mm-257
> There's more to my work than just arguing on Usenet
Yes, but that's your crowning achievement.
, so I'd like to
> point out that my paper Advanced Polynomial Factorization is
sla
> to be published:
> See http://www.megasociety.net/NoesisHighlights.html
How nice that the cranks have gotten together and put up a
website.
> The Mega Foundation is an organization of high IQ people,
and I'm glad
> to be associa with them. To learn further about the
organization
> you can use Google, or see:
> Why a group like the Mega Foundation?
> http://www.ultrahiq.org/Mega/WhyMega.htm
> I hope at least some of you will appreciate that often the
most
> important ideas in history have to get past people limi by
their
> lack of imagination and their prejudices, who act against
scientific
> progress.
Not any more, now that we have the Internet . . .
> What I want you to see is that there's more to me than
Usenet, so that
> you can begin to understand that the revolution I'm giving
you a
> chance to be a part of is bigger than the small-minded
people who
> continually throughout history work to halt progress.
Is there a publication party? Free drinks? Are we all invi?
===
Subject: Re: Key Core Error Argument
there is also a proof of the isomorphism
of deductive proofs with inductive ones,
which may perhaps be amenable to combining the two forms
into a tautology, or necklace.
> before the final link (you missed the green, dood !-)
--ils duces d'Enron!
http://www.wlym.com/covers/7101contents.png
===
Subject: Re: Key Core Error Argument
wow, how embarrassing for you. please,
don't tell me how many that I'm on --
I'm trying to get out of here, anyway!
> I am now at 2710, but I am only around in this newsgroup
since
> january 1988.
--ils duces d'Enron!
http://larouchepub.com/radio/index.html
===
Subject: Re: Key Core Error Argument
>>[snip umpteenth repeat of previous flawed argument]
>>I thought you were going to just post your proof elsewhere
and stop
>>cluttering up this newsgroup. You've given a URL, now go
away.
>Such a remarkably silly posting.
>What's the definition of insanity again?
>>Do the same thing over and over, and expecting different
>>results... such as acceptance of your proof.
> Praying would also constitute as insanity, then. Yes, I
think you're
> right.
Hmmm... in that case, since I pray, I must be insane... unless
it works.
*shrug*
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Key Core Error Argument
>>
>>[snip umpteenth repeat of previous flawed argument]
>>
>>I thought you were going to just post your proof elsewhere
and stop
>>cluttering up this newsgroup. You've given a URL, now go
away.
>Such a remarkably silly posting.
>What's the definition of insanity again?
>>Do the same thing over and over, and expecting different
results... such
>>as acceptance of your proof.
>But you see there's talk and there's substance. My proof is
>substance, and in fact, there is no error in it.
>>Except in step 6.
>If you disagree, then point out an error in what follows.
>>I have. You never responded.
> Oh yeah, I've looked over your stuff and it's wrong.
> Hey, I just assumed that Twentyman was a pseudonym, and some
poster
> poin out that it's an actual name, which I verified, so
sorry.
And I told you a long time ago you could look up my website on
Google.
Or you could have just asked.
> Well given that you're giving your real name and I assumed
you didn't,
> I'll rethink the possibility that you're serious.
> Now then, what you have said so far isn't correct, so why
don't you go
> think about it, and if you're really, really sure that
you've found
> some problem with my argument, go ahead and repost and I'll
try to
> check it out, and maybe comment.
*Where* do you disagree with what I said?
Your argument seems to be:
1) Let g_1(x) = 5 a_1(x) + 7
2) The constant term is 7, which is divisible by 7.
3) a_1(0) = 0, so 5 a_1(0) is divisible by 0.
4) Therefore, for all x, g_1(x) is divisible by 7 and, more
specifically, a_1(x) is divisible by 7.
I disagree with step 4.
Do you feel I have misrepresen your argument, or that the
above is
correct? If I have misrepresen your argument, how?
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Key Core Error Argument
>
>> Actually it is, as Dik Winter is trying to find a way that
7, 7 and
22
>> become 1, 1 and 22 based on a *varying* x.
That is, he's trying to make the change in *constants*
dependent on
a
>> variable.
>>
>>Eh?
I want readers to understand that his behavior is crank, while
I
guess
>> many of you may sympathize with his strong desire for me to
be
wrong,
>> remember, it's not about people as the math didn't just
decide to
>> change.
What you should be sympathetic to, is the truth.
>>
>>You claim that having P(x) = g1(x).g2(x).g3(x) with g1(0) =
g2(0) = 7
>>and g3(0) = 22; the *only* way to divide P(x) by 49 is by
dividing
g1(x)
>>and g2(x) by 7 and g3(x) by 1. Because now g1(0)/7 = 1,
g2(0)/7 = 1
and
>>g3(0)/1 = 22.
>>
>>I claim there are other ways to do that. Have w1(x), w2(x),
w3(x),
such
>>that w1(0) = w2(0) = 7, w3(0) = 1, w1(x).w2(x).w3(x) = 49.
Because now
>>g1(0)/w1(0) = 1, g2(0)/w2(0) = 1 and g3(0)/w3(0) = 22.
>>
>>Where do I change constants?
>The basic algebra is that if you have 7 and divide it by
*something*
>and get 1, then you divided by 7.
>And no tricks will change that fact, and it's simply crank
behavior to
>try and act like there's some complica way you can divide 7
to get
>1, without actually just dividing it by 7.
>>Since nobody is claiming to be doing anything else, your
>>diatribe seems a bit pointless.
>>At x=0: You divide by 7, Dik divides by 7
>>At x<>0: You divide by 7, Dik divides by something other
than 7
>>The two factorizations are thus different. But the constant
>>terms depend only on what happens at x=0. Thus the constant
>>terms are the same.
>> -William Hughes
> That's the kind of odd illogic which shows a crank, and I'll
explain
> quickly why your approach is specious.
> So I have the factors (a_1(x) + 7) and people like yourself
and Dik
> Winter want desperately to believe that some variable factor
of 49
> divides through so that the factor itself has a varying
factor of 7.
> I've poin out that the constant term of its corollary factor
is 1.
> So let's pick x=9 and see what happens, as then you have
> a_1(9) + 7
This is just a number.
> and you want a_1(9) to have some factor in common with 7
that is NOT
> 7, so divide that factor off, and notice that unless it's 7,
you don't
> have 1 as the constant term.
> Like, what if it were sqrt(7)?
> Then you'd have
> a_1(9)/sqrt(7) + sqrt(7)
Ok.
> where clearly the constant term is not now 1, it's sqrt(7).
Huh? There are two constants added together. Why is sqrt(7)
special?
> Now that's basic but it takes a crank to argue against the
basics.
Ummm... why is it that *you* are the only one who seems to
know the
basics?
> The cranks try to push the idea that you can divide off
something
> other than 7 because they desperately want you to believe
something
> false, and they hope that people won't realize that they
*still* have
> that constant 7 sitting there like a rock.
> If you divide off anything other than 7, you can't get 1 as
the
> result.
> So, yes, in fact, these posters arguing with me, are trying
to find
> some way to get around the fact that 7 divided by something
to get 1,
> means that the something IS 7.
You are dealing with seperate cases and don't seem to realize
it.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Key Core Error Argument
> So let's pick x=9 and see what happens, as then you have
a_1(9) + 7
and you want a_1(9) to have some factor in common with 7 that
is NOT
> 7, so divide that factor off, and notice that unless it's 7,
you
don't
> have 1 as the constant term.
Like, what if it were sqrt(7)?
Then you'd have
a_1(9)/sqrt(7) + sqrt(7)
where clearly the constant term is not now 1, it's sqrt(7).
> But a_1(9) is not a_1(0).
> Of course it's not, but what I'm pointing out is the strange
attempt
> to ignore the constancy of the constant term 7, as even if
you move
> from a_1(0) = 0, it's still there. I like to say it's
sitting there
> like a rock.
> So when you have something like a_1(9), some posters want to
act like
> maybe the constant term can change, as if it has one value
at x=0, but
> starts bouncing all over the place if it doesn't, but how
can 7 just
> change?
The constant term of (a_1(x) + 7)/w(x) does not depend on the
value
of x.
The constant term of (a_1(x) + 7)/w(x) is the same whether x
is 0, -1,
100 or 6.023e23.
In particular the constant term of (a_1(x) + 7)/w(x) does not
depend
on the value of (a_1(9) + 7)/w(9).
The constant term of (a_1(x) +7)/w(x) is (a_1(0) + 7)/w(0)
The constant term of (a_1(x) +7)/w(x) is 7/w(0)
If w(0) is 7 the constant term of (a_1(x) +7)/w(x) is 1
-William Hughes
===
Subject: Re: Key Core Error Argument
> That's the kind of odd illogic which shows a crank, and I'll
explain
> quickly why your approach is specious.
Has anyone else noticed how James seems to have fixa on the
word
crank lately? I suppose that's his version of I know you are,
but
what am I?.
It's always funny when he picks up on a word or phrase (like
can't
eat a math proof or now crank) and wears it out with
repetition.
Apparently once an idea works its way through all that thick
bone into
the small cavity within, it takes a while to find its way out
again...
--
===
Subject: Re: Key Core Error Argument
> That's the kind of odd illogic which shows a crank, and I'll
explain
> quickly why your approach is specious.
> Has anyone else noticed how James seems to have fixa on the
word
> crank lately? I suppose that's his version of I know you
are, but
> what am I?.
He gave an interesting definition of the word crank some time
ago:
------------8<--------------
Let me explain to you how a person can legitimately be
considered a
crank.
Cranks get a notion in their heads for supposedly solving
something
which they believe is extraordinary, and then despite all the
evidence
against, they refuse to let go of their notion. Logic and
rationality
don't matter to them, as they are completely unreasonable. As
you
can't reason with them, there's no point in talking with them,
as it
won't make any difference.
Yet, the evidence against me is that I DO let go of wrong
ideas,
showing myself to be reasonable. And I DO admit to having used
definitions wrong, as I'm an amateur, and this is my hobby. And
while
I've said some challenging things about mathematicians,
considering
the many scandals in human history (including recent ones like
Enron),
is it more reasonable to proclaim mathematicians saints?
------------8<--------------
> It's always funny when he picks up on a word or phrase (like
can't
> eat a math proof or now crank) and wears it out with
repetition.
> Apparently once an idea works its way through all that thick
bone into
> the small cavity within, it takes a while to find its way
out again...
> --
> Wayne Brown (HPCC #1104) | When your tail's in a crack, you
improvise
> fwbrown@bellsouth.net | if you're good enough. Otherwise you
give
> | your pelt to the trapper.
> e^(i*pi) = -1 -- Euler | -- John Myers Mye
===
Subject: Re: Key Core Error Argument
So let's pick x=9 and see what happens, as then you have
a_1(9) + 7
and you want a_1(9) to have some factor in common with 7 that
is
NOT
> 7, so divide that factor off, and notice that unless it's 7,
you
don't
> have 1 as the constant term.
Like, what if it were sqrt(7)?
Then you'd have
a_1(9)/sqrt(7) + sqrt(7)
where clearly the constant term is not now 1, it's sqrt(7).
But a_1(9) is not a_1(0).
Of course it's not, but what I'm pointing out is the strange
attempt
> to ignore the constancy of the constant term 7, as even if
you move
> from a_1(0) = 0, it's still there. I like to say it's
sitting there
> like a rock.
So when you have something like a_1(9), some posters want to
act like
> maybe the constant term can change, as if it has one value
at x=0, but
> starts bouncing all over the place if it doesn't, but how
can 7 just
> change?
What I want to point out is the reality that *fear* is the best
> explanation for why people would believe something so
basically
> stupid, after all 7 is CONSTANT, and not a variable
dependent on x.
Still posters seem quite comfortable questioning basics and I
guess
> there are readers who go along with them.
But basically they're questioning that for 7/x = 1, x = 7.
Remember people, the constant 7 is a *constant* and doesn't
have one
> value at x=0, only to suddenly change just because you use
another
> value!!!
> The way I read it, they are questioning that in (a_1(y) +
7)/x = 1, x
must
> always be 7.
> If y is not 0, you can't use the constant term tricks that
depend on
y
> being 0.
That's stupid. The constant term once found is distinguished
by being
constant.
All the trick is doing is finding it.
So let me give you the example that I've used elsewhere which
is
a_1(x) + 7, which has a constant term that is 7. Do you
understand
what it means for it to be constant?
Here's a test.
If I have x=11, then I necessarily have a_1(11) + 7, right?
Notice the constant term is STILL there, do you understand?
Now then, if I now divide P(11) by 49, what should the
constant term
be?
If you answer honestly I'll be shocked.
===
Subject: Re: Key Core Error Argument
>So let me give you the example that I've used elsewhere which
is
>a_1(x) + 7, which has a constant term that is 7. Do you
understand
>what it means for it to be constant?
But by your definition, the constant term of b(x) = a_1(x)+7
would be
the value of b(0) = a_1(0) + 7.
Who says a_1(0) has to be 7? What if a_1(x) = sqrt(x^2+1) +
0.5(x^3-5)?
>Here's a test.
>If I have x=11, then I necessarily have a_1(11) + 7, right?
>Notice the constant term is STILL there, do you understand?
Yes, but if a_1(0) is divisible by 7, e.g. a_1(x) =
sqrt(x+49), who
says that a_1(11) is still divisible by 7?
>Now then, if I now divide P(11) by 49, what should the
constant term
>be?
I dunno. What's the constant term of sqrt(x+49)/7 + 7/7?
What's the value at x=0?
- Randy
===
Subject: Re: Key Core Error Argument
If y is not 0, you can't use the constant term tricks that
depend on
y
> being 0.
> That's stupid. The constant term once found is distinguished
by being
> constant.
> All the trick is doing is finding it.
> So let me give you the example that I've used elsewhere
which is
> a_1(x) + 7, which has a constant term that is 7. Do you
understand
> what it means for it to be constant?
> Here's a test.
> If I have x=11, then I necessarily have a_1(11) + 7, right?
> Notice the constant term is STILL there, do you understand?
> Now then, if I now divide P(11) by 49, what should the
constant term
> be?
> If you answer honestly I'll be shocked.
Let me try again. (a_1(y) + 7)/x is not the same as 7/x,
unless a_1(y) =
0.
===
Subject: Re: Key Core Error Argument
If y is not 0, you can't use the constant term tricks that
depend
on y
> being 0.
That's stupid. The constant term once found is distinguished
by being
> constant.
All the trick is doing is finding it.
So let me give you the example that I've used elsewhere which
is
> a_1(x) + 7, which has a constant term that is 7. Do you
understand
> what it means for it to be constant?
Here's a test.
If I have x=11, then I necessarily have a_1(11) + 7, right?
Notice the constant term is STILL there, do you understand?
Now then, if I now divide P(11) by 49, what should the
constant term
> be?
If you answer honestly I'll be shocked.
> Let me try again. (a_1(y) + 7)/x is not the same as 7/x,
unless a_1(y) =
0.
Ok, I can go from there Richard Henry, and note that
necessarily
a_1(y) has *some* factor in common with 7, right?
So let's call that factor f, now dividing 49 from P(y) will
divide f
off from a_1(y) + 7, understand?
Now then, you have a_1(y)/f + 7/f, and if f does not equal 7,
what
does that tell you about the *constant* term Richard Henry?
Necessarily, you have 7/f left as the constant term for a
factor of
P(11)/49, and if 7/f does not equal 1, you have a
contradiction.
Understand?
The trick is to FIND the constant term, as you know that it's
not
affec by the value of x, and it sits there like a rock, unaffec
by the value of x, and none of your protestations against
mathematical
reality will change that fact.
===
Subject: Re: Key Core Error Argument
If y is not 0, you can't use the constant term tricks that
depend on y
> being 0.
That's stupid. The constant term once found is distinguished by
being
> constant.
All the trick is doing is finding it.
So let me give you the example that I've used elsewhere which
is
> a_1(x) + 7, which has a constant term that is 7. Do you
understand
> what it means for it to be constant?
Here's a test.
If I have x=11, then I necessarily have a_1(11) + 7, right?
Notice the constant term is STILL there, do you understand?
Now then, if I now divide P(11) by 49, what should the
constant term
> be?
If you answer honestly I'll be shocked.
> Let me try again. (a_1(y) + 7)/x is not the same as 7/x,
unless a_1(y)
= 0.
> Ok, I can go from there Richard Henry, and note that
necessarily
> a_1(y) has *some* factor in common with 7, right?
> So let's call that factor f, now dividing 49 from P(y) will
divide f
> off from a_1(y) + 7, understand?
f will depend upon y.
> Now then, you have a_1(y)/f + 7/f, and if f does not equal
7, what
> does that tell you about the *constant* term Richard Henry?
You have a_1(y)/f(y) + 7/f(y) and you could have f(0)=7
without every
other value of f(y) being equal to 7.
> Necessarily, you have 7/f left as the constant term for a
factor of
> P(11)/49, and if 7/f does not equal 1, you have a
contradiction.
> Understand?
> The trick is to FIND the constant term, as you know that
it's not
> affec by the value of x, and it sits there like a rock,
unaffec
> by the value of x, and none of your protestations against
mathematical
> reality will change that fact.
>
===
Subject: Re: Key Core Error Argument
If y is not 0, you can't use the constant term tricks that
depend
on y
> being 0.
That's stupid. The constant term once found is distinguished by
being
> constant.
All the trick is doing is finding it.
So let me give you the example that I've used elsewhere which
is
> a_1(x) + 7, which has a constant term that is 7. Do you
understand
> what it means for it to be constant?
Here's a test.
If I have x=11, then I necessarily have a_1(11) + 7, right?
Notice the constant term is STILL there, do you understand?
Now then, if I now divide P(11) by 49, what should the
constant term
> be?
If you answer honestly I'll be shocked.
Let me try again. (a_1(y) + 7)/x is not the same as 7/x,
unless a_1(y)
= 0.
> Ok, I can go from there Richard Henry, and note that
necessarily
> a_1(y) has *some* factor in common with 7, right?
Why?
> So let's call that factor f, now dividing 49 from P(y) will
divide f
> off from a_1(y) + 7, understand?
No. Why?
> Now then, you have a_1(y)/f + 7/f, and if f does not equal
7, what
> does that tell you about the *constant* term Richard Henry?
I do not understand the question. Please rephrase it.
> Necessarily, you have 7/f left as the constant term for a
factor of
> P(11)/49, and if 7/f does not equal 1, you have a
contradiction.
I don't follow your logic to say necessarily.
> Understand?
No.
> The trick is to FIND the constant term, as you know that
it's not
> affec by the value of x, and it sits there like a rock,
unaffec
> by the value of x, and none of your protestations against
mathematical
> reality will change that fact.
The constant term you refer to is found by zeroing out the
other terms
by
setting x to zero. When x is not zero, and those other terms
are not
necessarily zero, you can not always extend the properties
found for the
constant term to the entire expression.
===
Subject: Re: Key Core Error Argument
> If y is not 0, you can't use the constant term tricks that
depend
> on y
> being 0.
That's stupid. The constant term once found is distinguished by
being
> constant.
All the trick is doing is finding it.
So let me give you the example that I've used elsewhere which
is
> a_1(x) + 7, which has a constant term that is 7. Do you
understand
> what it means for it to be constant?
Here's a test.
If I have x=11, then I necessarily have a_1(11) + 7, right?
Notice the constant term is STILL there, do you understand?
Now then, if I now divide P(11) by 49, what should the constant
term
> be?
If you answer honestly I'll be shocked.
Let me try again. (a_1(y) + 7)/x is not the same as 7/x, unless
a_1(y)
> = 0.
Ok, I can go from there Richard Henry, and note that
necessarily
> a_1(y) has *some* factor in common with 7, right?
> Why?
Because a_1(y) + 7 has a factor in common with 7, right?
> So let's call that factor f, now dividing 49 from P(y) will
divide f
> off from a_1(y) + 7, understand?
> No. Why?
Look at my previous answer Richard Henry.
> Now then, you have a_1(y)/f + 7/f, and if f does not equal
7, what
> does that tell you about the *constant* term Richard Henry?
> I do not understand the question. Please rephrase it.
If the constant term of a_1(y) + 7 is 7, since a_1(0) = 0, and
now you
divide through by some number I've called f, what is the
constant term
of the new expression?
> Necessarily, you have 7/f left as the constant term for a
factor of
> P(11)/49, and if 7/f does not equal 1, you have a
contradiction.
> I don't follow your logic to say necessarily.
Well I rephrased above so maybe that has changed, has it
Richard
Henry?
> Understand?
> No.
> The trick is to FIND the constant term, as you know that
it's not
> affec by the value of x, and it sits there like a rock,
unaffec
> by the value of x, and none of your protestations against
mathematical
> reality will change that fact.
> The constant term you refer to is found by zeroing out the
other terms
by
> setting x to zero. When x is not zero, and those other terms
are not
> necessarily zero, you can not always extend the properties
found for the
> constant term to the entire expression.
The property of the constant term is being constant because
it's a
number.
For instance, with f(x) = x+ 2, the constant term is 2, as
notice that
at x=0, f(0) = 2, but the constant term just sits there like a
rock,
as you vary x, do you understand that Richard Henry?
So you can't assume that if you change x the constant term
changes,
understand?
===
Subject: Re: Key Core Error Argument
> ...
> Actually it is, as Dik Winter is trying to find a way that
7, 7
and 22
> become 1, 1 and 22 based on a *varying* x.
> That is, he's trying to make the change in *constants*
dependent on a
> variable.
> Eh?
> I want readers to understand that his behavior is crank,
while
I guess
> many of you may sympathize with his strong desire for me to
be
wrong,
> remember, it's not about people as the math didn't just
decide
to
> change.
> What you should be sympathetic to, is the truth.
> You claim that having P(x) = g1(x).g2(x).g3(x) with g1(0) =
g2(0)
= 7
> and g3(0) = 22; the *only* way to divide P(x) by 49 is by
dividing
g1(x)
> and g2(x) by 7 and g3(x) by 1. Because now g1(0)/7 = 1,
g2(0)/7 =
1 and
> g3(0)/1 = 22.
> I claim there are other ways to do that. Have w1(x), w2(x),
w3(x), such
> that w1(0) = w2(0) = 7, w3(0) = 1, w1(x).w2(x).w3(x) = 49.
Because now
> g1(0)/w1(0) = 1, g2(0)/w2(0) = 1 and g3(0)/w3(0) = 22.
> Where do I change constants?
The basic algebra is that if you have 7 and divide it by
*something*
> and get 1, then you divided by 7.
And no tricks will change that fact, and it's simply crank
behavior
to
> try and act like there's some complica way you can divide 7
to
get
> 1, without actually just dividing it by 7.
