mm-979

Subject: Re: Simple idea, mathematics and common-sense
> ab = 1,
> where again Œa¹ and Œb¹ are *irrational* the mathematicians
have a
> label for Œb¹ which is algebraic integer, but Œa¹ cannot be
an
> algebraic integer.
> So they don¹t have a label for it!
> You know how important naming is with human beings, and it
makes it
> that much harder for me to explain these ideas without a
label.
> I¹ve named numbers like Œa¹ objects
Take b = sqrt(2). Œb¹ is *irrational*. Thus a=1/sqrt(2) is an
object.
- William Hughes
===
Subject: Re: Simple idea, mathematics and common-sense
In sci.logic, James Harris
<jstevh@msn.com>
>> In sci.logic, James Harris
>> <jstevh@msn.com>
>> I¹ve had a time explaining some *very* simple mathematical
ideas that
>> lead to a few complexities, but it¹s been fruitful to
explain, or try
>> to explain, as I work to Žgure out why these simple ideas
either
>> excite derision, anger or confusion, and I think I have it
Žgured
>> out.
>>
>> I¹m sure many of you are put off by mathematics, so I
assure you up
>> front that what I¹ll be talking about will mostly be
*very* simple,
>> and there will only be a few slightly complicated things
at the end.
>>
>> First of all, I¹m going to talk about a case where
mathematicians gave
>> up because they couldn¹t see something, and assumed that
because they
>> couldn¹t see something it didn¹t exist!
>>
>> You know how with simple quadratics like x^2 + 3x + 2,
it¹s easy
>> enough to see factors of 2 in the roots?
>>
>> I mean, it¹s just (x^2 + 3x + 2) = (x+2)(x+1), and there
they are.
>>
>> However, if it¹s something like x^2 + 7x + 2, you can use
the
>> quadratic formula and get the roots to Žnd
>>
>> x = (-7 +/- sqrt(41))/2
>>
>> and who can see factors of 2 in that thing?
>> I think part of the problem here -- and I¹ll admit to not
being
>> entirely certain where the prime failure occurs -- is that,
>> if one has an integer such as 6, one can uniquely factor
it into
>> primes:
>> 6 = 2 * 3
> That¹s not the issue here.
> It¹s simple enough, but I have seen *posters* like this
person come
> forward and confuse the issue enough times that I think I¹d
better
> step in and make sure that the REAL issue isn¹t easily
obscured.
> P(x) = (x+8a)(x+b), where ab=1, and consider 8a+b an integer
> If a=b=1, notice you have (x+8)(x+1), which is one of two
basic
> possibilities with integers.
It and (a,b) = (-1,-1) are about the only ones in the integer
realm.
(a,b) = (-1,-1) gives you (x-8)(x-1), and of course 8a+b = -9
is an integer.
> Another possibility is a=1/2, b=4, which is the second basic
> possibility.
> That is, Œa¹ here can be a fraction, like 1/2, or it can be
more like
> an integer than a fraction, like actually being 1 with
*integers* in
> the other case.
Uh, OK. I fail to see your momentous point here, but you¹re
right
so far.
> Remember that making an issue between integers
> versus irrationals is a major part of the logical
> mistake that mathematicians made. I say
> that just because human beings LOVE being able to
> count something out on their Žngers, it¹s not a
> mathematical constraint!
Erm, that sentence made no sense at all.
> However, current mathematical dogma is that *if* Œa¹ and
Œb¹ are
> irrational then the Žrst type possibility is eliminated so
it
> must be the second possibility.
Well, if Œb¹ is irrational then it certainly isn¹t an integer.
An algebraic integer, perhaps, but not an integer. (One
example thereof: sqrt(2).)
> Part of the problem is that if you imagine the Žrst
> type possibility with
> ab = 1,
> where again Œa¹ and Œb¹ are *irrational* the mathematicians
have a
> label for Œb¹ which is algebraic integer, but Œa¹ cannot be
an
> algebraic integer.
a can most certainly be an algebraic integer if b is a unit.
Take b = 2 - sqrt(3), for example. 1/b = 1 / (2 - sqrt(3))
= (2 + sqrt(3)) / ((2 + sqrt(3)) * (2 - sqrt(3)))
= (2 + sqrt(3)) / (4 - 3) = 2 + sqrt(3).
Presto.
It¹s a little weird, but that¹s algebraic integers for you.
> So they don¹t have a label for it!
Are we supposed to level everything for your convenience?
As it so happens, there is a label for a, namely, the
reciprocal of an algebraic integer (if b is a non-unit).
As it is, I can fully characterize all integers and
reciprocals
of integers satisfying your equation. All one need do is
plug them in:
ab = 1
8a+b = n, for some integer n.
Therefore, 8a + 1/a = n
8a^2 - n*a + 1 = 0
a = (n +/- sqrt(n^2 - 32))/16
Presto. Two sets of numbers with a 1-1 mapping into Z, the
set of all integers. Or, if you prefer, you can combine the
sets and establish a 1-1 mapping with various tricks.
No doubt you¹ve done this already but it¹s an obvious
direction.
> You know how important naming is with human beings, and it
makes it
> that much harder for me to explain these ideas without a
label.
> I¹ve named numbers like Œa¹ objects, but while
mathematicians
> successfully trash my work or dismiss it, you can see that
using the
> name I¹ve given might not resolve things.
OK, so the set of all algebraic objects is the set W = {w:1/w
is an
algebraic integer}. No problem but I fail to see how that
helps
you any.
At best, it¹s a subset of all algebraic numbers. It¹s not
even a ring.
> That¹s it. That¹s the issue as what mathematicians teach in
this area
> doesn¹t follow from mathematics. It doesn¹t follow from
logic or any
> axioms. It¹s just some human notion that has settled into
dogma.
> Some mathematicians come to the logically specious
conclusion because
> they can¹t stick the label algebraic integer on it for that
reason
> Œa¹ is in no way an integer like 1 but is instead more like
a fraction
> like 1/2.
> There is nothing in mathematics to support that conclusion.
> It¹s just a human preference attached to the label algebraic
> integer.
> James Harris
--
#191, ewill3@earthlink.net
It¹s still legal to go .sigless.
===
Subject: Re: Simple idea, mathematics and common-sense
> In sci.logic, James Harris
> <jstevh@msn.com>
<snipYou know how with simple quadratics like x^2 + 3x + 2,
it¹s easy
> enough to see factors of 2 in the roots?
I mean, it¹s just (x^2 + 3x + 2) = (x+2)(x+1), and there they
are.
However, if it¹s something like x^2 + 7x + 2, you can use the
> quadratic formula and get the roots to Žnd
x = (-7 +/- sqrt(41))/2
and who can see factors of 2 in that thing?
>
> I think part of the problem here -- and I¹ll admit to not
being
> entirely certain where the prime failure occurs -- is that,
> if one has an integer such as 6, one can uniquely factor it
into
> primes:
>
> 6 = 2 * 3
> That¹s not the issue here.
So what _is_ the issue here?
> It¹s simple enough, but I have seen *posters* like this
person come
> forward and confuse the issue enough times that I think I¹d
better
> step in and make sure that the REAL issue isn¹t easily
obscured.
By all means, clarify. I¹ve lost track of what you¹re even
claiming
anymore.
> P(x) = (x+8a)(x+b), where ab=1, and consider 8a+b an integer
> If a=b=1, notice you have (x+8)(x+1), which is one of two
basic
> possibilities with integers.
> Another possibility is a=1/2, b=4, which is the second basic
> possibility.
> That is, Œa¹ here can be a fraction, like 1/2, or it can be
more like
> an integer than a fraction, like actually being 1 with
*integers* in
> the other case.
... more like an integer than a fraction...? To be more
precise, I
assume you mean that there are two possibilities (whose order
you
momentarily switch in the above paragraph): Given that Œb¹ is
a
(rational) integer, and a,b obey the above equations, then
either
(i) Œa¹ is a rational number which is an integer.
(ii) Œa¹ is a rational number which is not an integer.
> Remember that making an issue between integers versus
irrationals is a
> major part of the logical mistake that mathematicians made.
I say
> that just because human beings LOVE being able to count
something out
> on their Žngers, it¹s not a mathematical constraint!
> However, current mathematical dogma is that *if* Œa¹ and
Œb¹ are
> irrational then the Žrst type possibility is eliminated so
it must be
> the second possibility.
Here, it seems that you¹re saying: Given that neither a nor b
are
rational numbers, then if b is an algebraic integer, then the
two
possibilities are (by analogy):
(i) a is an algebraic integer.
(ii) a is an irrational complex number which is _not_ an
algebraic
integer.
And _you_ assert that, according to current mathematical
dogma, the
second case must hold.
> Part of the problem is that if you imagine the Žrst type
possibility
> with
> ab = 1,
> where again Œa¹ and Œb¹ are *irrational* the mathematicians
have a
> label for Œb¹ which is algebraic integer, but Œa¹ cannot be
an
> algebraic integer.
> So they don¹t have a label for it!
This part of the problem is not a problem. Since a is a root
of the
polynomial:
8a^2 - 17a + 1 = 0
which has integer coefŽcients, the standard label is algebraic
number; see, for example:
http://en.wikipedia.org/wiki/Algebraic_number
Note that, just as every integer is also a rational number,
every
algebraic integer is an algebraic number. And just as not
every
rational number is an integer, not every algebraic number is
an
algebraic integer.
Using this terminology, we can restate the two possibilities
as:
(i) Œa¹ is an algebraic number which is an algebraic integer.
(ii) Œa¹ is an algebraic number which is _not_ an algebraic
integer.
You assert that some mathematicians claim to have proven that
(ii) is
the only possibility if Œb¹ is an irrational algebraic
integer.
> You know how important naming is with human beings, and it
makes it
> that much harder for me to explain these ideas without a
label.
Hopefully this will make your argument easier.
> I¹ve named numbers like Œa¹ objects, but while
mathematicians
> successfully trash my work or dismiss it, you can see that
using the
> name I¹ve given might not resolve things.
Since I have no idea what you mean by like in this context, it
doesn¹t resolve things for me at all: objects may, or may
not, be the
same as algebraic numbers. However, clearly Œa¹ is an
algebraic
number.
> That¹s it. That¹s the issue as what mathematicians teach in
this area
> doesn¹t follow from mathematics. It doesn¹t follow from
logic or any
> axioms. It¹s just some human notion that has settled into
dogma.
Huh? _What¹s_ it? There is a standard label for numbers of
this
type. More precisely, there¹s a deŽnition by which we can
determine
whether or not a number deserves this label. Œa¹ meets the
criteria. Where¹s the problem?
> Some mathematicians come to the logically specious
conclusion because
> they can¹t stick the label algebraic integer on it for that
reason
> Œa¹ is in no way an integer like 1 but is instead more like
a fraction
> like 1/2.
This sentence doesn¹t parse; did you mean:
Some mathematicians come to the false conclusion that, since
Œa¹ is
not an algebraic integer, Œa¹ is not like a rational integer.
Instead,
it is more like a rational number.
If so, could you be more speciŽc? What does like an integer
and
more like a rational number mean here? As it stands, this
statement
could mean just about anything; as a result, it means nothing.
> There is nothing in mathematics to support that conclusion.
_What_ conclusion?
The closest I can come to is that you mean:
There are certain parallels between the algebraic integers and
algebraic numbers on the one hand, and the (rational)
integers and the
rational numbers on the other hand. For example, every
integer is an
algebraic integer, and every rational is an algebraic number;
furthermore, every integer is a rational number, while every
algebraic
integer is an algebraic number.
Given the example equations above, some mathematicians come
to the
false conclusion that, since Œa¹ does not meet the deŽnition
of
algebraic integer, Œa¹ is not an algebraic integer; instead
it is an
algebraic number.
_This_ is clearly a false statement (i.e., it is _not_ a false
conclusion). Œa¹ has the property meeting the deŽnition of
algebraic
number. The since... part of this statement is unnecessary;
Œa¹ has
this property since it is the root of a polynomial with
integer
coefŽcents. For that matter, Œb¹ has the same property, and
so is
also an algebraic number (as well as being an algebraic
integer).
What¹s the problem here?
> It¹s just a human preference attached to the label algebraic
> integer.
_What_ is just a human preference...?
> James Harris
===
Subject: Re: Simple idea, mathematics and common-sense
> It¹s simple enough, but I have seen *posters* like this
person come
> forward and confuse the issue enough times that I think I¹d
better
> step in and make sure that the REAL issue isn¹t easily
obscured.
Actually, what JSh is trying to do is obscure the real issue,
that JSH
cannot deal with the intellectual requirements of serious
mathematics.
Mathematical results are not proved by analogy, which seems
to be JSH¹s
only method of attack, but by logic.
> P(x) = (x+8a)(x+b), where ab=1, and consider 8a+b an integer
> If a=b=1, notice you have (x+8)(x+1), which is one of two
basic
> possibilities with integers.
> Another possibility is a=1/2, b=4, which is the second basic
> possibility.
> That is, Œa¹ here can be a fraction, like 1/2, or it can be
more like
> an integer than a fraction, like actually being 1 with
*integers* in
> the other case.
JSH seems to think that there is a gradual transition between
a number
being an integer and being a non-integral fraction. Can he
give an
example of some number that is part way in between to
illustrate his
theory of gradations between the two?
> Remember that making an issue between integers versus
irrationals is a
> major part of the logical mistake that mathematicians made.
I say
> that just because human beings LOVE being able to count
something out
> on their Žngers, it¹s not a mathematical constraint!
That is no more an error that designating some integers as
even and
others as not even. There are clear deŽnitions of what it
means to be a
rational number. The distinction mathematicians make is based
on whether
a number satisŽes that deŽnition or not.
> However, current mathematical dogma is that *if* Œa¹ and
Œb¹ are
> irrational then the Žrst type possibility is eliminated so
it must be
> the second possibility.
If a and b are irrational then it eliminates the possibility
that they
are rational. So what?
> Part of the problem is that if you imagine the Žrst type
possibility
> with
> ab = 1,
> where again Œa¹ and Œb¹ are *irrational* the mathematicians
have a
> label for Œb¹ which is algebraic integer, but Œa¹ cannot be
an
> algebraic integer.
> So they don¹t have a label for it!
How about non-algebraic-integer? or, to be more precise, an
algebraic
number which is not an algebraic integer.
> You know how important naming is with human beings, and it
makes it
> that much harder for me to explain these ideas without a
label.
Then create one by giving a clear deŽnition. You have used
object as
a label without ever deŽning it clearly enough for anyone to
understand
what you are talking about.
In mathematics, such labels are created all the time, but
they must be
clear and understandable
> I¹ve named numbers like Œa¹ objects, but while
mathematicians
> successfully trash my work or dismiss it, you can see that
using the
> name I¹ve given might not resolve things.
Since you have never given a clear enough deŽnition of what
you mean by
an object for anyone, except you, to distinguish between
objects and
non-objects, no one else can understand what they are.
It is not mathematicians who make your work trash, it is you.
And if
your object name won¹t resolve what you see as a problem, why
introduce it at all?
> That¹s it. That¹s the issue as what mathematicians teach in
this area
> doesn¹t follow from mathematics. It doesn¹t follow from
logic or any
> axioms. It¹s just some human notion that has settled into
dogma.
What is that? I see no mathematical issue here, just pointless
ramblings.
> Some mathematicians come to the logically specious
conclusion because
> they can¹t stick the label algebraic integer on it for that
reason
> Œa¹ is in no way an integer like 1 but is instead more like
a fraction
> like 1/2.
What logically specious conclusion?
> There is nothing in mathematics to support that conclusion.
What conclusion?
> It¹s just a human preference attached to the label algebraic
> integer.
One could replace each occurrence of algebraic integer with
root of a monic polynomial with integer coefŽcients, but
humans do
prefer the shorter phrase.
===
Subject: Re: Simple idea, mathematics and common-sense
>[...]
>That¹s not the issue here.
>It¹s simple enough, but I have seen *posters* like this
person come
>forward and confuse the issue enough times that I think I¹d
better
>step in and make sure that the REAL issue isn¹t easily
obscured.
Oh good. I¹ve been reading these ab = 1 posts and I have
no idea whatever what the issue you¹re talking about is.
>P(x) = (x+8a)(x+b), where ab=1, and consider 8a+b an integer
>If a=b=1, notice you have (x+8)(x+1), which is one of two
basic
>possibilities with integers.
>Another possibility is a=1/2, b=4, which is the second basic
>possibility.
Right...
>That is, Œa¹ here can be a fraction, like 1/2, or it can be
more like
>an integer than a fraction, like actually being 1 with
*integers* in
>the other case.
>Remember that making an issue between integers versus
irrationals is a
>major part of the logical mistake that mathematicians made.
Huh? What issue do we make, and what¹s incorrect about it?
> I say
>that just because human beings LOVE being able to count
something out
>on their Žngers, it¹s not a mathematical constraint!
>However, current mathematical dogma is that *if* Œa¹ and Œb¹
are
>irrational then the Žrst type possibility is eliminated so
it must be
>the second possibility.
>Part of the problem is that if you imagine the Žrst type
possibility
>with
>ab = 1,
>where again Œa¹ and Œb¹ are *irrational* the mathematicians
have a
>label for Œb¹ which is algebraic integer, but Œa¹ cannot be
an
>algebraic integer.
>So they don¹t have a label for it!
_That¹s_ the problem? The fact that there¹s (supposedly) no
label for something indicates that mathematics is _wrong_?
You¹re really sounding like an idiot. If you want to say
mathematicians are all wrong you need to come up with
something they actually say and show that it¹s false.
>You know how important naming is with human beings, and it
makes it
>that much harder for me to explain these ideas without a
label.
That¹s because of your complete lack of mathematical
competence.
When the rest of us have a new concept in mind we have no
problem
giving a _precise_ description of exactly what we mean. (With
or without a new label attached, depending on whether the
concept is going to come up again.)
>I¹ve named numbers like Œa¹ objects,
But you¹ve _never_ given a coherent version of the deŽnition
of
object - until you do that the things you say about objects
are meaningless.
Just for fun, why don¹t you supply a _proof_ that the a above
actually _is_ an object?
> but while mathematicians
>successfully trash my work or dismiss it, you can see that
using the
>name I¹ve given might not resolve things.
>That¹s it. That¹s the issue as what mathematicians teach in
this area
>doesn¹t follow from mathematics. It doesn¹t follow from
logic or any
>axioms. It¹s just some human notion that has settled into
dogma.
What _is_ it that mathematicians teach that you¹re referring
to here?
>Some mathematicians come to the logically specious
conclusion because
>they can¹t stick the label algebraic integer on it for that
reason
>¹a¹ is in no way an integer like 1 but is instead more like
a fraction
>like 1/2.
>There is nothing in mathematics to support that conclusion.
And no mathematician has ever _stated_ that conclusion,
either.
Name someone who, while he was speaking careful mathematics,
said something about numbers being like integers or not like
integers. (Without giving a careful deŽnition.)
>It¹s just a human preference attached to the label algebraic
>integer.
>James Harris
************************
David C. Ullrich
===
Subject: Re: Simple idea, mathematics and common-sense
> You¹re really sounding like an idiot. If you want to say
> mathematicians are all wrong you need to come up with
> something they actually say and show that it¹s false.
John
===
Subject: Re: Simple idea, mathematics and common-sense
>>In sci.logic, James Harris
>><jstevh@msn.com>
>I¹ve had a time explaining some *very* simple mathematical
ideas that
>lead to a few complexities, but it¹s been fruitful to
explain, or try
>to explain, as I work to Žgure out why these simple ideas
either
>excite derision, anger or confusion, and I think I have it
Žgured
>out.
>I¹m sure many of you are put off by mathematics, so I assure
you up
>front that what I¹ll be talking about will mostly be *very*
simple,
>and there will only be a few slightly complicated things at
the end.
>First of all, I¹m going to talk about a case where
mathematicians gave
>up because they couldn¹t see something, and assumed that
because they
>couldn¹t see something it didn¹t exist!
>You know how with simple quadratics like x^2 + 3x + 2, it¹s
easy
>enough to see factors of 2 in the roots?
>I mean, it¹s just (x^2 + 3x + 2) = (x+2)(x+1), and there
they are.
>However, if it¹s something like x^2 + 7x + 2, you can use the
>quadratic formula and get the roots to Žnd
>x = (-7 +/- sqrt(41))/2
>and who can see factors of 2 in that thing?
>>I think part of the problem here -- and I¹ll admit to not
being
>>entirely certain where the prime failure occurs -- is that,
>>if one has an integer such as 6, one can uniquely factor it
into
>>primes:
>>6 = 2 * 3
> That¹s not the issue here.
> It¹s simple enough, but I have seen *posters* like this
person come
> forward and confuse the issue enough times that I think I¹d
better
> step in and make sure that the REAL issue isn¹t easily
obscured.
> P(x) = (x+8a)(x+b), where ab=1, and consider 8a+b an integer
> If a=b=1, notice you have (x+8)(x+1), which is one of two
basic
> possibilities with integers.
You might also be able to have a = -1, b = -1.
> Another possibility is a=1/2, b=4, which is the second basic
> possibility.
Note that ab is not equal to 1 here.
You probably meant one of the following
a = 1/4, b = 4
a = 1/8, b = 8
You may also have
a = -1/4, b = -4
a = -1/8, b = -8
> That is, Œa¹ here can be a fraction, like 1/2, or it can be
more like
> an integer than a fraction, like actually being 1 with
*integers* in
> the other case.
So, you¹re saying that a pair (a,b) of rational numbers, with
the
property that their product is 1, and the sum 8a + b is
integral,
can either be both integral, or b can be an integer dividing
8,
with a being its reciprocal. I suppose the second case
subsumes
the Žrst case.
> Remember that making an issue between integers versus
irrationals is a
> major part of the logical mistake that mathematicians made.
I say
> that just because human beings LOVE being able to count
something out
> on their Žngers, it¹s not a mathematical constraint!
This is where you make the mistake of over-generalization.
The rational
integers are indeed distinguished, and can be identiŽed in
any number
Želd: take the additive group generated by 1. This has
nothing to do
with Žngers. In fact, the same process identiŽes the elements
of a
distinguished ring within any ring with identity, and of a
distinguished
Želd within any Žnite Želd.
> However, current mathematical dogma is that *if* Œa¹ and
Œb¹ are
> irrational then the Žrst type possibility is eliminated so
it must be
> the second possibility.
By the Žrst type possibility, you¹re referring to both a and
b being
integral, I assume.
If by integral, you¹re using the mathematical deŽnition
(viz., the
roots of monic polynomials over the integers), then it¹s a
derived fact
that the a¹s solving
ab = 1,
8a + b = k, a rational integer
are found by solving a quadratic equation:
8a^2 - ka + 1 = 0
and so when the polynomial is irreducible over Z, the
equation has no
rational solutions, in particular no solutions in the
rational integers.
In the other case, i.e., the polynomial factors as the
product of linear
polynomials with integral coefŽcients, well you do get
rational
solutions. That includes both cases (a,b) both integral as
well as
(a,b) not both integral.
> Part of the problem is that if you imagine the Žrst type
possibility
> with
> ab = 1,
> where again Œa¹ and Œb¹ are *irrational* the mathematicians
have a
> label for Œb¹ which is algebraic integer, but Œa¹ cannot be
an
> algebraic integer.
> So they don¹t have a label for it!
Is that what this is all about? Having a nice label? That¹s
not a
problem of mathematics.
> You know how important naming is with human beings, and it
makes it
> that much harder for me to explain these ideas without a
label.
No, naming is a convenience, not a necessity. It¹s a fact
that you
cannot possibly name all subsets of the real numbers using a
countable
alphabet and countable word length. We can¹t even name all
reals, using
a Žnite alphabet and Žnite word length. Should we all then
give up on
mathematics due to the impossibility of naming everything?
> I¹ve named numbers like Œa¹ objects, but while
mathematicians
> successfully trash my work or dismiss it, you can see that
using the
> name I¹ve given might not resolve things.
You¹re still going on about names. How sad.
> That¹s it. That¹s the issue as what mathematicians teach in
this area
> doesn¹t follow from mathematics. It doesn¹t follow from
logic or any
> axioms. It¹s just some human notion that has settled into
dogma.
Dogma. An established opinion. Is that the best you can do?
That the
whole issue is that mathematicians *deŽned something*, and
DOGMATICALLY
stick to that deŽnition?
Naming is via *convention*. You know, those parties in Vegas,
where
Is Bob lapel stickers, naked girls jumping out of big cakes,
and dozens
of manufacturers¹ reps giving out pens with company logos and
stupid-ass
software on funny-shaped CDs?
> Some mathematicians come to the logically specious
conclusion because
> they can¹t stick the label algebraic integer on it for that
reason
> Œa¹ is in no way an integer like 1 but is instead more like
a fraction
> like 1/2.
Look, it *could* have happened that the word algebraic
integer would
have been deŽned as all those numbers that žoat my boat, and
then
something different would have occurred, I don¹t care what.
Classes
of mathematical objects are found and named because of their
possession
of useful qualities. If they don¹t turn out to be interesting
enough,
then people stop studying them, but that¹s not the case for
the ring
of algebraic integers.
Just because you can¹t grasp that, or don¹t like the fact
that you can¹t
seem to come up with an *algebraic* shortcoming suffered by
that ring,
well, that¹s not our problem, then, is it?
> There is nothing in mathematics to support that conclusion.
I see no conclusion in your text, other than I sure don¹t
like it,
because I don¹t like it, and if you like it, then there¹s no
mathematical reason for it.
You fail to note that the ring of algebraic integers serves
much the
same role in the Želd of algebraic numbers as the integers
serves
in the rationals: both Želds are formed from their
corresponding sub-
rings by taking the Želd of quotients (oh, I¹m sorry, you¹re
an
algebra illiterate, aren¹t you?). The term integral as in
satisŽes
a monic polynomial equation serves as a true generalization
of the term
algebraic, meaning satisŽes a polynomial equation, which I¹ll
note
you aren¹t complaining about.
