mm-1080
===
Subject: Re: JSH:Understanding constant terms
> Where am I going wron ?
> Ill explain.
Yes,I think I understand you now.
To recap
A g_i(x) is any function that produces an algebraic integer
when x is an algebraic integer
The roots r1,r2 of x^2 -2x +2 are akgebraic integers
The roots, s1,s2,s3 of x^3 -7x +5 are likewise
algebraic integers so
g_1(x)= r1*x^(1/2) + s2*x + 2
g_2(x)= s2*x^(1/2) + r3
give algebaric integers when x is an algebraic
integer.
Their product P(x)is a polynomial in x^(1/2)
and g_1(0)*g_2(0) = 2 r3 is an algebraic integer.
So where does that get us ?
===
Subject: Re: JSH:Understanding constant terms
>If and only if the gs are monic polys
> No. If and only if the values g_i(0) are algebraic
integers. This will
> certainly happen if the g_i are monic polynomials with
integer
> coefficients, but it can happen in other cases as well (for
example,
> if the g_i are ANY polynomials with integer constant terms).
Dont the g_i, polys in x, also have to be
algebraic integers whenever x is an algebraic
integer ? So, say 6x^2 +(pi)x + 2 couldnt
be a g_1 even though its constant term
is a rational integer.
===
Subject: Re: JSH:Understanding constant terms
>>If and only if the gs are monic polys
>> No. If and only if the values g_i(0) are algebraic
integers. This will
>> certainly happen if the g_i are monic polynomials with
integer
>> coefficients, but it can happen in other cases as well (for
example,
>> if the g_i are ANY polynomials with integer constant
terms).
>Dont the g_i, polys in x, also have to be
>algebraic integers whenever x is an algebraic
>integer ?
Im sorry. Was that English?
You were assuming that the g_i(x) were polynomials, and that
the
product of the g_i(x) was equal to P(x). Then your statement
if and
only if the gs are monic polys was about the statement that
each
g_i(0) divides, in the algebraic integers, P(0).
But that is false. For each g_i(0) to divide P(0) in the ring
of all
algebraic integers, it is necessary and sufficient for each of
the
values g_i(0) to be algebraic integers. For certainly we need
each
g_i(0) to be an algebraic integer. And if they all are, then
clearly
each of them divides P(0) in the ring of all algebraic
integers.
You dont need the g_i(0) to be monic; you seemed to be led
astray by
the fact that g_i(0) is plus or minus the product of the
roots of
g_i(x). Certainly if g_i(x) is a monic polynomial with
algebraic
integer coefficients, then g_i(0) will be an algebraic
integer; but it
is possible for g_i(0) to be an algebraic integer without all
roots of
g_i(x) to be algebraic integer.
> So, say 6x^2 +(pi)x + 2 couldnt
>be a g_1 even though its constant term
>is a rational integer.
Doesnt matter. Your assertion was about the g_i(0) dividing
P(0) in
the algebraic integers. For that statement, you dont need
the g_i(x)
to be monic (note that P(x) was not monic either).
--
Its not denial. Im just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: JSH:Understanding constant terms
>>If and only if the gs are monic polys
>>
>> No. If and only if the values g_i(0) are algebraic
integers. This will
>> certainly happen if the g_i are monic polynomials with
integer
>> coefficients, but it can happen in other cases as well (for
example,
>> if the g_i are ANY polynomials with integer constant
terms).
>Dont the g_i, polys in x, also have to be
>algebraic integers whenever x is an algebraic
>integer ?
> Im sorry. Was that English?
I cant say that Ive found your
prose particularly pristine.
Are you a non-native English speaker ?
> You were assuming that the g_i(x) were polynomials, and
that the
> product of the g_i(x) was equal to P(x). Then your
statement if and
> only if the gs are monic polys was about the statement
that each
> g_i(0) divides, in the algebraic integers, P(0).
According to Harris, the g_i(x) are algebraic integers
whenever x is ( his algebraic integer functions).
Clearly g_i(0) has to be an algebraic
integer and so the constant term of g_i(x)has to be an
algerbaic integer.
You say that the gi can be ANY polynominal.
This is clearly not the case. For example pi*x +1.
Are you saying that p1*1 +1 is an alegebraic integer ?
> But that is false. For each g_i(0) to divide P(0) in the
ring of all
> algebraic integers, it is necessary and sufficient for each
of the
> values g_i(0) to be algebraic integers.
Yes,
> For certainly we need each
> g_i(0) to be an algebraic integer. And if they all are,
then clearly
> each of them divides P(0) in the ring of all algebraic
integers.
> You dont need the g_i(0) to be monic; you seemed to be 
led
astray by
> the fact that g_i(0) is plus or minus the product of the
roots of
> g_i(x).
No, that was not the problem.
Certainly if g_i(x) is a monic polynomial with algebraic
> integer coefficients, then g_i(0) will be an algebraic
integer; but it
> is possible for g_i(0) to be an algebraic integer without
all roots of
> g_i(x) to be algebraic integer.
Yes.
> So, say 6x^2 +(pi)x + 2 couldnt
>be a g_1 even though its constant term
>is a rational integer.
> Doesnt matter.
Im afraid it does matter if you are using the g_i(x)
as defined by Harris.
===
Subject: Re: JSH:Understanding constant terms
days. My association with the Department is that of an
alumnus.
>If and only if the gs are monic polys
>
> No. If and only if the values g_i(0) are algebraic
integers. This
will
> certainly happen if the g_i are monic polynomials with
integer
> coefficients, but it can happen in other cases as well (for
example,
> if the g_i are ANY polynomials with integer constant terms).
>Dont the g_i, polys in x, also have to be
>>algebraic integers whenever x is an algebraic
>>integer ?
>> Im sorry. Was that English?
>I cant say that Ive found your
>prose particularly pristine.
I do try to avoid the use of cute abbreviations like polys,
though.
>Are you a non-native English speaker ?
Indeed. English is, alas, my fourth language. It often shows.
>> You were assuming that the g_i(x) were polynomials, and
that the
>> product of the g_i(x) was equal to P(x). Then your
statement if and
>> only if the gs are monic polys was about the statement
that each
>> g_i(0) divides, in the algebraic integers, P(0).
>According to Harris, the g_i(x) are algebraic integers
>whenever x is ( his algebraic integer functions).
>Clearly g_i(0) has to be an algebraic
>integer and so the constant term of g_i(x)has to be an
>algerbaic integer.
>You say that the gi can be ANY polynominal.
Please note that I was not addressing Jamess argument. I 
was
addressing your restriction to the g_i being polynomials, and
your
assertion, ->under that further restriction<-, that each
g_i(0) would
divide P(0) in the ring of all algebraic integers if and only
if the
g_i(x) were monic. ->That<- assertion is false.
>This is clearly not the case. For example pi*x +1.
>Are you saying that p1*1 +1 is an alegebraic integer ?
No. I am saying that if g(x)=pi*x + 1, then it is entirely
possible
for g(0) to divide P(0) in the ring of all algebraic
integers, EVEN
THOUGH g(x) is not a monic polynomial.
Hell: take g1(x) = 2x+1 and g2(x) = x-1. Then
g1(x)*g2(x) = P(x) = 2x^2 - x - 1.
g1(0) and g2(0) both divide P(0) in the ring of all algebraic
integers, EVEN THOUGH g1(x) is not monic.
You claimed the divisibility conclusion would hold if and
only if the
gs are monic poly[nomial]s.
The if is true. The only if is false.
--
Its not denial. Im just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: JSH:Understanding constant terms
>If and only if the gs are monic polys
>
> No. If and only if the values g_i(0) are algebraic
integers. This
will
> certainly happen if the g_i are monic polynomials with
integer
> coefficients, but it can happen in other cases as well (for
example,
> if the g_i are ANY polynomials with integer constant terms).
>Dont the g_i, polys in x, also have to be
>>algebraic integers whenever x is an algebraic
>>integer ?
>>
>> Im sorry. Was that English?
>I cant say that Ive found your
>prose particularly pristine.
> I do try to avoid the use of cute
> abbreviations like polys, though.
Cute ? Poly used to be a widespr. abb.
for polytechnic. Ive never heard
it reviled on the basis of cuteness before.
But then you learn many strange things on sci.math.
>Are you a non-native English speaker ?
> Indeed. English is, alas, my fourth language. It often
shows.
>> You were assuming that the g_i(x) were polynomials, and
that the
>> product of the g_i(x) was equal to P(x). Then your
statement if and
>> only if the gs are monic polys was about the statement
that each
>> g_i(0) divides, in the algebraic integers, P(0).
>According to Harris, the g_i(x) are algebraic integers
>whenever x is ( his algebraic integer functions).
>Clearly g_i(0) has to be an algebraic
>integer and so the constant term of g_i(x)has to be an
>algerbaic integer.
>You say that the gi can be ANY polynominal.
> Please note that I was not addressing Jamess argument. I
was
> addressing your restriction to the g_i being polynomials,
and your
> assertion, ->under that further restriction<-, that each
g_i(0) would
> divide P(0) in the ring of all algebraic integers if and
only if the
> g_i(x) were monic. ->That<- assertion is false.
Yes, I agree
>This is clearly not the case. For example pi*x +1.
>Are you saying that p1*1 +1 is an alegebraic integer ?
> No. I am saying that if g(x)=pi*x + 1, then it is entirely
possible
> for g(0) to divide P(0) in the ring of all algebraic
integers, EVEN
> THOUGH g(x) is not a monic polynomial.
Yes, but we now know that the g_i(x)
must meet another condition.
What I said prev. was incorrect,
but your statement that ANY polyno.
would do is crassly passe in the light
of new revelations.
===
Subject: Re: JSH:Understanding constant terms
days. My association with the Department is that of an
alumnus.
[.snip.]
> You were assuming that the g_i(x) were polynomials, and
that the
> product of the g_i(x) was equal to P(x). Then your
statement if and
> only if the gs are monic polys was about the statement
that each
> g_i(0) divides, in the algebraic integers, P(0).
>According to Harris, the g_i(x) are algebraic integers
>>whenever x is ( his algebraic integer functions).
>>Clearly g_i(0) has to be an algebraic
>>integer and so the constant term of g_i(x)has to be an
>>algerbaic integer.
>You say that the gi can be ANY polynominal.
>> Please note that I was not addressing Jamess argument. I
was
>> addressing your restriction to the g_i being polynomials,
and your
>> assertion, ->under that further restriction<-, that each
g_i(0) would
>> divide P(0) in the ring of all algebraic integers if and
only if the
>> g_i(x) were monic. ->That<- assertion is false.
>Yes, I agree
>>This is clearly not the case. For example pi*x +1.
>>Are you saying that p1*1 +1 is an alegebraic integer ?
>> No. I am saying that if g(x)=pi*x + 1, then it is entirely
possible
>> for g(0) to divide P(0) in the ring of all algebraic
integers, EVEN
>> THOUGH g(x) is not a monic polynomial.
>Yes, but we now know that the g_i(x)
>must meet another condition.
>What I said prev. was incorrect,
>but your statement that ANY polyno.
>would do is crassly passe in the light
>of new revelations.
So... I addressed your comment as written. Since that
comment, you
have now changed your mind about what the g_i(x) should be;
based on
that change of heart, you complain that my correction was
wrong,
because it failed to meet your newfound and not-previously
asserted
conditions on the g_i(x).
How... interesting...
--
Its not denial. Im just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: JSH:Understanding constant terms
> [.snip.]
> You were assuming that the g_i(x) were polynomials, and
that the
> product of the g_i(x) was equal to P(x). Then your
statement if
and
> only if the gs are monic polys was about the statement
that each
> g_i(0) divides, in the algebraic integers, P(0).
>According to Harris, the g_i(x) are algebraic integers
>>whenever x is ( his algebraic integer functions).
>>Clearly g_i(0) has to be an algebraic
>>integer and so the constant term of g_i(x)has to be an
>>algerbaic integer.
>You say that the gi can be ANY polynominal.
>>
>> Please note that I was not addressing Jamess argument. I
was
>> addressing your restriction to the g_i being polynomials,
and your
>> assertion, ->under that further restriction<-, that each
g_i(0) would
>> divide P(0) in the ring of all algebraic integers if and
only if the
>> g_i(x) were monic. ->That<- assertion is false.
>Yes, I agree
>>This is clearly not the case. For example pi*x +1.
>>Are you saying that p1*1 +1 is an alegebraic integer ?
>>
>> No. I am saying that if g(x)=pi*x + 1, then it is entirely
possible
>> for g(0) to divide P(0) in the ring of all algebraic
integers, EVEN
>> THOUGH g(x) is not a monic polynomial.
>Yes, but we now know that the g_i(x)
>must meet another condition.
>What I said prev. was incorrect,
>but your statement that ANY polyno.
>would do is crassly passe in the light
>of new revelations.
> So... I addressed your comment as written. Since that
comment, you
> have now changed your mind about what the g_i(x) should be;
based on
> that change of heart, you complain that my correction was
wrong,
> because it failed to meet your newfound and not-previously
asserted
> conditions on the g_i(x).
> How... interesting...
The disingenuousness of your attempted
exculpation is breathtaking.
I am unversed in exegesising the
pronouncements of the Supreme Coprolocutor.
You, however, analyse His productions with
monomaniacal intensity. Therefore,
to claim that the conditions on the g_i(x),
although new to me, were unknown to you
is sophistry of the most egregious kind.
A simple admission of your trivial error would bring
the catharsis and ultimate peace of mind
you yearn for.
===
Subject: Re: JSH:Understanding constant terms
days. My association with the Department is that of an
alumnus.
[.snip.]
>> So... I addressed your comment as written. Since that
comment, you
>> have now changed your mind about what the g_i(x) should
be; based on
>> that change of heart, you complain that my correction was
wrong,
>> because it failed to meet your newfound and not-previously
asserted
>> conditions on the g_i(x).
>> How... interesting...
>The disingenuousness of your attempted
>exculpation is breathtaking.
>I am unversed in exegesising the
>pronouncements of the Supreme Coprolocutor.
>You, however, analyse His productions with
>monomaniacal intensity. Therefore,
>to claim that the conditions on the g_i(x),
>although new to me, were unknown to you
>is sophistry of the most egregious kind.
Sigh. I addressed YOUR comment. I noted an error in YOUR
comment,
based on YOUR hypothesis. You acknowledge that error, but
complain
that I noted it. I did not address anything the original
poster said.
Presumably, you are either joking, or dense.
I prefer to assume the former, though it seems hard to ignore
the
latter possibility.
--
Its not denial. Im just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: JSH:Understanding constant terms
> [.snip.]
>> So... I addressed your comment as written. Since that
comment, you
>> have now changed your mind about what the g_i(x) should
be; based on
>> that change of heart, you complain that my correction was
wrong,
>> because it failed to meet your newfound and not-previously
asserted
>> conditions on the g_i(x).
>>
>> How... interesting...
>The disingenuousness of your attempted
>exculpation is breathtaking.
>I am unversed in exegesising the
>pronouncements of the Supreme Coprolocutor.
>You, however, analyse His productions with
>monomaniacal intensity. Therefore,
>to claim that the conditions on the g_i(x),
>although new to me, were unknown to you
>is sophistry of the most egregious kind.
> Sigh. I addressed YOUR comment. I noted an error in YOUR
comment,
> based on YOUR hypothesis. You acknowledge that error, but
complain
> that I noted it. I did not address anything the original
poster said.
Typical
No, I wasnt complaining that you noted it at all.
My soul was suffused with the light that you
brought unto me. I wasnt saying that you
explicitly addressed anything the OP said,
but that you must have known implicitly
what the OP said about the g_i(x) due to
your regular enlightening intercourse with the OP.
Perhaps a ßeeting fit of insanity overcame
you as you pondered these deep matters causing
you to cast aside your usual intellectual caution
and make that rash and impetuous statement about
ANY polynomial.
> Presumably, you are either joking, or dense.
Joking ? Gay persißage ? Badinage of an ironic nature ?
Egad Sir,I would rather ritually disembowel
myself than make light of your
forensic quest for Truth.
> I prefer to assume the former, though it seems hard to
ignore the
> latter possibility.
Your sharpness of mind is matched only by your
generosity of spirit.
===
Subject: Re: JSH:Understanding constant terms
<coi0o6$rg0$1@agate.berkeley.edu>
<cokldp$1o3s$1@agate.berkeley.edu>
Discussion, linux)
> No, I wasnt complaining that you noted it at all.
> My soul was suffused with the light that you
> brought unto me.
Jesus, give it a rest and put away the thesaurus. Were
already
ing impressed. At least I know I am.
--
Jesse F. Hughes
To be honest, I dont have enough interest in math to spend
the time
it would take to clean up the mess that I believe has been
created in
the past 100 or so years. -- Curt Welch lets the world down.
===
Subject: Re: JSH:Understanding constant terms
>If and only if the gs are monic polys
> No. If and only if the values g_i(0) are algebraic
integers. This
will
> certainly happen if the g_i are monic polynomials with
integer
> coefficients, but it can happen in other cases as well (for
example,
> if the g_i are ANY polynomials with integer constant terms).
>Dont the g_i, polys in x, also have to be
>>algebraic integers whenever x is an algebraic
>>integer ?
> Im sorry. Was that English?
>I cant say that Ive found your
>prose particularly pristine.
> I do try to avoid the use of cute
> abbreviations like polys, though.
> Cute ? Poly used to be a widespr. abb.
Um Davidson, widespr. and abb. are not words in the English
language.
If you think they are acceptable abbreviations in the English
language you
are clearly not qualified to be asking others rudely if they
are a
non-native English speaker.
> for polytechnic. Ive never heard
So you were saying Dont the g_i, polytechnics in x, also
have to be
...?
KeithK
> it reviled on the basis of cuteness before.
> But then you learn many strange things on sci.math.
>Are you a non-native English speaker ?
> Indeed. English is, alas, my fourth language. It often
shows.
>> You were assuming that the g_i(x) were polynomials, and
that the
>> product of the g_i(x) was equal to P(x). Then your
statement if
and
>> only if the gs are monic polys was about the statement
that each
>> g_i(0) divides, in the algebraic integers, P(0).
>According to Harris, the g_i(x) are algebraic integers
>whenever x is ( his algebraic integer functions).
>Clearly g_i(0) has to be an algebraic
>integer and so the constant term of g_i(x)has to be an
>algerbaic integer.
>You say that the gi can be ANY polynominal.
> Please note that I was not addressing Jamess argument. I
was
> addressing your restriction to the g_i being polynomials,
and your
> assertion, ->under that further restriction<-, that each
g_i(0) would
> divide P(0) in the ring of all algebraic integers if and
only if the
> g_i(x) were monic. ->That<- assertion is false.
> Yes, I agree
>This is clearly not the case. For example pi*x +1.
>Are you saying that p1*1 +1 is an alegebraic integer ?
> No. I am saying that if g(x)=pi*x + 1, then it is entirely
possible
> for g(0) to divide P(0) in the ring of all algebraic
integers, EVEN
> THOUGH g(x) is not a monic polynomial.
> Yes, but we now know that the g_i(x)
> must meet another condition.
> What I said prev. was incorrect,
> but your statement that ANY polyno.
> would do is crassly passe in the light
> of new revelations.
===
Subject: Re: JSH:Understanding constant terms
>If and only if the gs are monic polys
> No. If and only if the values g_i(0) are algebraic integers.
This
> will
> certainly happen if the g_i are monic polynomials with
integer
> coefficients, but it can happen in other cases as well (for
> example,
> if the g_i are ANY polynomials with integer constant terms).
>Dont the g_i, polys in x, also have to be
>>algebraic integers whenever x is an algebraic
>>integer ?
> Im sorry. Was that English?
>I cant say that Ive found your
>prose particularly pristine.
> I do try to avoid the use of cute
> abbreviations like polys, though.
> Cute ? Poly used to be a widespr. abb.
> Um Davidson, widespr. and abb. are not words
in the English language.
Yes, thats wherefore me chosed they.
> If you think they are acceptable abbreviations in the
English language
you
> are clearly not qualified to be asking others rudely if they
are a
> non-native English speaker.
Me not it were which the one say
first bad bad things about my spoke Brit-talk was.
You cogitate big deep, you get it much quickly.
===
Subject: Re: JSH:Understanding constant terms
>If and only if the gs are monic polys
> No. If and only if the values g_i(0) are algebraic integers.
This
> will
> certainly happen if the g_i are monic polynomials with
integer
> coefficients, but it can happen in other cases as well (for
> example,
> if the g_i are ANY polynomials with integer constant terms).
>Dont the g_i, polys in x, also have to be
>>algebraic integers whenever x is an algebraic
>>integer ?
> Im sorry. Was that English?
>I cant say that Ive found your
>prose particularly pristine.
> I do try to avoid the use of cute
> abbreviations like polys, though.
> Cute ? Poly used to be a widespr. abb.
> Um Davidson, widespr. and abb. are not words
> in the English language.
> Yes, thats wherefore me chosed they.
> If you think they are acceptable abbreviations in the
English language
you
> are clearly not qualified to be asking others rudely if they
are a
> non-native English speaker.
> Me not it were which the one say
> first bad bad things about my spoke Brit-talk was.
> You cogitate big deep, you get it much quickly.
Me humbled muchly your by delightf. reply.
:)
KeithK
===
Subject: Re: JSH:Understanding constant terms
<cob9p0$1tpj$1@agate.berkeley.edu>>If and only if the gs 
are
monic polys
>> No. If and only if the values g_i(0) are algebraic
integers. This will
>> certainly happen if the g_i are monic polynomials with
integer
>> coefficients, but it can happen in other cases as well (for
example,
>> if the g_i are ANY polynomials with integer constant
terms).
>Dont the g_i, polys in x, also have to be
>algebraic integers whenever x is an algebraic
>integer ? So, say 6x^2 +(pi)x + 2 couldnt
>be a g_1 even though its constant term
>is a rational integer.
The g_i Ôs are not necessarily polynomials. In the 
specific
case JSH
considers later in his posting, the a_i(x) are the roots of a
cubic
whose coefficients are functions of x. They could be found
explicitly
using Cardanos formula: complex expressions involving cube-
and
square-roots of polynomials in x. (They are algebraic
integers whenever
x is, because the cubic involved is monic.)
--
David Hartley
===
Subject: Re: JSH:Understanding constant terms
>>If and only if the gs are monic polys
> No. If and only if the values g_i(0) are algebraic
integers. This will
>> certainly happen if the g_i are monic polynomials with
integer
>> coefficients, but it can happen in other cases as well (for
example,
>> if the g_i are ANY polynomials with integer constant
terms).
>Dont the g_i, polys in x, also have to be
>algebraic integers whenever x is an algebraic
>integer ? So, say 6x^2 +(pi)x + 2 couldnt
>be a g_1 even though its constant term
>is a rational integer.
> The g_i Ôs are not necessarily polynomials. In the 
specific
case JSH
> considers later in his posting, the a_i(x) are the roots of
a cubic
> whose coefficients are functions of x. They could be found
explicitly
> using Cardanos formula: complex expressions involving
cube- and
> square-roots of polynomials in x. (They are algebraic
integers whenever
> x is, because the cubic involved is monic.)
Apparently, all that is required of the g_i(x)
is that when x is an algebraic integer
g_1(x) is also an algebraic integer.
So something of the form A*log(x) + pi + tan(x)
would not do.
As the sums, products and roots of algebraic
integers are algebraic integers then, say, g_1(x)
A x^(3/5) + Bx^(1/2) + C, where A,B,C
are algebraic integers would be an algebaraic integer
if x is an algebraic integer. But this is
still a polynomial in fractional powers of x.
Non-polynomial seems a bit misleading.
===
Subject: Re: JSH:Understanding constant terms
...
> Apparently, all that is required of the g_i(x)
> is that when x is an algebraic integer
> g_1(x) is also an algebraic integer.
> So something of the form A*log(x) + pi + tan(x)
> would not do.
> As the sums, products and roots of algebraic
> integers are algebraic integers then, say, g_1(x)
> A x^(3/5) + Bx^(1/2) + C, where A,B,C
> are algebraic integers would be an algebaraic integer
> if x is an algebraic integer. But this is
> still a polynomial in fractional powers of x.
> Non-polynomial seems a bit misleading.
What if g_i(x) are the roots of g^2 + x.g + 1? I would
certainly
call that non-polynomial. And what do you think of sqrt(x +
1)?
--
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: JSH:Understanding constant terms
> ...
> > Apparently, all that is required of the g_i(x)
> > is that when x is an algebraic integer
> > g_1(x) is also an algebraic integer.
> > So something of the form A*log(x) + pi + tan(x)
> > would not do.
> > As the sums, products and roots of algebraic
> > integers are algebraic integers then, say, g_1(x)
> > A x^(3/5) + Bx^(1/2) + C, where A,B,C
> > are algebraic integers would be an algebaraic integer
> > if x is an algebraic integer. But this is
> > still a polynomial in fractional powers of x.
> > Non-polynomial seems a bit misleading.
> What if g_i(x) are the roots of g^2 + x.g + 1? I would
certainly
> call that non-polynomial. And what do you think of sqrt(x +
1)?
I dont know what your definition of polynomial 
or
non-polynomial
might be.
So I can only guess as to why you think your first example
is a non-polynomial.
A polynomial is finite sum of terms of the form Ax^k, where A
is an algebraic integer, x a variable and k a positive
rational number
How are you defining polynomial and non-polynomial ?
===
Subject: Re: JSH:Understanding constant terms
...
> > Non-polynomial seems a bit misleading.
>
> What if g_i(x) are the roots of g^2 + x.g + 1? I would
certainly
> call that non-polynomial. And what do you think of sqrt(x +
1)?
> I dont know what your definition of 
polynomial or
non-polynomial
> might be.
> So I can only guess as to why you think your first example
> is a non-polynomial.
In wat way is -x + sqrt(x^2 - 4) a polynomial?
> A polynomial is finite sum of terms of the form Ax^k, where 
A
> is an algebraic integer, x a variable and k a positive
> rational number
> How are you defining polynomial and non-polynomial ?
The same.
--
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: JSH:Understanding constant terms
> Consider a polynomial P(x), with n factors such that
> g_1(x) g_2(x)...g_n(x) = P(x)
> and g_1(x), g_2(x),...,g_n(x) are algebraic integer
functions.
> That is, for an algebraic integer x, and natural number 
Ôa
where
> 1<=a<=n, g_a(x) is an algebraic integer.
> Now consider setting x=0, and notice that gives
> g_1(0) g_2(0)...g_n(0) = P(0)
> and notice there is no dependency on x, as the constant
term is
> defined by terms where x is equal to 0, and 
thats not a
trick.
> Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have
> c_1 c_2...c_n = P(0)
> and the cs are necessarily algebraic integers and factors
of the
> constant term P(0).
> Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and
so on.
> Then I can substitute and have
> (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x).
> Now let
> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
> then
> P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)
> when the as are roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> so let
> g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) =
5a_3(x) + 7
> and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by
g_3(0).
> Setting x = 0 with the cubic defining the as 
gives
> a^3 - 3a^2 = 0, so two of the as are 0, and one is 3.
> Since indices are arbitrary let the first two equal 0, and I
have
> c_1 = 7, c_2 = 7, and c_3 = 22
> which is consistent with the constant term of P(x), which
is 1078.
> But dividing P(x) by 49, gives
> P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
> which means that 7 is divided from the constant terms as
well, and
> only two of the constant terms, c_1 and c_2 have 7 as a
factor, so
> necessarily 7 is divided from the factors where the
constant terms are
> 7.
> Now if you continue the analysis the conclusion that two of
the
> factors of P(x) have 7 as a factor leads to the conclusion
that two of
> the roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> have 7 as a factor, but you can rather easily prove that
with integer
> x, if the cubic is irreducible over rationals, then that
will not be
> true in the ring of algebraic integers.
> Algebraic integers are roots of monic polynomials with
integer
> coefficients.
> For a number to be an algebraic integer, it must be the
root of some
> monic polynomial that has integer coefficients.
> Therefore, you can conclude from the mathematics that for a
given x,
> when the cubic defining the as is irreducible 
over
rationals then its
> roots do not have 7 as a factor in the ring of algebraic
integers.
Which is, of course, an out-and-out contradiction, right?
So unless mathematics is inconsistent, there must be something
wrong. What is it?
It could be that algebraic number theory is wrong. Or possibly
Galois theory. Or both.
Or it could be that you have a logical error somewhere in what
you have said above.
You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That
certainly
seems OK. I dont disagree with it.
Then you noted that 49 = 7*7 must be factored out of the
expression
(r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22).
Lets look in detail at that troublesome third term,
r_3(x) + 22.
The objective here is to examine whether I can divide a
nonunit
factor of 7 out of this expression, with the result after the
factoring being an algebraic integer.
Let w be a nonunit factor of 7.
You note that 22 is coprime to 7. Therefore 22/w cannot be an
algebraic integer. So that is potentially a problem.
But recall that r_3(x) + 22 = 5 a_3(x) + 7.
Suppose I divide the latter expression through by w: I get
5 a_3(x)/w + 7/w.
The term 7/w is not a problem here because I was assuming that
w is a nonunit factor of 7. Thus 7/w *is* an algebraic
integer.
Now, if such a w can be chosen so that a_3(x)/w is also an
algebraic integer, I conclude that in the expression
5 a_3(x)/w + 7/w
all the coefficients are algebraic integers, and the sum is
an algebraic integer.
Can such a w be defined ?
If so, it is going to have to be a function of x: because when
x = 0, w has to be 1. Whereas, for most other integers x, the
polynomial that you mention above is irreducible, and w is a
nonunit
factor of 7.
Here is how w_1(x), w_2(x), and w_3(x) need to be defined:
w_1(x) = GCD(a_1(x), 7)
w_2(x) = GCD(a_2(x), 7)
w_3(x) = GCD(a_3(x), 7).
Here GCD denotes the Ôgreatest common divisor 
function. It
is defined in the ring of algebraic integers by a theorem of
Dedekind.
Obviously these definitions imply that all of
(5 a_1(x) + 7)/w_1(x),
(5 a_2(x) + 7)/w_2(x), and
(5 a_3(x) + 7)/w_3(x)
are algebraic integers.
Now the crucial thing to show is that
w_1(x) * w_2(x) * w_3(x) = 49.
Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in
general
not by any additional factors of 7. Why? Because
2401 x^3 - 147 x^2 + 3x
is coprime to 7 [except when x itself is divisible by 7], and
-49*(2401 x^3 - 147 x^2 + 3x)
is the constant term of the polynomial that the as satisfy,
as
you note above.
Since w_1(x), w_2(x), and w_3(x) are all greatest common
divisors
of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their
product
must equal 49, as desired.
What this shows is that it IS possible to divide 49 out of
(r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22)
in such a way that each factor is an algebraic integer.
But what about the constant terms ?
It is *not required* that the constant terms be respected.
That
is, even though it is true that
(7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22,
(as you can immediately check), there is NO REASON to require
that each
of these three terms be algebraic integers; and of course,
22/w_3(x)
is not.
Again, why? Isnt the product of the constant terms equal to
the
constant term of the product ???
Yes. But here is the last key fact. The constant term of
(r_2(x) + 22)/w_3(x)
is NOT what you think it is. You think it must be
22/w_3(x).
It isnt. You need to remember YOUR OWN DEFINITION of 
constant
term. It is instead
22/w_3(0).
So then you note that w_3(0) = 1, and you have no
contradiction.
Everything hangs together.
Again: you defined Ôconstant term 
in a perfectly reasonable
way,
and you should have stuck with your definition. Instead you 
got
confused and assumed that whatever is in the POSITION of the
constant
term *is* the constant term. But here, 22/w_3(x) is not even
constant,
because by its definition, w_3(x) is not constant.
Nora B.
> James Harris
> http://mathforprofit.blogspot.com/
===
Subject: Re: JSH:Understanding constant terms
> Consider a polynomial P(x), with n factors such that
>
> g_1(x) g_2(x)...g_n(x) = P(x)
>
> and g_1(x), g_2(x),...,g_n(x) are algebraic integer
functions.
>
> That is, for an algebraic integer x, and natural number 
Ôa
where
> 1<=a<=n, g_a(x) is an algebraic integer.
>
> Now consider setting x=0, and notice that gives
>
> g_1(0) g_2(0)...g_n(0) = P(0)
>
> and notice there is no dependency on x, as the constant
term is
> defined by terms where x is equal to 0, and 
thats not a
trick.
>
> Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have
>
> c_1 c_2...c_n = P(0)
>
> and the cs are necessarily algebraic integers and factors
of the
> constant term P(0).
>
> Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and
so on.
>
> Then I can substitute and have
>
> (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x).
>
> Now let
>
> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
>
> then
>
> P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)
>
> when the as are roots of
>
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>
> so let
>
> g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) =
5a_3(x) + 7
>
> and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by
g_3(0).
>
> Setting x = 0 with the cubic defining the as 
gives
>
> a^3 - 3a^2 = 0, so two of the as are 0, and one is 3.
>
> Since indices are arbitrary let the first two equal 0, and I
have
>
> c_1 = 7, c_2 = 7, and c_3 = 22
>
> which is consistent with the constant term of P(x), which
is 1078.
>
> But dividing P(x) by 49, gives
>
> P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
>
> which means that 7 is divided from the constant terms as
well, and
> only two of the constant terms, c_1 and c_2 have 7 as a
factor, so
> necessarily 7 is divided from the factors where the
constant terms are
> 7.
>
> Now if you continue the analysis the conclusion that two of
the
> factors of P(x) have 7 as a factor leads to the conclusion
that two of
> the roots of
>
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>
> have 7 as a factor, but you can rather easily prove that
with integer
> x, if the cubic is irreducible over rationals, then that
will not be
> true in the ring of algebraic integers.
>
> Algebraic integers are roots of monic polynomials with
integer
> coefficients.
>
> For a number to be an algebraic integer, it must be the
root of some
> monic polynomial that has integer coefficients.
>
> Therefore, you can conclude from the mathematics that for a
given x,
> when the cubic defining the as is irreducible 
over
rationals then its
> roots do not have 7 as a factor in the ring of algebraic
integers.
>
> Which is, of course, an out-and-out contradiction, right?
There is the appearance of contradiction, which is readily
resolved by
not giving the ring of algebraic integers a special position.
So then, just because the roots do not have 7 as a factor in
the ring
of algebraic integers, its not significant.
> So unless mathematics is inconsistent, there must be
something
> wrong. What is it?
If you overrate the ring of algebraic integers, then you can
make
arguments that are wrong.
Essentially, you have to understand that the requirement that
a number
be the root of a monic polynomial with integer coefficients is
a
meaningless technicality, with no real mathematical
importance.
There is no weight to the requirement mathematically that a
number be
the root of a monic polynomial with integer coefficients.
Thats what follows.
> It could be that algebraic number theory is wrong. Or
possibly
> Galois theory. Or both.
It turns out to be a problem in algebraic number theory,
which has
lead to a mis-use of Galois Theory.
Its not even really complicated, but there are reasons for
people to
get emotional over the issue, as its a mistake at the
foundations of
the discipline of mathematics.
Its an error in core.
> Or it could be that you have a logical error somewhere in
what
> you have said above.
You can continue with that assumption. I have no problem
defending
the math I outlined in my original post.
Ultimately, it boils down to constants being constant.
Specifically 7 and 22 are constants, and behave like 
constants.
> You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That
certainly
> seems OK. I dont disagree with it.
Good.
> Then you noted that 49 = 7*7 must be factored out of the
expression
> (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22).
> Lets look in detail at that troublesome third term,
Its not troublesome. Its asymmetrical with 
regard to the
rest.
The asymmetry is what blocks you out of the ring of algebraic
integers, which requires a lot of symmetry.
> r_3(x) + 22.
> The objective here is to examine whether I can divide a
nonunit
> factor of 7 out of this expression, with the result after
the
> factoring being an algebraic integer.
Now youre going to cheat, and Im going to 
explain how you
cheat
before you do it, so readers can see how it works.
A number cannot be an algebraic integer if it is not the root
of some
monic polynomial with integer coefficients.
You say nonunit above as if that applies globally, when
youre going
to rely on non-unit status in the ring of algebraic integers.
So consider a unit function u(x), which is not a unit in the
ring of
algebraic integers.
Now you can multiply with that unit, and then assert that you
have a
non-unit in the ring of algebraic integers, to claim that you
can do
that multiplication.
But its only a non-unit in the ring of algebraic integers
because
its not the root of some monic polynomial with integer
coefficients.
Ok readers, here we go, pay careful attention...
> Let w be a nonunit factor of 7.
Notice that the poster didnt say, in the ring of algebraic
integers.
Its important that you note that the claim is in the ring 
of
algebraic integers.
> You note that 22 is coprime to 7. Therefore 22/w cannot be
an
> algebraic integer. So that is potentially a problem.
The poster is apparently acknowledging that 22 properly is
coprime to
7. And is in fact coprime to 7 in the the ring of algebraic
integers.
> But recall that r_3(x) + 22 = 5 a_3(x) + 7.
> Suppose I divide the latter expression through by w: I get
> 5 a_3(x)/w + 7/w.
> The term 7/w is not a problem here because I was assuming
that
> w is a nonunit factor of 7. Thus 7/w *is* an algebraic
integer.
Remember readers, non-unit in the ring of algebraic integers!
> Now, if such a w can be chosen so that a_3(x)/w is also an
> algebraic integer, I conclude that in the expression
> 5 a_3(x)/w + 7/w
> all the coefficients are algebraic integers, and the sum is
> an algebraic integer.
> Can such a w be defined ?
> If so, it is going to have to be a function of x: because
when
> x = 0, w has to be 1. Whereas, for most other integers x,
the
> polynomial that you mention above is irreducible, and w is
a nonunit
> factor of 7.
> Here is how w_1(x), w_2(x), and w_3(x) need to be defined:
> w_1(x) = GCD(a_1(x), 7)
>
> w_2(x) = GCD(a_2(x), 7)
> w_3(x) = GCD(a_3(x), 7).
> Here GCD denotes the Ôgreatest common divisor 
function. It
> is defined in the ring of algebraic integers by a theorem of
> Dedekind.
> Obviously these definitions imply that all of
> (5 a_1(x) + 7)/w_1(x),
> (5 a_2(x) + 7)/w_2(x), and
> (5 a_3(x) + 7)/w_3(x)
> are algebraic integers.
Notice also that you have the implication that the factors do
not have
a constant term, which has to be made explicit in a bit.
Essentially the ws the poster is trying to use are unit
factors, but
are not units in the ring of algebraic integers by a
technicality that
they are not roots of a monic polynomial with integer
coefficients.
Its a neat trick when you think about it that requires
relying on the
very problem that has been outlined to try and use the ring of
algebraic integers to disprove that there is a problem with
that ring!
> Now the crucial thing to show is that
> w_1(x) * w_2(x) * w_3(x) = 49.
> Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in
general
> not by any additional factors of 7. Why? Because
> 2401 x^3 - 147 x^2 + 3x
> is coprime to 7 [except when x itself is divisible by 7],
and
> -49*(2401 x^3 - 147 x^2 + 3x)
> is the constant term of the polynomial that the as
satisfy, as
> you note above.
> Since w_1(x), w_2(x), and w_3(x) are all greatest common
divisors
> of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their
product
> must equal 49, as desired.
> What this shows is that it IS possible to divide 49 out of
> (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22)
> in such a way that each factor is an algebraic integer.
Well, yes, its possible.
> But what about the constant terms ?
> It is *not required* that the constant terms be respected.
That
Notice the poster doesnt even try at this point, simply
resorting to
hand-waving by asserting that the constant terms dont need
to be
respected!