> Since nobody is claiming to be doing anything else, your
> diatribe seems a bit pointless.
> At x=0: You divide by 7, Dik divides by 7
At x<>0: You divide by 7, Dik divides by something other than 7
The two factorizations are thus different. But the constant
> terms depend only on what happens at x=0. Thus the constant
> terms are the same.
-William Hughes
> That's the kind of odd illogic which shows a crank, and I'll
explain
> quickly why your approach is specious.
> So I have the factors (a_1(x) + 7) and people like yourself
and Dik
> Winter want desperately to believe that some variable factor
of 49
> divides through so that the factor itself has a varying
factor of 7.
> I've poin out that the constant term of its corollary factor
is 1.
> So let's pick x=9 and see what happens, as then you have
> a_1(9) + 7
> This is nonsense. The constant term of (a1(x) +7)/w(x) is
> (a1(0) +7)/w(0). The value (a_1(9) + 7)/w(9) has no bearing.
Since the factor is a_1(x) + 7, if x=9, it is a_1(9) + 7 as I
said.
> At x = 9 Dik divides by w1(9)
The constant term doesn't change with x, because it is
constant.
> The constant term does not depend on what Dik divides
> by at x=9.
Do you understand that in general the factor is a_1(x) + 7?
Now then, if you understand that, do you understand that at
x=9, it is
a_1(9)+7?
Maybe if that's gotten out of the way you can see the rest.
For othemy point is that the constant term is separate and
distinct from that part which varies, and doesn't change with
it.
So if I say a_1(9) + 7 with the factor, it has *some* factor
of 7,
right?
Well, I already know that if I divide P(x) by 49, the constant
terms
of the factors are 1, 1 and 22.
So if a_1(9) has any other factor than 7 itself, you have a
contradiction.
===
Subject: Re: Key Core Error Argument
> ...
> Actually it is, as Dik Winter is trying to find a way that 7,
7 and 22
> become 1, 1 and 22 based on a *varying* x.
> That is, he's trying to make the change in *constants*
dependent on a
> variable.
> Eh?
> I want readers to understand that his behavior is crank,
while I guess
> many of you may sympathize with his strong desire for me to
be wrong,
> remember, it's not about people as the math didn't just
decide to
> change.
> What you should be sympathetic to, is the truth.
> You claim that having P(x) = g1(x).g2(x).g3(x) with g1(0) =
g2(0) = 7
> and g3(0) = 22; the *only* way to divide P(x) by 49 is by
dividing g1(x)
> and g2(x) by 7 and g3(x) by 1. Because now g1(0)/7 = 1,
g2(0)/7
= 1 and
> g3(0)/1 = 22.
> I claim there are other ways to do that. Have w1(x), w2(x),
w3(x), such
> that w1(0) = w2(0) = 7, w3(0) = 1, w1(x).w2(x).w3(x) = 49.
Because now
> g1(0)/w1(0) = 1, g2(0)/w2(0) = 1 and g3(0)/w3(0) = 22.
> Where do I change constants?
> The basic algebra is that if you have 7 and divide it by
*something*
> and get 1, then you divided by 7.
> And no tricks will change that fact, and it's simply crank
behavior to
> try and act like there's some complica way you can divide 7
to
get
> 1, without actually just dividing it by 7.
Since nobody is claiming to be doing anything else, your
> diatribe seems a bit pointless.
> At x=0: You divide by 7, Dik divides by 7
At x<>0: You divide by 7, Dik divides by something other than 7
The two factorizations are thus different. But the constant
> terms depend only on what happens at x=0. Thus the constant
> terms are the same.
-William Hughes
That's the kind of odd illogic which shows a crank, and I'll
explain
> quickly why your approach is specious.
So I have the factors (a_1(x) + 7) and people like yourself
and Dik
> Winter want desperately to believe that some variable factor
of 49
> divides through so that the factor itself has a varying
factor of 7.
I've poin out that the constant term of its corollary factor
is 1.
So let's pick x=9 and see what happens, as then you have
a_1(9) + 7
> This is nonsense. The constant term of (a1(x) +7)/w(x) is
> (a1(0) +7)/w(0). The value (a_1(9) + 7)/w(9) has no bearing.
> Since the factor is a_1(x) + 7, if x=9, it is a_1(9) + 7 as
I said.
>
> At x = 9 Dik divides by w1(9)
> The constant term doesn't change with x, because it is
constant.
> The constant term does not depend on what Dik divides
> by at x=9.
> Do you understand that in general the factor is a_1(x) + 7?
> Now then, if you understand that, do you understand that at
x=9, it is
> a_1(9)+7?
> Maybe if that's gotten out of the way you can see the rest.
The factor is (a_1(x) +7)
The value of the factor is (a_1(x) +7)
The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7
At x=0
The value of (a_1(x) +7) is (a_1(0) + 7)
The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7
At x=1
The value of (a_1(x) +7) is (a_1(1) + 7)
The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7
At x=2
The value of (a_1(x) +7) is (a_1(2) + 7)
The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7
At x=9
The value of (a_1(x) +7) is (a_1(9) + 7)
The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7
Note that the value of (a_1(x) + 7) depends on x
Note that the constant term of (a_1(x) +7)
does not depend on x
Note that the constant term of (a_1(x) + 7) does not
depend on the value (a_1(9) + 7)
Now let's add the factor w(x)
The factor is (a_1(x)/w(x) + 7/w(x))
The value of the factor is (a_1(x)/w(x) +7/w(x))
The constant term of (a_1(x)/w(x) +7/w(x)) is
(a_1(0)/w(0) + 7/w(0)) = 7/w(0)
At x=0
The value of (a_1(x)/w(x) +7/w(x)) is (a_1(0)/w(0) + 7/w(0))
The constant term of (a_1(x)/w(x) +7/w(x)) is
(a_1(0)/w(0) + 7/w(0)) = 7/w(0)
At x=1
The value of (a_1(x)/w(x) +7/w(x)) is (a_1(1)/w(1) + 7/w(1))
The constant term of (a_1(x)/w(x) +7/w(x)) is
(a_1(0)/w(0) + 7/w(0)) = 7/w(0)
At x=2
The value of (a_1(2)/w(2) +7/w(2)) is (a_1(2)/w(2) + 7/w(2))
The constant term of (a_1(x)/w(x) +7/w(x)) is
(a_1(0)/w(0) + 7/w(0)) = 7/w(0)
At x=9
The value of (a_1(x)/w(x) +7/w(x)) is (a_1(9)/w(9) + 7/w(9))
The constant term of (a_1(x)/w(x) +7/w(x)) is
(a_1(0)/w(0) + 7/w(0)) = 7/w(0)
Note that the constant term does not depend on the value of x.
Note that the constant term does not depend on the
value (a_1(9)/w(9) + 7/w(9))
You are trying to do something like the following:
The constant term can be read off directly from
the factor.
The constant term of (a_1(x) +7) is 7, the bit
without the x
What can the constant term of (a_1(x)/w(x) + 7/w(x)) be?
It can't be a_1(x)/w(x) because a_1(x) depends on x
Therefore the constant term must be 7/w(x)
Therefore if the constant term is 1, then w(x)=7
The problem is that (a_1(x)/w(x) + 7/w(x)) is not a polynomial,
so you cannot read the constant term directly.
The constant term of (a_1(x)/w(x) + 7/w(x)) is
(a_1(0)/w(0) + 7/w(0)) = 7/w(0).
If the constant term is 1 then w(0)=7.
We do not know the value of w(x) for any
other value of x.
- William Hughes
===
Subject: Re: Key Core Error Argument
> ...
> Actually it is, as Dik Winter is trying to find a way that
7, 7 and 22
> become 1, 1 and 22 based on a *varying* x.
> That is, he's trying to make the change in *constants*
dependent on a
> variable.
> Eh?
> I want readers to understand that his behavior is crank,
while I guess
> many of you may sympathize with his strong desire for me to
be wrong,
> remember, it's not about people as the math didn't just
decide to
> change.
> What you should be sympathetic to, is the truth.
> You claim that having P(x) = g1(x).g2(x).g3(x) with g1(0) =
g2(0) = 7
> and g3(0) = 22; the *only* way to divide P(x) by 49 is by
dividing g1(x)
> and g2(x) by 7 and g3(x) by 1. Because now g1(0)/7 = 1,
g2(0)/7 = 1 and
> g3(0)/1 = 22.
> I claim there are other ways to do that. Have w1(x), w2(x),
w3(x), such
> that w1(0) = w2(0) = 7, w3(0) = 1, w1(x).w2(x).w3(x) = 49.
Because now
> g1(0)/w1(0) = 1, g2(0)/w2(0) = 1 and g3(0)/w3(0) = 22.
> Where do I change constants?
> The basic algebra is that if you have 7 and divide it by
*something*
> and get 1, then you divided by 7.
> And no tricks will change that fact, and it's simply crank
behavior to
> try and act like there's some complica way you can divide 7
to get
> 1, without actually just dividing it by 7.
> Since nobody is claiming to be doing anything else, your
> diatribe seems a bit pointless.
> At x=0: You divide by 7, Dik divides by 7
> At x<>0: You divide by 7, Dik divides by something other
than 7
> The two factorizations are thus different. But the constant
> terms depend only on what happens at x=0. Thus the constant
> terms are the same.
> -William Hughes
That's the kind of odd illogic which shows a crank, and I'll
explain
> quickly why your approach is specious.
So I have the factors (a_1(x) + 7) and people like yourself
and Dik
> Winter want desperately to believe that some variable factor
of 49
> divides through so that the factor itself has a varying
factor of
7.
I've poin out that the constant term of its corollary factor is
1.
So let's pick x=9 and see what happens, as then you have
a_1(9) + 7
This is nonsense. The constant term of (a1(x) +7)/w(x) is
> (a1(0) +7)/w(0). The value (a_1(9) + 7)/w(9) has no bearing.
> Since the factor is a_1(x) + 7, if x=9, it is a_1(9) + 7 as
I said.
>
> At x = 9 Dik divides by w1(9)
> The constant term doesn't change with x, because it is
constant.
> The constant term does not depend on what Dik divides
> by at x=9.
> Do you understand that in general the factor is a_1(x) + 7?
> Now then, if you understand that, do you understand that at
x=9, it is
> a_1(9)+7?
> Maybe if that's gotten out of the way you can see the rest.
> The factor is (a_1(x) +7)
> The value of the factor is (a_1(x) +7)
> The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7
> At x=0
> The value of (a_1(x) +7) is (a_1(0) + 7)
> The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7
> At x=1
> The value of (a_1(x) +7) is (a_1(1) + 7)
> The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7
> At x=2
> The value of (a_1(x) +7) is (a_1(2) + 7)
> The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7
> At x=9
> The value of (a_1(x) +7) is (a_1(9) + 7)
> The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7
> Note that the value of (a_1(x) + 7) depends on x
> Note that the constant term of (a_1(x) +7)
> does not depend on x
> Note that the constant term of (a_1(x) + 7) does not
> depend on the value (a_1(9) + 7)
> Now let's add the factor w(x)
> The factor is (a_1(x)/w(x) + 7/w(x))
> The value of the factor is (a_1(x)/w(x) +7/w(x))
> The constant term of (a_1(x)/w(x) +7/w(x)) is
> (a_1(0)/w(0) + 7/w(0)) = 7/w(0)
You keep writing a ratio because apparently you think a ratio
is more
powerful or mysterious, capable of doing something that it
can't.
Now then, if you admit that a_1(x)/w(x) is an algebraic
function, it
can be replaced by f(x), and if you admit that 7/w(x) is an
algebraic
function, it can be replaced by g(x), so then you have f(x) +
g(x).
But the constant term of f(x) + g(x) is 1, so let h(x) + 1 =
f(x) +
g(x), to isolate constant terms as before.
It's that result h(x) + 1 that you're terrified of, which is
why you
keep writing ratios as if algebra is mysticism!
You are a crank William Hughes who has seized upon ratios as a
way to
promote your illogical position.
The important factorization is
(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) =
300125 x^3 - 18375 x^2 - 360 x + 22
where you you have a clear separation between constant terms
and
varying terms, which is why 1(1)(22) equals the constant term
of
300125 x^3 - 18375 x^2 - 360 x + 22.
What you're trying to do is *hide* the simple reality by using
ratios!!!
It's *crank* behavior and anti-math, as you're trying to
refute a
logical, mathematical position by attempting to fool people
using a
dodge.
===
Subject: Re: Key Core Error Argument
[snip long boring discussion, look upthread if interes]
> You keep writing a ratio because apparently you think a
ratio is more
> powerful or mysterious, capable of doing something that it
can't.
> Now then, if you admit that a_1(x)/w(x) is an algebraic
function, it
> can be replaced by f(x), and if you admit that 7/w(x) is an
algebraic
> function, it can be replaced by g(x), so then you have f(x)
+ g(x).
> But the constant term of f(x) + g(x) is 1, so let h(x) + 1 =
f(x) +
> g(x), to isolate constant terms as before.
Which line contains the error?
1. The constant term of (h(x) + 1) is 1
2. (h(x) + 1) = (f(x) + g(x))
(Noting that f(x) = a_1(x)/w(x), g(x) = 7/w(x)
3. (h(x) + 1) = (a_1(x)/w(x) + 7/w(x))
4. The constant term of a_1(x)/w(x) + 7/w(x) is 1
-William Hughes
===
Subject: Re: Key Core Error Argument
>Which line contains the error?
Actually, I think most of JSH's proofs can be characterized by
the
following anecdote I think I heard from my old high-school math
teacher about one professor or another:
A lecture involved a somewhat algebraically heavy calculation
that
filled the entire blackboard while time was running out for
the day.
Since the desired answer did not appear at the end of the
calculation,
the professor simply ended his lecture by circling the entire
calculation, commenting that the error was contained somewhere
in
this region and giving the assignment to find it.
===
Subject: Re: Key Core Error Argument
...
> So let's pick x=9 and see what happens, as then you have
>
> a_1(9) + 7
>
> This is nonsense. The constant term of (a1(x) +7)/w(x) is
> (a1(0) +7)/w(0). The value (a_1(9) + 7)/w(9) has no bearing.
>
> Since the factor is a_1(x) + 7, if x=9, it is a_1(9) + 7 as
I said.
Yes, and the constant term of a_1(9) + 7 is a_1(9) + 7, by
your own
definition of constant term, and not 7 as you seem to assume
now.
> At x = 9 Dik divides by w1(9)
>
> The constant term doesn't change with x, because it is
constant.
Well? The constant term of a1(x) + 7 is 7 (by your definition,
because
we have to look at the part that does not vary with x. The
constant
term of a1(9) + 7 is *not* 7, or are you meaning that a1(9)
varies
with x? Strange. You accused us of assuming varying constants,
now
you do it your self. a1(9) is *constant*.
> So if I say a_1(9) + 7 with the factor, it has *some* factor
of 7,
> right?
Oh yes. gcd(a1(9) + 7, 7). Which is *not* necessarily 7.
> Well, I already know that if I divide P(x) by 49, the
constant terms
> of the factors are 1, 1 and 22.
>
> So if a_1(9) has any other factor than 7 itself, you have a
> contradiction.
What is the contradiction? 7 is *not* the constant term of
a1(9) + 7,
7 is the constant term of a1(x) + 7. By your *own* definition.
The
constant term (by your own definition) of (a1(x) + 7)/w(x) is
1 for
*all* functions w such that w(0) = 7. That is your *own*
definition.
--
===
Subject: Re: Ex(~x=x), counterpart theory, and contingent
identity
> I assume that you guys are more mature than He.
> One of you guys is Correy!!!!!
--John
===
Subject: Re: Ex(~x=x), counterpart theory, and contingent
identity
> In another post I sugges that John Correy's ideas about
non-reflexive
> identity might have a rational formulation with respect to
something
> called counterpart theory.
(I think categoreal would be the right term here--in my humble
> opinion, the apparent self-contradictions associa with John's
> intuitions arise from vagueness and ambiguity associa with
standard
> presuppositions rather than irrational error on his part. For
the
> record, Langholm's investigation of determinability and
> indeterminability in first-order contexts includes incoherent
formulas.
> Exclusion negation is not informationally well-behaved.)
The link,
http://www.sussex.ac.uk//Users/muralir/kct_final.pdf
>
<>((x=x) & ~(x=x))
is discussed as well as what is actually done in counterpart
theory
to
> exclude such an incoherent result.
>
>
mitch
>
http://www.sussex.ac.uk//Users/muralir/kct_final.pdf
>
<>((x=x) & ~(x=x))
is discussed as well as what is actually done in counterpart
theory
to
> exclude such an incoherent result.
>
>
mitch
The tie-in with contingent identity is as asser by (1).
(1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y)))
That is, identical(x,y) are necessarily identical if, ond only
if,
> N(x=x) and N(y=y). Thus, gran that the Morning Star and the
> Evening Star are each necessarily self-identical, if the
Morning Star
> and the Evening Star are identical, they are necessarily
identical.
> Hence the Morning Star and the Evening Star are necessarily
> identical. The same is not true, however, for Benjamin
Franklin
> and the inventor of bifocals. For although Benjamin Franklin
is
> necessarily self-identical, the inventor of bifocals is not
> necessarily self-identical. This is why Benjamin Franklin
> and the inventor of bifocals, although identical are not
> necessarily identical.
--John
> I am not sure that my remarks are welcome here, I have had
enough of
the
> remarks from (G. Frege)(II)
> : stupid asshole, ing bitch, and all of that childish
rhetoric, I
cannot
> reply to him at all.
> I assume that you guys are more mature than He.
> As to John's claim that: (John Correy = John Correy) is
necessarily true,
I
> disagree.
> It is not sufficient to say, x=y <-> AF(Fx <-> Fy).
> Rather, it is sufficient to say, x=y <->. E!x & E!y &AF( Fx
<-> Fy).
> After all, (ix: x=John Correy) is (John Correy).
> E!(John Correy) is just as doubtful as E!(The poster who
claims that
> ~Ax(x=x)).
> We cannot assume that E!(John Correy) any more than we can
assume
> E!(Vulcan)!
> (x=y) -> [](x=y) and E!x -> [](E!x), are consequences of
Leibnitz's Law.
> There are no contingent existences nor contingent identities.
> Exisence and Identity (and Membership) are analytic
properties.
> Can we assume that: []Exists(George W. Bush)?
> I don't think so, do you?
> Surely, the existence of, George W. Bush, is contingent.
> There cannot be an assumption that all names of purpor
physical
entities
> refer!
> Santa dosen't work, Vulcan doesn't work, Pegasus dosen't
work, etc.
> For although Benjamin Franklin is necessarily
self-identical, the
inventor
> of bifocals is not
> necessarily self-identical.
> Benjamin Franklin = Benjamin Franklin, is not necessarily
true.
> (the inventor of bifocals)=(the inventor of bifocals), is
also not
> necessarily true, imo.
> Necessary identity and existence, applies to
logical/mathematical
objects,
> not to emprical objects, imho.
> Witt
(1) states necessary and sufficient conditions for the
necessity of
(material) identity:
(1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y)))
If identicals x and y are necessarily self-identical, then--and
only then--is their identity a necessary one. Beyond this,
(1) states nothing more: from (1) it neither follows that
John Correy is necessarily self-identical nor that John
Correy is contingently self-identical--or indeed that John
Correy is self-identical at all. (1) does not say which of
the foregoing is the case.
We can sit around and argue about whether you or I are
necessarily
self-identical. However, although logicians do argue about such
matters--Who else would bother?--it is not as logicians that
they
argue but as metaphysicians, or as pataphysicians, or as what
have
you?
So, when I claim that (e.g.) Benjamin Franklin is necessarily
self-identical while the inventor of bifocals is not, my main
warrant for this claim is that if Benjamin Franklin is
necessarily
self-identical but the inventor of bifocals is not, then
Benjamin
Franklin and the inventor of bifocals are not necessarily
identical
(although they are identical). In other words, I take the
necessary
self-identity of the former and the contingent self-identity
of the
latter to constitute, together with (1), an *explanation* for
the
contingent identity of Benjamin Franklin and the inventor of
bifocals.
To this you might object that these would be also be
contingently
identical if both Benjamin Franklin and the inventor of
bifocals were
contingently self-identical. To which I would respond that
identities
involving what linguistically orien analytical osophers refer
to as rigid designatoare identities whose terms are necessarily
self-identical; whereas identities involving what such osophers
refer to as non-rigid designatoare identities whose terms are
contingently self-identical. Therefore, gran that I take rigid
and
non-rigid designation as the linguistic marks of necessary and
contingent self-identity--putting the cart back behind the
horse,
rather than approaching the matter bass ackwards as it is
fashionable to do these days--and gran that I take
Benjamin Franklin and John Correy to be 'rigid' designators
and 'the inventor of bifocals' to be 'non-rigid', I conclude
that the contingent identity of Benjamin Franklin and the
inventor
of bifocals has as its basis the contingent self-identity of
the
inventor of bifocals, while Benjamin Franklin is necessarily
self-identical.
As to whether physical or mathematical objects are contingently
self-identical or necessarily so, some sort of metaphysical
argument
(rather than a logical one) warranting one or the other of
these
conclusions would have to be made. I suspect that mathematical
properties are both essential in, and necessary to,
mathematical
objects--but this is an intuition and nothing more.
John
PS It won't surprise me if Paul Holbach or G. Frege bring in
talk
about 'scope', which I think is only peripherally relevant to
discussions of necessary and contingent identity.
===
Subject: Re: Ex(~x=x), counterpart theory, and contingent
identity
> (1) states necessary and sufficient conditions for the
necessity of
> (material) identity:
> (1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y)))
> If identicals x and y are necessarily self-identical,
then--and
> only then--is their identity a necessary one. Beyond this,
> (1) states nothing more: from (1) it neither follows that
> John Correy is necessarily self-identical nor that John
> Correy is contingently self-identical--or indeed that John
> Correy is self-identical at all. (1) does not say which of
> the foregoing is the case.
> We can sit around and argue about whether you or I are
necessarily
> self-identical. However, although logicians do argue about
such
> matters--Who else would bother?--it is not as logicians that
they
> argue but as metaphysicians, or as pataphysicians, or as
what have
> you?
> So, when I claim that (e.g.) Benjamin Franklin is necessarily
> self-identical while the inventor of bifocals is not, my main
> warrant for this claim is that if Benjamin Franklin is
necessarily
> self-identical but the inventor of bifocals is not, then
Benjamin
> Franklin and the inventor of bifocals are not necessarily
identical
> (although they are identical). In other words, I take the
necessary
> self-identity of the former and the contingent self-identity
of the
> latter to constitute, together with (1), an *explanation*
for the
> contingent identity of Benjamin Franklin and the inventor of
bifocals.