> It¹s just a human preference attached to the label algebraic
> integer.
It¹s a *name*, a *deŽnition*. You don¹t happen to like it,
and you
have been going on for over a year about something that you
think makes
a mathematical difference. Note that every time you have
thought you had
found a shortcoming of the set of algebraic integers, by
their lack of
numbers with certain divisibility properties, you have been
found to
have been wrong. You have taken the liberty of insulting,
threatening,
pleading, smearing your own fecal work all over your face and
prancing
about like a fool, and all to no avail.
Here¹s a suggestion: Žnd a new name for your preferred set of
numbers.
Algebrizzle Integizzle
That¹ll sure work. Now, deŽne them, and *prove* that they
have the
properties you claim they do. So far, your batting average at
that
second, more important, task has been pretty low.
Zero, if I divide successes by attempts.
> James Harris
Dale.
===
Subject: Re: Simple idea, mathematics and common-sense
<snipIt¹s simple enough, but I have seen *posters* like this
person come
> forward and confuse the issue enough times that I think I¹d
better
> step in and make sure that the REAL issue isn¹t easily
obscured.
> P(x) = (x+8a)(x+b), where ab=1, and consider 8a+b an integer
> If a=b=1, notice you have (x+8)(x+1), which is one of two
basic
> possibilities with integers.
> Another possibility is a=1/2, b=4, which is the second basic
> possibility.
This must be a typo, since ab isn¹t 1.
Rick
===
Subject: Re: Simple idea, mathematics and common-sense
In sci.math, James Harris
<jstevh@msn.com>
>> <snip>
>>
>> It¹s simple enough, but I have seen *posters* like this
person come
>> forward and confuse the issue enough times that I think
I¹d better
>> step in and make sure that the REAL issue isn¹t easily
obscured.
>>
>> P(x) = (x+8a)(x+b), where ab=1, and consider 8a+b an
integer
>>
>> If a=b=1, notice you have (x+8)(x+1), which is one of two
basic
>> possibilities with integers.
>>
>> Another possibility is a=1/2, b=4, which is the second
basic
>> possibility.
>> This must be a typo, since ab isn¹t 1.
> Just a dumb mistake. That should be b=2, or a=1/4.
> Still, beyond my dumb mistakes there¹s the real issue of
*two*
> possibilities, where one is that Œa¹ is a unit in a ring
where -1 and
> 1 are the only integer units.
Well, if a is an integer, then a is deŽnitely a unit in a ring
where -1 and 1 are the only integer units -- mostly because
b can¹t be non-integral in the sum 8a+b, and therefore must be
either +1 or -1, so a is equal to b in that case and 8a+b is
either
+9 or -9.
That¹s not much of a result.
Of course, if you want to generate the ring Z[(n +/- sqrt(n^2
- 32))/16],
feel free. (b = n - 8a is also in the ring. One also has some
issues because a^2, a^3, etc. have to be in the ring, too.)
For what
it¹s worth, you¹ve deŽned a meta-generator as well: a method
by
which one can spit out rings given an arbitrary integer Œn¹.
Some of
the rings, e.g., n = 9, are merely Z. Others are subsets
of Q (e.g., n = 6), R (|n| > 5) or C (|n| <= 5).
I suspect that the units may not be what you expect. One case
I
can readily identify is with n = 8; one unit in the resulting
ring will be 3 - 2*sqrt(2).
I¹d have to work to see if I can Žnd such a ring where 1 and
-1
are the only integer units, with n != 9 or -9 (since in those
case we¹re only dealing with Z). Ultimately, it should reduce
to solving the equation
n^2 - m^2 = 32
n and m integers. I for one would suspect that the number of
solutions for this equation are limited, since, if r > 0 is
selected to be a positive integer,
(r+1)^2 - r^2 = 2*r + 1
and therefore if n > 15 no solutions are possible. Therefore,
I can enumerate the solutions without too much difŽculty,
taking sign into account.
(6,2) : a = 1/2, b = 2, 8a+b = 6, P(x) = x^2 + 6a + 8
(6,-2) : a = 1/4, b = 4, 8a+b = 6, P(x) = x^2 + 6a + 8
(-6,2) : a =-1/4, b =-4, 8a+b =-6, P(x) = x^2 - 6a + 8
(-6,-2): a =-1/2, b =-2, 8a+b =-6, P(x) = x^2 - 6a + 8
(9,7) : a = 1, b = 1, 8a+b = 9, P(x) = x^2 + 9x + 8
(9,-7) : a = 1/8, b = 8, 8a+b = 9, P(x) = x^2 + 9x + 8
(-9,7) : a =-1/8, b =-8, 8a+b =-9, P(x) = x^2 - 9x + 8
(-9,-7): a = -1, b =-1, 8a+b =-9, P(x) = x^2 - 9x + 8
and that¹s it; that¹s all possible equations in the integer
realm.
For n != 6, -6, 9, -9, things get rather more interesting.
We can of course continue with our nomenclature:
(n, sqrt(n^2 - 32))
and attempt to Žnd units in the generated ring
Z[(n +/- sqrt(n^2 - 32))/16]. (I¹ve already noted one example
where that¹s possible, namely n = 8.)
For an arbitrary n other than 6, -6, 9, -9, if we can Žnd
integers
c and d such that
c + d(n +/- sqrt(n^2 - 32))/16
is a unit in our new ring, then it¹s clear that 1 and -1 are
not the only units.
The actual analysis of all these groups looks like a bit of
a schlog, as there are at least 8 cases to consider. The
easy one is when n = 0 (mod 16); the ring is equivalent
in that case to Z[sqrt(n^2 - 32)/16] or Z[sqrt(n^2/256 -
1/8)];
one unit in this ring is 4 * (n/16 - sqrt(n^2/256 - 1/8)),
with reciprocal 2 * (n/16 + sqrt(n^2/256 - 1/8)).
For the more general case, if we multiply
(c + d*(n + sqrt(n^2 - 32))/16) * (c - d*(n - sqrt(n^2 -
32))/16)
we get (c + d/16)^2 - d^2*(n^2 - 32)/256, or
((16c + d)^2 - d^2*n^2 - d^2*32)/256 =
(256c^2 + 32cd - d^2*(n^2 + 31))/256.
In order to have a pair of units we have to Žnd (c,d) such
that
(256c^2 + 32cd - d^2*(n^2 + 31)) = +256 or -256.
with d != 0. (If d = 0 we just get +1 or -1 by
setting c to +16 or -16.)
Of course these aren¹t the only possibilities for units. One
could, for instance, try to solve the pair of 4-variable
Diophantine equations generated by
(c + d*(n + sqrt(n^2 - 32))/16) * (e - f*(n - sqrt(n^2 -
32))/16) = 1
or
(c + d*(n + sqrt(n^2 - 32))/16) * (e - f*(n - sqrt(n^2 -
32))/16) = -1
To me that looks like the complicated way of doing it, but
certainly possible. Perhaps someone else can think of
a simpler method of either;
[1] Žnding an n, such that the generated ring
Z[(n +/- sqrt(n^2 - 32))/16], only has the two units -1 and
+1,
and n != 6, -6, 9, -9 (since I¹ve already accounted for
those),
or
[2] proving that no such n is possible.
I suspect it¹s [2] but can¹t prove it without a lot of
schlogging.
[rest snipped]
--
#191, ewill3@earthlink.net
It¹s still legal to go .sigless.
===
Subject: Re: Simple idea, mathematics and common-sense
>
> <snip
>
>
> It¹s simple enough, but I have seen *posters* like this
person come
> forward and confuse the issue enough times that I think I¹d
better
> step in and make sure that the REAL issue isn¹t easily
obscured.
The real issue is whether JSH is really confused or only
pretending.
>
> P(x) = (x+8a)(x+b), where ab=1, and consider 8a+b an integer
>
> If a=b=1, notice you have (x+8)(x+1), which is one of two
basic
> possibilities with integers.
>
> Another possibility is a=1/2, b=4, which is the second basic
> possibility.
Not with a and b integers, and not with a*b = 1.
>
>
> This must be a typo, since ab isn¹t 1.
>
> Just a dumb mistake. That should be b=2, or a=1/4.
> Still, beyond my dumb mistakes there¹s the real issue of
*two*
> possibilities, where one is that Œa¹ is a unit in a ring
where -1 and
> 1 are the only integer units.
What do you mean by a unit in a ring where -1 and 1 are the
only
integer units?
In at least one sense, 1 and -1 are the only *integer* units
in any
subring of the complex Želd as they are the only (rational)
integers
that are units. Is that what you mean?
Until you make what you want to say unambiguously clear,
nobody can
possibly agree with you.
> Sure I know, yeah, I make a dumb mistake and people jump on
it as if
> that proves that everything I say is wrong.
When everything that you claim is based on that mistake, then
everything
you claim must be redone without mistakes in order to be
validated.
> Losers. Why can¹t you people just play straight up?
You never play straight up.
If your rules do not require straight up play, you should not
complain
that others use your rules?
If your rules do require straight up play, then follow them
yourself
before you complain that others don¹t.
> Why can¹t you
> worry about the real ideas

We do, but then you bitch them up.
> versus the very human tendency to fail,
In your case, that tendency transcends mere humanity.
> including in ways so dumb as to have b=4 like I did?
> It amazes me how often over the years posters have jumped
on *every*
> single little mistake as if it¹s proof that everything I
say is wrong.
A chain is only as strong as its weakest link. If one link is
žawed,
the whole chain is žawed and will fail.
> I guess for those people I¹d have to be Jesus Christ or
something.
Unless JC was more of a mathematician than repute has it,
that would not
be nearly enough.
> As if unless I¹m a perfect man, nothing I say matters.
All you have to be willing to do is to acknowledge and repair
your
errors, as actual mathematicians do, rather than dissing
everyone who
refuses to overlook them.
That is a long way short of required perfection, but a lot
closer to
perfection than you have been these last 8 years.
> Such odd people they are, why do they succeed so much in
confusing the
> issue?
By being so clear on the issues that only people with
crosswired brains,
like you, can still manage to mess things up.
===
Subject: Re: Simple idea, mathematics and common-sense
> Still, beyond my dumb mistakes there¹s the real issue of
*two*
> possibilities, where one is that Œa¹ is a unit in a ring
where -1 and
> 1 are the only integer units.
You have set the condition that ab = 1. For the problem you
posed, Œb¹ is an
algebraic integer
and Œa¹ is not an algebraic integer. There is nothing
whatsoever unusual or
noteworthy about
this. For *any* non-unit algebraic integer Œb¹, there *is* a
reciprocal Œa¹
such that ab = 1,
but not in the ring of algebraic integers. For all these
cases Œa¹ would
still be an algebraic
number. So what? Is it your position that since ab = 1 and
since Œb¹ is an
algebraic integer
that Œa¹ *should* be an algebraic integer? Come to think of
it, that may the
root cause of all
the contention in these recent threads!
You seem to Žnd some necessity for Œa¹ and Œb¹ to be units
whenever one of
them is an
algebraic integer and their product is 1. Is *that* your
position? Is that
the source of all
the *should be an algebraic integer* assertions you have
made?? Is that why
you are reaching
out for other rings, to make units of all reciprocal pairs???
--
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: Simple idea, mathematics and common-sense
> <snip
> It¹s simple enough, but I have seen *posters* like this
person come
> forward and confuse the issue enough times that I think I¹d
better
> step in and make sure that the REAL issue isn¹t easily
obscured.
>
> P(x) = (x+8a)(x+b), where ab=1, and consider 8a+b an integer
>
> If a=b=1, notice you have (x+8)(x+1), which is one of two
basic
> possibilities with integers.
>
> Another possibility is a=1/2, b=4, which is the second basic
> possibility.
> This must be a typo, since ab isn¹t 1.
Just a dumb mistake. That should be b=2, or a=1/4.
Still, beyond my dumb mistakes there¹s the real issue of *two*
possibilities, where one is that Œa¹ is a unit in a ring
where -1 and
1 are the only integer units.
Sure I know, yeah, I make a dumb mistake and people jump on
it as if
that proves that everything I say is wrong.
Losers. Why can¹t you people just play straight up? Why can¹t
you
worry about the real ideas versus the very human tendency to
fail,
including in ways so dumb as to have b=4 like I did?
It amazes me how often over the years posters have jumped on
*every*
single little mistake as if it¹s proof that everything I say
is wrong.
I guess for those people I¹d have to be Jesus Christ or
something.
As if unless I¹m a perfect man, nothing I say matters.
Such odd people they are, why do they succeed so much in
confusing the
issue?
James Harris
===
Subject: Re: Simple idea, mathematics and common-sense
> P(x) = (x+8a)(x+b), where ab=1, and consider 8a+b an integer
> If a=b=1, notice you have (x+8)(x+1), which is one of two
basic
> possibilities with integers.
> Another possibility is a=1/2, b=4, which is the second basic
> possibility.
So there are *two* possibilities, according to you. a=b=1,
and a=1/2, b=4.
OK, proceed.
> That is, Œa¹ here can be a fraction, like 1/2, or it can be
more like
> an integer than a fraction, like actually being 1 with
*integers* in
> the other case.
Are you saying that Œ1¹ is more like an integer than a
fraction? Hmmm. I
believe it *is* an integer.
> Remember that making an issue between integers versus
irrationals is a
> major part of the logical mistake that mathematicians made.
Irrationals? So far you have two cases. a=b=1 and a=1/2, b =
4. No
irrationals here. Could you provide an example of this logical
mistake that mathematicians made? So far, I only see your
mistake in
equating rational numbers with irrational numbers.
> I say
> that just because human beings LOVE being able to count
something out
> on their Žngers, it¹s not a mathematical constraint!
I¹d be surprised if you could count past 21 with your pants
off.
> However, current mathematical dogma is that *if* Œa¹ and
Œb¹ are
> irrational then the Žrst type possibility is eliminated so
it must be
> the second possibility.
Huh? If Œa¹ and Œb¹ are irrational, then *both* possibilities
you gave are
eliminated. Your two possibilities involve integers and
rational numbers only.
> Part of the problem is that if you imagine the Žrst type
possibility
> with
> ab = 1,
> where again Œa¹ and Œb¹ are *irrational* the mathematicians
have a
> label for Œb¹ which is algebraic integer, but Œa¹ cannot be
an
> algebraic integer.
Why is that a problem? Œb¹ is the reciprocal of Œa¹. Unless
they are units
in the ring of algebraic integers, only one of them can be
an algebraic integer. In that case the other is an algebraic
number.
> So they don¹t have a label for it!
Yes, they do. I just gave it. Listen carefully: ALGEBRAIC
NUMBER.
> You know how important naming is with human beings, and it
makes it
> that much harder for me to explain these ideas without a
label.
It is much easier if the proper label for *you* is attached.
Namely, CRANK,
TROLL or CRACKPOT.
> I¹ve named numbers like Œa¹ objects, but while
mathematicians
> successfully trash my work or dismiss it, you can see that
using the
> name I¹ve given might not resolve things.
There is nothing to resolve. You have made a number of
serious errors,
gaffs, mistakes, nonsequitors, fallacies and blunders.
> That¹s it. That¹s the issue as what mathematicians teach in
this area
> doesn¹t follow from mathematics. It doesn¹t follow from
logic or any
> axioms. It¹s just some human notion that has settled into
dogma.
There is no dogma associated with mathematics. Everything
that is taught or
discussed is subject to review by anyone. Your material
has been subject to review and found to be false. Get a life.
> Some mathematicians come to the logically specious
conclusion because
> they can¹t stick the label algebraic integer on it for that
reason
> Œa¹ is in no way an integer like 1 but is instead more like
a fraction
> like 1/2.
Fractions are *rational numbers*. They have a distinct label
that is
important in some contexts. Just because *you* do not know the
correct terminology for classifying numbers does not mean
that others are
similarly ignorant.
> There is nothing in mathematics to support that conclusion.
What conclusion?
> It¹s just a human preference attached to the label algebraic
> integer.
Are you a *human*? Do you attach labels by *human
preference*? Guess what?
Mathematicians attach labels by *identiŽcation*, not
human preference. Algebraic integers form a distinct Želd.
Membership of
its elements is well deŽned. The proper label for you is
*idiot* -- and that¹s an identiŽcation, not an insult.
> James Often in error, but never in doubt. Harris
--
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: Simple idea, mathematics and common-sense
>Part of the problem is that if you imagine the Žrst type
possibility
>with
>ab = 1,
>where again Œa¹ and Œb¹ are *irrational* the mathematicians
have a
>label for Œb¹ which is algebraic integer, but Œa¹ cannot be
an
>algebraic integer.
>So they don¹t have a label for it!
Apart from algebriac number you mean? Do yuu even know what a
fraction
Želd is?
Your ignorance of these objects doesn¹t mean the mathematics
is
defective.
>You know how important naming is with human beings, and it
makes it
>that much harder for me to explain these ideas without a
label.
>I¹ve named numbers like Œa¹ objects, but while mathematicians
>successfully trash my work or dismiss it, you can see that
using the
>name I¹ve given might not resolve things.
>That¹s it. That¹s the issue as what mathematicians teach in
this area
>doesn¹t follow from mathematics. It doesn¹t follow from
logic or any
>axioms. It¹s just some human notion that has settled into
dogma.
>Some mathematicians come to the logically specious
conclusion because
>they can¹t stick the label algebraic integer on it for that
reason
>¹a¹ is in no way an integer like 1 but is instead more like
a fraction
>like 1/2.
>There is nothing in mathematics to support that conclusion.
>It¹s just a human preference attached to the label algebraic
>integer.
>James Harris
So now the issue is that we¹re misusing names?
===
Subject: Re: Simple idea, mathematics and common-sense
>The problem here is with a possibility not logically
eliminated where
>given
eh?
>x^2 + (8a + b)x + 8, ab=1, and (8a + b) an integer
>it¹s quite possible that Œa¹ is a member of a ring where -1
and 1 are
>the only integer units, though it is itself not an algebraic
integer.
Yes, many such exist.
If you had the slightest idea about anything other than the
simple
high school maths you¹ve been taught (hey, you said i could
insult you
all i like) you¹d know about something about these splitting
Želds
you started briežy citing.
Take A as the ring of algebraic integers, just adjoin a to it
if you
feel like doing so.
>So try to insult me all you like matt grime, and try to
obscure the
>issue all you like, as Arturo Magidin has successfully done
for so
>many years.
>Maybe later you¹ll go somewhere to laugh to yourself at the
ignorance
>of other readers, but the mathematics will not change. So
what if you
>can help other mathematicians get lost?
I don¹t think we¹re lost by the math, just bewildered by your
bizarre
bastardizations of it. In fact every mathematical claim
you¹ve made
has, with some difŽculty because of your inability to write a
coherent argument, been refuted. Or am I missing one of the
posts that
didn¹t have to get an ŒOOPS! I¹m wrong again¹ follow up?
Ideally of
course you¹d be the one to get lost.
>The mathematical truth remains as it has always been.
>James Harris
===
Subject: Re: Simple idea, mathematics and common-sense
>And I know what algebraic numbers are, but 1/2 is an
algebraic number
>along with 1.
why Œbut? Œ
>The problem here is with a possibility not logically
eliminated where
>given
>x^2 + (8a + b)x + 8, ab=1, and (8a + b) an integer
>it¹s quite possible that Œa¹ is a member of a ring where -1
and 1 are
>the only integer units, though it is itself not an algebraic
integer.
possible? what the hell¹s your point? you¹ve explained this
somewhere
else and there have been criticisms of your reasoning. What
is your
point? All we can say is that not every integer can be
written as the
summ8a+b where ab=1 and both are in the algebraic integers.
AND?
>So try to insult me all you like matt grime, and try to
obscure the
>issue all you like, as Arturo Magidin has successfully done
for so
>many years.
why is my name in quotation marks?
>Maybe later you¹ll go somewhere to laugh to yourself at the
ignorance
>of other readers, but the mathematics will not change. So
what if you
>can help other mathematicians get lost?
I don¹t laugh at ignorance, unless that person despite
repeated
instruction and clariŽcation carries on in the same vein
insisting
they are right, when perhaps ignorance isn t¹ the correct
word.
>The mathematical truth remains as it has always been.
>James Harris
===
Subject: Re: Simple idea, mathematics and common-sense
>
>
>
>Part of the problem is that if you imagine the Žrst type
possibility
>with
>
>ab = 1,
>
>where again Œa¹ and Œb¹ are *irrational* the mathematicians
have a
>label for Œb¹ which is algebraic integer, but Œa¹ cannot be
an
>algebraic integer.
>
>So they don¹t have a label for it!

>
> Apart from algebriac number you mean? Do yuu even know what
a fraction
> Želd is?
>
> Your ignorance of these objects doesn¹t mean the
mathematics is
> defective.
>
> Sly insults will get you nowhere here, as I have myself
noted that I¹m
> NOT a mathematician!
> And I know what algebraic numbers are, but 1/2 is an
algebraic number
> along with 1.
> The problem here is with a possibility not logically
eliminated where
> given
> x^2 + (8a + b)x + 8, ab=1, and (8a + b) an integer
> it¹s quite possible that Œa¹ is a member of a ring where -1
and 1 are
> the only integer units, though it is itself not an
algebraic integer.
Jimmy, you keep referring to some ring where 1 and -1 are the
only units.
Now the ring of integers is such a ring, but you have never
given any
examples of such a ring other than the ring of integers.
Could you please give us an example of such a ring other than
the ring
of integers, just so we know what sort of beast you are
talking about.
> So try to insult me all you like matt grime, and try to
obscure the
> issue all you like, as Arturo Magidin has successfully done
for so
> many years.
Arturo has clariŽed a lot of issues to the point where your
could no
longer maintain your illogical claims about them.
That sort of obscuring is what we need more of, i.e., making
clear
where your obscure bits are just nonsense..
> Maybe later you¹ll go somewhere to laugh to yourself at the
ignorance
> of other readers, but the mathematics will not change.
At least you will still be unable to change mathematics,
though it does,
in fact, change slowly from the advances made by myriad true
mathematicians.
> So what if you can help other mathematicians get lost?
Says JSH who needs a GPS to Žnd his ....
> The mathematical truth remains as it has always been.
Despite JSH¹s best efforts to alter it.
> James Harris
===
Subject: Re: Simple idea, mathematics and common-sense
>Part of the problem is that if you imagine the Žrst type
possibility
>with
>ab = 1,
>where again Œa¹ and Œb¹ are *irrational* the mathematicians
have a
>label for Œb¹ which is algebraic integer, but Œa¹ cannot be
an
>algebraic integer.
>So they don¹t have a label for it!
> Apart from algebriac number you mean? Do yuu even know what
a fraction
> Želd is?
> Your ignorance of these objects doesn¹t mean the
mathematics is
> defective.
Sly insults will get you nowhere here, as I have myself noted
that I¹m
NOT a mathematician!
And I know what algebraic numbers are, but 1/2 is an
algebraic number
along with 1.
The problem here is with a possibility not logically
eliminated where
given
x^2 + (8a + b)x + 8, ab=1, and (8a + b) an integer
it¹s quite possible that Œa¹ is a member of a ring where -1
and 1 are
the only integer units, though it is itself not an algebraic
integer.
So try to insult me all you like matt grime, and try to
obscure the
issue all you like, as Arturo Magidin has successfully done
for so
many years.
Maybe later you¹ll go somewhere to laugh to yourself at the
ignorance
of other readers, but the mathematics will not change. So
what if you
can help other mathematicians get lost?
The mathematical truth remains as it has always been.
James Harris
===
Subject: Re: Simple idea, mathematics and common-sense
> Now, if we go into the realm of algebraic integers,
> though, I¹m not even sure what a prime is therein, although
there
> are an awful lot of units.
In fact, in the ring of algebraic integers there are no
primes, because,
for example, for each algebraic integer p, the equation x^2 -
p = 0
shows that p factors into +- sqrt(p), which are also
algebraic integers.
===
Subject: Re: Simple idea, mathematics and common-sense
In sci.logic, Virgil
<virgil@COMCAST.com>
>> Now, if we go into the realm of algebraic integers,
>> though, I¹m not even sure what a prime is therein,
although there
>> are an awful lot of units.
> In fact, in the ring of algebraic integers there are no
primes, because,
> for example, for each algebraic integer p, the equation x^2
- p = 0
> shows that p factors into +- sqrt(p), which are also
algebraic integers.
An excellent point, and a good codiŽcation of the issue I was
trying to make earlier (although p in my case was an integer
prime,
but obviously it doesn¹t matter too much).
No wonder James is confused...
--
#191, ewill3@earthlink.net
It¹s still legal to go .sigless.
===
Subject: Re: Simple idea, mathematics and common-sense
> Now, if we go into the realm of algebraic integers,
> though, I¹m not even sure what a prime is therein, although
there
> are an awful lot of units.
> In fact, in the ring of algebraic integers there are no
primes, because,

> for example, for each algebraic integer p, the equation x^2
- p = 0
> shows that p factors into +- sqrt(p), which are also
algebraic integers.
And as far as units are concerned. Yes there is an awful lot.
Nora
Baron has shown (amongst others I think) that the algebraic
integer units
are dense in the complex numbers.
--
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: Simple idea, mathematics and common-sense
> It doesn¹t matter how much mathematicians try to deny the
obvious if
> people actually *check* what I¹m saying.
People have been checking your claims for years, but Žnding
any of the
the rare and tiny nuggets of truth in your claims will never
make anyone
rich or famous.
===
Subject: Re: Simple idea, mathematics and common-sense
>
> <SNIP
> Yes, I can talk it all out rigorously and in a heavily
mathematical
> format,
>
> James,
>
> I have been following your threads for some time now and I
have seen on

> more occasions than I can count a request for you to do
just this.
> Really?
As you seem to have no connection to what is actually real,
rather than
what merely strokes your ego, yes, really.
> I seem to recall that when you have deigned to acknowledge
these
requests
> it wasn¹t ...[talked] out rigourously and in a heavily
mathematical
> format it was more of a diatribe against mathematicians and
the evil
> society that they control.