The problem is that if you have a constant term that is 7,
another
that is 7, and one thats 22, then to get rid of the 
7s, you
have to
divide out by 7, but thats algebra thats 
inconvenient to
the poster!
So instead the poster just TELLS you that the math doesnt
care.
Lets watch to see what else this poster tries on you.
> is, even though it is true that
> (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22,
> (as you can immediately check), there is NO REASON to
require that each
> of these three terms be algebraic integers; and of course,
22/w_3(x)
> is not.
>
> Again, why? Isnt the product of the constant terms equal
to the
> constant term of the product ???
> Yes. But here is the last key fact. The constant term of
> (r_2(x) + 22)/w_3(x)
> is NOT what you think it is. You think it must be
> 22/w_3(x).
> It isnt. You need to remember YOUR OWN DEFINITION of
constant
> term. It is instead
> 22/w_3(0).
> So then you note that w_3(0) = 1, and you have no
contradiction.
> Everything hangs together.
Notice that the poster has asserted that the constant term is
constant
at x=0, but every where else its actually a function of x.
So the full assertion is that the constant term is NOT
CONSTANT, but
is instead a function of x.
Remember, when people try to fight mathematics they have to at
some
point rely on something that is just wacky. Here you can see
that
ultimately the poster is trying to get you to believe that a
constant
is in fact a function of x.
But given g_1(x) g_2(x)...g_n(x) = P(x), where the n factors
are
algebraic integer functions, then necessarily g_1(0)
g_2(0)...g_n(0) =
P(0), and notice, you dont have any functions of x, as x 
has
been set
to 0.
So the poster is using unit factors, which by a technicality
are not
units in the ring of algebraic integers, unless you do wish
to believe
that mathematics is inconsistent, and so what? Your belief
would be
wrong.
> Again: you defined Ôconstant 
term in a perfectly reasonable
way,
> and you should have stuck with your definition. Instead you
got
> confused and assumed that whatever is in the POSITION of
the constant
> term *is* the constant term. But here, 22/w_3(x) is not
even constant,
> because by its definition, w_3(x) is not constant.
> Nora B.
Well, I say youre using unit factors, which because they 
are
not
roots of a monic polynomial with integer coefficients are not
units in
the ring of algebraic integers.
How do you answer?
James Harris
===
Subject: Re: JSH:Understanding constant terms
> Consider a polynomial P(x), with n factors such that
> g_1(x) g_2(x)...g_n(x) = P(x)
> and g_1(x), g_2(x),...,g_n(x) are algebraic integer
functions.
> That is, for an algebraic integer x, and natural number 
Ôa
where
> 1<=a<=n, g_a(x) is an algebraic integer.
> Now consider setting x=0, and notice that gives
> g_1(0) g_2(0)...g_n(0) = P(0)
> and notice there is no dependency on x, as the constant
term is
> defined by terms where x is equal to 0, and 
thats not a
trick.
> Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have
> c_1 c_2...c_n = P(0)
> and the cs are necessarily algebraic integers and factors
of the
> constant term P(0).
> Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and
so on.
> Then I can substitute and have
> (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x).
> Now let
> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
> then
> P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)
> when the as are roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> so let
> g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) =
5a_3(x) + 7
> and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by
g_3(0).
> Setting x = 0 with the cubic defining the as 
gives
> a^3 - 3a^2 = 0, so two of the as are 0, and one is 3.
> Since indices are arbitrary let the first two equal 0, and I
have
> c_1 = 7, c_2 = 7, and c_3 = 22
> which is consistent with the constant term of P(x), which
is 1078.
> But dividing P(x) by 49, gives
> P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
> which means that 7 is divided from the constant terms as
well, and
> only two of the constant terms, c_1 and c_2 have 7 as a
factor, so
> necessarily 7 is divided from the factors where the
constant terms
are
> 7.
> Now if you continue the analysis the conclusion that two of
the
> factors of P(x) have 7 as a factor leads to the conclusion
that two
of
> the roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> have 7 as a factor, but you can rather easily prove that
with integer
> x, if the cubic is irreducible over rationals, then that
will not be
> true in the ring of algebraic integers.
> Algebraic integers are roots of monic polynomials with
integer
> coefficients.
> For a number to be an algebraic integer, it must be the
root of some
> monic polynomial that has integer coefficients.
> Therefore, you can conclude from the mathematics that for a
given x,
> when the cubic defining the as is irreducible 
over
rationals then
its
> roots do not have 7 as a factor in the ring of algebraic
integers.
> Which is, of course, an out-and-out contradiction, right?
> There is the appearance of contradiction, which is readily
resolved by
> not giving the ring of algebraic integers a special
position.
> So then, just because the roots do not have 7 as a factor
in the ring
> of algebraic integers, its not significant.
> So unless mathematics is inconsistent, there must be
something
> wrong. What is it?
> If you overrate the ring of algebraic integers, then you
can make
> arguments that are wrong.
> Essentially, you have to understand that the requirement
that a number
> be the root of a monic polynomial with integer coefficients
is a
> meaningless technicality, with no real mathematical
importance.
> There is no weight to the requirement mathematically that a
number be
> the root of a monic polynomial with integer coefficients.
> Thats what follows.
> It could be that algebraic number theory is wrong. Or
possibly
> Galois theory. Or both.
> It turns out to be a problem in algebraic number theory,
which has
> lead to a mis-use of Galois Theory.
> Its not even really complicated, but there are reasons 
for
people to
> get emotional over the issue, as its a mistake at the
foundations of
> the discipline of mathematics.
> Its an error in core.
> Or it could be that you have a logical error somewhere in
what
> you have said above.
> You can continue with that assumption. I have no problem
defending
> the math I outlined in my original post.
> Ultimately, it boils down to constants being constant.
> Specifically 7 and 22 are constants, and behave like
constants.
> You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That
certainly
> seems OK. I dont disagree with it.
> Good.
> Then you noted that 49 = 7*7 must be factored out of the
expression
> (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22).
> Lets look in detail at that troublesome third term,
> Its not troublesome. Its asymmetrical with 
regard to the
rest.
> The asymmetry is what blocks you out of the ring of
algebraic
> integers, which requires a lot of symmetry.
> r_3(x) + 22.
> The objective here is to examine whether I can divide a
nonunit
> factor of 7 out of this expression, with the result after
the
> factoring being an algebraic integer.
> Now youre going to cheat, and Im going to 
explain how you
cheat
> before you do it, so readers can see how it works.
> A number cannot be an algebraic integer if it is not the
root of some
> monic polynomial with integer coefficients.
> You say nonunit above as if that applies globally, when
youre going
> to rely on non-unit status in the ring of algebraic
integers.
> So consider a unit function u(x), which is not a unit in
the ring of
> algebraic integers.
> Now you can multiply with that unit, and then assert that
you have a
> non-unit in the ring of algebraic integers, to claim that
you can do
> that multiplication.
> But its only a non-unit in the ring of algebraic integers
because
> its not the root of some monic polynomial with integer
coefficients.
> Ok readers, here we go, pay careful attention...
> Let w be a nonunit factor of 7.
> Notice that the poster didnt say, in the ring of 
algebraic
integers.
> Its important that you note that the claim is in the ring
of
> algebraic integers.
> You note that 22 is coprime to 7. Therefore 22/w cannot be
an
> algebraic integer. So that is potentially a problem.
> The poster is apparently acknowledging that 22 properly is
coprime to
> 7. And is in fact coprime to 7 in the the ring of algebraic
integers.
> But recall that r_3(x) + 22 = 5 a_3(x) + 7.
> Suppose I divide the latter expression through by w: I get
> 5 a_3(x)/w + 7/w.
> The term 7/w is not a problem here because I was assuming
that
> w is a nonunit factor of 7. Thus 7/w *is* an algebraic
integer.
> Remember readers, non-unit in the ring of algebraic
integers!
> Now, if such a w can be chosen so that a_3(x)/w is also an
> algebraic integer, I conclude that in the expression
> 5 a_3(x)/w + 7/w
> all the coefficients are algebraic integers, and the sum is
> an algebraic integer.
> Can such a w be defined ?
> If so, it is going to have to be a function of x: because
when
> x = 0, w has to be 1. Whereas, for most other integers x,
the
> polynomial that you mention above is irreducible, and w is
a nonunit
> factor of 7.
> Here is how w_1(x), w_2(x), and w_3(x) need to be defined:
> w_1(x) = GCD(a_1(x), 7)
> w_2(x) = GCD(a_2(x), 7)
> w_3(x) = GCD(a_3(x), 7).
> Here GCD denotes the Ôgreatest common divisor 
function. It
> is defined in the ring of algebraic integers by a theorem of
> Dedekind.
> Obviously these definitions imply that all of
> (5 a_1(x) + 7)/w_1(x),
> (5 a_2(x) + 7)/w_2(x), and
> (5 a_3(x) + 7)/w_3(x)
> are algebraic integers.
> Notice also that you have the implication that the factors
do not have
> a constant term, which has to be made explicit in a bit.
> Essentially the ws the poster is trying to use are unit
factors, but
> are not units in the ring of algebraic integers by a
technicality that
> they are not roots of a monic polynomial with integer
coefficients.
> Its a neat trick when you think about it that requires
relying on the
> very problem that has been outlined to try and use the ring
of
> algebraic integers to disprove that there is a problem with
that ring!
> Now the crucial thing to show is that
> w_1(x) * w_2(x) * w_3(x) = 49.
> Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in
general
> not by any additional factors of 7. Why? Because
> 2401 x^3 - 147 x^2 + 3x
> is coprime to 7 [except when x itself is divisible by 7],
and
> -49*(2401 x^3 - 147 x^2 + 3x)
> is the constant term of the polynomial that the as
satisfy, as
> you note above.
> Since w_1(x), w_2(x), and w_3(x) are all greatest common
divisors
> of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their
product
> must equal 49, as desired.
> What this shows is that it IS possible to divide 49 out of
> (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22)
> in such a way that each factor is an algebraic integer.
> Well, yes, its possible.
> But what about the constant terms ?
> It is *not required* that the constant terms be respected.
That
> Notice the poster doesnt even try at this point, simply
resorting to
> hand-waving by asserting that the constant terms dont 
need
to be
> respected!
> The problem is that if you have a constant term that is 7,
another
> that is 7, and one thats 22, then to get rid of the 
7s,
you have to
> divide out by 7, but thats algebra thats 
inconvenient to
the poster!
> So instead the poster just TELLS you that the math doesnt
care.
> Lets watch to see what else this poster tries on you.
> is, even though it is true that
> (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22,
> (as you can immediately check), there is NO REASON to
require that each
> of these three terms be algebraic integers; and of course,
22/w_3(x)
> is not.
> Again, why? Isnt the product of the constant terms equal
to the
> constant term of the product ???
> Yes. But here is the last key fact. The constant term of
> (r_2(x) + 22)/w_3(x)
> is NOT what you think it is. You think it must be
> 22/w_3(x).
> It isnt. You need to remember YOUR OWN DEFINITION of
constant
> term. It is instead
> 22/w_3(0).
> So then you note that w_3(0) = 1, and you have no
contradiction.
> Everything hangs together.
> Notice that the poster has asserted that the constant term
is constant
> at x=0, but every where else its actually a function of 
x.
What nonsense! Nora Baron said no such thing!!
She said YOU fail to see that 22/w_3(x) is NOT the constant
term of
(r_2(x) + 22)/w_3(x).
Why? because w_3(x) is a function of x.
> So the full assertion is that the constant term is NOT
CONSTANT, but
> is instead a function of x.
The full assertion is that 22/w_3(x) is NOT CONSTANT, but is
instead a
function of x. (DUH)
KeithK
> Remember, when people try to fight mathematics they have to
at some
> point rely on something that is just wacky. Here you can
see that
> ultimately the poster is trying to get you to believe that
a constant
> is in fact a function of x.
> But given g_1(x) g_2(x)...g_n(x) = P(x), where the n
factors are
> algebraic integer functions, then necessarily g_1(0)
g_2(0)...g_n(0) =
> P(0), and notice, you dont have any functions of x, as x
has been set
> to 0.
> So the poster is using unit factors, which by a
technicality are not
> units in the ring of algebraic integers, unless you do wish
to believe
> that mathematics is inconsistent, and so what? Your belief
would be
> wrong.
> Again: you defined Ôconstant 
term in a perfectly reasonable
way,
> and you should have stuck with your definition. Instead you
got
> confused and assumed that whatever is in the POSITION of
the constant
> term *is* the constant term. But here, 22/w_3(x) is not
even constant,
> because by its definition, w_3(x) is not constant.
> Nora B.
> Well, I say youre using unit factors, which because they
are not
> roots of a monic polynomial with integer coefficients are
not units in
> the ring of algebraic integers.
> How do you answer?
> James Harris
===
Subject: Re: JSH:Understanding constant terms
> Consider a polynomial P(x), with n factors such that
>
> g_1(x) g_2(x)...g_n(x) = P(x)
>
> and g_1(x), g_2(x),...,g_n(x) are algebraic integer
functions.
>
> That is, for an algebraic integer x, and natural number 
Ôa
where
> 1<=a<=n, g_a(x) is an algebraic integer.
>
> Now consider setting x=0, and notice that gives
>
> g_1(0) g_2(0)...g_n(0) = P(0)
>
> and notice there is no dependency on x, as the constant
term is
> defined by terms where x is equal to 0, and 
thats not a
trick.
>
> Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have
>
> c_1 c_2...c_n = P(0)
>
> and the cs are necessarily algebraic integers and factors
of the
> constant term P(0).
>
> Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and
so on.
>
> Then I can substitute and have
>
> (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x).
>
> Now let
>
> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
>
> then
>
> P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)
>
> when the as are roots of
>
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>
> so let
>
> g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) =
5a_3(x) + 7
>
> and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by
g_3(0).
>
> Setting x = 0 with the cubic defining the as 
gives
>
> a^3 - 3a^2 = 0, so two of the as are 0, and one is 3.
>
> Since indices are arbitrary let the first two equal 0, and I
have
>
> c_1 = 7, c_2 = 7, and c_3 = 22
>
> which is consistent with the constant term of P(x), which
is 1078.
>
> But dividing P(x) by 49, gives
>
> P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
>
> which means that 7 is divided from the constant terms as
well, and
> only two of the constant terms, c_1 and c_2 have 7 as a
factor, so
> necessarily 7 is divided from the factors where the
constant terms
are
> 7.
>
> Now if you continue the analysis the conclusion that two of
the
> factors of P(x) have 7 as a factor leads to the conclusion
that two
of
> the roots of
>
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>
> have 7 as a factor, but you can rather easily prove that
with integer
> x, if the cubic is irreducible over rationals, then that
will not be
> true in the ring of algebraic integers.
>
> Algebraic integers are roots of monic polynomials with
integer
> coefficients.
>
> For a number to be an algebraic integer, it must be the
root of some
> monic polynomial that has integer coefficients.
>
> Therefore, you can conclude from the mathematics that for a
given x,
> when the cubic defining the as is irreducible 
over
rationals then
its
> roots do not have 7 as a factor in the ring of algebraic
integers.
>
>
> Which is, of course, an out-and-out contradiction, right?
> There is the appearance of contradiction, which is readily
resolved by
> not giving the ring of algebraic integers a special
position.
> So then, just because the roots do not have 7 as a factor
in the ring
> of algebraic integers, its not significant.
> So unless mathematics is inconsistent, there must be
something
> wrong. What is it?
>
> If you overrate the ring of algebraic integers, then you
can make
> arguments that are wrong.
> Essentially, you have to understand that the requirement
that a number
> be the root of a monic polynomial with integer coefficients
is a
> meaningless technicality, with no real mathematical
importance.
> There is no weight to the requirement mathematically that a
number be
> the root of a monic polynomial with integer coefficients.
> Thats what follows.
> It could be that algebraic number theory is wrong. Or
possibly
> Galois theory. Or both.
>
> It turns out to be a problem in algebraic number theory,
which has
> lead to a mis-use of Galois Theory.
> Its not even really complicated, but there are reasons 
for
people to
> get emotional over the issue, as its a mistake at the
foundations of
> the discipline of mathematics.
> Its an error in core.
> Or it could be that you have a logical error somewhere in
what
> you have said above.
>
> You can continue with that assumption. I have no problem
defending
> the math I outlined in my original post.
> Ultimately, it boils down to constants being constant.
> Specifically 7 and 22 are constants, and behave like
constants.
> You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That
certainly
> seems OK. I dont disagree with it.
>
> Good.
> Then you noted that 49 = 7*7 must be factored out of the
expression
>
> (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22).
>
> Lets look in detail at that troublesome third term,
> Its not troublesome. Its asymmetrical with 
regard to the
rest.
> The asymmetry is what blocks you out of the ring of
algebraic
> integers, which requires a lot of symmetry.
> r_3(x) + 22.
>
> The objective here is to examine whether I can divide a
nonunit
> factor of 7 out of this expression, with the result after
the
> factoring being an algebraic integer.
> Now youre going to cheat, and Im going to 
explain how you
cheat
> before you do it, so readers can see how it works.
> A number cannot be an algebraic integer if it is not the
root of some
> monic polynomial with integer coefficients.
> You say nonunit above as if that applies globally, when
youre going
> to rely on non-unit status in the ring of algebraic
integers.
> So consider a unit function u(x), which is not a unit in
the ring of
> algebraic integers.
> Now you can multiply with that unit, and then assert that
you have a
> non-unit in the ring of algebraic integers, to claim that
you can do
> that multiplication.
> But its only a non-unit in the ring of algebraic integers
because
> its not the root of some monic polynomial with integer
coefficients.
> Ok readers, here we go, pay careful attention...
> Let w be a nonunit factor of 7.
>
> Notice that the poster didnt say, in the ring of 
algebraic
integers.
> Its important that you note that the claim is in the ring
of
> algebraic integers.
> You note that 22 is coprime to 7. Therefore 22/w cannot be
an
> algebraic integer. So that is potentially a problem.
>
> The poster is apparently acknowledging that 22 properly is
coprime to
> 7. And is in fact coprime to 7 in the the ring of algebraic
integers.
> But recall that r_3(x) + 22 = 5 a_3(x) + 7.
>
> Suppose I divide the latter expression through by w: I get
>
> 5 a_3(x)/w + 7/w.
>
> The term 7/w is not a problem here because I was assuming
that
> w is a nonunit factor of 7. Thus 7/w *is* an algebraic
integer.
>
> Remember readers, non-unit in the ring of algebraic
integers!
> Now, if such a w can be chosen so that a_3(x)/w is also an
> algebraic integer, I conclude that in the expression
>
> 5 a_3(x)/w + 7/w
>
> all the coefficients are algebraic integers, and the sum is
> an algebraic integer.
>
> Can such a w be defined ?
>
> If so, it is going to have to be a function of x: because
when
> x = 0, w has to be 1. Whereas, for most other integers x,
the
> polynomial that you mention above is irreducible, and w is
a nonunit
> factor of 7.
>
> Here is how w_1(x), w_2(x), and w_3(x) need to be defined:
>
> w_1(x) = GCD(a_1(x), 7)
>
> w_2(x) = GCD(a_2(x), 7)
>
> w_3(x) = GCD(a_3(x), 7).
>
> Here GCD denotes the Ôgreatest common divisor 
function. It
> is defined in the ring of algebraic integers by a theorem of
> Dedekind.
>
> Obviously these definitions imply that all of
>
> (5 a_1(x) + 7)/w_1(x),
>
> (5 a_2(x) + 7)/w_2(x), and
>
> (5 a_3(x) + 7)/w_3(x)
>
> are algebraic integers.
>
> Notice also that you have the implication that the factors
do not have
> a constant term, which has to be made explicit in a bit.
> Essentially the ws the poster is trying to use are unit
factors, but
> are not units in the ring of algebraic integers by a
technicality that
> they are not roots of a monic polynomial with integer
coefficients.
> Its a neat trick when you think about it that requires
relying on the
> very problem that has been outlined to try and use the ring
of
> algebraic integers to disprove that there is a problem with
that ring!
> Now the crucial thing to show is that
>
> w_1(x) * w_2(x) * w_3(x) = 49.
>
> Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in
general
> not by any additional factors of 7. Why? Because
>
> 2401 x^3 - 147 x^2 + 3x
>
> is coprime to 7 [except when x itself is divisible by 7],
and
>
> -49*(2401 x^3 - 147 x^2 + 3x)
>
> is the constant term of the polynomial that the as
satisfy, as
> you note above.
>
> Since w_1(x), w_2(x), and w_3(x) are all greatest common
divisors
> of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their
product
> must equal 49, as desired.
>
> What this shows is that it IS possible to divide 49 out of
>
> (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22)
>
> in such a way that each factor is an algebraic integer.
>
> Well, yes, its possible.
> But what about the constant terms ?
>
> It is *not required* that the constant terms be respected.
That
> Notice the poster doesnt even try at this point, simply
resorting to
> hand-waving by asserting that the constant terms dont 
need
to be
> respected!
> The problem is that if you have a constant term that is 7,
another
> that is 7, and one thats 22, then to get rid of the 
7s,
you have to
> divide out by 7, but thats algebra thats 
inconvenient to
the poster!
> So instead the poster just TELLS you that the math doesnt
care.
> Lets watch to see what else this poster tries on you.
> is, even though it is true that
>
> (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22,
>
> (as you can immediately check), there is NO REASON to
require that each
> of these three terms be algebraic integers; and of course,
22/w_3(x)
> is not.
>
> Again, why? Isnt the product of the constant terms equal
to the
> constant term of the product ???
>
> Yes. But here is the last key fact. The constant term of
>
> (r_2(x) + 22)/w_3(x)
>
> is NOT what you think it is. You think it must be
>
> 22/w_3(x).
>
> It isnt. You need to remember YOUR OWN DEFINITION of
constant
> term. It is instead
>
> 22/w_3(0).
>
> So then you note that w_3(0) = 1, and you have no
contradiction.
> Everything hangs together.
>
> Notice that the poster has asserted that the constant term
is constant
> at x=0, but every where else its actually a function of 
x.
The poster has asserted that the contant term is 22/w_3(0).
The poster has asserted explicitely that the constant term
is not 22/w_3(x). 22/w_3(0) is not does not vary with x, it
has the same value if x=0, x=10, x=pi x = whatever.
-William Hughes
===
Subject: Re: JSH:Understanding constant terms
> Consider a polynomial P(x), with n factors such that
>
> g_1(x) g_2(x)...g_n(x) = P(x)
>
> and g_1(x), g_2(x),...,g_n(x) are algebraic integer
functions.
[snip to save space ...]
>
> Which is, of course, an out-and-out contradiction, right?
> There is the appearance of contradiction, which is readily
resolved by
> not giving the ring of algebraic integers a special
position.
No, its not the *appearance* of a contradiction. It is
just an outright contradiction - unless someone, somewhere has
made a mistake.
> So then, just because the roots do not have 7 as a factor
in the ring
> of algebraic integers, its not significant.
> So unless mathematics is inconsistent, there must be
something
> wrong. What is it?
>
> If you overrate the ring of algebraic integers, then you
can make
> arguments that are wrong.
Overrate ??? The set of algebraic integers has a clear,
unambiguous definition. There is a textbook proof that it
forms a ring. It is not a field; it is simply a natural
extension of the integers. It has some key properties: for
example, every algebraic number can be written in the form
a/n, where a is an algebraic integer and n is an ordinary
integer. And it has the property that any two elements have
a greatest common divisor. Unlike the integers, it does NOT
have the property of factorization into primes. Which of
these characteristics would you describe as overrating
the ring of algebraic integers?
> Essentially, you have to understand that the requirement
that a number
> be the root of a monic polynomial with integer coefficients
is a
> meaningless technicality, with no real mathematical
importance.
I disagree. If you throw that out, you lose some of the
interesting properties that the ring has. Dedekind devoted
intense study to it for good reason.
> There is no weight to the requirement mathematically that a
number be
> the root of a monic polynomial with integer coefficients.
A matter of opinion, clearly, rather than a matter of fact.
> Thats what follows.
> It could be that algebraic number theory is wrong. Or
possibly
> Galois theory. Or both.
>
> It turns out to be a problem in algebraic number theory,
which has
> lead to a mis-use of Galois Theory.
> Its not even really complicated, but there are reasons 
for
people to
> get emotional over the issue, as its a mistake at the
foundations of
> the discipline of mathematics.
> Its an error in core.
The error is in your non-rigorous application of your own
definition of constant term, as shown below.
> Or it could be that you have a logical error somewhere in
what
> you have said above.
>
> You can continue with that assumption. I have no problem
defending
> the math I outlined in my original post.
> Ultimately, it boils down to constants being constant.
> Specifically 7 and 22 are constants, and behave like
constants.
> You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That
certainly
> seems OK. I dont disagree with it.
>
> Good.
> Then you noted that 49 = 7*7 must be factored out of the
expression
>
> (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22).
>
> Lets look in detail at that troublesome third term,
> Its not troublesome. Its asymmetrical with 
regard to the
rest.
Spare me the semantics.
> The asymmetry is what blocks you out of the ring of
algebraic
> integers, which requires a lot of symmetry.
> r_3(x) + 22.
>
> The objective here is to examine whether I can divide a
nonunit
> factor of 7 out of this expression, with the result after
the
> factoring being an algebraic integer.
> Now youre going to cheat, and Im going to 
explain how you
cheat
> before you do it, so readers can see how it works.
> A number cannot be an algebraic integer if it is not the
root of some
> monic polynomial with integer coefficients.
No disagreement there. Thats the *definition*.
> You say nonunit above as if that applies globally, when
youre going
> to rely on non-unit status in the ring of algebraic
integers.
Sorry that I did not qualify every statement with in the
algebraic integers. I thought it was clearly implied. In
any case, just to be definite, what I meant was: nonunit *in
the
ring of algebraic integers*.
> So consider a unit function u(x), which is not a unit in
the ring of
> algebraic integers.
Now its your turn to be imprecise, apparently. What is a
unit function ???
> Now you can multiply with that unit,
A unit in what ring? Any nonzero complex number is a unit in
SOME ring; many are nonunits in others. The word unit in the
context you are using it here is not well-defined. You must
specify: in what ring?
> and then assert that you have a
> non-unit in the ring of algebraic integers, to claim that
you can do
> that multiplication.
> But its only a non-unit in the ring of algebraic integers
because
> its not the root of some monic polynomial with integer
coefficients.
> Ok readers, here we go, pay careful attention...
> Let w be a nonunit factor of 7.
>
> Notice that the poster didnt say, in the ring of 
algebraic
integers.
OK, Ill say it: Let w be a nonunit factor of 7 in the ring
of algebraic integers. I think you know that is what I meant
in the first place.
> Its important that you note that the claim is in the ring
of
> algebraic integers.
Sure.
> You note that 22 is coprime to 7. Therefore 22/w cannot be
an
> algebraic integer. So that is potentially a problem.
>
> The poster is apparently acknowledging that 22 properly is
coprime to
> 7.
*In the ring of algebraic integers*. That is clearly implied.
> And is in fact coprime to 7 in the the ring of algebraic
integers.
> But recall that r_3(x) + 22 = 5 a_3(x) + 7.
>
> Suppose I divide the latter expression through by w: I get
>
> 5 a_3(x)/w + 7/w.
>
> The term 7/w is not a problem here because I was assuming
that
> w is a nonunit factor of 7. Thus 7/w *is* an algebraic
integer.
>
> Remember readers, non-unit in the ring of algebraic
integers!
No quarrel with that. Do tell us when you find where I start
cheating.
> Now, if such a w can be chosen so that a_3(x)/w is also an
> algebraic integer, I conclude that in the expression
>
> 5 a_3(x)/w + 7/w
>
> all the coefficients are algebraic integers, and the sum is
> an algebraic integer.
>
> Can such a w be defined ?
>
> If so, it is going to have to be a function of x: because
when
> x = 0, w has to be 1. Whereas, for most other integers x,
the
> polynomial that you mention above is irreducible, and w is
a nonunit
> factor of 7.
>
> Here is how w_1(x), w_2(x), and w_3(x) need to be defined:
>
> w_1(x) = GCD(a_1(x), 7)
>
> w_2(x) = GCD(a_2(x), 7)
>
> w_3(x) = GCD(a_3(x), 7).
>
> Here GCD denotes the Ôgreatest common divisor 
function. It
> is defined in the ring of algebraic integers by a theorem of
> Dedekind.
>
> Obviously these definitions imply that all of
>
> (5 a_1(x) + 7)/w_1(x),
>
> (5 a_2(x) + 7)/w_2(x), and
>
> (5 a_3(x) + 7)/w_3(x)
>
> are algebraic integers.
>
> Notice also that you have the implication that the factors
do not have
> a constant term, which has to be made explicit in a bit.
Wrong! Here is where you make your mistake! You define
constant term of a function G(x) to be G(0). So here, the
constant term of, for example, (r_3(x) + 22)/w_3(x), is
(r_3(0) + 22)/w_3(0) = r_3(0)/w_3(0) + 22/w_3(0).
No problem at all with *existence* of a constant term !!!
YOUR problem is, the constant term is not what you think it
is,
or want it to be!
> Essentially the ws the poster is trying to use are unit
factors,
They most certainly are not: not in the ring of algebraic
integers. Is THIS where you think I am cheating? Dividing
algebraic integers by other (nonunit) algebraic integers? Is
that cheating? Remember, I am NOT claiming that 22/w_3(x) is
an algebraic integer (unless x = 0).
> but
> are not units in the ring of algebraic integers by a
technicality that
> they are not roots of a monic polynomial with integer
coefficients.
Its not a technicality, unless you regard the very phrase
that
defines the ring to be a technicality !
> Its a neat trick when you think about it that requires
relying on the
> very problem that has been outlined to try and use the ring
of
> algebraic integers to disprove that there is a problem with
that ring!
The ring has certain properties that I described above, one of
which is the existence of greatest-common-divisors of any two
of
its elements. It is NOT a neat trick or cheating in any way to
make use of known properties! The integers, for example, have
the property of unique factorization into primes. Would it be
*cheating* to use that property to prove a theorem ???
> Now the crucial thing to show is that
>
> w_1(x) * w_2(x) * w_3(x) = 49.
>
> Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in
general
> not by any additional factors of 7. Why? Because
>
> 2401 x^3 - 147 x^2 + 3x
>
> is coprime to 7 [except when x itself is divisible by 7],
and
>
> -49*(2401 x^3 - 147 x^2 + 3x)
>
> is the constant term of the polynomial that the as
satisfy, as
> you note above.
>
> Since w_1(x), w_2(x), and w_3(x) are all greatest common
divisors
> of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their
product
> must equal 49, as desired.
>
> What this shows is that it IS possible to divide 49 out of
>
> (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22)
>
> in such a way that each factor is an algebraic integer.
>
> Well, yes, its possible.
concede everything.
> But what about the constant terms ?
>
> It is *not required* that the constant terms be respected.
That
> Notice the poster doesnt even try at this point, simply
resorting to
> hand-waving by asserting that the constant terms dont 
need
to be
> respected!
Keep reading. I explain very clearly what I meant.
> The problem is that if you have a constant term that is 7,
another
> that is 7, and one thats 22, then to get rid of the 
7s,
you have to
> divide out by 7, but thats algebra thats 
inconvenient to
the poster!
> So instead the poster just TELLS you that the math doesnt
care.
> Lets watch to see what else this poster tries on you.
> is, even though it is true that
>
> (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22,
>
> (as you can immediately check), there is NO REASON to
require that each
> of these three terms be algebraic integers; and of course,
22/w_3(x)
> is not.
>
> Again, why? Isnt the product of the constant terms equal
to the
> constant term of the product ???
>
> Yes. But here is the last key fact. The constant term of
>
> (r_2(x) + 22)/w_3(x)
>
> is NOT what you think it is. You think it must be
>
> 22/w_3(x).
>
> It isnt. You need to remember YOUR OWN DEFINITION of
constant
> term. It is instead
>
> 22/w_3(0).
>
> So then you note that w_3(0) = 1, and you have no
contradiction.
> Everything hangs together.
>
> Notice that the poster has asserted that the constant term
is constant
> at x=0, but every where else its actually a function of 
x.
Read the following very, very carefully, and try to absorb
something.
I am using YOUR definition of constant term. The constant term
of
r_3(x)/w_3(x) + 22/w_3(x)
is NOT 22/w_3(x). By YOUR OWN DEFINITION, it is
22/w_3(0).
Got that?
> So the full assertion is that the constant term is NOT
CONSTANT, but
> is instead a function of x.
You are missing the point so incredibly thoroughly. The
constant
term in question is 22/w_3(0). It IS CONSTANT. Yes, 22/w_3(x)
is NOT CONSTANT. But also, by your definition, it is NOT THE
CONSTANT
TERM.
You are deeply confused on this point. As William Hughes has
pointed out in another thread, your thinking is based on the
idea
that you can look at a function, and BY INSPECTION tell what
the
constant term is. You say, the constant term of
r_3(x) + 22
is obviously 22. Thats correct. Then you consider
r_3(x)/w_3(x) + 22/w_3(x)
and you think: hey, 22/w_3(x) is in the POSITION where the
constant term ought to be, so it must BE the constant term.
In other words, you just throw YOUR OWN DEFINITION out the
window, and declare what you believe to be the constant term
BY INSPECTION. And its not. You should have used your own
definition consistently and RIGOROUSLY throughout. When you
deviated from it, you got into trouble and led yourself down
the path to this erroneous conclusion.
THIS is what has led you to think there is a contradiction,
which in your addled logic leads you to believe there is some
kind of problem with the ring of algebraic integers.
Again, just to be really, really clear about this. The
constant term of (r_3(x) + 22)/w_3(x) is NOT what you
think it is, i.e., it is NOT
22/w_3(x).
No, by YOUR OWN DEFINITION, the constant term is
22/w_3(0).
What you need to do is RIGOROUSLY USE that definition. It
does not lead you down the path to the wrong conclusion which
you have followed previously.
> Remember, when people try to fight mathematics they have to
at some
> point rely on something that is just wacky. Here you can
see that
> ultimately the poster is trying to get you to believe that
a constant
> is in fact a function of x.
> But given g_1(x) g_2(x)...g_n(x) = P(x), where the n
factors are
> algebraic integer functions, then necessarily g_1(0)
g_2(0)...g_n(0) =
> P(0), and notice, you dont have any functions of x, as x
has been set
> to 0.
> So the poster is using unit factors,
because you know that 22 and w_3(x) are coprime in the ring of
algebraic integers. That is a CORRECT FACT. However, it is not
important. The important thing here is that, when you divide
the WHOLE
EXPRESSION (r_3(x) + 22) by w_3(x), the result IS an algebraic
integer. That is what you require in the first place. There is
no reason that the individual terms must be algebraic
integers.
Heres an analogy in integers. Note that
15 = 13 + 2.
Divide both sides by 3. On the left, I get 5,
an integer. On the right I get
13/3 + 2/3.
Neither of these is an integer. So what? Do I
conclude that the right side is not divisible by 3? No!!!
The SUM of the two parts still equals 5, an integer. The fact
that 2/3 is not an integer is NOT A PROBLEM. You can add
two nonintegers and get an integer. Similarly, you can add
two non-algebraic-integers and get an algebraic integer. Not
a problem at all.
> which by a technicality are not
> units in the ring of algebraic integers,
Saying the DEFINING PROPERTY of the ring is a technicality is
utterly absurd. You must see that at least.
> unless you do wish to believe
> that mathematics is inconsistent, and so what? Your belief
would be
> wrong.
I agree, or at least I agree that what we have discussed here
does not imply that mathematics is inconsistent. If you were
right, it would. But because you have failed to use YOUR OWN
DEFINITION of constant term, you are not right, and
mathematics
is still safe.
> Again: you defined Ôconstant 
term in a perfectly reasonable
way,
> and you should have stuck with your definition. Instead you
got
> confused and assumed that whatever is in the POSITION of
the constant
> term *is* the constant term. But here, 22/w_3(x) is not
even constant,
> because by its definition, w_3(x) is not constant.
>
> Nora B.
> Well, I say youre using unit factors, which because they
are not
> roots of a monic polynomial with integer coefficients are
not units in
> the ring of algebraic integers.
> How do you answer?
As above: 22/w_3(x) is not an algebraic integer. I dont 
claim
it is, and there is no reason it SHOULD be. It is NOT the
constant
term of
(r_3(x) + 22)/w_3(x),
unless x = 0.
Or do you actually still think 22/w_3(x) IS the constant term,
in spite of your own definition ?? Try it - use YOUR OWN
DEFINITION
of constant term - tell me what you get!
Nora B.
> James Harris
===
Subject: Re: JSH:Understanding constant terms
...
> You note that 22 is coprime to 7. Therefore 22/w cannot be
an
> algebraic integer. So that is potentially a problem.
>
>
> The poster is apparently acknowledging that 22 properly is
coprime to
> 7.
> *In the ring of algebraic integers*. That is clearly
implied.
Actually, with most reasonable definitions of coprime, in any
ring that
contains both 7 and 22, they are coprime.
Standard definition: a and b are coprime if p and q in the
ring can be
found such that p*a + q*b = 1.
Alternative definition: a and b are coprime if the have no
common
non-unit factor in the ring.
With both definitions 7 and 22 are coprime regardless of the
ring
involved.
I have no idea what James means with properly coprime.
--
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: JSH:Understanding constant terms
> If you overrate the ring of algebraic integers, then you
can make
> arguments that are wrong.
And *that* is precisely what you have done. You have imposed
demands on the
ring of algebraic integers which are
inconsistent with their properties.
> Essentially, you have to understand that the requirement
that a number
> be the root of a monic polynomial with integer coefficients
is a
> meaningless technicality, with no real mathematical
importance.
That requirement defines a perfectly valid ring. Whether it is
important or
not depends on whether it is useful
in some mathematical context or not. In your application it
is useless. That
doesnt speak to the interests of
others.
> There is no weight to the requirement mathematically that a
number be
> the root of a monic polynomial with integer coefficients.
No, of course not. Rational numbers may not meet that
requirement. If your
requirements are not satisfied by
algebraic integers, dont use them.
[snip irrelevant and redundant exposition of fallacious
arguments]
--
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: JSH:Understanding constant terms
Discussion, linux)
> It turns out to be a problem in algebraic number theory,
which has
> lead to a mis-use of Galois Theory.
> Its not even really complicated, but there are reasons 
for
people to
> get emotional over the issue, as its a mistake at the
foundations of
> the discipline of mathematics.
> Its an error in core.
Er, when did algebraic number theory or Galois theory become
part of
the foundations of mathematics?
I missed that memo. Probably, they dropped me from the
mailing list
when I moved to a philosophy section thats part of a
management
department in a technical university.
--
Ive been thinking about my problems with getting any kind 
of
admission that my math arguments showing the core error in
mathematics
are correct, so Ive gone to marketing books.
-- James S. Harris, on when mathematics isnt enough
===
Subject: Re: final proof Einstein 1905 is crap.
> Nice one, Mike.
> Androcles
I tried hard to incorporate every known formal/informal
fallacy
exactly like the originator of the thread has done. :)
But speaking about fallacies, I think thw Ônon 
sequitur is
the most
serious one, that is the conclusion does not follow from the
premises,
a disconected idea.
Mike
===
Subject: Re: final proof Einstein 1905 is crap.
>> Nice one, Mike.
>> Androcles
> I tried hard to incorporate every known formal/informal
fallacy
> exactly like the originator of the thread has done. :)
> But speaking about fallacies, I think thw Ônon 
sequitur is
the most
> serious one, that is the conclusion does not follow from
the premises,
> a disconected idea.