> To this you might object that these would be also be
contingently
> identical if both Benjamin Franklin and the inventor of
bifocals were
> contingently self-identical. To which I would respond that
identities
> involving what linguistically orien analytical osophers refer
> to as rigid designatoare identities whose terms are
necessarily
> self-identical; whereas identities involving what such
osophers
> refer to as non-rigid designatoare identities whose terms are
> contingently self-identical. Therefore, gran that I take
rigid and
> non-rigid designation as the linguistic marks of necessary
and
> contingent self-identity--putting the cart back behind the
horse,
> rather than approaching the matter bass ackwards as it is
> fashionable to do these days--and gran that I take
> Benjamin Franklin and John Correy to be 'rigid' designators
> and 'the inventor of bifocals' to be 'non-rigid', I conclude
> that the contingent identity of Benjamin Franklin and the
inventor
> of bifocals has as its basis the contingent self-identity of
the
> inventor of bifocals, while Benjamin Franklin is necessarily
> self-identical.
> As to whether physical or mathematical objects are
contingently
> self-identical or necessarily so, some sort of metaphysical
argument
> (rather than a logical one) warranting one or the other of
these
> conclusions would have to be made. I suspect that
mathematical
> properties are both essential in, and necessary to,
mathematical
> objects--but this is an intuition and nothing more.
>
> John
> PS It won't surprise me if Paul Holbach or G. Frege bring in
talk
> about 'scope', which I think is only peripherally relevant to
> discussions of necessary and contingent identity.
Let me bring in a quote:
One must distinguish between the claim that identity sentences
are
contingent and the claim that the identity relation itself is
contingent. For the relation to be contingent, there need to
be things
between which it holds merely contingent. For it to be
necessary, it
has to be that if the relation obtains between things, it
obtains
between those very things of necessity. [...] One can
consistently say
that there are contingent identity sentences, though the
relation
itself is necessary. Thus one could say that The first
Postmaster-General of the US was the inventor of bifocal
lenses. is
contingent and is an identity sentence, but that if we
consider the
object, x, which is in fact referred to by the first
Postmaster-General of the US and the object, y, which is in
fact
referred to by the inventor of bifocal lenses, it is necessary
that
x is identical to y.
[Sainsbury, M. (1995). osophical Logic. In A. C. Grayling
(Ed.),
/osophy. A Guide Through the Subject/ (pp. 61-122). Oxford:
Oxford
University Press. (p. 93)]
PH
===
Subject: Re: Ex(~x=x), counterpart theory, and contingent
identity
> (1) states necessary and sufficient conditions for the
necessity of
> (material) identity:
> (1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y)))
> If identicals x and y are necessarily self-identical,
then--and
> only then--is their identity a necessary one. Beyond this,
> (1) states nothing more: from (1) it neither follows that
> John Correy is necessarily self-identical nor that John
> Correy is contingently self-identical--or indeed that John
> Correy is self-identical at all. (1) does not say which of
> the foregoing is the case.
> We can sit around and argue about whether you or I are
necessarily
> self-identical. However, although logicians do argue about
such
> matters--Who else would bother?--it is not as logicians that
they
> argue but as metaphysicians, or as pataphysicians, or as
what have
> you?
> So, when I claim that (e.g.) Benjamin Franklin is necessarily
> self-identical while the inventor of bifocals is not, my main
> warrant for this claim is that if Benjamin Franklin is
necessarily
> self-identical but the inventor of bifocals is not, then
Benjamin
> Franklin and the inventor of bifocals are not necessarily
identical
> (although they are identical). In other words, I take the
necessary
> self-identity of the former and the contingent self-identity
of the
> latter to constitute, together with (1), an *explanation*
for the
> contingent identity of Benjamin Franklin and the inventor of
bifocals.
> To this you might object that these would be also be
contingently
> identical if both Benjamin Franklin and the inventor of
bifocals were
> contingently self-identical. To which I would respond that
identities
> involving what linguistically orien analytical osophers refer
> to as rigid designatoare identities whose terms are
necessarily
> self-identical; whereas identities involving what such
osophers
> refer to as non-rigid designatoare identities whose terms are
> contingently self-identical. Therefore, gran that I take
rigid and
> non-rigid designation as the linguistic marks of necessary
and
> contingent self-identity--putting the cart back behind the
horse,
> rather than approaching the matter bass ackwards as it is
> fashionable to do these days--and gran that I take
Benjamin Franklin and John Correy to be 'rigid' designators
> and 'the inventor of bifocals' to be 'non-rigid', I conclude
> that the contingent identity of Benjamin Franklin and the
inventor
> of bifocals has as its basis the contingent self-identity of
the
> inventor of bifocals, while Benjamin Franklin is necessarily
> self-identical.
> As to whether physical or mathematical objects are
contingently
> self-identical or necessarily so, some sort of metaphysical
argument
> (rather than a logical one) warranting one or the other of
these
> conclusions would have to be made. I suspect that
mathematical
> properties are both essential in, and necessary to,
mathematical
> objects--but this is an intuition and nothing more.
>
> John
> PS It won't surprise me if Paul Holbach or G. Frege bring in
talk
> about 'scope', which I think is only peripherally relevant to
> discussions of necessary and contingent identity.
> Let me bring in a quote:
> One must distinguish between the claim that identity
sentences are
> contingent and the claim that the identity relation itself is
> contingent. For the relation to be contingent, there need to
be things
> between which it holds merely contingent. For it to be
necessary, it
> has to be that if the relation obtains between things, it
obtains
> between those very things of necessity. [...] One can
consistently say
> that there are contingent identity sentences, though the
relation
> itself is necessary. Thus one could say that The first
> Postmaster-General of the US was the inventor of bifocal
lenses. is
> contingent and is an identity sentence, but that if we
consider the
> object, x, which is in fact referred to by the first
> Postmaster-General of the US and the object, y, which is in
fact
> referred to by the inventor of bifocal lenses, it is
necessary that
> x is identical to y.
> [Sainsbury, M. (1995). osophical Logic. In A. C. Grayling
(Ed.),
> /osophy. A Guide Through the Subject/ (pp. 61-122). Oxford:
Oxford
> University Press. (p. 93)]
> PH
This sounds so much like what Kripke says, either in
Identity and Necessity or in _Naming and Necessity_, that I
hope
Sainsbury ci him.
Of course, what Kripsbury says represents the Party Line on
contingent identity: There are *statements* of contingent
identity but no instances of contingent identity itself.
Oops! Before I forget, let me forestall the inevitable
bUllrich-ism:
Duh. You're not saying anything *new* you know.
'AxAy(x=y -> (N(x=x & y=y) <-> N(x=y)))' is a theorem of
standard
quantified modal logic with identity. (Giggle)
--John
===
Subject: Re: Ex(~x=x), counterpart theory, and contingent
identity
> In another post I sugges that John Correy's ideas about
non-reflexive
> identity might have a rational formulation with respect to
something
> called counterpart theory.
(I think categoreal would be the right term here--in my humble
> opinion, the apparent self-contradictions associa with John's
> intuitions arise from vagueness and ambiguity associa with
standard
> presuppositions rather than irrational error on his part. For
the
> record, Langholm's investigation of determinability and
> indeterminability in first-order contexts includes incoherent
formulas.
> Exclusion negation is not informationally well-behaved.)
The link,
http://www.sussex.ac.uk//Users/muralir/kct_final.pdf
>
<>((x=x) & ~(x=x))
is discussed as well as what is actually done in counterpart
theory
to
> exclude such an incoherent result.
>
>
mitch
>
http://www.sussex.ac.uk//Users/muralir/kct_final.pdf
>
<>((x=x) & ~(x=x))
is discussed as well as what is actually done in counterpart
theory
to
> exclude such an incoherent result.
>
>
mitch
The tie-in with contingent identity is as asser by (1).
(1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y)))
That is, identical(x,y) are necessarily identical if, ond only
if,
> N(x=x) and N(y=y). Thus, gran that the Morning Star and the
> Evening Star are each necessarily self-identical, if the
Morning Star
> and the Evening Star are identical, they are necessarily
identical.
> Hence the Morning Star and the Evening Star are necessarily
> identical. The same is not true, however, for Benjamin
Franklin
> and the inventor of bifocals. For although Benjamin Franklin
is
> necessarily self-identical, the inventor of bifocals is not
> necessarily self-identical. This is why Benjamin Franklin
> and the inventor of bifocals, although identical are not
> necessarily identical.
--John
>
>
I am not sure that my remarks are welcome here, I have had
enough of
the
> remarks from (G. Frege)(II)
> : stupid asshole, ing bitch, and all of that childish
rhetoric, I
cannot
> reply to him at all.
I assume that you guys are more mature than He.
As to John's claim that: (John Correy = John Correy) is
necessarily
true, I
> disagree.
> It is not sufficient to say, x=y <-> AF(Fx <-> Fy).
> Rather, it is sufficient to say, x=y <->. E!x & E!y &AF( Fx
<-> Fy).
After all, (ix: x=John Correy) is (John Correy).
E!(John Correy) is just as doubtful as E!(The poster who
claims that
> ~Ax(x=x)).
We cannot assume that E!(John Correy) any more than we can
assume
> E!(Vulcan)!
(x=y) -> [](x=y) and E!x -> [](E!x), are consequences of
Leibnitz's
Law.
There are no contingent existences nor contingent identities.
> Exisence and Identity (and Membership) are analytic
properties.
Can we assume that: []Exists(George W. Bush)?
I don't think so, do you?
Surely, the existence of, George W. Bush, is contingent.
There cannot be an assumption that all names of purpor physical
entities
> refer!
Santa dosen't work, Vulcan doesn't work, Pegasus dosen't work,
etc.
>
For although Benjamin Franklin is necessarily self-identical,
the
inventor
> of bifocals is not
> necessarily self-identical.
Benjamin Franklin = Benjamin Franklin, is not necessarily true.
> (the inventor of bifocals)=(the inventor of bifocals), is
also not
> necessarily true, imo.
Necessary identity and existence, applies to
logical/mathematical
objects,
> not to emprical objects, imho.
Witt
> (1) states necessary and sufficient conditions for the
necessity of
> (material) identity:
> (1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y)))
> If identicals x and y are necessarily self-identical,
then--and
> only then--is their identity a necessary one. Beyond this,
> (1) states nothing more: from (1) it neither follows that
> John Correy is necessarily self-identical nor that John
> Correy is contingently self-identical--or indeed that John
> Correy is self-identical at all. (1) does not say which of
> the foregoing is the case.
> We can sit around and argue about whether you or I are
necessarily
> self-identical. However, although logicians do argue about
such
> matters--Who else would bother?--it is not as logicians that
they
> argue but as metaphysicians, or as pataphysicians, or as
what have
> you?
Good point. That is one of the reasons that I have always
known that there
is a osophical element involved with
discussions of identity.
> So, when I claim that (e.g.) Benjamin Franklin is necessarily
> self-identical while the inventor of bifocals is not, my main
> warrant for this claim is that if Benjamin Franklin is
necessarily
> self-identical but the inventor of bifocals is not, then
Benjamin
> Franklin and the inventor of bifocals are not necessarily
identical
> (although they are identical). In other words, I take the
necessary
> self-identity of the former and the contingent self-identity
of the
> latter to constitute, together with (1), an *explanation*
for the
> contingent identity of Benjamin Franklin and the inventor of
bifocals.
> To this you might object that these would be also be
contingently
> identical if both Benjamin Franklin and the inventor of
bifocals were
> contingently self-identical. To which I would respond that
identities
> involving what linguistically orien analytical osophers refer
> to as rigid designatoare identities whose terms are
necessarily
> self-identical; whereas identities involving what such
osophers
> refer to as non-rigid designatoare identities whose terms are
> contingently self-identical. Therefore, gran that I take
rigid and
> non-rigid designation as the linguistic marks of necessary
and
> contingent self-identity--putting the cart back behind the
horse,
> rather than approaching the matter bass ackwards as it is
> fashionable to do these days--and gran that I take
> Benjamin Franklin and John Correy to be 'rigid' designators
> and 'the inventor of bifocals' to be 'non-rigid', I conclude
> that the contingent identity of Benjamin Franklin and the
inventor
> of bifocals has as its basis the contingent self-identity of
the
> inventor of bifocals, while Benjamin Franklin is necessarily
> self-identical.
Personally, I still have some doubts about self-identical and
not
self-identical. But I am now fairly certain that if
one accepts Frege's argument for the definition of number as
attaching to an
object, one should logically accept this
distinction.
> As to whether physical or mathematical objects are
contingently
> self-identical or necessarily so, some sort of metaphysical
argument
> (rather than a logical one) warranting one or the other of
these
> conclusions would have to be made. I suspect that
mathematical
> properties are both essential in, and necessary to,
mathematical
> objects--but this is an intuition and nothing more.
> It won't surprise me if Paul Holbach or G. Frege bring in
talk
> about 'scope', which I think is only peripherally relevant to
> discussions of necessary and contingent identity.
Well, when you claim that you can define scope so that
self-identity is
always implicit, you must deal with a Kantian
possibility--...
The logical determination of a concept by reason
is based upon a disjunctive syllogism, in which the
major premiss contains a logical division (the division
of the sphere of a universal concept), the minor
premiss limiting this sphere to a certain part, and
the conclusion determining the concept by means
of this part.
--Immanuel Kant
Critique of Pure Reason A577/605
...--namely, the non-self-identical.
Of course, the concept is still a fiction in Kantian
epistemology. However,
it is also the reason for the apparent
complexity of his ideas--he does not trivially assert
self-identity as
self-evident.
:-)
mitch
===
Subject: Re: Ex(~x=x), counterpart theory, and contingent
identity
> Well, when you claim that you can define scope so that
self-identity >is
always implicit, you must deal with a Kantian
> possibility--...
> The logical determination of a concept by reason
> is based upon a disjunctive syllogism, in which the
> major premiss contains a logical division (the division
> of the sphere of a universal concept), the minor
> premiss limiting this sphere to a certain part, and
> the conclusion determining the concept by means
> of this part.
> --Immanuel Kant
> Critique of Pure Reason A577/605
I'm not sure how this cashes out where the logic of identity is
concerned. One might construe the classical Ax[x=x <-> Ey(x=y)]
as characterizing by exclusion the relation between
self-identity and
identity-with-something: no self-identical is an
identical-with-nothing,
and no identical-with-something is a non-self-identical.
In respect of the foregoing, ~AxEy(x=y) might be the minor
premise
and ~Ax(x=x) the conclusion.
> ...--namely, the non-self-identical.
> Of course, the concept is still a fiction in Kantian
epistemology.
> However, it is also the reason for the apparent
> complexity of his ideas--he does not trivially assert
> self-identity as self-evident.
--John
===
Subject: Re: Ex(~x=x), counterpart theory, and contingent
identity
> I am not sure that my remarks are welcome here
That may make two of us.
> Necessary identity and existence, applies to
logical/mathematical
objects,
> not to emprical objects, imho.
The problem then becomes how one obtains criteria by which to
distinguish.
Just so you understand why I pay so much attention to Kant, I
will quote a
passage from The Amphiboly of Concepts of Reflection. In part,
it is also
an
answer to Frege's somewhat naive statement,
Perhaps Kant uses the word 'object' in a rather
different sense;
Before we leave the Transcendental Analytic we
must add some remarks which, although in themselves
not of special importance, might nevertheless be
regarded as requisite for the completeness of the
system. The supreme concept with which it is customary
to begin a transcendental osophy is the division
into the possible and the impossible. But, since all
division presupposes a concept to be divided, a still
higher one is required, and this is the concept of an
object in general, taken problematically, without its
having been decided whether it is something or nothing.
As the categories are the only concepts which refer
to objects in general, the distinguishing of an object,
whether it is something or nothing, will proceed
according to the order and under the guidance of
the categories.
1. To the concepts of all, many, and one there
that is, none. Thus, the object of a concept to
which no assignable intuition whatsoever corresponds
is equal to nothing. That is, it is a concept without
an object (ens rationis), like noumena, which cannot
be reckoned among the possibilities, although they
must not for that reason be declared to be also
impossible; or like certain new fundamental forces,
which though entertained in thought without
self-contradiction are yet also in our thinking
unsuppor by any example from experience,
and are therefore not to be coun as possible.
2.Reality is something; negation is nothing, namely
a concept of the absence of an object, such as
shadow, cold (nihil privativum)
3. The mere form of intuition, without substance,
is in itself no object, but the merely formal condition
of an object (as appearance), as pure space and
pure time (ens imaginarium). These are indeed
something, as forms of intuition, but are not
themselves objects which are to intui.
4. The object of a concept which contradicts
itself is nothing, because the concept is nothing,
is the impossible, e.g., a two-sided rectilinear
figure (nihil negativum)
The table of this division of the concept of
nothing would therefore have to be drawn up
as follows. (The corresponding division of
something follows directly from it.).
[begin fixed width]
1
[quantity]
Empty concept without object
ens rationis
2 3
[quality] [relation]
Empty object of a concept Empty intuition without object
nihil privativum ens imaginarium
4
[modality]
Empty object without concept
nihil negativum
[end fixed width]
We see that the ens rationis (1) is distinguished
from the nihil negativum (4) in that the fomer is
not to be coun among the possibilities because
it is a mere fiction (although not self-contradictory),
whereas the latter is opposed to possibility in that
empty concepts. On the other hand, the nihil
privativum (2) and the ens imaginarium (3) are
empty data for concepts. If light were not
given to the senses, we could not represent
darkness, and if extended beings were not perceived
we could not represent space. Negation and
the mere form of intuition, in the absence of
something real are not objects.
It would seem that Kant is in agreement with your humble
opinion. However,
he
uses the necessity ascribed to mathematical objects to ground
the
predictive
coherence observed in the empirical realm as objective
knowledge.
Frege used a non-self contradictory concept (ens rationis) as
the foundation
of
his definition of number. I suspect that John's notion of not
self-identical
corresponds to a notion of absence of an object (nihil
privativum) since
absolute metalinguistic notions of identity permit us to apply
the concept
to
objects absent from the domain of discourse.
:-)
mitch
===
Subject: Re: Ex(~x=x), counterpart theory, and contingent
identity
>
I am not sure that my remarks are welcome here
> That may make two of us.
> Necessary identity and existence, applies to
logical/mathematical
objects,
> not to emprical objects, imho.
> The problem then becomes how one obtains criteria by which to
distinguish.
Empirical objects are physical, logical objects are not
(physical).
Of course, we distinguish material objects by their
materialism, and,
abstract objects by their abstractness.
Physical objects are those which our senses commit us to!
If we can admit physical objects into our universe of
understanding
we can explain 'applied mathematics'.
> Just so you understand why I pay so much attention to Kant,
I will quote
a
> passage from The Amphiboly of Concepts of Reflection. In
part, it is
also
an
> answer to Frege's somewhat naive statement,
> Perhaps Kant uses the word 'object' in a rather
> different sense;
> Before we leave the Transcendental Analytic we
> must add some remarks which, although in themselves
> not of special importance, might nevertheless be
> regarded as requisite for the completeness of the
> system. The supreme concept with which it is customary
> to begin a transcendental osophy is the division
> into the possible and the impossible. But, since all
> division presupposes a concept to be divided, a still
> higher one is required, and this is the concept of an
> object in general, taken problematically, without its
> having been decided whether it is something or nothing.
> As the categories are the only concepts which refer
> to objects in general, the distinguishing of an object,
> whether it is something or nothing, will proceed
> according to the order and under the guidance of
> the categories.
> 1. To the concepts of all, many, and one there
> that is, none. Thus, the object of a concept to
> which no assignable intuition whatsoever corresponds
> is equal to nothing. That is, it is a concept without
> an object (ens rationis), like noumena, which cannot
> be reckoned among the possibilities, although they
> must not for that reason be declared to be also
> impossible; or like certain new fundamental forces,
> which though entertained in thought without
> self-contradiction are yet also in our thinking
> unsuppor by any example from experience,
> and are therefore not to be coun as possible.
> 2.Reality is something; negation is nothing, namely
> a concept of the absence of an object, such as
> shadow, cold (nihil privativum)
> 3. The mere form of intuition, without substance,
> is in itself no object, but the merely formal condition
> of an object (as appearance), as pure space and
> pure time (ens imaginarium). These are indeed
> something, as forms of intuition, but are not
> themselves objects which are to intui.
> 4. The object of a concept which contradicts
> itself is nothing, because the concept is nothing,
> is the impossible, e.g., a two-sided rectilinear
> figure (nihil negativum)
> The table of this division of the concept of
> nothing would therefore have to be drawn up
> as follows. (The corresponding division of
> something follows directly from it.).
> [begin fixed width]
> 1
> [quantity]
> Empty concept without object
> ens rationis
> 2 3
> [quality] [relation]
> Empty object of a concept Empty intuition without object
> nihil privativum ens imaginarium
> 4
> [modality]
> Empty object without concept
> nihil negativum
> [end fixed width]
> We see that the ens rationis (1) is distinguished
> from the nihil negativum (4) in that the fomer is
> not to be coun among the possibilities because
> it is a mere fiction (although not self-contradictory),
> whereas the latter is opposed to possibility in that
> empty concepts. On the other hand, the nihil
> privativum (2) and the ens imaginarium (3) are
> empty data for concepts. If light were not
> given to the senses, we could not represent
> darkness, and if extended beings were not perceived
> we could not represent space. Negation and
> the mere form of intuition, in the absence of
> something real are not objects.
> It would seem that Kant is in agreement with your humble
opinion.
However, he
> uses the necessity ascribed to mathematical objects to
ground the
predictive
> coherence observed in the empirical realm as objective
knowledge.
> Frege used a non-self contradictory concept (ens rationis)
as the
foundation of
> his definition of number. I suspect that John's notion of not
self-identical
> corresponds to a notion of absence of an object (nihil
privativum) since
> absolute metalinguistic notions of identity permit us to
apply the
concept
to
> objects absent from the domain of discourse.
mitch
I have not read Kant, yet, so I miss much of what you say.
Witt
===
Subject: Re: Ex(~x=x), counterpart theory, and contingent
identity
> Physical objects are those which our senses commit us to!
Close enough for a start.
:-)
mitch
===
Subject: Re: Ex(~x=x), counterpart theory, and contingent
identity
> Frege used a non-self contradictory concept (ens rationis)
as the
foundation of
> his definition of number. I suspect that John's notion of not
self-identical
> corresponds to a notion of absence of an object (nihil
privativum) since
> absolute metalinguistic notions of identity permit us to
apply the concept
to
> objects absent from the domain of discourse.
Defined parameters of usage, however, do not permit such
application. That
is why
it is so easy to win arguments on this matter.