> Hey, there¹s this silly error, and I¹ve explained it,
But you keep persisting in that error even after you have
explained it
and publicly apologized for it.
How is it that you now claim that it was not really an error
after:
(1) persisting in that error for months despite clear proofs
by many
posters that it was an error.
(2) your own acknowledgement that it was an error.
Only by isolating yourself from reality.
> mathematicians seem either willing to ignore it, or engage
in rather
> vicious personal attacks or arguments which I see as
designed to hide
> the issue.
You have a great talent both for seeing what is not there and
for not
seeing what is there, and there is no health in you.
> I have seen you claim that clearly stated deŽnitions are
wrong, that
(as
> you did in this post) there is some inherent ambiguity and
that
> professional mathematicians (as well as the very capable
amatuers) are
too
> stupid to understand but I have *NEVER* seen you post
anything that even
I
> would consider to be rigourous.
> Well I¹m tired, so I¹m pushing more on to people like
yourself, so
> consider
Is this another of your dream fantasies and non-sequitur
mathematics
coming up?
> P(x) = (x + 8a)(x + b), where ab=1, and (8a + b) is an odd
integer.
> You should have the expertise to play with that and Žgure
out the
> error in core yourself!
The error in the core exists only in the core of James
Harris, whom
To the rest of us, James Harris is merely a prime example of
what not to
be.
> Having now said that you can do this, when can I expect to
see it?
When your delusion generator gets into high gear, you are
capable of
seeing anything.
> Play with P(x), and see if you¹re smart enough to see the
mistake in
> reasoning with current teaching about the ring of algebraic
integers.
We already see mistakes in reasoning aplenty, but they are
all in your
reasoning.
> Why Žght against humanity, against progress and truth?
We don¹t know why you are compelled to do so. Perhaps it was
due to your
being badly potty trained, assuming, of course, you have been
potty
trained at all.
> James Harris
===
Subject: Re: Simple idea, mathematics and common-sense
<4037AF19.1020602@pacbell.net>
Discussion, linux)
> that of direct algebraic manipulation.
You forgot appeal to Dedekind (or Gauss or anyone whose name
is
mathematically sacred). This is such a powerful proof
principle that
it doesn¹t even require reading anything by these authors or
anything
about these authors aside from popular math histories.
--
Jesse F. Hughes
Radicals are interesting because they were considered
Œradical¹ by
modern mathematics depends on. --Another JSH history lesson
===
Subject: Re: Question regarding Archimedes¹ use of inŽnity to
Žnd a
volume
> I watched the Nova program InŽnite Secrets last
> September, and I have been wondering ever since
> about a part of his Palimpsest that they only briežy
> mentioned. At one point, they found that Archimedes
> had developed a way to use inŽnite slices to calculate
> the volume of an odd shape, but the program does not
> explain what this method is. I have been struggling
> to Žnd an answer to this question, but have had no
> success. Does anyone here happen to know what
> reasoning Archimedes used to develop his method and
> what that method was?
Archimedes was a great inventor and very able in geometry
and reckoning. Here is a good reference regarding his
volume computation:
http://www.mathpages.com/home/kmath343.htm
Archimedes knew about the two faces of invention:
(1) How to play with thoughts and analogies and combining
known facts.
(2) How to rigorously present the results: as rockhard
formulas as well as by elegant proofs.
In a letter to a friend he revealed his Method, which
is a more informal description of (1). This Method was
widely known to exist in some copies but all of them seemed
to have been lost. Until in 1907 a copy was found in Russia.
This was nearly unreadable due to the fact that somebody
had been using the paper again (reused folia = palimpsest).
interesting Žnding:
http://www.wired.com/news/school/0,1383,39870,00.html
To reconstruct the original document, Walters set up a
competition to search for a team of scientists able to
decipher the work. We had lots of applications for the
job globally and we whittled it down to two, Noel said.
The two winners were the Rochester Institute of Technology
(RIT) and Johns Hopkins University.
Both teams were given Žve pages of the palimpsest to
work from. The Žve pages had a variety of problems --
faint text, mold -- one had a forgery on it, Noel said.
The manuscript was so badly damaged that traditional
methods of imaging, using visible or ultraviolet light,
proved ineffective.
of mathematics at the U. S. Military Academy at West Point:
http://www.ihes.fr/~ilan/sawit.html where he describes, how
he got near to that funny thing. Unfortunately the link is
broken. If you like to search any further, I give you the
information at the end of that HTML-File:
MAA Online is edited by Fernando Q. Gouv.90a
(fqgouvea@colby.edu).
Last modiŽed: Wed Oct 28 17:57:21 -0500 1998
Or if you are interested, I can send it to you by e-mail.
Rainer Rosenthal
r.rosenthal@web.de
===
Subject: Re: Question regarding Archimedes¹ use of inŽnity to
Žnd a
volume
> I watched the Nova program InŽnite Secrets last September,
and I
> have been wondering ever since about a part of his
Palimpsest that
> they only briežy mentioned. At one point, they found that
Archimedes
> had developed a way to use inŽnite slices to calculate the
volume of
> an odd shape, but the program does not explain what this
method is. I
> have been struggling to Žnd an answer to this question, but
have had
> no success. Does anyone here happen to know what reasoning
Archimedes
> used to develop his method and what that method was?
He divided the volume into inŽnitesmial slices and used his
lever
prinicples to change those slices into more managable slices
by
balancing the original slices with the new slices.
> When I saw the
> program, I was taking multivariable calculus and
Differential
> Equations, so you have an idea of where I am in my math
education.
> Please, does anyone know the answer?
> Renee
Look for the Great Books of the Western World. This is a 50
or so
volume of translations into English of classic books of the
western world.
One volume is partially devoted to the works of Archimedes.
These volumes may be on the reference shelf of your library.
-- Bill Hale
===
Subject: Re: Question regarding Archimedes¹ use of inŽnity to
Žnd a
volume
Renee Reavis
> I watched the Nova program InŽnite Secrets last September,
and I
> have been wondering ever since about a part of his
Palimpsest that
> they only briežy mentioned. At one point, they found that
Archimedes
> had developed a way to use inŽnite slices to calculate the
volume of
> an odd shape, but the program does not explain what this
method is. I
> have been struggling to Žnd an answer to this question, but
have had
> no success. Does anyone here happen to know what reasoning
Archimedes
> used to develop his method and what that method was? When I
saw the
> program, I was taking multivariable calculus and
Differential
> Equations, so you have an idea of where I am in my math
education.
> Please, does anyone know the answer?
The keyword is method of exhaustion. Google found this:
http://www.newton.dep.anl.gov/newton/askasci/1995/math/
MATH015.HTM
Very ofŽcial-looking :)
LH
===
Subject: Question regarding Archimedes¹ use of inŽnity to Žnd
a volume
I watched the Nova program InŽnite Secrets last September,
and I
have been wondering ever since about a part of his Palimpsest
that
they only briežy mentioned. At one point, they found that
Archimedes
had developed a way to use inŽnite slices to calculate the
volume of
an odd shape, but the program does not explain what this
method is. I
have been struggling to Žnd an answer to this question, but
have had
no success. Does anyone here happen to know what reasoning
Archimedes
used to develop his method and what that method was? When I
saw the
program, I was taking multivariable calculus and Differential
Equations, so you have an idea of where I am in my math
education.
Please, does anyone know the answer?
Renee
===
Subject: 2 and only 2 geometries where Euclidean is like
Newton¹s absolute
time and absolute space
I am going to have to revamp File 103 on FLT, and File 120 of
3 and
only 3 geometries and File 125 of two proofs of the Riemann
Hypothesis in my website of www.iw.net/~a_plutonium/
I did not do much mathematics after 1997 and recently when I
reviewed
my Riemann Hypothesis proof I realized that it is the p-adics
that are
on the 1/2 Real line which means that lines are curved when
out at
inŽnity. There are no straightlines. I have to change and
revise my
Poincare Conjecture proof also.
But directly, I have to toss out my early 1990s proof of 3
and only 3
geometries
because it is really 2 and only 2 geometries. A lot of
revision and I
can hack it if I go slowly.
What does this say about physic? It says alot. I was sort of
uncomfortable in the early 1990s with the statement of 3 and
only 3
geometries idea. Because physics is dominated not by
threesome but
This is basic and fundamental in physics and so why should
mathematics
be cloaked and dressed in triality when physics is duality.
That would mean that Riemannian and Lobachevskian are the
only 2
geometries where the zeroness of Euclidean žat space is not a
geometry. Zeroness is contained in Riemannian geometry as
well as
Lobachevskian.
If the Riemann Hypothesis has all the Natural Numbers on the
1/2 Real
Line and if the NaturalNumbers are really the P-adics, and
since the
p-adics curve then there exists no straight lines. There
exists no
Euclidean Geometry and Euclidean is just a human mental
construct with
no physical substance. The same as Newtonian absolute space
and
absolute time is just a human dreamed up mental construct.
I see alot of work ahead in revising that website of mine.
Archimedes Plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
(www.iw.net/~a_plutonium) website of the science of AP under
revision
what used to be my old science website
www.newphys.se/elektromagnum/physics/LudwigPlutonium from
years 1993
===
Subject: Re: bound on partitions of subsets
> Given a set S of n elements, i need to choose a subset S¹ of
> these elements and then choose a partitioning of S¹. Can
someone tell
> me a compact asymptotic upper bound on the number of ways
this
> can be done? Anything that is better than (2^n)*(B_n). {B_n
is the
> nth bell number}. Essentially a bound on the series sigma
(nC_i B_i)
This number is B_{n+1}. Take your set S and form another S u
{*}
by adding an extra element. Form any partition of your larger
set,
and the cast out the part containing *; this yields a
partition of
a subset of S.
--
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: Isn¹t it beyond obvious ?
Hi Edward Green,
In closing, you mentioned, I don¹t much like Hawking. .
Our tastes obviously diverge then, no crime there.
You started by saying,
Jeff ... I doubt that either of us
understands Godel¹s work well enough to know
just how well it may be extended in its application
outside its original statement .
Should I try to prove to you that I can apply
Godel¹s incompleteness theorem to physics
similar to the way that Hawking
( and so many others ) have done ?
Can you prove that all the Premises of physics
are perfectly complete and consistent ...
and that science can just kick back and relax ?
Isn¹t it simply beyond obvious that
physics and mathematics are both founded
on well tested assumptions ?
===
Subject: Re: An AB¹ + A¹B Kernel in the Matrix and Algebraic
Riccati
Control-Filter Equations
> Should it surprise us to Žnd fg¹ + f¹g, AB¹ U A¹B, AB¹ + A¹B
> to have the same form for functions, sets, and matrices
respect-
> ively? Let¹s look again at zz* or ww* = (x + iy)(x - iy) =
> x^2 + y^2, and as I mentioned in a recent posting, z and z*
or
> w and w* are režections around the real (x) axis. The
transpose
> A¹ of matrix A is also a režection of the matrix. The
> complement A¹ of a set A is an outward or inward režection
> from A to the rest of the Universe. Is the derivative f¹ a
> režection of f in any sense?
No. The derivative has properties such as the Leibniz identity
not shared by complementation. The only link you have
demonstrated
between them is that you use the same notation for both.
You talk about režections but mean involutions --- operations
which when repeated yield the identity. You ask if the
(operation of)
derivative is one. The simple answer is no. It is not the
case that f¹¹ = f
in general. The proof of this should be within even your
abilities.
--
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: An AB¹ + A¹B Kernel in the Matrix and Algebraic
Riccati
Control-Filter Equations
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1NBLxl05338;
The Riccati Differential Equations for scalars is:
1) dy/dt = A(t) + B(t)y + C(t)y^2
with A(t), B(t), C(t) scalar functions of time. However, in
optimal
control theory including Kalman Žlter-predictors we get Matrix
Riccati Equations including for Linear Quadratic Regulators
(LQR)
and Optimal Linear Filtering for the Covariance Matrix P. The
Matrix Riccati Differential Equation has the form:
2) dP/dt = PA¹ + AP +/- PBP -/+ Q
where this time prime (Œ) denotes transpose of a matrix and P
is
symmetric so that P¹ = P and AP¹ = AP. There is also a steady
state or asymptotic time-independent version of (2) called the
Algebraic Matrix Riccati Equation (ARE), with both continuous
and
discrete versions, in which dP/dt = 0.
This would all be of little surprise if the Riccati Equation
were
not the fundamental equation of Growth-Expansion-Contraction
and
similarly of Rare Event Theory (RET) and Logic-Based
Probability
(LBP). For those who do not have the time to examine my long
series of postings to geometry-research and math-history-list
and
researchmathematics@yahoogroups.com and superstringtheory.com
(which at the last time I looked was under a virus and/or
being
moved to a different server), I¹ll be glad to explain it
later,
or you could try to Žgure it out as an exercise.
I call PA¹ + AP (which is PA¹ + AP¹ since P is symmetric) the
kernel of the Riccati Equations in the sense that it is the
key
term involving P on the right-hand-side of the equation other
than the symmetric PBP term. It is what B(t)y of the scalar
Riccati equation becomes for matrices in these control
scenarios.
Should it surprise us to Žnd fg¹ + f¹g, AB¹ U A¹B, AB¹ + A¹B
to have the same form for functions, sets, and matrices
respect-
ively? Let¹s look again at zz* or ww* = (x + iy)(x - iy) =
x^2 + y^2, and as I mentioned in a recent posting, z and z* or
w and w* are režections around the real (x) axis. The
transpose
A¹ of matrix A is also a režection of the matrix. The
complement A¹ of a set A is an outward or inward režection
from A to the rest of the Universe. Is the derivative f¹ a
režection of f in any sense?
Let¹s think of that as an exercise, and I¹ll give a hint. We
know that Aristotle (and if I recall even Plato) regarded the
Potential and Actual as somewhat režections of each other.
Is there any sense in which change of a function can be
thought
of as the kinetic or dynamic part of whatever the function
represents as the potential part?
Osher Doctorow
===
Subject: Re: JSH: Reply, reply, reply
>[...]
>What Harris said was that you¹d be back, and sure enough,
there you
>are.
True. So? I¹ve never taken issue with his claims to be in
control
of the rest of us (given that he¹s ordered some of us not to
reply and also ordered us to reply it¹s hard to see how we
could
behave in a way that could not be construed as giving him just
what we wants. Why would anyone care about that?)
>earle
************************
David C. Ullrich
===
Subject: Re: JSH: Reply, reply, reply
>Come on, reply to my posts!!! Come why don¹t you say
something else,
>huh?
>Try to point out some mistake Nora Baron you stupid shit!
>Hey, you lapdog David Ullrich, yeah, I called you a LAPDOG
Ullrich!!!
>Aren¹t you mad now? Hey, you know, you HAVE to reply right,
or I¹ll
>call your school and try to get you Žred, right?
>Ullrich the university professor cursing in the muck, with
no shame.
>REPLY TO MY POSTS YOU CUR!
>Hey Dik Winter, you fucking shit, reply again! Why don¹t you
throw
>more of my old arguments on a webpage, you stupid fuck!!!
>Yeah, you¹ll all be back want you? Along with C. Bond and
all the
>rest.
>lapdogs
>You are curs, my , my little traveling circus of obsessive
>repliers.
>I spit on you.
>But you¹ll be back, now won¹t you?
> Totally bizarre. When people reply to your posts your curse
them
> and tell them not to. When you make a post or two and people
> don¹t reply immediately, possibly because they¹re at work or
> something, you complain about that.
> The strangest thing I¹ve ever seen. And considering that
I¹ve
> been reading _your_ posts for years that¹s saying something.
>James Harris
> ************************
> David C. Ullrich
*
What Harris said was that you¹d be back, and sure enough,
there you
are.
earle
*
===
Subject: Re: Decomposing higher-order differential equations
> Hi all.
> There is, apparently, a theorem somewhere that states that
> a higher order linear differential equation can be separated
> into a system of coupled Žrst-order differential equations.
> I guess (I¹m not a mathematician) this has to do with the
> Fourier transform and that a n¹th order differential
operator
> takes the form of an n¹th order polynomial in Fourier space,
> and can be factorized into a product of n 1st order
polynomials.
It¹s simpler than that: a_n d^nx/dt^n + ... + a1 dx/dt + a0 x
= 0 can
be rewritten as n coupled Žrst order equations as follows:
dx_n/dt = - (a_{n-1} x_{n-1} + ... + a_1 x_1 + a_0 x_0 ) / a_n
dx_{n-1}/dt = x_n
dx_{n-2}/dt = x_{n-1}
...
dx_1/dt = x_2
dx_0/dt = x_1
In fact, dx^n/dt^n = f(x, dx/dt, ... dx^(n-1)/dt^(n-1), t)
can be
rewritten in this form for any (not necessarily linear) f:
dx_n/dt = f(x_0, x_1, ... , x_{n-1}, t)
dx_{i}/dt = x_{i+1}, 0 <= i <= n - 1
(If f depends explicitly on t, one usually adds dx_{n+1}/dt =
1,
x_{n+1}(0) = 0 to remove such dependence.)
--
P.A.C. Smith
The vast majority of Iraqis want to live in a peaceful, free
world.
And we will Žnd these people and we will bring them to
justice.
===
Subject: Decomposing higher-order differential equations
Hi all.
There is, apparently, a theorem somewhere that states that
a higher order linear differential equation can be separated
into a system of coupled Žrst-order differential equations.
I guess (I¹m not a mathematician) this has to do with the
Fourier transform and that a n¹th order differential operator
takes the form of an n¹th order polynomial in Fourier space,
and can be factorized into a product of n 1st order
polynomials.
Does this theorem have a name I can search for? Is it
described
or mentioned in some text that is accessible for an ad hoc
maths
user like myself?
Rune
===
Subject: Re: Goldberg dual
Don¹t change the total edge length. Let the radius change.
Area
changes and volume changes, too.
> in other words, the fullerenes.
> your guess is ill-posed, since the fullerenes are all
> of 3-way vertices, their duals will be trigonated --
> but the lengths for duals is not the same, although
> it¹s the same *number* of edges.
> it¹s interesting that the fullerenes can thus be modelled
> with trigona, and that Bucky never noticed that!
> A Goldberg polyhedron has all hexagonal faces except for 12
pentagons
> or 6 fourgons or 4 trigons, and are multi-symmetrical. A
soccor ball
> is a Goldberg polyhedron. Goldberg-like is the same but
asymmetrical.
> My guess is this. For 3(N-2) edge elements whose lengths
are variable,
> and where N is is an integer greater that 2, ther is a
maximum
> diameter polyhedron that can be contructed and a minimum
diameter
> polyhedron that can be constructed. The maximim diameter
polyhedron
> that can be built with 3(N-2) edges is a Goldberg
Polyhedron like
> structure. The minimum diameter sphere will be an
omnitriangulated
> icosahedral arrangement of the edge elements.
===
Subject: Re: Decomposing higher-order differential equations
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1NFSFi25579;
>Hi all.
>There is, apparently, a theorem somewhere that states that
>a higher order linear differential equation can be separated
>into a system of coupled Žrst-order differential equations.
>I guess (I¹m not a mathematician) this has to do with the
>Fourier transform and that a n¹th order differential operator
>takes the form of an n¹th order polynomial in Fourier space,
>and can be factorized into a product of n 1st order
polynomials.
>Does this theorem have a name I can search for? Is it
described
>or mentioned in some text that is accessible for an ad hoc
maths
>user like myself?
>Rune
Hi Rune,
if
F(y,y¹,...,y^(n-1),y^(n)) = 0 (*)
is the n-th order differential equation, you can set
y_0 = y, y_1 = y¹,...,y_(n-1) = y^(n-1)
and write equation (*) equivalently as a system of Žrst order
differential equations in the form
y_0¹ = y_1
y_1¹ = y_2
...
y_(n-2)¹ = y_(n-1)
F(y_0,y_1,...,y_(n-1),y_(n-1)¹) = 0.
Is it this what you meant with your question ?
Best wishes
Torsten.
===
Subject: Re: Core error proof, simpler, shorter
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1NFqIV28085;
>It turns out you can prove that there¹s an error in core
with rather
>basic math, using a quadratic:
*Incapable*
>That means that Œa¹ *cannot* be an algebraic integer, as
algebraic
>integers cannot be roots of primitive non-monic polynomials
>irreducible over Q!
So because algebraic integers are all roots of monic
polynimials
it follows that they cannot be roots of non-monic polynomials
?
Ruhahaa, Harristotelian !
Like if all people Žt in a small car it follows that
they do not Žt in a non-small (=large) car ?
Mathematicians are *incapable* ? *Incapable* ?
Maurice
>But notice that Œb¹ IS an algebraic integer as it¹s the root
of a
>monic polynomial with integer coefŽcient.
>Solving 8a^2 - 17a + 1 = 0 with the quadratic formula gives
>a = (17 +/- sqrt(257))/16
>but it¹s clear that *only* one of those roots can be like
1/8 before,
>while the other is like 8 from before, but *neither* is an
algebraic
>integer!
>So what¹s the core error?
>The assumptions of some mathematicians would mean that
>P(x) = (x+8a)(x+b),
>is impossible in an ring where Œa¹ and Œb¹ have properties
like
>integers, because it¹s impossible in the ring of algebraic
integers,
>if ab = 1 and (8a + b) is an integer, when the result is a
polynomial
>irreducible over Q that has Œa¹ as a root!
>See the odd illogic that has been behind posters arguing
with me?
>Clearly there isn¹t anything special about that case besides
our
>inability to *look* at the roots because of the
irreducibility, unlike
>with integers!
>It¹s an integer prejudice, which is kind of funny now, but
also silly.
>It turns out that there *has* to be a ring beyond algebraic
integers,
>which I call the object ring, where there¹s no problem.
>Now you can Žght the core error for all you¹re worth, and
you¹ll just
>be silly.
>The error has been in mathematics long enough.
>James Harris
===
Subject: Re: Statistical independence test for continuous
variables
>> I need to know what are the available statistical
independence tests
>> for continuous variables. I know that the correlation is a
test for
>> linear dependence for continuous variables, but I was
wondering if
>> there were others.
>Both Kendall¹s tau and Spearman¹s rho are sensitive to an
overall
>monotone relation (which may be nonlinear) between two
variables.
>Beyond that, you pretty much need to specify the form of the
relation
>that you¹re looking for. The tighter the speciŽcation, the
more
>powerful the test can be.
>Terminological note: correlations are not tests.
The analog of the Kolmogorov-Smirnov and the
Cramer-von Mises tests are universal tests
in any number of variables. Under the null
hypothesis, they are distribution free.
However, I am not familiar with a tabulation
of their signiŽcance levels. One could also
consider other tests of a similar nature.
However, there are ways of using random signiŽcance
levels to get a test of the desired size. Take k
random samples from the null hypothesis, and if
j = (k+1)*alpha, rejecting if the observed sample
is in the upper j is at level alpha. For the above
tests, if the sample is large, sqrt(n)*T, T the
test statistic, has a limiting distribution, so
that if the sample size is large, one might want
to scale using smaller samples for null data.
--
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: Statistical independence test for continuous
variables
> I need to know what are the available statistical
independence tests
> for continuous variables. I know that the correlation is a
test for
> linear dependence for continuous variables, but I was
wondering if
> there were others.
Both Kendall¹s tau and Spearman¹s rho are sensitive to an
overall
monotone relation (which may be nonlinear) between two
variables.
Beyond that, you pretty much need to specify the form of the
relation
that you¹re looking for. The tighter the speciŽcation, the
more
powerful the test can be.
Terminological note: correlations are not tests.
===
Subject: Cyclotomic Želds
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1NGNqJ31231;
Hi everyone,
Does anyone know of an example of a cyclotomic extension of
the rationals
which is cyclic but not obtained by adjoining a primitive p^m
root of unity
for some prime p? I would be interested in any examples or
references that
you could provide.
Mark
===
Subject: Re: Explaining restricted choice
> Many people (even some fairly good maths graduate students)
struggle a
> bit with the concept of restricted choice. This principle
explains
> (to take the most common illustration) why, in two-child
families, if
> you ask a daughter the gender of her sibling, she will
answer male
> 2/3 of the time. [I¹m making commonsense mathematical
simplifying
> assumptions here. In reality, she would, of course, be
likely to
> answer something like Why are you asking me that for? etc.
And the
> assumption that each birth has a 50/50 probability of being
female is
> open to question, too.]
How did you go about calculating the 2/3? I¹m not aware of
anything in
biology that says that the genders of subsequent children are
dependent, but I don¹t think that¹s what¹s giving you that
number
anyway. Assuming that it¹s 50/50, we have that the
probability of a
boy being the younger sibling, given that the Žrst sibling is
a girl,
is just the probability of a boy being the younger sibling,
which is
1/2, since the two events in our model are independent.
Are you saying that reality somehow adjusts itself to keep the
expected value near 50/50? That by restricting the Žrst birth
to be a
girl, the second one becomes more likely a boy?
> Here is one example which, I think, clariŽes the concept.
Consider
> two experiments, A and B, which both involve 100 coin
tosses. In A,
> the result of the Žrst toss is predetermined to be heads,
but the 99
> other tosses have the usual 50% heads/ 50% tails
probabilities. In B,
> all 100 coins are tossed in the usual way but a restriction
is imposed
> that the trial is rejected and the experiment repeated if
all 100
> coins land tails. If all 100 are tails, the experiment
keeps getting
> repeated until at least 1 of the 100 coins are heads. The
Žnal
> sample of 100 coin tosses is considered the result of B.
> In A, the expected number of heads is 1 + 99/2 = 50.5. In
B, there is
> no restriction in any practical sense because the
rejections are
> astronomically unlikely. Therefore, in B, the expected
number of
> heads is very, very close to 50.
Unfortunately, by your own calculations it isn¹t quite 50.
What are we
to draw from this example? That small restrictions lead to
small
changes in expected value? You use conditional expectation in
A and a
completely revised experiment in B, so it¹s a bit unclear
what you¹re
driving at.