> Mike
Actually the commonest fallacy in Special Relativity is to
call a
conclusion
a postulate via the fallacy of circularity, and that is
pretty serious.
The velocity of light is c in all inertial frames is bandied
about as
if it were
common knowledge, but stems from V = (c+w)/(1+ w/c), and that
was
a conclusion Einstein drew in section 5 with the help of the
equations
of transformation developed in section 3 (i.e. the Lorentz
transformations)
from which he extracts sqrt(1-v^2/c^2).
Ref. http://www.fourmilab.ch/etexts/einstein/specrel/www/
The idiots claim to be able to derive the LTs from a
conclusion that was
itself derived from the LT and circularity is thus
established.
In fact the original postulate ( which by definition of
postulate it
cannot be)
is light is always propagated in empty space with a definite
velocity c
which is independent of the state of motion of the emitting
body.
The reason it cannot be a postulate is that it is not
intuitive, it is
testable,
and it has no experimental data to support it. It is an
hypothesis. That
it reaches absurd conclusions indicates it to be a failed
hypothesis.
Reading Einsteins paper carefully, it becomes obvious he 
was
perpetrating a hoax. He begins with Galilean Relativity,
using a magnet
and a conductor as an example, carefully avoiding any explicit
description but vaguely referring to laws that hold good,
claims
that although it is irreconcilable with his second hypothesis
this is
only apparent , further claims that the velocity of light in
our
theory
plays the part, physically, of an infinitely great velocity 
and
produces
a mathematical fiction,
xi = x/sqrt(1-v^2/c^2) ..... (recall that x 
= x-vt)
The reason Einstein so carefully avoided stating the PoR is
that
he makes use of it when he says
But the ray moves relatively to the initial point of k, when
measured
in the stationary system, with the velocity c-v...
so he must be careful not to be too explicit in the
introduction and
preceding sections.
To conclude
It follows, further, that the velocity of light c cannot be
altered by
composition with a velocity less than that of light. For this
case we
obtain V = (c+w)/(1+w/c) = c.
denies him the use of the ray moves [at] c-v... without which
he
cannot derive the Lorentz transforms.
Shubert makes the same fallacy of circularity in his rather
long-winded
but actually quite simple minded derivation, although he now
admits
pulling sqrt(1-v^2/c^2) from thin air. He is
nothingimportant, of
course, although he likes to think he is everythingimportant.
Egomania
is treatable, I understand, although it may take the services
of a
trained psychiatrist.
Androcles
===
Subject: Re: final proof Einstein 1905 is crap.
> Actually the commonest fallacy in Special Relativity is to
call a
> conclusion
> a postulate via the fallacy of circularity, and that is
pretty serious.
Well, you are hetting into really dark domains here. Before
getting as
dvanced as SR, the use of circularity was enforced by those
who
missinterpreted Newton, or actually desired to present
Newtons
metaphysics as a science. To be explicit:
Newton used a modified version of Keplers third 
law in order
to
derive the law of universal gravitation. Please note that
without such
law, which was purely empirical, the derivation of LUG is
impossible.
Then, in mechanics books you see the famous Kepler problem
and the
derivation of Keplers third law using the LUG as given!
But of course, if the conclusion is true, all premised in the
deduction process must be true otherwise the deduction was
unsound in
the first place.
The same of course hold with Einstein and some claims the
constancy of
c can be derived , etc.
Let me tell you what it boild down to: Math infidels with no
understanding of the principles of logic the Greeks
established and
Aristotle put in a massive organized works called the
Organon, have
taken over and their misunderstanding and paranoia has
resulted in the
chaos we see in science todat with inductions presented as
deductions,
deductions as induction, abductions as inductions and
deductions.
Question of the day: how to you distinguish between paranoid
and
genious? Often, both sound as equally ntelligent.
My answer: Paranoids are easy to spot is you understand
formal and
informal fallacies and the ways they can be concealed in what
appears
to be a sound inference. Watch for:
Non sequitur
Circularity
affirmation of the consequent
denial of the antecedant
false analogy
red herring
faulty causality
faulty generalization
Straw man
etc.
I belive Einstein has concealed all of the above fallacies in
what is
a masterpiece of paranoid thinking to result in a circular
theory
where all predictions will match observations but no
prediction will
result that will alter the fundamental basis of the theory
which is:
paranoia.
Mike
===
Subject: Re: final proof Einstein 1905 is crap.
Mike:
>Newton used a modified version of Keplers 
third law in order
to
>derive the law of universal gravitation. Please note that
without such
>law, which was purely empirical, the derivation of LUG is
impossible.
Newton did not derive his law of gravitation, nor can it be
derived
from keplers laws. Newton deduced a general principle from
keplers
laws and if that principle was to stand a chance of being
correct,
the first requirement is that keplers laws must 
be derivable
from
that principle.
>Then, in mechanics books you see the famous Kepler problem
and the
>derivation of Keplers third law using the LUG as given!
Keplers laws _can_ be derived from newtons 
law of gravity.
Its
called consistency, not circularity. Newtons law if gravity
_cannot_
be derived from keplers laws. Keplers laws 
say nothing
about forces
or potentials or even why the planetary motion should be
described
by his laws. By contrast, the 1/r potential newton proposed
can be
used to show that stable orbits only occur for forces of the
form
r^n, where n = +/- 2.
>But of course, if the conclusion is true, all premised in the
>deduction process must be true otherwise the deduction was
unsound in
>the first place.
>The same of course hold with Einstein and some claims the
constancy of
>c can be derived , etc.
The assertion that `c is a constant cannot be derived, 
since
obviously
its possible to create a self-consistent theory of 
mechanics
in which
c -> infty, called galilean relativity. Distinguishing
between special
relativity and galilean relativity is an experimental issue
and since
galilean relativity makes predictions that are incompatible
with what
is observed, experiment has determined that there exists some
velocity,
`c, which is finite. Electromagnetic theory 
predicts that `c
and the
speed of light must be the same.
[..]
>I belive Einstein has concealed all of the above fallacies
in what is
>a masterpiece of paranoid thinking to result in a circular
theory
>where all predictions will match observations but no
prediction will
>result that will alter the fundamental basis of the theory
which is:
>paranoia.
Special relativity has been used to make quite a number of
predictions.
For example, the standard model is based on special
relativity. If
relativity was wrong, the standard model would have had zero
chance of
correctly predicting the mass ratio between the W and Z or
the myriad
of other predictions it make correctly. While there is more
to the
standard model than relativity, relativity is the foundation
upon which
the standard model is constructed.
===
Subject: Re: final proof Einstein 1905 is crap.
>> Actually the commonest fallacy in Special Relativity is to
call a
>> conclusion
>> a postulate via the fallacy of circularity, and that is
pretty
>> serious.
> Well, you are hetting into really dark domains here. Before
getting as
> dvanced as SR, the use of circularity was enforced by those
who
> missinterpreted Newton, or actually desired to present
Newtons
> metaphysics as a science. To be explicit:
> Newton used a modified version of Keplers 
third law in
order to
> derive the law of universal gravitation. Please note that
without such
> law, which was purely empirical, the derivation of LUG is
impossible.
> Then, in mechanics books you see the famous Kepler problem
and the
> derivation of Keplers third law using the LUG as given!
> But of course, if the conclusion is true, all premised in
the
> deduction process must be true otherwise the deduction was
unsound in
> the first place.
Hold on a moment. Keplers third law is empirical, 
youve
said.
Thats originating in or based on observation or experience.
Keplers third law (however modified) is
The ratio of the squares of the revolutionary periods for two
planets
is equal to the ratio of the cubes of their semimajor axes:
(P1/P2)^2 = (R1/R2)^3
Are you saying this is NOT what is observed?
All I see here is that a simple equation has been determined
to describe
observation.
> The same of course hold with Einstein and some claims the
constancy of
> c can be derived , etc.
That is a very different matter. Nobody has ever observed
light is
always propagated in empty space with a definite velocity c
which is
independent of the state of motion of the emitting body.
There is no empirical foundation to it.
In fact the empirical data contradicts it, aside from any
logical
deduction.
> Let me tell you what it boild down to: Math infidels with no
> understanding of the principles of logic the Greeks
established and
> Aristotle put in a massive organized works called the
Organon, have
> taken over and their misunderstanding and paranoia has
resulted in the
> chaos we see in science todat with inductions presented as
deductions,
> deductions as induction, abductions as inductions and
deductions.
Youll get no argument from me on that.
> Question of the day: how to you distinguish between
paranoid and
> genious? Often, both sound as equally ntelligent.
> My answer: Paranoids are easy to spot is you understand
formal and
> informal fallacies and the ways they can be concealed in
what appears
> to be a sound inference. Watch for:
> Non sequitur
> Circularity
> affirmation of the consequent
> denial of the antecedant
> false analogy
> red herring
> faulty causality
> faulty generalization
> Straw man
> etc.
> I belive Einstein has concealed all of the above fallacies
in what is
> a masterpiece of paranoid thinking to result in a circular
theory
> where all predictions will match observations but no
prediction will
> result that will alter the fundamental basis of the theory
which is:
> paranoia.
To you he was paranoid.
You are looking into the cause of his ill-concieved ideas and
attributing
it to a form of mental illness.
To me he was a huckster.
Without getting into psychology or psychiatry, fields in which
I do not
claim expertise, for maybe thieves and murderers suffer from
mental
abberations and it can all be blamed on an abusive childhood,
nevertheless
I want them locked away in a correctional institution or be
taken
out of society if it can be justly determined that they are
guilty of
crime.
It is too late to correct Einstein the man, hes lived and
died.
We are still obligated to eradicate the consequences of his
fraud,
whatever the cause.
Therefore I have to say your analysis is actually --- Non
sequitur.
But then, so is mine. Whatever the cause, it is our duty to
promote
freedom of thought in the next generation of budding young
physicists
and to dissuade them from the indoctrinators that infest this
newsgroup.
Androcles.
> Mike
===
Subject: Re: final proof Einstein 1905 is crap.
> Actually the commonest fallacy in Special Relativity is to
call a
>> conclusion
>> a postulate via the fallacy of circularity, and that is
pretty
>> serious.
>> Well, you are hetting into really dark domains here.
Before getting as
> dvanced as SR, the use of circularity was enforced by those
who
> missinterpreted Newton, or actually desired to present
Newtons
> metaphysics as a science. To be explicit:
> Newton used a modified version of Keplers 
third law in
order to
> derive the law of universal gravitation. Please note that
without such
> law, which was purely empirical, the derivation of LUG is
impossible.
> Then, in mechanics books you see the famous Kepler problem
and the
> derivation of Keplers third law using the LUG as given!
> But of course, if the conclusion is true, all premised in
the
> deduction process must be true otherwise the deduction was
unsound in
> the first place.
> Hold on a moment. Keplers third law is empirical, 
youve
said.
> Thats originating in or based on observation or 
experience.
> Keplers third law (however modified) is
> The ratio of the squares of the revolutionary periods for
two planets
> is equal to the ratio of the cubes of their semimajor axes:
> (P1/P2)^2 = (R1/R2)^3
> Are you saying this is NOT what is observed?
> All I see here is that a simple equation has been
determined to describe
> observation.
> The same of course hold with Einstein and some claims the
constancy of
> c can be derived , etc.
> That is a very different matter. Nobody has ever observed
light is
> always propagated in empty space with a definite velocity c
which is
> independent of the state of motion of the emitting body.
> There is no empirical foundation to it.
> In fact the empirical data contradicts it, aside from any
logical
> deduction.
Not that hard to do!
Use the standard apparatus for testing light velocity, the
two,
toothed wheels, form which c can be calculated from the
difference in
their rotation and separation when a beam is shone through.
Now stick
it on a rocket, and test the sunlight speed when the rocket is
travelling away/towards, the sun.
Probably cost a few million on the next interplanetary run-
FAR too
expensive for the DHR budget! (make that too likely to
embarass)
> Let me tell you what it boild down to: Math infidels with no
> understanding of the principles of logic the Greeks
established and
> Aristotle put in a massive organized works called the
Organon, have
> taken over and their misunderstanding and paranoia has
resulted in the
> chaos we see in science todat with inductions presented as
deductions,
> deductions as induction, abductions as inductions and
deductions.
> Youll get no argument from me on that.
> Question of the day: how to you distinguish between
paranoid and
> genious? Often, both sound as equally ntelligent.
> My answer: Paranoids are easy to spot is you understand
formal and
> informal fallacies and the ways they can be concealed in
what appears
> to be a sound inference. Watch for:
> Non sequitur
> Circularity
> affirmation of the consequent
> denial of the antecedant
> false analogy
> red herring
> faulty causality
> faulty generalization
> Straw man
> etc.
> I belive Einstein has concealed all of the above fallacies
in what is
> a masterpiece of paranoid thinking to result in a circular
theory
> where all predictions will match observations but no
prediction will
> result that will alter the fundamental basis of the theory
which is:
> paranoia.
> To you he was paranoid.
> You are looking into the cause of his ill-concieved ideas
and
> attributing
> it to a form of mental illness.
> To me he was a huckster.
> Without getting into psychology or psychiatry, fields in
which I do not
> claim expertise, for maybe thieves and murderers suffer
from mental
> abberations and it can all be blamed on an abusive
childhood,
> nevertheless
> I want them locked away in a correctional institution or be
taken
> out of society if it can be justly determined that they are
guilty of
> crime.
> It is too late to correct Einstein the man, hes lived and
died.
> We are still obligated to eradicate the consequences of his
fraud,
> whatever the cause.
> Therefore I have to say your analysis is actually --- Non
sequitur.
> But then, so is mine. Whatever the cause, it is our duty to
promote
> freedom of thought in the next generation of budding young
physicists
> and to dissuade them from the indoctrinators that infest
this newsgroup.
> Androcles.
> Mike
===
Subject: Re: final proof Einstein 1905 is crap.
>> Actually the commonest fallacy in Special Relativity is to
call a
> conclusion
> a postulate via the fallacy of circularity, and that is
pretty
> serious.
> Well, you are hetting into really dark domains here. Before
getting
>> as
>> dvanced as SR, the use of circularity was enforced by
those who
>> missinterpreted Newton, or actually desired to present
Newtons
>> metaphysics as a science. To be explicit:
> Newton used a modified version of Keplers 
third law in
order to
>> derive the law of universal gravitation. Please note that
without
>> such
>> law, which was purely empirical, the derivation of LUG is
>> impossible.
> Then, in mechanics books you see the famous Kepler problem
and the
>> derivation of Keplers third law using the LUG as given!
> But of course, if the conclusion is true, all premised in
the
>> deduction process must be true otherwise the deduction was
unsound
>> in
>> the first place.
>> Hold on a moment. Keplers third law is empirical, 
youve
said.
>> Thats originating in or based on observation or
experience.
>> Keplers third law (however modified) is
>> The ratio of the squares of the revolutionary periods for
two
>> planets
>> is equal to the ratio of the cubes of their semimajor axes:
>> (P1/P2)^2 = (R1/R2)^3
>> Are you saying this is NOT what is observed?
>> All I see here is that a simple equation has been
determined to
>> describe
>> observation.
> The same of course hold with Einstein and some claims the
constancy
>> of
>> c can be derived , etc.
>> That is a very different matter. Nobody has ever observed
light is
>> always propagated in empty space with a definite velocity c
which is
>> independent of the state of motion of the emitting body.
>> There is no empirical foundation to it.
>> In fact the empirical data contradicts it, aside from any
logical
>> deduction.
> Not that hard to do!
> Use the standard apparatus for testing light velocity, the
two,
> toothed wheels, form which c can be calculated from the
difference in
> their rotation and separation when a beam is shone through.
Now stick
> it on a rocket, and test the sunlight speed when the rocket
is
> travelling away/towards, the sun.
> Probably cost a few million on the next interplanetary run-
FAR too
> expensive for the DHR budget! (make that too likely to
embarass)
The easiest way is Roemers method and Cassini. All it takes
is
data gathering. The DHRs would say time was increasing and
decreasing during the orbit, but it wouldnt be all that
convincing.
Androcles
> Let me tell you what it boild down to: Math infidels with no
>> understanding of the principles of logic the Greeks
established and
>> Aristotle put in a massive organized works called the
Organon, have
>> taken over and their misunderstanding and paranoia has
resulted in
>> the
>> chaos we see in science todat with inductions presented as
>> deductions,
>> deductions as induction, abductions as inductions and
deductions.
>> Youll get no argument from me on that.
> Question of the day: how to you distinguish between
paranoid and
>> genious? Often, both sound as equally ntelligent.
> My answer: Paranoids are easy to spot is you understand
formal and
>> informal fallacies and the ways they can be concealed in
what
>> appears
>> to be a sound inference. Watch for:
> Non sequitur
>> Circularity
>> affirmation of the consequent
>> denial of the antecedant
>> false analogy
>> red herring
>> faulty causality
>> faulty generalization
>> Straw man
>> etc.
>> I belive Einstein has concealed all of the above fallacies
in what
>> is
>> a masterpiece of paranoid thinking to result in a circular
theory
>> where all predictions will match observations but no
prediction
>> will
>> result that will alter the fundamental basis of the theory
which
>> is:
>> paranoia.
> To you he was paranoid.
>> You are looking into the cause of his ill-concieved ideas
and
>> attributing
>> it to a form of mental illness.
>> To me he was a huckster.
>> Without getting into psychology or psychiatry, fields in
which I do
>> not
>> claim expertise, for maybe thieves and murderers suffer
from mental
>> abberations and it can all be blamed on an abusive
childhood,
>> nevertheless
>> I want them locked away in a correctional institution or
be taken
>> out of society if it can be justly determined that they
are guilty of
>> crime.
>> It is too late to correct Einstein the man, hes lived 
and
died.
>> We are still obligated to eradicate the consequences of
his fraud,
>> whatever the cause.
>> Therefore I have to say your analysis is actually --- Non
sequitur.
>> But then, so is mine. Whatever the cause, it is our duty
to promote
>> freedom of thought in the next generation of budding young
physicists
>> and to dissuade them from the indoctrinators that infest
this
>> newsgroup.
>> Androcles.
>> Mike
===
Subject: Re: final proof Einstein 1905 is crap.
> The same of course hold with Einstein and some claims the
constancy
of
> c can be derived , etc.
>
> That is a very different matter. Nobody has ever observed
light is
> always propagated in empty space with a definite velocity c
which is
> independent of the state of motion of the emitting body.
Androcles is being a retard, as per usual.
> There is no empirical foundation to it.
More of Androcles not reading the literature, as per usual.
> In fact the empirical data contradicts it, aside from any
logical
> deduction.
More of Androcles mistakening his suppositions for reality,
as per
usual.
> Not that hard to do!
> Use the standard apparatus for testing light velocity, the
two,
> toothed wheels, form which c can be calculated from the
difference in
> their rotation and separation when a beam is shone through.
Now stick
> it on a rocket, and test the sunlight speed when the rocket
is
> travelling away/towards, the sun.
> Probably cost a few million on the next interplanetary run-
FAR too
> expensive for the DHR budget! (make that too likely to
embarass)
Since you have no idea what goes into designing a mission
package,
such ignorant statements come as no surprise.
Why dont you read the literature? Have you ever considered
the
possibility that you are wrong?
===
Subject: Re: final proof Einstein 1905 is crap.
Jim Greenfield:
>Not that hard to do!
>Use the standard apparatus for testing light velocity, the
two,
>toothed wheels, form which c can be calculated from the
difference in
>their rotation and separation when a beam is shone through.
Now stick
>it on a rocket, and test the sunlight speed when the rocket
is
>travelling away/towards, the sun.
>Probably cost a few million on the next interplanetary run-
FAR too
>expensive for the DHR budget! (make that too likely to
embarass)
Ill be happy to perform the experiment if you pay for it.
My guess is that you arent willing to back up your beliefs
with cash.
===
Subject: Re: final proof Einstein 1905 is crap.
[snip
> Hold on a moment. Keplers third law is empirical, 
youve
said.
> Thats originating in or based on observation or 
experience.
> Keplers third law (however modified) is
> The ratio of the squares of the revolutionary periods for
two planets
> is equal to the ratio of the cubes of their semimajor axes:
> (P1/P2)^2 = (R1/R2)^3
> Are you saying this is NOT what is observed?
> All I see here is that a simple equation has been
determined to describe
> observation.
Actually thats the observed law. Newton did not use that
because it
will gets you nowhere. Newton used kmT^2 = R^3 where k is a
constant.
This is not mentioned in most books. The constant k ends up
part of G
in the LUG. This is what Newton did, he assumed that in the
two-body
problem:
k(m1+m2)T^2 = (R1+R2)^3=R^3
He then assumed that m1>m2 to get:
k1m1T^2 ~ R^3
and he assumed again that this hold for all planets so that:
km1T1^2 = R1^3
km2T2^2 = R2^3
divide and you get Keplers observed law!
This series of assumptions by Newton hides the hypothesis
that one of
the bodies is fixed in absolute space and is immovable, and
that is
the SUN. He actually states that hypothesis I:
HYPOTHESIS I.
That the center of the system of the world is immovable.
This is acknowledged by all, while some contend that the
earth, others
that the sun is fixed in that center. Let us see what may from
hence
follow.
PROPOSITION XI. THEOREM XI.
That the common center of gravity of the earth, the sun, and
all the
planets, is immovable.
LOL
Without such hypothesis, derivation of the LUG is impossible.
Now, math infidels such as Newton and Einstein did not realize
that
unless bodies trajectories are fixed in space by a magical
Ôspirit
(actually Newton believed in such spirit) their gravitational
dynamics
lead to a collapse of the universe in a single point over
time. This
is why Einstein wanted to include the cosmological constant
to prevent
that.
Yet, everybody is deceived to believe that the derived
theories
explain gravitation but is deprived of the underline
assumptions that
basically invoke magic.
Most books attempt to circumvent the above using hand waiving
arguments and Kolker eats them for breakfast. Yet, what is,
is.
Mike
===
Subject: Re: final proof Einstein 1905 is crap.
> [snip
>> Hold on a moment. Keplers third law is empirical, 
youve
said.
>> Thats originating in or based on observation or
experience.
>> Keplers third law (however modified) is
>> The ratio of the squares of the revolutionary periods for
two
>> planets
>> is equal to the ratio of the cubes of their semimajor axes:
>> (P1/P2)^2 = (R1/R2)^3
>> Are you saying this is NOT what is observed?
>> All I see here is that a simple equation has been
determined to
>> describe
>> observation.
>> Actually thats the observed law. Newton did not use that
because it
> will gets you nowhere. Newton used kmT^2 = R^3 where k is a
constant.
> This is not mentioned in most books. The constant k ends up
part of G
> in the LUG. This is what Newton did, he assumed that in the
two-body
> problem:
> k(m1+m2)T^2 = (R1+R2)^3=R^3
> He then assumed that m1>m2 to get:
> k1m1T^2 ~ R^3
> and he assumed again that this hold for all planets so that:
> km1T1^2 = R1^3
> km2T2^2 = R2^3
> divide and you get Keplers observed law!
> This series of assumptions by Newton hides the hypothesis
that one of
> the bodies is fixed in absolute space and is immovable, and
that is
> the SUN. He actually states that hypothesis I:
> HYPOTHESIS I.
> That the center of the system of the world is immovable.
> This is acknowledged by all, while some contend that the
earth, others
> that the sun is fixed in that center. Let us see what may
from hence
> follow.
Einstein does the same thing, calling it the stationary
system.
However, Newton is careful to call an hypothesis by its name,
but Einstein calls his own light speed hypothesis a postulate.
> PROPOSITION XI. THEOREM XI.
> That the common center of gravity of the earth, the sun,
and all the
> planets, is immovable.
> LOL
> Without such hypothesis, derivation of the LUG is
impossible.
Granted that Newton envisaged a heliocentric universe,
nevertheless
the solar system can be treated as having a common centre of
gravity.
He didnt know of the existence of Pluto, the orbit of which
is Sol
dependent.
Moreover, any perturbation of the orbits of the planets by
the nearest
star are neglible. There is a considerable difference between
Pluto
being circa 4 light hours from Sol and circa 4 light years
from Proxima
Centauri, particularly when R^3 is taken into account.
(24 * 365.25 )^3 ~= 5,000,000.
> Now, math infidels such as Newton and Einstein did not
realize that
> unless bodies trajectories are fixed in space by a magical
Ôspirit
> (actually Newton believed in such spirit) their
gravitational dynamics
> lead to a collapse of the universe in a single point over
time.
A trajectory cannot be fixed in space by 
definition of
trajectory,
which implies motion. I Ôve missed the point you are
attempting to make.
Until such times as our solar system enters the vicinity of a
massive
body such as another star, Ill go along with 
Newtons
description,
it is an adequate approximation.
> This
> is why Einstein wanted to include the cosmological constant
to prevent
> that.
> Yet, everybody is deceived to believe that the derived
theories
> explain gravitation but is deprived of the underline
assumptions that
> basically invoke magic.
Action at a distance IS magic, in that we do not understand
it.
However, we do observe it and Newton gives a good
approximation that
will adequately predict the future positions of the planets,
even though
it
will ultimately be chaotic.
> Most books attempt to circumvent the above using hand
waiving
> arguments and Kolker eats them for breakfast. Yet, what is,
is.
Why bother with Kolker? Hes just another parrot, repeating
what
hes read. Hes not capable of actually 
thinking, few are.
There are
three distinct categories of people that write to this
newsgroup.
1) Parrots that can only repeat what theyve read.
2) Egocentrics that want to promote their own pet theory.
3) Rational thinkers.
There are three mutually exclusive concepts that compete.
a) the velocity of light is medium dependent
b) the velocity of light is source dependent
c) the velocity of light is observer dependent
Kolker is a 1c)
Androcles.
> Mike
===
Subject: Re: final proof Einstein 1905 is crap.
> Newton used a modified version of Keplers 
third law in
order to
> derive the law of universal gravitation. Please note that
without such
> law, which was purely empirical, the derivation of LUG is
impossible.
Not true. A central force with a 1/r potential will give you
all of the
Keplarian laws. It is quite true that Keplers third law
-suggest- to
Newton that the inverse square law is correct, but Kepler
independent
ways of deriving the famous inversesquare law plus all of
Keplers
empirical laws exist as I have indicated. Refere to
Goldsteins book on
Mechanics for the mathematical details.
> Then, in mechanics books you see the famous Kepler problem
and the
> derivation of Keplers third law using the LUG as given!
Not true, as I have shown.
> Let me tell you what it boild down to: Math infidels with no
> understanding of the principles of logic the Greeks
established and
> Aristotle put in a massive organized works called the
Organon, have
> taken over and their misunderstanding and paranoia has
resulted in the
> chaos we see in science todat with inductions presented as
deductions,
> deductions as induction, abductions as inductions and
deductions.
Aristotles logic is a restricted subset of 
first order logic.
Furthermore the syllogism is not the ideal form of
mathematical proof
structure. The Stoics* developed a form of logic** that is
much closer
to first order logic than Aristotle. Unfortunately, most of
their works
in logic have not survived. For details refer to -The
Development of
Logic- by Kneale and Kneale (hom et ux).
> I belive Einstein has concealed all of the above fallacies
in what is
> a masterpiece of paranoid thinking to result in a circular
theory
> where all predictions will match observations but no
prediction will
> result that will alter the fundamental basis of the theory
which is:
> paranoia.
Utter nonsense. SR is potentially falsifiable. Later ways of
expressing
SR by means of Minkowski geometry show how stragight forward
SR is. The
group of transformation (Lorentz Group) which is the bases of
SR were
established first and independently by Lorentz and Poincare so
there are
no mathematical mysteries here.
Bob Kolker
* especially Chrysippus
** modus ponens
> Mike
===
Subject: Re: final proof Einstein 1905 is crap.
>
> Newton used a modified version of Keplers 
third law in
order to
> derive the law of universal gravitation. Please note that
without such
> law, which was purely empirical, the derivation of LUG is
impossible.
> Not true. A central force with a 1/r potential will give
you all of the
> Keplarian laws. It is quite true that Keplers third law
-suggest- to
> Newton that the inverse square law is correct, but Kepler
independent
> ways of deriving the famous inversesquare law plus all of
Keplers
> empirical laws exist as I have indicated. Refere to
Goldsteins book on
> Mechanics for the mathematical details.
>
> Then, in mechanics books you see the famous Kepler problem
and the
> derivation of Keplers third law using the LUG as given!
> Not true, as I have shown.
>
> Let me tell you what it boild down to: Math infidels with no
> understanding of the principles of logic the Greeks
established and
> Aristotle put in a massive organized works called the
Organon, have
> taken over and their misunderstanding and paranoia has
resulted in the
> chaos we see in science todat with inductions presented as
deductions,
> deductions as induction, abductions as inductions and
deductions.
> Aristotles logic is a restricted subset of 
first order
logic.
> Furthermore the syllogism is not the ideal form of
mathematical proof
> structure. The Stoics* developed a form of logic** that is
much closer
> to first order logic than Aristotle. Unfortunately, most of
their works
> in logic have not survived. For details refer to -The
Development of
> Logic- by Kneale and Kneale (hom et ux).
>
> I belive Einstein has concealed all of the above fallacies
in what is
> a masterpiece of paranoid thinking to result in a circular
theory
> where all predictions will match observations but no
prediction will
> result that will alter the fundamental basis of the theory
which is:
> paranoia.
> Utter nonsense. SR is potentially falsifiable. Later ways of
expressing
> SR by means of Minkowski geometry show how stragight
forward SR is. The
> group of transformation (Lorentz Group) which is the bases
of SR were
> established first and independently by Lorentz and Poincare
so there are
> no mathematical mysteries here.
Just the biggie!
HOW does photon A, emitted from a source stationary ref us,
KNOW how
fast B, emitted from a moving source at that location, is
travelling?????????????
Jim G
c=c+v
> Bob Kolker
> * especially Chrysippus
> ** modus ponens
>
>
> Mike
===
Subject: Re: final proof Einstein 1905 is crap.
Jim Greenfield:
>Just the biggie!
>HOW does photon A, emitted from a source stationary ref us,
KNOW how
>fast B, emitted from a moving source at that location, is
>travelling?????????????
Why would a photon have to know that?
===
Subject: Re: final proof Einstein 1905 is crap.
> Newton used a modified version of Keplers 
third law in
order to
> derive the law of universal gravitation. Please note that
without such
> law, which was purely empirical, the derivation of LUG is
impossible.
> Not true. A central force with a 1/r potential will give
you all of the
> Keplarian laws. It is quite true that Keplers third law
-suggest- to
> Newton that the inverse square law is correct, but Kepler
independent
> ways of deriving the famous inversesquare law plus all of
Keplers
> empirical laws exist as I have indicated. Refere to
Goldsteins book on
> Mechanics for the mathematical details.
> Then, in mechanics books you see the famous Kepler problem
and the
> derivation of Keplers third law using the LUG as given!
> Not true, as I have shown.
> Let me tell you what it boild down to: Math infidels with no
> understanding of the principles of logic the Greeks
established and
> Aristotle put in a massive organized works called the
Organon, have
> taken over and their misunderstanding and paranoia has
resulted in the
> chaos we see in science todat with inductions presented as
deductions,
> deductions as induction, abductions as inductions and
deductions.
> Aristotles logic is a restricted subset of 
first order
logic.
> Furthermore the syllogism is not the ideal form of
mathematical proof
> structure. The Stoics* developed a form of logic** that is
much closer
> to first order logic than Aristotle. Unfortunately, most of
their works
> in logic have not survived. For details refer to -The
Development of
> Logic- by Kneale and Kneale (hom et ux).
> I belive Einstein has concealed all of the above fallacies
in what is
> a masterpiece of paranoid thinking to result in a circular
theory
> where all predictions will match observations but no
prediction will
> result that will alter the fundamental basis of the theory
which is:
> paranoia.
> Utter nonsense. SR is potentially falsifiable.
Not for religious fanatics like Mike (aka Bill Smith,
Undeniable, Eleatis
etc...)
> Later ways of expressing
> SR by means of Minkowski geometry show how stragight
forward SR is. The
> group of transformation (Lorentz Group) which is the bases
of SR were
> established first and independently by Lorentz and Poincare
so there are
> no mathematical mysteries here.
You are wasting your time with this idiot.
Dirk Vdm
===
Subject: Re: final proof Einstein 1905 is crap.
> Shubert makes the same fallacy of circularity in his rather
long-winded
> but actually quite simple minded derivation, although he
now admits
> pulling sqrt(1-v^2/c^2) from thin air.
I missed his derivation but have noted here that Tom 
Roberts
derivation
involves hand-waving when he reaches his penultimate step:
the evaluation
of
a criterion, value one for Newton, over or under for SR and
... polar
coordinates? (Dont remember because I didnt 
actually read
that part.) The
Newton part was automatic, the SR parts logic trail
disappeared and he
just
said, well make it this particular value and we have SR.
SR-cult cretins here have agreed Einsteins 1916/1962 
(lets
hear it from
the idiots) did really idiotic things. AE 1905 was just as
bad but less
obvious.
Ive seen a number of so-called derivations that started 
with
the
conclusion.
eleaticus
He is nothingimportant, of
> course, although he likes to think he is
everythingimportant. Egomania
> is treatable, I understand, although it may take the
services of a
> trained psychiatrist.
> Androcles
===
Subject: Re: final proof Einstein 1905 is crap.
>> Shubert makes the same fallacy of circularity in his rather
>> long-winded
>> but actually quite simple minded derivation, although he
now admits
>> pulling sqrt(1-v^2/c^2) from thin air.
> I missed his derivation but have noted here that Tom
Roberts
> derivation
> involves hand-waving when he reaches his penultimate step:
the
> evaluation of
> a criterion, value one for Newton, over or under for SR and
... polar
> coordinates? (Dont remember because I 
didnt actually read
that
> part.) The
> Newton part was automatic, the SR parts logic trail
disappeared and
> he just
> said, well make it this particular value and we have SR.
Ive carved Roberts up big time in the thread
Roberts considers Einstein paper to be irrelevant verbiage.
He really is a sad case of fanaticism.
Androcles
> SR-cult cretins here have agreed Einsteins 1916/1962
(lets hear it
> from
> the idiots) did really idiotic things. AE 1905 was just as
bad but
> less
> obvious.
> Ive seen a number of so-called derivations that started
with the
> conclusion.
> eleaticus
> He is nothingimportant, of
>> course, although he likes to think he is
everythingimportant.
>> Egomania
>> is treatable, I understand, although it may take the
services of a
>> trained psychiatrist.
>> Androcles
===
Subject: Re: final proof Einstein 1905 is crap.
eleaticus:
>Ive seen a number of so-called derivations that started
with the
>conclusion.
on a regular basis.
hello.....doctor~
The set of all complex numbers with the usual topology
is both separable and 2nd countable.
------------------------------------------------
this is true.
because, f : C -> R X R, f(x+yi) = (x,y)
so, C and R^2 is homeomorphic.
and countable basis of R^2 is
{U X V | U=(a,b),V=(c,d), a,b,c,d in Q}
so, 2nd countable. so, separable....
um.......right ??
thank you very much for your advice.
> ------------------------------------------------
> this is true.
> because, f : C -> R X R, f(x+yi) = (x,y)
> so, C and R^2 is homeomorphic.
> and countable basis of R^2 is
> {U X V | U=(a,b),V=(c,d), a,b,c,d in Q}
This set is not a basis. but this set is dense in R^2 and
countable.
Therefore, this set shows that R^2 is separable.
Furthermore, Since R^2 is metrizable, R^2 is 2nd countable.
(for its known
that a metrizable and separable space is second countable.)
> so, 2nd countable. so, separable....
> um.......right ??
> thank you very much for your advice.
pointed out.
I misreaded the set as something else.
mina_worlds argument is right.
>>------------------------------------------------
>>this is true.
>>because, f : C -> R X R, f(x+yi) = (x,y)
>>so, C and R^2 is homeomorphic.
>>and countable basis of R^2 is
>>{U X V | U=(a,b),V=(c,d), a,b,c,d in Q}
> This set is not a basis. but this set is dense in R^2 and
countable.
> Therefore, this set shows that R^2 is separable.
> Furthermore, Since R^2 is metrizable, R^2 is 2nd countable.
(for its
known
> that a metrizable and separable space is second countable.)
>>so, 2nd countable. so, separable....
>>um.......right ??
>>thank you very much for your advice.
How is this set not a basis? this is the set of all products
of open
intervals with rational endpoints. if B1 is a basis for X and
B2 is a
basis for Y then the set {AxB|A is in B1,B is in B2} is a
basis for XxY.
>>------------------------------------------------
>>this is true.
>>because, f : C -> R X R, f(x+yi) = (x,y)
>>so, C and R^2 is homeomorphic.
>>and countable basis of R^2 is
>>{U X V | U=(a,b),V=(c,d), a,b,c,d in Q}
> This set is not a basis. but this set is dense in R^2 and
countable.
> Therefore, this set shows that R^2 is separable.
> Furthermore, Since R^2 is metrizable, R^2 is 2nd countable.
(for its
known
> that a metrizable and separable space is second countable.)
Perhaps its been too long and my recollection is faulty, 
but
if (x,y) is in R^2, and U is an open set of the plane
containing
(x,y), then isnt there a product (a,b)x(c,d) of intervals
with
rational endpoints, within U, that contains (x,y)? Surely, if
there is a ball around (x,y) with positive radius within U,
there is a box with rational vertices as well.
Doesnt that count as forming a basis?
>>so, 2nd countable. so, separable....
>>um.......right ??
>>thank you very much for your advice.
Dale
> hello.....doctor~
> The set of all complex numbers with the usual topology
> is both separable and 2nd countable.
> ------------------------------------------------
> this is true.
> because, f : C -> R X R, f(x+yi) = (x,y)
> so, C and R^2 is homeomorphic.
> and countable basis of R^2 is
> {U X V | U=(a,b),V=(c,d), a,b,c,d in Q}
> so, 2nd countable. so, separable....
Your proof of second countability is correct. Separability
means the
existence of a countable dense subset. For R^2, Q^2 naturally
embedded is
both dense and countable.
Igor
===
Subject: Function naming
Im doing some study on a perticular type of functions (in 
my
spare time
just for fun). The problem is however that I dont know how
these functions
are called in English. Could someone provide me with the
correct english
term and point me perhaps to some more info on this topic.
The functions are of the form;
y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2
This function has two input variables and has an order of
two. The number
of
variables and order can vary accordingly ofcourse. Is there a
general term
for this kind of functions?
===
Subject: Re: Function naming
> Im doing some study on a perticular type of functions (in
my spare time
> just for fun). The problem is however that I dont know 
how
these
functions
> are called in English. Could someone provide me with the
correct english
> term and point me perhaps to some more info on this topic.
> The functions are of the form;
> y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2
> This function has two input variables and has an order of
two. The number
of
> variables and order can vary accordingly ofcourse. Is there
a general
term
> for this kind of functions?
Darius,
This is a quadratic (what you call order 2 but in English is
usually called degree two) polynomial in two variables. If
you vary
the number of variables (inputs) then they are called
polynomials in
several variables. It is impossible to tell your level of
mathematical education, so it is not clear where to direct
you for
further information.
Achava
===
Subject: Re: Function naming
function group as I described polynomials. I have some
knowledge of
math,
but I was always under the impression that a polynome was
defined as a one
variable function, and that these were merely a special case
of the
multivariable case. Probably has to do with my understanding
;-)
Achava Nakhash, the Loving Snake <achava@hotmail.com> schreef
in
bericht
> Im doing some study on a perticular type of functions (in
my spare
time
> just for fun). The problem is however that I dont know 
how
these
functions
> are called in English. Could someone provide me with the
correct
english
> term and point me perhaps to some more info on this topic.