But, these questions are important because they go back to
presuppositions
about
*knowing* or *believing that we know* a domain of discourse.
Any reference
to
truth-in-the-sense-of-what-really-is-true is an implicit
reference to the
class
universe, the set universe, or the intended interpretation.
These
are all
definite descriptions without formal sense.
Perhaps everyone should solve the problem by not using them.
Apparently, a large part of osophy involves complaints about
people not
using
words they way that they were meant to be used. Damn people.
:-)
mitch
===
Subject: Oneness of a number
I am trying to build fractals genera by the principal of the
oneness of
a
number.
For the definition of the oneness of a number, view the
internet, or
below.*
I have been unseccesful in creating a pattern around the
oneness although I
have struck on some intriguing details. These mainly resolve
around the
issue of, if we take the oneness of the oneness of a number,
and keep doing
that, how long does it take 'til we reach 1? (note that for 5
this never
happens) I think I might be able to find a pattern in that
since, if I call
this the recursive oneness loop length (if anybody can think a
better
name, tell me), then, the recursive oneness loop length of a
number like
20000000 is still only 15. I am trying to make a large-scale
2d map of it,
but so far it is still pure chaos, even if the range is so
limi.
Does anybody know any other numerical iterations like the
oneness that
create a chaotic result? I think this is very interesting..
* For the oneness of a number 'n' :
if(n is even) n = n/2;
if(n is odd) n = 3*n + 1;
Keep doing this until n = 1. The amount of necessary steps for
this is
called the oneness of n.
--
Quaternion
===
Subject: Re: Math dependency logic REVISED
... but Usenet was *way* to small to contain it.
> P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078
> but the correct polynomial is
> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078.
--ils duces d'Enron!
http://www.movisol.org/
http://members.tripod.com/~american_almanac/
===
Subject: Re: Math dependency logic REVISED
Ain't it touchin' the lengths to which Logikoi will go to bail
> one another out? See Camaraderie of the Experts
> I did go to that link. Here's how it begins.
>
> Lonely, are you?
Not at all. But don't fail to read Camaraderie of the Experts
at
It explains why you Boyz go to such great lengths to fend off
assaults on one or another's 'expertise'.
But you'll find nothing there about bum-fuzzling. Sorry.
> Anyway, the lengths the Logikoi will go to bail each other
out of
> *what*? Was there something threatening in this thread?
Oooh, were
what with
> my mortgage[1] and all, we can't have that.
--John
===
Subject: Re: Math dependency logic REVISED
>
Ain't it touchin' the lengths to which Logikoi will go to bail
>> one another out? See Camaraderie of the Experts
>
>I did go to that link. Here's how it begins.
>
>>
>
>Lonely, are you?
> If not for the way _he_ tends to speculate on people's
personal lives
> when he can't refute their arguments I'd say this wasn't
very nice,
> pointing out what a pathetic character he must be, forced to
talk to
> himself like this in public where everyone can see.
>Anyway, the lengths the Logikoi will go to bail each other
out of
>*what*? Was there something threatening in this thread?
> I know _I've_ been terrified of the possible consequences. I
mean
> of course everything he's said here has been nonsense, but
> regardless, what if someone at OSU found out that there was
> someone on the internet saying bad things about me?
> (Giggle. Worse yet: You may not have noticed, but he often
> quotes me saying wild things like everything is equal to
> itself. What if someone at OSU found out I was promulgating
> that sort of heresy? I can just picture it, once that
post-tenure
> review he mentioned elsewhere is implemen: There's a
> committee meeting in Whitehurst. Ullrich says everything is
> equal to itself? Off with his head.)
Of more concern to this committee might be your UseNet stalking
of JSH.
Searched Groups for JSH OR James OR Harris
author:ullrich@math.okstate.edu. Results 1 - 10 of about 2,700.
Search took 0.70 seconds.
Then again, you could always plead insanity...
http://www.fetchfido.co.uk/sound_files/giggle.wav
--John
===
Subject: Re: Math dependency logic REVISED
>
>Ain't it touchin' the lengths to which Logikoi will go to bail
>one another out?
> Giggle. What's fascinating is the consistency with which you
reply
> with non-sequiturs when it becomes apparent even to you that
> you've been making an idiot of yourself.
C1 AxAy[x=y -> Az(z in x <-> z in y)]
C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)]
C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in
A)Classification
C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] (Weak
Extensionality)
Exhibit of proof of Ex~(x=x) from C1-C4 and someone will point
out
the error.
Laud as you will valid reasoning from true premises, this is
something you don't know how to do. Imagine! A Ph.D. in
mathematics
who. instead of pointing out that Ex~(x=x) follows from C3,C4
alone, DENIES that Ex~(x=x) follows from these premises at
all!!!
--John
You concentra on mathematics because it is predictable, because
there is always a right answer you can check in the back of
the book,
because you like following very precise rules, because it
allows
you to escape from everyday life into a world that has nothing
to
do with everyday life, and because mathematics does not require
the creativity that you completely lack.
Keith Devlin, Commencement address to the mathematics
graduating class
of UC Berkeley, May 23, 1997
A performance system is designed to work in a defined task
domain,
accepting particular goals and seeking to reach them by some
kind of
highly selective search. The system must be told what goal is
to be
reached and must be given a description of the structure and
characteristics of the task domain in which it is to operate:
its
problem space. [...] In contrast, a learning system, is
capable of
acquiring a problem space, in whole or part, by interacting
with the
external environment and without being instruc about it
directly.
A performance system is designed to work in a defined task
domain,
accepting particular goals and seeking to reach them by some
kind of
highly selective search. The system must be told what goal is
to be
reached and must be given a description of the structure and
>characteristics of the task domain in which it is to operate:
its
problem space. [...] In contrast, a learning system, is
capable of
acquiring a problem space, in whole or part, by interacting
with the
external environment and without being instruc about it
directly.
Herbert Simon <http://www.isi.edu/~shen/book-foreword.html>
===
Subject: Re: Math dependency logic REVISED
>
Ain't it touchin' the lengths to which Logikoi will go to bail
>> one another out? See Camaraderie of the Experts
>
>I did go to that link. Here's how it begins.
>
>>
>
>Lonely, are you?
> If not for the way _he_ tends to speculate on people's
personal lives
> when he can't refute their arguments I'd say this wasn't
very nice,
> pointing out what a pathetic character he must be, forced to
talk to
> himself like this in public where everyone can see.
>Anyway, the lengths the Logikoi will go to bail each other
out of
>*what*? Was there something threatening in this thread?
> I know _I've_ been terrified of the possible consequences. I
mean
> of course everything he's said here has been nonsense, but
> regardless, what if someone at OSU found out that there was
> someone on the internet saying bad things about me?
> (Giggle.
Nothing makes a pig happier than a roll in the wallow, as this
giggling
porker knows!
http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg&
width=314&height=4
00#
http://www.fetchfido.co.uk/sound_files/giggle.wav
===
Subject: Re: Math dependency logic REVISED
<giggle>
Nothing makes a pig happier than a roll in the wallow, as
this giggling porker knows!
http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg&
width=314&height=4
00#
http://www.fetchfido.co.uk/sound_files/giggle.wav
===
Subject: Re: Math dependency logic REVISED
> <giggle>
>Nothing makes a pig happier than a roll in the wallow, as
>this giggling porker knows!
Excellent point. You finally decided to answer my question.
Lucky thing, too, because people were starting to think that
you were resorting to content-free nonsequiturs because you
didn't have an answer... giggle.
>http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg&
width=314&height=
400#
>http://www.fetchfido.co.uk/sound_files/giggle.wav
************************
===
Subject: Re: Math dependency logic REVISED
<giggle>
>
>Nothing makes a pig happier than a roll in the wallow, as
>this giggling porker knows!
> Excellent point. You finally decided to answer my question.
What question was that?
--John
Nothing makes a pig happier than a roll in the wallow, as
this giggling porker knows!
http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg&
width=314&height=4
00#
http://www.fetchfido.co.uk/sound_files/giggle.wav
===
Subject: Re: Math dependency logic REVISED
>
>> You yourself dispu that fact. Starting from C1-C4,
by applying correct reasoning, one deduces Ex~(x=x).
>>
Exhibit of proof of Ex~(x=x) from C1-C4 and someone will point
out
>> the error.
>
>So C1-C4 |- Ex~(x = x)? I assume that's right -- I won't
check.
>Ullrich said it didn't? Maybe, though if so, it's likely just
a minor
>error (perhaps caused by context?).
> Caused by context indeed. He quotes that thing I said really
> a lot, but he never quotes the post where I actually _said_
it,
> it's always a followup to a reply to a response. I'd like to
see
> the original, where I make that statement with no > at the
> start of the lines...
Exhibit of proof of Ex~(x=x) from C1-C4 and someone will point
out
the error.
No > at the start of the line! PURE, UNADULTERA BULLRICH!
>Where does this entail a denial of if you start with
something true
>and apply correct reasoning then the conclusion must be true?
>
>Are you just lying? Or, shall we be charitable and assume
that you're
>stupid?
> It _is_ quite remarkable how slow he's being about this
point,
> how it is that calling a proof incorrect, (correctly or
incorrectly)
> does not involve denying that if you start with something
true
> and apply correct reasoning the conclusion must be true.
Of course, you don't deny that if one starts with something
true and
apply correct reasoning, then the conclusion must be true. But
coming from you these are empty words, for you have demonstra
(on more
than one occasion!) your inability to reason from C1-C4 to
Ex~(x=x).
C1 AxAy[x=y -> Az(z in x <-> z in y)]
C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)]
C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] (Weak
Extensionality)
What sets you apart from a eunuch in a harem? You 'know how
it's done'--you've seen it every day--but you're unable to do
it yourself!
--John
===
Subject: Re: Math dependency logic REVISED
what bearing does this have on James' historically simple proof
of the very last theorem of Fermat?... is your exhibit an
example
of a Leftwing or a Rightwing proof, or
do you not believe in Devlin's metaphor?... if his metaphor is
serious,
I'm afraid to ask the party-affiliation of *any*
mathemeaticians!
note that there's a simple (1.5-page) isomorphism
of deductive & inductive mathematical proofs. just
for the concrete of it, I'll give an example
of a simple, unsolved problem, the Perfect Box:
take a rectangular box (or parallelipiped)
with 3 edges, 3 facial digonals and 1 interior diagonal, and
find a solution in integeor prove that there is none;
taht's 7 different lengths, although they're *dependent*
on only 3 of them, say the rectanular edges of the box.
(nothing more than the pythagorean theorem is required,
as far as algebra goes, but it's a nicety to *prove* it,
before using it. for the sake of clarity,
call the rectangular edges x, y, z, and the face diagonals a,
b, c,
where A=Y+Z, B=Z+X, C=X+Y (X=xx etc., and
D=dd is the skware of the interior diagonal) and
D = X+A = Y+B = Z+C = D.)
> Of course, you don't deny that if one starts with something
true and
> apply correct reasoning, then the conclusion must be true.
But
> coming from you these are empty words, for you have
demonstra (on more
> than one occasion!) your inability to reason from C1-C4 to
Ex~(x=x).
> C1 AxAy[x=y -> Az(z in x <-> z in y)]
> C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)]
> C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
> C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] (Weak
> Extensionality)
--ils duces d'Enron!
http://larouchepub.com/radio/index.html
===
Subject: Re: Math dependency logic REVISED
>> [...]
>Of course, you don't deny that if one starts with something
true and
>apply correct reasoning, then the conclusion must be true.
Well that's progress - for some reason you've decided not to be
entirely stupid today.
So of course I didn't deny that. So that example is totally
irrelevant regarding your (_repea_) statement that I was
_lying_ when I said nobody was denying that. Which brings
us back to the beginning: If I was lying when I said nobody
was denying that you should produce an example where
someone _has_ denied that. Or admit that you were being
either stupid or dishonest when you said I was lying.
Of course neither of those is going to happen, because
you have more concern for exposing evildoers than for
telling the truth.
> But
>coming from you these are empty words, for you have demonstra
(on more
>than one occasion!) your inability to reason from C1-C4 to
Ex~(x=x).
Oops. Back to the irrelvancies...
************************
===
Subject: Re: Math dependency logic REVISED
[...]
>
>Of course, you don't deny that if one starts with something
true and
>apply correct reasoning, then the conclusion must be true.
> Well that's progress - for some reason you've decided not to
be
> entirely stupid today.
> So of course I didn't deny that. So that example is totally
> irrelevant regarding your (_repea_) statement that I was
> _lying_ when I said nobody was denying that. Which brings
> us back to the beginning: If I was lying when I said nobody
> was denying that you should produce an example where
> someone _has_ denied that. Or admit that you were being
> either stupid or dishonest when you said I was lying.
> Of course neither of those is going to happen, because
> you have more concern for exposing evildoers than for
> telling the truth.
> But
>coming from you these are empty words, for you have demonstra
(on
more
>than one occasion!) your inability to reason from C1-C4 to
Ex~(x=x).
> Oops. Back to the irrelvancies...
When (with no evidence) *you* accuse someone of faulty
reasoning,
you expect to be taken seriously. But when glaring instances
of your own imperviousness to logic are ci and referenced, you
dismiss these as irrelevant!
Are you Rush Limbaugh? Do you do prescription drugs?
--John
C1 AxAy[x=y -> Az(z in x <-> z in y)]
C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)]
C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in
A)Classification
C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] (Weak
Extensionality)
Exhibit of proof of Ex~(x=x) from C1-C4 and someone will point
out
the error.
===
Subject: Re: Probability of a Run
>I've appended some code in VBA for the original algorithm and
for the
>approximation I mentioned.
>I make no great claims for the code but it should be adequate
for any
>investigations you might want to make.
CONGRATULATIONS IAN SMITH!
With a few lines of Basic code you have cut through mountains
of
mathematical horse and crea a program that gives quick answers
to what formerly was considered a difficult problem. This is
just
what I hoped would happen.
I por your code to BC7 for DOS and ran a few tests. The results
were exactly the same as what I got using the recursive
program or
through evaluating the coefficients of the generating function.
May I have your permission to compile it and put it on the web
site of
the Rancocas Valley Journal of Applied Mathematics, with full
credit
to yourself, of course?
Sam Allen
===
Subject: Re: Probability of a Run
>Your formula
>
>u(m + 1) = u(m) + (1 - u(m - n)) (1 - p) p^n
>
>is deducible from mine, for m > n. Just calculate
P(n,m+1)-P(n,m) in
>my notation, swap P(n,m) to u(m) and q to p and you'll get
the same. I
>had just about realized this couldn't be the difficult part
of the
>problem!
>
>I'm slightly puzzled as to why you said I want to calculate
the
>probability of this if you know the answer, unless you're not
a kid
>with a TI-83 or a home computer!
>
> Glad to explain! I am working on the subject of betting
strategies for
> my web site at
> http://www.cybcity.com/ranmath/start.htm
> A resort city close to me has been inunda by gambling
casinos. If
> everyone knew as much math as I do or as you do or had
studied
> psychology under B. F. Skinner as I have, they wouldn't
gamble, yet
> they do. I am trying to educate them. In a recent paper,
Edward O.
> expectation for a game with fixed expectation in the
slightest, yet I
> have found a counterexample. I have found a betting strategy
that will
> make the player's expectation WORSE, so I feel that the last
word on
> this subject has not yet been spoken.
> To calculate the player's expectation for game + strategy
one must
> take into account the probability of ruin and for the case
of the
> player who uses a Martingale betting strategy it is
necessary to refer
> to the Theory of Runs. It is a deep and difficult subject,
or at least
> it has been up to now. The above recurrence relation is not
mine. I
> got it from Burnside and Uspensky. My own attempts to derive
such
> things generally lead to wrong answers.
> I have the same mathematical machinery as Robert Israel does
but he
> knows how to use it better and even has helped me with it. I
don't
> feel I can tell my people that all they have to do is to
expand this
> complica rational function in powers of s and the
coefficients will
> be the required probabilities. Robert Israel can do it and
maybe I can
> do it but they can't do it. I want to give them something
they can
> use.
> I am studying everything you write but am a little behind. I
have
> gotten as far as your formula
>P(n,m) = 0 if m < n,
> = q^n.(1 + (1-q).( (m-n) - sum(i=n..m-n-1, P(n,i) ) ).
>
>The logic is simple. Either the first n trials are losses or
the first
> trial is a success and then we have n losses or we have no
runs of n
> or more in i trials followed by a success and then n losses
(for i =
> 1.. m-n-1). I think this enumerates all the possibile
combinations
> without duplication.
> and have the following observation:
> P(n,m) appears to be defined in terms of P(n,n) for the
range n > m 2n+2 and there seems to be no definition of P(n,n)
except a circular
> one. Setting m = n, the limits of the summation are n to -1
and the
> first term in the sum is P(n,n). Presumably the other terms
would have
> the value 0 because m < n.
> I construe the character between (m-n) and sum in your
formula as
> a minus sign. I assume that your formula is not intended to
be
> executable and will study your programming code. Have you
checked the
> output against results obtained in other ways?
I have no difficulty believing my explanations are difficult to
understand. My father was a very good teacher but I never
inheri
any of those skills.
was
beginning to think I was back at schoool.
I'll answer your last question first. The only results I could
obtain
were hand calculations for very small examples, the simple
case where
n=1 and m is any value and quite a few results from large
simulations.
I did say I'd tried to make sure the formula wasn't ludicrous
before I
put forward any reply.
When I said Burnside and Uspensky's formula is deducible from
mine, I
didn't make it clear that my definition is also deducible from
their's
(i.e. they are equivalent and I have not discovered anything
which was
not already known).
In my definition the summation is intended to be 0 if the
lower limit
exceeds to upper limit as is the case for P(n,n).
Although I have seen languages in which you could execute the
function
definitions, this wasn't really the intention here. Besides,
it would
not be an efficient algorithm if it worked directly from the
definition. The first program does work using this definition,
you'll
be reassured to know.
By writing
SP(n,m) = 0 if n > m
= sum(i=n..m, P(n,i)) otherwise
we can get rid of the summations.
Basically we calculate P(n,n) and SP(n,n), P(n,n+1) and
SP(n,n+1)...P(n,j) {using SP(n,j-n-1) which we have already
calcula} and SP(n,j)...P(n,m)
We don't really need all the values of SP at any given moment
and as
old habits die hard I economised on the space. Unfortunately,
it makes
a simple algorithm look complica.
The second algorithm works by undoing the recursive part of the
function definition. I'll rewrite the new definition as there
were
errors in the ranges for parts of the definition. I'll also
replace
(1-q) by p just to save space.
P(n,m)
= 0, if m < n
= q^n(1+p((m-n))), if m < 2n+1
= q^n(1+p((m-n)-q^n((m-2n)+p/2((m-2n-1)(m-2n))))), if m < 3n+2
=
q^n(1+p((m-n)-q^n((m-2n)+p/2((m-2n-1)(m-2n)/2)-q^n((m-3n-1)(m-
3n))+p/3((m-3n-
2)(m-3n-1)(m-3n))))))),
if m < 4n+3
...
This has been reorganised for calculation purposes as
P(n,m)
= 0, if m < n
= q^n(1+p(m-n)), if m < 2n+1
= q^n(1+p(m-n)(1-q^n(m-2n)/(m-n)(1+p/2(m-2n-1)))), if m < 3n+2
=
q^n(1+p(m-n)(1-q^n(m-2n)/(m-n)(1+p/2(m-2n-1)(1-q^n(m-3n-1)(m-
3n)/(m-2n)/(m-2n
-1)(1+p/3(m-3n-2)))))),
if m < 4n+3
...
The things to notice are that the new definition is exact. The
approx
in the function name in the code merely means that the
expression for
the m < 20n+19 case is similar to that for the case m < 21n+20
and you
don't really have to go all the way down to the bottom of the
expression to evaluate it accurately.
sense if you
compare this algorithm with that for 1-exp(-x) which can be
calcula
as
x(1-x/2(1-x/3(1-x/3(1-x/4...))))
This formula is ok until x gets too large and then its starts
returning nonsense.
It might be sensible to limit how big m can be - basically if
j (the
number of levels inside the brackets it starts the expression
from) is
too big you probably won't get a sensible answer. You might
also have
to wait a while due to an inefficient implementation of mnfunc
for
large values of i - I struggle to pick sensible names!
The code I've produced is in VBA because I find spreadsheets
convenient for holding disorganised calculations. The code is
pretty
simple so I don't think it will be hard to translate the code
into a
language of your choice.
Ian Smith
===
Subject: Re: JSH: Difficult social problem
congratulation!
> 7/x = 1 requires that x=7.
> It's an intriguing problem.
> http://mathforprofit.blogspot.com/
--ils duces d'Enron!
http://www.movisol.org/
http://members.tripod.com/~american_almanac/
===
Subject: Re: Difficult social problem
> It looks like I'm swinging at tissue paper with a
sledgehammer when it
> comes to getting acceptance of my work, as while I can get
initial
> contact with mathematicians they tend to run as soon as I
give them
> enough information to realize the implications of my work
and that I
> am correct.
Mr. Harris,
even if your work were correct (and it is not), it would be as
awe
inspiring
as my morning piss after waking up.
It would hardly be worth a mention anywhere, but hey, it has
only taken you
7 + years to form a wrong page of nonsense (I dare not refer
to it as
proof).
You are so full of yourself and don't even consider the
possibility that
you
could be wrong.
Get a reality check.
===
Subject: Re: Difficult social problem
> It looks like I'm swinging at tissue paper with a
sledgehammer when it
> comes to getting acceptance of my work, as while I can get
initial
> contact with mathematicians they tend to run as soon as I
give them
> enough information to realize the implications of my work
and that I
> am correct.
> Mr. Harris,
> even if your work were correct (and it is not), it would be
as awe
inspiring
> as my morning piss after waking up.
And people that's math society.
That's how math people *really* are, so forget the movies.
Keep that image of a math person taking his morning piss.
===
Subject: Re: Difficult social problem
[...]
> Mr. Harris,
even if your work were correct (and it is not), it would be as
> awe inspiring as my morning piss after waking up.
> And people that's math society.
> That's how math people *really* are, so forget the movies.
> Keep that image of a math person taking his morning piss.
>
James, how soon you forget:
: David Ullrich is a ing piece of dog.
:
: I think it's funny that I can call a professor at Oklahoma
State
: University a ing piece of dog knowing that he'll keep
: replying in my threads.
:
: You see, he has to keep replying pushing the same old lies.
:
: He's stuck. He's trapped in something that he can't get out
of,
: so it doesn't matter what I call him, or what I say about
him,
: he has to come back.
:
: You see I'm the person who has the correct math argument, so
: posters like David Ullrich or are *compelled*
: to reply out of fear that if they go away, then I'll get some
: people who'll pay attention to the truth.