> In my opinion, this extreme example clariŽes the restricted
choice
> concept.
Not especially, to me anyway. Stating formally what
restricted choice
is, without using examples, might be helpful.
> Paul Epstein
===
Subject: Explaining restricted choice
Many people (even some fairly good maths graduate students)
struggle a
bit with the concept of restricted choice. This principle
explains
(to take the most common illustration) why, in two-child
families, if
you ask a daughter the gender of her sibling, she will answer
male
2/3 of the time. [I¹m making commonsense mathematical
simplifying
assumptions here. In reality, she would, of course, be likely
to
answer something like Why are you asking me that for? etc.
And the
assumption that each birth has a 50/50 probability of being
female is
open to question, too.]
Here is one example which, I think, clariŽes the concept.
Consider
two experiments, A and B, which both involve 100 coin tosses.
In A,
the result of the Žrst toss is predetermined to be heads, but
the 99
other tosses have the usual 50% heads/ 50% tails
probabilities. In B,
all 100 coins are tossed in the usual way but a restriction
is imposed
that the trial is rejected and the experiment repeated if all
100
coins land tails. If all 100 are tails, the experiment keeps
getting
repeated until at least 1 of the 100 coins are heads. The Žnal
sample of 100 coin tosses is considered the result of B.
In A, the expected number of heads is 1 + 99/2 = 50.5. In B,
there is
no restriction in any practical sense because the rejections
are
astronomically unlikely. Therefore, in B, the expected number
of
heads is very, very close to 50.
In my opinion, this extreme example clariŽes the restricted
choice
concept.
Paul Epstein
===
Subject: Re: hi....integral problem...
> hello........every body.......
> i saw a integral.
> int [sqrt{1+4(x^2)}] / {c-(x^2)} dx , c is constant.
> ------------------------
> um......possible??
> let me advice, please....thank you.
As usual, I frown at trig substitutions where they can be
avoided:
Classical rationalization:
set
2*x + sqrt(1+4*x^2) = u, so that
-2*x + sqrt(1+4*x^2) = 1/u
x = (u^2-1)/(4*u)
sqrt(1+4*x^2) = (u^2+1)/(2*u)
dx/du = (u^2+1)/(4*u^2)
and you get a rational function to integrate.
(Still lot of work, though.)
A rationalization that takes advantage of the situation:
set x = (v-2)/(4*sqrt(v-1)) , so one of the inverses is
v = (2*x + sqrt(4*x^2+1))^2
then you get it rational and simpler than by the classical
method.
For checking purposes, you should get the transformed
integrand
-1/v - 4*(4*c+1)/(v^2 - 2*(8*c+1)*v + 1)
so the form of the second integral depends on the value of c;
discuss separately c=0, c=-1/4, -1/4<c<0, c>0, c<-1/4.
Looks like more than you asked for; where on earth does such
an
integral arise?
===
Subject: Re: hi....integral problem...
> hello........every body.......
> i saw a integral.
> int [sqrt{1+4(x^2)}] / {c-(x^2)} dx , c is constant.
> ------------------------
> um......possible??
I think so Have you tried substituting x = 1/2 tan(t). Or
perhaps x =
u/(1-u^2)?
-Michael.
===
Subject: hi....integral problem...
hello........every body.......
i saw a integral.
int [sqrt{1+4(x^2)}] / {c-(x^2)} dx , c is constant.
------------------------
um......possible??
let me advice, please....thank you.
===
Subject: Re: problems that are NP hard but not NP?
3QLpj-NoP*NzsIC,boYU]bQ]H¹<P%JD/,.J>y<#4ga3$21:
> Yes. Any decision problem that is so hard that it isn¹t
even in NP
> is NP-hard, but not NP-complete.
While this is the right intuition, I don¹t think it¹s
technically
correct. E.g. a Parity-P-complete problem is unlikely to be
in NP, but
is also unlikely to be NP-hard.
On the other hand, if a problem is complete for a strict
superset of NP,
such as, say, PSPACE (assuming NP!=PSPACE), then it¹s NP-hard
but not
NP-complete.
NP-hardness also arises in situations where for technical
reasons we
don¹t know that a language belongs in NP. For instance, the
Euclidean
traveling salesman problem, with integer coordinates on the
points -- if
we are given a potential solution (a speciŽc tour), we don¹t
know in
polynomial time how to test that the tour has length at most
K, because
the length is a messy sum of square roots and for all we can
prove it
requires exponential precision to compare such values.
--
David Eppstein http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer
Science
===
Subject: Re: problems that are NP hard but not NP?
>I know from deŽnition that NP complete is NP and NP hard.
>But I still don¹t know what¹s the difference between NP
complete and NP
hard.
>Are there problems that are NP hard but not NP complete,
i.e., NP hard but
not NP?
Yes. Any decision problem that is so hard that it isn¹t even
in NP
is NP-hard, but not NP-complete. Also, NP and NP-complete
usually refer to
classes of languages, i.e. decision problems whose answers
are either
yes
(meaning membership in the language) or no (meaning not in the
language).
NP-hard problems are allowed by some authors to be either
languages or
function problems, where the answer is allowed to be a
number, or a set,
or whatever.
Examples of NP-hard but not NP-complete problems:
The travelling salesman optimization problem. Given a
weighted graph, what
is the minimum distance of a path through the graph that
visits all
vertices?
This is a function problem, so it isn¹t in NP because it
isn¹t a decision
problem. The associated decision problem, i.e. is there a
path of distance
at most k for some k, is NP-complete.
The Boolean circuit equivalence problem. Given two
combinational Boolean
circuits, do they compute the same function? This decision
problem isn¹t
in NP (unless the polynomial hierarchy collapses), but is it
in NP^NP,
i.e., NP with an NP oracle. It is NP-hard by reduction from,
say, SAT.
The halting problem. Given a Turing machine, will it halt?
This is a
decsion
problem that isn¹t in NP, or in any other class of the
polynomial hierarchy

since it is undecideable. It is NP-hard: given some language
L in NP, you
can construct a machine M for a particular instance I of L
such that the
machine will halt iff I is in L.
--
Daniel A. Jim.8enez djimenez@cs.utexas.edu
Assistant Professor djimenez@cs.rutgers.edu
Department of Computer Science
Rutgers University
===
Subject: problems that are NP hard but not NP?
I know from deŽnition that NP complete is NP and NP hard.
But I still don¹t know what¹s the difference between NP
complete and NP
hard.
Are there problems that are NP hard but not NP complete,
i.e., NP hard but
not NP?
===
Subject: Re: Collatz Conjecture : Symmetry question.
>Second: The two divmod¹s can be combined into one.
> How? My thinking was to do the divmod 3 Žrst since it will
> fail the test two out of three times allowing the divmod 2
> test to be skipped.
> Well, if n==1 mod 3 and n==0 mod 2, that is the same as
n==4 mod 6.
> -Michael.
Ok, I thought you were refering to the nested if statements,
which if
combined would still have done two mod functions. The mod 6
thing never
occurred to me.
I changed
p3 = divmod(n,3)
if (p3[1]==1):
p2 = divmod(n,2)
if (p2[1]==0):
g.write(str((n-1)/3)+¹n¹)
to
p3 = divmod(n,6)
if (p3[1]==4):
g.write(str((n-1)/3)+¹n¹)
and found that it saves about half a second (1.4%) at Level
64 where the
input Žle is 22 MB. By Level 69, where the input Žle is 76
MB, the
difference is up to about 3.5 seconds (4%). I didn¹t try it,
but I would
expect that by the time you get to level 84, where the input
Žle is
3.3 GB and the run time in hours, the savings would be
signiŽcant.
===
Subject: Re: Collatz Conjecture : Symmetry question.
>Second: The two divmod¹s can be combined into one.
> How? My thinking was to do the divmod 3 Žrst since it will
> fail the test two out of three times allowing the divmod 2
> test to be skipped.
Well, if n==1 mod 3 and n==0 mod 2, that is the same as n==4
mod 6.
-Michael.
===
Subject: Re: Explaining restricted choice
>> Many people (even some fairly good maths graduate
students) struggle a
>> bit with the concept of restricted choice. This principle
explains
>> (to take the most common illustration) why, in two-child
families, if
>> you ask a daughter the gender of her sibling, she will
answer male
>> 2/3 of the time.
> <snip>
>How did you go about calculating the 2/3? I¹m not aware of
anything in
>biology that says that the genders of subsequent children are
>dependent, but I don¹t think that¹s what¹s giving you that
number
>anyway.
> <snipThink conditional probability.
> With 2 children there are 4 equally likely possibilities.
> BB
> BG
> GB
> GG
> We know that there is at least 1 girl. In 2 of the 3
remaining
> cases the other sibling is a boy.
It¹s reasonable to ask for the OP¹s deŽnition. It isn¹t
enough to show that there exists a deŽnition under which the
calculation makes sense. (For example, if the OP meant that
we¹re to choose a daughter randomly from all the daughters in
two-child families and ask her about her sibling, then his
calculation is wrong.)
It sounds as though the OP is referring to what is sometimes
called reduction of the sample space to compute conditional
probabilities. But the event whose conditional probability
is required must still be properly deŽned, and here it isn¹t.
--r.e.s.
===
Subject: Re: Explaining restricted choice
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1NIHIV09314;
>> Many people (even some fairly good maths graduate
students) struggle a
>> bit with the concept of restricted choice. This principle
explains
>> (to take the most common illustration) why, in two-child
families, if
>> you ask a daughter the gender of her sibling, she will
answer male
>> 2/3 of the time.
<snip>
>How did you go about calculating the 2/3? I¹m not aware of
anything in
>biology that says that the genders of subsequent children are
>dependent, but I don¹t think that¹s what¹s giving you that
number
>anyway.
<snip>
Think conditional probability.
With 2 children there are 4 equally likely possibilities.
BB
BG
GB
GG
We know that there is at least 1 girl. In 2 of the 3 remaining
cases the other sibling is a boy.
===
Subject: Re: What is standard deviation?
> Your deŽnition of SD as Root Mean Square Deviation is well
understood.
> Yet I do not see why the degrees of deviation of two sets
of values must
be compared by way of SD deŽned in the above way.
> In other words, What is intrinsically inappropriate if we
were to try to
express SD as either of the following?
> - Mean Absolute Deviation
> - Mean Root Sum Square Deviation
There is nothing intrinsically inappropriate about expressing
the variation
of a data set as the Mean Absolute Deviation, but it wouldn¹t
be the SD.
The SD is deŽned as the Root Mean Squared Deviation. Asking
why you
can¹t deŽne SD = MAD is about the same as asking why you
can¹t call the
numbers {2,3,5,7,11,13,17...} ODD numbers instead of prime
numbers.
You could, but no one would understand you.
As to why the SD is used so frequently, it turns out that
this quantity has
some theoretical statistical properties that make it
convenient to work
with.
It is a natural when paired with least squares estimation.
Minimizing a sum
of squares leads to a mathematical problem that can be solved
by calculus.
The MAD is naturally paired with a different kind of
estimation where
you try to minimize the sum of the absolute deviations.
Nothing wrong with
that, but it is much harder to do analytically because the
absolute value
function is not differentiable at 0.
HTH.
===
Subject: Re: What is standard deviation?
> Your deŽnition of SD as Root Mean Square Deviation is well
understood.
> Yet I do not see why the degrees of deviation of two sets
of values must
be
> compared by way of SD deŽned in the above way.
> In other words, What is intrinsically inappropriate if we
were to try to
> express SD as either of the following?
> - Mean Absolute Deviation
> - Mean Root Sum Square Deviation
> I may sound irrational but as a non-Math student, I am just
trying to get
to
> offering any help.
> Peter
Given values a_1, ..., a_n and probabilities p_1, ... , p_n
(nonnegative adding to 1) the mean is
m = a_1*p_1 + a_n*p_n .
Now for the deviation... For any c, let
f(c) = ((a_1-c)^2*p_1 + ... + (a_n-c)^2*p_n)^(1/2) ,
this is a minimum when c is m. Exponents other than 2 do not
have this
property. They are not minimized at the mean, but at some
other value.
These other exponents may be appropriate when you are using
median or
mode or harmonic mean, rather than ordinary (arithmetic) mean.
[your alternatives are exponent 1/2 and 1.]
===
Subject: Re: What is standard deviation?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1NH2gT02725;
Your deŽnition of SD as Root Mean Square Deviation is well
understood.
Yet I do not see why the degrees of deviation of two sets of
values must be
compared by way of SD deŽned in the above way.
In other words, What is intrinsically inappropriate if we
were to try to
express SD as either of the following?
- Mean Absolute Deviation
- Mean Root Sum Square Deviation
Peter
===
Subject: A Nonlinear Solution
A NONLINEAR SOLUTION
Jon Giffen
eqn of plane pi[1]:
a[1]x[1]+a[2]x[2]+a[3]x[3]+a[4]x[4]+a[5]x[5]+a[6]x[6]+a[7]x[7
]+a[0]=0
N[1]=(a[1],a[2],a[3],a[4],a[5],a[6],a[7]) normal
T=(t,t^2,t^3,t^4,t^5,t^6,t^7) space curve
N[1]*T + a[0]=0 7th degree polynomial or
a[1]t+a[2]t^2+a[3]t^3+a[4]t^4+a[5]t^5+a[6]t^6+a[7]t^7+a[0]=0
Due to symmetry between components of the space curve, the
result is
eqn of plane p[2]:
( a[3]t^2)x[1]
+( -a[3]t+a[4]t^2)x[2]
+(a[3]-a[4]t+a[5]t^2)x[3]
+(a[4]-a[5]t+a[6]t^2)x[4]
+(a[5]-a[6]t+a[7]t^2)x[5]
+(a[6]-a[7]t )x[6]
+(a[7] )x[7]
+ a[0]+a[1]t+a[2]t^2 = 0
The normal to pi[2] is,
N[2]
=
( a[3]t^2,
-a[3]t+a[4]t^2,
a[3]-a[4]t+a[5]t^2,
a[4]-a[5]t+a[6]t^2,
a[5]-a[6]t+a[7]t^2,
a[6]-a[7]t ,
a[7] )
Solve N[2]*N=-a[0] for t using the quadradic formula:
(a[1]a[3]+a[2]a[4]+a[3]a[5]+a[4]a[6]+a[5]a[7])t^2
+(a[2]a[3]+a[3]a[4]+a[4]a[5]+a[5]a[6]+a[6]a[7])t
+ a[3]^2+a[4]^2+a[5]^2+a[6]^2+a[7]^2+a[0] = 0
Sub this t in N[2] for the intersection of N[2] with the
plane pi[1]. Call this point P. Let Q=(-a[0]/|N|^2)N.
pi[1] intersects pi[2] along the line L[1]. The line
through P and Q is line L[2]. T is then the intersecion
of L[1] and L[2].
===
Subject: Re: Solution to a Nonlinear Equation
> A NONLINEAR SOLUTION
> Jon Giffen
> eqn of plane pi[1]:
>
a[1]x[1]+a[2]x[2]+a[3]x[3]+a[4]x[4]+a[5]x[5]+a[6]x[6]+a[7]x[7
]+a[0]=0
> N[1]=(a[1],a[2],a[3],a[4],a[5],a[6],a[7]) normal
> T=(t,t^2,t^3,t^4,t^5,t^6,t^7) space curve
> N[1]*T + a[0]=0 7th degree polynomial or
> a[1]t+a[2]t^2+a[3]t^3+a[4]t^4+a[5]t^5+a[6]t^6+a[7]t^7+a[0]=0
Your case for having a solution to a 7th degree polynomial
would be much
stronger if you would actually solve one.
--
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: Solution to a Nonlinear Equation
A NONLINEAR SOLUTION
Jon Giffen
eqn of plane pi[1]:
a[1]x[1]+a[2]x[2]+a[3]x[3]+a[4]x[4]+a[5]x[5]+a[6]x[6]+a[7]x[7
]+a[0]=0
N[1]=(a[1],a[2],a[3],a[4],a[5],a[6],a[7]) normal
T=(t,t^2,t^3,t^4,t^5,t^6,t^7) space curve
N[1]*T + a[0]=0 7th degree polynomial or
a[1]t+a[2]t^2+a[3]t^3+a[4]t^4+a[5]t^5+a[6]t^6+a[7]t^7+a[0]=0
Due to symmetry between components of the space curve, the
result is
eqn of plane p[2]:
( a[3]t^2)x[1]
+( -a[3]t+a[4]t^2)x[2]
+(a[3]-a[4]t+a[5]t^2)x[3]
+(a[4]-a[5]t+a[6]t^2)x[4]
+(a[5]-a[6]t+a[7]t^2)x[5]
+(a[6]-a[7]t )x[6]
+(a[7] )x[7]
+ a[0]+a[1]t+a[2]t^2 = 0
The normal to pi[2] is,
N[2]
=
( a[3]t^2,
-a[3]t+a[4]t^2,
a[3]-a[4]t+a[5]t^2,
a[4]-a[5]t+a[6]t^2,
a[5]-a[6]t+a[7]t^2,
a[6]-a[7]t ,
a[7] )
Solve N[2]*N=-a[0] for t using the quadradic formula:
(a[1]a[3]+a[2]a[4]+a[3]a[5]+a[4]a[6]+a[5]a[7])t^2
+(a[2]a[3]+a[3]a[4]+a[4]a[5]+a[5]a[6]+a[6]a[7])t
+ a[3]^2+a[4]^2+a[5]^2+a[6]^2+a[7]^2+a[0] = 0
Sub this t in N[2] for the intersection of N[2] with the
plane pi[1]. Call this point P. Let Q=(-a[0]/|N|^2)N.
pi[1] intersects pi[2] along the line L[1]. The line
through P and Q is line L[2]. T is then the intersecion
of L[1] and L[2].
===
Subject: Re: 2 problems for you to solve
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1NKe6C22222;
>> CHALLENGE 1:
>> Can anybody prove/disprove:
>> Let a > 0, b > 0 and c > 0 integers, a different from b,
and d > 0 with
d^2 = a^2 + b^2.
>> If the discriminant of the equation is not 0 and c > 2d,
then the cubic
equation t^3 + (d^2 - c^2) t^2 + 4 a^2 b^2 c^2 = 0 has no
integral roots.
>Notice about the cubic :
>- The fact that c > 2d implies three real roots, two
positive and one
>negative (that can be deduced from the stationary points).
>- Not all roots can be integers, because, if we call them
r1, r2 and
>r3, we have
>r3 = - r1 r2 / (r1 + r2) since the 1st-deg coefŽcient of the
cubic is
>zero, and this can only be an integer if r1 = r2 (impossible
with a
>nonzero discriminant).
>- Let r1 be an integer root and r2 a non-integer one. Then,
from the
>2nd-deg
>coefŽcient of the cubic and the previous equation for r3, we
arrive
>at the
>equation (r1^3 - r2^3) / (r1^2 - r2^2) = c^2 - d^2. But the
LHS cannot
>be an integer by construction (since r1 is integral and r2
is not).
>Thus the cubic cannot have any integer roots.
>Dimitris
Rgerading your Žrst statement:
r1 r2 /(r1 + r2) may be an integer even if r1 and r2 are
different:
Put r1= 21 and r2 = 28. The r1 + r2 = 49 and r1 r2 = 49 12 so
r1 + r2 divides r1 r2. ???
Kent Holing
===
Subject: Re: question about exact Žrst order ODE
> ....
> -1/x dx + 1/y dy = 0
> or M(x,y) = -1/x and N(x,y) = 1/y
> Since dM/dy = dN/dx, this is an exact equation.
> But I could have split the original equation up as
> -y dx + x dy = 0
> In this case, M(x,y) = -y and N(x,y) = x, and now the
partials are not
> equal
> dM/dy = -1 whereas dN/dx = 1
> so now the equation appears to not be exact....
The term to look up in your text book is integrating factors.
Ken Pledger.
===
Subject: Re: question about exact Žrst order ODE
X-SessionID: 8kt_b-5923-45-12120@news.uchicago.edu
X-Hash-Info: post-Žlter,v:1.4
X-Hash: ce8d79e3 b5e0f3b0 1ffb74ad 91d59879 01f55e4e
>I have a question about exact Žrst order differential
equations. As a
>simple example, consider
>dy/dx = y/x
>Of course, I can solve this equation by separation of
variables, or I
>could recognize that this is a homogeneous equation, etc.
Instead, I
>want to look at it in the context of exact equations.
>I can write this as
>M(x,y) dx + N(x,y) dy = 0
>which for this particular example is
>-1/x dx + 1/y dy = 0
>or M(x,y) = -1/x and N(x,y) = 1/y
>Since dM/dy = dN/dx, this is an exact equation.
>But I could have split the original equation up as
>-y dx + x dy = 0
>In this case, M(x,y) = -y and N(x,y) = x, and now the
partials are not
>equal
>dM/dy = -1 whereas dN/dx = 1
>so now the equation appears to not be exact. Am I right that
>exactness depends on how you factor the original equation?
Yes.
Consider that all the various factorizations differ one form
another
only by a global multiplicative factor. In another words, if
you¹ve
one factorization of the form
1) M(x,y) dx + N(x,y) dy = 0
Than any other factorization will be of the form
2) M¹(x,y) dx + N¹(x,y) dy = P(x,y)*(M(x,y) dx + N(x,y) dy) =
0
Now, assume that the equation in form (1) is exact, i.e.
you¹ve
dM/dy = dN/dX
So, what about form (2)? In order for it to be exact, you
must have
d/dy(PM) = d/dx(PN)
i.e.
P*dM/dy + M*dP/dy = P*dN/dx + N*dP/dx
and using the the fact that M anN satisfy the exactness
condition,
this yields
(*) M*dP/dy = N*dP/dx
Now, this is by no means an identity, just a differential
equation for
P, so if you pick an arbitrary P, relationship (*) does not
need to be
satisŽed. In other words, the exactness of (1) does not
translate
into exactness of (2). So, yes, the exactness does depend on
the
factorization.
Mati Meron | When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the
same
===
Subject: Re: question about exact Žrst order ODE
> Am I right that exactness depends on how you factor the
original
equation?
Yes.
===
Subject: question about exact Žrst order ODE
I have a question about exact Žrst order differential
equations. As a
simple example, consider
dy/dx = y/x
Of course, I can solve this equation by separation of
variables, or I
could recognize that this is a homogeneous equation, etc.
Instead, I
want to look at it in the context of exact equations.
I can write this as
M(x,y) dx + N(x,y) dy = 0
which for this particular example is
-1/x dx + 1/y dy = 0
or M(x,y) = -1/x and N(x,y) = 1/y
Since dM/dy = dN/dx, this is an exact equation.
But I could have split the original equation up as
-y dx + x dy = 0
In this case, M(x,y) = -y and N(x,y) = x, and now the
partials are not
equal
dM/dy = -1 whereas dN/dx = 1
so now the equation appears to not be exact. Am I right that
exactness depends on how you factor the original equation?
===
Subject: Re: Graph Isomorphism
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1NLBpO25315;
>Of course I agree that a planar graph may easily have
>exponentially many embeddings. Consequently, checking each
>embedding separately will not make for a polynomial time
>isomorphism algorithm. I was thinking more along the lines of
>breaking up the graph into 3-connected pieces and dealing
with
>each of these separately. I have not worked out the details
and
>am not completely sure if anyone has....
Yes, they have--this was hot stuff thirty years ago. The
pinnacle
of the art was presented in
J. E. Hopcroft and J. K. Wong. Linear time algorithm for
isomorphism of planar graphs (preliminary report). In
Conference Record of Sixth Annual ACM Symposium on Theory of
Computing, pages 172-184, Seattle, Washington, 30 April-2 May
1974.
though I thought the n log n algorithms were easier to
understand
and not so expensive if you are going to generate a canonical
form.
The canonical form requires sorting, so I thought it couldn¹t
possibly be done in linear time. Now I don¹t know if
radix-sort
techniques might be able to do even that.
Dan Hoey
haoyuep@aol.com
===
Subject: Re: Tea with Sarfatti
> Isn¹t he worried about rejection?
> I don¹t think there¹s any worry about that. He¹s his own
biggest fan.
He¹s his only fan.
===
Subject: Re: Prime factorization of a Gaussian integer
> Hi everyone
> I need some help with this:
> Prove that every gaussian integer has a canonic prime
factorization on
> the form z=ep(1)p(2)...p(n) where e is an unit in Z[i] and
p(1) stands
> for p index 1.
> All help is very appreciated
> Thnx in advance
> Pierre
Actually I Žgured this one out myself. So sorry if I spent
anyone¹s time in
vain.
===
Subject: Prime factorization of a Gaussian integer
Hi everyone
I need some help with this:
Prove that every gaussian integer has a canonic prime
factorization on
the form z=ep(1)p(2)...p(n) where e is an unit in Z[i] and
p(1) stands
for p index 1.
All help is very appreciated
Thnx in advance
Pierre
===
I have the pleasure to inform you that the International
conference
Payment Modality, TUNISAIR
Reduction, Flights, Transfer Airport-Hotel)
are now online at <http://www.universites.tn/setit/>
Med Salim BOUHLEL
Chairman SETIT
===
Subject: Re: Old Arguments, Dik Winter and Rick Decker
> Rick Decker¹s page is especially egregious in that he
maintains it on
> the Hamilton College website. Now I¹ve contacted Hamilton
College to
> complain about his webpage, but they trust him.
> Why is that especially egregious? Dik Winter¹s page is at
CWI.
> That¹s a national research institute. Hamilton College is
just a
> private liberal arts school.
> I¹d go for Dik. Bring down CWI. Besides, you already know
that
> Hamilton College trusts Rick. Maybe CWI doesn¹t trust Dik.
You
> haven¹t even checked.
> And, James, it would be *international*! Think CNN! Think
Oprah!
> Think Hard Copy! (Is that show still on?)
Next, James will try to bring down Google, and any other
search engines
that allow one to Žnd out what James did yesterday.