> The functions are of the form;
> y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2
> This function has two input variables and has an order of
two. The
number of
> variables and order can vary accordingly ofcourse. Is there
a general
term
> for this kind of functions?
> Darius,
> This is a quadratic (what you call order 2 but in English is
> usually called degree two) polynomial in two variables. If
you vary
> the number of variables (inputs) then they are called
polynomials in
> several variables. It is impossible to tell your level of
> mathematical education, so it is not clear where to direct
you for
> further information.
> Achava
===
Subject: Re: Function naming
>function group as I described polynomials. I have some
knowledge of
math,
>but I was always under the impression that a polynome was
defined as a one
>variable function, and that these were merely a special case
of the
>multivariable case. Probably has to do with my understanding
;-)
(Please dont post upside down.)
A good resource for definitions is
http://mathworld.wolfram.com --
the definition at http://mathworld.wolfram.com/Polynomial.html
says
A polynomial is a mathematical expression involving a sum of
powers
in one or more variables multiplied by coefficients.
I understand that this wouldnt help you when you 
didnt know
the
term, but when you do want a definition of a 
specific term (or
facts
about a term or concept) its a great starting point.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
A: Maybe because some people are too annoyed by top-posting.
Q: Why do I not get an answer to my question(s)?
A: Because it messes up the order in which people normally
read text.
Q: Why is top-posting such a bad thing?
===
Subject: Re: Function naming
>function group as I described polynomials. I have some
knowledge of
math,
>but I was always under the impression that a polynome was
defined as a one
>variable function, and that these were merely a special case
of the
>multivariable case. Probably has to do with my understanding
;-)
>Im doing some study on a perticular type of functions (in
my spare time
>just for fun). The problem is however that I dont know how
these
>
>functions
>are called in English. Could someone provide me with the
correct english
>term and point me perhaps to some more info on this topic.
>The functions are of the form;
>y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2
>This function has two input variables and has an order of
two. The
>
>number of
>variables and order can vary accordingly ofcourse. Is there
a general
>
>term
>for this kind of functions?
>
To be more specific, I would call it a bivariate quadratic
function.
--
Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Re: Function naming
> function group as I described polynomials. I have some
knowledge of
math,
> but I was always under the impression that a polynome was
defined as a
one
> variable function, and that these were merely a special
case of the
> multivariable case. Probably has to do with my
understanding ;-)
You can call them polynomials indeed, but in most cases this
is (again
indeed) as involving a single variable. To alleviate you
could use
polynomials with multiple variables for the group of
functions. (I
have seen the term multinomial used for it, but this creates a
different kind of confusion.)
--
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: Function naming
> Im doing some study on a perticular type of functions (in
my spare time
> just for fun). The problem is however that I dont know 
how
these
functions
> are called in English. Could someone provide me with the
correct english
> term and point me perhaps to some more info on this topic.
> The functions are of the form;
> y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2
> This function has two input variables and has an order of
two. The number
of
> variables and order can vary accordingly ofcourse. Is there
a general
term
> for this kind of functions?
Its a quadratic (= your order of two) polynomial in two
indeterminates (= your input variables). If you dont want 
to
be tied
to order and number of indeterminates just call it a
polynomial.
Actually _as functions_ these things are better called
polynomial
functions rather than just polynomials, but that may not
matter to you.
You may also want to Google binary quadratic form.
===
Subject: Re: Function naming
> I dont know how these functions
>are called in English. Could someone provide me with the
correct english
>term and point me perhaps to some more info on this topic.
>The functions are of the form;
>y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2
This particular one is a quadratic in two variables. (The last
_three_ terms are all second degree.)
>This function has two input variables and has an order of
two. The number
of
>variables and order can vary accordingly ofcourse. Is there
a general term
>for this kind of functions?
More generally, a polynomial of degree N in M variables.
degree of 1 -- linear
degree of 2 -- quadratic
degree of 3 -- cubic
degree of 4 -- quartic
degree of 5 -- quintic
degree of 6 -- sextic
etc.
But above quadratic the special terms are progressively less
used.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
A: Maybe because some people are too annoyed by top-posting.
Q: Why do I not get an answer to my question(s)?
A: Because it messes up the order in which people normally
read text.
Q: Why is top-posting such a bad thing?
===
Subject: Re: Function naming
> Im doing some study on a perticular type of functions (in
my spare time
> just for fun). The problem is however that I dont know 
how
these
> functions are called in English. Could someone provide me
with the
correct
> english term and point me perhaps to some more info on this
topic.
> The functions are of the form;
> y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2
> This function has two input variables and has an order of
two. The number
> of variables and order can vary accordingly ofcourse. Is
there a general
> term for this kind of functions?
Id call it quadratic.
--
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: Function naming
>> Im doing some study on a particular type of function (in
my spare time
>> just for fun). The problem is however that I dont know
how these
>> functions are called in English. Could someone provide me
with the
>> correct English term and point me perhaps to some more
info on
>> this topic?
>> The functions are of the form;
>> y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2
>> This function has two input variables and has an order of
two.
>> The number of variables and order can vary accordingly of
course.
The functions are polynomial functions in several variables.
>> Is there a general term for these kind of functions?
> Id call it quadratic.
The example you gave is of a quadratic polynomial function,
whose graph is a quadric surface.
===
Subject: Re: prime numbers and 111111.....
> Hi there.
> When doing a math problem involving numbers with only 1s,
like 11 and
> 1111111 etc (which could be any base really), i noticed
that n digit 1s
> divide by m digit 1s if and only if m is a factor of n.
> But I guess this is reducable to prime numbers and is not
sagnificant.
> Martin Johansen
Similar to the previous respondent,
A number of repeatable digits, 1s, e.g. 11111 in any base b
is
representable as the sum of n terms of the geometric
progression
(b^n-1)/(b - 1) = b^0 + b^1 + b^2 + .. b^(n-1)
The polynomial Ôf(b)n defined as
f(b)n = b^0 + b^1 + b^2 + .. b^(n-1)
i.e. such that
f(b)n * (b-1) = (b^n-1)
is a cyclotomic polynomial, see Wolfram. It has some very
interesting
factor
properties. Most importantly, if n is composite, the
polynomial is
composite. Furthermore, it will only have factors of the form
2ln+1 for odd
n.
e.g. if n = 5, 11111 = 41 * 271, both are of the form 2*l*5 +
1, for 41 l
=
4, for 271 l = 27
You can see that a number x such as x=11111 is always such
that, when n=5,
base 10, 2n|(x-1) and you will get to your interesting
observation for some
n. f(b)n is one hell of an interesting polynomial.
Richard Miller
===
Subject: Re: third order stationarity behavior of random
process
>The first order stationarity behavior of random process is
that all
first
>order pdf/pmfs are time-independent;
>The second order stationarity behavior of random process is
that all
joint
>pdf/pmfs of any two samples are only dependent on the
difference of the
two
>sampling times, i.e., (t1-t2).
>>I dont think those are correct. At least 
theyre not
consistent with
>>the definitions Ive seen. For example, in Cox 
& Miller, The
Theory
>>of Stochastic Processes:
>> ... a process is stationary of order p if all moments up
to order p
have
>> the stationarity property.
>> It depends on how one defines moments. If one only uses
>> products of the random variables, it is woefully
inadequate.
>> Moments might not exist; this does not affect stationarity.
>Inadequate for what? Are you denying that this is the
definition
>of stationary of order p in Cox & Miller? Similarly in
Bartlett,
>An Introduction to Stochastic Processes, section 6.1:
> More generally, a process with finite moments will be called
stationary
> to the rth order if all the moments of order r or less are
invariant
> under translation.
>Or Grimmett and Stirzaker, Probability and Random Processes,
sec. 8.2:
> X = {X(t): t >= 0} is weakly (or second-order or
covariance) stationary
if
> E(X(t_1)) = E(X(t_2))
> and
> cov(X(t_1),X(t_2)) = cov(X(t_1+h),X(t_2+h))
> for all t_1, t_2 and h > 0.
The term weakly is very important here. Just as
uncorrelated is much weaker than independent, so is
covariance stationarity much weaker than stationary.
Also, for Gaussian processes, the two are equivalent
in both cases, again because joint normal
distributions are determined by their moments of order
1 and 2. It is this form of stationarity which is
needed for the Loeve-Karhunen representation.
Stationary of order k can be defined as above, or as
joint distributions at k or fewer time points are
preserved under time translation.
--
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: Symbolic solution of quadratic matrix equations
I need to find a symbolic solution of a quadratic matrix
equation
X^2 * A + X * B + A = 0
where A is the transpose of (kxk) nonsingular matrix A and 
B
is a
(kxk) symmetric matrix. Ive found some papers about numeric
solution
of such kind of matrix equation but I couldnt 
find anything
about the
symbolic solution. If somebody helps me, I will appreciate a
lot.
===
Subject: Re: Symbolic solution of quadratic matrix equations
>I need to find a symbolic solution of a quadratic matrix
equation
> X^2 * A + X * B + A = 0
>where A is the transpose of (kxk) nonsingular matrix A and
B is a
>(kxk) symmetric matrix. Ive found some papers about 
numeric
solution
>of such kind of matrix equation but I couldnt 
find anything
about the
>symbolic solution. If somebody helps me, I will appreciate a
lot.
This is not going to be easy, except in the case
where all the matrices commute. You can treat it
as a system of k^2 quadratic polynomials in the entries
of X, and use try Groebner basis techniques, but Id guess
its likely to be rather complicated unless k is very small.
A 2 x 2 example should work pretty well, but even 3 x 3
might be very ugly.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: Symbolic solution of quadratic matrix equations
>>I need to find a symbolic solution of a quadratic matrix
equation
>> X^2 * A + X * B + A = 0
>>where A is the transpose of (kxk) nonsingular matrix A 
and
B is a
>>(kxk) symmetric matrix.
>This is not going to be easy, except in the case
>where all the matrices commute. You can treat it
>as a system of k^2 quadratic polynomials in the entries
>of X, and use try Groebner basis techniques, but Id guess
>its likely to be rather complicated unless k is very 
small.
>A 2 x 2 example should work pretty well, but even 3 x 3
>might be very ugly.
Of course Robert is correct; this is going to be a mess.
But it looks like there IS some structure to the answer;
I dont have an explanation for it but I offer my 
findings
so someone else can explain the theory making this work.
I tried some 2x2 and 3x3 examples, talking to Maple like this:
X:=matrix(3,3,[z,y,x,w,v,u,t,s,r]);
A:=matrix(3,3,[seq(rand() mod 100, i=1..9)]);
B:=matrix(3,3,[seq(rand() mod 100, i=1..9)]);
for i to 3 do for j to i-1 do B[i,j]:=B[j,i]:od:od:print(B);
evalm(X&*X&*A + X&*B + transpose(A));
eq:={seq(seq(%[i,j],j=1..3),i=1..3)}; lprint(%);
and then to Magma like this:
Q:=RationalField();
P<z,y,x,w,v,u,t,s,r>:=PolynomialRing(Q,9);
I:=ideal<P| ... nine quadratic polynomials not shown ... >;
Groebner(I);
Write(SeeMe,I);
The 2x2 case is finished right away; the 3x3 case takes a
minute or two.
The 3x3 solution looks like this: each of z, y, x, ... can be
expressed
as a degree-19 polynomial in r with coefficients which are
ratios of
roughly 150-digit numbers. So there is one solution matrix X
for
each value of r. And there are 20 possible values of r,
namely the
roots of a polynomial of degree 20 with integer coefficients.
(In the 2x2 case, replace 20 by 6.)
The only surprise (to me) is that the polynomial always seems
to factor:
in the 2x2 case its the product of a quadratic and a 
quartic
with
coefficients in the 7- and 15-digit range, respectively.
In the 3x3 case theres a degree-8 factor (60-digit
coefficients)
and a degree-12 factor (90-digit coefficients). Moreover, 
these
polynomials are special: the quartic has a dihedral Galois
group,
and the two factors in the 3x3 case have galois groups of
order just 48.
So it appears there are two types of solutions to the original
quadratic equation, and that at least for these low degrees,
the different solutions can be obtained by some easy root
extractions.
I admit I dont understand why there is this much of a
pattern to
what should be a more random-looking polynomial.
dave
===
Subject: Re: Symbolic solution of quadratic matrix equations
In answer to Daves question below, I believe I can shed a
little more
light on this. Rewrite the matrix equation as
A.X^2 + B.X + C = 0 (1)
Let k and v be an eigenvalue and eigenvector of X so X.v = k.v
Then multiplying (1) by v gives
(A.k^2 + B.k + C).v = 0 (2)
so that
det(A.k^2 + B.k + C) = 0 (3)
This is then a polynomial equation of degree 2.n for the n
eigenvalues
of X. Now we take n solutions k_i of (3) we can then solve
(2) for
v_i.
Then X is given by the eigen decomposition formula
X = V.K.V^-1
where
K = diagonal matrix k_i
V = matrix of vectors v_i
Which n solutions of (3) should be chosen I dont know. 
Maybe
they all
give valid solutions, maybe only some of them do ...
>>I need to find a symbolic solution of a quadratic matrix
equation
>> X^2 * A + X * B + A = 0
>>where A is the transpose of (kxk) nonsingular matrix A 
and
B is a
>>(kxk) symmetric matrix.
>This is not going to be easy, except in the case
>where all the matrices commute. You can treat it
>as a system of k^2 quadratic polynomials in the entries
>of X, and use try Groebner basis techniques, but Id guess
>its likely to be rather complicated unless k is very 
small.
>A 2 x 2 example should work pretty well, but even 3 x 3
>might be very ugly.
> Of course Robert is correct; this is going to be a mess.
> But it looks like there IS some structure to the answer;
> I dont have an explanation for it but I offer my 
findings
> so someone else can explain the theory making this work.
> I tried some 2x2 and 3x3 examples, talking to Maple like
this:
> X:=matrix(3,3,[z,y,x,w,v,u,t,s,r]);
> A:=matrix(3,3,[seq(rand() mod 100, i=1..9)]);
> B:=matrix(3,3,[seq(rand() mod 100, i=1..9)]);
> for i to 3 do for j to i-1 do B[i,j]:=B[j,i]:od:od:print(B);
> evalm(X&*X&*A + X&*B + transpose(A));
> eq:={seq(seq(%[i,j],j=1..3),i=1..3)}; lprint(%);
> and then to Magma like this:
> Q:=RationalField();
> P<z,y,x,w,v,u,t,s,r>:=PolynomialRing(Q,9);
> I:=ideal<P| ... nine quadratic polynomials not shown ... >;
> Groebner(I);
> Write(SeeMe,I);
> The 2x2 case is finished right away; the 3x3 case takes a
minute or two.
> The 3x3 solution looks like this: each of z, y, x, ... can
be expressed
> as a degree-19 polynomial in r with coefficients which are
ratios of
> roughly 150-digit numbers. So there is one solution matrix
X for
> each value of r. And there are 20 possible values of r,
namely the
> roots of a polynomial of degree 20 with integer 
coefficients.
> (In the 2x2 case, replace 20 by 6.)
> The only surprise (to me) is that the polynomial always
seems to factor:
> in the 2x2 case its the product of a quadratic and a
quartic with
> coefficients in the 7- and 15-digit range, respectively.
> In the 3x3 case theres a degree-8 factor (60-digit
coefficients)
> and a degree-12 factor (90-digit coefficients). Moreover,
these
> polynomials are special: the quartic has a dihedral Galois
group,
> and the two factors in the 3x3 case have galois groups of
order just 48.
> So it appears there are two types of solutions to the
original
> quadratic equation, and that at least for these low degrees,
> the different solutions can be obtained by some easy root
extractions.
> I admit I dont understand why there is this much of a
pattern to
> what should be a more random-looking polynomial.
> dave
===
Subject: Re: basic covering space question
at 05:50 PM, nojb@fibertel.com.ar (Nicolas Ojeda Baer) said:
>Let X -> B, Y -> B be two covering maps. Then X x_B Y -> B
is a
>covering map. (X x_B Y is the fiber product of X and Y). I
know this
>is basic, but I cant seem to get it right.
Consider these questions:
1. What is the inverse image of a point in B under X x_B Y ->
B?
2. How do you define the topology of X x_B Y?
Then apply the definition of a covering map.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
<http://patriot.net/~shmuel>
Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply
to
domain Patriot dot net user shmuel+news to contact me. Do not
===
Subject: Re: In what sense does this vector converge to zero?
===
>Subject: In what sense does this vector converge to zero?
None; convergence is a property of sequences.
>Let M be an nxn symmetric positive semidefinite matrix.
Suppose v
>is an n-tuplet that satisfy 2 conditions (only) in the limit
that n
>goes to infinity:
> v v -> 1
> v M v -> 0.
The above has no meaning. You need to replace it with
statements about
sequences and their limits. Presumably you mean that {vn} and
{Mn} are
sequences such that vn and Mn are an n-tuplet and an n*n
symmetric
positive semidefinite matrix and
vn vn -> 1
vn Mn vn -> 0.
Similarly, you dont have a definition for w but 
rather a
sequence
{wn}, wn=Mn vn. In order to talk about limits youl need to
imbed all
of your vectors into a common vector space and specify what
topology
youre using on it, at which point you can ask whether {wn}
has a
limit.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
<http://patriot.net/~shmuel>
Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply
to
domain Patriot dot net user shmuel+news to contact me. Do not
===
Subject: Cardinality of an Infinite set, which is smaller than
Alef 0?
Without the Axiom of Choice, can anyone prove that Alef 0, is
the
smallest cardinality of an infinite set?
Please note: Assuming that for any pair of non empty sets, the
smaller set can be mapped one-to-one to a subset of the larger
set, is equivalent to the Axiom of Choice.
http://math.vanderbilt.edu/~schectex/ccc/choice.html
===
Subject: Re: Cardinality of an Infinite set, which is smaller
than Alef 0?
>Without the Axiom of Choice, can anyone prove that Alef 0,
is the
>smallest cardinality of an infinite set?
No; is is the only smallest cardinality of an infinite
set, in every other infinite cardinality must have a
smaller cardinality. But there can be incomparable
infinite cardinalities.
>Please note: Assuming that for any pair of non empty sets,
the
>smaller set can be mapped one-to-one to a subset of the
larger
>set, is equivalent to the Axiom of Choice.
That is part of the definition. What is equivalent ot
the Axiom of Choice is that of two sets, on is larger
than the other.
--
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: Cardinality of an Infinite set, which is smaller
than Alef 0?
> Without the Axiom of Choice, can anyone prove that Alef 0,
is the
> smallest cardinality of an infinite set?
It depends what you mean.
On the one hand if a set A has cardinality less than or equal
to
aleph_0, then by definition there is an injection f:A ->
omega. This
means A is wellorderable. If it is also infinite it must have
cardinality aleph_0.
On the other hand without AC there may be sets which are not
comparable
with aleph_0 in cardinality. So aleph_0 is minimal among
infinite
cardinals but not necessarily the minimum without AC. It is
the minimum
among the cardinals of infinite wellorderable sets.
===
Subject: Re: Cardinality of an Infinite set, which is smaller
than Alef 0?
>Without the Axiom of Choice, can anyone prove that Alef 0,
is the
>smallest cardinality of an infinite set?
One does not need AC to prove the following: Any subset of a
countably
infinite set is countable.
However, without the axiom of choice, one cannot prove that
an infinite
set necessarily has a countably infinite subset. A set is
Dedekind
infinite if and only if it has a countably 
infinite subset, and
it
consistent with ZF+not AC that there exist infinite Dedekind
finite sets.
Thus, countably infinite is a minimal infinite 
cardinality, but
without
AC it is not necessarily the only one.
Note that aleph0 has a very precise definition to many
authors: It is
the set of all finite ordinals; the axiom of 
infinity
guarantees its
existence. Thus, one does not need the axiom of choice to
show that if
an ordinal (or cardinal) a < aleph0, then a is finite.
--
Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Re: Cardinality of an Infinite set, which is smaller
than Alef 0?
>Without the Axiom of Choice, can anyone prove that Alef 0,
is the
>smallest cardinality of an infinite set?
> One does not need AC to prove the following: Any subset of
a countably
> infinite set is countable.
> However, without the axiom of choice, one cannot prove that
an infinite
> set necessarily has a countably infinite subset. A set is
Dedekind
> infinite if and only if it has a countably 
infinite subset,
and it
> consistent with ZF+not AC that there exist infinite Dedekind
finite
sets.
I have always been massively confused by this, and I figure
this is as good
a time to ask as any, since it has been explicitly stated
here. The obvious
question is, why cant I just pick a countable series of
elements from the
infinite set? What is it thats preventing me 
from doing that?
Let A be the infinite set, then we know that there exists a
member x of the
set. Put that x into a subset S. Then we know that there
exists a y in A
such that y != x, else it would be finite. Put that into a
subset S.
Continuing, finding, for instance, a z such that z != x, y,
you have a
countable set.
Is it the statements we know that there exists a member ...
of the set
A
that prevent this construction?
> Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Re: Cardinality of an Infinite set, which is smaller
than Alef 0?
>Without the Axiom of Choice, can anyone prove that Alef 0,
is the
>smallest cardinality of an infinite set?
>
> One does not need AC to prove the following: Any subset of a
countably
> infinite set is countable.
> However, without the axiom of choice, one cannot prove that
an infinite
> set necessarily has a countably infinite subset. A set is
Dedekind
> infinite if and only if it has a countably 
infinite subset,
and it
> consistent with ZF+not AC that there exist infinite Dedekind
finite
> sets.
> I have always been massively confused by this, and I figure
this is as
good
> a time to ask as any, since it has been explicitly stated
here. The
obvious
> question is, why cant I just pick a countable series of
elements from
the
> infinite set? What is it thats preventing me 
from doing
that?
> Let A be the infinite set, then we know that there exists a
member x of
the
> set. Put that x into a subset S. Then we know that there
exists a y in A
> such that y != x, else it would be finite. Put that into a
subset S.
> Continuing, finding, for instance, a z such that z != x, y,
you have a
> countable set.
> Is it the statements we know that there exists a member ...
of the set
A
> that prevent this construction?
> Stephen J. Herschkorn sjherschko@netscape.net
shows that this construction is impossible without the axiom
of choice?
===
Subject: Re: Cardinality of an Infinite set, which is smaller
than Alef 0?
>>Without the Axiom of Choice, can anyone prove that Alef 0,
is the
>>smallest cardinality of an infinite set?
.....................
>> Let A be the infinite set, then we know that there exists a
member x of
>the
>> set. Put that x into a subset S. Then we know that there
exists a y in A
>> such that y != x, else it would be finite. Put that into a
subset S.
>> Continuing, finding, for instance, a z such that z != x, y,
you have a
>> countable set.
>> Is it the statements we know that there exists a member
... of the set
A
>> that prevent this construction?
>shows that this construction is impossible without the axiom
of choice?
It at most requires the countable axiom of choice, and I
believe it is weaker than that.
--
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: Cardinality of an Infinite set, which is smaller
than Alef 0?
|shows that this construction is impossible without the axiom
of choice?
When Paul Cohen proved that if ZF is consistent, then so is
ZF+the axiom
of choice is false, along with that he also proved that if ZF
is
consistent
then so is ZF+there exists a set S which is infinite but has
no countably
infinite subset. The proof is in his famous book if 
youre
curious.
Here the definition of infinite is that S is 
infinite if there
is a
family F
of
subsets of S that contains the empty set, and such that if X
is in F and
s is in S, then the union of X with {s} is in F too. This
corresponds to
the
usual definition of finite which is most closely 
related to the
principle
of mathematical induction on the natural numbers. Sometimes
authors
give a different definition of infinite which is 
called Dedekind
infinite
which is equivalent to containing a countably infinite subset
(S is called
Dedekind infinite if there exists a 1-1 correspondence between
S and
a proper subset of S.) Assuming the axiom of choice the two
definitions
are equivalent. While not assuming the axiom of choice I
believe the
first definition I gave is preferred.
Keith Ramsay
===
Subject: Re: Cardinality of an Infinite set, which is smaller
than Alef 0?
>>
>However, without the axiom of choice, one cannot prove that
an infinite
>set necessarily has a countably infinite subset. A set is
Dedekind
>infinite if and only if it has a countably 
infinite subset,
and it
>consistent with ZF+not AC that there exist infinite Dedekind
finite
>
>>sets.
>>
>>I have always been massively confused by this, and I figure
this is as
>>
>good
>>a time to ask as any, since it has been explicitly stated
here. The
>>
>obvious
>>question is, why cant I just pick a countable series of
elements from
the
>>infinite set? What is it thats preventing me 
from doing
that?
>>
>shows that this construction is impossible without the axiom
of choice?
start, though an explicit model of ZF where there exists a
Dedekind set
is not given there.
--
Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Re: Cardinality of an Infinite set, which is smaller
than Alef 0?
>>However, without the axiom of choice, one cannot prove that
an infinite
>>set necessarily has a countably infinite subset. A set is
Dedekind
>>infinite if and only if it has a countably 
infinite subset,
and it
>>consistent with ZF+not AC that there exist infinite Dedekind
finite
>>
>sets.
>I have always been massively confused by this, and I figure
this is as
good
>a time to ask as any, since it has been explicitly stated
here. The
obvious
>question is, why cant I just pick a countable series of
elements from the
>infinite set? What is it thats preventing me 
from doing that?
>Let A be the infinite set, then we know that there exists a
member x of
the
>set. Put that x into a subset S. Then we know that there
exists a y in A
>such that y != x, else it would be finite. Put that into a
subset S.
>Continuing, finding, for instance, a z such that z != x, y,
you have a
>countable set.
>Is it the statements we know that there exists a member ...
of the set
A
>that prevent this construction?
Logical conjunction is a binary operation, so you can use
induction to
form a subset of any arbitrary *finite* size. Without the
axiom of
choice, a problem arises here because you want to conjunct
infinitely
many times.
--
Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Re: Cardinality of an Infinite set, which is smaller
than Alef 0?
...
> However, without the axiom of choice, one cannot prove that
an
infinite
> set necessarily has a countably infinite subset. A set is
Dedekind
> infinite if and only if it has a countably 
infinite subset,
and it
> consistent with ZF+not AC that there exist infinite Dedekind
finite
> sets.
> I have always been massively confused by this, and I figure
this is as
good
> a time to ask as any, since it has been explicitly stated
here. The
obvious
> question is, why cant I just pick a countable series of
elements from
the
> infinite set? What is it thats preventing me 
from doing
that?
My gut feeling is that it is the axiom of choice that allows
you to do
it. If there is no axiom of choice you can not pick such a
countable
series of elements.
--
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: Cardinality of an Infinite set, which is smaller
than Alef 0?
>Without the Axiom of Choice, can anyone prove that Alef 0,
is the
>smallest cardinality of an infinite set?
>
> One does not need AC to prove the following: Any subset of a
countably
> infinite set is countable.
> However, without the axiom of choice, one cannot prove that
an infinite
> set necessarily has a countably infinite subset. A set is
Dedekind
> infinite if and only if it has a countably 
infinite subset,
and it
> consistent with ZF+not AC that there exist infinite Dedekind
finite
> sets.
> I have always been massively confused by this, and I figure
this is as
good
> a time to ask as any, since it has been explicitly stated
here. The
obvious
> question is, why cant I just pick a countable series of
elements from
the
> infinite set? What is it thats preventing me 
from doing
that?
> Let A be the infinite set, then we know that there exists a
member x of
the
> set. Put that x into a subset S. Then we know that there
exists a y in A
> such that y != x, else it would be finite. Put that into a
subset S.
> Continuing, finding, for instance, a z such that z != x, y,
you have a
> countable set.
> Is it the statements we know that there exists a member ...
of the set
A
> that prevent this construction?
> Stephen J. Herschkorn sjherschko@netscape.net
The problem is that to choose these countably many elements,
you use
at least the countable AC. In various (topos) models of set
theory,
you can even have infinite (that is non-finite) 
subsets of finite
sets. In fact, there are a number of distinct definitions of
finite.
A regular morass, that disappears under AC.
===
Subject: Re: Cardinality of an Infinite set, which is smaller
than Alef 0?
>Without the Axiom of Choice, can anyone prove that Alef 0,
is the
>smallest cardinality of an infinite set?
>
> One does not need AC to prove the following: Any subset of a
countably
> infinite set is countable.
> However, without the axiom of choice, one cannot prove that
an infinite
> set necessarily has a countably infinite subset. A set is
Dedekind
> infinite if and only if it has a countably 
infinite subset,
and it
> consistent with ZF+not AC that there exist infinite Dedekind
finite
> sets.
> I have always been massively confused by this, and I figure
this is as
good
> a time to ask as any, since it has been explicitly stated
here. The
obvious
> question is, why cant I just pick a countable series of
elements from
the
> infinite set? What is it thats preventing me 
from doing
that?
> Let A be the infinite set, then we know that there exists a
member x of
the
> set. Put that x into a subset S. Then we know that there
exists a y in A
> such that y != x, else it would be finite. Put that into a
subset S.
> Continuing, finding, for instance, a z such that z != x, y,
you have a
> countable set.
> Is it the statements we know that there exists a member ...
of the set
A
> that prevent this construction?
> Stephen J. Herschkorn sjherschko@netscape.net
Youre making infinitely many choices. 
Basically, you are
assuming a
choice function for the nonempty subsets of A.
===
Subject: Re: Cardinality of an Infinite set, which is smaller
than Alef 0?
>>Without the Axiom of Choice, can anyone prove that Alef 0,
is the
>>smallest cardinality of an infinite set?
>> One does not need AC to prove the following: Any subset of
a countably
>> infinite set is countable.
>> However, without the axiom of choice, one cannot prove
that an infinite
>> set necessarily has a countably infinite subset. A set is
Dedekind
>> infinite if and only if it has a countably 
infinite subset,
and it
>> consistent with ZF+not AC that there exist infinite
Dedekind finite
>sets.
>I have always been massively confused by this, and I figure
this is as
good
>a time to ask as any, since it has been explicitly stated
here. The
obvious
>question is, why cant I just pick a countable series of
elements from the
>infinite set? What is it thats preventing me 
from doing that?
>Let A be the infinite set, then we know that there exists a
member x of
the
>set. Put that x into a subset S. Then we know that there
exists a y in A
>such that y != x, else it would be finite. Put that into a
subset S.
>Continuing, finding, for instance, a z such that z != x, y,
you have a
>countable set.
So how do you keep choosing the elements? You can choose a
finite number of them, but ... .
>Is it the statements we know that there exists a member ...
of the set
A
>that prevent this construction?
>> Stephen J. Herschkorn sjherschko@netscape.net
--
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: Determinants and commuting matrices (old Hoffman and
Kunze
Exercise)
The last exercise in section 5.4 of Hoffman and Kunzes
Linear Algebra
asks the reader to prove that, given 4 commuting n x n
matrices A, B,
C, and D, the determinant of the 2n x 2n matrix
A B
C D
is given by det(AD - BC).
Im feeling dense, but I dont see why this 
should be. Any
insight
would be appreciated...
===
Subject: Re: Determinants and commuting matrices (old Hoffman
and Kunze
Exercise)
> The last exercise in section 5.4 of Hoffman and Kunzes
Linear Algebra
> asks the reader to prove that, given 4 commuting n x n
matrices A, B, C,
> and D, the determinant of the 2n x 2n matrix
> A B
> C D
> is given by det(AD - BC).
> Im feeling dense, but I dont see why this 
should be. Any
insight would
> be appreciated...
If they all commute (including AD-BC), then they have the
same invariant
subspaces. Try to find a basis where the matrix [ A, B; C, D ]
becomes
block-diagonal. The determinant of a block diagonal matrix is
the product
of determinants of each block.
Hope this helps.
Igor
===
Subject: Re: Determinants and commuting matrices (old Hoffman
and Kunze
Exercise)
posting-account=
y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g
Perhaps it can be done a bit more directly.
det ( A, B ; C, D ) = det(A)*det(C)*det(I,A^-1*B; I, C^-1*D)
= det (A)*det(C)*det(C^-1*D - A^-1*B)
= det(A) * det( D - A^-1 * B * C )
= det(AD - BC)
det-ails of why commutativity justifies these steps left to
the reader.
===
Subject: Re: Determinants and commuting matrices (old Hoffman
and Kunze
Exercise)
> Perhaps it can be done a bit more directly.
> det ( A, B ; C, D ) = det(A)*det(C)*det(I,A^-1*B; I, C^-1*D)
I dont see how whats below follows from 
whats above. How
do you reduce
the determinant of a 2nx2n matrix to the determinant of an
nxn matirx?
You also get into trouble if A or C is not invertible.
Igor
> = det (A)*det(C)*det(C^-1*D - A^-1*B)
> = det(A) * det( D - A^-1 * B * C )
> = det(AD - BC)
> det-ails of why commutativity justifies these steps left to
the reader.
===
Subject: Re: Determinants and commuting matrices (old Hoffman
and Kunze
Exercise)
posting-account=
y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g
Apologies if this turns into a double-post; Google Groups 2
beta
kicked back the first attempt.
det( I, M; I, N) = det( I, M; 0, N-M)
Subtract the upper rows from the lower ones.
To generalize slightly what you pointed out, determinant of a
block
triangular matrix is the product of the determinants of the
diagonal
blocks.
Yes, you get in trouble if A or C is not invertible, but this
can be
handled by a generic perturbation argument. Add a small
multiple of
the identity matrix, sufficient to avoid the singularity, then
take the
limit as the multiple tends to zero.
===
Subject: Re: Determinants and commuting matrices (old Hoffman
and Kunze
Exercise)
posting-account=
y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g
I M
I N
Subtract the upper rows from the lower ones.
Determinant of a block triangular matrix is ....
Yes, you get in trouble if A or C is not invertible, but this
can be
handled by a generic perturbation argument. Add a small
multiple of
the identity matrix, sufficient to avoid the singularity, then
take the
limit as the multiple tends to zero.
===
Subject: Re: Determinants and commuting matrices (old Hoffman
and Kunze
Exercise)
> The last exercise in section 5.4 of Hoffman and Kunzes
Linear Algebra
> asks the reader to prove that, given 4 commuting n x n
matrices A, B,
> C, and D, the determinant of the 2n x 2n matrix
> A B
> C D
> is given by det(AD - BC).
> Im feeling dense, but I dont see why this 
should be.
Any insight would be appreciated...
If you had a block matrix of the form
| P Q | =M
| Z R |
where Z is the appropriate sized
zero matrix, then any term of the det of
M expanded about the 1st column of P
must contain a zero term if it contains
an element or elements of Q.
So DetM comprises priducts of the form
product n elements of P times product n elements
of R i.e. Det(P)Det(R) = Det(PR).
If you multiply
|A B| by | I -(A^(-1)*B |
|C D| | Z I |
The determinant of the 2nx2n matrix is
unaltered.
The product equals
| A Z |
| C D - CA^(-1)B|
So Det(M) = det(A)Det(D - CA^(-1)B
= Det(AD -AC A^(-1)B)
= Det(AD -CB) for matrices that commute.
This assumes that A is invertible but if
both A and D were not invertible Det(M)=0
I dont know if this would be the approach
envisaged for the excercise
===
Subject: Re: Determinants and commuting matrices (old Hoffman
and Kunze
Exercise)
posting-account=
y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g
> This assumes that A is invertible but if
> both A and D were not invertible Det(M)=0
Consider A = D = 0, B = C = I.
I prefer a perturbation argument to overcome singularity of
(for
example) A.
===
Subject: Re: Determinants and commuting matrices (old Hoffman
and Kunze
Exercise)
> This assumes that A is invertible but if
> both A and D were not invertible Det(M)=0
> Consider A = D = 0, B = C = I.
Yes, but you are still assuming that
the matrices on a diagonal are invertible.
You would then swap columns or choose a
different multiplying matrix.
In any case, it seems that you would have to assume
that at least one matrix was non-singualar.
> I prefer a perturbation argument to overcome
singularity of (for
> example) A.
Yes, but I suppose you are going beyond
elementary methods.
I am not familiar with the textbook
from which the excercise was taken.
Knowing what topics had been covered
might suggest how the problem should be
solved using commutivity alone or with
a peturbation argument if that had been covvered.
===
Subject: Re: Determinants and commuting matrices (old Hoffman
and Kunze
Exercise)
> This assumes that A is invertible but if
> both A and D were not invertible Det(M)=0
> Consider A = D = 0, B = C = I.
Yes, but you are still assuming that
the matrices on a diagonal are invertible.
You would then swap columns or choose a
different multiplying matrix.
In any case, it seems that you would have to assume
that at least one matrix was non-singualar.
> I prefer a perturbation argument to overcome
singularity of (for
> example) A.
Yes, but I suppose you are going beyond
elementary methods.
I am not familiar with the textbook
from which the excercise was taken.
Knowing what topics had been covered
might suggest how the problem should be
solved using commutivity alone or with
a peturbation argument if that had been covvered.
===
Subject: Re: Determinants and commuting matrices (old Hoffman
and Kunze
Exercise)
posting-account=JrQSJw0AAAA6UtXIbhl36iP_cLZ-hNeT
Just to clarify what tools a reader of the textbook might
reasonably
bring to bear on the exercise, perturbation arguments and even
invariant subspaces have not yet been introduced. Topics
covered thus
far include elementary row operations, vector spaces, bases,
and the
idea of dimension, linear transformations and their
representation by
matrices, dual spaces and dual bases, elementary polynomial
algebra
(ideals of polynomials and some simple results depending
therefrom),
and determinants of matrices. No canonical forms of any sort
have been
discussed, though similarity insofar as it relates to changes
of bases
has been covered.
My suspicion, after reading through this thread, is that the
solution
theyre looking for involves something along the lines of 
first
assuming that both A and D are invertible, proving the result
based on
something similar I.M. Davidsons line of reasoning, and 
then
showing
that it still holds even if A and/or D are both singular.
===
Subject: Re: Determinants and commuting matrices (old Hoffman
and Kunze
Exercise)
>> This assumes that A is invertible but if
>> both A and D were not invertible Det(M)=0
>> Consider A = D = 0, B = C = I.
>Yes, but you are still assuming that
>the matrices on a diagonal are invertible.
>You would then swap columns or choose a
>different multiplying matrix.
>In any case, it seems that you would have to assume
>that at least one matrix was non-singualar.
>> I prefer a perturbation argument to overcome
>singularity of (for
>> example) A.
>Yes, but I suppose you are going beyond
>elementary methods.
Not really. The determinant of a matrix being a
polynomial in its elements is continuous. As the
determinant of A+tI is a polynomial in t, it must
be non-singular for all non-zero t which are
sufficiently small, and hence A+tI is non-singular.
Now pass to the limit as t -> 0.
--
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: Is x^n - x^{n - 1} - ... - x - 1 irreducible?
>For n = 1, 2, ... let f_n(x) = x^n - x^{n - 1} - ... - x - 1
>= (x^{n + 1} - 2x^n + 1)/(x - 1). Is f_n always irreducible
>over Z?
Leafing further through my sheaf of photocopies 
(Ive a bad
habit of collecting stuff and not reading it), I found that
this has been posed and solved as an American Mathematical
Monthly problem:
Elementary Problem E 3008 (Vol. 90, No. 7, Aug. - Sep., 1983)
<http://links.jstor.org/sici?sici=0002-9890%28198308%2F09%
2990%3A7%3C482%3AE
PE%3E2.0.CO%3B2-B>
Solution of Elementary Problem E 3008 (Vol. 96, No. 2, Feb.,
1989)
<http://links.jstor.org/sici?sici=0002-9890%28198902%2996%
3A2%3C155%3AE%3E2.