:
: So David Ullrich, the math professor at Oklahoma State
University,
: is demeaned by me as the piece of ing dog he is, and he
: *has* to keep coming back.
:
:
:
===
Subject: Re: JSH: Difficult social problem
...
> Given that mathematicians as a group are both capable of a
high degree
> of irrationality, and fearful to the extent that they will
avoid
> troubling results that challenge their viewpoint, how do you
break
> through with a spectacular result?
You could douse yourself in gasoline and set yourself on fire.
That'd
get their attention.
Gib
===
Subject: Re: Difficult social problem
> It looks like I'm swinging at tissue paper with a
sledgehammer when it
> comes to getting acceptance of my work...
Does this mean that Southwest Journal has rejec your paper,
JSH?
Narcissism is anti-social, Harris. So much for your social
problem.
I've seen lots of cases of vanity going to war against
reality. Reality
always wins those conflicts. You're no different. How long
your string of
defeats will continue, and how much they will cost you, is up
to you. Give
it up.
LH
===
Subject: Re: JSH: Difficult social problem
> It looks like I'm swinging at tissue paper with a
sledgehammer when it
> comes to getting acceptance of my work, as while I can get
initial
> contact with mathematicians they tend to run as soon as I
give them
> enough information to realize the implications of my work
and that I
> am correct.
Perhaps if you learned some of the standard terminology and
techniques,
you'd have better luck communicating. It's like going to
France,
refusing to learn any French, and complaining that the waiters
won't
bring you soup. And believe me, if you don't speak any French,
the
waiters in France won't bring you soup.
> Extrapolating that tendency I come to the conclusion that in
general
> mathematicians are far less likely to accept a result that
goes
> against their beliefs, and more likely to simply avoid it
than even
> the general population.
Actually most mathematicians are pretty quick to recognize your
crankitude. What impressses me is the ongoing politeness and
patience
shown you by many professional mathematicians.
> And also mathematicians seem to be strangely capable of
ignoring even
> basic mathematical logic, like with my current discussions
where
> people are fighting a necessary conclusion from the simple
result that
> 7/x = 1 requires that x=7.
You make these simple statements, but they are not really rela
to the
rest of your argument.
> Given that mathematicians as a group are both capable of a
high degree
> of irrationality, and fearful to the extent that they will
avoid
> troubling results that challenge their viewpoint, how do you
break
> through with a spectacular result?
You've been at this for six or seven years now. If you made a
committment to spend the next six months learning some basic
algebra --
working through, say, Herstein's Topics in Algebra, asking
questions
here when you don't understand something -- you'd be amazed at
how much
better your relationship with the math community will become.
===
Subject: Re: JSH: Difficult social problem
>And believe me, if you don't speak any French, the
>waiters in France won't bring you soup.
Have you considered the possibility that the reason waiters
don't
bring you soup might *not* be that you don't speak French?
-- Richard
--
Spam filter: to mail me from a .com/.net site, put my surname
in the
headers.
FreeBSD rules!
===
Subject: Re: JSH: Difficult social problem
Discussion, linux)
> Perhaps if you learned some of the standard terminology and
techniques,
> you'd have better luck communicating. It's like going to
France,
> refusing to learn any French, and complaining that the
waiters won't
> bring you soup. And believe me, if you don't speak any
French, the
> waiters in France won't bring you soup.
Gosh. That means that when I visi Marseilles this summer, the
whole place was crawling with imposters posing as waiters and
bringing
me soup. That's a little freaky.
--
Now I'm informing all of you that the people arguing against
me are
EVIL, yes they are real, live EVIL people as mathematics is
that
important, so it's important enough for Evil itself to send
minions
like them. -- on Evil's interest in Alg. Number Theory
===
Subject: Re: JSH: Difficult social problem
>It looks like I'm swinging at tissue paper with a
sledgehammer when it
>comes to getting acceptance of my work, as while I can get
initial
>contact with mathematicians they tend to run as soon as I
give them
>enough information to realize the implications of my work and
that I
>am correct.
Giggle. That's a very amusing way to put it. Yes, you can get
initial
contact with mathematicians, especially if they're not familiar
with the name . Anyone with a PC can do that these
days. And yes, they run when you give them enough information
about your work. But it's not because they realize you're
correct...
>Extrapolating that tendency I come to the conclusion that in
general
>mathematicians are far less likely to accept a result that
goes
>against their beliefs, and more likely to simply avoid it
than even
>the general population.
>And also mathematicians seem to be strangely capable of
ignoring even
>basic mathematical logic, like with my current discussions
where
>people are fighting a necessary conclusion from the simple
result that
>7/x = 1 requires that x=7.
You're either a total moron or a blatant liar. Nobody has dispu
this fact.
>Given that mathematicians as a group are both capable of a
high degree
>of irrationality, and fearful to the extent that they will
avoid
>troubling results that challenge their viewpoint, how do you
break
>through with a spectacular result?
>It's an intriguing problem.
>
>http://mathforprofit.blogspot.com/
************************
===
Subject: Re: JSH: Difficult social problem
Question: What does one get when one crosses David Ullrich
with the progeny of a moebius strip, a parrot and a pooch?
Answer: A one-sided, two-dimensional lapdog that covets
crackers.
===
Subject: Re: JSH: Difficult social problem
> It looks like I'm swinging at tissue paper with a
sledgehammer when it
> comes to getting acceptance of my work, as while I can get
initial
> contact with mathematicians they tend to run
You might have better luck if you stop using
as the salutaion in your e-mails.
>
> http://mathforprofit.blogspot.com/
===
Subject: Re: JSH: Difficult social problem
> It looks like I'm swinging at tissue paper with a
sledgehammer when it
> comes to getting acceptance of my work, as while I can get
initial
> contact with mathematicians they tend to run
> You might have better luck if you stop using
> as the salutaion in your e-mails.
<giggle> I think that's the funniest thing I've read in one of
his threads.
>
> http://mathforprofit.blogspot.com/
===
Subject: Re: PDE to ODE in Maple
|>> PDE1a :=
|>convert(PDE1,fddiff,order=[1,0],forward=[1,0],indexletters=[
j,none]):
|>PDE1a;
|> d nu (x[j + 1] - x[j])
|>
|> -- x[j] = --------------------
|>
|> dt h[r]
I don't know what this convert(..., fddiff) is. Is it defined
in one of
your utilities? There doesn't seem to be any fddiff mentioned
in any of
the Maple 9 help files. I presume that, since the left side
didn't
evaluate to 0, this must be using the inert form Diff rather
than diff.
|>> xeqns := [seq(PDE1b,j=2..n_x-1)];
|>> S := convertsys(xeqns, [], [seq(x[j],j=2..n_x-1)], t, x,
xp ):
|>Error, (in DEtools/convertsys) invalid system of differential
|>equations
I think it would work if you put back the t dependence into
your
system, e.g.
> xeqns:= subs(seq(x[j] = x[j](t), j=2..n_x-1),xeqns);
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: Factor with Modulus option question
coefficients.
Ok. That seems to explain a little better what it is trying to
do.
again.
Dana
> In the program Mathematica, one can factor an equation
Modulus
a
Prime
> number. However, I am having a hard time trying to
understand how to
> interpret the output. Could someone offer a simple
explanation of
> For example, the following normally cannot be factored any
further.
>
Factor[5 + x^2]
5 + x^2
However, if we factor this using the option Modulus a Prime
number,
we
get
> a different output.
> Factor[5 + x^2, Modulus -> 3]
(1 + x)*(2 + x)
> ..or
> 2 + 3*x + x^2
I have been having a hard time understanding just what the
output is
trying
> to tell me.
> I understand how Mod[7, 3] returns 1, but I do not see the
relationship
> Calculating mod 3, Mathematica maps
> ..., -6,-3,0,3,6,9, .... to 0
> ..., -5,-2,1,4,7,10, ... to 1
> ..., -4,-1,2,5,8,11, ... to 2
> so you asked for the factors of x^2+2, when reduced mod 3.
> It gave you the factors 1+x and 2+x.
> multiplied together gives 2+3x+x^2, but note that 3 is the
same as 0.
> so the result is 2+x^2. That's what you asked to factor.
===
Subject: Autioning papers for coauthor credit
Hello everyone,
I am going to aution my papeon ebay. The highest bidder will
own the
coauthor rights to my papers.
Initially, this is the first paper, i am going to run the
trial on.
http://www.wikiworld.com/KooMar/ulat.pdf
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=2979558376#
ebayphotohosti
ng
-suresh
===
Subject: Re: Autioning papers for coauthor credit
> I am going to aution my papeon ebay. The highest bidder will
own the
> coauthor rights to my papers.
I should think the above provides ample grounds for instant
ex-communication from the academic priesthood. It certainly
gives a whole new meaning to the market for ideas.
But, I guess it would not be without interest:
* The Ebay Price Ranking of Journals
Presumably, a paper that has been accep for publication
would sell for considerably more than a mere draft; we would
then
have a price ranking of journals instead of the ubiquitous
quality rankings (citation impacts etc) that never seem to
agree
with each other. This scenario could be quite easily played
out,
and might even make a nice paper in experimental economics ...
* Academics seeking tenure and in need of some more papers
could obtain
them efficiently on ebay, rather than muscling their own
students
into providing co-authorship.
* Or, taking the inverse case, students who believe that
'names'
sell papers could shop around for a co-author to adorn their
paper
without even having to attend an expensive US grad school.
Delightfully poor taste
Colin
______________________________
Dr Colin Rose
mathStatica Pty Ltd
Email: colin@mathStatica.com
Web: www.mathStatica.com
______________________________
===
Subject: Re: Autioning papers for coauthor credit
In some branches of science it is common to list as coauthor
the one
who obtained the funding for the lab where the work was done.
There may be some conventions like: the last-lis author is the
one
the words, etc...
But not in mathematics.
Yet.
===
Subject: Autioning papers on ebay
I am going to aution my papeon ebay. The highest bidder will
own the
coauthor rights to my papers.
Initially, this is the first paper, i am going to run the
trial on
http://www.wikiworld.com/KooMar/ulat.pdf
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=2979558376#
ebayphotohosti
ng
I am going to aution every paper on this page
https://sourceforge.net/project/showfiles.php?group_id=61183
Each paper(pdf file, not zip file) will start at a initial bid
of $600.
-suresh
===
Subject: evaluating inequalities
Does anyone know of a computer algebra package that can
evaluate
inequalities
in a general way. That is, one that could state that:
2^n > n^2 for all n > 4
is true.
Don
===
Subject: Re: evaluating inequalities
All inequalities?
Your particular example can be solved by
trying to
solve(2^n-n^2=0), which has 3 roots n=4, n=2 and
n = - ((2 * lambert_w(((log(2))/2)))/(log(2)))
which is about n=-0.76666.
Thus one can deduce that the expression does not change sign
after
n=4. try n=5, when 32>25 so it is positive.
Thus the statement below can be proved, if you can state it
as given above.
I do not know if there is a program to do exactly this, but
I've
described how one might write it.
> Does anyone know of a computer algebra package that can
evaluate
> inequalities
> in a general way. That is, one that could state that:
> 2^n > n^2 for all n > 4
> is true.
> Don
===
Subject: Re: Is the Functional Paradigm suitable for Math
Heavy problems?
> Ever looked at the ICFP contest results?
No I haven't. Having found
http://www.dtek.chalmers.se/groups/icfpcontest/,
I will take a gander.
--
www.indiegamedesign.com
Brandon Van Every Seattle, WA
Desperation is the motherer of Invention. - Robert Prestridge
===
Subject: W32.HLLW.Torvel.b virus->Fw: Returned mail--
More virus being spewed to the world.
Here is the (castra) evidence.
...
><HTML><HEAD></HEAD><BODY>
><iframe src=3Dcid:WCGmWkmPPaIpW height=3D0 width=3D0></iframe>
></BODY></HTML>
>--mMckiDTIqhDpoluAgbLhFEYiVVdPydWL
> name=SMSSfl.exe
>Content-ID: <WCGmWkmPPaIpW>
>TVpQAAIAAAAEAA8A//
8AALgAAAAAAAAAQAAaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
A
<<>>
>--mMckiDTIqhDpoluAgbLhFEYiVVdPydWL--
===
Subject: Re: Your Research: Not So Revolutionary After All
I am certain that if I would ever hire a person like you I
> would fall into a serious error as such a person is obviously
> not a team player and would make constant attempts to break
> communication within the company.
That's strange. I thought that Vladimir Bondarenko hasn't been
hired by
Maplesoft exactly for that reason... Hmm. Maybe, he is a
Mathematica team
player?
Alec
===
Subject: Re: Universal set theory and three-valued logic
Originator: israel@math.ubc.ca (Robert Israel)
> : In my formulation, I identify each set with its
characteristic
> : function from sets to {T,F,bot}. Thus given s:set, the
membership of
> : an element x can be tes with s(x). In the general case,
this is a
> : partial function, returning bot for some values. I call
such sets
> : partial sets, and sets whose characteristic function is in
{T,F}
> : total sets. Every set of ZF and NF is a total set, with
this
theory
> : admitting a strictly larger class of sets than either.
> In some ways, this is the structure of a topoi, or at least
a functor from
a
> topoi category to something similar. Is the approach meant
to be
categorial
> (which, by the way, is a good thing in my eyes -- I'm just
curious)?
Then,
> your partial classification appears to mainly distinguish
the topoi of
sets
> from some of the many other topoi.
My goal is to show evidence that a particular type theory is
reasonable, by equating it to some existing system. An
axiomatic set
theory was my first hope due to the relative simplicity, but a
topos
will do.
> The categorial study of paradox is becoming a large field
these days, and
it
> appears you may be repeating some of the work already done
(which can be
> soooo frustrating sometimes!). I don't mean to assume any
level of
study,
> but perhaps I might suggest that, if you haven't, you should
check out
some
> online I can suggest.
Do you have any pointers to research on topoi coinciding to a
set
theory with a universal set, or more generally to any
topos-centric
analysis of paradoxes due to nonwellfoundedness?
> I could make one suggestion, it would be to not restrict
yourself to your
> trivalent logic. Any Heyting algebra is possible, and
expands your
research
> into the much more fruitful world that all topoi present. In
fact,
because
> of natural distinctions that present themselves between
propositions that
> take on the middle value, trivalent theories are often
looked at only
as
> summarisations of a more natural infinitely valent theory.
Good
resources
> for this can be found in intuitionist discussions, but it is
more
general.
In my setting, I am mainly interes in mechanically analyzing
sets
to determine whether they are inhabi and, if so, whether they
are
singletons. A trivalent logic seems sufficient for this: any
finer
structure can be projec down for my purposes.
> Also, may I ask why you pos to comp.lang.functional? This
intrigues
me
> because some of my own research has been around the
evaluation of the
lambda
> calculus and proof / evaluation theory in the context of non
standard
> logics, but I do not see this approach explicitly sta in
your message.
I'm developing a programming language with enough expressive
power
that one may easily express the major paradoxes from set
theory, both
those due to recursion (the liar paradox) and those due to
nonwellfoundedness or reflection (Russell's paradox). I have a
compiler that analyzes programs and assigns to each term a set
cardinality approximations classifying its potential
inhabitance, as
well as an indicator of the decidability of this
classification.
The general idea is that single-valued terms correspond to
runtime-executable parts of programs, while more general
(potentially
uninhabi or multivalued terms) may be bundled up into
extensional
types and reasoned about within the limits of the compiler's
resolution rules. This all works to the extent of being able to
express the paradoxes, recognize their undecidability, and
mechanically determine that, for example, certain concrete
terms do or
don't belong to Russell's set. But though it tends to work
well in
mechanical terms, it lacks a theoretical basis.
Tim Sweeney
===
Subject: This Week's Finds in Mathematical Physics (Week 201)
Originator: baez@math-cl-n01.math.ucr.edu (John Baez)
Originator: israel@math.ubc.ca (Robert Israel)
Also available at http://math.ucr.edu/home/baez/week201.html
This Week's Finds in Mathematical Physics - Week 201
John Baez
Lately James Dolan and I have been studying number theory. I
used to
*hate* this subject: it seemed like a massive waste of time.
Newspape
magazines and even lots of math books seem to celebrate the
idea of people
slaving away for centuries on puzzles whose only virtue is
that they're
easy to state but hard to solve. For example: are any odd
numbers the
sum of all their divisors? Are there infinitely many pairs of
primes
that differ by 2? Is every even number bigger than 2 a sum of
two primes?
Are there any positive integer solutions to
x^n + y^n = z^n
for n > 2? My response to all these was: WHO CARES?!
Sure, it's noble to seek knowledge for its own sake. But
working on a
math problem just because it's *hard* is like trying to drill
a hole in
a concrete wall with your nose, just to prove you can! If you
succeed,
I'll be impressed - but I'll still wonder why you didn't put
all that
energy into something more interesting.
Now my attitude has changed, because I'm beginning to see that
behind
these silly hard problems there lurks an actual *theory*, full
of deep
ideas and interesting links to other branches of mathematics,
including
mathematical physics. It just so happens that now and then
this theory
happens to crack another hard nut.
I'd known for a while that something like this must be true:
after all,
when Andrew Wiles proved Fermat's Last Theorem, even the
newspapers
admit this was just a spinoff of something more important,
namely
a special case of the Taniyama-Shimura Conjecture. They said
this had
something to do with elliptic curves and modular forms, which
are very
nice geometrical things that show up all over in complex
analysis and
string theory. Unfortunately, the actual statement of this
conjecture
seemed impenetrable - it didn't resonate with things I
understood.
In fact, the Taniyama-Shimura Conjecture is part of a big
*network* of
problems that are more interesting but harder to explain than
the flashy
ones I lis above: problems like the Extended Riemann
Hypothesis, the
Weil Conjecture (now solved), the Birch-Swinnerton-Dyer
Conjecture,
and bigger projects like the Langlands Program and developing
the theory
of motives. And these problems rest on top of a solid
foundation of
beautiful stuff that's already known, like Galois theory and
class field
theory, and stuff about modular forms and L-functions.
As I'm gradually beginning to understand little bits of these
things,
I'm getting really exci about number theory... so I'm dying to
*explain*
some of it! But where to start? I have to start with something
basic that
underlies all the fancy stuff. Hmm, I think I'll start with
Galois theory.
As you may have heard, Galois inven group theory in the
process of
showing you can't solve the quintic equation
ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0
by radicals. In other words, he showed you can't solve this
equation
by means of some souped-up version of the quadratic formula
that just
involves taking the coefficients a,b,c,d,e,f and adding,
subtracting,
multiplying, dividing and taking nth roots.
The basic idea is something like this. In general, a quintic
equation
has 5 solutions - and there's no best one, so your formula has
got to
be a formula for all five. And there's a puzzle: how do you
give one
formula for five things?
Well, think about the quadratic formula! It has that plus or
minus
in it, which comes from taking a square root. So, it's really
a formula
for *both* solutions of the quadratic equation. If there were
a formula
for the quintic that worked like this, we'd have to get all 5
solutions
from different choices of nth roots in this formula.
Galois showed this can't happen. And the way he did it used
*symmetry*!
Roughly speaking, he showed that the general quintic equation
is completely
symmetrical under permuting all 5 solutions, and that this
symmetry group -
the group of permutations of 5 things - can't be built up from
the symmetry
groups that arise when you take nth roots.
The moral is this: you can't solve a problem if the answer has
some
symmetry, and your method of solution doesn't let you write
down an
answer that has this symmetry!
An old example of this principle is the medieval puzzle called
Buridan's
Ass. Placed equidistant between two equally good piles of hay,
this
donkey starves to death because it can't make up its mind
which alternative
is best. The problem has a symmetry, but the donkey's method
of solution
doesn't, so it's stuck.
Buridan's ass would also get stuck if you asked it for *the*
solution
to the quadratic equation. Galois proof of the unsolvability
of the
quintic by radicals is just a more sophistica variation on
this theme.
(Of course, you *can* solve the quintic if you strengthen your
methods.)
A closely rela idea is Curie's principle, named after Marie's
husband Pierre. This says that if your problem has a symmetry
and
it is a unique solution, the solution must be symmetrical.
For example, if some physical system has rotation symmetry and
it has a
unique equilibrium state, this state must be rotationally
invariant.
Now, in the case of a ferromagnet below its Curie temperature,
the
equilibrium state is *not* rotationally invariant: the little
magnetized
electrons line up in some specific direction! But this doesn't
contradict Curie's principle, since there's not a unique
equilibrium
state - there are lots, since the electrons can line up in any
direction.
Physicists use the term spontaneous symmetry breaking when any
*one*
solution of a symmetric problem is not symmetrical, but the
whole set
of them is. This is precisely what happens with the quintic,
or even
the quadratic equation.
While these general ideas about symmetry apply to problems of
all sorts,
their application to number theory kicks in when we apply them
to *fields*.
A field is a gadget where you can add, subtract, multiply and
divide
by anything nonzero, and a bunch of familiar laws of
arithmetic hold,
which I won't bore you with here. The three most famous fields
are the
rational numbers Q, the real numbers R, and the complex
numbers C.
However, there are lots of other interesting fields.
Number theorists are especially fond of algebraic number
fields. An
algebraic number is a solution to a polynomial equation whose
coefficients
are rational numbers. You get an algebraic number field by
taking the
field of rational numbethrowing in finitely many algebraic
numbeand
then adding, subtracting, multiplying and dividing them to get
more numbers
until you've got a field.
For example, we could take the rationals, throw in the square
root of 2,
and get a field consisting of all numbers of the form
a + b sqrt(2)
where a and b are rational. Notice: if we add, multiply,
subtract or
divide two numbers like this, we get another number of this
form.
So this is really a field - and it's called Q(sqrt(2)), since
we use
round parentheses to denote the result of taking a field and
extending
it by throwing in some extra numbers.
More generally, we could throw in the square root of any
integer n,
and get an algebraic number field called Q(sqrt(n)),
consisting of all
numbers
a + b sqrt(n)
where a and b are rational. If sqrt(n) is rational then this
field is
just Q, which is boring. Otherwise, we call it a quadratic
number
field.
Even more generally, we could take the rationals and throw in a
solution of any quadratic equation with rational coefficients.
But
it's easy to see that this doesn't give anything beyond fields
like
Q(sqrt(n)). And that's the real reason we call these the
quadratic
number fields.
There are also cubic number fields, and quartic number fields,
and quintic number fields, and so on. And othetoo, where we
throw in solutions to a whole bunch of polynomial equations!
Now, it turns out you can answer lots of classic but rather
goofy-sounding
number theory puzzles like which integers are a sum of two
squares?
by converting them into questions about algebraic number
fields.