James, what you have posted is now a part of history, and you
cannot
expunge it.
All you do by bitching about those web sites is make more
people aware
of them, and widen the population of those who are familiar
with your
errors.
===
Subject: Re: Old Arguments, Dik Winter and Rick Decker
Discussion, linux)
> Rick Decker¹s page is especially egregious in that he
maintains it on
> the Hamilton College website. Now I¹ve contacted Hamilton
College to
> complain about his webpage, but they trust him.
Why is that especially egregious? Dik Winter¹s page is at CWI.
That¹s a national research institute. Hamilton College is
just a
private liberal arts school.
I¹d go for Dik. Bring down CWI. Besides, you already know that
Hamilton College trusts Rick. Maybe CWI doesn¹t trust Dik. You
haven¹t even checked.
And, James, it would be *international*! Think CNN! Think
Oprah!
Think Hard Copy! (Is that show still on?)
--
But in our enthusiasm, we could not resist a radical overhaul
of the
system, in which all of its major weaknesses have been
exposed,
analyzed, and replaced with new weaknesses.
-- Bruce Leverett (presumably with apologies to Ambrose
Bierce)
===
Subject: Re: Cyclotomic Želds
> Does anyone know of an example of a cyclotomic extension of
the
> rationals which is cyclic, but not obtained by adjoining a
primitive
> p^m root of unity for some prime p?
By cyclotomic extension, do you mean Q(exp(2 pi i / n)) for
some n?
Then each s in the Galois group takes z to z^j, where z =
exp(2 pi i / n)
and j is relatively prime to n. So we can identify the Galois
group
with the multiplicative group of the integers (mod n). It¹s
known
that this group is cyclic if and only if n is a power of an
odd prime,
twice a power of an odd prime, 2, or 4. So the only hope is
to take
n twice a power of an odd prime. But this gives the same
extension as
taking n to be the power of the odd prime.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Cyclotomic Želds
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1NJUqS16095;
Sorry about the lack of line breaks. Let¹s try that again.
Does anyone know of an example of a cyclotomic extension of
the
rationals which is cyclic, but not obtained by adjoining a
primitive
p^m root of unity for some prime p?
Mark
===
Subject: Re: Simply connected domain (cross-posting on
sci.math.research)
> How do I show that a punctured disc is not simply
connected? It looks
> obvious any contour around the hole is not homotopic to a
contour not
> enclosing the hole, in particular a point. But how do I
show it formally,
> using a topological / analytical argument (preferably
utilizing homotopic
> maps)?
> Ryan
For the sake of deŽniteness, let¹s consider the punctured
plane
C* = C {0}
The following approaches work for C*; they can easily be
modiŽed
for a punctured disc.
There are two approaches that are worth considering:
Topological approach: show that the map
z |---> exp(z)
maps C to C*, and is a covering map: every element of z in C
has a neighborhood U that is mapped homeomorphically to a
neighborhood of exp(z) in C*. Further, the set of points
mapped
to a given w in C* is of the form {z + N*2 pi i | N in Z}.
Since C is connected, C* cannot be simply connected, for a
simply
connected space has only one connected covering space; the
covering
map in this case is a (global) homeomorphism.
Analytical approach: DeŽne the 1-form w by:
w = -y/(x^2 + y^2) dx + x/(x^2 + y^2) dy.
Show dw = 0 (that is w is a closed differential), and then
show that if you integrate around the unit circle in the
positive (counterclockwise) direction, the result is 2 pi.
This shows that the punctured plane is not simply-connected,
for a simply-connected space has the property that the
integral of any closed 1-form over a closed loop is zero.
Dale.
===
Subject: Simply connected domain (cross-posting on
sci.math.research)
How do I show that a punctured disc is not simply connected?
It looks
obvious any contour around the hole is not homotopic to a
contour not
enclosing the hole, in particular a point. But how do I show
it formally,
using a topological / analytical argument (preferably
utilizing homotopic
maps)?
Ryan
===
Subject: Re: Combinatorial graph theory question?
> Hi all, can somebody help me with this question?
> Given a bipartite graph G = (X,Y,E) with |X| = 40 and |Y| =
80 such
> that for any subset Z of X, we have that |R(Z)| >= 2|Z|
(where R(Z) is
> the range of Z), show that for each element x of X we can
construct a
> pairing {x, y_1, y_2} where no two x share a y_1 or y_2.
That is, we
Sounds like you need Hall¹s Marriage Theorem. See any good
book
on combinatorics or graph theory.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Combinatorial graph theory question?
Hi all, can somebody help me with this question?
Given a bipartite graph G = (X,Y,E) with |X| = 40 and |Y| =
80 such
that for any subset Z of X, we have that |R(Z)| >= 2|Z|
(where R(Z) is
the range of Z), show that for each element x of X we can
construct a
pairing {x, y_1, y_2} where no two x share a y_1 or y_2. That
is, we
===
Subject: Is there a standard term for semi-inverse of a 2x2
matrix?
Given a 2x2 real matrix A = [[a,b],[c,d]] with d nonzero,
deŽne its
semi-inverse B to be [[|A|,b],[-c,1]]/d, |A| being the
determinant
ad-bc. This has the the property that if A[x,y] = [x¹,y¹],
then
B[x,y¹] = [x¹,y]. That is, B is the inverse of A with respect
to only
one of the two dimensions.
Does this matrix operation have a standard name? Semi-inverse
seems
to be in use already with the meaning of half a solution for
the case
when there¹s no inverse, similar to pseudo-inverse. The above
is
semi as in half the dimensions rather than semi as in
half-baked.
Vaughan Pratt
--
Don¹t contact me at pratt@boole.stanford.edu, substitute cs
for boole
instead.
===
Subject: Re: Fourier Transform of Cumulative Distribution
>Suppose F(k) is Fourier Transform of f(x).
>Let g(x) be antiderivative of f(x). What is G(k) -- the
Fourier
>Transform -- of g(x)?
>For example,
>Dirac Delta function delta(x-x_0) has Fourier Transform
e^(-2*pi*i*k*x_0)
>while
>Heviside step function has Fourier Transform
[delta(k)-i/(pi*k)]/2
>What is connection between F,f,g and G in general case?
These Fourier transforms must be understood in the sense of
distributions.
Formally, one antiderivative g of f is the convolution f*H
where H is the
Heaviside step function. So its Fourier transform is
G(k) = F(k)(delta(k)-i/(pi k))/2. Caution is needed in
interpreting this,
because the product of two tempered distributions is not
necessarily
a tempered distribution.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2

===
Subject: Fourier Transform of Cumulative Distribution
Suppose F(k) is Fourier Transform of f(x).
Let g(x) be antiderivative of f(x). What is G(k) -- the
Fourier
Transform -- of g(x)?
For example,
Dirac Delta function delta(x-x_0) has Fourier Transform
e^(-2*pi*i*k*x_0)
while
Heviside step function has Fourier Transform
[delta(k)-i/(pi*k)]/2
What is connection between F,f,g and G in general case?
===
Subject: Re: Splitting Field
Timothy Murphy <tim@birdsnest.maths.tcd.ie> ha scritto nel
messaggio
> I¹d like to know a quick way to Žnd a splitting Želd for
f(x)=x^5-1
in
> Z_3[x]. A solution is 1. Then setting x+I=c (where I=(g(x)),
> g(x)=1+x+x^2+x^3+x^4), we can consider c in Z_3[x]/I as a
root.
> How to determine other roots?
> I may have misunderstood the question,
> but g(x) is irreducible in F_3 (the Želd with 3 elements).
> (It doesn¹t have a linear factor x+/-1 ,
> so if it were not irreducible it would be the product of 2
quadratics.
> These would have to be different, since gcd(f(x),f¹(x)) =
gcd(x^5-1,x^4)
=
1
> It follows that g(x) | x^{3^2} - 1 = x^9 - 1 => x^5 - 1 |
x^9 - 1,
> which is easily seen not to be the case.)
> It follows that F_3[x]/(g(x)) = F_{3^4},
> and this is a splitting Želd for g(x).
> If c is one root of g(x) then the others are c^3, c^{3^2}
and c^{3^3}.
> Note that all Žnite extensions of a Žnite Želd are normal
(or galois).
> Thus if an irreducible polynomial over F_p has one root in
k > F_p
> it must factorise completely in k.
> So the concept of splitting Želd is more or less trivial
> in the case of Žnite Želds.
> --
===
Subject: Re: Splitting Field
> I¹d like to know a quick way to Žnd a splitting Želd for
f(x)=x^5-1 in
> Z_3[x]. A solution is 1. Then setting x+I=c (where I=(g(x)),
> g(x)=1+x+x^2+x^3+x^4), we can consider c in Z_3[x]/I as a
root.
> How to determine other roots?
I may have misunderstood the question,
but g(x) is irreducible in F_3 (the Želd with 3 elements).
(It doesn¹t have a linear factor x+/-1 ,
so if it were not irreducible it would be the product of 2
quadratics.
These would have to be different, since gcd(f(x),f¹(x)) =
gcd(x^5-1,x^4) =
1
It follows that g(x) | x^{3^2} - 1 = x^9 - 1 => x^5 - 1 | x^9
- 1,
which is easily seen not to be the case.)
It follows that F_3[x]/(g(x)) = F_{3^4},
and this is a splitting Želd for g(x).
If c is one root of g(x) then the others are c^3, c^{3^2} and
c^{3^3}.
Note that all Žnite extensions of a Žnite Želd are normal (or
galois).
Thus if an irreducible polynomial over F_p has one root in k
> F_p
it must factorise completely in k.
So the concept of splitting Želd is more or less trivial
in the case of Žnite Želds.
--
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: Splitting Field
William Hale <hale@tulane.edu> ha scritto nel messaggio
> You can use the same approach that is used for determining
the
> splitting Želd of x^5-1 over the rationals Q.
I need a quicker one.
===
Subject: Re: Splitting Field
> I¹d like to know a quick way to Žnd a splitting Želd for
f(x)=x^5-1 in
> Z_3[x]. A solution is 1. Then setting x+I=c (where I=(g(x)),
> g(x)=1+x+x^2+x^3+x^4), we can consider c in Z_3[x]/I as a
root.
> How to determine other roots?
First, did you prove that g(x) is irreducible over Z_3? (It
is).
You can use the same approach that is used for determining the
splitting Želd of x^5-1 over the rationals Q.
-- Bill Hale
===
Subject: Splitting Field
I¹d like to know a quick way to Žnd a splitting Želd for
f(x)=x^5-1 in
Z_3[x]. A solution is 1. Then setting x+I=c (where I=(g(x)),
g(x)=1+x+x^2+x^3+x^4), we can consider c in Z_3[x]/I as a
root.
How to determine other roots?
TIA
===
Subject: Mean distance to an ellipse from points within it
The mean distance to a circle from points within it can be
shown to be 1/3
of its radius, an easy result which is fairly well known.
Last month, in
geometry.research, someone asked about the mean distance to
an ellipse from
points within it. That mean distance, M, can be expressed as
an integral in
terms of the lengths, a and b, of the semiaxes. Details of
the messy
calculation are omitted here. The result expresses the mean
distance in
closed form in terms of K and E, the complete elliptic
integrals of the
Žrst and second kinds, resp. With e^2 = 1 - (b/a)^2, we get
* M = 4 b/(3 pi) ( E(e^2) - 1/2 (1-e^2) K(e^2) )
[Lest there be any notational confusion, I¹m using
essentially the same
conventions used in
<http://functions.wolfram.com/EllipticIntegrals/>.]
The mean distance can be approximated nicely by an expression
in the form
# 4 b/(3 pi) ( 1 - (1-pi/4) (b/a)^p )
where p is a suitable positive constant. Note that,
regardless of p,
# gives M precisely at both extremes of eccentricity, e = 0
and e = 1.
If we wish # to approximate the mean distance as accurately
as possible
for small eccentricities, it can be shown that p = pi/(8 - 2
pi), which is
about 1.83 . Using that value for p, # provides a lower bound
for the mean
distance, with its worst relative error being roughly -0.0021
.
If we wish # to approximate the mean distance as accurately
as possible
for large eccentricities, it can be shown that p = 2 . Using
that value for
p, # provides an upper bound for the mean distance, with its
worst relative
error being roughly 0.0064 .
If we wish # to approximate the mean distance as accurately
as possible
over all eccentricities, then the value of p, between pi/(8 -
2 pi) and 2,
which minimizes max|relative error| can be determined
numerically. This
gives p = 1.86906..., and # then provides |relative error| <
0.00087 .
Example: For an ellipse having semiaxes lengths 10 and 5, Žnd
the mean
distance to it from points within it.
Using *, we get M = 20/(3 pi) ( E(3/4) - 1/8 K(3/4)) =
1.9979071667...
Using # with p = 1.86906, we get 1.9974 as an approximation
for M, and so
relative error is about -0.00025 .
Has anyone seen a closed form, an approximation, or bounds
for the mean
distance previously?
David W. Cantrell
===
Subject: Re: a Žbred link question
>Well, I can¹t quite see why the automorphism should be
expected
>to come in a reasonable way from the embedding.
If the automorphism doesn¹t come in a reasonable way from the
embedding, then there¹s no reason to expect that the space you
construct from the automorphism (that is, the mapping cylinder
completed by the binding) should be the 3-sphere which was the
target of the embedding. And, in fact, it isn¹t in your case.
Yes, you started with a link L in S^3, and a Seifert surface
A for
that link, but you didn¹t actually *use* the embedding of A
in S^3
to construct the automorphism. You¹d have had a similar
problem
if you¹d started with a link that *is* Žbered and a Seifert
surface
for it that *isn¹t* a Žber surface; you might Žnd it
illuminating
to think about the simplest example of that, namely, the
unknot O
and a maximally-uncomplicated Seifert surface F for O of
genus 1
(in other words, a Heegard torus in S^3 with an open disk
removed).
Why can¹t you construct g:F->F in this case?
>I¹m not so clear
>on your alternate explanation either. I mean, the
construction of
>my Žrst paragraph should give me some sort of 3-manifold for
any
>link L and g in the mapping class group of A, should it not?
For
>g=Id, this is obviously S^1 x S^2, so for g a dehn twist, i
feel
>(or at least i interpret Lickorish to the effect that) the
result
>should be cutting S^1 x S^2 along some surface and gluing
back
>with a dehn twist. Actually, now that I put it that way, it
>seems that all I am doing is a dehn twist around one of the
>S^2¹s, i.e. nothing at all, so I still get S^1 x S^2. Sound
>good? But I still would like to understand what you meant.
>link to Agora.
Lee Rudolph
===
Subject: Re: a Žbred link question
Well, I can¹t quite see why the automorphism should be
expected
to come in a reasonable way from the embedding. I¹m not so
clear
on your alternate explanation either. I mean, the
construction of
my Žrst paragraph should give me some sort of 3-manifold for
any
link L and g in the mapping class group of A, should it not?
For
g=Id, this is obviously S^1 x S^2, so for g a dehn twist, i
feel
(or at least i interpret Lickorish to the effect that) the
result
should be cutting S^1 x S^2 along some surface and gluing back
with a dehn twist. Actually, now that I put it that way, it
seems that all I am doing is a dehn twist around one of the
S^2¹s, i.e. nothing at all, so I still get S^1 x S^2. Sound
good? But I still would like to understand what you meant.
link to Agora.
>>consider a Žbred link L in S3, with Žber A, and
>>f:S3L -> S1,
>>so to reconstruct S3, take the mapping torus
>>(A x I)/~, where (p,1)~(g(p),0) for some automorphism
g:A->A, and
>>then add in the Œbinding¹, i.e. a neighborhood of the link,
>>which amounts to the quotient (p,t)~(p,s) for all p on the
>>boundary of A, and all s,t in I. then it seems to me that
>>i can consider the inverse images of [0,/pi],[pi,2pi],
>>quotient their boundaries as above to get a kind of Heegaard
>>splitting of S^3 into two handlebodies and paste them
together by
>>curves given by g.
> This is all OK so far, if properly interpreted; but you¹ve
> set yourself up to misinterpret it too easily, and in fact,
> that¹s what you do next:
>>The speciŽc example of this that is
>>giving me trouble is L = two unlinked unknots, so A is an
>>untwisted annulus, and g given by a dehn twist along one of
>>the Žbers.
> Since (as you point out below) this L is *not* a Žbered
link,
> this cannot be an example of your construction properly
> interpreted. What¹s gone wrong is that your automorphism g
> of the annulus does not, in fact, come in any reasonable way
> from the embedding of A as an untwisted (and unknotted)
annulus
> in S^3. Or, putting it another way, the identiŽcation of
> A x [0,1-epsilon] with a 1-sided collar of A in S^3, for any
> epsilon > 0, does *not* yield (by passing to the limit as
> epsilon goest to 0) a *continuous* map g:A->A.
>>then my heegaard splitting is genus one, and if
>>i suppose my dehn twist is performed along say f^{-1}(0),
>>i should end up with a meridional disk of one solid torus
>>pasted to a (1,1) curve on the other (i.e. the meridian with
>>the dehn twist), and so the result is S3, and so L is
>>Žbred. the problem is that i have reason to believe that
>>L shouldnt be a Žbred link, mainly because the complement
>>S3L is a solid torus with another solid torus hollowed out,
>>where the hollowed-out torus is contained in a three ball in
>>the Žrst solid torus, and i cant imagine this space as the
>>mapping torus of an annulus. any ideas, help would be
greatly
> You should go look at Gabai¹s paper Detecting Žbred links
in S^3$,
> Gottingen¹s marvelous Agora: search for his name on
> http://134.76.163.65/agora_docs/217324TABLE_OF_CONTENTS.html
> To get a handle on what¹s happening with your unlink L, you
could
> also look at my recent paper with Mikami Hirasawa,
> http://arxiv.org/abs/math.GT/0311134 . L is the next best
> thing to Žbered, in a reasonable sense; and that makes it a
> useful tool for some purposes.
> Lee Rudolph
===
Subject: Re: a Žbred link question
>consider a Žbred link L in S3, with Žber A, and
>f:S3L -> S1,
>so to reconstruct S3, take the mapping torus
>(A x I)/~, where (p,1)~(g(p),0) for some automorphism
g:A->A, and
>then add in the Œbinding¹, i.e. a neighborhood of the link,
>which amounts to the quotient (p,t)~(p,s) for all p on the
>boundary of A, and all s,t in I. then it seems to me that
>i can consider the inverse images of [0,/pi],[pi,2pi],
>quotient their boundaries as above to get a kind of Heegaard
>splitting of S^3 into two handlebodies and paste them
together by
>curves given by g.
This is all OK so far, if properly interpreted; but you¹ve
set yourself up to misinterpret it too easily, and in fact,
that¹s what you do next:
>The speciŽc example of this that is
>giving me trouble is L = two unlinked unknots, so A is an
>untwisted annulus, and g given by a dehn twist along one of
>the Žbers.
Since (as you point out below) this L is *not* a Žbered link,
this cannot be an example of your construction properly
interpreted. What¹s gone wrong is that your automorphism g
of the annulus does not, in fact, come in any reasonable way
from the embedding of A as an untwisted (and unknotted)
annulus
in S^3. Or, putting it another way, the identiŽcation of
A x [0,1-epsilon] with a 1-sided collar of A in S^3, for any
epsilon > 0, does *not* yield (by passing to the limit as
epsilon goest to 0) a *continuous* map g:A->A.
>then my heegaard splitting is genus one, and if
>i suppose my dehn twist is performed along say f^{-1}(0),
>i should end up with a meridional disk of one solid torus
>pasted to a (1,1) curve on the other (i.e. the meridian with
>the dehn twist), and so the result is S3, and so L is
>Žbred. the problem is that i have reason to believe that
>L shouldnt be a Žbred link, mainly because the complement
>S3L is a solid torus with another solid torus hollowed out,
>where the hollowed-out torus is contained in a three ball in
>the Žrst solid torus, and i cant imagine this space as the
>mapping torus of an annulus. any ideas, help would be greatly
You should go look at Gabai¹s paper Detecting Žbred links in
S^3$,
Gottingen¹s marvelous Agora: search for his name on
http://134.76.163.65/agora_docs/217324TABLE_OF_CONTENTS.html
To get a handle on what¹s happening with your unlink L, you
could
also look at my recent paper with Mikami Hirasawa,
http://arxiv.org/abs/math.GT/0311134 . L is the next best
thing to Žbered, in a reasonable sense; and that makes it a
useful tool for some purposes.
Lee Rudolph
===
Subject: a Žbred link question
consider a Žbered link L in S^3, with Žber A, and
f:S^3L -> S^1,
so to reconstruct S^3, take the mapping torus
(A x I)/~, where (p,1)~(g(p),0) for some automorphism g:A->A,
and
then add in the Œbinding¹, i.e. a neighborhood of the link,
which amounts to the quotient (p,t)~(p,s) for all p on the
boundary of A, and all s,t in I. then it seems to me that
i can consider the inverse images of [0,/pi],[pi,2pi],
quotient their boundaries as above to get a kind of Heegaard
splitting into two handlebodies and paste them together by
curves given by g. The speciŽc example of this that is
giving me trouble is L two unlinked unknots, so A is an
untwisted annulus, and g given by a dehn twist along one of
the Žbers. then my heegaard splitting is genus one, and if
i suppose my dehn twist is performed along say f^{-1}(0),
i should end up with a meridional disk of one solid torus
pasted to a (1,1) curve on the other (i.e. the meridian with
the dehn twist), and so the result is S^3, and so L is
Žbred. the problem is that i have reason to believe that
L shouldnt be a Žbred link, mainly because the complement
S^3L is a solid torus with another solid torus hollowed out,
where the hollowed-out torus is contained in a three ball in
the Žrst solid torus, and i cant imagine this space as the
mapping torus of an annulus. any ideas, help would be greatly
appreciated
===
Subject: Re: series clariŽcation
> The text also says its convenient to write p(x) in the form:
> * p(x) = c0 + c1(x-x0) + c2(x-x0)^2+...+ cn(x-x0)^n
> versus p(x) = c0 + c1x + c2x^2 +...+cnx^n
> Is the only reason for writing the approx. polynomial p(x)
in the form
(*)
> so that we get zeros when we plug in x0 into p(x) and its
derivatives to
> Žnd the constants c0...cn?
No, consider log x.
You can use *, but not with x0 = 0.
===
Subject: series clariŽcation
My book shows how to derive the Maclaurin __polynomial__, and
leaves it as
an exercise to do the taylor polynomial.
So from my understanding, the taylor polynomial will
approximate some
function f(x) at x0. In contrast, the Maclaurin polynomial
approximates
f(x) at x0 =0.
The text also says its convenient to write p(x) in the form:
* p(x) = c0 + c1(x-x0) + c2(x-x0)^2+...+ cn(x-x0)^n
versus p(x) = c0 + c1x + c2x^2 +...+cnx^n
Is the only reason for writing the approx. polynomial p(x) in
the form (*)
so that we get zeros when we plug in x0 into p(x) and its
derivatives to
Žnd the constants c0...cn?
===
Subject: Re: Motivation for e; was : Re: Transcendental
numbers other than
pi and e?
> In sci.math, Stephen J. Herschkorn
>
<herschko@rutcor.rutgers.edu<40399C5B.7030104@
rutcor.rutgers.edu>:
>>[Btw, e arises when solving dy/dx = y, and in other ways.]
>Yes, I knew that, but that¹s still not a geometrically
tangible idea,
>the way pi is.
>>Here are three ways to motivate the concept of e.
> 4) The area under 1/t from 1 to x turns out to have many
odd properties;
> in fact it looks a lot like a logarithm. Turns out it is a
> logarithm, with base e.
[ as another poster said, the x value for which the area is 1
is x = e]
> Still not horribly geometric, I know, but somehow it¹s a
very natural
> (ahem) result.
My contribution to some geometric meaning of e :
Let a right triangle A(0),B(0),C(0), A(0)B(0) = 1 horizontal,
B(0)C(0) = 1 vertical and a distance d = A(0)B(0)/n = 1/n
C1
/|
/ |
/ |
/ |
C0 + |
/| |
// | |
/ / | |
/ / | |
/ / | |
/ / | |
/_____/______|____|
A0 A1 B0 B1
Translate segment A(0)B(0) by distance d to segment A(1)B(1)
A(1)C(0) intersects the vertical from B(1) at C(1)
This gives a new triangle A(1),B(1),C(1) with height
h(1) = B(1)C(1)
Repeat the same step to get A(i+1),B(i+1),C(i+1) from
A(i),B(i),C(i):
Translate segment A(i)B(i) by d to A(i+1)B(i+1).
A(i+1)C(i) intersects vertical from B(i+1) at C(i+1).
After n steps, A(n)=B(0) because d=1/n and h(n)=B(n)C(n).
When n increases, h(n) -> e
Proof :
Thales gives h(i+1)/h(i)=1/(1-1/n)
that is h(n) = h(n)/h(0) = ( 1/(1-1/n) )^n
Ln(h(n)) = n*Ln( 1/(1-1/n) ) ~ n*Ln(1+1/n) ~ n*1/n -> 1
and h(n) -> e when n -> oo
--
philippe
(chephip at free dot fr)
===
Subject: Re: Motivation for e; was : Re: Transcendental
numbers other than
pi and e?
In sci.math, Stephen J. Herschkorn
<herschko@rutcor.rutgers.edu>
<40399C5B.7030104@rutcor.rutgers.edu>:
>>
>
>[Btw, e arises when solving dy/dx = y, and in other ways.]
>
>>Yes, I knew that, but that¹s still not a geometrically
tangible idea,
>>the way pi is.
> Here are three ways to motivate the concept of e.
> 1) Compound interest: The effective annual yield for a
nominal annual
> rate r compounded n times annually is (1 + r/n)^n. As n
grows
> arbitrariliy large, the effective yield approaches e^r.
> 2) With some effort, for positive a, one can show that if
f(x) =
> a^x, then f is differentiable everywhere and f¹(x) is
proportional
> to f(x). e is the unique value of a which renders as unity
the
> constant of proportionality. Geometircally tangible? Try
drawing the
> graph of such a function.