0.CO%3B2-L>
Editorial comment. M. J. DeLeon noted that the result
of the problem is contained in the following theorem of
Alfred Brauer: if a_1, a_2, ..., a_n are integers with
a_1 >= a_2 >= ... > = a_n > 0, then the polynomial
x^n - a_1x^{n - 1} - a_2x^{n - 2} - ... - a_n is
irreducible over the irrationals. Cf. Alfred Brauer,
On algebraic equations with all but one root in the
interior of the unit circle, Math. Nachrichten, 4
(1950/51), 250--257.
The positive root of the polynomial considered in this
problem (or more generally of the polynomial considered
in Brauers Theorem) is an example of what is known as a
PV-number, which is a real algebraic integer greater than
1 all of whose conjugates lie inside the unit circle. Cf.
J. W. S. Cassells, _An Introduction to Diophantine Approx-
imation_, Cambridge University Press, 1957, Chapter VIII.
--
Angus Rodgers
Contains mild peril
===
Subject: Re: Is x^n - x^{n - 1} - ... - x - 1 irreducible?
> is irreducible over the irrationals.
a misprint, right?
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Is x^n - x^{n - 1} - ... - x - 1 irreducible?
>> is irreducible over the irrationals.
>a misprint, right?
I have discovered a whole new field of mathematics,
disproving the ravings of those upstarts Galois and
Cantor. Bow down and worship me! Woe to those who
dare to suggest that my Word is not Law! Death to
the infidel Edgar!
--
Angus Rodgers
Contains mild peril
===
Subject: Re: Is x^n - x^{n - 1} - ... - x - 1 irreducible?
>> For n = 1, 2, ... let f_n(x) = x^n - x^{n - 1} - ... - x -
1
>> = (x^{n + 1} - 2x^n + 1)/(x - 1). Is f_n always
irreducible over Z?
> Before anyone wastes too much time on this: the answer is
yes
> (although I dont know why).
> Ernst S. Selmer, On the irreducibility of certain
trinomials,
> Math. Scand. 4 (1956), 287--302.
> I actually had a photocopy of this paper lying around
(because
> of a question on irreducibility that came up in sci.math a
few
> years back), but hadnt got around to reading it.
> An important criterion, typical of one approach to the
problem,
> is given by Perron [8]: The polynomial (with integer
coefficients)
> x^n + a_1x^{n - 1} + a_2x^{n - 2} + ... + a_{n - 1}x + a_n
> is irreducible if
> |a_1| > 1 + |a_2| + |a_3| + ... + |a_n|.
> Applied to f(x) = x^n + ax +- 1, where we substitute x =
1/z,
> this shows that f(x) is irreducible for |a| >= 3. When |a| =
> 2 and f(+-1) [is not equal to] 0, we can still conclude
> irreducibility according to Perron. When |a| = 2 and f(x)
has
> a rational factor x + 1 or x - 1, the second factor of f(x)
> will be irreducible (this is not contained among Perrons
> statements, but follows easily from his method).
> I dont have access to Perrons paper, but 
the reference is:
> 8. O. Perron, Neue Kriterien fuer die Irreduzibilitat
> algebraischer Gleichungen, J. reine angew. Math. 132 (1907),
> 288--307.
This was discussed on sci.math.research on 1998/03/28, see
[1].
For an extensive treatment of irreducibility criteria see
Filasetas online book [2].
--Bill Dubuque
[2]
http://www.math.sc.edu/~filaseta/gradcourses/Math788F/
latexbook/
===
Subject: Re: Is x^n - x^{n - 1} - ... - x - 1 irreducible?
<posted & mailed> For an extensive treatment of
irreducibility criteria see
> Filasetas online book [2].
> [2]
http://www.math.sc.edu/~filaseta/gradcourses/Math788F/
latexbook/
I wasnt able to access this.
--
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: Is x^n - x^{n - 1} - ... - x - 1 irreducible?
>> For an extensive treatment of irreducibility criteria see
>> Filasetas online book [2].
>> [2]
http://www.math.sc.edu/~filaseta/gradcourses/Math788F/
latexbook/
> I wasnt able to access this.
In the past it was accessible without a password,
so perhaps if you email Filaseta hell tell you
how to access it.
--Bill Dubuque
===
Subject: Re: Is x^n - x^{n - 1} - ... - x - 1 irreducible?
> Ernst S. Selmer, On the irreducibility of certain
trinomials,
> Math. Scand. 4 (1956), 287--302.
> An important criterion, typical of one approach to the
problem,
> is given by Perron [8]: The polynomial (with integer
coefficients)
> x^n + a_1x^{n - 1} + a_2x^{n - 2} + ... + a_{n - 1}x + a_n
> is irreducible if
> |a_1| > 1 + |a_2| + |a_3| + ... + |a_n|.
>[...] I dont have access to Perrons paper 
[...]
... but of course theres Google:
<http://mathforum.org/epigone/sci.math.research/khilnußang/
6fo20m$6mh$1@nnr
p1.dejanews.com>
(Both references note the necessity of the additional
condition
that a_n is nonzero.)
--
Angus Rodgers
Contains mild peril
===
Subject: JSH: Being me
I know, many of you probably think its horrible being me,
with all
these people calling me name, putting up nasty webpages, and
spending
so much time talking bad about me.
Some of you are in your own little world and maybe still
think Im
wrong, while some of you realize that Im right and STILL
wouldnt
want to be me considering how much opposition my results have
faced
and are likely to face.
But hey, its actually fun!
Like I havent just done arguing with people on math or even
just math
research as I have an open source project on SourceForge, and
I have
made some friends (believe it or not) in a few Internet
communities.
I *thought* I could use the Internet in a groundbreaking way
to
introduce major results but here I am WAITING ON A JOURNAL,
when I
said in the past that I wouldnt even use journals!
:-)
Live and learn.
In any event, I also get to go over my own mathematical work,
and
enjoy talking about it as you might have noticed with me
starting new
threads--yet again--going over the arguments in different
ways.
Different looks. Perspective.
But, on to something practical, as Ive been looking at
BitTorrent
clients while Im doing downloads, and I see the download
speeds
hopping all over the place as the clients--I guess--use
various
algorithms to try and figure out whats the best 
connection to
make.
Anyone here know anything about those algorithms? Im
thinking maybe
I might make my own BitTorrent client, or go in and fiddle
with the
algorithms on the one I have as I have Azureus. But its an
idle
thought, so why not toss it on sci.math?
Any of you know anything about the algorithms being used in
BitTorrent
clients?
James Harris
===
Subject: Re: JSH: Being me
> I know, many of you probably think its horrible being me,
with all
> these people calling me name, putting up nasty webpages,
and spending
> so much time talking bad about me.
I have to be honest, if being you is horrible, those are not
the causes
I would have listed for it. You receive very little
name-calling, the
websites mainly say you are obstinately wrong. Ive seen
people treated
far worse in the public schools. Youve got nothing on some
of the
politicians out there.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: JSH: Being me
>I know, many of you probably think its horrible being me,
with all
>these people calling me name, putting up nasty webpages, and
spending
>so much time talking bad about me.
Why in the world would you think anyone cares?
And just out of curiosity, who are you adressing here? This
happens every once in a while, you seem to be talking to all
your other readers, the ones who are not part of the vast
conspiracy. What makes you think that they exist?
>Some of you are in your own little world and maybe still
think Im
>wrong, while some of you realize that Im right and STILL
wouldnt
>want to be me considering how much opposition my results
have faced
>and are likely to face.
Like, what gives you the idea that there are people out there
who think youre right? Nobody ever _says_ they think 
youre
right.
>But hey, its actually fun!
>Like I havent just done arguing with people on math or 
even
just math
>research as I have an open source project on SourceForge,
and I have
>made some friends (believe it or not) in a few Internet
communities.
That _is_ hard to believe. Why do you keep coming back here?
>I *thought* I could use the Internet in a groundbreaking way
to
>introduce major results but here I am WAITING ON A JOURNAL,
when I
>said in the past that I wouldnt even use journals!
Youve also said in the past that the internet 
didnt matter,
all that mattered was the fact that you were about to be
published
in a journal.
Given just about anything youve ever said, 
youve said
something
directly contraditory in the past.
************************
David C. Ullrich
===
Subject: Re: JSH: Being me
> Like, what gives you the idea that there are people out
there
> who think youre right? Nobody ever _says_ they think 
youre
> right.
Wasnt there that guy from the genius club? What ever
happened to him?
===
Subject: Re: JSH: Being me
<icckq09v5fdmh2oumkv2smk8qbvf24rhkd@4ax.com>
<Ypsqd.102306$SW3.41170@fed1read01>
posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK
> Like, what gives you the idea that there are people out
there
> who think youre right? Nobody ever _says_ they think 
youre
> right.
> Wasnt there that guy from the genius club? What ever
happened to
him?
I dont recall Quinn Tyler-Jackson ever saying Harris was
right,
he objected to how Harris was treated.
I dont recall A. Beckwith ever saying Harris was right,
he just tried to steal the credit for his paper.
===
Subject: Re: JSH: Being me
Discussion, linux)
> Like, what gives you the idea that there are people out
there
> who think youre right? Nobody ever _says_ they think 
youre
> right.
Just think.
Sci.math has *lots* of readers. What are the odds that
theyre *all*
wrong about Jamess work?
--
So I speak before a crowd of the damned, cursed to be unloved
throughout time, with only their hatred and bile to comfort
them now,
having betrayed what should have been their one true lover:
Mathematics. -- James Harris reaches a bit
===
Subject: Re: JSH: Being me
> Like, what gives you the idea that there are people out
there
> who think youre right? Nobody ever _says_ they think 
youre
> right.
> Just think.
> Sci.math has *lots* of readers. What are the odds that
theyre *all*
> wrong about Jamess work?
Well, I think sci.math readers are stupid.
Besides, I do remind you that Im just here 
goofing off, while
I wait.
It makes me no never mind to play with you people and see
just what I
can see.
I trace out your neural pathways this way.
Like, tomorrow I will come to see who replies to this post,
and read
information bounced around in their heads.
I do this enough that I can build a map, and a model, then I
simply
test the model.
If it fails to respond as you would, then I test again, until
it
responds as you do.
So I build a world and then I can simply test that world,
moving the
pieces in it as I see fit. And then I know how you will move,
as I
see fit.
I am mostly done.
James Harris
===
Subject: Re: JSH: Being me
> Well, I think sci.math readers are stupid.
> Besides, I do remind you that Im just here 
goofing off,
while I wait.
> It makes me no never mind to play with you people and see
just what I
> can see.
> I trace out your neural pathways this way.
> Like, tomorrow I will come to see who replies to this post,
and read
> information bounced around in their heads.
> I do this enough that I can build a map, and a model, then
I simply
> test the model.
> If it fails to respond as you would, then I test again,
until it
> responds as you do.
> So I build a world and then I can simply test that world,
moving the
> pieces in it as I see fit. And then I know how you will
move, as I
> see fit.
> I am mostly done.
> James Harris
It appears you have managed to elevate your delusions of
grandeur to
delusions of divinity. Keep it
up. There are special places for you -- even here on earth!
--
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: JSH: Being me
>
> Like, what gives you the idea that there are people out
there
> who think youre right? Nobody ever _says_ they think 
youre
> right.
>
> Just think.
>
> Sci.math has *lots* of readers. What are the odds that
theyre *all*
> wrong about Jamess work?
> Well, I think sci.math readers are stupid.
You read sci.math, right?
===
Subject: Re: JSH: Being me
> I know, many of you probably think its horrible being me,
with all
> these people calling me name, putting up nasty webpages,
and spending
> so much time talking bad about me.
The unswerving self-belief (in the face of a mountain of
contradictory
evidence) must be some comfort, I imagine.
> Some of you are in your own little world
Project much?
> and maybe still think Im
> wrong, while some of you realize that Im right and STILL
wouldnt
> want to be me considering how much opposition my results
have faced
> and are likely to face.
> But hey, its actually fun!
Must have an almost narcotic effect, that level of
self-belief.
> Like I havent just done arguing with people on math or
even just math
> research as I have an open source project on SourceForge,
and I have
> made some friends (believe it or not) in a few Internet
communities.
You should be careful making friends on the Internet James,
theres a
lot of weirdos out there.
> I *thought* I could use the Internet in a groundbreaking
way to
> introduce major results but here I am WAITING ON A JOURNAL,
when I
> said in the past that I wouldnt even use journals!
Youll be waiting a while, Id wager.
> :-)
> Live and learn.
If only you would!
> In any event, I also get to go over my own mathematical
work, and
> enjoy talking about it as you might have noticed with me
starting new
> threads--yet again--going over the arguments in different
ways.
You? Starting new threds? Cant say Id 
spotted any, no...
> Different looks. Perspective.
> But, on to something practical, as Ive been looking at
BitTorrent
> clients while Im doing downloads, and I see the download
speeds
> hopping all over the place as the clients--I guess--use
various
> algorithms to try and figure out whats the 
best connection
to make.
> Anyone here know anything about those algorithms? Im
thinking maybe
> I might make my own BitTorrent client, or go in and fiddle
with the
> algorithms on the one I have as I have Azureus. But its 
an
idle
> thought, so why not toss it on sci.math?
> Any of you know anything about the algorithms being used in
BitTorrent
> clients?
You have Internet access, learn how to use it. Oh wait, I
forgot,
youre pathologically incapacle of learning.
--
Larry Lard
Replies to group please
===
Subject: Re: JSH: Being me
> I know, many of you probably think its horrible being me,
> But hey, its actually fun!
Would you describe it as . . . bliss?
===
Subject: Re: JSH: Being me
> But, on to something practical, as Ive been looking at
BitTorrent
> clients while Im doing downloads, and I see the download
speeds
> hopping all over the place as the clients--I guess--use
various
> algorithms to try and figure out whats the 
best connection
to make.
> Anyone here know anything about those algorithms? Im
thinking maybe
> I might make my own BitTorrent client, or go in and fiddle
with the
> algorithms on the one I have as I have Azureus. But its 
an
idle
> thought, so why not toss it on sci.math?
> Any of you know anything about the algorithms being used in
BitTorrent
> clients?
Sure. What do you want to know? Its a much more naive
algorithm
than you seem to think it is.
http://www.bittorrent.com/protocol.html
Ôcid Ôooh
===
Subject: Re: JSH: Being me
jstevh@msn.com says...
> I know, many of you probably think its horrible being me,
with all
> these people calling me name, putting up nasty webpages,
and spending
> so much time talking bad about me.
> Some of you are in your own little world and maybe still
think Im
> wrong, while some of you realize that Im right and STILL
wouldnt
> want to be me considering how much opposition my results
have faced
> and are likely to face.
> But hey, its actually fun!
> Like I havent just done arguing with people on math or
even just math
> research as I have an open source project on SourceForge,
and I have
> made some friends (believe it or not) in a few Internet
communities.
> I *thought* I could use the Internet in a groundbreaking
way to
> introduce major results but here I am WAITING ON A JOURNAL,
when I
> said in the past that I wouldnt even use journals!
> :-)
> Live and learn.
> In any event, I also get to go over my own mathematical
work, and
> enjoy talking about it as you might have noticed with me
starting new
> threads--yet again--going over the arguments in different
ways.
> Different looks. Perspective.
> But, on to something practical, as Ive been looking at
BitTorrent
> clients while Im doing downloads, and I see the download
speeds
> hopping all over the place as the clients--I guess--use
various
> algorithms to try and figure out whats the 
best connection
to make.
> Anyone here know anything about those algorithms? Im
thinking maybe
> I might make my own BitTorrent client, or go in and fiddle
with the
> algorithms on the one I have as I have Azureus. But its 
an
idle
> thought, so why not toss it on sci.math?
> Any of you know anything about the algorithms being used in
BitTorrent
> clients?
> James Harris
Bi-Polar much?
===
Subject: Re: JSH: Being me
Frightening thought.
===
Subject: Re: JSH: Being me
> Frightening thought.
Must be hell in there.
Dirk Vdm
===
Subject: Simple Group Theory Question - AGAIN
I cant seem to find any way to solve these 
problems!
Let G be a cyclic group of order p when p is prime.
Find all the automorphisims of G.
===
Subject: Re: Simple Group Theory Question - AGAIN
>I cant seem to find any way to solve these 
problems!
>Let G be a cyclic group of order p when p is prime.
>Find all the automorphisims of G.
First, can you tell us which elements of G _generate_ G?
************************
David C. Ullrich
===
Subject: Re: Simple Group Theory Question - AGAIN
> I cant seem to find any way to solve these 
problems!
> Let G be a cyclic group of order p when p is prime.
> Find all the automorphisims of G.
An automorphism f:G -> G is a homomorphism which maps
generators to
generators. So fix a generator g so that G = <g>. Since p is a
prime, g |-> g^n is an automorphism for 0<n and n less than
or equal
to p-1.
Youll want to *prove* all of this before you turn it in.
Ôcid Ôooh
===
Subject: A question about ßoor(e^x)
Let n(x) be the least integer such that:
1+x+x^2/2+...+x^n/n!>=ßoor(e^x)
So n(1)=2, n(2)=5, n(3)=9, n(4)=10, ect.
What is the maximum value of n(x) on the interval (1,2)? At
what value(s)
of x
is n(x) largest?
Rich
===
Subject: Re: A question about ßoor(e^x)
>Let n(x) be the least integer such that:
>1+x+x^2/2+...+x^n/n!>=ßoor(e^x)
>So n(1)=2, n(2)=5, n(3)=9, n(4)=10, ect.
I think youre off by 1 on all of these, e.g.
ßoor(e^3) = 20 and sum_{j=0}^8 3^j/j! > 20.
>What is the maximum value of n(x) on the interval (1,2)? At
what value(s)
of x
>is n(x) largest?
n(ln(k)) does not exist if k is an integer > 1, and n(x) ->
infinity
as x -> ln(k)+.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: A question about ßoor(e^x)
>>Let n(x) be the least integer such that:
>>1+x+x^2/2+...+x^n/n!>=ßoor(e^x)
>>So n(1)=2, n(2)=5, n(3)=9, n(4)=10, ect.
>I think youre off by 1 on all of these, e.g.
>ßoor(e^3) = 20 and sum_{j=0}^8 3^j/j! > 20.
Yes. I was thinking of n(x) as the number of terms of e^x
needed to
compute
ßoor(e^x) and did not make the adjustment when I posted the
(slightly)
different question.
>>What is the maximum value of n(x) on the interval (1,2)? At
what
value(s)
>of x
>>is n(x) largest?
>n(ln(k)) does not exist if k is an integer > 1, and n(x) ->
infinity
>as x -> ln(k)+.
Rich
===
Subject: Re: A question about ßoor(e^x)
> Let n(x) be the least integer such that:
> 1+x+x^2/2+...+x^n/n!>=ßoor(e^x)
> So n(1)=2, n(2)=5, n(3)=9, n(4)=10, ect.
> What is the maximum value of n(x) on the interval (1,2)? At
what value(s)
of x
> is n(x) largest?
How about x = ln(3)?
Asger.
===
Subject: adjoining elements to rings
posting-account=rtZaKw0AAAD069Dyt1QHY9YKfVai02Jb
What happens when we adjoin the inverse of 2 to the ring Z_12
i.e. what
is the field Z_12[x]/(2x-1) isomorphic to? I have reason to
believe
that it is simply the field of 3 elements, because of the
following:
In Z_12[x]/(2x-1) we have that 12=0 and 2x-1=0. Therefore
12x-6=0 ==>
6=0. Similarly we have 6x-3=0 ==> 3=0. This makes me believe
that we
are talking about F_3 here.
However, I cannot prove this. I tried setting up the
following maps:
j:Z ---> Z_12
pi:Z_12 ---> Z_12[x]/(2x-1)
i:Z ---> Z_3
If ker i = ker (pi o j) then we have that Z_12[x]/(2x-1) is
isomorphic
to Z_3 by the first isomorphism theorem. I know that
ker j = 12Z
ker pi = (2x-1)
ker i = 3Z
But for some reason I cannot compute the kernel of pi o j. Is
this even
the right track to go on?
Similarly I am trying to describe Z[i]/(2+i). Since 2+i=0 we
must have
5=0, which leads me to believe that Z[i]/(2+i) is simply Z_5.
I know
that 2+i is prime in Z[i] and therefore the quotient should
be a field.
I tried a similar approach to the above but did not get far.
I must be going in the wrong direction for these problems.
Can someone
tell me what is going on?
===
Subject: Re: adjoining elements to rings
>What happens when we adjoin the inverse of 2 to the ring
Z_12 i.e. what
>is the field Z_12[x]/(2x-1) isomorphic to? I have reason to
believe
>that it is simply the field of 3 elements, because of the
following:
>In Z_12[x]/(2x-1) we have that 12=0 and 2x-1=0. Therefore
12x-6=0 ==>6=0. Similarly we have 6x-3=0 ==> 3=0. This makes
me believe that we
>are talking about F_3 here.
>However, I cannot prove this. I tried setting up the
following maps:
>j:Z ---> Z_12
>pi:Z_12 ---> Z_12[x]/(2x-1)
>i:Z ---> Z_3
>If ker i = ker (pi o j) then we have that Z_12[x]/(2x-1) is
isomorphic
>to Z_3 by the first isomorphism theorem. I know that
>ker j = 12Z
>ker pi = (2x-1)
>ker i = 3Z
>But for some reason I cannot compute the kernel of pi o j.
Is this even
>the right track to go on?
Well, you know that the kernel contains (3). Since (3) is a
maximal
ideal of Z, the only possibilities are for the kernel to be
all of Z,
or else to be just equal to (3). So the question then
becomes: is 1 in
the kernel?
This will happen if and only if 1 = 0 (mod (2x-1)) in
Z_{12}[x]. That
is, if and only if there exists a polynomial f(x) in
Z_{12}[x] such
that f(x)(2x-1) = 1.
Write f(x) = a_n*x^n + ... + a_1*x + a_0 with a_i in Z. Then
(2x-1)f(x) = 2a_n*x^{n+1} + (2a_{n-1}-a_n)x^n + ...
(2a_0-a_1)x - a_0.
So you must have a_0 = - 1 (mod 12)
2a_{i-1} - a_i = 0 (mod 12) for i=1,...,n
2a_n = 0 (mod 12).
So, from a_0 = -1, we have -2 - a_1 = 0 (mod 12), or a_1 = -2.
Then 2a_1 - a_2 = 0 (mod 12), so -4 = a_2 (mod 12).
2a_2 - a_3 = 0 (mod 12) so a_3 = -8 (mod 12).
Continuing in this way, a_{i} = -2^i (mod 12).
Since we also need 2a_n = 0 (mod 12), that means that
-2^{n+1}=0 (mod
12). But this never happens. So no such polynomial exists.
Therefore,
1 is not in the kernel of the map. Since the kernel already
contains
3, and is not everything, the kernel of pi o j must be (3).
HOWEVER: you have not proven that Z_{12}[x]/(2x-1) is
isomorphic to
F_3; youve only shown that it CONTAINS F_3; you would also
have to
prove that the induced map from Z is surjective in order to
prove
that. (I mean, what if it is some other field of
characteristic 3?)
Slightly easier: you have already shown that 3 is in (2x-1),
so the
quotient Z_{12}[x]/(2x-1) factors through Z_{3}[x]/(2x-1).
And since
Z_3[x]/(2x-1) is isomorphic to Z_3, you are done.
>Similarly I am trying to describe Z[i]/(2+i).
This is a bit simpler since Z[i] is not only a domain, but a
UFD.
> Since 2+i=0 we must have
>5=0, which leads me to believe that Z[i]/(2+i) is simply
Z_5. I know
>that 2+i is prime in Z[i] and therefore the quotient should
be a field.
Its maximal, and therefore the quotient should be a 
field
(prime and
maximal are equivalent in Z[i], but not in general).
>I tried a similar approach to the above but did not get far.
Yes, it is isomorphic to Z/5Z. Certainly, it is a field a
characteristic 5, since a similar approach to the above will
establish
that the kernel of the map Z -> Z[i] -> Z[i]/(2+i) is just
(5), by
noting that 1 is not in (2+i). But you would still have to
show that
this map is surjective, which is not too hard: you can easily
show
that every element of Z[i] can be written as (a+bi)(2+i) + r,
with r =
0, 1, 2, 3, or 4.
This because 5 = (2+i)(2-i). So, given x+yi, with x and y in
Z, then
we have x-2y = (x+yi) - y(2+i). Now let x-2y = 5t + r, with t
an
integer, r=0,1,2,3, or 4; Then
x - 2y = (2t-ti)(2+i) + r
So
(2t-ti)(2+i) + r = (x+yi)-y(2+i)
hence
(x+yi) = (2t+y - ti)(2+i) + r.
Thus, every element in Z[i]/(2+i) is congruent modulo 2+i to
0, 1, 2,
3, 4, or 5, so the map Z->Z[i]->Z[i]/(2+i) is surjective,
kernel (5),
and you are done.
--
Its not denial. Im just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: adjoining elements to rings
<codke1$2idh$1@agate.berkeley.edu>
posting-account=rtZaKw0AAAD069Dyt1QHY9YKfVai02Jb
>What happens when we adjoin the inverse of 2 to the ring
Z_12 i.e.
what
>is the field Z_12[x]/(2x-1) isomorphic to? I have reason to
believe
>that it is simply the field of 3 elements, because of the
following:
>In Z_12[x]/(2x-1) we have that 12=0 and 2x-1=0. Therefore
12x-6=0
>6=0. Similarly we have 6x-3=0 ==> 3=0. This makes me believe
that we
>are talking about F_3 here.
>However, I cannot prove this. I tried setting up the
following maps:
>j:Z ---> Z_12
>pi:Z_12 ---> Z_12[x]/(2x-1)
>i:Z ---> Z_3
>If ker i = ker (pi o j) then we have that Z_12[x]/(2x-1) is
isomorphic
>to Z_3 by the first isomorphism theorem. I know that
>ker j = 12Z
>ker pi = (2x-1)
>ker i = 3Z
>But for some reason I cannot compute the kernel of pi o j.
Is this
even
>the right track to go on?
> Well, you know that the kernel contains (3). Since (3) is a
maximal
> ideal of Z, the only possibilities are for the kernel to be
all of Z,
> or else to be just equal to (3). So the question then
becomes: is 1
in
> the kernel?
> This will happen if and only if 1 = 0 (mod (2x-1)) in
Z_{12}[x]. That
> is, if and only if there exists a polynomial f(x) in
Z_{12}[x] such
> that f(x)(2x-1) = 1.
> Write f(x) = a_n*x^n + ... + a_1*x + a_0 with a_i in Z. Then
> (2x-1)f(x) = 2a_n*x^{n+1} + (2a_{n-1}-a_n)x^n + ...
(2a_0-a_1)x -
a_0.
> So you must have a_0 = - 1 (mod 12)
> 2a_{i-1} - a_i = 0 (mod 12) for i=1,...,n
> 2a_n = 0 (mod 12).
> So, from a_0 = -1, we have -2 - a_1 = 0 (mod 12), or a_1 =
-2.
> Then 2a_1 - a_2 = 0 (mod 12), so -4 = a_2 (mod 12).
> 2a_2 - a_3 = 0 (mod 12) so a_3 = -8 (mod 12).
> Continuing in this way, a_{i} = -2^i (mod 12).
> Since we also need 2a_n = 0 (mod 12), that means that
-2^{n+1}=0 (mod
> 12). But this never happens. So no such polynomial exists.
Therefore,
> 1 is not in the kernel of the map. Since the kernel already
contains
> 3, and is not everything, the kernel of pi o j must be (3).
> HOWEVER: you have not proven that Z_{12}[x]/(2x-1) is
isomorphic to
> F_3; youve only shown that it CONTAINS F_3; you would 
also
have to
> prove that the induced map from Z is surjective in order to
prove
> that. (I mean, what if it is some other field of
characteristic 3?)
> Slightly easier: you have already shown that 3 is in
(2x-1), so the
> quotient Z_{12}[x]/(2x-1) factors through Z_{3}[x]/(2x-1).
And since
> Z_3[x]/(2x-1) is isomorphic to Z_3, you are done.
>Similarly I am trying to describe Z[i]/(2+i).
> This is a bit simpler since Z[i] is not only a domain, but
a UFD.
> Since 2+i=0 we must have
>5=0, which leads me to believe that Z[i]/(2+i) is simply
Z_5. I know
>that 2+i is prime in Z[i] and therefore the quotient should
be a
field.
> Its maximal, and therefore the quotient should be a 
field
(prime and
> maximal are equivalent in Z[i], but not in general).
>I tried a similar approach to the above but did not get far.
> Yes, it is isomorphic to Z/5Z. Certainly, it is a field a
> characteristic 5, since a similar approach to the above will
establish
> that the kernel of the map Z -> Z[i] -> Z[i]/(2+i) is just
(5), by
> noting that 1 is not in (2+i). But you would still have to
show that
> this map is surjective, which is not too hard: you can
easily show
> that every element of Z[i] can be written as (a+bi)(2+i) +
r, with r
=
> 0, 1, 2, 3, or 4.
> This because 5 = (2+i)(2-i). So, given x+yi, with x and y
in Z, then
> we have x-2y = (x+yi) - y(2+i). Now let x-2y = 5t + r, with
t an
> integer, r=0,1,2,3, or 4; Then
> x - 2y = (2t-ti)(2+i) + r
> So
> (2t-ti)(2+i) + r = (x+yi)-y(2+i)
> hence
> (x+yi) = (2t+y - ti)(2+i) + r.
> Thus, every element in Z[i]/(2+i) is congruent modulo 2+i
to 0, 1, 2,
> 3, 4, or 5, so the map Z->Z[i]->Z[i]/(2+i) is surjective,
kernel (5),
> and you are done.
> --
> Its not denial. Im just very selective 
about
> what I accept as reality.
> --- Calvin (Calvin and Hobbes)
> Arturo Magidin
> magidin@math.berkeley.edu
first.
While messing with the second I came across a question that I
could not
answer. Do you get a different object if you adjoin i to C? I
know that
R[x]/(x^2-1)=C, but does C[x]/(x^2+1)=C or do you get
something else? I
am guessing you get something else since a ring mod an ideal
is a field
if an only if the ideal is maximal... however (x^2+1) is not
maximal in
C[x]. But what do you get?
===
Subject: Re: adjoining elements to rings
<codke1$2idh$1@agate.berkeley.edu>
posting-account=rtZaKw0AAAD069Dyt1QHY9YKfVai02Jb
>What happens when we adjoin the inverse of 2 to the ring
Z_12 i.e.
what
>is the field Z_12[x]/(2x-1) isomorphic to? I have reason to
believe
>that it is simply the field of 3 elements, because of the
following:
>In Z_12[x]/(2x-1) we have that 12=0 and 2x-1=0. Therefore
12x-6=0
>6=0. Similarly we have 6x-3=0 ==> 3=0. This makes me believe
that we
>are talking about F_3 here.
>However, I cannot prove this. I tried setting up the
following maps:
>j:Z ---> Z_12
>pi:Z_12 ---> Z_12[x]/(2x-1)
>i:Z ---> Z_3
>If ker i = ker (pi o j) then we have that Z_12[x]/(2x-1) is
isomorphic
>to Z_3 by the first isomorphism theorem. I know that
>ker j = 12Z
>ker pi = (2x-1)
>ker i = 3Z
>But for some reason I cannot compute the kernel of pi o j.
Is this
even
>the right track to go on?
> Well, you know that the kernel contains (3). Since (3) is a
maximal
> ideal of Z, the only possibilities are for the kernel to be
all of Z,
> or else to be just equal to (3). So the question then
becomes: is 1
in
> the kernel?
> This will happen if and only if 1 = 0 (mod (2x-1)) in
Z_{12}[x]. That
> is, if and only if there exists a polynomial f(x) in
Z_{12}[x] such
> that f(x)(2x-1) = 1.
> Write f(x) = a_n*x^n + ... + a_1*x + a_0 with a_i in Z. Then
> (2x-1)f(x) = 2a_n*x^{n+1} + (2a_{n-1}-a_n)x^n + ...
(2a_0-a_1)x -
a_0.
> So you must have a_0 = - 1 (mod 12)
> 2a_{i-1} - a_i = 0 (mod 12) for i=1,...,n
> 2a_n = 0 (mod 12).
> So, from a_0 = -1, we have -2 - a_1 = 0 (mod 12), or a_1 =
-2.
> Then 2a_1 - a_2 = 0 (mod 12), so -4 = a_2 (mod 12).
> 2a_2 - a_3 = 0 (mod 12) so a_3 = -8 (mod 12).
> Continuing in this way, a_{i} = -2^i (mod 12).
> Since we also need 2a_n = 0 (mod 12), that means that
-2^{n+1}=0 (mod
> 12). But this never happens. So no such polynomial exists.
Therefore,
> 1 is not in the kernel of the map. Since the kernel already
contains
> 3, and is not everything, the kernel of pi o j must be (3).
> HOWEVER: you have not proven that Z_{12}[x]/(2x-1) is
isomorphic to
> F_3; youve only shown that it CONTAINS F_3; you would 
also
have to
> prove that the induced map from Z is surjective in order to
prove
> that. (I mean, what if it is some other field of
characteristic 3?)
> Slightly easier: you have already shown that 3 is in
(2x-1), so the
> quotient Z_{12}[x]/(2x-1) factors through Z_{3}[x]/(2x-1).
And since
> Z_3[x]/(2x-1) is isomorphic to Z_3, you are done.
>Similarly I am trying to describe Z[i]/(2+i).
> This is a bit simpler since Z[i] is not only a domain, but
a UFD.
> Since 2+i=0 we must have
>5=0, which leads me to believe that Z[i]/(2+i) is simply
Z_5. I know
>that 2+i is prime in Z[i] and therefore the quotient should
be a
field.
> Its maximal, and therefore the quotient should be a 
field
(prime and
> maximal are equivalent in Z[i], but not in general).
>I tried a similar approach to the above but did not get far.
> Yes, it is isomorphic to Z/5Z. Certainly, it is a field a
> characteristic 5, since a similar approach to the above will
establish
> that the kernel of the map Z -> Z[i] -> Z[i]/(2+i) is just
(5), by
> noting that 1 is not in (2+i). But you would still have to
show that
> this map is surjective, which is not too hard: you can
easily show
> that every element of Z[i] can be written as (a+bi)(2+i) +
r, with r
=
> 0, 1, 2, 3, or 4.
> This because 5 = (2+i)(2-i). So, given x+yi, with x and y
in Z, then
> we have x-2y = (x+yi) - y(2+i). Now let x-2y = 5t + r, with
t an
> integer, r=0,1,2,3, or 4; Then
> x - 2y = (2t-ti)(2+i) + r
> So
> (2t-ti)(2+i) + r = (x+yi)-y(2+i)
> hence
> (x+yi) = (2t+y - ti)(2+i) + r.
> Thus, every element in Z[i]/(2+i) is congruent modulo 2+i
to 0, 1, 2,
> 3, 4, or 5, so the map Z->Z[i]->Z[i]/(2+i) is surjective,
kernel (5),
> and you are done.
> --
> Its not denial. Im just very selective 
about
> what I accept as reality.
> --- Calvin (Calvin and Hobbes)
> Arturo Magidin
> magidin@math.berkeley.edu
first.
While messing with the second I came across a question that I
could not
answer. Do you get a different object if you adjoin i to C? I
know that
R[x]/(x^2-1)=C, but does C[x]/(x^2+1)=C or do you get
something else? I
am guessing you get something else since a ring mod an ideal
is a field
if an only if the ideal is maximal... however (x^2+1) is not
maximal in
C[x]. But what do you get?
===
Subject: Re: adjoining elements to rings
days. My association with the Department is that of an
alumnus.
>first.
>While messing with the second I came across a question that
I could not
>answer. Do you get a different object if you adjoin i to C?
Depends what you mean by adjoin i! If you add the complex
number i
to a ring that already has it, then you just get the same ring
back. If you mean taking quotients of certain polynomial
rings, thats
something else:
>I know that
>R[x]/(x^2-1)=C, but does C[x]/(x^2+1)=C or do you get
something else? I
>am guessing you get something else since a ring mod an ideal
is a field
>if an only if the ideal is maximal... however (x^2+1) is not
maximal in
>C[x].
Exactly! So C[x]/(x^2+1) CANNOT be a field.
> But what do you get?
Notice that (x^2+1) = (x+i)(x-i); that (x+i) + (x-i) = 1. So
the
that
C[x]/(x^2+1) is isomoprhic to ( C[x]/(x+i) ) x ( C[x]/(x-i) )
which in turn is isomorphic to C x C, the product of two
copies of the complex numbers.
--
Its not denial. Im just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Prove this bound on integral
Hi
Let f(z)=(z+i)/(z-2i). Prove that |Int( f(z)) | <= sqrt(2)
where the
integral is taken over the contour C which is the line z=t ,
t in [-1,1].
So I used the estimate lemma (also called fundamental lemma).
So we first bound |f(z)|.
We notice that |z|=|t|<=1 so |z|<=1 , which implies -|z|>= -1.
|z+i|<=1+1 = 2. and |z-2i|>= |2i|-|z| >= 2-1 = 1.
so |z+i|/|z-2i| <= 2, i.e |f(z) | <=2.
The arclength of z=t in [-1,1] is 2.
So using the estimate lemma we get |int(f(z) dz) | <= 2*
arclength = 2* 2 =
4.
I cant seem to improve to lower the bound to sqrt(2), any
ideas?
===
Subject: Re: Prove this bound on integral
>Let f(z)=(z+i)/(z-2i). Prove that |Int( f(z)) | <= sqrt(2)
where the
>integral is taken over the contour C which is the line z=t ,
t in [-1,1].
>So I used the estimate lemma (also called fundamental lemma).
>So we first bound |f(z)|.
>We notice that |z|=|t|<=1 so |z|<=1 , which implies -|z|>=
-1.
>|z+i|<=1+1 = 2. and |z-2i|>= |2i|-|z| >= 2-1 = 1.
>so |z+i|/|z-2i| <= 2, i.e |f(z) | <=2.
>The arclength of z=t in [-1,1] is 2.
>So using the estimate lemma we get |int(f(z) dz) | <= 2*
arclength = 2* 2 =
4.
>I cant seem to improve to lower the bound to sqrt(2), any
ideas?
You can get sharper bounds than the triangle inequality in
both cases.
Draw a picture and look at |z + i| for z in [-1,1]. Ditto for
|z- 2i|.
Other than that, you are on the right track.
--Lynn
===
Subject: Re: Prove this bound on integral (Lynn)
So following your hint, we write z as z=x+iy, since z is in
[-1,1]
then z is real so z=x with x in [-1,1]. Now we consider |z+i|.
|z+i|=|x+i|= sqrt(x^2 + 1), this is maximum when x=1 so we get
|z+i|<= sqrt(1+1)= sqrt(2).
Now |z-2i|= |x-2i| = sqrt(x^2+4), this is minimm when x=0 so
we get
|z-2i| >= 2. Then we apply estimate lemma and were done.
Correct?
===
Subject: Re: Prove this bound on integral (Lynn)
>So following your hint, we write z as z=x+iy, since z is in
[-1,1]
>then z is real so z=x with x in [-1,1]. Now we consider
|z+i|.
>|z+i|=|x+i|= sqrt(x^2 + 1), this is maximum when x=1 so we
get
>|z+i|<= sqrt(1+1)= sqrt(2).