And the good part is, the resulting questions are connec to
all sorts
of other topics in math - they're not just glorified mental
gymnastics!
So, from a modern viewpoint, a bunch of classic number theory
puzzles are
secretly just tricks to get certain kinds of people interes in
algebraic
number fields.
But right now I *don't* want to explain how we can use
algebraic number
fields to solve classic but goofy-sounding number theory
puzzles.
In fact, I want to downplay the whole puzzle aspect of number
theory.
Instead, I hope you're reeling with horror at thought of this
vast
complica wilderness of fields containing Q but contained in C.
First there's a huge infinite thicket of algebraic number
fields...
and then, there's an ever scarier jungle of fields that contain
transcendental numbers like pi and e! I won't even talk about
*that*
jungle, it's so dark and scary. Physicists usually zip
straight past
this whole wilderness and work with C.
But in fact, if you stop and carefully examine all the
algebraic number
fields and how they sit inside each other, you'll find some
incredibly
beautiful patterns. And these patterns are turning out to be
rela to
Feynman diagrams, topological quantum field theory, and so
on...
However, before we can talk about all that, we need to
understand the
basic tool for analyzing how one field fits inside another:
Galois theory!
A function from a field to itself that preserves addition,
subtraction,
multiplication and division is called an automorphism. It's
just
a *symmetry* of the field. But now, suppose we have a field K
which
contains some smaller field k. Then we define the Galois group
of K
over k to be the group of all automorphisms of K that act as
the
identity on k. We call this group
Gal(K/k)
for short.
The classic example, familiar to all physicists, is the Galois
group of
the complex numbeC, over the real numbeR. This group has two
elements: the identity transformation, which leaves everything
alone, and
complex conjugation, which switches i and -i. Since the only
group with 2
elements is Z/2, we have
Gal(C/R) = Z/2
Where does complex conjugation come from? It comes from the
fact that
C from R by throwing in a solution of the quadratic equation
x^2 = -1.
We say C is a quadratic extension of R. But as soon as we
throw in one
solution of this equation, we inevitably throw in another,
namely its
negative - and there's no way to tell which is which. And
complex
conjugation is the symmetry that switches them!
Note: we know that i and -i are different, but we can't tell
which is
which! This sounds a bit odd at first. It's a bit hard to
explain
precisely in ordinary language, which is part of why Galois
had to invent
group theory. But it's fun to try to explain it in plain
English...
so let me try. The complex numbers have two solutions to
x^2 = -1.
By convention, one of them is called i, and the other is called
-i. Having made this convention, there's never any problem
telling
them apart. But we could reverse our convention and nothing
would
go wrong. For example, if the ghost of Galois waf into your
office
wherever there had been i, everything in these books would
still be
true!
Here's another way to think about it. Suppose we meet some
extraterrestrials and find that they too have developed the
complex
numbers by taking the real numbers and adjoining a square root
of -1,
only they call it @. Then there would be no way for us to tell
if
their @ was our i or our -i. All we can do is choose an
arbitrary
convention as to which is which.
Of course, if they put their @ in the lower halfplane when
drawing the
complex plane, we might feel like calling it -i... but here we
are
secretly making use of a convention for matching their complex
plane with
ouand the *other* convention would work equally well! If they
drew
their real line *vertically* in the complex plane, it would be
more
obvious that we need a convention to match their complex plane
with ou
and that there are two conventions for doing this, both
perfectly
self-consistent.
If you've studied enough physics, this extraterrestrial
scenario
should remind you of those thought experiments where you're
trying to
explain to some alien civilization the difference between left
and
right... by means of radio, say, where you're *not* allowed to
refer
to specific objects you both know - so it's cheating to say
imagine
you're on Earth looking at the Big Dipper and the handle is
pointing
down; then Arcturus is to the right.
If the laws of physics didn't distinguish between left and
right,
you couldn't explain the difference between left and right
without
cheating like this, so the laws of physics would have a
symmetry
group with two elements: the identity and the transformation
that
switches left and right. As it turns out, the laws of physics
*do*
distinguish between left and right - see week73 for more on
that.
But that's another story. My point here is that the Galois
group of C
over R is a similar sort of thing, but built into the very
fabric of
mathematics! And that's why complex conjugation is so
important.
I could tell you a nice long story about how complex
conjugation is
rela to charge conjugation (switching matter and antimatter)
and
also time reversal (switching past and future). But I won't!
Here's another example of a Galois group that physicists
should like.
Let C(z) be the field of rational functions in one complex
variable z -
in other words, functions like
f(z) = P(z)/Q(z)
where P and Q are polynomials in z with complex coefficients.
You can
add, subtract, multiply and divide rational functions and get
other
rational functions, so they form a field. And they contain C
as a
subfield, because we can think of any complex number as a
*constant*
function. So, we can ask about the Galois group of C(z) over C.
What's it like?
It's the Lorentz group!
To see this, it's best to think of rational functions as
functions not
on the complex plane but on the Riemann sphere - the complex
plane
together with one extra point, the point at infinity. The only
conformal transformations of the Riemann sphere are fractional
linear
transformations:
az + b
T(z) = ------
cz + d
So, the only symmetries of the field of rational functions that
act as the identity on constant functions are those coming from
fractional transformations, like this:
f |-> fT where fT(z) = f(T(z))
If you don't follow my reasoning here, don't worry - the
details aren't
hard to fill in, but they'd be distracting here.
The last step is to check that the group of fractional linear
transformations is the same as the Lorentz group. You can do
this
algebraically, but you can also do it geometrically by
thinking of the
Riemann sphere as the heavenly sphere: that imaginary sphere
the stars
look like they're sitting on. The key step is to check this
remarkable
fact:
if you shoot past the earth near the speed of light, the
constellations
will
look distor by a Lorentz transformation - but if you draw lines
connecting
the staall the *angles* between these lines will remain the
same; only
their *lengths* will get messed up!
Moreover, it's obvious that if you rotate your head, both
angles and
lengths
on the heavenly spehere are preserved. So, any rotation or
Lorentz boost
gives an angle-preserving transformation of the heavenly
sphere - that is,
a conformal transformation! And this must be a fractional
linear
transformation.
Summarizing, the Galois group of C(z) over C is the Lorentz
group, or
more precisely, its connec component, SO_0(3,1):
Gal(C(z)/C) = SO_0(3,1).
We've talked about the Galois group of C(z) over C and the
Galois group
of C over R. What about the Galois group of C(z) over R?
Unsurprisingly,
this is the group of transformations of the Riemann sphere
genera by
fractional linear transformations *and* complex conjugation.
And
physically,
this corresponds to taking the connec component of the Lorentz
group
and throwing in *time reversal*! So you see, complex
conjugation is
rela
to time reversal. But I promised not to go into that....
I've been talking about Galois groups that physicists should
like, but
you're probably wondering where the number theory went! Well,
it's
all part of the same big story. In number theory we're
especially
interes in Galois groups like
Gal(K/k)
where K is some algebraic number field and k is some subfield
of K.
For starteconsider this example:
Gal(Q(sqrt(n))/Q)
where sqrt(n) is irrational. I've already hin at what this
group is!
Q(sqrt(n)) has sqrt(n) in it, so it also has -sqrt(n) in it,
and there's
an automorphism that switches these two while leaving all the
rational
numbers alone, namely
a + b sqrt(n) |-> a - b sqrt(n) (a,b in Q)
So, we have:
Gal(Q(sqrt(n)))/Q) = Z/2
just like the Galois group of C over R.
To get some bigger Galois groups, let's take Q and throw in a
primitive
nth root of unity. Hmm, I may need to explain what that means.
There
are n different nth roots of 1 - but unlike the two square
roots of -1,
these are not all crea equal! Only some are primitive.
For example, among the 4th roots of unity we have 1 and -1,
which are
actually square roots of unity, and i and -i, which aren't. A
primitive
nth root of unity is an nth root of 1 that's not an kth root
for any
k < n. If you take all the powers of any primitive nth root of
unity,
you get *all* the nth roots of unity. So, if we take some
primitive nth
root of unity, call it
1^{1/n}
for lack of a better name, and extend the rationals by this
number,
we get a field
Q(1^{1/n})
which contains all the nth roots of unity. Since the nth roots
of unity
are evenly distribu around the unit circle, this sort of field
is called
a cyclotomic field, for the Greek word for circle cutting. In
fact,
one can apply Galois theory to this field to figure out which
regular
n-gons one can construct with a ruler and compass!
But what's the Galois group
Gal(Q(1^{1/n})/Q)
like? Any symmetry in this group must map 1^{1/n} to some root
of unity,
say 1^{m/n} - and once you know which one, you completely know
the
symmetry. But actually, this symmetry must map 1^{1/n} to some
*primitive*
root of unity, so m has to be relatively prime to n. Apart
from that,
though, anything goes - so the size of
Gal(Q(1^{1/n})/Q)
is just the number of m less than n that are relatively prime
to n. And
if you think about it, these numbers relatively prime to n are
just the
same as elements of Z/n that have multiplicative inverses! So
if you think
some more, you'll see that
Gal(Q(1^{1/n})/Q) = (Z/n)*
where (Z/n)* is the multiplicative group of Z/n - that is, the
elements of Z/n that have multiplicative inverses, made into a
group
via multiplication!
This group can be big, but it's still abelian. Can we get some
nonabelian
Galois groups from algebraic number fields?
Sure! Let's say you take some polynomial equation with rational
coefficients,
take *all* its solutions, throw them into the rationals - and
keep adding,
subtracting, multiplying and dividing until you get some field
K. This K
is called the splitting field of your polynomial.
But here's the interesting thing: if you pick your polynomial
equation at
random, the chances are really good that it has n different
solutions if
the polynomial is of degree n, and that *any* permutation of
these
solutions comes from a unique symmetry of the field K. In
other words:
barring some coincidence, all roots are crea equal! So in
general we
have
Gal(K/Q) = S_n
where S_n is the group of all permutations of n things.
Sometimes of course the Galois group will be smaller, since
our polynomial
could have repea roots or, more subtly, algebraic relations
between
roots - as in the cyclotomic case we just looked at.
But, we can already start to see how to prove the
unsolvability of the
general quintic! Pick some random 5th-degree polynomial, let K
be its
splitting field, and note
Gal(K/Q) = S_5
Then, show that if we build up an algebraic number field by
starting
with Q and repealy throwing in nth roots of numbers we've
already got,
we just can't get S_5 as its Galois group over the rationals!
We've
already seen this in the case where we throw in a square root
of n, or
an nth root of 1. The general case is a bit more work. But
instead of
giving the details, I'll just mention a good textbook on
Galois theory for
beginners:
1) Ian Stewart, Galois Theory, 3rd edition, Chapman and Hall,
New York,
Ian Stewart is famous as a popularizer of mathematics, and it
shows
here - he has nice discussions of the history of the famous
problems
solved by Galois theory, and a nice demystification of the
Galois'
famous duel. But, this is a real math textbook - so you can
really
learn Galois theory from it! Make sure to get the 3rd edition,
since
it has more examples than the earlier ones.
Having given Ian Stewart the dirty work of explaining Galois
theory in
the usual way, let me say some things that few people admit in
a first
course on the subject.
So far, we've looked at examples of a field k contained in
some bigger
field K, and worked out the group Gal(K/k) consisting of all
automorphisms
of K that fix everything in k.
But here's the big secret: this has NOTHING TO DO WITH FIELDS!
It works
for ANY sort of mathematical gadget! If you've got a little
gadget k
sitting in a big gadget K, you get a Galois group Gal(K/k)
consisting of
symmetries of the big gadget that fix everything in the little
one.
But now here's the cool part, which is also very general. Any
subgroup
of Gal(K/k) gives a gadget containing k and contained in K:
namely, the
gadget consisting of all the elements of K that are fixed by
everything
in this subgroup.
And conversely, any gadget containing k and contained in K
gives a
subgroup of Gal(K/k): namely, the group consisting of all the
symmetries
of K that fix every element of this gadget.
This was Galois' biggest idea: we call this a GALOIS
CORRESPONDENCE.
It lets us use *group theory* to classify gadgets contained in
one
and containing another. He applied it to fields, but it turns
out
to be useful much more generally.
Now, it would be great if the Galois corresondence were always
a perfect
1-1 correspondence between subgroups of Gal(K/k) and gadgets
containing
k and contained in K. But, it ain't true. It ain't even true
when we're
talking about fields!
However, that needn't stop us. For example, we can restrict
ourselves
to cases when it *is* true. And this is where the Fundamental
Theorem of
Galois Theory comes in! It's easiest to state this theorem
when k and K
are algebraic number fields, so that's what I'll do. In this
case, there's
a 1-1 correspondence between subgroups of Gal(K/k) and
extensions of k
contained in K if:
i) K is a finite extension of k. In other words, K is a
finite-dimensional
vector space over k.
ii) K is a normal extension of k. In other words, if a
polynomial with
coefficients in k has no roots in k, but one root in K, then
all its roots
are in K.
For general fields we also need another condition, namely that
K be a
separable extension of k. But this is automatic for algebraic
number
fields, so let's not worry about it.
At this point, if we had time, we could work out a bunch of
Galois groups
and see a bunch of patterns. Using these, we could see why you
can't
solve the general quintic using radicals, why you can't
trisect the angle
or double the cube using ruler-and-compass constructions, and
why you can
draw a regular pentagon using ruler and compass, but not a
regular
heptagon.
Basically, to prove something is impossible, you just show
that some number
can't possibly lie in some particular algebraic number field,
because it's
the root of a polynomial whose splitting field has a Galois
group that's
fancier than the Galois group of that algebraic number field.
For example, ruler-and-compass constructions produce distances
that lie in
itera quadratic extensions of the rationals - meaning that you
just
keep throwing in square roots of stuff you've got. Doubling
the cube
requires getting your hands on the cube root of 2. But the
Galois group
of the splitting field of
x^3 = 2
has size divisible by 3, while an itera quadratic extension
has a Galois
group whose size is a power of 2. Using the Galois
correspondence, we see
there's no way to stuff the former field into the latter.
But you can read about this in any good book on Galois theory,
so I'd
rather
dive right into that thicket I was hinting at earlier: the
field of ALL
algebraic numbers! The roots of any polynomial with
coefficients in this
field again lie in this field, so we say this field is
algebraically
closed. And since it's the smallest algebraically closed field
containing
Q, it's called the algebraic closure of Q, or Qbar for short -
that is,
Q
with a bar over it.
This field Qbar is huge. In particular, it's an
infinite-dimensional vector
space over Q. So, condition i) in the Fundamental Theorem of
Galois Theory
doesn't hold. But that's no disaster: when this happens, we
just need to
put a topology on the group Gal(K/k) and set up the Galois
correspondence
using
*closed* subgroups of Gal(K/k). Using this trick, every
algebraic number
field
corresponds to some closed subgroup of Gal(Qbar/Q).
So, for people studying algebraic number fields,
Gal(Qbar/Q)
is like the holy grail. It's the symmetry group of the
algebraic numbe
and the key to how all algebraic number fields sit inside each
other!
But alas, this group is devilishly complica. In fact, it has
literally
driven men mad. One of my grad students knows someone who had
a breakdown
and went to the mental hospital while trying to understand
this group!
(There may have been other reasons for his breakdown, too, but
as readers
of E. T. Bell's book Men in Mathematics know, the facts should
never
get
in the way of a good anecdote.)
If Gal(Qbar/Q) were just an infinitely tangled thicket, it
wouldn't be so
tantalizing. But there are things we can understand about it!
To describe
these, I'll have to turn up the math level a notch...
First of all, an extension K of a field k is called abelian if
Gal(K/k)
is an abelian group. Abelian extensions of algebraic number
fields can be
understood using something called class field theory. In
particular, the
Kronecker-Weber theorem says that every finite abelian
extension of Q is
contained in a cyclotomic field. So, they all sit inside a
field called
Qcyc, which is gotten by taking the rationals and throwing in
*all* nth
roots of unity for *all* n. Since
Gal(Q(1^{1/n})/Q) = (Z/n)*
we know from Galois theory that Gal(Qcyc/Q) must be a big
group containing
all the groups (Z/n)* as closed subgroups. It's easy to see
that (Z/n)* is
a quotient group of (Z/m)* if m is divisible by n; this lets
us take the
inverse limit of all the groups (Z/m)* - and that's
Gal(Qcyc/Q). This
inverse limit is also the multiplicative group of the ring Z^,
the inverse
limit of all the rings Z/n. Z^ is also called the profinite
completion of
the integeand I urge you to play around with it if you never
have!
It's a cute gadget.
In short:
Gal(Qcyc/Q) = Z^*
and if we stay inside Qcyc, we're in a zone where the pattern
of algebraic
number fields can be understood. This stuff was worked out by
people like
Weber, Kronecker, Hilbert and Takagi, with the final keystone,
the Artin
reciprocity theorem, laid in place by Emil Artin in 1927. In a
certain
sense Qcyc is to Qbar as homology theory is to homotopy
theory: it's all
about *abelian* Galois groups, so it's manageable.
People now use Qcyc as a kind of base camp for further
expeditions into
the depths of Qbar. In particular, since
Q is contained in Qcyc and Qcyc is contained in Qbar
we get an exact sequence of Galois groups:
1 -> Gal(Qbar/Qcyc) -> Gal(Qbar/Q) -> Gal(Qcyc/Q) -> 1
So, to understand Gal(Qbar/Q) we need to understand
Gal(Qcyc/Q),
Gal(Qbar/Qcyc) and how they fit together! The last two steps
are not
so easy. Shafarevich has conjectured that Gal(Qbar/Qcyc) is the
profinite completion of a free group, say F^. This would give
1 -> F^ -> Gal(Qbar/Q) -> Z^* -> 1
but I have no idea how much evidence there is for
Shafarevich's conjecture,
or how much people know or guess about this exact sequence.
More recently, Deligne has turned attention to a certain
motivic
version
of Gal(Qbar/Q), which is a proalgebraic group scheme. This
sort of group
has a *Lie algebra*, which makes it more tractable. And there
are a bunch
of fascinating conjectures about this Lie algebra is rela to
the Riemann
zeta function at odd numbeConnes and Kreimer's work on Feynman
diagrams,
Drinfeld's work on the Grothendieck-Teichmueller group, and
more!
I really want to understand this stuff better - right now,
it's a complete
muddle in my mind. When I do, I will report back to you. For
now, though,
let me give you some references.
For two very nice but very different introductions to
algebraic number
fields, try these:
2) H. P. F. Swinnerton-Dyer, A Brief Guide to Algebraic Number
Theory,
Cambridge U. Press, Cambridge 2001.
3) Juergen Neukirch, Algebraic Number Theory, trans. Norbert
Schappacher,
Springer, Berlin, 1986.
Both assume you know some Galois theory or at least can fake
it.
Neukirch's book is good for the all-important analogy between
Galois
groups and fundamental groups, which I haven't even touched
upon here!
Swinnerton-Dyer's book has the virtue of brevity, so you can
see the
forest for the trees. Both have a friendly, slightly chatty
style that
I like.
For Shafarevich's conjecture, try this:
4) K. Iwasawa, On solvable extensions of algebraic number
fields,
Ann. Math. 58 (1953) 548-572.
For Deligne's motivic analogue, try this:
5) Pierre Deligne, Le groupe fondamental de la droite
projective
moins trois points, in Galois Groups over Q, MSRI Publications
16 (1989),
79-313.
This stuff has a lot of relationships to 3d topological
quantum field
theory, braided monoidal categories, and the like... and it
all goes
back to the Grothendieck-Teichmueller group. To learn about
this group
6) Leila Schneps, The Grothendieck-Teichmueller group: a
survey,
in The Grothendieck Theory of Dessins D'Enfants, London Math.
Society
Notes 200, Cambridge U. Press, Cambridge 1994, pp. 183-204.
To hear and watch some online lectures on this material, try:
7) Leila Schneps, The Grothendieck-Teichmuller group and
fundamental
groups of moduli spaces, MSRI lecture available at
http://www.msri.org/publications/ln/msri/1999/vonneumann/
schneps/1/
Grothendieck-Teichmueller group and Hopf algebras,
MSRI lecture available at
http://www.msri.org/publications/ln/msri/1999/vonneumann/
schneps/2/
For a quick romp through many mindblowing ideas which touches
on this
material near the end:
8) Pierre Cartier, A mad day's work: from Grothendieck to
Connes
and Kontsevich - the evolution of concepts of space and
symmetry,
http://www.ams.org/joursearch/index.html
For even more mindblowing ideas along these lines:
9) Jack Morava, The motivic Thom isomorphism, talk at the
Newton Institute,
December 2002, also available at math.AT/0306151
Quote of the week:
Paris, 1 June - A deplorable duel yesterday has deprived the
exact
sciences of a young man who gave the highest expectations, but
whose
celebra precosity was lately overshadowed by his political
activities.
The young Evariste Galois... was fighting with one of his old
friends,
a young man like himself, like himself a member of the Society
of
Friends of the People, and who was known to have figured
equally in
a political trial. It is said that love was the cause of the
combat.
The pistol was the chosen weapon of the adversaries, but
because of
their old friendship they could not bear to look at one
another and
left their decision to blind fate. - Le Precursor, June 4 1832
--------------------------------------------------------------
---------
mathematics and physics, as well as some of my research
papecan be
obtained at
http://math.ucr.edu/home/baez/
For a table of contents of all the issues of This Week's
Finds, try
http://math.ucr.edu/home/baez/twf.html
A simple jumping-off point to the old issues is available at
http://math.ucr.edu/home/baez/twfshort.html
If you just want the latest issue, go to
http://math.ucr.edu/home/baez/this.week.html
===
Subject: Re: This Week's Finds in Mathematical Physics (Week
201)
Originator: baez@math-cl-n01.math.ucr.edu (John Baez)
Originator: israel@math.ubc.ca (Robert Israel)
>And this is where the Fundamental Theorem of
>Galois Theory comes in! It's easiest to state this theorem
when k and K
>are algebraic number fields, so that's what I'll do. In this
case,
there's
>a 1-1 correspondence between subgroups of Gal(K/k) and
extensions of k
>contained in K if:
>i) K is a finite extension of k. In other words, K is a
finite-dimensional
>vector space over k.
>ii) K is a normal extension of k. In other words, if a
polynomial with
>coefficients in k has no roots in k, but one root in K, then
all its roots
>are in K.
In part ii) I was trying to say if a polynomial with
coefficients in
k is irreducible in k, but has a root in K, then all its roots
are in K
without introducing the term irreducible. Joseph Silverman
kindly
poin out that the above doesn't quite do it!
While I'm at it, I'd like to thank Kevin Buzzard, Noam Elkies
and
Joseph Silverman for emailing me nice replies to my post The
secret
inner meaning of modular forms. There are indeed some good
general
reasons for a sequence of natural numbers a_n to be the
coefficients
of the q-expansion of a modular form! Someday I'll talk about
them
in This Week's Finds.