> 3) Letting f(x) = log_a (x) (logariithm to the base a) for
a > 0,
> f¹(x) is proportional to 1/x. e is the unique value of a
which
> renders as unity the constant of proportionality.
4) The area under 1/t from 1 to x turns out to have many odd
properties;
in fact it looks a lot like a logarithm. Turns out it is a
logarithm, with base e.
Obviously this is 3), but coming from the reverse direction.
If one takes L(a) = integral (t = 1, a) dt/t, then
L(ab) = L(a) + integral(t = a, ab) dt/t
= L(a) + integral(u = 1, b) (dt/du) (du/(au))
(where t = au)
= L(a) + integral(u = 1, b) (adu/au)
= L(a) + integral(u = 1, b) (du/u)
= L(a) + L(b).
Similar methods show L(1/a) = -L(a). Of course L(1) = 0.
5) If we hypothesize a function f(x) = a0 + a1*x + a2*x^2 +
... + an*x^n +
...,
the usual Maclaurin series, and f¹(x) = f(x) and f(0) = 1,
then we get the inŽnite series of identities
a0 = 1 (the initial condition)
a0 = a1
a1 = 2a2
a2 = 3a3
...
a{n-1} = n(an)
...
and ultimately the absolutely convergent series we all know
and love, exp(x) = 1 + x + x^2/2! + x^3/3!+ x^4/4! + ... +
x^n/n! + ...
exp(a+b) can probably be shown, with a *lot* of schlogging, to
be exp(a)exp(b), and ultimately e = exp(1). (One of the tools
that might be of assistance is the binomial expansion.)
Still not horribly geometric, I know, but somehow it¹s a very
natural
(ahem) result.
--
#191, ewill3@earthlink.net
It¹s still legal to go .sigless.
===
Subject: Motivation for e; was : Re: Transcendental numbers
other than pi
and e?
>>
>>[Btw, e arises when solving dy/dx = y, and in other ways.]
>>
>Yes, I knew that, but that¹s still not a geometrically
tangible idea,
>the way pi is.
Here are three ways to motivate the concept of e.
1) Compound interest: The effective annual yield for a
nominal annual
rate r compounded n times annually is (1 + r/n)^n. As n grows
arbitrariliy large, the effective yield approaches e^r.
2) With some effort, for positive a, one can show that if
f(x) =
a^x, then f is differentiable everywhere and f¹(x) is
proportional
to f(x). e is the unique value of a which renders as unity the
constant of proportionality. Geometircally tangible? Try
drawing the
graph of such a function.
3) Letting f(x) = log_a (x) (logariithm to the base a) for a
> 0,
f¹(x) is proportional to 1/x. e is the unique value of a which
renders as unity the constant of proportionality.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Having trouble with some math that deals with
x,y coordinates
> x¹ = x + d * cos a
> y¹= y + d * sin a
===
Subject: Re: Having trouble with some math that deals with
x,y coordinates
> Basicly, I have a point x,y
> Also, I have an angle from 0 to 360 a
> And I have a distance d
> What I want to do is to start at point x,y.
> Face the angle a
> Move distance d
> and compute the new point
> Anyone know the formula/math for this?
Suppose (x,y) is the origin (0,0). The ray starting at the
origin and
making angle a with the x-axis consists of the points (d *
cos a, d *
sin a) where d is the distance from the origin. If you start
instead at
arbitrary (x,y) you translate everything accordingly. If
(x¹,y¹) is the
Žnal point, then
x¹ = x + d * cos a
y¹= y + d * sin a
===
Subject: Having trouble with some math that deals with x,y
coordinates
Basicly, I have a point x,y
Also, I have an angle from 0 to 360 a
And I have a distance d
What I want to do is to start at point x,y.
Face the angle a
Move distance d
and compute the new point
Anyone know the formula/math for this?
===
Subject: Re: Analysis help
<l6em3059rofmdpmidi77ci9lk9ssprum6t@4ax.com>
>>
>>if f is differentiable on an interval containing zero and if
>>lim(x->0)f¹(x) = L
>>then L = f¹(0).
>>
>> Hint: What does the Mean Value Theorem tell you about
>> (f(x) - f(0))/x when x is near 0?
>That f¹(0) exists and equals L.
>>
>> if f differentiable on (a,b)0, for some a < 0 < b and if
>> lim(x->0)f¹(x) = L
>> then L = f¹(0).
>>
>> is true? No it¹s not - I _gave_ you a counterexample.
>>
>> Let f(x) = -1 for x < 0, f(x) = 1 for x > 0. Are you
>> claiming that f is not differentiable on (a,b){0}
>> for some a < 0 < b, that lim(x->0)f¹(x) does not
>> equal 0, or that f¹(0) = 0?
>>
>Ok, it¹s a counterexample. f¹(0) may not exist.
>Mean value theorem requires f¹ over all of (a,b).
> Actually there _is_ a very interesting/useful
> version of the result that does not require that
> you assume that f¹(0) exists. But it does require
> a hypothesis that does not appear in your version...
Continuity of f over (a,b)
===
Subject: Re: Analysis help
>>
>>if f is differentiable on an interval containing zero and if
>>lim(x->0)f¹(x) = L
>>then L = f¹(0).
>>
>> Hint: What does the Mean Value Theorem tell you about
>> (f(x) - f(0))/x when x is near 0?
>That f¹(0) exists and equals L.
>
>So the premise can be reduced to
>f differentiable on (a,b)0, for some a < 0 < b
>
> Really? Let f(x) = -1 for x < 0, f(x) = 1 for x > 0.
>>Yes, the (top line) of the premise can be reduced to ...
>> if f differentiable on (a,b)0, for some a < 0 < b and if
>> lim(x->0)f¹(x) = L
>> then L = f¹(0).
>> is true? No it¹s not - I _gave_ you a counterexample.
>> Let f(x) = -1 for x < 0, f(x) = 1 for x > 0. Are you
>> claiming that f is not differentiable on (a,b){0}
>> for some a < 0 < b, that lim(x->0)f¹(x) does not
>> equal 0, or that f¹(0) = 0?
>Ok, it¹s a counterexample. f¹(0) may not exist.
>Mean value theorem requires f¹ over all of (a,b).
Actually there _is_ a very interesting/useful
version of the result that does not require that
you assume that f¹(0) exists. But it does require
a hypothesis that does not appear in your version...
************************
David C. Ullrich
===
Subject: Re: Analysis help
===
Subject: Re: Analysis help
> if f differentiable on (a,b)0, for some a < 0 < b and if
> lim(x->0)f¹(x) = L
> then L = f¹(0).
> is true? No it¹s not - I _gave_ you a counterexample.
> Let f(x) = -1 for x < 0, f(x) = 1 for x > 0. Are you
> claiming that f is not differentiable on (a,b){0}
> for some a < 0 < b, that lim(x->0)f¹(x) does not
> equal 0, or that f¹(0) = 0?
>>Ok, it¹s a counterexample. f¹(0) may not exist.
>>Mean value theorem requires f¹ over all of (a,b).
>The mean value theorem was used in this case on [0,b] and
[a,0],
>and those don¹t require f¹(0) to exist. But what else does
the
>mean value theorem require?
f continuous on [a,b]
----
===
Subject: Re: Analysis help
>if f is differentiable on an interval containing zero and if
>lim(x->0)f¹(x) = L
>then L = f¹(0).
>>Hint: What does the Mean Value Theorem tell you about
>>(f(x) - f(0))/x when x is near 0?
> there exists a c in (x,0) or (0,x) such that
> f¹(c) = (f(x) - f(0))/x
> so there is a sequence of c_n¹s in the interval (and not
equal to 0)
> that approach 0, so (f(c_n) - f(0))/c_n -> f¹(0)
> and thus f¹(c_n) -> f¹(0)
> so, since lim(x->0) f¹(x) = L, f¹(c_n) -> L, therefore L =
f¹(0).
> is that correct?
Yes, although I would rather put it as a delta-epsilon
argument.
> Based on the exercise¹s location in the book (understanding
analysis
> by abbott) i don¹t think he intends you to use the MVT. so
do you see
> another way to do it?
No, and I would be surprised if it turned out that there is
another
approach which can be left as an exercise.
Jose Carlos Santos
===
Subject: Re: Analysis help
>> if f differentiable on (a,b)0, for some a < 0 < b and if
>> lim(x->0)f¹(x) = L
>> then L = f¹(0).
>> is true? No it¹s not - I _gave_ you a counterexample.
>> Let f(x) = -1 for x < 0, f(x) = 1 for x > 0. Are you
>> claiming that f is not differentiable on (a,b){0}
>> for some a < 0 < b, that lim(x->0)f¹(x) does not
>> equal 0, or that f¹(0) = 0?
>Ok, it¹s a counterexample. f¹(0) may not exist.
>Mean value theorem requires f¹ over all of (a,b).
The mean value theorem was used in this case on [0,b] and
[a,0],
and those don¹t require f¹(0) to exist. But what else does the
mean value theorem require?
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Analysis help
<a46k305cdguhgj961u6qpu7f14lpe6g1td@4ax.com>
>
>if f is differentiable on an interval containing zero and if
>lim(x->0)f¹(x) = L
>then L = f¹(0).
>
> Hint: What does the Mean Value Theorem tell you about
> (f(x) - f(0))/x when x is near 0?
>>That f¹(0) exists and equals L.
>>
>>So the premise can be reduced to
>>f differentiable on (a,b)0, for some a < 0 < b
>>
>> Really? Let f(x) = -1 for x < 0, f(x) = 1 for x > 0.
>Yes, the (top line) of the premise can be reduced to ...
> if f differentiable on (a,b)0, for some a < 0 < b and if
> lim(x->0)f¹(x) = L
> then L = f¹(0).
> is true? No it¹s not - I _gave_ you a counterexample.
> Let f(x) = -1 for x < 0, f(x) = 1 for x > 0. Are you
> claiming that f is not differentiable on (a,b){0}
> for some a < 0 < b, that lim(x->0)f¹(x) does not
> equal 0, or that f¹(0) = 0?
Ok, it¹s a counterexample. f¹(0) may not exist.
Mean value theorem requires f¹ over all of (a,b).
===
Subject: Re: Analysis help
> if f differentiable on (a,b)0, for some a < 0 < b and if
> lim(x->0)f¹(x) = L
> then L = f¹(0).
> is true? No it¹s not - I _gave_ you a counterexample.
> Let f(x) = -1 for x < 0, f(x) = 1 for x > 0. Are you
> claiming that f is not differentiable on (a,b){0}
> for some a < 0 < b, that lim(x->0)f¹(x) does not
> equal 0, or that f¹(0) = 0?
What if we changed the function and deŽned it at 0,
say...f(0) = 0...it¹s still differentiable on (a,b){0), and
lim(x->0)f¹(x) = 0 right?
but since it¹s not continuous at 0, f¹(0) does not exist (and
does not
equal 0!).
===
Subject: Re: Analysis help
>i am trying to prove the following:
>if f is differentiable on an interval containing zero and if
>lim(x->0)f¹(x) = L
>then L = f¹(0).
> Hint: What does the Mean Value Theorem tell you about
> (f(x) - f(0))/x when x is near 0?
there exists a c in (x,0) or (0,x) such that
f¹(c) = (f(x) - f(0))/x
so there is a sequence of c_n¹s in the interval (and not
equal to 0)
that approach 0, so (f(c_n) - f(0))/c_n -> f¹(0)
and thus f¹(c_n) -> f¹(0)
so, since lim(x->0) f¹(x) = L, f¹(c_n) -> L, therefore L =
f¹(0).
is that correct?
Based on the exercise¹s location in the book (understanding
analysis
by abbott) i don¹t think he intends you to use the MVT. so do
you see
another way to do it?
i¹m happy to have A solution, but i¹m curious to know what
the author
was thinking, i have spent way too much time on this.
===
Subject: Re: Analysis help
> i am trying to prove the following:
> if f is differentiable on an interval containing zero and if
> lim(x->0)f¹(x) = L
> then L = f¹(0).
> I have tried a number of tricks, contradiction, etc. but
nothing seems
> to get me anywhere. Can someone give me a hint as to how a
proof like
> this might go?
> This is not homework, btw.
It is one of my favourite theorems in Real Analysis; the
proof uses the
Mean Value Theorem: for h not zero but still in the interval,
(f(0+h)-f(0))/h = f¹(0+theta*h) where 0<theta<1.
Then you take limit as h goes to 0. You can make it
epsilon-delta
ofŽcial, if needed.
===
Subject: Re: Analysis help
>
>if f is differentiable on an interval containing zero and if
>lim(x->0)f¹(x) = L
>then L = f¹(0).
>
> Hint: What does the Mean Value Theorem tell you about
> (f(x) - f(0))/x when x is near 0?
>>That f¹(0) exists and equals L.
>>
>>So the premise can be reduced to
>>f differentiable on (a,b)0, for some a < 0 < b
>> Really? Let f(x) = -1 for x < 0, f(x) = 1 for x > 0.
>Yes, the (top line) of the premise can be reduced to ...
Meaning that
if f differentiable on (a,b)0, for some a < 0 < b and if
lim(x->0)f¹(x) = L
then L = f¹(0).
is true? No it¹s not - I _gave_ you a counterexample.
Let f(x) = -1 for x < 0, f(x) = 1 for x > 0. Are you
claiming that f is not differentiable on (a,b){0}
for some a < 0 < b, that lim(x->0)f¹(x) does not
equal 0, or that f¹(0) = 0?
************************
David C. Ullrich
===
Subject: Re: Analysis help
<5kuj305vf1sk1umiqfc012f0rm3qh55ofr@4ax.com>
>>
>>if f is differentiable on an interval containing zero and if
>>lim(x->0)f¹(x) = L
>>then L = f¹(0).
>>
>> Hint: What does the Mean Value Theorem tell you about
>> (f(x) - f(0))/x when x is near 0?
>That f¹(0) exists and equals L.
>So the premise can be reduced to
>f differentiable on (a,b)0, for some a < 0 < b
> Really? Let f(x) = -1 for x < 0, f(x) = 1 for x > 0.
Yes, the (top line) of the premise can be reduced to ...
===
Subject: Re: Analysis help
>>if f is differentiable on an interval containing zero and if
>>lim(x->0)f¹(x) = L
>>then L = f¹(0).
>> Hint: What does the Mean Value Theorem tell you about
>> (f(x) - f(0))/x when x is near 0?
>That f¹(0) exists and equals L.
>So the premise can be reduced to
>f differentiable on (a,b)0, for some a < 0 < b
Really? Let f(x) = -1 for x < 0, f(x) = 1 for x > 0.
************************
David C. Ullrich
===
Subject: Re: Analysis help
<c1c7og$9pe$1@nntp.itservices.ubc.ca>
>if f is differentiable on an interval containing zero and if
>lim(x->0)f¹(x) = L
>then L = f¹(0).
> Hint: What does the Mean Value Theorem tell you about
> (f(x) - f(0))/x when x is near 0?
That f¹(0) exists and equals L.
So the premise can be reduced to
f differentiable on (a,b)0, for some a < 0 < b
===
Subject: Re: Analysis help
>i am trying to prove the following:
>if f is differentiable on an interval containing zero and if
>lim(x->0)f¹(x) = L
>then L = f¹(0).
Hint: What does the Mean Value Theorem tell you about
(f(x) - f(0))/x when x is near 0?
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Analysis help
i am trying to prove the following:
if f is differentiable on an interval containing zero and if
lim(x->0)f¹(x) = L
then L = f¹(0).
I have tried a number of tricks, contradiction, etc. but
nothing seems
to get me anywhere. Can someone give me a hint as to how a
proof like
this might go?
This is not homework, btw.
===
Subject: Re: Transcendental numbers other than pi and e?
>Yes, I knew that, but that¹s still not a geometrically
tangible idea,
>the way pi is.
e is the continous 2. JustiŽcation follows.
Take the difference equation:
x_(k+1) - x_k = x_k
<=> x_(k+1) = 2 x_k
<=> x_k = x_0 * 2^k
Now consider the equivalent differential equation:
f¹(x) = f(x)
<=> f(x) = C * e^x
The number e arrises when we take a discrete form and use a
limit to
extend it to the continuum. In here the process takes us from
2 -> e.
--
I¹m not interested in mathematics that might have anything
to do with reality. -- Russell Easterly, in sci.math
===
Subject: Re: Transcendental numbers other than pi and e?
>>
>>[Btw, e arises when solving dy/dx = y, and in other ways.]
>>
>Yes, I knew that, but that¹s still not a geometrically
tangible idea,
>the way pi is.
Still not geometrically, but does it mean more to you when we
express this
by saying that f(x) = e^x is an eigenfunction of the
differential operator
corresponding to the eigenvalue 1? That is, there are
inŽnite-dimensional
vector spaces; the inŽnitely differentiable functions on R is
one such
space. Even in inŽnite-dimesional spaces, linear
transformations
(differentiation here) have eigenvalues and eigenvectors.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Transcendental numbers other than pi and e?
> That is, the characteristic word of an irrational in (0,1)
(which
> always consists of 0¹s and 1¹s only) can be interpreted as
the
> base-b representation of a transcendental number. So for
each base
> b >= 2, the theorem neatly gives uncountably many
transcendentals,
> one for each irrational in (0,1).
> (The Fibonacci word 1011010110110..., or Rabbit Sequence,
which is
> the characteristic word of the golden mean¹s fractional
part, is the
> best-known example. Just this single word represents a
countable
> inŽnity of transcendentals, corresponding to b = 2,3,4,...)
I forgot to add something for Steven O., who might like this
geometrical interpretation of these transcendental numbers ...
Take a piece of graph paper ruled into squares, and deŽne x-
and y-axes. Through the origin draw a staight line whose slope
s is irrational, with 0 < s < 1. Now for the part of the line
in the upper righthand quadrant, note the points where the
line
intersects the graph paper rulings, as you move away from the
origin. These occur on a sequence of vertical and horizontal
rulings, and the 0/1 sequence you get by coding vertical = 0
and horizontal = 1 is the characteristic word of t = s/(s+1).
So pick any irrational slope in (0,1) and Žnd the
corresponding
0/1 sequence, say a,b,c,d,... Then for any integer base b >
=2,
say binary, the number represented by 0.abcd... is
transcendental.
--r.e.s.
===
Subject: Re: Transcendental numbers other than pi and e?
>
> Are there other known, established transcendental numbers
besides pi
> and e? If so, I¹d be curious to know what they are called,
and if
> they can be characterized in a way that would be clear to my
> mathematically simple brain. (For example, I¹ve never found
it easy
> to understand were e comes from, but the idea that pi is
the ratio
> between a circle¹s circumŽrence and diameter is clear
enough.)
> It seems that those which arise naturally are few, or at
least only a
> few are _known_ to be transcendental.
> Liouville constructed an inŽnity of transcendental numbers
using
> continued fractions. Cantor proved that almost all real
numbers are
> transcendental.
I just happen to be studying parts of the book by Allouche &
Shallit
(Automatic Sequences), who prove a very interesting theorem
attributed
to Bohmer. Paraphrasing ...
Let t be an irrational in the unit interval, and let
t_n = žoor((n+1)t) - žoor(nt) for n = 1,2,3,...
Let b be an integer >= 2. Then sum(t_n / b**n, n>=1)
is transcendental.
That is, the characteristic word of an irrational in (0,1)
(which
always consists of 0¹s and 1¹s only) can be interpreted as the
base-b representation of a transcendental number. So for each
base
b >= 2, the theorem neatly gives uncountably many
transcendentals,
one for each irrational in (0,1).
(The Fibonacci word 1011010110110..., or Rabbit Sequence,
which is
the characteristic word of the golden mean¹s fractional part,
is the
best-known example. Just this single word represents a
countable
inŽnity of transcendentals, corresponding to b = 2,3,4,...)
--r.e.s.
===
Subject: Re: Transcendental numbers other than pi and e?
>>
>[Btw, e arises when solving dy/dx = y, and in other ways.]
> Yes, I knew that, but that¹s still not a geometrically
tangible idea,
> the way pi is.
> I appreciate the other replies on the thread, and I did go
to
> Wolfram¹s site and some of the others. It seems there are
dozens of
> special mathematical functions that generate
transcendentals. But pi
> seems to be the only one that pops up from some picture
that is both
> simple and of obvious relevance.
Let f(x) be pretty nearly any function from high school math.
Graph y = f(x) from x = 0 to x = 1 and compute the arc length
of that graph.
Chances are you get a transcendental number. Pi is just the
case
f(x) = sqrt(1 - x^2), up to some rational multiplier.
In particular, are lengths of ellipses, parabolas, and
hyperbolas
tend to be transcendental.
As for e, draw the hyperbola xy = 1 and the vertical line x =
1
and ask yourself how far to the right of that vertical line do
you have to go before the area between the hyperbola, the
x-axis,
and x = 1 accumulates to 1. The answer is e.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Transcendental numbers other than pi and e?
>>
>>[...]
>>
>>
>Hmm, aren¹t the degree measures of the acute angles of a
right
>triangle with integer sides transcendental? This is a sincere
>question - I don¹t know if that is actually known.
>
>easy : arctan p/q is transcendental if p and q are integers
and p/q is not
0
>or 1...
1) Proof or reference?
2) But I made a claim about (arctan p/q) / pi. Is this still
known to
be transcendental?
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Transcendental numbers other than pi and e?
>> Are there other known, established transcendental numbers
besides
>> pi and e? If so, I¹d be curious to know what they are
called, and
>> if they can be characterized in a way that would be clear
to my
>> mathematically simple brain. (For example, I¹ve never
found it
>> easy to understand were e comes from, but the idea that pi
is
>> the ratio between a circle¹s circumŽrence and diameter is
clear
>> enough.)
>>
>>
> It seems that those which arise naturally are few, or at
least
> only a few are _known_ to be transcendental.
>> Hmm, aren¹t the degree measures of the acute angles of a
right
>> triangle with integer sides transcendental? This is a
sincere
>> question - I don¹t know if that is actually known.
easy : arctan p/q is transcendental if p and q are integers
and p/q is not
0
or 1...
> There are 2^{aleph_0} transcendental numbers, betcha no
more than
> aleph_0 are known. To me, aleph_0 out of 2^{aleph_0} is few
If (a_i) is a sequence of 1 and 2 (and tehre is 2^(aleph_0)
of those, x=sum
(a_n /10^(n!)) is trancendental
===
Subject: Re: Transcendental numbers other than pi and e?
> Are there other known, established transcendental numbers
besides pi
> and e? If so, I¹d be curious to know what they are called,
and if
> they can be characterized in a way that would be clear to my
> mathematically simple brain. (For example, I¹ve never found
it easy
> to understand were e comes from, but the idea that pi is
the ratio
> between a circle¹s circumŽrence and diameter is clear
enough.)
> Well, the sine of 1 degree is a transcendental number. The
common
> logarithm of 4 is also a transcendental number.
> Transcendental numbers are, by deŽnition, numbers that can
not be
> derived from integers solely by arithmetic operations and
root
> extractions.
Surely there are algebraic numbers that cannot be derived
from integers
solely by arithmetic operations and root extractions? The
unsolvability
of the general quintic in radicals shows this.
--
G.C.
===
Subject: Re: Transcendental numbers other than pi and e?
>>Are there other known, established transcendental numbers
besides pi
>>and e? If so, I¹d be curious to know what they are called,
and if
>>they can be characterized in a way that would be clear to my
>>mathematically simple brain. (For example, I¹ve never found
it easy
>>to understand were e comes from, but the idea that pi is
the ratio
>>between a circle¹s circumŽrence and diameter is clear
enough.)
>>
>>
>It seems that those which arise naturally are few, or at
least only
a
>few are _known_ to be transcendental.
> Hmm, aren¹t the degree measures of the acute angles of a
right triangle
> with integer sides transcendental? This is a sincere
question - I don¹t
> know if that is actually known.
There are 2^{aleph_0} transcendental numbers, betcha no more
than
aleph_0 are known. To me, aleph_0 out of 2^{aleph_0} is few
--
G.C.
===
Subject: Re: Transcendental numbers other than pi and e?
> Are there other known, established transcendental numbers
besides pi
> and e? If so, I¹d be curious to know what they are called,
and if
> they can be characterized in a way that would be clear to my
> mathematically simple brain. (For example, I¹ve never found
it easy
> to understand were e comes from, but the idea that pi is
the ratio
> between a circle¹s circumŽrence and diameter is clear
enough.)
> Well, the sine of 1 degree is a transcendental number.
False: it is the imaginary part of the 90th root of i, hence
algebraic.
> The common
> logarithm of 4 is also a transcendental number.
> Transcendental numbers are, by deŽnition, numbers that can
not be
> derived from integers solely by arithmetic operations and
root
> extractions.
Incomplete or vague: some roots of polynomials of degree 5 or
more with
integer coefŽcients cannot be obtained by radicals, if that¹s
what you
mean by root extraction. (Galois classiŽed these polynomials.)
Algebraic: roots of polynomials with integer coefŽcients.
Transcendental: real or complex, but not algebraic.
> Michael
===
Subject: Re: Transcendental numbers other than pi and e?
> Are there other known, established transcendental numbers
besides pi
> and e? If so, I¹d be curious to know what they are called,
and if
> they can be characterized in a way that would be clear to my
> mathematically simple brain. (For example, I¹ve never found
it easy
> to understand were e comes from, but the idea that pi is
the ratio
> between a circle¹s circumŽrence and diameter is clear
enough.)
Well, the sine of 1 degree is a transcendental number. The
common
logarithm of 4 is also a transcendental number.
Transcendental numbers are, by deŽnition, numbers that can
not be
derived from integers solely by arithmetic operations and root
extractions.
Michael
===
Subject: Re: Transcendental numbers other than pi and e?
> Arturo, perhaps you mean:
>
> L=sum_{i=0,+oo} 1/10^{i!}
>
> ?