>Now |z-2i|= |x-2i| = sqrt(x^2+4), this is minimm when x=0 so
we get
>|z-2i| >= 2. Then we apply estimate lemma and were done.
>Correct?
I think you know it is correct. :-)
--Lynn
===
Subject: Re: Bible Codes Resurrected?
>
> As a follow up, I notified Art Levitt, the author of the
website
that
> I pointed to, of your objection. He contacted Haralick and
it
appears
> that Haralick indeed was unaware of the other lists that you
> mentioned.
>
> He said, I will take a look to see if there are any
additional
> web site.
>
> This is a good example of the quality of information that
gets around
> on this subject.
>
> Of course Haralick is aware of the other lists because they
are
> described in my paper published in Statistical Science. He
also
> http://cs.anu.edu.au/~bdm/dilugim/StatSci/emanuel_rep.html
> So what you report is nonsense.
>
> I directly quoted an email that Haralick cced me. It
appears to me
> that he only made a mistake by forgetting about those extra
lists
> he already used. I wouldnt infer from this that he
deliberately
> ignored these lists because he knew that they would produce
bad
> results - after all, what purpose would that serve him, as
it would
> only be a matter of time when you or someone else would try
it out and
> disprove him. And maybe they produce good results? I too
would like to
> see him perform the experiment with the lists on your
website.
> They will only provide a few extra appellations. Anyway,
note that
> I was responding to a specific claim and did not say that
Haralick
> should have used or not used any particular data. Rather,
the whole
> idea of his experiment is broken. A little on this below.
> More interestingly, Haralick has even collected
appellations himself
> that are not in the two lists he used.
>
> ones?
> Other way around. Haralick knows lots of valid appellations
that are
> not used in his experiment. These include appellations he
has collected
> himself for his own (other) experiments. They also include
many that
> everyone knows. For example, the original experiment used
appellations
> of the form Rabbi Yakov, but not Rav Yakov. As you know, the
latter
> is an extremely common expression; why is it not used?
There are many
> more examples. Why is the much smaller and more irregular
set of
> appellations that work in War and Peace the correct set of
additional
> appellations? And why is it correct to only add
appellations when
> (according to our hypothesis) a major problem with the
original data was
> the selective inclusion of doubtful appellations? (There
are even two
> that nobody ever found a single example of in the
literature.)
That is for certain true that there are many more possible
appellations, at least twice as many as you pointed out in the
Rabbi, Rav example.
> And thats only the start of the problems with this
appallingly bad
> paper of Haralick.
>
> What are some other problems? I appreciate the work that
you have done
> the subjectivity of the lists. If there are ßaws in the
Haralick
> Ill summarise what Haralick did. There are two sets of
data and
> a third constructed from them.
> D1: data used by WRR for Genesis
> D2: data used by BMMK (McKay & al.) for War and Peace
> D3. the union of D1 and D2
> Note that D1 and D2 have a very large overlap.
> D1 works well in Genesis and very poorly in War and Peace.
> D2 works well in War and Peace and fairly poorly in Genesis
> The question is: how does D3 behave?
> Using the analysis method of WRR, D3 works worse in Genesis
> than D1 did, and worse in War and Peace than D2 did.
I did not know this. I wish Haralick would have mentioned
this in his
paper. I got the impression from reading his paper that the
reason
that he used a different metric was because it was more
accurate, not
that it did not work on the combined lists. This fact 
doesnt
only
prove that the experimental methodology of WRR was ßawed as
you
showed in MBBK; it shows that the only way to argue the
thesis of WRR
is to do what Witztum does on his website, which is to try to
convince
everyone that the WRR list is superior than your list, which
is
completely subjective.
> Somehow or other, Haralick found a different method of
analysis
> for which D3 works about the same or slightly better in
Genesis
> than D1 did, whereas D3 still works worse in War and Peace
> than D2 did. (Actually he found 4 new methods of analysis,
some
> of which behave this way and some of which do not.)
> Haralick claims this proves something. I think it only
proves that
> wishful thinking is strong medicine.
> The fact is that the result is extremely sensitive to both
the data
> and the method of analysis. Seemingly tiny changes (like
adding
> or deleting a single word, or replacing a mathematical
expression
> by another with the same properties) can make the result
jump
> up or down dramatically. Factors of 10 or more are nothing.
> I know of tiny changes to the analysis method (far far more
minor
> than the changes between WRR and Haralick) that make the
> result for D2 in War and Peace 100 times better than the
result
> for D1 in Genesis. What does this prove? Nothing. What does
> it prove if Haralick can come up with a method showing some
> other behaviour? Nothing.
> Besides this, Haralick has not established a significance
level for
> his results. How likely is it for his observations to occur
by chance?
> He doesnt even ask the question. (I have reason to 
believe
that
> they are not very unlikely.) And lets not get started on
the errors
> in his new methods.
> Brendan.
realize how much the choice of the metric has such a big
effect on the
results of the experiment, as I have never tested it out
myself. This
fact and the fact that there are probably infinitely many
reasonable
compactness measures casts a lot of doubt on the methodology
of
Haralicks experiment, implying that there still is wiggle
room in the
choice of a metric - even though Haralick only tested 32
metrics on
list one, someone else doing the experiment might test
another 32
possible metrics on list one and get drastically different
results.
Haralick would have to give an explanation for the choice of
32
metrics that he tested - and lets not go down that road
again as
weve been there before regarding the choice of the
appellations.
Regardless of these critisisms, I would still like to give
Professor
Haralick the courtesy of a chance to respond to your
critisisms
(assuming that he is reading this thread) and would be
interested in
hearing the results of his same experiment adding all of the
lists
posted on your website (that Avi Norowitz mentioned before).
Anyway, Im sure that you also have criticisms of the other
paper
(only 4 pages) regarding the phrase Cursed is bin Laden and
revenge
belongs to the Messiah found in an ELS in the Bible? I and
probably
others reading this thread would probably be interested in
hearing
them.
Craig
===
Subject: Partial products
It is possible that the following problem has been solved
already by
someone, but I couldnt find any refence with 
Google.
Given is a set of n=2^m numbers {x(0),...x(n-1)}. It is
required
to compute all partial products P(0) to P(n-1), such that each
includes n-1 numbers from the set:
P(0) = 1 *x(1)*...*x(n-1)
P(1) =x(0)* 1 *...*x(n-1)
...
P(n-1)=x(0)*x(1)*...* 1
The only operation permited is multiplication. I need an
efficient
algorithm to generate these partial products using minimum
number
of multiplications.
My use of the above is in a factorization algorithm.
A.D.
===
Subject: Re: Partial products
> It is possible that the following problem has been solved
already by
> someone, but I couldnt find any refence with 
Google.
> Given is a set of n=2^m numbers {x(0),...x(n-1)}. It is
required
> to compute all partial products P(0) to P(n-1), such that
each
> includes n-1 numbers from the set:
> P(0) = 1 *x(1)*...*x(n-1)
> P(1) =x(0)* 1 *...*x(n-1)
> ...
> P(n-1)=x(0)*x(1)*...* 1
> The only operation permited is multiplication. I need an
efficient
> algorithm to generate these partial products using minimum
number
> of multiplications.
> My use of the above is in a factorization algorithm.
> A.D.
I hoped that this problem would attract some attention from
the
readers of this group. In the meantime I made some progress
on solving
it.
Obviously, naive direct method requires n*(n-1) multiplication
operations.
Desired approach will pre-compute some values and combine
them to
obtain the products P(i) using minimum number of operations.
The pre-computed values can be stored in a 2-dimensional
matrix. Then
partial products P(i) can be obtained with minimum number of
additional operations by cleverly combining the matrix
elements.
Let pre-computed value be stored in a matrix d(m,2^m) in the
following
manner:
for i=0 to n-1
d(0,i)=x(i)
next i
Ô
for j=1 to m-1
k=2^(m-j)-1
for i=0 to k
d(j,i)=d(j-1,2*i)*d(j-1,2*i+1)
next i
next j
The filling up of the matrix d(j,i) needs n-2 operations. The
following step will combine the elements from the matrix into
the
products P(i).
For example take m=4, n=16. One gets the following:
P(0)=d(0,1)*d(1,1)*d(2,1)*d(3,1) which uses 3 operations.
However P(1)
needs only 1 operation if d(0,1) only is replaced in the
above product
with d(0,0). It can be shown that P(2) will need 2
operations, P(3)
again 1 operation, P(4) 3 operations, and so on ... After
counting all
the operations I found that the total is 3*(16-2) including
those
needed to calculate d(j,i) products.
It looks that optimum method needs 3*(n-2) operations, versus
n*(n-1)
operation needed by direct inefficient method. For n=2^7 the
difference is quite significant, 378 versus 16256 operations!
What I still need is an algorithm to combine the elements
from matrix
d(m,2^m) to obtain P(i)s.
Any suggestion is appreciated.
A.D.
===
Subject: Re: Partial products
This best way to deal with this is the Forward-Backward
Algorithm.
we construct to arrays:
1st array is: x(0) x(0)*x(1) x(0)*x(1)*x(2), .........
2st array is: x(n-1) x(n-1)*x(n-2) x(n-1)*x(n-2)*x(n-3),
...........
It takes O(n) to construct these 2 arrays.
every partial product is the product from 2 number in 1st and
2st array.
and it takes constant time to locate the 2 numbers,
so the whole process takes O(n).
BTW, what are you going to make use of this algorithm? I
would like to know
it.
===
Subject: Re: infinitesimal calculation ?
> I also understand that because the function 1/0 is not
defined in the
> standard |R, It cannot be defined in the extended |R if we
assume
> true the standardisation axiom. Therefore, what does happen
if we are
> not taking this axiom true?
> Does anybody knows if the non standard analysis is self
consistent
> without the standardisation axiom? (It helps me to
understand the
> embedding constraints of the standard sets/properties into
the new
> one)
It must be -- there obviously exist models in which the axiom
holds,
and since it is a non-logical axiom (i.e. not a tautology or
first-order validity) there must be models in which it
doesnt. This,
of course, is assuming that it is independent from the other
axioms
(in the sense that you cant derive it from the other two).
> However, I keep learning on the extended algebra and the
possible
> representations we can build on these new (for me ; ) sets.
May be
> some one has a pointer towards a condensed lecture on this
aspect?
> (more condensed than keisler?).
You may want to look at Abraham Robinsons book on the 
topic,
or
Handbook of Analysis and Its Foundations by Eric Schechter if
condensed is what youre looking for.
> One of the questions concerning the representations is if
there exists
> an isomorphism to the elements of the representation or a
class of
> elements ? e.g. infinite sequence of standard real numbers.
Well, you cant have an order-isomorphism, since the
non-standard
reals are non-Archimedean. For the same reason, you cant
have an
(in my cursory examinations), the non-standard reals are
something
like the direct product of three continuum sized groups with
some
relations and a multiplicative structure thrown in. So the
best I can
think of is *maybe* a set theoretic isomorphism, which only
requires
the cardinalities of the involved sets to be equal. I dont
know what
the cardinality of the set of non-standard reals is, but I
suspect it
is that of the continuum.
Ôcid Ôooh
===
Subject: Re: JSH: Understanding polynomials
...
[context]
> > > which is a simple example of non-polynomial
factorization as
you
> > > have factored x-1 into non-polynomial factors.
> >
> > You have factored a polynomial in x^{1/2} into two
polynomials in
> > x^{1/2}.
> > So that is a polynomial factorisation.
> Harris stated that x^{1/2} + 1 is not a polynomial, which
is obviously
Is it false? Depends. I clearly stated (see the explanation
below) that
it is a polynomial in some contexts, but not in other
contexts.
> Your explanation: It is a polynomial in x^{1/2}. Which is
true, but
silly
> and irrelevant.
And I added that it is not a polynomial in x.
But that is the point. In the context it is *not* irrelevant.
In a way
(x - 1) = (x^{1/2} + 1)(x^{1/2} - 1) is a polynomial
factorisation. When
you can provide a polynomial factorisation in such a way
(where the
polynomial variable is 0 for x = 0), you can visibly inspect
the constant
terms. When you can *not* do that, you can *not* make such
visible
inspections. (And I may note in addition that constant term
is used
normally in the context of polynomials.)
And JSH tried to explain why in his factorisation of P(x)
into:
(5 a1(x) + 7)(5 a2(x) + 7)(5 a3(x) + 7)
there was no clear visible connection between the constant
term of P(x)
and the 7s in the factorisation, by arguing that there are
fractional
exponents involved in the a_i(x), and so it was not a
polynomial
factorisation. The explanation was wrong for two reasons.
First, in
the a_i(x) there are no fractional exponents of x involved.
Second,
if there had (positive) fractional exponents been involved
here, there
*would* have been a clear visible connection. If the a_i(x)
could
be written as polynomials in some x^{1/p} for some natural p,
the
constant term of P(x) must be 343.
This is reminiscent to an error JSH made quite a few years
ago when
where ... stands for a (possibly infinite) series of
a_rs.x^r.y^s
with rational r and s.
--
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: what determines upper bound on eigenvalues of a
covariance
matrix?
>>Suppose M is an nxn covariance matrix of n normally
distributed
random
>>variables.
>>Thus, M is symmetric and positive semidefinite.
>>Let z be the largest eigenvalue of M. In the limit that the
size n
of
>>M goes to infinity,
>>what are necessary and sufficient conditions such that:
> i) z/n -> 0?
>> ii) 0 < z/n < infinity?
>>
>> That will always be true except in the trivial case where
M = 0.
>>
>> Im afraid not so, Robert. Let e be a random variable 
with
mean zero
>> and unit variance.
>> Let X(n) be the n-tuplet whose every component equals e.
>> Then let M(n) = var(X(n)). In this case, M is the nxn
square matrix
>> whose every
>> element is 1. The largest eigenvalue of M in this case is
z=n. Thus,
>> z/n=1 is strictly
>> non-zero for all n.
>>
>> How is that not so? Last time I looked, 0 < 1 < infinity.
>(ii) is true, but (i) is not so.
>Roderick
The that I was referring to was (ii), not (i).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: induction vs Cantor II
Sorry, forgot sci.math
> Summary from the first thread:
> L_1 is a list of reals. This creates an implied
> mapping (injection) F_1 from the naturals to the reals.
> D_n is the Cantor number, CN (anti-diagonal number),
> that can be formed using the mapping F_n.
> Let L_n+1 be a list of reals by inserting D_n into L_n
> at row 2n.
> Row i in L_n+1 = Row i in L_n if i < 2n.
> Row i in L_n+1 = Row i-1 in L_n if i > 2n.
> This process is clearly an inductive process.
> It creates a new mapping for each natural
> number n.
> For all j > n, D_n is in L_j at row 2n.
> The union of all the L_n, (a countable set) contains
> all the D_n. Lets call the union CR (Cantor reals).
> Theres one problem. Someone is bound to come along
> and claim a bijection exists between N and CR and they
> are bound to claim that there is no CN to prove that
> its not a bijection N<->R.
> We can stop them by acting like 3-year-old babies and
> call them names, we can deny its true by utilizing the
> results of Cantors proof, or we can just prove that
> they are wrong by coming up with the CN. I think the
> best way is to actually support Cantors proof and
> show how to construct the CN.
> Someone is bound to claim this is a bijection:
> Each n can be associated with a unique real by using
> the real it is mapped to with F_n. This is the same
> real as the real its maps to with F_i for all i > n/2.
> They will then try to claim that the CN associated
> with that bijection exists in that bijection. They
> are going to try and claim that their bijection
> includes its own CN!
> Im not sure how to construct the CN in this case.
> If you can help me, You can post a response. I dont
> like a bunch of garbage in my e-mail but if you think
> you know me, e-mail me.
> GMN-75-23335.
> P.S. Im not looking for anything but the CN so I can
> prove those anti-Cantor people wrong.
===
Subject: Re: induction vs Cantor II
Poker Joker says...
>Sorry, forgot sci.math
>> Summary from the first thread:
>> L_1 is a list of reals. This creates an implied
>> mapping (injection) F_1 from the naturals to the reals.
>> D_n is the Cantor number, CN (anti-diagonal number),
>> that can be formed using the mapping F_n.
>> Let L_n+1 be a list of reals by inserting D_n into L_n
>> at row 2n.
>> Row i in L_n+1 = Row i in L_n if i < 2n.
>> Row i in L_n+1 = Row i-1 in L_n if i > 2n.
Okay. So the original list L_1 is
r_1
r_2
r_3
.
.
.
L_2 is the list (where d_1 is the anti-diagonal number of L_1)
r_1
d_1
r_2
r_3
.
.
.
L_3 is the list (where d_2 is the anti-diagonal number of L_2)
r_1
d_1
r_2
d_2
r_3
.
.
.
In the limit, we have L_infinity which will look like the
following
r_1
d_1
r_2
d_2
r_3
d_3
r_4
d_4
r_5
d_5
.
.
.
This list contains all the elements of L_1 plus all the
anti-diagonal numbers d_1, d_2, etc. If you use Cantors
diagonalization procedure on *this* list, you get a new
number d_infinity that is not equal to r_1, nor d_1, nor
r_2, nor d_2, etc.
d_infinity is guaranteed to be different from r_1 in the
first decimal place, different from d_1 in the second
decimal place, different from r_2 in the third place,
different from d_2 in the fourth place, etc.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: induction vs Cantor II
Poker Joker says...
>> Summary from the first thread:
>> L_1 is a list of reals. This creates an implied
>> mapping (injection) F_1 from the naturals to the reals.
>> D_n is the Cantor number, CN (anti-diagonal number),
>> that can be formed using the mapping F_n.
>> Let L_n+1 be a list of reals by inserting D_n into L_n
>> at row 2n.
>> Row i in L_n+1 = Row i in L_n if i < 2n.
>> Row i in L_n+1 = Row i-1 in L_n if i > 2n.
>> This process is clearly an inductive process.
>> It creates a new mapping for each natural
>> number n.
>> For all j > n, D_n is in L_j at row 2n.
>> The union of all the L_n, (a countable set) contains
>> all the D_n. Lets call the union CR (Cantor reals).
>> Theres one problem. Someone is bound to come along
>> and claim a bijection exists between N and CR and they
>> are bound to claim that there is no CN to prove that
>> its not a bijection N<->R.
Yes, there is a bijection between N and CR, if by CR
you mean the set of all reals r such that r is in L_n
for some natural number n. You define a merged list
L_merged as follows:
1. Let positions 1, 3, 5, 7, etc. in L_merged be filled in
with elements of L_1.
2. Let positions 2, 6, 10, 14, etc. in L_merged be filled in
with elements of L_2.
3. Let positions 4, 12, 20, 28, etc. in L_merged be filled in
with elements of L_3.
In general, real number i of list L_j will appear at position
2^{j-1} * (2i - 1) of list L_merged. For example, real number
1
of list L_1 will appear at position 2^0 * (2-1) = 1. Real
number 2
of list L_1 will appear at position 2^0 * (4-1) = 3. Real
number
3 of list L_3 will appear at position 2^2 * (6-1) = 20. etc.
Then you perform the regular Cantor diagonalization procedure
to
come up with a real that is not in the list L_merged (and
therefore,
it is not in any of the lists L_1, L_2, etc.)
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: induction vs Cantor II
>> D_n is the Cantor number, CN (anti-diagonal number),
>> that can be formed using the mapping F_n.
Some additional explanation is in order here.
D_n is really any number that can be generated by
any method as long as the construction gaurantees
that D_n is not already in L_n. For example, D_n
may match one of the reals in the list along the
diagonal. Thats okay as long as the construction
gaurantees a unique real.
Note that the construction itself can change at each
iteration. The only real restriction is that at each
iteration, we describe the construction that we are
using and that it generates a unique real.
I dont see anything that restricts any given real from
showing up on the list.
===
Subject: Re: induction vs Cantor II
Poker Joker says...
> D_n is the Cantor number, CN (anti-diagonal number),
> that can be formed using the mapping F_n.
>Some additional explanation is in order here.
>D_n is really any number that can be generated by
>any method as long as the construction gaurantees
>that D_n is not already in L_n. For example, D_n
>may match one of the reals in the list along the
>diagonal. Thats okay as long as the construction
>gaurantees a unique real.
Okay, so lets say that a function f is a diagonalizing
function
if for any infinite list of reals L, f(L) returns a real that
is
not on list L. One such diagonalizing function is the one
Cantor
described: f(L) = that real whose nth digit is equal to d+1
(mod 10)
where d = the nth digit of the nth real of list L.
If you pick one fixed diagonalizing function f, then we can
let L_1 be
our original list, and then let L_2 be the result of adding
f(L_1) to
L_1, and let L_3 be the result of adding f(L_2) to L_2, etc.
Then
if we let L_merged be the merged list of L_1, L_2, L_3, etc.
then
f(L_merged) will produce a real that is not on any of the
lists.
--
Daryl McCullough
Ithaca, NY
===
Subject: (image processing) how to match the perceived energy
for two
images...
Hi all,
This question is regarding two images.
One image is the original image; another image is obtained by
proforming
downsampling, upsampling, filtering, color-space conversion,
gamma
conversion, etc, on the original image.
After all of these steps, I want the two images to have same
perceived
energy... (or maybe I should say that I want the two images
to appear
similar viewing from several meters away... ) currently the
processed image
is a lot darker than the original image. I should defintely
scale up the
second image in order to make it brighter and perceived
similar to the
original one... it is just a question about the scaling
factor, after so
many linear and non-linear steps, I have already lost track
of the scaling
factor...
Now in order to match the two images and make them look
similar when viewing
from several meters away, I think I want to be able to take
any N x N block
from both images Red plane, compare the the mean of two
blocks, and make the
second one scale-up to the original image with the same mean.
The problem with my scheme is that if I choose N larger, then
two adjacent
different blocks may have two different scaling-factor, and
hence it has
visible artifacts...
If I choose N small, say N=1, then it is in fact I am
cheating by making the
second image point by point the same as the first one, then I
lose all
information of my processing of the second image...
So are there any good schemes that can let me retain the
processed
information of the second image, while scale up its
brightness and make
their perceived energy the same when viewing at several
meters distance?
===
Subject: Re: Operator ambiguity, Escultura
> As this example shows, in the domain of complex numbers,
> the square root of the product is not equal to the product
of the
> square roots and the same for quotients.
And how do you define the square root in the domain of complex
numbers?
> As a consequence of the Fundamental Theorem of Algebra,
> every complex number has exactly two complex square roots.
In any field, every element has at most two square roots.
Besides,
you dont need the Fundamental Theorem of Algebra in order 
to
prove
that every complex complex number has a square root.
Jose Carlos Santos
===
Subject: Re: Operator ambiguity, Escultura
>> As this example shows, in the domain of complex numbers,
>> the square root of the product is not equal to the product
of the
>> square roots and the same for quotients.
> And how do you define the square root in the domain of
complex
> numbers?
A square root of a complex number a+bi is a solution of the
equation
z^2 = a+ib.
>> As a consequence of the Fundamental Theorem of Algebra,
>> every complex number has exactly two complex square roots.
> In any field, every element has at most two square roots.
Besides,
> you dont need the Fundamental Theorem of Algebra in order
to prove
> that every complex complex number has a square root.
===
Subject: Re: Operator ambiguity, Escultura
>As this example shows, in the domain of complex numbers,
>the square root of the product is not equal to the product
of the
>square roots and the same for quotients.
>>And how do you define the square root in the domain of
complex
>>numbers?
> A square root of a complex number a+bi is a solution of the
equation
> z^2 = a+ib.
reproduced above) something about the square root. Now, I
know very
well what _a_ square root is, thank you very much, but when
you use the
expression the square root you are making a choice: you are
choosing
one among the two square roots of a complex number (different
from 0).
Only after that you can state that in the domain of complex
numbers,
the square root of the product is not equal to the product of
the square
roots.
So, I repeat my question: how do you define the square root in
the
domain of complex numbers?
Jose Carlos Santos
===
Subject: Re: Operator ambiguity, Escultura
Originator: richard@cogsci.ed.ac.uk (Richard Tobin)
>Only after that you can state that in the domain of complex
numbers,
>the square root of the product is not equal to the product
of the square
>roots.
>So, I repeat my question: how do you define the square root
in the
>domain of complex numbers?
Any definition of which root is *the* root will have the
problem
stated, so the statement is true regardless of which
definition you
use.
-- Richard
===
Subject: Re: Operator ambiguity, Escultura
>>Only after that you can state that in the domain of complex
numbers,
>>the square root of the product is not equal to the product
of the square
>>roots.
>>So, I repeat my question: how do you define the square root
in the
>>domain of complex numbers?
> Any definition of which root is *the* root will have the
problem
> stated, so the statement is true regardless of which
definition you
> use.
I know that. I just wanted to know which definition was N.
Silver
the square root of the product is not equal to the product of
the square
roots. Note that he did not add regardless of which definition
you
use.
Jose Carlos Santos
===
Subject: Re: Operator ambiguity, Escultura
I think that the error is in the following step:
sqrt(1/-1) = 1/i
Its a bit like saying
sqrt((-3)(-3))=-3
And then proclaiming that you have taken the positive square
root.
Clearly this is incorrect.
> I would like to pull out and highlight something
interesting that E.
> E. Escultura posted a few days ago, which Id guess 
hes
probably
> talked about many times before, but I just noticed it and
think its
> neat.
> First some more preamble as *by convention* as has been
noted when I
> brought up the subject of operator ambiguity before,
sqrt(x) is taken
> to be positive.
> So, by the convention, sqrt(4) = 2, and thats good as,
-2(-2) = 4, so
> if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2,
and 4 = 0,
> which is not good.
> Naively then, you may believe that you can just say, take
the positive
> of the square root but as Escultura showed, that doesnt
work:
> i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1.
Contradiction.
> You see, the ambiguity in the square root operator still
remains,
> despite the convention.
> It doesnt work to just try and always take the positive 
as
> Esculturas example shows so clearly.
> Who has the resolution? Im curious as to whether or not
any of you
> think you can answer.
> James Harris
===
Subject: Re: Operator ambiguity, Escultura
>I would like to pull out and highlight something interesting
> at E.E. Escultura posted a few days ago, which Id guess
> hes probably talked about many times before, but I just
> noticed it and think its neat.
You just ache to show mathematics is not on firm foundations,
dont you?! Because you are stumped at times does not mean
that mathematics is contradictory. On the contrary, it means
you are not working hard enough, do not know enough, or
are not mathematically sophisticated enough to overcome
your mathematical difficulties.
This elementary stuff has been worked over many, many
times before by those much smarter than you or I.
It would be prudent to give up the ghost now. You have
not a snowballs chance of ever winning in this forum.
===
Subject: Re: Kick classic in shins, win prizes
> Misner, Thorne and Wheeler have undoubtedly each
accomplished more
> than I might hope to should I live another 50 years. (Ob
kowtow).
> But one thing they had _not_ accomplished as of 1973 was to
edit their
> massive textbook free of casual carelessness.
> My complaint is with Box 2.3 -- Differentials.
> In this box we are made acquainted with the exterior
derivative or
> gradient Df (boldface small-d italic-f), which is
contrasted with
> the bad old differential df (italic-d italic-f). The new
model is
> superior to the old in every way, and comes with a lifetime
powertrain
> warranty! Lets see what it can do...
> A symbol v (bold-v) represents a particular infinitesimally
long
> displacement of the point P (script-P), which is the
argument of the
> function f = f(P). An inner product <Df,v> is formed and
set equal to
> a symbol @_v f (partial d subscript bold-v, italic-f),
which is the
> change of f in going from the tail of v to its tip. Mighty
like a
> differential, no? In fact we can identify the old notation
with the
> new term-by-term:
> <Df,v> = @_v f == <grad f,dX> = df
> Each slot on the rhs and lhs is occupied by the same
object, but with
> a different symbol (grad f is the gradient, dX the
infinitesimal
> displacement of the argument, df the infinitesimal change in
the
> function). Notice in particular that df and Df are _not_
mapped
> to each other, but are distinct objects.
> Ok. Now lets go back and catalogue the sophistry enlisted
in a
> casual effort to calumniate the old notation:
> Df ... is a more rigorous version of the elementary concept
of
> differential i.e., df.
> So they claim. Too bad the two symbols simply represent
different
> objects in relabled but otherwise identical equations;
neither more or
> less elementary or rigorous than the other.
> In elementary textbooks...
> Nice touch. ;-)
> ...one is presented with differential df as representing Ôan
> infinitesimal change in the function f(P) 
associated with
some
> infinitesimal displacment of the point P; but one will
recall that the
> displacement of P is left arbitrary, albeit infinitesimal.
Thus df
> represents a change in f in some unspecified direction.
> Heavens! This means that df is itself a variable or a
function of
> its argument (the change in P), as _well_ as an
infinitesimal! What
> are we to do!
> But this is precisely what the exterior derivative Df
represents...
> Nope. Df represents an _engine_ to _calculate_ such a
thing, given an
> infinitesimal displacement as input, same as the gradient.
In fact,
> it _is_ the gradient, as they just finished telling us.
Seems they
> forgot.
> Thus Df, before it has been pierced <their favorite imagery
for
> forming the inner product> to give a number, represents the
change in
> f in an unspecified direction.
> Same comment. Df represents a linear machine (a tensor, as
they well
> should know) for calculating the change in an arbitrary
direction,
> given an input. Nor is it not some brave new object never
before
> seen, it is the gradient.
> And now get ready for the coup-de-grace..
> The only failing of the textbook presentation...
> Forgot to add elementary to exclude the book we are reading?
> ...then, was its suggestion that Df was a scalar or a
number; the
> explicit recognition of the need for specifying a
directiong v to
> reduce Df to a number <Df,v> shows that in fact Df is a
1-form, the
> gradient of f.
> It takes time to appreciate such confusion. It really must
be savored:
> and particularly the interesting claim that the textbook
> presentation suggests that Df <is> a scalar or number. How
on
> earth could it! _Df_ (bold-d f) is part of a _new_ notation
--
> which in truth looks confusingly similar to a part of the
old notation
> (df) which it does not map to.
> The bad old textbook could only have been making a
suggestion about
> _df_ , not Df, and it would have been perfectly correct to
do so: df
> is indeed a scalar, albeit a variable one -- a concept we
might
> perhaps recall -- and is equivalent to the weird @_v f
symbol in the
> new notation, not to Df. This would possibly account for df
acting
> more like a number or scalar, and Df acting more like a
1-form or
> tensor? And we havent made the variable 
infinitesimal any
less
> variable by relabeling it to include an argument in its
symbol,
> anymore than relabeling y as y(x) makes y a definite number.
> The symbols Df and df represent distinct objects in
different notation
> schemes, and attempting to identify them leads to the
unsettling idea
> that somebody must have been confused somewhere -- surely
it must have
> been the feeble, unrigorous (and conveniently absent)
authors of this
> bad old notation? But the confusion instead lies in the mind
> attempting to make the identification.
LOL, MTWs dual prose is certainly interesting,
had to try it once.
While component based notation may have some
faults, I can always get out my ruler and clock,
and figure out where the I am.
Understandly, mathematical physicists seek notational
concistion, especially to eliminate the unimportant
distractions, and thats good.
I guess its up to the new students to determine what
fits best, maybe differential algebra will evolve legs,
and out run tensor analysis.
Ken S. Tucker
===
Subject: Re: Kick classic in shins, win prizes
<...> LOL, MTWs dual prose is certainly interesting,
> had to try it once.
> While component based notation may have some
> faults, I can always get out my ruler and clock,
> and figure out where the I am.
> Understandly, mathematical physicists seek notational
> concistion, especially to eliminate the unimportant
> distractions, and thats good.
> I guess its up to the new students to determine what
> fits best, maybe differential algebra will evolve legs,
> and out run tensor analysis.
Huh! I was almost afraid to go back and look at this. I
expected at
least a few how dare you criticize MTW, you lickspittles. 
But
nothing...
===
Subject: Orthogonal complement in L^2(M)
Hi all,
Let M be a measure space and consider the space L^2(M) of all
square-integrable functions from M into the reals. A proof
from a book
that I was reading uses the following fact implicitely: if E
is a closed
subspace of L^2(M), then its orthogonal complement is
different form
{0}, unless, of course, E = L^2(M). My question is: is this
obvious?
I know that L^2(M) is a Hilbert space and that in any Hilbert
space
that statement is true, but the authors of the book make no
mention at
all to Hilbert spaces (they dont appear at the subject
index).
So, is the statement obvious for Hilbert spaces of the form
L^2(M)?
I know that every Hilbert space is isomorphic to a Hilbert
space of
that type, but there could be an easier prove in this context.
Jose Carlos Santos
===
Subject: Re: Orthogonal complement in L^2(M)
>Let M be a measure space and consider the space L^2(M) of all
>square-integrable functions from M into the reals. A proof
from a book
>that I was reading uses the following fact implicitely: if E
is a closed
>subspace of L^2(M), then its orthogonal complement is
different form
>{0}, unless, of course, E = L^2(M). My question is: is this
obvious?
>I know that L^2(M) is a Hilbert space and that in any
Hilbert space
>that statement is true, but the authors of the book make no
mention at
>all to Hilbert spaces (they dont appear at the subject
index).
>So, is the statement obvious for Hilbert spaces of the form
L^2(M)?
>I know that every Hilbert space is isomorphic to a Hilbert
space of
>that type, but there could be an easier prove in this
context.
I cant think of anything easier than this, which 
doesnt use
the fact that its L^2(M).
Let y be a member of the Hilbert space not in E, and d =
dist(y,E).
There is a sequence x_n in E with ||x_n - y|| -> d. Use the
parallelogram
law to show that x_n is a Cauchy sequence, so by completeness
it
has a limit x with ||x - y|| = d, and x is in E. Then x - y
will be in
the orthogonal complement of E.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: Orthogonal complement in L^2(M)
>>Let M be a measure space and consider the space L^2(M) of
all
>>square-integrable functions from M into the reals. A proof
from a book
>>that I was reading uses the following fact implicitely: if
E is a closed
>>subspace of L^2(M), then its orthogonal complement is
different form
>>{0}, unless, of course, E = L^2(M). My question is: is this
obvious?
>>I know that L^2(M) is a Hilbert space and that in any
Hilbert space
>>that statement is true, but the authors of the book make no
mention at
>>all to Hilbert spaces (they dont appear at the subject
index).
>>So, is the statement obvious for Hilbert spaces of the form
L^2(M)?
>>I know that every Hilbert space is isomorphic to a Hilbert
space of
>>that type, but there could be an easier prove in this
context.
> I cant think of anything easier than this, which 
doesnt
use
> the fact that its L^2(M).
> Let y be a member of the Hilbert space not in E, and d =
dist(y,E).
> There is a sequence x_n in E with ||x_n - y|| -> d. Use the
parallelogram
> law to show that x_n is a Cauchy sequence, so by
completeness it
> has a limit x with ||x - y|| = d, and x is in E. Then x - y
will be in
> the orthogonal complement of E.
that there was something missing here.
Jose Carlos Santos
===
Subject: Re: Orthogonal complement in L^2(M)
>Let M be a measure space and consider the space L^2(M) of all
>square-integrable functions from M into the reals. A proof
from a book
>that I was reading uses the following fact implicitely: if E
is a closed
>subspace of L^2(M), then its orthogonal complement is
different form
>{0}, unless, of course, E = L^2(M). My question is: is this
obvious?
>I know that L^2(M) is a Hilbert space and that in any
Hilbert space
>that statement is true, but the authors of the book make no
mention at
>all to Hilbert spaces (they dont appear at the subject
index).
>So, is the statement obvious for Hilbert spaces of the form
L^2(M)?
>I know that every Hilbert space is isomorphic to a Hilbert
space of
>that type, but there could be an easier prove in this
context.
>> I cant think of anything easier than this, which 
doesnt
use
>> the fact that its L^2(M).
>> Let y be a member of the Hilbert space not in E, and d =
dist(y,E).
>> There is a sequence x_n in E with ||x_n - y|| -> d. Use the
parallelogram
>> law to show that x_n is a Cauchy sequence, so by
completeness it
>> has a limit x with ||x - y|| = d, and x is in E. Then x -
y will be in
>> the orthogonal complement of E.
>that there was something missing here.
No, surely its as Chapman suggests - the authors of a book
titled
Representations of Compact Lie Groups are just assuming the
reader already knows a little basic real analysis.
>Jose Carlos Santos
************************
David C. Ullrich
===
Subject: Re: Orthogonal complement in L^2(M)
> Hi all,
> Let M be a measure space and consider the space L^2(M) of
all
> square-integrable functions from M into the reals. A proof
from a book
> that I was reading uses the following fact implicitely: if
E is a closed
> subspace of L^2(M), then its orthogonal complement is
different form
> {0}, unless, of course, E = L^2(M). My question is: is this
obvious?
Its true: whether its obvious is a matter of 
opinion.
> I know that L^2(M) is a Hilbert space and that in any
Hilbert space
> that statement is true, but the authors of the book make no
mention at
> all to Hilbert spaces (they dont appear at the subject
index).
> So, is the statement obvious for Hilbert spaces of the form
L^2(M)?
> I know that every Hilbert space is isomorphic to a Hilbert
space of
> that type, but there could be an easier prove in this
context.
Seems unlikely.
Which book is it? Is it a physics book?
--
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: Orthogonal complement in L^2(M)
>>Let M be a measure space and consider the space L^2(M) of
all
>>square-integrable functions from M into the reals. A proof
from a book
>>that I was reading uses the following fact implicitely: if
E is a closed
>>subspace of L^2(M), then its orthogonal complement is
different form
>>{0}, unless, of course, E = L^2(M). My question is: is this
obvious?
> Its true: whether its obvious is a matter 
of opinion.
>>I know that L^2(M) is a Hilbert space and that in any
Hilbert space
>>that statement is true, but the authors of the book make no
mention at
>>all to Hilbert spaces (they dont appear at the subject
index).
>>So, is the statement obvious for Hilbert spaces of the form
L^2(M)?
>>I know that every Hilbert space is isomorphic to a Hilbert
space of
>>that type, but there could be an easier prove in this
context.
> Seems unlikely.
> Which book is it? Is it a physics book?
Not at all. Its Representations of Compact Lie Groups by
Theodor
Broecker and Tammo tom Dieck. It appears at the end of the
proof of
the theorem 2.6 (chapter III), which states that given a
compact
self-adjoint automorphism of L^2(M), the direct sum of the
eigenspaces
is dense and, furthermore, given r > 0, the direct sum of
those
eigenspaces whose eigenvalue has absolute value greater or
equal than r
is finite-dimensional (in other words, the spectral theorem
for compact
self-adjoint operators).
Jose Carlos Santos
===
Subject: Re: Orthogonal complement in L^2(M)
Originator: grubb@lola
>Let M be a measure space and consider the space L^2(M) of all
>square-integrable functions from M into the reals. A proof
from a book
>that I was reading uses the following fact implicitely: if E
is a closed
>subspace of L^2(M), then its orthogonal complement is
different form
>{0}, unless, of course, E = L^2(M). My question is: is this
obvious?
>> Its true: whether its obvious is a matter 
of opinion.
>I know that L^2(M) is a Hilbert space and that in any
Hilbert space
>that statement is true, but the authors of the book make no
mention at
>all to Hilbert spaces (they dont appear at the subject
index).
>So, is the statement obvious for Hilbert spaces of the form
L^2(M)?
>I know that every Hilbert space is isomorphic to a Hilbert
space of
>that type, but there could be an easier prove in this
context.
>> Seems unlikely.
>> Which book is it? Is it a physics book?