===
Subject: Re: A problem of quadrilaterals
Originator: israel@math.ubc.ca (Robert Israel)
> ABCD is a convex quadrilateral (a normal one, where the
sides don't
> intersect and none of the interior angles exceeds 180
degrees). You
> are given that angles A and B sum up to greater than 180
degrees.
> Prove that:
> AC + BD > AD + BC
> i.e. sum of diagonals exceeds sum of sides AD and BC.
> I think I have a proof, but it is not a simple one. I
believe that
> something simple and elegant should be able to do it.
This is the theorem of Ptolemy (equality for points on a
circle) and his
inequality in the general case. Proofs you find here
http://planetmath.org/encyclopedia/
ProofOfPtolemysInequality.html
http://matholymp.com/TUTORIALS/Ptolemy.pdf
--
Roland Franzius
===
Subject: Moreau's necklace-counting function
Originator: israel@math.ubc.ca (Robert Israel)
In _Witt_Vectors_and_the_Algebra_of_Necklaces_, by
N. Metropolis and G.-C. Rota, reprin in the book
_Gian-Carlo_Rota_on_Combinatorics_, we find Moreau's
necklace-counting function
M(n,c) = (1/n) * SUM { mu(n/d) c^d : d divides n },
where mu is the classic Moebius function. It is sta that
this is the number of necklaces of n beads when each bead can
be of any of c colors. I think the paper gives a bijective
proof, which it says is more natural than earlier proofs, but
I haven't read that far yet.
Two questions:
(1) Metropolis & Rota attribute this result to Colonel Moreau
of the French army, but the paper's bibliography lists no one
by that name, and the list of biographical sketches of
mathematicians
at <http://www-gap.dcs.st-and.ac.uk/~history/Indexes/M.html>
lists no on by that name. Who was Moreau and on what grounds
can this be attribu this to him?
(2) I hope I'm being temporarily horribly stupid on this one.
Suppose four beads, chosen from among three colors. Then we
have
1. All four red.
2. All four blue.
3. All four green.
4. 3 R, 1 G.
5. 3 R, 1 B.
6. 3 B, 1 G.
7. 3 B, 1 R.
8. 3 G, 1 R.
9. 3 G, 1 B.
10. 2 R, 2 G, identical colors adjacent.
11. 2 R, 2 G, identical colors opposite.
12 & 13. Like 10 & 11 above; 2 B.
14 & 15. Like 10 & 11 above; 2 G.
16. 2 R adjacent, 1 B, 1 G.
17. 2 R opposite, 1 B, 1 G.
18 & 19. Like 16 & 17 above, 2 B.
20 & 21. Like 16 & 17 above, 2 G.
In #16-21, I assumed it doesn't matter in which order the
singleton colors appear, since one can pick the necklace up
off the table and turn it over, reversing clockwise and
counterclockwise, and it's still the same necklace.
But Moreau's necklace-counting function gives
(1/4) * ( mu(4/1) 3^1 + mu(4/2) 3^2 + mu(4/4) 3^4)
= (1/4) * ( [0 * 3] + [-1 * 9] + [1 * 81] )
= 18 =/= 21.
Next I found via a Google search a working draft of a book
by Frank Ruskey of the University of Victoria in British
Columbia. I can't tell what the URL is since clicking on
it opens up a postscript viewer rather than a web browser.
One page 213 he says ... a necklace is an equivalence class
of strings under rotation ... [although] we expect to be able
to pick up a necklace and turn it over ... we stick with
mathematical tradition and regard 001101 as a different
necklace than 001011 ... That makes my count of 21 too small.
But two pages later he gives that same formula, which seems to
lead to 18.
Please tell me I am stupidly missing something.
Mike Hardy
===
Subject: Re: Moreau's necklace-counting function
Originator: israel@math.ubc.ca (Robert Israel)
> In _Witt_Vectors_and_the_Algebra_of_Necklaces_, by
> N. Metropolis and G.-C. Rota, reprin in the book
> _Gian-Carlo_Rota_on_Combinatorics_, we find Moreau's
> necklace-counting function
> M(n,c) = (1/n) * SUM { mu(n/d) c^d : d divides n },
> where mu is the classic Moebius function. It is sta that
> this is the number of necklaces of n beads when each bead can
> be of any of c colors.
[ ..... ]
> (2) I hope I'm being temporarily horribly stupid on this one.
> Suppose four beads, chosen from among three colors. Then we
have
[ ..... ]
> Please tell me I am stupidly missing something.
OK, I was right, in that I was wrong. What it says is
necklaces that are asymetric under rotations. That rules out
six of my 21 necklaces, and then we add three more because
(unnaturally, given the name necklace) we are forbidden to
pick the necklace up off the table and turn it over, and call
it the same necklace (unless it has that kind of symmetry),
so we have to add three more that are reflections of three
that I lis (the others are the same as their reflections).
Once all this is done, Moreau's formula gives the right
answer.
This still leaves my other question: who was Moreau and
what did he write about this? -- Mike Hardy
===
Subject: Re: Moreau's necklace-counting function
Originator: israel@math.ubc.ca (Robert Israel)
>> In _Witt_Vectors_and_the_Algebra_of_Necklaces_, by
>> N. Metropolis and G.-C. Rota, reprin in the book
>> _Gian-Carlo_Rota_on_Combinatorics_, we find Moreau's
>> necklace-counting function
>> M(n,c) = (1/n) * SUM { mu(n/d) c^d : d divides n },
> This still leaves my other question: who was Moreau and
>what did he write about this? -- Mike Hardy
Googling for [Moreau necklace counting] I get a postscript
file:
Cycle counting for isomorphism types of endofunctions
by Volker Strehl
Informatik I, Universitat Erlangen-Nurnberg, FRG
which includes the following reference:
C. Moreau, Sur les permutations circulaire distinct, Nouv.
Ann. Math.,
11:309--311, 1872.
It also references the Metropolis-Rota paper.
There is also a Colonel Moreau mentioned prominantly on chess
sites.
Apparently he hos a tournament in Monte Carlo in 1903.
Dan
--
Dan Luecking Department of Mathematical Sciences
University of Arkansas Fayetteville, Arkansas 72701
luecking at uark dot edu
===
Subject: Re: Moreau's necklace-counting function
Originator: israel@math.ubc.ca (Robert Israel)
> This still leaves my other question: who was Moreau and
> what did he write about this? -- Mike Hardy
http://www1.informatik.uni-erlangen.de/tree/Research/reports/
rep91_5.ps
with the reference (for the necklace formula you gave):
C. Moreau. Sur les permutations circulaires distincts. Nouv.
Ann. Math.,
11:309-314, 1872.
Mitch Harris
===
Subject: Re: Partially Ordered Hausdorff
Originator: israel@math.ubc.ca (Robert Israel)
I think you need more. Standard reference for this
stuff is Topology and Order, L. Nachbin, van Nostrand
Mathematical Studies, 1965.
> Let S be a partially ordered set with a Hausdorff topology
> which has a base of convex sets.
> Let <= be the order of S and let L = { (x,y) | x <= y }.
> Can it been shown L is closed in SxS ?
> Or is it necessary to know more about the topology or
> base sets to conclude L is closed?
> The converse is easy and doesn't require convex base sets.
> Assume L is closed and x /= y. Take (wlog) not x <= y.
> Then (x,y) in open (SxS)L.
> Some open U,V with (x,y) in UxV subset (SxS)L
> U,V disjoint. Otherwise: some z in U,V
> (z,z) in UxV subset (SxS)L; not z <= z
> ----
===
Subject: Re: computability with random Oracles
Originator: israel@math.ubc.ca (Robert Israel)
> Here is another way of stating the question, which is
perhaps clearer:
> define a preorder -> (Turing reducibility) on the set 2^N of
subsets
> of N (the naturals) by letting O->P when there exists a
Turing machine
> that computes (the characteristic function of) P given (the
> characteristic function of) O as an Oracle. So recursive
subsets of N
> are minimal elements for this preorder: P is recursive iff
{O | O->P}
> is all of 2^N. Now I assume that P is such that {O | O->P}
has
> measure 1 and I ask whether P is, actually, recursive.
Yes, P must be computable (recursive) in this case. If P is
not computable
then the set of all sets in which P is computable has measure
0. This is
proved in Gerald Sacks' book Degrees of Unsolvability (Annals
of
Mathematical Studies 55, Princeton University Press, 1963, pp.
154-156).
The proof goes like this: Suppose that the set of all sets in
which P is
computable has nonzero measure. By countable additivity, there
is some
Turing machine T for which the set S of all sets O such that T
computes P
with oracle O has measure m>0. Let r be a rational s.t.
r/2<m<r. There is
a subset A of 2^N such that A is a finite union of basic open
sets of 2^N,
the measure of A is at most r, and the measure of the
intersection of A
and S is greater than r/2. For a natural number n and i = 0 or
1, let
B^n_i be the set of all finite strings F such that T(n) halts
with oracle
F and gives the answer i, and let C^n_i be the subset of 2^N
consisting of
all extensions of elements of B^n_i. The B^n_i are uniformly
computably
enumerable, and (identifying P with its characteristic
function) if P(n)=i
then the intersection of C^n_i with A has measure greater than
r/2, while
the intersection of C^n_{1-i} with A has measure less than
r/2. So to
compute P(n), it suffices to wait until we see the measure of
the
intersection of C^n_i with A exceed r/2 for some i, which we
can do
effectively, since A is a finite union of basic open sets.
When this
happens, we know that P(n)=i.
Denis
===
Subject: mesh network question
Originator: israel@math.ubc.ca (Robert Israel)
I am going crazy trying to find a formula/equation for the
equivalent
resistance asross diagonally opposite ends of a resistive
mesh. Can
someone help.
X
o--+--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+
| | | | | | |
Rv Rv Rv | | Rv Rv
| | | | | | |
+--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+
| | | | | | |
Rv Rv Rv | | Rv Rv
| | | | | | |
. . . .(m rows) .--Rh--+--Rh--+
| | | | | | |
Rv Rv Rv | | Rv Rv
| | | | | | |
+--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+
| | | | | | |
Rv Rv Rv | | Rv Rv
| | | | | | |
+--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+--o Y
Imagine a rectangular mesh with resistors at every small
segment. The
horizontal segment resistances are Rh and vertical ones are
Rv. Need
to get a formula to calculate the effective resistance between
X and
Y. Any help welcome!
===
Subject: Re: mesh network question
Originator: israel@math.ubc.ca (Robert Israel)
> I am going crazy trying to find a formula/equation for the
> equivalent resistance across diagonally opposite ends of a
> resistive mesh. Can someone help.
> X
> o--+--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+
> | | | | | | |
> Rv Rv Rv | | Rv Rv
> | | | | | | |
> +--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+
> | | | | | | |
> Rv Rv Rv | | Rv Rv
> | | | | | | |
> . . . .(m rows) .--Rh--+--Rh--+
> | | | | | | |
> Rv Rv Rv | | Rv Rv
> | | | | | | |
> +--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+
> | | | | | | |
> Rv Rv Rv | | Rv Rv
> | | | | | | |
> +--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+--o Y
> Imagine a rectangular mesh with resistors at every small
segment.
> The horizontal segment resistances are Rh and vertical ones
are Rv.
> Need to get a formula to calculate the effective resistance
between
> X and Y. Any help welcome!
For one ohm resistothe resistance can be found hypothetically
by
dividing the number of spanning trees into the number of
spanning
trees there would be with the ends shor together. There is a
simple formula for the number of spanning trees, but not for
the
shor variety as far as I know.
Trees = 1/(mn) prod_{i=0 to n-1 and j=0 to m-1 but not both
zero}
(4 - 2cos(Pi i/n) - 2cos(Pi j/m))
The wye-delta formula can be used repealy. It is easier to use
in the conductance form since parallel conductances can be
added.
Each vertex can be elimina by replacing each pair of resistors
to it by a resistor that skips the vertex. A vertex surrounded
by 4 resistors becomes 6 resistors that bypass the vertex since
there are 6 ways to choose 2 out of the 4 resistors.
G_bypass = G_1 G_2 / (G_1 + G_2 + ... + G_n) for n resistor
vertex.
===
Subject: Re: Complex Darboux Theorem
Originator: israel@math.ubc.ca (Robert Israel)
> I can't figure out what a complex Darboux Theorem would even
_state_,
> much less whether it's true. Maybe we're talking about
different
> results: The Darboux Theorem I have in mind is the fact that
if
> f : (a,b) -> R is differentiable then the derivative f'
satisfies the
> intermediate value property. (What _is_ the intermediate
value
> property in the complex case?)
You are talking about different results. The relevant Darboux
theorem
is that for a symplectic manifold with symplectic form omega
(omega is a
closed 2-form that gives a nondegenerate pairing on each
tangent space),
there exist local coordinates (x1,...,xn,y1,...,yn) such that
omega=sum_i (dx_i ^ dy_i). This terminology is standard in
differential
geometry.
David
===
Subject: An Inequality with Dirichlet Series
Epigone-thread: salmerljong
Originator: israel@math.ubc.ca (Robert Israel)
Hi. Is it true that for a real valued function f(x),
sum_{m=1}^infty m^{-r}(sum_{d|m} f(d))^2 <=
zeta^3(r)sum_{m=1}^infty f(m)/m^r?
More generally, is it true that, if we define F_i(m) =
sum_{d|m}
f_i(d) for functions f_i, i = 1...h,
sum_{m} m^{-r} F_1(m)F_2(m)...F_h(m)
converges iff
sum_{m} m^{-r} f_1(m)f_2(m)...f_h(m)
converges?
Brad
[I'd prefer you just respond on this board, but if you'd like
to
===
Subject: Re: Reports on an alleged solution to the P vs NP
problem
Originator: israel@math.ubc.ca (Robert Israel)
Prof. Kim yang-gon doesn't stop to contact media.......
In SBS (Korean Major Broadcasting System) Program Seven Days ,
Prof. Kim Yang-gon has said (with his own voice)
P means Class of existences of ordinary-organizing process
NP means Class of All Matters in the Universe
Then NP Includes P
In the set -NP but not P- , there are GOD and UFO
It is theoretically possible to prove existence of GOD and UFO
So I wanna to say -we can prove existence of GOD by mathematics
This isn't Joke. It was his own voice.
Now I believe he doesn't understand P/NP Problem.
===
Subject: x=?;y=?;z=?
x + y + z = 120
5x + 2y + 0.1z = 120
Can't you help me ?
===
Subject: Re: x=?;y=?;z=?
> x + y + z = 120
> 5x + 2y + 0.1z = 120
> Can't you help me ?
You need some further restriction in order to get a unique
solution.
The nature of the equations suggests that you might want to
reatrict x,y
and z to integers , or even positive integer values.
If this is the case, then clearly z must be a multiple of 10,
and a
little judicious trial and error work sould solve it.
If the variables are not constrained to be integethen any
solutions
will have to contain a parameter, which could be any of the
original
variables, expressing the other two in terms of that one.
===
Subject: Re: x=?;y=?;z=?
> x + y + z = 120
> 5x + 2y + 0.1z = 120
> Can't you help me ?
You need a 3rd equation for this. Best you can do here is find
the equation
of the intersection line.
===
Subject: Re: =?ISO-8859-1?B?eD0/O3k9Pzt6PT8=?=
>> x + y + z = 120
>> 5x + 2y + 0.1z = 120
>> Can't you help me ?
> You need a 3rd equation for this. Best you can do here is
find the
equation
> of the intersection line.
[...]
Waaal. Meebe the point is to find the solution with 2
equations --
e.g. x and y in terms of z?
Interestingly, std Gausian elimination can solve that case,
too.
===
Subject: Re: x=?;y=?;z=?
17X
13Y
90Z
> x + y + z = 120
5x + 2y + 0.1z = 120
Can't you help me ?
>
You need a 3rd equation for this. Best you can do here is find
the
equation
> of the intersection line.
>
===
Subject: Re: x=?;y=?;z=?
> 17X
> 13Y
> 90Z
> x + y + z = 120
5x + 2y + 0.1z = 120
Can't you help me ?
>
You need a 3rd equation for this. Best you can do here is find
the
> equation
> of the intersection line.
>
>
If I were solving this, here's how I'd think about it.
2 Equations, 3 variables means that you need one more equation
or one of
the
variables given.
Let z=0. This gives
x+y=120
5x+2y=120
-2x-2y=-240
5x+2y=120
3x=-120
x=-40
y=160
So one point is (-40,160,0)
Now let z=50
x+y=70
5x+2y=115
-2x-2y=-140
5x+2y=115
3x=-25
x=-25/3
y=235/3
Another point is (-25/3,235/3,50)
Now we can use 3D geometry
Find the Normal Vector
n=(-95/3,245/3,-50)
And then resolve into parametric equations
x=-25/3-95/3t
y=235/3+245/3t
z=50-50t
It doesn't get much nicer than that, unfortunately.
===
Subject: Re: x=?;y=?;z=?
Outlook Express
can't do this!
> 17X
> 13Y
> 90Z
No, you have a solution.
As David Morgan said, you the solution is the interesction of
two planes &
is a line.
If you had wan a Diophantine solution you should have said so.
gtoomey
===
Subject: Re: Combination lock
> Hi everyone.
> I've just bought a combination lock and was wondering how
many
combinations
> it has - and more importantly how secure stuff secured using
this lock
> really is. I had a 3 number combnation lock and someone
flicked through
> them all and guessed it (only 999 combinations) which I
experimen with
> and it takes a meer 8 minutes.
> I guess that this is a mathematics problem so would like to
appeal to
your
> good nature in finding the answer.
> The new lock is as follows:
> It has 10 push buttons on the front, numbered 0 up to and
including 9.
Each
> button has two positions, in or out.
> All numbers can be pushed in or out, in any order or
combination.
> It has a preset to unlock it - 5 numbers must be pushed in,
the remaining
> five left out - these have to be pushed in, but not in any
particular
order.
> So essentially to unlock it, 5 out of the 10 numbers must be
pushed in and
5
> must not.
> Given that the order the numbers are pushed in does not
matter, and that
we
> know there are 5 numbers that must be pushed in to unlock
it. What's the
> maximum possible number of combinations of buttons one would
have to try
in
> order to open the lock.
> Also If the number of buttons that must be pushed in can be
any number
other
> than 5 (this may be the case with these locks) how many
combinations are
> there then?
Number of buttons Number of combinations
to be pushed
0 1
1 10/1 = 10
2 (10*9)/(2*1) = 45
3 (10*9*8)/(3*2*1) = 120
4 (10*9*8*7)/(4*3*2*1) = 210
5 (10*9*8*7*6)/(5*4*3*2*1) = 252
6 210 = N(4)
7 120 = N(3)
8 45 = N(2)
9 10 = N(1)
10 1 = N(0)
----
1024
As a whole there are 1024=2^10 possible combinations of
IN/OUT. Thus
your new lock is not better than your old one. In fact it is
probably
worse because it takes less time to push buttons than to choose
numbers on a wheel.
--
Horst
===
Subject: Re: Rieman Sums to Integrals
> Hey to all, quick question about Rieman Sums
Wouldn't the following Summation:
lim(n->infinity) (1/n)Sum(1/(k/n)) where Sum is k=1 to n
equal the intergarl
Intergral(1/x)dx from 0 to 1?
Why not?
> I mean, decomposing the Summation we would get
> length of each interval = 1/n
> start of integral = k/n where k = 1, n=infinity = 0
> end of integral = k/n where k=n, n=infinity =1
> and there are n intervals in total
Only in the sense that both are undefined.
> The limit does not exist (as a real number) and
> the integral does not exist (as a real number).
However, that is accidental. A divergent integral can have
the right-hand endpoint Riemann sums converging to a real
number.
Not if the function is positive and decreasing.
===
Subject: Re: Round off error
> I was wondering if there is some sort of test that can be
run on a
> numerical algorithm to quantify round off error. Needless to
say, the
> algorithm is stable.
> For me, this comes up as follows: I run finer and finer
meshes on a
> problem and observe some particular value of the solution.
If the exact
> value is y_ex, the compu values approach and then diverge
from y_ex.
> I am wondering if I can attribute this divergence to round
off errors
> entirely. Is there some other definitive test that will
establish this ?
Yes, there is a practical test (will require some work) which
goes
something like this:
Let S(t;h) be the system model, then
-- generate system response r(t) on [0,T] at h
-- generate system response p(t) on [0,T] at h/2
-- calculate some measure (d) between r(t) & p(t)
Continue with successive step reductions until a decreasing
sequence {d} reverses the trend; that is, when h crosses the
optimal
step implied by the saddle of an inherent error model,
e(h) = ah + 1/bh
In practice, a decreasing sequence {d} between successive
responses
will indicate that the solution converges uniformly on [0,T]
toward an
optimum with a min error e(h).
--
Dr.B.Voh
------------------------------------------------------
Applied Algorithms http://sdynamix.com
===
Subject: Re: what are the relatioship between linear/nonlinear
optimization
and convex optimization?
Cc: mizhael@yahoo.com
> Convex optimization seems hot today... anybody can tell me
what is the
> relationship and difference between linear/nonlinear,
discrete, and
convex
> optimization?
Linear problems: linear objective function and linear
constraints
non-linear problems: nonlinear objective function and/or
nonlinear
constraints
Discrete problems: the variables are constrained to be integer
numbers
(or some other finite/countable set)
Convex problems: convex objective function and the constraints
form a
convex feasible set. Linear problems are a class of convex
problems.
Linear problems are generally the easiest to solve.
I know my explanation wasn't very detailed but I hope it will
help you a
bit.
Fernando G. del Cueto.
===
Subject: a question rela to doing Jacobian for the
transformation
I'm an electrical engineering student doing some derivations
that involve Jacobian for transformation, and I have
encountered the
following problem, which leads me to suspect that the
expression I
obtained after performing the Jacobian transformation is
incorrect. I very
much appreciate whoever can give me some help and assistance
on this.
Below is the detailed derivation so that you know how I
arrived at my
current
result, and my question comes after it. I truly appreciate your
correspondence and please send your reply to
eyh5@ece.cornell.edu .
I have a function f(m,n) that describes the joint probability
density function (jpdf) of two random variables m and n, and
is written as
the following:
f(m,n)=-1/(2*b)^4*m*n-1/(2*b)^3*m+1/(2*b)^3*n+1/(2*b), [1]
where b is a non-zero positive constant (it's just used as a
parameter
here), and the ranges of m and n are:
0 <= m <= 2*b, -2*b <= n <= 0 [2]
Now, I define two new variables x and y such that
x=sqrt(m^2+n^2), y=arctan(n/m) [3]
You might recognize this is very much like the change from
Cartesian to
polar coordinate system (in fact, the way m and n are defined
indicates
we are operating in the fourth quadrant of the Cartesian
system). So,
to solve for m and n in terms of x and y, we can write the
following:
m=x*cos(y), n=x*sin(y) [4]
What I want to do is to find the joint pdf of x and y, deno as
g(x,y).
