area, with very little theory or
methodology
> worked out, which came from noticing that there was this
factorization
> (jk - Tk + sT)(jk + Tk + sT) = T^4
> where unlike with my x^2 + ax - T example, you get
> k = (-jsT +/- T^2 sqrt(j^2 - T^2 + s^2))/(j^2 - T^2)
> where the square root gives an integer dependent NOT on the
factors of
> T, but on the factors of T^2-s^2, which is rather
extraordinary, and YOU
> get to pick s.
> I just thought about this a few weeks ago, and have made
only initial
> efforts at exploring it.
> The reason it might be a big deal is that public key
encryption depends
> on finding large numbers with only two large prime factors
specially
> chosen so that their product is hard to factor, BUT maybe
it's possible
> to shift to another large number that will have many prime
factors.
> My hope is that some number theorists will show interest.
>> One way to explain would be to illustrate with an example.
Say I take
>> T = 178765164017621646307.
>> Trying to factor T^2-s^2 for odd s>0, I find that for s=15
>> T^2-s^2 =
8*11*13*307*347*373*50153*124471*42027701*326147*8215783
>> and indeed, factoring T^2-s^2 is easier than factoring T.
Now how do we
>> proceed?
>> Feel free to use a smaller example; or give an algorithm,
I'll compute
>> the example.
> I know it may be hard for some of you to handle this
situation as
> cryptology is so well-worked, and there are so many
theories that have
> been put together with so many methodologies, but at this
point it's at
> the ground stage.
> There is no well-worked out theory to lean on, and no
full-scale
> methodology that I'm aware of, but just an idea.
> If it's too early for you, fine, 
don't worry about it.
> My hope is that some number theorists might get interested
in this
> development.
> Algebraically it seems to be a way to factor by using a
surrogate
> factorization which leads to a factoring of your target.
> If it can be made to work, then that would probably be
significant.
> James Harris
James, this problem is extremely trivial. To factor n, a
product of two
primes:
(1) Let t<>n be some number such that 1<gcd(t,n)<n.
(2) Compute gcd(t,n) using Euclid's algorithm.
(3) Clearly gcd(t,n) must be one of n's two prime factors.
Use this to
find the other prime factor.
* A note on picking t: If you don't already know the two
prime factors,
use magic. If you do know the two prime factors (p and q),
pick something
like 7p.
--
Lance Lamboy
I tell them the truth and they think it's hell. ~ Harry S.
Truman
===
Subject: Re: Surrogate factoring
> For some time it has been known that given a target
integer T, you 
can
> factor it using something like
> x^2 + ax - T = 0,
> which can be solved by the quadratic formula to give
> x = (-a +/- sqrt(a^2 + 4T))/2
> and if you can find an 'a' that 
is an integer that works,
then 
there's
> a good chance you'll have non-trivially factored T, since
 The hard part here is finding a 
'a' that works. Note that
your 
requirement
> is stronger than the requirement in more modern methods
where an 
attempt
> is made to find an integer 'a' 
such that 'a^2 = b^2 mod T'
for some 
'b'.
 So now, factoring T is dependent on the factorization of
T^2-s^2, 
and
> the method does work.
 Yes, the method is quite old, although mostly performed
with s = 1. 
See
> Knuth, the Art of Computer Programming. In many cases it
relies on 
other
> numbers to be factored, and these new numbers can also be
quite large.
 Apparently the OP is talking about Pollard's.
The OP is a poster called James Harris... But I did not know
that the
methods Knuth describes are Pollard's. At least, 
Pollard's
method does
not rely on factoring 'n - 1'.
> Unfortunately there are several posters from the sci.math
newsgroup
> who typically just make ßuff posts which I guess are meant
to confuse
> the issue.
> Who knows what motivates these people but they throw up a
lot of
> distractions, and don't often really seem to know much
about the
> subject at hand.
Oh, so be it. I have been involved in factoring numbers some
10 years
ago. (Probably more, I disremember.) I know how MPQS works,
how other
methods work (NFS). Do *you* know how MPQs works? Have a look
at the
people that cracked the first RSA challenge.
> This Dik Winter is one of the worst of the lot, and he also
has
> several webpages where he copied old ßawed arguments of
mine from
> Usenet and put them on webpages with his own commentary, of
course.
> I told him to take off my Usenet posts, he refuses to do so.
I have told you *why* I refuse to do so. As long as you
continue
reasoning as you did at that time, I see no reason to pull
them off
the net. Moreover, they provide an excellent example of the
behaviour of some of the people who do not understand.
> Anyone out there understand that kind of behavior?
> I find it curious.
Yes, just you. Attacking the person, not the content.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland, 
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Surrogate factoring
> As for why this idea is of any interest at all, consider
that popular
> Internet public key encryption methods depend on picking
large primes p_1
> and p_2 which are carefully chosen such that T = p_1 p_2 is
hard to
> factor.
Actually (for RSA, anyway), I don't believe the primes are
carefully
chosen anymore. Randomly chosen is good enough. It used to
be
recommended that you use so-called strong primes, but
advances in 
general
purpose factoring algorithms have made it so that products of
strong primes
aren't harder to factor than products of random primes.