>Not at all. Its Representations of Compact Lie Groups by
Theodor
>Broecker and Tammo tom Dieck. It appears at the end of the
proof of
>the theorem 2.6 (chapter III), which states that given a
compact
>self-adjoint automorphism of L^2(M), the direct sum of the
eigenspaces
>is dense and, furthermore, given r > 0, the direct sum of
those
>eigenspaces whose eigenvalue has absolute value greater or
equal than r
>is finite-dimensional (in other words, the spectral theorem
for compact
>self-adjoint operators).
I would assume that knowledge of Hilbert spaces and the fact
that
L^2(M) is a Hilbert space would be a prerequisite for such a
book.
In fact, I would assume that elementary facts about Hilbert
spaces would
be used almost without comment.
--Dan Grubb
===
Subject: Re: Orthogonal complement in L^2(M)
>Let M be a measure space and consider the space L^2(M) of all
>square-integrable functions from M into the reals. A proof
from a book
>that I was reading uses the following fact implicitely: if E
is a closed
>subspace of L^2(M), then its orthogonal complement is
different form
>{0}, unless, of course, E = L^2(M). My question is: is this
obvious?
>> Its true: whether its obvious is a matter 
of opinion.
>I know that L^2(M) is a Hilbert space and that in any
Hilbert space
>that statement is true, but the authors of the book make no
mention at
>all to Hilbert spaces (they dont appear at the subject
index).
>So, is the statement obvious for Hilbert spaces of the form
L^2(M)?
>I know that every Hilbert space is isomorphic to a Hilbert
space of
>that type, but there could be an easier prove in this
context.
>> Seems unlikely.
>> Which book is it? Is it a physics book?
> Not at all. Its Representations of Compact Lie Groups by
Theodor
> Broecker and Tammo tom Dieck. It appears at the end of the
proof of
> the theorem 2.6 (chapter III), which states that given a
compact
> self-adjoint automorphism of L^2(M), the direct sum of the
eigenspaces
> is dense and, furthermore, given r > 0, the direct sum of
those
> eigenspaces whose eigenvalue has absolute value greater or
equal than r
> is finite-dimensional (in other words, the spectral theorem
for compact
> self-adjoint operators).
Im sure these guys regard that as assumed knowledge.
--
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: Orthogonal complement in L^2(M)
>Which book is it? Is it a physics book?
>>Not at all. Its Representations of Compact Lie Groups by
Theodor
>>Broecker and Tammo tom Dieck. It appears at the end of the
proof of
>>the theorem 2.6 (chapter III), which states that given a
compact
>>self-adjoint automorphism of L^2(M), the direct sum of the
eigenspaces
>>is dense and, furthermore, given r > 0, the direct sum of
those
>>eigenspaces whose eigenvalue has absolute value greater or
equal than r
>>is finite-dimensional (in other words, the spectral theorem
for compact
>>self-adjoint operators).
> Im sure these guys regard that as assumed knowledge.
I did not reply yesterday because I did not have the book at
hand then,
but now that I have it I can confirm that youre 
right
(together with
David C. Ullrich and Daniel Grubb): the authors are assuming
that
knowledge. In fact, a few pages before that theorem they state
(concerning the space H of all continuous functions from a
compact Lie
group G into the complex numbers, equiped with the norm
derived from
the standard scalar product) that [c]ompletion of H for this
norm yelds
the Hilbert space H^ = L^2(G) of square integrable functions.
So, some
basic facts concerning HIlbert spaces are assumed here.
Jose Carlos Santos
===
Subject: Re: The-State-of-the-Art in Mathematics
Hey, Escultura,
I disagree.
When you say dichotomy, do you mean trichotomy?
What are your vague numbers? What is your definition of a real
number? Do you consider the nonstandard constructions of the
surreal
numbers, or hyperreals or colonies of germs and germinal
functions?
You appear to promote geometry.
Im getting somewhat defensive about all this. I would like
to not
share with you.
Please go off about the real numbers. You think they are
decimals?
Write the decimal form of square root of two.
You claim to solve all paradoxes? What are they? Where are
paradoxes
these days? The ßaw is axiomatization, but not the root
probabilistic ßaw.
What do you think about iota? Not you, Ullrich. Ullrich is a
troll.
Please explain Banach-Tarski, and the bleeding points on the
line.
The real numbers are complete.
The state of the art in mathematics is symplectic integration
and
geometric algebra, and not manifold theory, although there
are some
fantastic combinatorial results out in the polynomial realm.
Mathematics tends to lead physics by some time.
I encourage you to go off about the real numbers. Please
explain your
opinion of what is wrong with contemporary standard
treatments of the
real numbers. This sci.math group is full of reviewers.
The real number is a scalar.
Ross Finlayson
===
Subject: The function x^x
Hi.
Does anyone know what the graph of the function f(x) =
Imag(x^x) for x
< 0, that is, the imaginary part of x^x (e^(x ln(x)) with
principal
branch of
ln(x)), looks like? I havent really been able to get a good
idea of
its behavior from looking at a few points.
===
Subject: Re: The function x^x
> Does anyone know what the graph of the function f(x) =
Imag(x^x) for
> x < 0, that is, the imaginary part of x^x (e^(x ln(x)) with
principal
> branch of ln(x)), looks like?
Mark Meyerson has written something about the graph of x^x.
Have a look
at http://www.usna.edu/MathDept/mdm/homepage.html
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: The function x^x
> Hi.
> Does anyone know what the graph of the function f(x) =
Imag(x^x) for x
> < 0, that is, the imaginary part of x^x (e^(x ln(x)) with
principal
> branch of
> ln(x)), looks like? I havent really been able to get a
good idea of
> its behavior from looking at a few points.
The negative axis, taken as a branch cut for the principal
branch of the
complex function log, can be parametrized as:
x=r*exp(i*pi), 0<=r<+inf.
Plug x in exp(x*log(x)) to get:
exp(x*log(x))=exp(r*(-1)*(r+i*pi))=
exp(-r^2)*exp(-i*r*pi)
Separate real and imaginary parts,
Im(x^x)=exp(-r^2)*sin(-r*pi)
Now graph that for 0<=r<+inf.
--
I. N. G. --- http://users.forthnet.gr/ath/jgal/
===
Subject: Re: The function x^x
> Hi.
> Does anyone know what the graph of the function f(x) =
Imag(x^x) for x
> < 0, that is, the imaginary part of x^x (e^(x ln(x)) with
principal
> branch of
> ln(x)), looks like? I havent really been able to get a
good idea of
> its behavior from looking at a few points.
> The negative axis, taken as a branch cut for the principal
branch of the
> complex function log, can be parametrized as:
> x=r*exp(i*pi), 0<=r<+inf.
> Plug x in exp(x*log(x)) to get:
> exp(x*log(x))=exp(r*(-1)*(r+i*pi))=
I suppose that, instead of (r+i*pi), you meant to have
(log(r) + i*pi).
etc.
> exp(-r^2)*exp(-i*r*pi)
> Separate real and imaginary parts,
> Im(x^x)=exp(-r^2)*sin(-r*pi)
> Now graph that for 0<=r<+inf.
With the change noted above, you would have gotten
r^(-r)*sin(-r*pi),
which one could graph, as you said, for 0<=r<+inf. I suggest
writing
the answer as
Im(x^x) = (-x)^x * sin(pi*x) for x <= 0.
David
===
Subject: Re: The function x^x
>Hi.
>Does anyone know what the graph of the function f(x) =
Imag(x^x) for x
>< 0, that is, the imaginary part of x^x (e^(x ln(x)) with
principal
>branch of
>ln(x)), looks like? I havent really been able to get a 
good
idea of
>its behavior from looking at a few points.
>>The negative axis, taken as a branch cut for the principal
branch of the
>>complex function log, can be parametrized as:
>>x=r*exp(i*pi), 0<=r<+inf.
>>Plug x in exp(x*log(x)) to get:
>>exp(x*log(x))=exp(r*(-1)*(r+i*pi))=
> I suppose that, instead of (r+i*pi), you meant to have
(log(r) + i*pi).
> etc.
and my head wasnt working properly :-)
>>exp(-r^2)*exp(-i*r*pi)
>>Separate real and imaginary parts,
>>Im(x^x)=exp(-r^2)*sin(-r*pi)
>>Now graph that for 0<=r<+inf.
> With the change noted above, you would have gotten
r^(-r)*sin(-r*pi),
> which one could graph, as you said, for 0<=r<+inf. I
suggest writing
> the answer as
> Im(x^x) = (-x)^x * sin(pi*x) for x <= 0.
> David
--
I. N. G. --- http://users.forthnet.gr/ath/jgal/
===
Subject: Re: Infinite number of football teams
<coa9tb$g4g$1@news3.infoave.net>
<coas0c$qnu$1@lust.ihug.co.nz>There is an infinite number of
football teams that play in a
>single-elimination tournament. All even numbered teams play
an odd
number
>team in the first round. The winners eliminate the losers
then they in
turn
>play one another and eliminate another half. This goes on
hundreds, if
not
>thousands, of times. When is there less than an infinite
amount of teams?
If
>there is always an infinite amount of teams no matter how
many rounds
are
>played, how do we know since an infinite amount has been
eliminated?
> Suppose that the higher-numbered team always wins. Then the
teams in
each
> round go like this:
> (1) 1 2 3 4 5 6 7 8 9 10 11 12 ...
> (2) 2 4 6 8 10 12 ...
> (3) 4 8 12 ...
> ...
Let m_0 = 1 and m_j be the smallest numbered undefeated team
of the j-th match.
The winner of the tournament is lim(j->oo) m_j unless (m_j)_j
has
no limit, as in the example above, when the tournament is a
draw.
A limited option giving the tax payers a big break for not
having to fund
infinitely many stadiums, is to let m_0 = 1 and m_(j+1) be the
first team
with team number > m_j to defeat m_j. Again lim(j->oo) m_j is
the winner
unless (m_j)_j has no limit resulting in a drawn tournament.
===
Subject: Discontinuous everywhere functions in Maple?
Is there anyway to define an expression or function in Maple
that is
discontinuous everywhere, say the characteristic of Q? I
looked in the
index
for some way to input notation x in R and x in R Q but
couldnt
find
it. Id imagine there would be extreme computational
difficulties in
determining irrationality of numbers.
Im basically trying to see if it can do Lebesgue 
integration
and how far
it
will actually go.
Jason
===
Subject: Re: Discontinuous everywhere functions in Maple?
> Is there anyway to define an expression or function in Maple
that is
> discontinuous everywhere, say the characteristic of Q? I
looked in the
index
> for some way to input notation x in R and x in R Q but
couldnt
find
> it. Id imagine there would be extreme computational
difficulties in
> determining irrationality of numbers.
Try this:
f:=x->if type(x,rational) then 1 else 0 end if;
OTOH, the comp.soft-sys.math.maple newsgroup is the
appropriate place
for questions concerning Maple.
Jose Carlos Santos
===
Subject: Re: Discontinuous everywhere functions in Maple?
> Try this:
> f:=x->if type(x,rational) then 1 else 0 end if;
Hmm, Ive encountered this thing before ...
Question: who is the inventor of this function. And for what
purpose?
Han de Bruijn
===
Subject: Re: Discontinuous everywhere functions in Maple?
>> Try this:
>> f:=x->if type(x,rational) then 1 else 0 end if;
> Hmm, Ive encountered this thing before ...
> Question: who is the inventor of this function. And for
what purpose?
Its called the Dirichlet function. I dont 
know its original
purpose,
but it is frequently encountered as a standard example of a
function that
is Lebesgue integrable but not Riemann integrable. In fact,
its an
example of what is called a simple function in measure
theory, since it
takes on only finitely many values, each on a measurable set.
--
Dave Seaman
Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: Discontinuous everywhere functions in Maple?
>Try this:
>f:=x->if type(x,rational) then 1 else 0 end if;
>>Hmm, Ive encountered this thing before ...
>>Question: who is the inventor of this function. And for
what purpose?
> Its called the Dirichlet function. [ ... ]
> is Lebesgue integrable but not Riemann integrable.
Maybe youre in the position then to answer another 
question:
> Can somebody enumerate all different integration techniques?
> Ive heard now about Lebesgue integration, Stieltjes
integral ..
And Riemann integrable too.
> As a physicist, I always calculated integrals the common
way,
> by employing the main theorem of calculus. Or numerically .
> Whats the beef with these definitions? And 
how many are
there?
I posted this in a thread about Cantors Theory but got no
answer.
Han de Bruijn
===
Subject: Re: Discontinuous everywhere functions in Maple?
>>Try this:
>>f:=x->if type(x,rational) then 1 else 0 end if;
>Hmm, Ive encountered this thing before ...
>Question: who is the inventor of this function. And for what
purpose?
>> Its called the Dirichlet function. [ ... ]
>> is Lebesgue integrable but not Riemann integrable.
This is too trivial an example; it can be changed on a set
of measure 0 to become Riemann integrable. One can produce
a function on the unit interval which only takes on the
values 0 and 1, which is Lebesgue measurable, and for which
the Lebesgue integral can be easily computed, which cannot
be so modified, as there is no interval where it does not
take on both values on a set of positive measure.
>Maybe youre in the position then to answer another 
question:
>> Can somebody enumerate all different integration
techniques?
>> Ive heard now about Lebesgue integration, Stieltjes
integral ..
There are limits of these, such as Cauchy and Denjoy.
And there is the Riemann-Hellinger integral, which has
many applications.
I doubt if there is an easy answer to this; for probability,
usually what is wanted is the Lebesgue-type integral for an
arbitrary probability measure. But there are such things
as the somewhat Riemann-Stieltjes integrals with respect to
white noise, which are much more irregular, and have more
stringent conditions. There are extensions of this as well.
>And Riemann integrable too.
>> As a physicist, I always calculated integrals the common
way,
>> by employing the main theorem of calculus. Or numerically .
As to how to calculate an integral, use anything that works.
But the Fundamental Theorem of Calculus states that the
Riemann integral, which does not involve the notion of
derivative, can be calculated by antidifferentiation.
>> Whats the beef with these definitions? And 
how many are
there?
The main idea of integral is limit of sum. There are
may ways of passing to the limit. The integral sign
itself is an old s.
--
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: Discontinuous everywhere functions in Maple?
> [ ... snip ... ] But there are such things
> as the somewhat Riemann-Stieltjes integrals with respect to
> white noise, which are much more irregular, and have more
> stringent conditions. There are extensions of this as well.
How about smoothing the irregular (possibly discrete) signal,
by convoluting it with Gauss distributions and then
integrating
the result. It seems that Ill just need a Riemann
integration,
with no significant deviations from results obtained 
otherwise.
I could do it via the Fourier domain as well.
Any remarks on that kind of procedure?
Han de Bruijn
===
Subject: Re: Discontinuous everywhere functions in Maple?
>> [ ... snip ... ] But there are such things
>> as the somewhat Riemann-Stieltjes integrals with respect to
>> white noise, which are much more irregular, and have more
>> stringent conditions. There are extensions of this as well.
>How about smoothing the irregular (possibly discrete) signal,
>by convoluting it with Gauss distributions and then
integrating
>the result. It seems that Ill just need a Riemann
integration,
>with no significant deviations from results obtained
otherwise.
>I could do it via the Fourier domain as well.
If this could be done, it would have been. As it is, it makes
a difference which point is used to multiply the length of
the interval. If W is the Wiener process, do dW is white
noise, the Ito integral from a to b, int_a^b W(t) dW(t), is
(W^2(b) - W^2(a) - b + a)/2, while the Rubin-Fisk-Stratonovich
integral is what one would expect. This is because the
Riemann type integral int_a^b (dW(t))^2 = b-a.
--
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: Discontinuous everywhere functions in Maple?
>> Its called the Dirichlet function. [ ... ]
>> is Lebesgue integrable but not Riemann integrable.
> Maybe youre in the position then to answer another
question:
>> Can somebody enumerate all different integration
techniques?
>> Ive heard now about Lebesgue integration, Stieltjes
integral ..
> And Riemann integrable too.
In addition to the references that Dik T. Winter provided,
you might look
at
<http://mathworld.wolfram.com/Measure.html>. The Lebesgue
integral is
integration with respect to Lebesgue measure. The concept of
a measure
can
be generalized, as explained on that page, and when you
consider
integration
with respect to arbitrary measures you get pretty much all
the types of
integrals.
A little further generalization is also possible. See
<http://mathworld.wolfram.com/GeneralizedFunction.html>.
--
Dave Seaman
Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: Discontinuous everywhere functions in Maple?
...
> Can somebody enumerate all different integration techniques?
> Ive heard now about Lebesgue integration, Stieltjes
integral ..
> And Riemann integrable too.
See <http://mathworld.wolfram.com/Integral.html>, especially
the section
see also at the bottom, and the quote in the first paragraph.
It will
also answer your other questions.
--
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: Discontinuous everywhere functions in Maple?
> See <http://mathworld.wolfram.com/Integral.html>,
especially the section
> see also at the bottom, and the quote in the first
paragraph. It
will
> also answer your other questions.
Indeed! Quoted from this page:
it appears that cases where these methods [i.e.,
generalizations
of the Riemann integral] are applicable and Riemanns
[definition
of the integral] is not are too rare in physics to repay the
extra
difficulty.
Well, that explains why, as a physicist, I didnt need
anything but
Han de Bruijn
===
Subject: Re: Discontinuous everywhere functions in Maple?
>> Is there anyway to define an expression or function in
Maple that is
>> discontinuous everywhere, say the characteristic of Q? I
looked in the
index
>> for some way to input notation x in R and x in R Q but
couldnt find
>> it. Id imagine there would be extreme computational
difficulties in
>> determining irrationality of numbers.
>Try this:
>f:=x->if type(x,rational) then 1 else 0 end if;
Yes, if you only give f numeric inputs.
But type only looks at the surface, so for example
type((sqrt(2)+1)^2 + (sqrt(2)-1)^2), rational) will return
false.
Somewhat better is to use is, which will make some effort
(Im not sure how much) to see whether the expression can be
simplified
to a rational:
is((sqrt(2)+1)^2 + (sqrt(2)-1)^2), rational) will return true.
On the other hand, even is is not infallible:
is((arctan(1/3)+arctan(1/2))/Pi, rational) will return false
although (arctan(1/3)+arctan(1/2))/Pi simplifies to 1/4.
>OTOH, the comp.soft-sys.math.maple newsgroup is the
appropriate place
>for questions concerning Maple.
Yes.
>>Im basically trying to see if it can do Lebesgue
integration and how
far
it
>>will actually go.
No, it cant do Lebesgue integration. All the integration
methods are meant for functions that are at least piecewise
smooth.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: A simple proof for the following?
/Carl
> Is there for the following statement a simple proof (or a
way to show
> that it is plausible)?
> Statement: Any 8-bit word can be stored as a 14-bit word,
with the
> 14-bit word having at least two zeros between every one.
> Carl
===
Subject: Re: topology...~!
> let X be a countably infinite set and
> T be a finite-complement topology on X.
> (b) show that every continuous real valued function on X
> is constant
> I thought it might be amusing to use the finite-complement
topology
> more directly, so heres an alternate proof.
> Let f: X ---> R be continuous and non-constant.
> Suppose a < b are in f(X), and let c = (a+b)/2.
> Actually, this appears to prove that you cant even have
> a non-constant continuous map into the rationals.
Heres ultimate generalization.
A space is hyper-connected when
for all open nonnul U,V, U/V /= nulset
(all open nonnul sets intersect)
Notice cofinite and cocountable spaces are hyper-connected.
hyper-connected X, Hausdorff Y, f in C(X,Y) ==> constant.
Otherwise some x,y with f(x) /= f(y)
some open U,V separate f(x), f(y)
open f^-1(U), f^-1(V) separate x,y
which contradicts X is hyper-connected
open U,V separate x,y when U,V disjoint, x in U, y in V
Thus for example a continuous function from a cocountable
space
into the complex rationals Q[i], is constant.
===
Subject: Re: topology...~!
> let X be a countably infinite set and
> T be a finite-complement topology on X.
> (b) show that every continuous real valued function on X
> is constant
> Actually, this appears to prove that you cant even have
> a non-constant continuous map into the rationals.
Minor correction made below regarding this generalization.
> A space is hyper-connected when
> for all open nonnul U,V, U/V /= nulset
> (all open nonnul sets intersect)
> Notice cofinite and cocountable spaces are hyper-connected.
> hyper-connected X, Hausdorff Y, f in C(X,Y) ==> f is
constant.
> Otherwise some x,y with f(x) /= f(y)
> some open U,V separate f(x), f(y)
> open f^-1(U), f^-1(V) separate x,y
> which contradicts X is hyper-connected
> open U,V separate x,y when U,V disjoint, x in U, y in V
> Thus for example a continuous function from a cocountable
space
> into the complex rationals Q[i], is constant.
===
Subject: Re: topology.......
> hello.....doctor~
> let X be a countably infinite set and
> T be a finite-complement topology on X.
> (a) show that X is compact and connected
> (b) show that every continuous real valued function on X
> is constant
> -----------------------------------------------------
> i can show (a).
> but i cant show (b).
> so, i need your advice.
Let F:X->R be continuous. Then F(X) is connected.
But X is countable, so F(X) is likewise countable,
and no non-singleton countable subset of R is
connected. F(X) is therefore a singleton, which
is to say, F is constant.
===
Subject: Re: Discontinuous convex function
>Has somebody got an example of a discontinuous convex
functional over a
>Banach space?
Depends on what it means to have an example - you can show
such
things exist using the Axiom of Choice.
>Pereger
************************
David C. Ullrich
===
Subject: Re: Discontinuous convex function
> Has somebody got an example of a discontinuous convex
functional over a
> Banach space?
> Pereger
A discontinuous linear functional will do, right?
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
There are a number of authors and a number of contributors
to this NG who claim that sampling of analogue
signals is represented by multiplying the incoming waveform
f(t) by a comb of Diracian Delta Functions of the form
d(t - T) and who then go on to claim that each such sample
gives rise to a contribution to the spectrum of the sampled
signal of f(T).e^(-sT).
As this claim, this apparently faulty meme, ought to be such a
fundamental part of DSP, its foundation stone in fact, it
should
be possible to prove this assertion by appeal to Diracs
properties
of his Delta Function, and to the Laplace transform. I think
that such
a proof should be a simple thing to be provided by that
body of authors and contributors if the claim were to be true.
In our training of the Laplacian method, we are presented
with a whole range of such derivations, the spectra for
d(t), u(t), t, sin(t), rect(t), sinc(t), etc etc etc. Why not
a similar derivation for f(t).d(t - T)? It cannot be a
difficult
matter for those who make the claim that sampling
is so represented!
As that body of authors and contributors seem unable
to provide such a proof and resort to side issues and
rather silly and childish ad hominem attacks when
challenged upon the matter the conclusion that I reach
is that the claim is false, and that that body of authors
and contributors hold a religious-like stance to the matter
and respond with the emotional maladjustment which is
the mark of all those who are the religious loonies of the
world today. (11/9 and the resultant war brought on
by the religionists Bush, Blair and Windsor being prime
examples of catastrophes brought on by emotional
maladjustment)
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
> There are a number of authors and a number of contributors
> to this NG who claim that sampling of analogue
> signals is represented by multiplying the incoming waveform
> f(t) by a comb of Diracian Delta Functions of the form
> d(t - T) and who then go on to claim that each such sample
> gives rise to a contribution to the spectrum of the sampled
> signal of f(T).e^(-sT).
> As this claim, this apparently faulty meme, ought to be
such a
> fundamental part of DSP, its foundation stone in fact, it
should
> be possible to prove this assertion by appeal to Diracs
properties
> of his Delta Function, and to the Laplace transform. I
think that such
> a proof should be a simple thing to be provided by that
> body of authors and contributors if the claim were to be
true.
> In our training of the Laplacian method, we are presented
> with a whole range of such derivations, the spectra for
> d(t), u(t), t, sin(t), rect(t), sinc(t), etc etc etc. Why
not
> a similar derivation for f(t).d(t - T)? It cannot be a
difficult
> matter for those who make the claim that sampling
> is so represented!
It appears that when you say spectrum you mean Laplace
Transform.
Assuming thats what you mean, then this is indeed very easy
to show.
The definition of the LT is
(i) L(f)(s) = int_0^infinity f(t) e^(-st) dt.
And the definition of delta(t-T) is that if g is continuous
(and T > 0) then
(ii) int_0^infinity g(t) delta(t-T) dt = g(T).
(Im calling it delta instead of d to avoid confusion
with the d in dt.))
Now (i) says that
L(f(t)delta(t - T))(s) = int_0^infinity f(t)delta(t-T) e^(-st)
dt,
and applying (i) with g(t) = f(t) e^(-st) shows that
int_0^infinity f(t)delta(t-T) e^(-st) dt = f(T) e^(-sT).
> As that body of authors and contributors seem unable
> to provide such a proof and resort to side issues and
> rather silly and childish ad hominem attacks when
> challenged upon the matter the conclusion that I reach
> is that the claim is false, and that that body of authors
> and contributors hold a religious-like stance to the matter
> and respond with the emotional maladjustment which is
> the mark of all those who are the religious loonies of the
> world today. (11/9 and the resultant war brought on
> by the religionists Bush, Blair and Windsor being prime
> examples of catastrophes brought on by emotional
> maladjustment)
Uh, right. Who is it whos been unable to give the proof
above? Are you sure its unable and not just unwilling?
I mean this is all covered in every book on differential
equations Ive ever seen...
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Up to the point quoted below, I agree with
input.
But. to evaluate the (unilateral) Laplace
transform, we either have to be able to
present f(t) as a function of exponentials,
in which case simple integration applies
together with the adding of exponents,
or we have to apply Integration by Parts.
(or else for a time domain product we
have to evaluate an s domain convolution)
You simply cannot make a simple suggestion
that g(t) = f(t).d(t - T) as you have done
below.
Integration by parts.....
int(UV) = U.int(v) - int[ dU.int(V) ]
which becomes even more unwieldy when
it needs to be int(UVW) as in evaluating
the LT of f(t).d(t - T)
> Now (i) says that
> L(f(t)delta(t - T))(s) = int_0^infinity f(t)delta(t-T)
e^(-st) dt,
> and applying (i) with g(t) = f(t) e^(-st) shows that
> int_0^infinity f(t)delta(t-T) e^(-st) dt = f(T) e^(-sT).
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>Up to the point quoted below, I agree with
>input.
>But. to evaluate the (unilateral) Laplace
>transform, we either have to be able to
>present f(t) as a function of exponentials,
>in which case simple integration applies
>together with the adding of exponents,
>or we have to apply Integration by Parts.
>(or else for a time domain product we
>have to evaluate an s domain convolution)
>You simply cannot make a simple suggestion
>that g(t) = f(t).d(t - T) as you have done
>below.
Uh, right. What I gave was a complete and correct
proof (well, it was extremely informal by mathematical
standards). You just saying I cant do what I did
doesnt make the proof invalid, sorry. The fact
that you dont understand something doesnt 
make
it wrong.
>Integration by parts.....
>int(UV) = U.int(v) - int[ dU.int(V) ]
>which becomes even more unwieldy when
>it needs to be int(UVW) as in evaluating
>the LT of f(t).d(t - T)
>> Now (i) says that
>> L(f(t)delta(t - T))(s) = int_0^infinity f(t)delta(t-T)
e^(-st) dt,
>> and applying (i) with g(t) = f(t) e^(-st) shows that
>> int_0^infinity f(t)delta(t-T) e^(-st) dt = f(T) e^(-sT).
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
The fact that you cannot do what you did _DOES_
make the proof invalid.
I understand fully what I am proposing.
If there is to be any criticism that there is a lack
of understanding then it seems to arise in your
own contribution.
Nevertheless, thank-you for your attempt at
assistance, wrong though it was.
> Uh, right. What I gave was a complete and correct
> proof (well, it was extremely informal by mathematical
> standards). You just saying I cant do what I did
> doesnt make the proof invalid, sorry. The fact
> that you dont understand something doesnt 
make
> it wrong.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>The fact that you cannot do what you did _DOES_
>make the proof invalid.
It would if you gave an explanation for _why_ I
cant do that, instead of just asserting it.
>I understand fully what I am proposing.
>If there is to be any criticism that there is a lack
>of understanding then it seems to arise in your
>own contribution.
>Nevertheless, thank-you for your attempt at
>assistance, wrong though it was.
I begin to understand the comments that you called
ad hominem a few posts up, although I havent seen them.
If you were interested in understanding this youd
ask me to explain why the steps in the proof were
correct instead of just asserting theyre invalid.
But youre somehow convinced that you must be right,
even though were talking about standard mathematical
facts that you can read in undergraduate textbooks.
The idea that you can be so certain youre right
and all those books are wrong is simply kooky.
>> Uh, right. What I gave was a complete and correct
>> proof (well, it was extremely informal by mathematical
>> standards). You just saying I cant do what I did
>> doesnt make the proof invalid, sorry. The fact
>> that you dont understand something doesnt 
make
>> it wrong.
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
I did give you an explanation, suggesting that
you must either integrate by parts or else
must resort to frequency-domain convolution.
>The fact that you cannot do what you did _DOES_
>make the proof invalid.
> It would if you gave an explanation for _why_ I
> cant do that, instead of just asserting it.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
> I did give you an explanation, suggesting that
> you must either integrate by parts or else
> must resort to frequency-domain convolution.
OK, lets integrate by parts:
L[f(t).delta(t-T)] = int_0^infinity 
exp(-st).f(t).delta(t-T).dt
using int_a^b udv = (uv)|_a^b - int_a^b v.du
choose u = exp(-st) => du = -s.exp(-st)dt
and,
dv = f(t).delta(t-T) dt
=> v = int_0^t x(u) delta(u-T)du
= 0 if t < T, or x(T) if t >= T
so, L[f(t).delta(t-T)] = [exp(-st).v(t)]_0^inf
- int_0^inf -s.exp(-st).v(t)dt
= 0 + s.int_T^inf x(T)exp(-st)dt
= [-x(T).exp(-st)]_T^inf
= x(T).exp(-sT)
hopefully that is satisfactory for you, excuse the sloppiness
Richard
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
The point at which I have stopped quoting you
is the point at which you seem to have made a
simple change of variable from t to u which
does not seem to support your change of limits
from a^b to 0^t (assuming that I have understood
your notation for expressing limits)
How have you calculated your change of limits?
> OK, lets integrate by parts:
> L[f(t).delta(t-T)] = int_0^infinity
exp(-st).f(t).delta(t-T).dt
> using int_a^b udv = (uv)|_a^b - int_a^b v.du
> choose u = exp(-st) => du = -s.exp(-st)dt
> and,
> dv = f(t).delta(t-T) dt
> => v = int_0^t x(u) delta(u-T)du
> = 0 if t < T, or x(T) if t >= T
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>I did give you an explanation, suggesting that
>you must either integrate by parts or else
>must resort to frequency-domain convolution.
Thats exactly the part that you simply _stated_
without proof. Its not true, by the way.
>>The fact that you cannot do what you did _DOES_
>>make the proof invalid.
>> It would if you gave an explanation for _why_ I
>> cant do that, instead of just asserting it.
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
.....[Pantomime mode ON].....
Oh, Yes! It is!
.....[Pantomime mode ON].....
> Thats exactly the part that you simply _stated_
> without proof. Its not true, by the way.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>.....[Pantomime mode ON].....
>Oh, Yes! It is!
>.....[Pantomime mode ON].....
Uh, right. Now why dont you give us a _proof_ of your
assertion that the only way to evaluate a Laplace transform
is integration by parts or convolution?
Its a funny thing. Ive been a mathematician 
for over 20
years, and this is the first time Ive ever seen 
anyone
insist that X is the _only_ way to accomplish Y.
>> Thats exactly the part that you simply _stated_
>> without proof. Its not true, by the way.
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
What other way is there to evaluate it that is not
based upon the two methods I described?
Your reputation as a debater is not enhanced by
your wont to sneer but without producing
counter-argument.
I have never asserted that X is the _only_ way
to accomplish Y
>.....[Pantomime mode ON].....
>Oh, Yes! It is!
>.....[Pantomime mode OFF].....
> Uh, right. Now why dont you give us a _proof_ of your
> assertion that the only way to evaluate a Laplace transform
> is integration by parts or convolution?
> Its a funny thing. Ive been a 
mathematician for over 20
> years, and this is the first time Ive ever 
seen anyone
> insist that X is the _only_ way to accomplish Y.
>> Thats exactly the part that you simply _stated_
>> without proof. Its not true, by the way.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>What other way is there to evaluate it that is not
>based upon the two methods I described?
I explained exactly how to evaluate this Laplace
transform correctly in my first post on the topic.
(I also just posted a more complicated proof,
using the fact that the delta function is in
some sense a limit of ordinary functions.)
>Your reputation as a debater is not enhanced by
>your wont to sneer but without producing
>counter-argument.
Guffaw.
>I have never asserted that X is the _only_ way
>to accomplish Y
Uh, yes you did. In your reply to my first post
you said in part
But. to evaluate the (unilateral) Laplace
transform, we either have to be able to
present f(t) as a function of exponentials,
in which case simple integration applies
together with the adding of exponents,
or we have to apply Integration by Parts.
(or else for a time domain product we
have to evaluate an s domain convolution)
which does say that this and that are the
only way to do something.
>>.....[Pantomime mode ON].....
>>Oh, Yes! It is!
>>.....[Pantomime mode OFF].....
>> Uh, right. Now why dont you give us a _proof_ of your
>> assertion that the only way to evaluate a Laplace transform
>> is integration by parts or convolution?
>> Its a funny thing. Ive been a 
mathematician for over 20
>> years, and this is the first time Ive ever 
seen anyone
>> insist that X is the _only_ way to accomplish Y.
> Thats exactly the part that you simply _stated_
> without proof. Its not true, by the way.
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
So first of all you say that I said, X is the _only_ way
to accomplish Y
Then you said that I said, this and that are the
only way to do something.
Youre not consistent.
In any case, I suggested three ways of proceeding
and certainly did not state that X is the _only_ way
to accomplish Y nor did I state this and that are the
only way to do something..
>I have never asserted that X is the _only_ way
>to accomplish Y
> Uh, yes you did. In your reply to my first post
> you said in part
> which does say that this and that are the
> only way to do something.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
.....[Pantomime mode ON].....
Oh, no! You didnt!
.....[Pantomime mode OFF].....
>What other way is there to evaluate it that is not
>based upon the two methods I described?
> I explained exactly how to evaluate this Laplace
> transform correctly in my first post on the topic.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Is not the same thing true of you, and therefore an
invalid comment to make?
> But youre somehow convinced that you must be right,
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>Is not the same thing true of you, and therefore an
>invalid comment to make?
No, because I gave an actual proof that I was right,
starting from the definitions.
>> But youre somehow convinced that you must be right,
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
If you say, No to the question posed below, then you
are saying that you are not convinced that you
must be right.
>Is not the same thing true of you, and therefore an
>invalid comment to make?
> No, because I gave an actual proof that I was right,
> starting from the definitions.
>> But youre somehow convinced that you must be right,
> ************************
> David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>If you say, No to the question posed below, then you
>are saying that you are not convinced that you
>must be right.
Uh, sorry, I didnt read carefully, in particular I 
didnt
realize that you were now quoting less than complete
sentences.
Yes, I am convinced Im right. The complete sentence
that I thought the is not the same true of you was
asking about was this:
But youre somehow convinced that you must be right,
even though were talking about standard mathematical
facts that you can read in undergraduate textbooks.
Yes, Im convinced Im right. No, 
its not true that
Im convinced Im right _in spite of_ the fact 
that
Im contradicting whats in hundreds of 
textbooks.
The person whos somehow convinced hes right 
and
all those books are wrong is you.
>>Is not the same thing true of you, and therefore an
>>invalid comment to make?
>> No, because I gave an actual proof that I was right,
>> starting from the definitions.
> But youre somehow convinced that you must be right,
> ************************
>> David C. Ullrich
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Therefore it is also true to say of you But youre somehow
convinced that you must be right, and therefore it was an
inappropriate thing for you to introduce.
Furthermore, if something that you read in one of your
textbooks
is your reply to a point of debate, then it is a simple
matter for
you to type it out. Far better than sneering at your
correspondent!
>If you say, No to the question posed below, then you
>are saying that you are not convinced that you
>must be right.
> Yes, I am convinced Im right. The complete sentence
> that I thought the is not the same true of you was
> asking about was this:
> But youre somehow convinced that you must be right,
> even though were talking about standard mathematical
> facts that you can read in undergraduate textbooks.
> Yes, Im convinced Im right. No, 
its not true that
> Im convinced Im right _in spite of_ the 
fact that
> Im contradicting whats in hundreds of 
textbooks.
> The person whos somehow convinced hes 
right and
> all those books are wrong is you.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
You just saying that it was a complete and correct
proof does not make the proof valid, sorry.
The fact that you dont understand something 
doesnt make
it right.
> Uh, right. What I gave was a complete and correct
> proof (well, it was extremely informal by mathematical
> standards). You just saying I cant do what I did
> doesnt make the proof invalid, sorry. The fact
> that you dont understand something doesnt 
make
> it wrong.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Hi there,
Im still a student but I recall studying the effect of delta
function
sampling
and I recall the proof by David. The way I understood the
proof of it was
that f(t).delta(t-T) = f(T) where T is often a constant delay
or shift.
You can say that
f(t) has been sampled at point T.
This occurs because the delta function is zero everywhere
except at T so
if you multiply the two, every point is zero except for the
one at T (where
delta(t-T) = 1).
This is a basic property of a delta function and can be
proved but Its
fairly obvious so im not going to.
Now you can evaluate f(t) at T, and use that in the laplace
transform like:
integral(from 0 to inf) of ( f(T).e^-(st) ) dt
if f(T) is a constant you can pull it out the front.
If you periodically sample a signal, e.g with a delta comb:
f(t).(sigma
(from n=0 to inf) of (delta( t - nT)))
function you get an array of values, such that sampled f(t) =
f(0) + f(T)
+
f(2T)........
since it is not continuous now, you can represent the signal
as f(kT).
Now you are looking at a starred laplace transform which is
no longer an
integral because it is not continuous. Instead
it is a summation which is multiplied by a complex variable:
F*(s) =
sigma(k=0 to inf) of ( f(kT).e^-(T.s.k) )
To complete the transform,
apply the proof of a geometric sum to get a neat ratio of
polynomials (in
basic problems).
The z-transform is more commonly used and is strongly related
to the
starred
laplace transform.
I hope that is what you were looking for and also that it is
all correct
(my
memory sometimes plays tricks on me)
Marc
Now that you
>Up to the point quoted below, I agree with
>input.
>But. to evaluate the (unilateral) Laplace
>transform, we either have to be able to
>present f(t) as a function of exponentials,
>in which case simple integration applies
>together with the adding of exponents,
>or we have to apply Integration by Parts.
>(or else for a time domain product we
>have to evaluate an s domain convolution)
>You simply cannot make a simple suggestion
>that g(t) = f(t).d(t - T) as you have done
>below.
> Uh, right. What I gave was a complete and correct
> proof (well, it was extremely informal by mathematical
> standards). You just saying I cant do what I did
> doesnt make the proof invalid, sorry. The fact
> that you dont understand something doesnt 
make
> it wrong.
>Integration by parts.....
>int(UV) = U.int(v) - int[ dU.int(V) ]
>which becomes even more unwieldy when
>it needs to be int(UVW) as in evaluating
>the LT of f(t).d(t - T)
>> Now (i) says that
> L(f(t)delta(t - T))(s) = int_0^infinity f(t)delta(t-T)
e^(-st) dt,
> and applying (i) with g(t) = f(t) e^(-st) shows that
> int_0^infinity f(t)delta(t-T) e^(-st) dt = f(T) e^(-sT).