>
> Actually, I meant L = sum_{n=0 to inŽnity} 10^{n!}... (-:
>>I think his point was your series does not converge.
>So, two typos...
True. But as far as I can see _this_ post was absolutely
perfect! Nice work.
************************
David C. Ullrich
===
Subject: Re: Transcendental numbers other than pi and e?
>> Arturo, perhaps you mean:
>>
>> L=sum_{i=0,+oo} 1/10^{i!}
>>
>> ?
>> Actually, I meant L = sum_{n=0 to inŽnity} 10^{n!}... (-:
>I think his point was your series does not converge.
So, two typos...
--
It¹s not denial. I¹m just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Transcendental numbers other than pi and e?
> Isn¹t natural log of 2 transcendental?? Or have I been
mistaken all
> these years?
It is (by Gelfond-Schneider theorem). Lots of other similar
numbers are (ln
(a) for all algebraic =/= 1, a^b for all and b algebraic, a
=/= 0 or
1, and b irrational, etc.) Conjectures are that most usual
numbers are
transcendental (like e+pi, or ln(2)/(sqrt(pi)+1),...but no
proof are yet
known...
>> Steven O.
<Steven@OpZZREMOVE_ALL_Zs_AND_ALL_BETWEEN_ZZComm.com>
> Are there other known, established transcendental numbers
besides pi
> and e?
>> Yes; of course, you probably want to avoid things like
k*pi or k*e,
>> where k is a rational...
> If so, I¹d be curious to know what they are called, and if
> they can be characterized in a way that would be clear to my
> mathematically simple brain. (For example, I¹ve never found
it easy
> to understand were e comes from, but the idea that pi is
the ratio
> between a circle¹s circumŽrence and diameter is clear
enough.)
>> There¹s Liouville¹s constant,
>> L = sum_{i=0 to inŽnity} 10^{n!};
>> The Gelfond-Schneider constant, 2^{sqrt(2)}.
>> Champernowne¹s constant, 0.12345678910111213... (list all
positive
>> integers in order)
>> And others. See for example,
>> http://mathworld.wolfram.com/TranscendentalNumber.html
>> Some of them arise naturally; other¹s arise when trying to
Žnd
>> transcendental numbers (either for the sake of it, or to
answer
>> questions speciŽcally about transcendental numbers, such as
>> Hilbert¹s 7th problem).
>> --
>> It¹s not denial. I¹m just very selective about
>> what I accept as reality.
>> --- Calvin (Calvin and Hobbes)
>> Arturo Magidin
>> magidin@math.berkeley.edu
===
Subject: Re: Transcendental numbers other than pi and e?
Isn¹t natural log of 2 transcendental?? Or have I been
mistaken all these
years?
>Are there other known, established transcendental numbers
besides pi
>and e?
> Yes; of course, you probably want to avoid things like k*pi
or k*e,
> where k is a rational...
> If so, I¹d be curious to know what they are called, and if
>they can be characterized in a way that would be clear to my
>mathematically simple brain. (For example, I¹ve never found
it easy
>to understand were e comes from, but the idea that pi is the
ratio
>between a circle¹s circumŽrence and diameter is clear
enough.)
> There¹s Liouville¹s constant,
> L = sum_{i=0 to inŽnity} 10^{n!};
> The Gelfond-Schneider constant, 2^{sqrt(2)}.
> Champernowne¹s constant, 0.12345678910111213... (list all
positive
> integers in order)
> And others. See for example,
> http://mathworld.wolfram.com/TranscendentalNumber.html
> Some of them arise naturally; other¹s arise when trying to
Žnd
> transcendental numbers (either for the sake of it, or to
answer
> questions speciŽcally about transcendental numbers, such as
Hilbert¹s
> 7th problem).
> --
> It¹s not denial. I¹m just very selective about
> what I accept as reality.
> --- Calvin (Calvin and Hobbes)
> Arturo Magidin
> magidin@math.berkeley.edu
===
Subject: Re: Transcendental numbers other than pi and e?
> Arturo, perhaps you mean:
L=sum_{i=0,+oo} 1/10^{i!}
?
> Actually, I meant L = sum_{n=0 to inŽnity} 10^{n!}... (-:
> I think his point was your series does not converge.
Right.
Liouville¹s constant is sum_{n=1 to inŽnity} 10^{-n!} .
David
===
Subject: Re: Transcendental numbers other than pi and e?
> Arturo, perhaps you mean:
> L=sum_{i=0,+oo} 1/10^{i!}
> ?
> Actually, I meant L = sum_{n=0 to inŽnity} 10^{n!}... (-:
I think his point was your series does not converge.
===
Subject: Re: Transcendental numbers other than pi and e?
>>Are there other known, established transcendental numbers
besides pi
>>and e? If so, I¹d be curious to know what they are called,
and if
>>they can be characterized in a way that would be clear to my
>>mathematically simple brain. (For example, I¹ve never found
it easy
>>to understand were e comes from, but the idea that pi is
the ratio
>>between a circle¹s circumŽrence and diameter is clear
enough.)
>>
>>
>It seems that those which arise naturally are few, or at
least only
a
>few are _known_ to be transcendental.
> Hmm, aren¹t the degree measures of the acute angles of a
right triangle
> with integer sides transcendental? This is a sincere
question - I don¹t
> know if that is actually known.
> --
> Stephen J. Herschkorn herschko@rutcor.rutgers.edu
Lindemann¹s Theorem produces plenty of transcendental
numbers. Here it is,
copied (Žlling in obvious omissions) from
100 Great Problems of Elementary Mathematics, Their History
and
Solution
by Heinrich Doerrie, Dover, ISBN 0-486-61348-8:
As usual, exp(z) means e^z.
[begin]
The [Žnite, non-empty] expression
A_1*exp(a_1) + A_2*exp(a_2) + ...
in which the [algebraic] coefŽcients A_j differ from zero and
in which
the exponents a_j are algebraic numbers differing from each
other, cannot
equal zero.
[end]
The algebraic numbers are allowed to be complex.
The transcendency of e follows from 1*exp(1) + A*exp(0) not
being zero for
any algebraic A.
The transcendency of pi follows from 1*exp(i*pi) + 1*exp(0) =
0
(if pi were algebraic then...)
Now, natural logarithms (use notation log rather than ln) of
(complex)
algebraic numbers other than 1 are transcendental (set up the
argument
imitating the above lines!). In particular, we learn in
complex analysis
that
arctan(x) = (1/(2*i)) * log ((1+i*x) / (1-i*x))
If a right triangle has all sides integer and one of its
acute angles is
theta then tan(theta) is a rational number, so its arctangent
(expressed
by logarithms, and in radians!) is transcendental. And who
uses angular
degrees in calculus anyway? =
Seriously, the degree measure of theta is simply
theta*(180/pi); it may be
easy to complete the answer to your degree inquiry, but I am
eager to
go
home now.
===
Subject: Re: Transcendental numbers other than pi and e?
>[Btw, e arises when solving dy/dx = y, and in other ways.]
Yes, I knew that, but that¹s still not a geometrically
tangible idea,
the way pi is.
I appreciate the other replies on the thread, and I did go to
Wolfram¹s site and some of the others. It seems there are
dozens of
special mathematical functions that generate transcendentals.
But pi
seems to be the only one that pops up from some picture that
is both
simple and of obvious relevance.
Stee O.
===
Subject: Re: Transcendental numbers other than pi and e?
Ć Arturo Magidin <magidin@math.berkeley.edu>
Ūēņįćå
óōļ
ķÜīłķį
> Actually, I meant L = sum_{n=0 to inŽnity} 10^{n!}... (-:
But wouldn¹t L above diverge? Or am I missing something
obvious?
> Sorry for the typo...
> Arturo Magidin
> magidin@math.berkeley.edu
--
Ioannis Galidakis
http://users.forthnet.gr/ath/jgal/
------------------------------------------
Eventually, _everything_ is understandable
===
Subject: Re: Transcendental numbers other than pi and e?
>>Are there other known, established transcendental numbers
besides pi
>>and e? If so, I¹d be curious to know what they are called,
and if
>>they can be characterized in a way that would be clear to my
>>mathematically simple brain. (For example, I¹ve never found
it easy
>>to understand were e comes from, but the idea that pi is
the ratio
>>between a circle¹s circumŽrence and diameter is clear
enough.)
>>
>It seems that those which arise naturally are few, or at
least only a
>few are _known_ to be transcendental.
Hmm, aren¹t the degree measures of the acute angles of a
right triangle
with integer sides transcendental? This is a sincere question
- I don¹t
know if that is actually known.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Transcendental numbers other than pi and e?
> Are there other known, established transcendental numbers
besides pi
> and e? If so, I¹d be curious to know what they are called,
and if
> they can be characterized in a way that would be clear to my
> mathematically simple brain. (For example, I¹ve never found
it easy
> to understand were e comes from, but the idea that pi is
the ratio
> between a circle¹s circumŽrence and diameter is clear
enough.)
> It seems that those which arise naturally are few, or at
least only a
> few are _known_ to be transcendental.
> Liouville constructed an inŽnity of transcendental numbers
using
> continued fractions. Cantor proved that almost all real
numbers are
> transcendental.
> [Btw, e arises when solving dy/dx = y, and in other ways.]
See Courant, R. and Robbins, H. What Is Mathematics?: An
Elementary
Approach to Ideas and Methods for Liouville¹s construction of
transcendental numbers.
--
G.C.
===
Subject: Re: Transcendental numbers other than pi and e?
> Are there other known, established transcendental numbers
besides pi
> and e? If so, I¹d be curious to know what they are called,
and if
> they can be characterized in a way that would be clear to my
> mathematically simple brain. (For example, I¹ve never found
it easy
> to understand were e comes from, but the idea that pi is
the ratio
> between a circle¹s circumŽrence and diameter is clear
enough.)
It seems that those which arise naturally are few, or at
least only a
few are _known_ to be transcendental.
Liouville constructed an inŽnity of transcendental numbers
using
continued fractions. Cantor proved that almost all real
numbers are
transcendental.
[Btw, e arises when solving dy/dx = y, and in other ways.]
--
G.C.
===
Subject: Re: Transcendental numbers other than pi and e?
Adjunct Assistant Professor at the University of Montana.
Ioannis <morpheus@olympus.mons> escribio:
> ģ Arturo Magidin <magidin@math.berkeley.edu>
[NonBreakingSpace].8b.96.87æ.8c .97.99.95
.93[CapitalThorn].92.9b.93.87
>> There¹s Liouville¹s constant,
>> L = sum_{i=0 to inŽnity} 10^{n!};
> Arturo, perhaps you mean:
> L=sum_{i=0,+oo} 1/10^{i!}
Actually, I meant L = sum_{n=0 to inŽnity} 10^{n!}... (-:
Sorry for the typo...
--
It¹s not denial. I¹m just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Transcendental numbers other than pi and e?
Ioannis <morpheus@olympus.mons> escribio:
> ģ Arturo Magidin <magidin@math.berkeley.edu>
[NonBreakingSpace].8b.96.87æ.8c .97.99.95
.93[CapitalThorn].92.9b.93.87
>> There¹s Liouville¹s constant,
>> L = sum_{i=0 to inŽnity} 10^{n!};
> Arturo, perhaps you mean:
> L=sum_{i=0,+oo} 1/10^{i!}
> Also, an easy consequence (should be) that one could derive
countably
> many transcendentals from the above construction :*)
Also
N(b) = Sum(1/b^2^n, n, 1, inf) for b integer >= 2 (I¹m not
sure for b = 2)
The prove is independent of Lioville¹s Th, but relatively
easy:
An Oceans of Zeros Proof That a Certain
Non-Liouville Number is Transcendental
M. J. Knight
American Mathematical Monthly, Vol. 98, No.
10. (Dec., 1991), pp. 947-949.
--
Ignacio Larrosa Canestro
A Coruna (Espana)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: Transcendental numbers other than pi and e?
Ć Arturo Magidin <magidin@math.berkeley.edu>
Ūēņįćå
óōļ
ķÜīłķį
> There¹s Liouville¹s constant,
> L = sum_{i=0 to inŽnity} 10^{n!};
Arturo, perhaps you mean:
L=sum_{i=0,+oo} 1/10^{i!}
?
Also, an easy consequence (should be) that one could derive
countably many
transcendentals from the above construction :*)
> Arturo Magidin
> magidin@math.berkeley.edu
--
Ioannis Galidakis
http://users.forthnet.gr/ath/jgal/
------------------------------------------
Eventually, _everything_ is understandable
===
Subject: Re: Transcendental numbers other than pi and e?
Adjunct Assistant Professor at the University of Montana.
>Are there other known, established transcendental numbers
besides pi
>and e?
Yes; of course, you probably want to avoid things like k*pi
or k*e,
where k is a rational...
> If so, I¹d be curious to know what they are called, and if
>they can be characterized in a way that would be clear to my
>mathematically simple brain. (For example, I¹ve never found
it easy
>to understand were e comes from, but the idea that pi is the
ratio
>between a circle¹s circumŽrence and diameter is clear
enough.)
There¹s Liouville¹s constant,
L = sum_{i=0 to inŽnity} 10^{n!};
The Gelfond-Schneider constant, 2^{sqrt(2)}.
Champernowne¹s constant, 0.12345678910111213... (list all
positive
integers in order)
And others. See for example,
http://mathworld.wolfram.com/TranscendentalNumber.html
Some of them arise naturally; other¹s arise when trying to Žnd
transcendental numbers (either for the sake of it, or to
answer
questions speciŽcally about transcendental numbers, such as
Hilbert¹s
7th problem).
--
It¹s not denial. I¹m just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Transcendental numbers other than pi and e?
Are there other known, established transcendental numbers
besides pi
and e? If so, I¹d be curious to know what they are called,
and if
they can be characterized in a way that would be clear to my
mathematically simple brain. (For example, I¹ve never found
it easy
to understand were e comes from, but the idea that pi is the
ratio
between a circle¹s circumŽrence and diameter is clear enough.)
Steve O.
===
Subject: Re: integration with respect to a function?
integral f(x) dg(x) ... this is a Stieltjes integral.
===
Subject: Re: integration with respect to a function?
>In reading a probability text I encountered this formula:
>INT(-oo,oo) F_Y (t-x) dF_X (x) = 1 if t >= 1, 0 otherwise. I
need to
>conclude something about F_Y, but I was wondering what
exactly is meant by
>the
>dF_X (x)
>in the integral. How should I interpret this?
F_X is almost surely a cumulative distribution function of a
random
variable X. This is a Stieltjes integral. It represents the
same
thing as E[F_Y(t - X)]. (This notation is a little bit
ambiguous,
since the subscript Y in the expectand is not a random
variable.)
By the way, this is the formula for the convolution of the
distributions F_X and F_Y.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: integration with respect to a function?
In reading a probability text I encountered this formula:
INT(-oo,oo) F_Y (t-x) dF_X (x) = 1 if t >= 1, 0 otherwise. I
need to
conclude something about F_Y, but I was wondering what
exactly is meant by
the
dF_X (x)
in the integral. How should I interpret this?
===
Subject: Re: I got low score on math test, please advise me
and take a look
don¹t you think that such rule should have a
> provision stateing that students need some
> performance feedback prior to the drop deadline?
How do you know that he didn¹t get any performance
> feedback before the deadline?
i believe he said that somewhere. now, can you answer
> my question - perhaps in a direct manner this time?
Fine. RULES ARE RULES.
do you always follow rules blind-folded, or are there
ocassions
> where you wonder if some rules are well-founded?
Non-sequitur, since I agree completely with this rule.
> i am not particularly questioning whether you agree with
that one
> rule. my speciŽc question to you, is whether you agree with
the idea
> that a withdrawal deadline should have a provision stating
that
> students need some performance feedback prior to the
deadline.
> a simple Œyes¹ or Œno¹ will sufŽce.
> I¹m not playing your lawyer games. I¹m not on trial, and
you¹re not
> cross-examining me. Good night.
i am deŽnitely NOT playing any games. i am simply trying to
determine
why you objected to my post - that¹s all.
> Doug
===
Subject: Re: I got low score on math test, please advise me
and take a look
don¹t you think that such rule should have a
> provision stateing that students need some
> performance feedback prior to the drop deadline?
How do you know that he didn¹t get any performance
> feedback before the deadline?
i believe he said that somewhere. now, can you answer
> my question - perhaps in a direct manner this time?
Fine. RULES ARE RULES.
do you always follow rules blind-folded, or are there
ocassions
> where you wonder if some rules are well-founded?
> Non-sequitur, since I agree completely with this rule.
> i am not particularly questioning whether you agree with
that one
> rule. my speciŽc question to you, is whether you agree with
the idea
> that a withdrawal deadline should have a provision stating
that
> students need some performance feedback prior to the
deadline.
> a simple Œyes¹ or Œno¹ will sufŽce.
I¹m not playing your lawyer games. I¹m not on trial, and
you¹re not
cross-examining me. Good night.
Doug
===
Subject: Re: I got low score on math test, please advise me
and take a look
> .
> .
> .
> don¹t you think that such rule should have a
> provision stateing that students need some
> performance feedback prior to the drop deadline?
How do you know that he didn¹t get any performance
> feedback before the deadline?
i believe he said that somewhere. now, can you answer
> my question - perhaps in a direct manner this time?
Fine. RULES ARE RULES.
> do you always follow rules blind-folded, or are there
ocassions
> where you wonder if some rules are well-founded?
> Non-sequitur, since I agree completely with this rule.
i am not particularly questioning whether you agree with that
one
rule. my speciŽc question to you, is whether you agree with
the idea
that a withdrawal deadline should have a provision stating
that
students need some performance feedback prior to the deadline.
a simple Œyes¹ or Œno¹ will sufŽce.
> How¹s that?
> does not seem to address the question i originally asked in
a
> direct manner.
> It certainly did. You didn¹t like the answer, but it was
direct.
> Doug
===
Subject: Re: I got low score on math test, please advise me
and take a look
> don¹t you think that such rule should have a
> provision stateing that students need some
> performance feedback prior to the drop deadline?
How do you know that he didn¹t get any performance
> feedback before the deadline?
i believe he said that somewhere. now, can you answer
> my question - perhaps in a direct manner this time?
> Fine. RULES ARE RULES.
> do you always follow rules blind-folded, or are there
ocassions
> where you wonder if some rules are well-founded?
Non-sequitur, since I agree completely with this rule.
> How¹s that?
> does not seem to address the question i originally asked in
a
> direct manner.
It certainly did. You didn¹t like the answer, but it was
direct.
Doug
===
Subject: Re: I got low score on math test, please advise me
and take a look
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1N1ahY25201;
.
.
.
> don¹t you think that such rule should have a
> provision stateing that students need some
> performance feedback prior to the drop deadline?
How do you know that he didn¹t get any performance
> feedback before the deadline?
>
> i believe he said that somewhere. now, can you answer
> my question - perhaps in a direct manner this time?
> Fine. RULES ARE RULES.
do you always follow rules blind-folded, or are there
ocassions
where you wonder if some rules are well-founded?
> How¹s that?
does not seem to address the question i originally asked in a
direct manner.
>Doug
===
Subject: Re: JSH
> There is an internet and use-net phenomenon known as a
troll. JSH is
> a troll. A troll makes deliberately provocative posts to
get people
> to reply so he/she can make lots more posts, etc., wasting
everyone¹s
> time. This is what JSH does and he even gets angry when
people don¹t
> reply to his posts, further evidence of his troll status.
There is
> also a name for people who compulsively reply to trolls:
troll fodder.
Not to sound rude, but do you think you¹re the Žrst person to
piece this
together?
And, after countless other similar attempts have failed over
the years, do
you think that *you* are the one who will succeed in ending
the madness?
===
Subject: Re: JSH
> There is an internet and use-net phenomenon known as a
troll. JSH is
> a troll. A troll makes deliberately provocative posts to
get people
> to reply so he/she can make lots more posts, etc., wasting
everyone¹s
> time. This is what JSH does and he even gets angry when
people don¹t
> reply to his posts, further evidence of his troll status.
There is
> also a name for people who compulsively reply to trolls:
troll fodder.
Hello Mudder. Hello fodder. I know he¹s got me. I see he¹s
got you too.
===
Subject: Re: JSH
> There is also a name for people who compulsively reply to
trolls: troll
fodder.
What about people that can¹t mind their own damned business?
Failing that, what¹s a good name for people who start threads
about
trolls?
Doug
===
Subject: JSH
There is an internet and use-net phenomenon known as a troll.
JSH is
a troll. A troll makes deliberately provocative posts to get
people
to reply so he/she can make lots more posts, etc., wasting
everyone¹s
time. This is what JSH does and he even gets angry when
people don¹t
reply to his posts, further evidence of his troll status.
There is
also a name for people who compulsively reply to trolls:
troll fodder.
===
Subject: Re: simple surfaces and tangents
> appreciate any help.
> 1. Let U={(u1,u2) in R^2| -pi<u1<pi, -pi<u2<pi}
> and deŽne:
> x(u1,u2)=((2+cos(u1))*cos(u2),(2+cos(u1))*sin(u2),sin(u1))
> Prove that x is a simple surface.
> Suppose:
> x(u1,u2)=((2+cos(u1))*cos(u2),(2+cos(u1))*sin(u2),sin(u1))
>
=y((v1,v2))=((2+cos(v1))*cos(v2),(2+cos(v1))*sin(v2),sin(v1))
that should be
x(u1,u2) = x(v1,v2)
and for injectivity this equality should impy
(u1,u2) = (v1,v2)
> then (2+cos(u1))*cos(u2)=(2+cos(v1))*cos(v2)
> 2cos(u2)+cos(u1)cos(u2)=2cos(v2)+cos(v1)cos(v2)
> 2[cos(u2)-cos(v2)]+cos(u1)cos(u2)-cos(v1)cos(v2)=0
> Stuck here..how do you show¹s its injective?
Show that u1 = v1 and u2 = v2
For example: from
x(u1,u2) = x(v1,v2)
you get 3 equations:
( (2+cos(u1))*cos(u2) , (2+cos(u1))*sin(u2) , sin(u1) )
= ( (2+cos(v1))*cos(v2) , (2+cos(v1))*sin(v2) , sin(v1) )
or
[1] { (2+cos(u1))*cos(u2) = (2+cos(v1))*cos(v2)
[2] { (2+cos(u1))*sin(u2) = (2+cos(v1))*sin(v2)
[3] { sin(u1) = sin(v1)
square [1] and [2] and add to eliminate u2 and v2 and
use [3] to Žnd that
cos(u1) = cos(v1).
Together with [3] this already implies u1 = v1.
Put this information in [1] and [2] to Žnd
cos(u2) = cos(v2)
sin(u2) = sin(v2)
implying
u2 = v2.
Carefully write down each step.
> Also when i compute the cross product of the derivative
> I get a messy expression, what¹s the usual argument or
> techinque in proving such expressions are 1-1?
> For the cross product I get
>
(-cos^2(u1)*cos(u2),-cos^2(u1)sin(u2),-sin(u1)cos(u1)cos^2(u2)
> -sin(u1)cos(u1)*sin(u2)*cos(u2))
If you do it a bit more carefully, you get this:
(2+cos(u1)) * ( cos(u1)cos(u2) , -cos(u1)sin(u2), -sin(u2) )
Verify this, and then calculate the length of this vector for
some
u1 and u2.
Can it be zero? Do you now understand why they took
the number 2? What would happen if they had taken 1/2?
Dirk Vdm
> how do you compute x1,x2 and n as functions of u1 and u2???
> 2. Prove that a(s) is a straight line iff all its tangent
> lines are parallel.
> Stuck on this one.
===
Subject: simple surfaces and tangents
appreciate any help.
1. Let U={(u1,u2) in R^2| -pi<u1<pi, -pi<u2<pi}
and deŽne:
x(u1,u2)=((2+cos(u1))*cos(u2),(2+cos(u1))*sin(u2),sin(u1))
Prove that x is a simple surface.
Suppose:
x(u1,u2)=((2+cos(u1))*cos(u2),(2+cos(u1))*sin(u2),sin(u1))
=y((v1,v2))=((2+cos(v1))*cos(v2),(2+cos(v1))*sin(v2),sin(v1))
then (2+cos(u1))*cos(u2)=(2+cos(v1))*cos(v2)
2cos(u2)+cos(u1)cos(u2)=2cos(v2)+cos(v1)cos(v2)
2[cos(u2)-cos(v2)]+cos(u1)cos(u2)-cos(v1)cos(v2)=0
Stuck here..how do you show¹s its injective?
Also when i compute the cross product of the derivative
I get a messy expression, what¹s the usual argument or
techinque in proving such expressions are 1-1?
For the cross product I get
(-cos^2(u1)*cos(u2),-cos^2(u1)sin(u2),-sin(u1)cos(u1)cos^2(u2)
-sin(u1)cos(u1)*sin(u2)*cos(u2))
how do you compute x1,x2 and n as functions of u1 and u2???
2. Prove that a(s) is a straight line iff all its tangent
lines are parallel.
Stuck on this one.
===
Subject: Re: Having trouble with some math that deals with
x,y coordinates
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1OEF7r11027;
>Basicly, I have a point x,y
>Also, I have an angle from 0 to 360 a
>And I have a distance d
>What I want to do is to start at point x,y.
>Face the angle a
>Move distance d
>and compute the new point
>Anyone know the formula/math for this?
How¹s your trig? Do sine and cosine ring any bells?
The increment in the x-direction is d cos a
The increment in the y-direction is d sin a
Now add the increments to the original position
to Žnd the new position
x_new = x_orig + d cos a
y_new = y_orig + d sin a
Of course, you should do a bit of checking Žrst.
E.g., make sure that d is non-negative
Good luck
phil
===
Subject: re:Axioms deŽning a Žnite Želd
You¹re using the - operation, which is not well deŽned
apriori and
its properties yet to be shown, so this argument is
incomplete. And
of course there¹s a lot more to be shown before it¹s a Želd.