Since this is a case of one function with two variables, I go
ahead and use the Jacobian for transformation by applying its
definition:
g(x,y)=f(m,n) * abs(J(x,y))
=f(x*cos(y), x*sin(y)) * abs(J(x,y)) [5]
where abs(k) denotes the absolute value of k, and J(x,y) is
the Jacobian.
f(x*cos(y), x*sin(y)) is pretty straightforward and is given
by:
f(x*cos(y), x*sin(y)) = [6]
-1/(2b)^4*x^2*sin(y)cos(y)-1/(2b)^3*x*(cos((y)-sin(y))+1/(2b)^2
J(x,y) can also be pretty easily found by using its
definition, and is
given as:
J(x,y) = x [7]
Note that by way of its definition in [3] above, x is always
positive. The
joint pdf of x and y is therefore:
g(x,y)=f(x*cos(y), x*sin(y)) *abs(x) [8]
=-1/(2b)^4*x^3*sin(y)*cos(y)-1/(2b)^3*x^2*(cos((y)-sin(y))+1/(
2b)^2*x
the ranges for x and y are:
0 <= x <= 2*sqrt(2)*b, 3*pi/2 <= y <= 2*pi [9]
Now I want to verify that I've arrived at the correct g(x,y)
(with the
correct bounds for x and y). To do that, I first take the
double integral
of Eq. [1], with the bounds of m and n as defined in [2]. The
result is
0.25. Now, I take the double integral of Eq. [8], with the
bounds of x and
y defined in [9], and the result obtained is 0.1852. There is
this 0.07
difference between the two.
I'm really puzzled by this discrepancy since, if I understand
correctly, the Jacobian for transformation is supposed to
preserve the
probability before and after. So I should expect 0.25 as the
result of
the double integral of [8], right? But it's not.
I went over my steps a couple times and am now pretty certain
as far
as carrying out the double-integral computations, I didn't
make any
mistakes. In fact, both my computation by hand and by Matlab
give me the
same result, 0.1852. And I also double-checked Eq. [1] to make
sure for
all values of m and n, f(m,n)>=0, which is a valid condition
as a pdf
(actually, this is a partial pdf for it occupies but one of
the four
quadrants; hence the value 0.25). So where could this error
arise from?
I would really appreciate if someone can take a look at my
derivations and see if there's anything wrong or that I've
overlooked when
doing the Jacobian for transformation. Please send your reply
to
-Ed
===
Subject: Re: a question rela to doing Jacobian for the
transformation
>
> I'm an electrical engineering student doing some derivations
>that involve Jacobian for transformation, and I have
encountered the
>following problem, which leads me to suspect that the
expression I
>obtained after performing the Jacobian transformation is
incorrect. I very
>much appreciate whoever can give me some help and assistance
on this.
>Below is the detailed derivation so that you know how I
arrived at my
current
>result, and my question comes after it. I truly appreciate
your
>correspondence and please send your reply to
eyh5@ece.cornell.edu .
>
> I have a function f(m,n) that describes the joint probability
>density function (jpdf) of two random variables m and n, and
is written as
>the following:
>
> f(m,n)=-1/(2*b)^4*m*n-1/(2*b)^3*m+1/(2*b)^3*n+1/(2*b), [1]
>
>where b is a non-zero positive constant (it's just used as a
parameter
>here), and the ranges of m and n are:
>
> 0 <= m <= 2*b, -2*b <= n <= 0 [2]
>
>Now, I define two new variables x and y such that
>
> x=sqrt(m^2+n^2), y=arctan(n/m) [3]
>
>You might recognize this is very much like the change from
Cartesian to
>polar coordinate system (in fact, the way m and n are defined
indicates
>we are operating in the fourth quadrant of the Cartesian
system). So,
>to solve for m and n in terms of x and y, we can write the
following:
>
> m=x*cos(y), n=x*sin(y) [4]
>
>What I want to do is to find the joint pdf of x and y, deno
as g(x,y).
>
> Since this is a case of one function with two variables, I go
>ahead and use the Jacobian for transformation by applying its
definition:
>
> g(x,y)=f(m,n) * abs(J(x,y))
!!!! obviously you have here the integral transformation rule
in mind
and mean abs(det(J(x,y))
> =f(x*cos(y), x*sin(y)) * abs(J(x,y)) [5]
>
>where abs(k) denotes the absolute value of k, and J(x,y) is
the Jacobian.
>f(x*cos(y), x*sin(y)) is pretty straightforward and is given
by:
>
> f(x*cos(y), x*sin(y)) = [6]
>
-1/(2b)^4*x^2*sin(y)cos(y)-1/(2b)^3*x*(cos((y)-sin(y))+1/(2b)^2
>J(x,y) can also be pretty easily found by using its
definition, and is
>given as:
>
> J(x,y) = x [7]
>
since x is positive this is abs(det(J(x,y)) o.k.
>Note that by way of its definition in [3] above, x is always
positive. The
>joint pdf of x and y is therefore:
>
>g(x,y)=f(x*cos(y), x*sin(y)) *abs(x) [8]
>
=-1/(2b)^4*x^3*sin(y)*cos(y)-1/(2b)^3*x^2*(cos((y)-sin(y))+1/(
2b)^2*x
>
>the ranges for x and y are:
>
> 0 <= x <= 2*sqrt(2)*b, 3*pi/2 <= y <= 2*pi [9]
>
>Now I want to verify that I've arrived at the correct g(x,y)
(with the
>correct bounds for x and y). To do that, I first take the
double integral
>of Eq. [1], with the bounds of m and n as defined in [2]. The
result is
>0.25. Now, I take the double integral of Eq. [8], with the
bounds of x and
>y defined in [9], and the result obtained is 0.1852. There is
this 0.07
>difference between the two.
>
> I'm really puzzled by this discrepancy since, if I understand
>correctly, the Jacobian for transformation is supposed to
preserve the
>probability before and after. So I should expect 0.25 as the
result of
>the double integral of [8], right? But it's not.
>
> I went over my steps a couple times and am now pretty
certain as far
>as carrying out the double-integral computations, I didn't
make any
>mistakes. In fact, both my computation by hand and by Matlab
give me the
>same result, 0.1852. And I also double-checked Eq. [1] to
make sure for
>all values of m and n, f(m,n)>=0, which is a valid condition
as a pdf
>(actually, this is a partial pdf for it occupies but one of
the four
>quadrants; hence the value 0.25). So where could this error
arise from?
>
> I would really appreciate if someone can take a look at my
>derivations and see if there's anything wrong or that I've
overlooked when
>doing the Jacobian for transformation. Please send your reply
to
but integration over a rectangle in (x,y) space means
integrating over
a section of a circle in (n,m) space and you wan a rectangle
in (n,m)
space.
hth
peter
===
Subject: Looking for: Gauss-Seidel preconditioner to be used
in PETSc
we use PETSc for solving our linear systems for matrices that
are stored
in the sparse-block-format BAIJ.
I am looking for a Gauss-Seidel preconditoner that works with
PETSc (in
order to compare the results with the
point-block-ilu('0')-method that
is available in PETSc as 'PCILU' -- the Gauss-Seidel
preconditioner
should respect the block structure of the matrix).
If somebody has already written that preconditioner, please
contact me!
That would save me a lot of time !!
Bernhard
===
Subject: Re: Looking for: Gauss-Seidel preconditioner to be
used in PETSc
> I am looking for a Gauss-Seidel preconditoner that works
with PETSc
Just modify the ILU factorisation code to not do fill-in; only
invert
the diagonal blocks. The reuse the ILU code for solving the
triangular
factors.
V.
--
email: lastname at cs utk edu
homepage: cs utk edu tilde lastname
In a recent thread [titled log(x+y) = f( log(x), log(y))] I no
languages to create a well-defined problem specification of the
problem under discussion. This got me interes in the use of
such
languages.
are in quite widespread use for the description of optimization
problems. I am now interes to know if there is any
corresponding
standard language in use for more general mathematical
problems - in
particular for ODEs and PDEs.
I note that several 'problem solving environments' or code
generators
exist for PDE problems (e.g. Citadel, PDELab, PIER, Falcon
etc.) but I
haven't been able to conclude if there is a common system for
the
problem definition. Maple and Mathematica also have their own
syntaxes
which do similar things.
Do any readers know of any 'standard' languages for describing
general
PDE problems (i.e. equation, domain, boundary conditions etc)?
Any
that are widely used / well established? Any
preferences/experiences?
Best wishes
andy
> are in quite widespread use for the description of
optimization
> problems. I am now interes to know if there is any
corresponding
> standard language in use for more general mathematical
problems - in
> particular for ODEs and PDEs.
> I note that several 'problem solving environments' or code
generators
> exist for PDE problems (e.g. Citadel, PDELab, PIER, Falcon
etc.) but I
> haven't been able to conclude if there is a common system
for the
> problem definition. Maple and Mathematica also have their
own syntaxes
> which do similar things.
> Do any readers know of any 'standard' languages for
describing general
> PDE problems (i.e. equation, domain, boundary conditions
etc)? Any
> that are widely used / well established? Any
preferences/experiences?
there is MINOPT for modeling problems in optimal control,
which allows one to specify ordinary differential equations,
but apparently it is not widely used at present.
I don't know enough about PDE solvebut probably each system
has its own environment; I never heard about analogues of
To become a standard, a modeling system must be successfully
interfaced to _many_different_ solvers; only then people will
change from what they already use to a standardized language.
Having such a general system would be very useful;
unfortunately it takes many man years to build one...
Arnold Neumaier
===
Subject: Re: Sparse Matrix, eigenvalue, eigenvector...
>Hy all,
>
>this is my problem: I have to find *all* eigenvalue and
eigenvector
>from very large sparse matrix (from 30000x30000 to
100000x100000
>elements). Matrix has value very condensed to the diagonal,
it's
>symmetric and it's stored in memory with skyline or CSR
method (I
>heaven't decide yet...)
>
>Which method can I use to solve my problem? Another
(sub)problem: is
> there a method to find eigenvalue and maintain the matrix
structure
>(in order not to occupy much of the memory)?
>
>
> Attilio Gelosa.
for doing an earthquake analysis of a building for sure you
don't need all
eigenvalues, you need the eigenvalues which correspond to the
frequency
spectrum
of an typical earthquake, that means all eigenvalues in an
interval.
-> arpack.
http://www.caam.rice.edu/software/ARPACK
having said this, how to proceed yourself: if you would need
indeed all of
them,
then separate eigenvalue/eigenvector computation.
Lanczos is not the way to go, since doing Lanczos for all
eigenvalues you
need to
use complete reorthogonalization, storing all Lanczos vectors
and this
alone
will render it impossible,
now to the next step.
since the matrix is symmetric, the pivot signs of the
LDL(transpose)
decomposition
of A-sI indicate the number of eigenvalues >s =s <s.
A possibility to compute all of them, but questionable.
This might become
instable with the exception of the tridiagonal case.
If indeed the matrix can be considered as narrow banded then:
use H.R. Schwarz bandwidth reduction which is done by applying
in a tricky
way
Givens rotations. you have to store all the Givens rotations
in the
form (i,j,c,s) (row numbers involvd, cosine, sine value).
(code is in the Handbook for automatic computation
(Wikinson-Reinsch,
Springers Yellow series), in ALGOL60, but easy to understand
and to
translate.
having come down to the tridiagonal form, you are happy:
LAPACK has an excellent efficient code for computing all
eigenvalues of
a large tridiagonal matrix using nes dissection. then, having
found the
eigenvalues, you do a rudimentary inverse iteration step to
get the
eigenvector of the tridiagonal matrix (no problems involved
here: gaussian
elimination with row pivoting. if some subdiagonal elements
are extremely
small
you could dissect again.) then, using the list of Givens
rotations stored,
transform back to an eigenvector of the original matrix. this
can be done
one by one, again no much memory access problems occur. Hence
main effort is
the bandwidth reduction. there will be fill in within the band!
(this is the bad part of the message: sparsity alone will not
help).
hth
peter
===
Subject: Re: Sparse Matrix, eigenvalue, eigenvector...
> I would suggest lanczos type methods for this. Still, it is
going
> to be difficult just to store that many vectors. Another
possible
> approach would be to compute all the eigenvalues with
lanczos, and
> then repeat the process to compute blocks of eigenvectors
using the
> precompu expansion vectors from the tridiagonal
representation.
I read something about Lanczos method... Probably I'm wrong
(...not
true: certainly I'm wrong!), but I understood that Lanczos
responses
are about a little whole of the eigenvectors...
Can you explain me something (or tell me about where i can
find docs)?
Attilio Gelosa
===
Subject: Re: Sparse Matrix, eigenvalue, eigenvector...
> This will mean your matrix of eigenvectors is 100000x100000
but full!
> How on earth are you going to utilize this information?
I'm making a feasibility study about heartquake verification of
concrete/steel building
> Probably you need an approach where you generate and use one
eigenvector
> after the other, to avoid storing them all?
...mmm... interesting solution type... So, there is a method to
extract one eigenvalue after another and study it
separately... Do you
know where can i find docs about it?
Attilio Gelosa
===
Subject: Re: Sparse Matrix, eigenvalue, eigenvector...
>>This will mean your matrix of eigenvectors is 100000x100000
but full!
>>How on earth are you going to utilize this information?
> I'm making a feasibility study about heartquake verification
of
> concrete/steel building
>>Probably you need an approach where you generate and use one
eigenvector
>>after the other, to avoid storing them all?
> ...mmm... interesting solution type... So, there is a method
to
> extract one eigenvalue after another and study it
separately... Do you
> know where can i find docs about it?
For general background on eigenvalue methods look at the books
by
Parlett or Golub & van Loan; there you can find the general
background needed. You can calculate blocks of eigenpairs by
the
subspace method; it gives you the k absolutely largest
eigenvalues
and their eigenvalues. So to get the k eigenvalues closest to
some
reference value s, you need to apply it to (A-sI)^{-1} in
place of A,
and transform the eigenvalues back to those for A; the
eigenvectors
will be unchanged.
To get all eigenvalues, you need to use many different s,
moving it
over the range of interest. I hope that all your eigenvalues
are
real; then you only need to search the real line; searching the
complex plane is much more expensive (and less reliable).
Arnold Neumaier
===
Subject: Re: Sparse Matrix, eigenvalue, eigenvector...
I have no experience with finding ALL eigenvalues of such large
matrices. But,
just from the standard knowledge about LR/QR algorithms:
Look for sparse QR or Cholesky factorizations suitable for your
matrices. If those do a decomposition in a reasonable time you
can try
to apply Cholesky-LR with Wilkinson shift (cubically
convergent) or QR
with Francis shift or similar. Since the shift is to the
diagonal only
you should able to use the symbolic factorization you did once
and
only redo the numerical one. Also, use deflation after an
eigenvalue
has been found.
Hans Mittelmann
--------------------------------------------------------------
--------------
----
> Hy all,
> this is my problem: I have to find *all* eigenvalue and
eigenvector
> from very large sparse matrix (from 30000x30000 to
100000x100000
> elements). Matrix has value very condensed to the diagonal,
it's
> symmetric and it's stored in memory with skyline or CSR
method (I
> heaven't decide yet...)
> Which method can I use to solve my problem? Another
(sub)problem: is
> there a method to find eigenvalue and maintain the matrix
structure
> (in order not to occupy much of the memory)?
> Attilio Gelosa.
===
Subject: Joint space multidimensional scaling in SPSS
I'm trying to perform a joint space mds in SPSS and my problem
is to
calculate the rectangular joint distance matrix (between cases
and
variables).
Could you please help me with this?
I would also appreciate it, if you could suggest me any useful
software/reading material about mds.
Dimitris Vatikiotis
===
Subject: Number combinations
Is there a mathmatical formula or computer program that can
generate
all 10,000 possible combinations of 1-9 using 4 digits at a
time (i.e
1,3,4,5 1,1,1,1,) repititions are okay?
--
tml
===
Subject: Re: Can't figure this out...
>> Common denominator:
>> (sin x)/(1+cos x)+(1+cos x)/(sin x)
>> = ((sin x)^2 + (1 + cos x)^2)/ ((sin x)(1+cos x))
>> = ((sin x)^2 + 1 +2(cos x) +(cos x)^2)/ ((sin x)(1+cos x))
>> = 2(1+ (cos x)) / ((sin x)(1+cos x))
>> = 2 / (sin x)
>> = 2 csc x
>> Constraint: cos(x)!=-1, to avoid division by 0.
> Would you not also need a constraint: sin x !=0, for the
same reason?
When sin(x)=0, cos(x) is either +1 or -1. If cos(x)=-1, the
left hand
side of the identity to prove is undefined, but the right hand
side is
infinite. When cos(x)=+1 both sides of the identity are
infinite,
which one COULD argue is still an acceptable identity. That
is, one
can allow division by 0, but not 0/0. I'd be happy, though,
with a
student who disallowed all multiples of pi for x, and not just
the odd
multiples.
--
Kevin Karplus karplus@soe.ucsc.edu
http://www.soe.ucsc.edu/~karplus
life member (LAB, Adventure Cycling, American Youth Hostels)
Effective Cycling Instructor #218-ck (lapsed)
Professor of Computer Engineering, University of California,
Santa Cruz
Undergraduate and Graduate Director, Bioinformatics
Affiliations for identification only.
--
tml
===
Subject: Re: Can't figure this out...
> What is necessary is that sin(x) != 0, that is, x != k(pi),
where k is
any
> integer. That is, all integer multiples of pi are excluded
as x values.
> This subsumes the constraint cos(x) != -1, which happens for
x being an
> *odd* integer multiple of pi.
> Notes: 1. != is used for is not equal to. 2. It's probably a
50-50
> chance that a typical trig or pre-calc book would not bother
mentioning
> any constraint for this trig identity, since both sides of
the (proposed)
> identity have the same constraint.
We have the statement:
(sin x)/(1+cos x)+(1+cos x)/(sin x)=2csc x
It's not where the individual expressions themselves are
defined, but where
the two _sides_ of the equation are defined that are relevent
in calling it
an identity. Both the left and right sides are defined over
the exact same
set. Many define identity to be consistent with this (where
both sides
defined there is equality). Neither of the aforementioned
restrictions
really needs to be made explicit, in such a context. The
question of
making
the restriction explicit or not, probably depends more on pet
peeve than
anything else. ...and for pet peeves, everyone has one but the
current
instructor's judgement is usually governing, so do whatever
he/she says, if
they have any insistence one way or the other.
That said, in some lines of reasoning eg establishing a
bidirectional
implication, it may be necessary to make such a restriction
explicit
without
misspeaking. This usually occurs where some act results in two
equations
that are not equivelent, such as where one is defined for
certain values
where another is not. In such cases, we may have an
implication (if) in
one
direction, but not in both (iff). I, for one, in my earlier
sugges
proof did not intend to communicate any bidirectional
implications,
just
one-way ones.
--
Darrell
--
tml
===
Subject: Re: handhelds in k-12 settings
> Has anyone personally had experience with handhelds in
mathematics
> classes in k-12.
> I'm interes in your experience.
Handheld...what? Computers?
Personally I'm beginning to favor not allowing calculators
until 11th
grade..
--
tml
===
Subject: Re: heron's formula
> I am a junior in highschool doing a geometry report on
Heron's
> Formula...if anyone is able to help me find resources I
thank you very
> much now! Please and thank you to all those we read this!
One of the skills you should develop that will stand you in
good stead is
learning to search for things on the web on your own. There
are many
search engines available. Perhaps the most popular is called
and you
will get a long list of websites that deal with heron's
formula. You
will have learned about heron's formula and you will learn a
little about
web searching in the bargain.
Rich
--
tml
===
Subject: Re: heron's formula
> I am a junior in highschool doing a geometry report on
Heron's
> Formula...if anyone is able to help me find resources I
thank you very
> much now! Please and thank you to all those we read this!
Try Google, The following has much information.
http://jwilson.ce.uga.edu/emt725/Heron/Heron.html
David
--
tml
===
Subject: Re: heron's formula
> I am a junior in highschool doing a geometry report on
Heron's
> Formula...if anyone is able to help me find resources I
thank you very
> much now! Please and thank you to all those we read this!
--
tml
===
Subject: Re: re-arranging formulaes
> what is the answer to this question s=1/2gtsquared you have
to make t
> the subject
Multiply both sides by 2
2s=gt^2
2s/g=t^2
t=sqrt(2s/g)
--
tml
===
Subject: Re: re-arranging formulaes
> what is the answer to this question s=1/2gtsquared you have
to make t
> the subject
That isn't a question; it's a formula.
You start with:
s = (1/2) * g * t^2
The first thing to do is get t by itself, so you multiply both
sides by
2/g. This gives you:
2 * s / g = t^2
Now you can take the square root of both sides for your answer:
t = sqrt(2s/g)
--
tml
===
Subject: Re: Grad school
> I myself have only a BA in mathematics, and have been
recently
> exploring obtaining a master's in mathematics. (Full-time
college
> teaching usually requires a subject master's, minimum. Still
plan to
> get a master's in education as well, though - the people in
k12 who
> have the power of hiring and firing usually want that, not a
subject
> master's. [There are plenty of online master's degrees in
education,
> but none that I know of in math.]
Are you aware of this program? It is an online/video-based MA
in
Teaching Math. The good news is that of the required courses
only 2 are
in education and the rest can be math courses.
http://www.uidaho.edu/eo/newhtml/progmath.htm
Rich
--
tml
===
Subject: Re: Grad school
>The student advisor of a math dept that is trying to become
recognized
>as a top-notch Ph.D program directly told me to stay away
from the
>program: The professors there won't think you're serious if
all you
>want is a Master's.
>So I'm sort of in the same boat. To obtain what you've talked
about -
>the personal attention by the professoetc. - I wondered about
>checking out schools where the Master's program is the
terminal
>program. Don't know whether it's a good idea. If it is,
anyone have
>some thoughts on good, reputable schools in this regard?
>Paul
The state college or state universities such as California
State University
CityName have bachelor and master programs, but no PhD
programs (or is this
not
completely correct?); The University schools, such as
University of
California at CityName have bachelor, master, and PhD
programs. Maybe this
forms a possible but somewhat incomplete way of judging which
sort of
institutions would want masters students or PhD students ---
This in
relation
to what Paul Tanner has sta.
G C
--
tml