> ************************
> David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>Hi there,
>Im still a student but I recall studying the effect of delta
function
>sampling
>and I recall the proof by David. The way I understood the
proof of it
was
>that f(t).delta(t-T) = f(T)
Thats not true. Whats true is that
f(t).delta(t-T) = f(T).delta(t-T).
> where T is often a constant delay or shift.
>You can say that
>f(t) has been sampled at point T.
>This occurs because the delta function is zero everywhere
except at T so
>if you multiply the two, every point is zero except for the
one at T
(where
>delta(t-T) = 1).
delta(t-T) = 1 when t = T is not true either.
Actually theres no such thing as the value of delta(t-T) 
when
t = T. The delta function is not a function. What it is is a
generalized function. What that means is a slightly long
story,
but the simplest way to think of it is probably this: 
Theres
no such thing as delta(t-T) _except_ when it appears under
an integral sign. And if g is a continuous function then
(*) int_-infinity^infinity g(t)delta(t-T) dt = 
g(T).
Really - (*) is a version of the _definition_ of what the
delta function really is.
Thats the way it has to be to make things like the Laplace
transform come out right. If you set g(t) = e^(-st) in (*)
you see that the Laplace transform of delta(t-T) is e^(-Ts),
which is what we want it to be. If on the other hand
it were true that delta(t-T) = 0 when t <> T and 1 when
t = T then that Laplace transform would be the integral
of a function that vanishes except at one point, and that
integral is _zero_.
(If we were talking about discrete variables instead of
continuous variables then what you say about delta(t-T)
would be exactly right.)
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
I have been doing more DSP lately (without sampling) and had
somehow
managed
to confuse the real-time version of the delta with the
discrete one
( delta(n) = { 1 if n = 0
{ 0 otherwise )
After reading both of your explanations I realised that my
working
was incorrect as it was a weird hybrid.
Now that I have revised the correct continuous time version,
I understand
Davids original working and also remember trying to do it
the way Airy
is doing it when I first learnt about laplace transforms. It
wasnt fun
lol.
Just to clarify to make sure im not off the rails again:
Airy, I dont understand why you cant call f(t).delta(t-T) =
g(t)? I
personally cannot see
anything wrong with it. g(t) itself will be undefined at T and
zero at t =
not T. But since it
is being integrated, the sifting theorem should still work
shouldnt it?
More importantly you should be able to go: g(t) = e^-(st).f(t)
now you have: int(from 0 to inf) of ( g(t).delta(t-T) ) dt
which is a basic sifting theorem problem which David
illustrated
previously.
Marc
>Hi there,
>Im still a student but I recall studying the effect of delta
function
>sampling
>and I recall the proof by David. The way I understood the
proof of it
was
>that f(t).delta(t-T) = f(T)
> Thats not true. Whats true is that
> f(t).delta(t-T) = f(T).delta(t-T).
> where T is often a constant delay or shift.
>You can say that
>f(t) has been sampled at point T.
>This occurs because the delta function is zero everywhere
except at T so
>if you multiply the two, every point is zero except for the
one at T
(where
>delta(t-T) = 1).
> delta(t-T) = 1 when t = T is not true either.
> Actually theres no such thing as the value of delta(t-T)
when
> t = T. The delta function is not a function. What it is is a
> generalized function. What that means is a slightly long
story,
> but the simplest way to think of it is probably this:
Theres
> no such thing as delta(t-T) _except_ when it appears under
> an integral sign. And if g is a continuous function then
> (*) int_-infinity^infinity g(t)delta(t-T) dt = 
g(T).
> Really - (*) is a version of the _definition_ of what the
> delta function really is.
> Thats the way it has to be to make things like the 
Laplace
> transform come out right. If you set g(t) = e^(-st) in (*)
> you see that the Laplace transform of delta(t-T) is e^(-Ts),
> which is what we want it to be. If on the other hand
> it were true that delta(t-T) = 0 when t <> T and 1 when
> t = T then that Laplace transform would be the integral
> of a function that vanishes except at one point, and that
> integral is _zero_.
> (If we were talking about discrete variables instead of
> continuous variables then what you say about delta(t-T)
> would be exactly right.)
> ************************
> David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
In the evaluation of integral transforms, the sifting
theory will work if it is present as one of the
integrals.
> Airy, I dont understand why you cant call f(t).delta(t-T)
= g(t)? I
> personally cannot see
> anything wrong with it. g(t) itself will be undefined at T
and zero at t
=
> not T. But since it
> is being integrated, the sifting theorem should still work
shouldnt it?
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
It doesnt agree with integration by parts.
> Airy, I dont understand why you cant call f(t).delta(t-T)
= g(t)? I
> personally cannot see
> anything wrong with it.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>It doesnt agree with integration by parts.
The integration by parts that youre doing is _wrong_.
The fact that integration by parts works is a theorem.
Theorems have hypotheses - whatever delta(t-T) is,
its not a continuous function.
>> Airy, I dont understand why you cant call 
f(t).delta(t-T)
= g(t)? I
>> personally cannot see
>> anything wrong with it.
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Show where it is wrong, or else resort
to infantile sneering.
Oh!!! Youre doing that already!
>It doesnt agree with integration by parts.
> The integration by parts that youre doing is _wrong_.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>Show where it is wrong, or else resort
>to infantile sneering.
Ok. Took me a minute to find the place
where you showed us the integration
by parts:
Certainly, if you integrate by parts, you would choose
int(f(t).d(t-T)) as the integrated bit to yield f(T), but when
I try this, I get 0!......
int(UV) = U.int(V) - int[dU.int(V)] giving.....
int(+/-inf)(f(t).d(t-T).e^(-st)) as .....
with f(t).d(t-T) as V and e^(-sT) as U .....
f(T).e^(-st) - int(e^(-st)/-s . f(T)).....
f(T).e^(-st) - -s.e^(-st)/-s . f(T).....
f(T).e^(-sT) - f(T).e^(-sT)....
0.
Your first step,
f(T).e^(-st) - int(e^(-st)/-s . f(T))
is already wrong. My guess is because of a confusion
over definite integrals versus antiderivatives, ie
indefinite integrals. The (definite) integral 
from
0 to infinity of f(t).d(t-T) is indeed f(T). But
what we need here is an antiderivative.
An antiderivative of f(t).d(t-T) is given by
F(t), where F(t) = 0 for t < T and f(T) for t > T.
So that line should be
F(t).e^(-st) - int(e^(-st)/-s . F(t)).
Golly, I assumed that you had some idea how to
do simple calculus - looking at this I realize
you took the derivative wrong. What we really
get is this:
F(t).e^(-st) - int(e^(-st)(-s) . F(t)).
Now when we put in the limits t = 0 to t = infinity
the first term vanishes since F(0) = 0 and e^(-st)
tends to 0 as t -> infinty (at least if s > 0; if s < 0
this integration by parts is not going to work.)
So out Laplace transform becomes
- int_0^infinity (e^(-st)/-s . F(t))
Since F(t) = 0 for t < T and f(T) for t > T
this equals
f(T) int_T^infinity (e^(-st)s ).
Again, its easy to evaluate the last integral;
int_T^infinity (e^(-st)s ) = e^(-sT), so we
finally get f(T)e^(-sT) for the Laplace transform.
>Oh!!! Youre doing that already!
So now weve found _two_ mistakes in your
integration by parts, confusing antiderivatives
and definite integrals and taking a very simple
derivative incorrectly.
>>It doesnt agree with integration by parts.
>> The integration by parts that youre doing is _wrong_.
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Dirac did not define his function a as generalised function
because
the concept did not exist in 1930.
The introduction of generalised functions and the Theory Of
Distributions is entirely irrelevant when no mathematics
resulting from that theory is applied.
The product of a generalised function is another generalised
function. No generalised functions appear in our DSPs.
> Actually theres no such thing as the value of delta(t-T)
when
> t = T. The delta function is not a function. What it is is a
> generalized function.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>Dirac did not define his function a as generalised function
because
>the concept did not exist in 1930.
I didnt say that Dirac used the term generalized function.
The delta function _is_ a generalized function, aka
distribution.
Diracs delta function was one of the first 
examples of the
_concept_, which was named and put on a rigorous mathematical
basis by Schwarz years later.
>The introduction of generalised functions and the Theory Of
>Distributions is entirely irrelevant when no mathematics
>resulting from that theory is applied.
>The product of a generalised function is another generalised
>function. No generalised functions appear in our DSPs.
>> Actually theres no such thing as the value of delta(t-T)
when
>> t = T. The delta function is not a function. What it is is
a
>> generalized function.
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
The Delta function was on a rigorous mathematical basis
when Dirac defined it, although he spoke to the contrary.
The idea of a function being the limit of another function
has always been entirely respectable and is found in
the evaluation of every derivative.
The introduction of generalised functions and the Theory Of
Distributions is entirely irrelevant when no mathematics
resulting from that theory is applied.
>Dirac did not define his function a as generalised function
because
>the concept did not exist in 1930.
> I didnt say that Dirac used the term generalized 
function.
> The delta function _is_ a generalized function, aka
distribution.
> Diracs delta function was one of the first 
examples of the
> _concept_, which was named and put on a rigorous
mathematical
> basis by Schwarz years later.
>The introduction of generalised functions and the Theory Of
>Distributions is entirely irrelevant when no mathematics
>resulting from that theory is applied.
>The product of a generalised function is another generalised
>function. No generalised functions appear in our DSPs.
>> Actually theres no such thing as the value of delta(t-T)
when
>> t = T. The delta function is not a function. What it is is
a
>> generalized function.
> ************************
> David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>The Delta function was on a rigorous mathematical basis
>when Dirac defined it, although he spoke to the contrary.
>The idea of a function being the limit of another function
>has always been entirely respectable and is found in
>the evaluation of every derivative.
Yes, but the delta function is _not_ the limit of
a sequence of ordinary functions, at least not in
the sense that one speaks of limits in calculus.
Its not a function at all.
But it is true that one can define the _action_
of a delta function as a limit - that is, the
linear functional defined by the delta function
is indeed a limit of the linear functionals defined
by a sequence of ordinary functions. If you want to
look at it that way its very easy to show from that
point of view that the Laplace transform is what it
is:
Fix T, and say f_n(t) = n for T < t < T + 1/n,
f_n(t) = 0 for other t. Then the f_n _do_ converge
to delta(t-T) in a certain sense - in particular
the Laplace transform of delta(t-T) _is_ the
limit of the Laplace transform of f_n as n -> infinity.
Lets say F_n is the LT of f_n. Then
F_n(s) = n int_T^{T+1/n} e^{-st} dt
= n (e^{-s(T+1/n)) - e^{-sT}) / (-s)
= e^{-sT} (1 - e^{-s/n}) / (s/n).
Its an easy calculus exercise to show that
(1 - e^{-s/n}) / (s/n) -> 1 as n -> infinity.
So the limit of F_n(s) is e^{-sT}.
>The introduction of generalised functions and the Theory Of
>Distributions is entirely irrelevant when no mathematics
>resulting from that theory is applied.
>>Dirac did not define his function a as generalised function
because
>>the concept did not exist in 1930.
>> I didnt say that Dirac used the term generalized 
function.
>> The delta function _is_ a generalized function, aka
distribution.
>> Diracs delta function was one of the first 
examples of the
>> _concept_, which was named and put on a rigorous
mathematical
>> basis by Schwarz years later.
>>The introduction of generalised functions and the Theory Of
>>Distributions is entirely irrelevant when no mathematics
>>resulting from that theory is applied.
>The product of a generalised function is another generalised
>>function. No generalised functions appear in our DSPs.
>> Actually theres no such thing as the value of delta(t-T)
when
> t = T. The delta function is not a function. What it is is a
> generalized function.
> ************************
>> David C. Ullrich
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Make your mind up - first you say that it is not
a function, then you say that it is the delta function.
.
> Its not a function at all.
> But it is true that one can define the _action_
> of a delta function as a limit
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
So you have argued a derivation, without using the
concept of generalised functions which were in any case
irrelevant, to arrive at a result that wasnt in dispute.
I wonder why?
> Its an easy calculus exercise to show that
> (1 - e^{-s/n}) / (s/n) -> 1 as n -> infinity.
> So the limit of F_n(s) is e^{-sT}.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>So you have argued a derivation, without using the
>concept of generalised functions which were in any case
>irrelevant, to arrive at a result that wasnt in dispute.
The result I arrived at was that the Laplace transform
of delta(t-T) is e^{-sT}. That wasnt in dispute?
>I wonder why?
Because youre too stupid or stubborn to agree with the
much simpler argument proving the same thing.
>> Its an easy calculus exercise to show that
>> (1 - e^{-s/n}) / (s/n) -> 1 as n -> infinity.
>> So the limit of F_n(s) is e^{-sT}.
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
.....[Pantomime mode ON].....
Oh, yes! It _IS_ a function.
.....[Pantomime mode OFF].....
And I have never defined it as a limit
of a sequence of functions any more
than I would define a derivative as
such a limit.
The limit employed is very much the same
limit as used in calculus; t > 0.
>The Delta function was on a rigorous mathematical basis
>when Dirac defined it, although he spoke to the contrary.
>The idea of a function being the limit of another function
>has always been entirely respectable and is found in
>the evaluation of every derivative.
> Yes, but the delta function is _not_ the limit of
> a sequence of ordinary functions, at least not in
> the sense that one speaks of limits in calculus.
> Its not a function at all.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
No, thats not true and is not supported in any way
by the properties of the Diracian Delta Function.
There are no simple multiplication properties
unless expressed under an integral sign.
What is true is that int(+/- inf)(f(t).d(t-T)) is f(T),
but the Delta is no longer present having been
integrated out.
> Thats not true. Whats true is that
> f(t).delta(t-T) = f(T).delta(t-T).
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>No, thats not true and is not supported in any way
>by the properties of the Diracian Delta Function.
>There are no simple multiplication properties
>unless expressed under an integral sign.
You really dont know nearly as much about any
of this as you think you do.
delta is a _measure_. The product of a measure
and a continuous function is a standard thing;
by _definition_ saying that f(t).delta(t-T) = f(T).delta(t-T)
means that if g is a continuous function then
(*) int g(t).f(t).delta(t-T) = int g(t).f(T).delta(t-T)
(where int is the integral from -infinity to 
infinity
in general, or from 0 to infinity here). And (*) is
true, because both sides equal g(T)f(T).
>What is true is that int(+/- inf)(f(t).d(t-T)) is f(T),
>but the Delta is no longer present having been
>integrated out.
Yes, thats true. Doesnt imply that 
f(t).delta(t-T) =
f(T).delta(t-T)
is false.
>> Thats not true. Whats true is that
>> f(t).delta(t-T) = f(T).delta(t-T).
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
You cannot prove something merely by defining
it to be correct.
> delta is a _measure_. The product of a measure
> and a continuous function is a standard thing;
> by _definition_ saying that f(t).delta(t-T) = 
f(T).delta(t-T)
> means that if g is a continuous function then
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
To use your own argument, just saying it, does
not prove it one way or the other.
> delta is a _measure_. The product of a measure
> and a continuous function is a standard thing;
> by _definition_ saying that f(t).delta(t-T) = 
f(T).delta(t-T)
> means that if g is a continuous function then
> (*) int g(t).f(t).delta(t-T) = int g(t).f(T).delta(t-T)
> (where int is the integral from -infinity to 
infinity
> in general, or from 0 to infinity here). And (*) is
> true, because both sides equal g(T)f(T).
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>To use your own argument, just saying it, does
>not prove it one way or the other.
Of course not. Here the question was about a _definition_.
I didnt claim to have proved I was giving a correct
definition.
The definition I gave _is_ correct, and perfectly
standard, your ignorance of the matter notwithstanding.
>> delta is a _measure_. The product of a measure
>> and a continuous function is a standard thing;
>> by _definition_ saying that f(t).delta(t-T) =
f(T).delta(t-T)
>> means that if g is a continuous function then
>> (*) int g(t).f(t).delta(t-T) = int g(t).f(T).delta(t-T)
>> (where int is the integral from -infinity to 
infinity
>> in general, or from 0 to infinity here). And (*) is
>> true, because both sides equal g(T)f(T).
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Your reputation will not be enhanced by your
tendency to resort to ad hominem remarks.
> The definition I gave _is_ correct, and perfectly
> standard, your ignorance of the matter notwithstanding.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Looking at the work done by Dirac, his Delta function is a
function
defined as a limit.
No problem there. All derivatives are so defined.
>No, thats not true and is not supported in any way
>by the properties of the Diracian Delta Function.
>There are no simple multiplication properties
>unless expressed under an integral sign.
> You really dont know nearly as much about any
> of this as you think you do.
> delta is a _measure_. The product of a measure
> and a continuous function is a standard thing;
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Shame on you for making personal remarks.
The conclusion about your university-style
email address is that you are a first-year man.
>No, thats not true and is not supported in any way
>by the properties of the Diracian Delta Function.
>There are no simple multiplication properties
>unless expressed under an integral sign.
> You really dont know nearly as much about any
> of this as you think you do.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
> No, thats not true and is not supported in any way
> by the properties of the Diracian Delta Function.
> There are no simple multiplication properties
> unless expressed under an integral sign.
> What is true is that int(+/- inf)(f(t).d(t-T)) is f(T),
> but the Delta is no longer present having been
> integrated out.
> Thats not true. Whats true is that
> f(t).delta(t-T) = f(T).delta(t-T).
Perhaps you should try to understand it a bit more
before saying that someone is making false statements.
The above means that (ie is equivalent with):
int_{-oo}^oo f(t).delta(t-T) dt = int_{-oo}^oo
f(T).delta(t-T) dt
The left handside simplifies to f(T). The right handside
simplies
to f(T). If you dont see the latter, define 
h(t)=1 (for all
t). Now
int_{-oo}^oo f(T).delta(t-T) dt = int_{-oo}^oo
f(T).h(t).delta(t-T) dt
= f(T) . int_{-oo}^oo h(t).delta(t-T) dt = f(T).h(T) = f(T)
since
h(T)=1.
Hence they are indeed equal.
Wilbert
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Multiplying one or both sides of an identity by 1
does not value to your argument.
>If you dont see the latter, define h(t)=1 (for 
all t). Now
> int_{-oo}^oo f(T).delta(t-T) dt = int_{-oo}^oo
f(T).h(t).delta(t-T) dt
> = f(T) . int_{-oo}^oo h(t).delta(t-T) dt = f(T).h(T) = f(T)
since
> h(T)=1.
> Hence they are indeed equal.
> Wilbert
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Is it necessarily true that if two integrals are
equal, then the functions under those integrals
are equal?
> No, thats not true and is not supported in any way
> by the properties of the Diracian Delta Function.
> There are no simple multiplication properties
> unless expressed under an integral sign.
> What is true is that int(+/- inf)(f(t).d(t-T)) is f(T),
> but the Delta is no longer present having been
> integrated out.
> Thats not true. Whats true is that
> f(t).delta(t-T) = f(T).delta(t-T).
> The above means that (ie is equivalent with):
> int_{-oo}^oo f(t).delta(t-T) dt = int_{-oo}^oo
f(T).delta(t-T) dt
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>Is it necessarily true that if two integrals are
>equal, then the functions under those integrals
>are equal?
Of course not.
>> No, thats not true and is not supported in any way
>> by the properties of the Diracian Delta Function.
>> There are no simple multiplication properties
>> unless expressed under an integral sign.
>> What is true is that int(+/- inf)(f(t).d(t-T)) is f(T),
>> but the Delta is no longer present having been
>> integrated out.
>> Thats not true. Whats true is that
>> f(t).delta(t-T) = f(T).delta(t-T).
>> The above means that (ie is equivalent with):
>> int_{-oo}^oo f(t).delta(t-T) dt = int_{-oo}^oo
f(T).delta(t-T) dt
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
Perhaps you should pay attention to the discussion rather than
resorting in the first instance to personal remarks?
> Perhaps you should try to understand it a bit more
> before saying that someone is making false statements.
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
>> No, thats not true and is not supported in any way
>> by the properties of the Diracian Delta Function.
>> There are no simple multiplication properties
>> unless expressed under an integral sign.
>> What is true is that int(+/- inf)(f(t).d(t-T)) is f(T),
>> but the Delta is no longer present having been
>> integrated out.
>> Thats not true. Whats true is that
> f(t).delta(t-T) = f(T).delta(t-T).
>>Perhaps you should try to understand it a bit more
>before saying that someone is making false statements.
>The above means that (ie is equivalent with):
>int_{-oo}^oo f(t).delta(t-T) dt = int_{-oo}^oo
f(T).delta(t-T) dt
Well actually no, thats not what it means. See my reply
to Airy.
>The left handside simplifies to f(T). The right handside
simplies
>to f(T). If you dont see the latter, define 
h(t)=1 (for all
t). Now
>int_{-oo}^oo f(T).delta(t-T) dt = int_{-oo}^oo
f(T).h(t).delta(t-T) dt
>= f(T) . int_{-oo}^oo h(t).delta(t-T) dt = f(T).h(T) = f(T)
since
>h(T)=1.
>Hence they are indeed equal.
>Wilbert
************************
David C. Ullrich
===
Subject: Re: Derivation Of The Spectrum Due To f(t).d(t -
T)?????
What you say below is not supported by the
properties of the Diracian Delta function. It has
to be under an integral sign to get the f(T).
> Im still a student but I recall studying the effect of
delta function
> sampling
> and I recall the proof by David. The way I understood the
proof of it
was
> that f(t).delta(t-T) = f(T) where T is often a constant
delay or shift.
> You can say that
> f(t) has been sampled at point T.
===
Subject: Re: Derivation Of The Spectrom Due To f(t).d(t -
T)?????
with as follows.....
You need to do int(+/-inf)(f(t).d(t-T).e^(-st)) to
determine your result.
You simply cannot say that f(t).d(t-T) yields f(T) unless
you do so under an integral. This arises from the fundamental
properties of the Diracian Delta function .
Certainly, if you integrate by parts, you would choose
int(f(t).d(t-T)) as the integrated bit to yield f(T), but when
I try this, I get 0!......
int(UV) = U.int(V) - int[dU.int(V)] giving.....
int(+/-inf)(f(t).d(t-T).e^(-st)) as .....
with f(t).d(t-T) as V and e^(-sT) as U .....
f(T).e^(-st) - int(e^(-st)/-s . f(T)).....
f(T).e^(-st) - -s.e^(-st)/-s . f(T).....
f(T).e^(-sT) - f(T).e^(-sT)....
0.
What I seek is a sound mathematical proof of the claim
that f(t).d(t-T) gives rise to a spectrum contribution
of f(T).e^-(st), and claiming that f(t).d(t-T) equals
f(T) is not sound. We use the properties of the Diracian
Delta Function in so many other aspects of signal
processing that it is just not right to pull-a-fast-one.
> Now you can evaluate f(t) at T, and use that in the laplace
transform
like:
> integral(from 0 to inf) of ( f(T).e^-(st) ) dt
> if f(T) is a constant you can pull it out the front.
===
Subject: Re: Infantile authours degrading this NG.
It is interesting the size of the amount of posts.
If I am wrong in what I say, it would be a simple
matter to ignore me and what I posit.
That so many people, people who would seek to
represent themselves as authorities, respond with
emotional postings that do not address the
points raised, seems to suggest that I am not wrong, and that
they are embarrassed about a basic failing in their
professed knowledge and so feel threatened.
the amount of posts, posts that are non-technical and
of an undesirable ad hominem style has reached such a
scale that I have binned them all this morning.
This leaves just one point that has not yet been addressed,
and
that is, that any mathematical analysis of a real system must,
to be respectable, deal with measurements of that system.
There is no system of which I am aware that has sampling
pulses whose measurements match those of the Diracian
Impulse...
1. Their amplitudes do not approach infinity.
2. Their areas do not approach unity.
3. In any case, the operation of f(t).d(t - T) is not
defined unless under an integral sign, and cannot be
evaluated unless under that integral sign.
4. The evaluation of the spectrum of the Diracian relies
on it being over all time, -oo^+oo. It is improper, therefore,
to attempt to evaluate other operations using the Diracian
with a reduced domain, and yet still rely on the spectrum
derivation.
> of posts that it has generated, it stands to reason that
you are
> unlikely to get a response suitable to you from them.
===
Subject: Re: Infantile authours degrading this NG.
> It is interesting the size of the amount of posts.
Airy
The expression Bollocks springs to mind as far as your
diatribes are
concerned.
Marco
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Infantile authours degrading this NG.
Why dont you and James Harris collaborate
and solve all the major outstanding problems.
>It is interesting the size of the amount of posts.
>If I am wrong in what I say, it would be a simple
>matter to ignore me and what I posit.
>That so many people, people who would seek to
>represent themselves as authorities, respond with
>emotional postings that do not address the
>points raised, seems to suggest that I am not wrong, and that
>they are embarrassed about a basic failing in their
>professed knowledge and so feel threatened.
>the amount of posts, posts that are non-technical and
>of an undesirable ad hominem style has reached such a
>scale that I have binned them all this morning.
>This leaves just one point that has not yet been addressed,
and
>that is, that any mathematical analysis of a real system
must,
>to be respectable, deal with measurements of that system.
>There is no system of which I am aware that has sampling
>pulses whose measurements match those of the Diracian
>Impulse...
>1. Their amplitudes do not approach infinity.
>2. Their areas do not approach unity.
>3. In any case, the operation of f(t).d(t - T) is not
>defined unless under an integral sign, and cannot be
>evaluated unless under that integral sign.
>4. The evaluation of the spectrum of the Diracian relies
>on it being over all time, -oo^+oo. It is improper,
therefore,
>to attempt to evaluate other operations using the Diracian
>with a reduced domain, and yet still rely on the spectrum
>derivation.
>> of posts that it has generated, it stands to reason that
you are
>> unlikely to get a response suitable to you from them.
===
Subject: Re: Infantile authours degrading this NG.
Classic crackpot. If nobody says youre wrong you must
be right. But if a large number of people say youre wrong
it follows you must be right as well.
Things must be very pleasant in your little world.
Btw, since you dont seem to have noticed my post
where I reply to your request to show _where_ your
integration by parts fails: When I said that
integration by parts does not apply because delta
is not a continuous I was assuming that you
were making one sort of moderately subtle error.
Hadnt looked at the actual argument you gave.
In case you didnt see that post, in your
integration by parts you simply do the basic
calculus wrong.
>It is interesting the size of the amount of posts.
>If I am wrong in what I say, it would be a simple
>matter to ignore me and what I posit.
>That so many people, people who would seek to
>represent themselves as authorities, respond with
>emotional postings that do not address the
>points raised, seems to suggest that I am not wrong, and that
>they are embarrassed about a basic failing in their
>professed knowledge and so feel threatened.
>the amount of posts, posts that are non-technical and
>of an undesirable ad hominem style has reached such a
>scale that I have binned them all this morning.
>This leaves just one point that has not yet been addressed,
and
>that is, that any mathematical analysis of a real system
must,
>to be respectable, deal with measurements of that system.
>There is no system of which I am aware that has sampling
>pulses whose measurements match those of the Diracian
>Impulse...
>1. Their amplitudes do not approach infinity.
>2. Their areas do not approach unity.
>3. In any case, the operation of f(t).d(t - T) is not
>defined unless under an integral sign, and cannot be
>evaluated unless under that integral sign.
>4. The evaluation of the spectrum of the Diracian relies
>on it being over all time, -oo^+oo. It is improper,
therefore,
>to attempt to evaluate other operations using the Diracian
>with a reduced domain, and yet still rely on the spectrum
>derivation.
>> of posts that it has generated, it stands to reason that
you are
>> unlikely to get a response suitable to you from them.
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
(1/root(2.PI.N)).e^(t^2/n^2) is pretty continuous.
: When I said that
> integration by parts does not apply because delta
> is not a continuous
===
Subject: Re: Infantile authours degrading this NG.
You claim to be a mathematician of 20 years standing.
I suggest that you do not try to be a teacher of mathematics.
Mathematics is an abstract subject for which your students
must be relaxed in mind in order to be able to take on
abstractions. Your emotive and insulting posts
will disrupt such a state of mind. In my time as a part-time
tutor of mathematics to adults I came across this time and
time again - people who have been put off mathematics, not
by the subject per se, but by the aggression and personal
attacks addressed to them by their teachers.
Your infantile tirade below illustrates well the persons
proscribed
Shame on you.
(I notice that you do not address the mathematical points
below)
> Classic crackpot. If nobody says youre wrong you must
> be right. But if a large number of people say youre wrong
> it follows you must be right as well.
> Things must be very pleasant in your little world.
> Btw, since you dont seem to have noticed my post
> where I reply to your request to show _where_ your
> integration by parts fails: When I said that
> integration by parts does not apply because delta
> is not a continuous I was assuming that you
> were making one sort of moderately subtle error.
> Hadnt looked at the actual argument you gave.
> In case you didnt see that post, in your
> integration by parts you simply do the basic
> calculus wrong.
>It is interesting the size of the amount of posts.
>If I am wrong in what I say, it would be a simple
>matter to ignore me and what I posit.
>That so many people, people who would seek to
>represent themselves as authorities, respond with
>emotional postings that do not address the
>points raised, seems to suggest that I am not wrong, and that
>they are embarrassed about a basic failing in their
>professed knowledge and so feel threatened.
>the amount of posts, posts that are non-technical and
>of an undesirable ad hominem style has reached such a
>scale that I have binned them all this morning.
>This leaves just one point that has not yet been addressed,
and
>that is, that any mathematical analysis of a real system
must,
>to be respectable, deal with measurements of that system.
>There is no system of which I am aware that has sampling
>pulses whose measurements match those of the Diracian
>Impulse...
>1. Their amplitudes do not approach infinity.
>2. Their areas do not approach unity.
>3. In any case, the operation of f(t).d(t - T) is not
>defined unless under an integral sign, and cannot be
>evaluated unless under that integral sign.
>4. The evaluation of the spectrum of the Diracian relies
>on it being over all time, -oo^+oo. It is improper,
therefore,
>to attempt to evaluate other operations using the Diracian
>with a reduced domain, and yet still rely on the spectrum
>derivation.
>> of posts that it has generated, it stands to reason that
you are
>> unlikely to get a response suitable to you from them.
> ************************
> David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
> In my time as a part-time tutor of mathematics to adults
the more of this thread I read the more I pity those poor
souls
===
Subject: Re: Infantile authours degrading this NG.
> You claim to be a mathematician of 20 years standing.
> I suggest that you do not try to be a teacher of
mathematics.
> Mathematics is an abstract subject for which your students
> must be relaxed in mind in order to be able to take on
> abstractions.
What qualifies a failed software engineer (remember
Westinghouse) to make
such a statement?
What are YOUR qualifications?
Did you ever actually get a degree?
===
Subject: Re: Infantile authours degrading this NG.
> You claim to be a mathematician of 20 years standing.
> I suggest that you do not try to be a teacher of
mathematics.
> Mathematics is an abstract subject for which your students
> must be relaxed in mind in order to be able to take on
> abstractions. Your emotive and insulting posts
> will disrupt such a state of mind. In my time as a part-time
> tutor of mathematics to adults
Sounds a bit far fetched to me. Was that before or after the
Westinghouse
affair?
===
Subject: Re: Infantile authours degrading this NG.
> You claim to be a mathematician of 20 years standing.
<personal attacks on others snipped>
Yet, he posted a detailed analysis of your errors in your
Ôintegration by
parts of the posited function.
If you stay focused on the problem at hand, you would see
that your mistakes
have been pointed out to you. Then you can learn from your
mistakes.
daestrom
===
Subject: Re: Infantile authours degrading this NG.
Unfortunately, although he claimed to have posted such, it was
masked by his over-riding urge to be insulting and was
therefore
not seen by me.
personal remarks, then I would be happy to respond. I have
challenged a number of posers in this respect, but all have
declined to
reply without their spleen aventing, which leads me to
question
that they had a genuine contribution to offer in the first
place.
Sic transit gloria Mundi.
have anything to do with childish outbursts.
There were no personal attacks for you to snip. Defence
in asserting the right of reply, maybe, but no attacks.
> You claim to be a mathematician of 20 years standing.
> <personal attacks on others snipped> Yet, he posted a
detailed analysis of your errors in your Ôintegration by
> parts of the posited function.
> If you stay focused on the problem at hand, you would see
that your
mistakes
> have been pointed out to you. Then you can learn from your
mistakes.
===
Subject: Re: Infantile authours degrading this NG.
Neener, and Neener
that...
> Unfortunately, although he claimed to have posted such, it
was
> masked by his over-riding urge to be insulting and was
therefore
> not seen by me.
Because of your over-riding urge to be insulted.
--
Checkmate
all rights reserved
===
Subject: Re: Infantile authours degrading this NG.
> Unfortunately, although he claimed to have posted such, it
was
> masked by his over-riding urge to be insulting and was
therefore
> not seen by me.
Well, here is what David C. Ullrich posted.... (lines with >>
are from Airy
R. Beans previous post, those preceded by 
Ô> are from David
C. Ullrichs
pose, my comments are embedded with Ô<*<)
>>Show where it is wrong, or else resort
>>to infantile sneering.
>Ok. Took me a minute to find the place
>where you showed us the integration
>by parts:
>>Certainly, if you integrate by parts, you would choose
>>int(f(t).d(t-T)) as the integrated bit to yield f(T), but
when
>>I try this, I get 0!......
>>int(UV) = U.int(V) - int[dU.int(V)] giving.....
>> int(+/-inf)(f(t).d(t-T).e^(-st)) as .....
>>with f(t).d(t-T) as V and e^(-sT) as U .....
>>f(T).e^(-st) - int(e^(-st)/-s . f(T)).....
>>f(T).e^(-st) - -s.e^(-st)/-s . f(T).....
>>f(T).e^(-sT) - f(T).e^(-sT)....
>>0.
>Your first step,
> f(T).e^(-st) - int(e^(-st)/-s . f(T))
>is already wrong. My guess is because of a confusion
>over definite integrals versus antiderivatives, ie
>indefinite integrals. The (definite) integral 
from
>0 to infinity of f(t).d(t-T) is indeed f(T). But
>what we need here is an antiderivative.
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>So that line should be
> F(t).e^(-st) - int(e^(-st)/-s . F(t)).
<*<daestrom snipped from David C. Ullrich quote>*>Now when we
put in the limits t = 0 to t = infinity
>the first term vanishes since F(0) = 0 and e^(-st)
>tends to 0 as t -> infinty (at least if s > 0; if s < 0
>this integration by parts is not going to work.)
>So out Laplace transform becomes
> - int_0^infinity (e^(-st)/-s . F(t))
>Since F(t) = 0 for t < T and f(T) for t > T
>this equals
> f(T) int_T^infinity (e^(-st)s ).
<*<daestrom: At this point DCU changes from 
Ôe^(-st)/-s to
Ôe^(-st)s. It
is apparent the Ô- was taken out front. But 
since the
beginning, dU should
have been Ô-e^(-st)s with the 
Ôs in the numerator, not
denominator. I
suspect this error came from ARBs original attempt at the
integration by
>Again, its easy to evaluate the last integral;
>int_T^infinity (e^(-st)s ) = e^(-sT), so we
>finally get f(T)e^(-sT) for the Laplace transform.
<*<daestrom: Snipped the rest of David C. Ullrichs post >*>
daestrom
===
Subject: Re: Infantile authours degrading this NG.
Where do you get that from?
The definite integral int -oo^+oo f(t).d(t-T)
gives us f(T), but there is no information
given to us about the indefinite integral
nor of the anti-derivative.
Indeed, the definite integral gives us f(T), which
is a constant function having no aspect of t in its
evaluation and which graphs as a horizontal
line from -oo to +oo. This comes from the
basic definitions of the Diracian.
Your claimed anti-derivative has a strong
dependence on t, and therefore has not
come from anything formally defined.
Are you making this up as you go along?
Well, here is what David C. Ullrich posted..
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>So that line should be
===
Subject: Re: Infantile authours degrading this NG.
>Where do you get that from?
Just explained that in a different post.
>The definite integral int -oo^+oo f(t).d(t-T)
>gives us f(T), but there is no information
>given to us about the indefinite integral
>nor of the anti-derivative.
>Indeed, the definite integral gives us f(T), which
>is a constant function having no aspect of t in its
>evaluation and which graphs as a horizontal
>line from -oo to +oo. This comes from the
>basic definitions of the Diracian.
True, and of no relevance whatever to the question
of what an antiderivative is.
Hint: Youre _really_ showing over and over that
you have no understanding of _basic_ calculus.
You keep kindly informing me that Im making
a laughingstock of myself, presumably youd
want to know how utterly ignorant youre
revealing yourself to be.
PS: Look up the definition of ad hominem.
Im not saying you must be wrong because 
youre
stupid, Im saying youre looking very stupid
because the things youre saying are wrong,
at such a basic level.
>Your claimed anti-derivative has a strong
>dependence on t, and therefore has not
>come from anything formally defined.
>Are you making this up as you go along?
>Well, here is what David C. Ullrich posted..
>>An antiderivative of f(t).d(t-T) is given by
>>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>>So that line should be
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
>>Where do you get that from?
>Just explained that in a different post.
>>The definite integral int -oo^+oo f(t).d(t-T)
>>gives us f(T), but there is no information
>>given to us about the indefinite integral
>>nor of the anti-derivative.
>>Indeed, the definite integral gives us f(T), which
>>is a constant function having no aspect of t in its
>>evaluation and which graphs as a horizontal
>>line from -oo to +oo. This comes from the
>>basic definitions of the Diracian.
>True, and of no relevance whatever to the question
>of what an antiderivative is.
>Hint: Youre _really_ showing over and over that
>you have no understanding of _basic_ calculus.
>You keep kindly informing me that Im making
>a laughingstock of myself, presumably youd
>want to know how utterly ignorant youre
>revealing yourself to be.
I tangled with him in another newsgroup. He is untrainable
and is proud of that fact.
/BAH
<snip>
/BAH
Subtract a hundred and four for e-mail.
===
Subject: Re: Infantile authours degrading this NG.
An ad hominem attack from you of no value to the
discussion.
Shame on you.
You should no better.
I do not keep kindly informing you that youre making
a laughingstock of myself. I have suggested that once
and once only.
Please get your facts correct.
You say that the things quoted below from me are true
and then you say that Im utterly ignorant.
You are inconsistent.
>Where do you get that from?
> Just explained that in a different post.
>The definite integral int -oo^+oo f(t).d(t-T)
>gives us f(T), but there is no information
>given to us about the indefinite integral
>nor of the anti-derivative.
>Indeed, the definite integral gives us f(T), which
>is a constant function having no aspect of t in its
>evaluation and which graphs as a horizontal
>line from -oo to +oo. This comes from the
>basic definitions of the Diracian.
> True, and of no relevance whatever to the question
> of what an antiderivative is.
> Hint: Youre _really_ showing over and over that
> you have no understanding of _basic_ calculus.
> You keep kindly informing me that Im making
> a laughingstock of myself, presumably youd
> want to know how utterly ignorant youre
> revealing yourself to be.
> PS: Look up the definition of ad hominem.
> Im not saying you must be wrong because 
youre
> stupid, Im saying youre looking very 
stupid
> because the things youre saying are wrong,
> at such a basic level.
>Your claimed anti-derivative has a strong
>dependence on t, and therefore has not
>come from anything formally defined.
>Are you making this up as you go along?
>Well, here is what David C. Ullrich posted..
>>An antiderivative of f(t).d(t-T) is given by
>>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>>So that line should be