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
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===
Subject: re:Axioms deŽning a Žnite Želd
I didn¹t realize that this was supposed to be a quiz. As far
as -c is
concerned, you can deŽne it as +(-c), where (-c) is the
element
which added to c gives 0. I am aware that to complete this
stage,
uniqueness of the additive inverse has to be shown.
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
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===
Subject: re:Axioms deŽning a Žnite Želd
The Žrst part is a quiz. The second is more of an open
challenge,
as I don¹t know how much you can cut down the axioms so that
they
still deŽne a Žnite Želd.
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
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===
Subject: re:Axioms deŽning a Žnite Želd
Finiteness is also important for showing that a*b = b*a.
I suspect that the a*1 = a axiom isn¹t essential. I might be
wrong.
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
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===
Subject: re:Axioms deŽning a Žnite Želd
To the best of recollection (this goes back over 50 years)
(1) - (7)
with a multiplicative inverse give a Želd (without any
Žniteness).
Replacing the multiplicative inverse by your (9) and
Žniteness makes
it work. I don¹t believe you can get away with any less.
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
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===
Subject: re:Axioms deŽning a Žnite Želd
By the classical deŽnition a Želd must be commutative,
otherwise
it¹s called a skew Želd or something like that. The guys who
developed the concept originally (Galois, Dirichlet, Caushy)
had only
commutative Želds in mind and I stick to their deŽnition.
As for a*1 = a, possibly it can be omitted and it¹s still
would be
possible to prove that there is a unity, perhaps different
from 1.
Alternatively, maybe we can replace this axiom with just 1*1
= 1.
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
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===
Subject: re:Axioms deŽning a Žnite Želd
a*b=b*a is not necessary for something to be a Želd. a*1=a is
part of
the deŽnition of 1, i.e. there exists an element (called 1) so
that a*1=a.
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
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===
Subject: Re: Cyclotomic Želds
> z=exp(2 pi i / n), where n is a power of an odd prime p.
Then there is
> a rational prime, namely p, which has a unique prime ideal
of Q(z)
> containing it.
That¹s not true. If q is a prime which is a primitive root
modulo p^k,
then q is inert in K = Q(exp(2 pi i/p^k)), that is q
generates a prime
ideal in the ring of integers of K. That¹s inŽnitely many q
for each p.
> Is this possible in any other case? That is, does there
> exist w = exp(2 pi i / m) where m is an odd integer which
is not a power
> of a prime and such that there is a prime q which is
contained in a
unique
> prime ideal of Q(w)? Just wondering,
I suppose what you really want is for the ideal generated by
p to be a
non-trivial power of a prime ideal. For this it is necessary
that p
ramiŽes, which is the case iff p | m. Also you need p to be
inert in
Q(exp(2 pi i / m¹)) where m = p^k m¹ and m¹ is prime to p.
Now one case
where this happens is where q = m¹ is prime and p is a
primitive root
modulo q. In particular if m = pq where p is a primitive root
modulo q
but q is not a primitive root modulo p, then p is the only
rational prime
which is a nontrivial power of a prime ideal. For instance,
take p = 3 and
q = 7. Then 3 is a square of a prime ideal in the integers of
Q(exp(2pi i/21)) but 7 is the sixth power of a product of two
distinct
prime ideals.
--
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: Cyclotomic Želds
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1OEHG812529;
z=exp(2 pi i / n), where n is a power of an odd prime p. Then
there is
a rational prime, namely p, which has a unique prime ideal of
Q(z)
containing it. Is this possible in any other case? That is,
does
there exist w = exp(2 pi i / m) where m is an odd integer
which is
not a power of a prime and such that there is a prime q which
is
contained in a unique prime ideal of Q(w)?
Just wondering,
Mark
>> Does anyone know of an example of a cyclotomic extension
of the
>> rationals which is cyclic, but not obtained by adjoining a
primitive
>> p^m root of unity for some prime p?
>By cyclotomic extension, do you mean Q(exp(2 pi i / n)) for
some n?
>Then each s in the Galois group takes z to z^j, where z =
exp(2 pi i / n)
>and j is relatively prime to n. So we can identify the
Galois group
>with the multiplicative group of the integers (mod n). It¹s
known
>that this group is cyclic if and only if n is a power of an
odd prime,
>twice a power of an odd prime, 2, or 4. So the only hope is
to take
>n twice a power of an odd prime. But this gives the same
extension as
>taking n to be the power of the odd prime.
>--
>Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: e is transcendental
>My,question has been resolved neatly and Laconicaly,
>within two lines:
>Re(e^[iPi]) = Re(-1+i[0]) = -1 AND
>Im(e^[iPi]) = Im(-1+i[0]) = 0
>Since there is acceptance to this,
>any more saying has no value.
>>Except to point out that it means that your original
conclusion that
>>exp(i pi) = 0 was not validly drawn from what preceded it.
>>David McAnally
>Yes because I was not either happy with:
>e^[ipi]+1=0 as given in my referenced book:
Well, that is a matter of taste, rather than an objective
observation of
whether the result is true or false. In the Želd of complex
numbers,
exp(i pi)+1 = 0 is true, whether one likes the equality or
not.
>.....However ,since the transcendental numbers formulated by
Liouville
>and G. Cantor,were somehow TECHNICAL, Charles Hermite proved
that
>number e ,the basis of the natural logarithms ,is not
algebraic That was
>important since it was being proved that it was
transcendental ,and
>related to pi via Euler¹s relationship
>e^[iPi]+1 =0
Euler¹s identity is distinct from any consideration of
whether or not e is
transcendental. Spivak gives in his book, Calculus, a proof
of the
transcendence of e which is completely unrelated to Euler¹s
identity. On
the other hand, the proof that pi is transcendental is very
closely
trelated to the fact that e^[i pi]+1 = 0.
>[ Eyler gave the general formula :
> e^[ix]=cosx+isinx in 1748 ,
> gave e and its value 2.718 ,and new since 1728
> the relationship:
> e^[iPi]+1 =0 ].......
>Panagiotis Stefanides
David McAnally
Despite anything you may have heard to the contrary,
the rain in Spain stays almost invariably in the hills.
===
Subject: Re: e is transcendental
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1OEFA111128;
>>My,question has been resolved neatly and Laconicaly,
>>within two lines:
>>Re(e^[iPi]) = Re(-1+i[0]) = -1 AND
>>Im(e^[iPi]) = Im(-1+i[0]) = 0
>>Since there is acceptance to this,
>>any more saying has no value.
>Except to point out that it means that your original
conclusion that
>exp(i pi) = 0 was not validly drawn from what preceded it.
>David McAnally
Yes because I was not either happy with:
e^[ipi]+1=0 as given in my referenced book:
.....However ,since the transcendental numbers formulated by
Liouville and G. Cantor,were somehow TECHNICAL,
Charles Hermite proved that number e ,the basis of the
natural logarithms
,is not algebraic
That was important since it was being proved that it was
transcendental ,and related to pi via Euler¹s relationship
e^[iPi]+1 =0
[ Eyler gave the general formula :
e^[ix]=cosx+isinx in 1748 ,
gave e and its value 2.718 ,and new since 1728
the relationship:
e^[iPi]+1 =0 ].......
Panagiotis Stefanides
===
Subject: Re: Pythagoreian triangles and gaussian integers
>Calculate all natural numbers h<100 so that there are more
than one
>pythagoreian triangle with the hypthenus h and calculate all
>pythagoreian triangles that correspond to these numbers h
[...]
> of P which isn¹t a prime (N=7)). But then how do I Žnd the
triangles.
> I know you went on to explain that but I just can¹t relate
that to the
> facts that I¹m given in the assignement. It says that
> A gaussian integer z is pythagoreian if it¹s of the form
N*w^2 where
> N is a natural number and w^2 is neither all real or all
imaginary.
> It then goes on to say that this gives a perfect way of
Žnding
> triangles. But how does it? I mean the number that I Žnd
isn¹t a
> gaussian integer so how should I do? I realy need help on
this one
Depending on your skill on Gaussian integers (Z[i])
2 methods for Žnding all triangles with a given hypotenus
That is all ways of x^2 + y^2 = h^2 for a given h.
Both start from prime decomposition of h^2
1) elementary method :
Use the Fibonacci relation
(a^2 + b^2)*(c^2 + d^2) = (a*c +/- b*d)^2 + (a*d -/+ b*c)^2
example with 5=1^2+2^2 and 13=2^2+3^2
we derive
5*13 = (1*2+2*3)^2 + (1*3-2*2)^2 = 8^2 + 1^2
and (1*2-2*3)^2 + (1*3+2*2)^2 = 4^2 + 7^2
add each prime factor one at a time in the product to get
the list of all sums of squares = h^2.
2) using Z[i]
Use 5=(2+i)*(2-i), 13=(3+2*i)*(3-2*i) and s.o.
So you have the decomposition in prime factors in Z[i] of h^2
h^2 = x^2 + y^2 = (x+i*y)*(x-i*y)
Hence x+i*y is a selected subset of prime factors from h^2
Just choose any possible subset and calculate x+i*y for every
subset
In the example of h=5*13
h^2 = (2+i)^2 * (2-i)^2 * (3+2*i)^2 * (3-2*i)^2
x+i*y = e * (2+i)^r * (2-i)^(2-r) * (3+2*i)^s * (3-2*i)^(2-s)
You let vary e (e=1,-1,i,-i is a unit), r=0,1,2, s=0,1,2.
(the exponents r, 2-r because the two factors x+i*y and x-i*y
have to be
conjugate)
In both methods, without special care, you get every solution
several
times.
Choose carefully the subsets so you get the solutions only
once.
In our case r=0,s=0 is the same as r=2,s=2 for example
(conjugate)
You have to choose only half the values for r and s (and Žx
e) :
r=0, s=(0,1,2)
r=1, s=(0,1)
That is 5 decompositions of 65^2=4225
63^2+16^2
39^2+52^2
33^2+56^2
25^2+60^2
65^2+0^2
Note that this method gives the trivial decomposition too
h^2 = h^2 + 0^2
I have a java script using Z[i] to list all decompositions of
z = x^2 + y^2 (french, but doesn¹t matter for numbers)
http://chephip.free.fr/exe001.html
--
philippe
(chephip at free dot fr)
===
Subject: Re: Pythagoreian triangles and gaussian integers
> This time I got really stuck on a certain problem. It goes
like this:
>
> A natural number H can be the hypothenus? in a pythagoreian
triangle
> if it contains any of the prime factors of the form 4n+1.
> If H=K*h where h=(p_1^k_1)(p_2^k_2)...p_n^k_n) is the
product of all
> primes on the form 4n+1 then the number N of triangles with
the
> hypothenus h os given by the formula N=(P-1)/2 where
> P=(2k_1+1)(2k_2+1)...(2k_n+1)
> Now here¹s the actual assignment:
> Calculate all natural numbers h<100 so that there are more
than one
> pythagoreian triangle with the hypthenus h and calculate all
> pythagoreian triangles that correspond to these numbers h
> Where are you getting stuck?
> Do you not know what a prime factor is?
> Do you know what a prime factor is, but not know what it
means
> to be of the form 4n + 1?
> Would it help if I ran through an example? Let H = 975.
> Then H = Kh where K = 3 and h = (5^2)(13)
> so n = 2, k_1 = 2, and k_2 = 1
> so P = (5)(3) = 15
> so N = (15 - 1) / 2 = 7
> so there should be 7 pythagorian triangles with hypotenuse
h = (5^2)(13).
> Or maybe you are getting stuck on Žnding the actual
triangles?
> The key here is that surely somewhere you¹ve been told that
> if h = (m^2 + n^2)r with m > n then it is the hypotenuse of
a triangle
> whose sides are 2mnr and (m^2 - n^2)r. So all you have to
do is Žnd
> all the ways of writing (5^2)(13) as (m^2 + n^2)r.
> It helps to note that 5 = 2^2 + 1^2 and 13 = 3^2 + 2^2.
> With some algebraic trickery (actually, this is where the
Gaussian
> integers of your subject header come in handy!) this leads
to
> 5^2 = 4^2 + 3^2 and (5)(13) = 8^2 + 1^2 = 7^2 + 4^2 and
similarly
> for (5^2)(13). Then don¹t forget that r can be 1, 5, 13,
25, or 65,
> and you should be getting there.
Actually there are a few things that I would really
appreciate if
someone could help me understand.
For a start, I¹m really confused on how to Žnd the triangles
that
correspond to a certain hypothenuse. For e.g. a problem is to
Žnd the
number h which corresponds to more than 4 triangles. Finding
this
number isn¹t a problem (just put in N=5,6 etc until you Žnd a
number
of P which isn¹t a prime (N=7)). But then how do I Žnd the
triangles.
I know you went on to explain that but I just can¹t relate
that to the
facts that I¹m given in the assignement. It says that
A gaussian integer z is pythagoreian if it¹s of the form
N*w^2 where
N is a natural number and w^2 is neither all real or all
imaginary.
It then goes on to say that this gives a perfect way of Žnding
triangles. But how does it? I mean the number that I Žnd
isn¹t a
gaussian integer so how should I do? I realy need help on
this one
Thnx
Pierre
===
Subject: Re: Pythagoreian triangles and gaussian integers
> This time I got really stuck on a certain problem. It goes
like this:
>
> A natural number H can be the hypothenus? in a pythagoreian
triangle
> if it contains any of the prime factors of the form 4n+1.
> If H=K*h where h=(p_1^k_1)(p_2^k_2)...p_n^k_n) is the
product of all
> primes on the form 4n+1 then the number N of triangles with
the
> hypothenus h os given by the formula N=(P-1)/2 where
> P=(2k_1+1)(2k_2+1)...(2k_n+1)
> Now here¹s the actual assignment:
> Calculate all natural numbers h<100 so that there are more
than one
> pythagoreian triangle with the hypthenus h and calculate all
> pythagoreian triangles that correspond to these numbers h
> Where are you getting stuck?
> Do you not know what a prime factor is?
> Do you know what a prime factor is, but not know what it
means
> to be of the form 4n + 1?
> Would it help if I ran through an example? Let H = 975.
> Then H = Kh where K = 3 and h = (5^2)(13)
> so n = 2, k_1 = 2, and k_2 = 1
> so P = (5)(3) = 15
> so N = (15 - 1) / 2 = 7
> so there should be 7 pythagorian triangles with hypotenuse
h = (5^2)(13).
> Or maybe you are getting stuck on Žnding the actual
triangles?
> The key here is that surely somewhere you¹ve been told that
> if h = (m^2 + n^2)r with m > n then it is the hypotenuse of
a triangle
> whose sides are 2mnr and (m^2 - n^2)r. So all you have to
do is Žnd
> all the ways of writing (5^2)(13) as (m^2 + n^2)r.
> It helps to note that 5 = 2^2 + 1^2 and 13 = 3^2 + 2^2.
> With some algebraic trickery (actually, this is where the
Gaussian
> integers of your subject header come in handy!) this leads
to
> 5^2 = 4^2 + 3^2 and (5)(13) = 8^2 + 1^2 = 7^2 + 4^2 and
similarly
> for (5^2)(13). Then don¹t forget that r can be 1, 5, 13,
25, or 65,
> and you should be getting there.
thnx for the help, cleared some things out. Actually now my
problem is
to Žnd those h<100. I got 5^2=25, 5*13=65 and 5*17=85. But
there
should be 2 more h<100.
===
Subject: Re: Pythagoreian triangles and gaussian integers
> This time I got really stuck on a certain problem. It goes
like this:
> A natural number H can be the hypothenus? in a pythagoreian
triangle
> if it contains any of the prime factors of the form 4n+1.
> If H=K*h where h=(p_1^k_1)(p_2^k_2)...p_n^k_n) is the
product of all
> primes on the form 4n+1 then the number N of triangles with
the
> hypothenus h os given by the formula N=(P-1)/2 where
> P=(2k_1+1)(2k_2+1)...(2k_n+1)
> Now here¹s the actual assignment:
> Calculate all natural numbers h<100 so that there are more
than one
> pythagoreian triangle with the hypthenus h and calculate all
> pythagoreian triangles that correspond to these numbers h
Where are you getting stuck?
Do you not know what a prime factor is?
Do you know what a prime factor is, but not know what it means
to be of the form 4n + 1?
Would it help if I ran through an example? Let H = 975.
Then H = Kh where K = 3 and h = (5^2)(13)
so n = 2, k_1 = 2, and k_2 = 1
so P = (5)(3) = 15
so N = (15 - 1) / 2 = 7
so there should be 7 pythagorian triangles with hypotenuse h
= (5^2)(13).
Or maybe you are getting stuck on Žnding the actual triangles?
The key here is that surely somewhere you¹ve been told that
if h = (m^2 + n^2)r with m > n then it is the hypotenuse of a
triangle
whose sides are 2mnr and (m^2 - n^2)r. So all you have to do
is Žnd
all the ways of writing (5^2)(13) as (m^2 + n^2)r.
It helps to note that 5 = 2^2 + 1^2 and 13 = 3^2 + 2^2.
With some algebraic trickery (actually, this is where the
Gaussian
integers of your subject header come in handy!) this leads to
5^2 = 4^2 + 3^2 and (5)(13) = 8^2 + 1^2 = 7^2 + 4^2 and
similarly
for (5^2)(13). Then don¹t forget that r can be 1, 5, 13, 25,
or 65,
and you should be getting there.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Pythagoreian triangles and gaussian integers
Hi everyone, again
This time I got really stuck on a certain problem. It goes
like this:
A natural number H can be the hypothenus? in a pythagoreian
triangle
if it contains any of the prime factors of the form 4n+1.
If H=K*h where h=(p_1^k_1)(p_2^k_2)...p_n^k_n) is the product
of all
primes on the form 4n+1 then the number N of triangles with
the
hypothenus h os given by the formula N=(P-1)/2 where
P=(2k_1+1)(2k_2+1)...(2k_n+1)
Now here¹s the actual assignment:
Calculate all natural numbers h<100 so that there are more
than one
pythagoreian triangle with the hypthenus h and calculate all
pythagoreian triangles that correspond to these numbers h
Excuse me for the language but it¹s a tricky part to
translate and I
got little time so. Don¹t really know if it¹s spelled
hypothenus or
even close to it but I mean the longest side of a triangle
(e.g 5 on
the Egyptian triangle)
Thnx a lot in advance
Pierre
===
Subject: Re: Ideas for course on great ideas in (theoretical)
CS?
> I was looking at Amazon for the book you have mentained,
but didn¹t Žnd
it.
> Could you please tell me who the writer is?
Here it is, along with a few related others. All can be found
at Amazon:
Great Ideas in Computer Science - 2nd Edition: A Gentle
Introduction
by Alan W. Bierman
Publisher: MIT Press; 2nd edition (March 1, 1997) ISBN:
0262522233
ACM Turing Award Lectures: The First Twenty Years 1966-1985
Publisher: Addison-Wesley Pub Co; 1st edition (July 1, 1991)
ISBN:
0201548852
by Jean-Luc Chabert (Editor), E. Barbin (Editor), R.
Aasnogorodski, V.
Malyshev
Publisher: Springer Verlag; (September 1999) ISBN: 3540633693
People & Ideas in Theoretical Computer Science (Springer
Series in
Discrete Mathematics and Theoretical Computer Science)
by Cristian Calude
Publisher: Springer Verlag; (October 1999)
ISBN: 981402113X
===
Subject: Re: Ideas for course on great ideas in (theoretical)
CS?
Hi.
You MUST talk also about the halting problem, it would be
nice to talk
about
the tiling problem as well.
I was looking at Amazon for the book you have mentained, but
didn¹t Žnd
it.
Could you please tell me who the writer is?
Gila
Chip Klostermeyer <klostermeyer@hotmail.com> ???
> I have been coerced into teaching a Honors course the Fall
> (mostly for non-CS freshman/sophomore Honors students). My
> idea was to do some Great Ideas/Problems/Puzzles etc. from
> computer science -- emphasis on theory/algorithms
> and related areas like graph theory/combinatorics.
> Of course, the honors college wants a syllabus in one week!
> I looked at the book, Great Ideas in CS and though a nice
> book, seems a bit light on the theory side ... given that
> I want to focus on theory to keep me interested. I saw the
> course/web site at CMU Great Ideas in Theoretical Computer
Science
> and may use that as a guide for some of the course. Examples
> of some things I might discuss (besides a couple weeks on
> basics/deŽnitions/history) include Towers of Hanoi,
Byzantine
> Generals, voting problems, maybe a gentle discussion
> of interactive proofs, prisoner¹s dilemma, game of life,
primality
> testing, graph coloring ... anything that can be discussed
in a
> day or so to folks with no CS background, yet which has some
> theory component to it ... stuff that is surprising or
counter-intuitive
> is all the better ;)
> Anyway, if anyone has any suggestions for material/topics
> that I might cover, I would most appreciate it. Any pointers
> would be accessible to students would be great (I have a
couple)
> would be great.
> (or a link to one) to comp.theory in a week or so.
> Chip Klostermeyer
===
Subject: Re: Ideas for course on great ideas in (theoretical)
CS?
The structure theorem:
Any program can be transformed to a program whose structure
is formed
from combining sequences, while-loops, and an if-then-else
structure.
The use of invariants, such as loop invariants, in
correctness proofs.
The structure theorem can be found proven in H. D. Mill et.
al.¹s
Structured Programming. Some neat algorithms and ideas along
these
lines are be in Dijkstra¹s A Discipline of Programming. See
also
Gries¹ Science of Programming.
By the way, I recently convinced a colleague of mine that
Conway¹s
Game of Life was a universal computer. I showed him memory,
an AND
gate, and a NOT gate. Maybe somthing like that would be neat
as
part of a lecture on the Church-Turing thesis.
--
Try
http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/
Bukharin.html
To solve Linear Programs: .../LPSolver.html
r c A game: .../Keynes.html
v s a Whether strength of body or of mind, or wisdom, or
i m p virtue, are found in proportion to the power or
wealth
e a e of a man is a question Žt perhaps to be discussed
by
n e . slaves in the hearing of their masters, but highly
@ r c m unbecoming to reasonable and free men in search of
d o the truth. -- Rousseau
===
Subject: Divisions algorithm for Z[i]
Hi
I posted here concerning this subject a few weeks ago and now
I just
want to see that I got things right. I¹m handing in a project
done in
mathematics but my teacher can¹t help me let alone say if I¹m
on the
right track with things so I kind of depend on you guys.
How would you prove that the remainder gets smaller every
time when
the divisions algorithm is applied to two gaussian integers
and then
show that the remainder is zero after a Žnite number of steps?
I really appreciate all the help that the people here have
been giving
me during this project
Thnx
Piere
===
Subject: Re: Stepwise regression
Gup is right, I was thinking of recursive estimation (which
you do
stepwise), sorry for the misinformation.
> [... moved top post to bottom...]
> I¹d like to know the advantages of a stepwise linear
regression. In
what
> way
> does it differentiate itself from a normal linear
regression?
> for one, if the parameters (a&b in y= ax+b) are not
constant then
this
> technique would stand a good chance of revealing this.
> may also be quicker if you have to do it in real time, not
sure about
that
> though.
> Rob¹s reply is pretty nonsensical.
> Stepwise regression is a way to put
> predictor variables into a regression model one at a time.
This is
> especially useful if you have lots of variables and want to
screen them,
> as only signiŽcant predictors are added to the model. (In
most
> computer implementations you get to decide what signiŽcant
means.)
> The typical model you strive for is parsimonious, that is,
it
contains
> only few predictors, and every predictor contributes
signiŽcantly to
> explaining the variation of the response.
> The downside to stepwise regression is that you leave the
selection
> of important variables to the computer, and all it has to
go by is
> the data. If you do not check that the resulting model is
plausible
> based on your own understanding of the problem, you can end
up
> with pretty wacky models.
> Consult an introductory stats text for further information
--- this is
> standard stuff, and every text should cover it.
===
Subject: Re: Stepwise regression
[... moved top post to bottom...]
> I¹d like to know the advantages of a stepwise linear
regression. In
what
> way
> does it differentiate itself from a normal linear
regression?
> for one, if the parameters (a&b in y= ax+b) are not
constant then this
> technique would stand a good chance of revealing this.
> may also be quicker if you have to do it in real time, not
sure about
that
> though.
Rob¹s reply is pretty nonsensical.
Stepwise regression is a way to put
predictor variables into a regression model one at a time.
This is
especially useful if you have lots of variables and want to
screen them,
as only signiŽcant predictors are added to the model. (In most
computer implementations you get to decide what signiŽcant
means.)
The typical model you strive for is parsimonious, that is, it
contains
only few predictors, and every predictor contributes
signiŽcantly to
explaining the variation of the response.
The downside to stepwise regression is that you leave the
selection
of important variables to the computer, and all it has to go
by is
the data. If you do not check that the resulting model is
plausible
based on your own understanding of the problem, you can end up
with pretty wacky models.
Consult an introductory stats text for further information
--- this is
standard stuff, and every text should cover it.
===
Subject: Re: Stepwise regression
> Z.Stunic @eudoramail.com> replied: to Michael Jærgensen
> Certainly, my problem is not so simply. Symbols aj and bj
denotes
> coefŽcients(or parameters) for one particular subsets from
original
> set of data marked B. Only some of xi,yi data is related to
for
> example a1 and b1 coefŽcients, other member of original set
is
> connected with to another pair of coefŽcients a2,b2 and so
far.
> Obviously, with index j=1,2,...=K we can specify one
particular subset
> inside of set B. Number of possible subsets inside original
ones B is
> a priore unknown. There are two special cases:
> 1) a1 = a2=..........aj =.......aK; b1 <> b2 <> b3
<>......bj <> bK ,
> and
> 2) a1 <> a2 <>........aj <>.....aK; b1 = b2 = b3 =......bj
= bK.
> Z.Stunic
If I understand you, and I¹m not so sure of that, you have a
bunch of
ordered pairs, some subsets of which you want to regress
linearly.
What is keeping you from using ordinary least squares on those
subsets?