mm-1399
===
Subject: Re: Poll: Are PCs Turing Machines?
> I wonder what people really think about this.
> Are PCs physical examples to Turing Machines? [*] 
> Please write only Yes/No to avoid discussion.
> --
> Eray
> [*] The alternative being they are not. There are some arguments that
> this is the case.
No, they are not Turing Machines.
-- 
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Poll: Are PCs Turing Machines?
> I wonder what people really think about this.
> Are PCs physical examples to Turing Machines? [*]
No.
===
Subject: Re: Poll: Are PCs Turing Machines?
> I wonder what people really think about this.
> Are PCs physical examples to Turing Machines? [*] 
Even if you change to to of the answer is no.
Rick
===
Subject: Re: Are PCs Turing Machines?
>I wonder what people really think about this.
> Are PCs physical examples to Turing Machines? [*]
> Please write only Yes/No to avoid discussion.
no 
===
Subject: Re: Are PCs Turing Machines?
> Are PCs physical examples to Turing Machines? [*]
> Please write only Yes/No to avoid discussion.
No. 
===
Subject: Re: Are PCs Turing Machines?
> Are PCs physical examples to Turing Machines? [*]
> Please write only Yes/No to avoid discussion.
> No.
Yes
===
Subject: Re: Are PCs Turing Machines?
>Are PCs physical examples to Turing Machines? [*]
>Please write only Yes/No to avoid discussion.
>>No.
> Yes
No!!
===
Subject: Cyclic subgroup
Hi again,
Im doing another Abstract Algebra problem and just want some feedback on 
my thinking.
Let G be a subgroup of order pq, where p and q are primes, not 
necessarily diffrent primes. Show that every real/genuine (dont know 
how to translate this) subgroup in G is cyclic.
Here is my thought, Lagranges says that if H is a subgroup in G, the 
order of H is a devider in the order of G.
So, any real number can be written as a multiplication of primes, and if 
  the order of G in the example has the order if pq, the subgroup H must 
have the order of either p or q (or 1?). Ever subgroup that has a prime 
as order is cyclic so any subgroup in G must be cyclic.
Am I on the right path here? Is there a nicer proof?
-- 
Stefan
===
Subject: Re: Cyclic subgroup
            days. My association with the Department is that of an alumnus.
>Im doing another Abstract Algebra problem and just want some feedback on 
>my thinking.
>Let G be a subgroup of order pq, where p and q are primes, not 
>necessarily diffrent primes. Show that every real/genuine (dont 
know 
>how to translate this) 
Proper. Means that the subgroup is not the trivial subgroup of one
element, and not the full group either.
>subgroup in G is cyclic.
>Here is my thought, Lagranges says that if H is a subgroup in G, the 
>order of H is a devider in the order of G.
>So, any real number
any positive integer
> can be written as a multiplication of primes, and if 
>  the order of G in the example has the order if pq, the subgroup H must 
>have the order of either p or q (or 1?).
Order 1 is excluded by asking that the subgroup be proper.
> Ever subgroup that has a prime 
>as order is cyclic so any subgroup in G must be cyclic.
>Am I on the right path here? Is there a nicer proof?
No. That's it.
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Cyclic subgroup
> Order 1 is excluded by asking that the subgroup be proper.
Not the way I heard it. You say proper if you want to exclude 
the whole group (but allow order 1), and non-trivial if you want 
to exclude order 1 (but allow the whole group).
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Cyclic subgroup
>>Im doing another Abstract Algebra problem and just want some feedback on 
>>my thinking.
>>Let G be a subgroup of order pq, where p and q are primes, not 
>>necessarily diffrent primes. Show that every real/genuine (dont 
know 
>>how to translate this) 
>Proper. Means that the subgroup is not the trivial subgroup of one
>element, and not the full group either.
>>subgroup in G is cyclic.
When I say H is a proper subgroup of G, I just mean that H is not equal
to G - I do not exclude the trivial subgroup. I woudl say proper 
nontrivial
subgroup for that.
My definition agrees with that on mathworld 
(see mathworld.wolfram.com/ProperSubgroup.html),
but I guess that won't help to convince anybody!
Derek Holt.
===
Subject: Re: Cyclic subgroup
            days. My association with the Department is that of an alumnus.
>Im doing another Abstract Algebra problem and just want some feedback on 
>my thinking.
>Let G be a subgroup of order pq, where p and q are primes, not 
>necessarily diffrent primes. Show that every real/genuine (dont 
know 
>how to translate this) 
>>Proper. Means that the subgroup is not the trivial subgroup of one
>>element, and not the full group either.
>subgroup in G is cyclic.
>When I say H is a proper subgroup of G, I just mean that H is not equal
>to G - I do not exclude the trivial subgroup. I woudl say proper 
nontrivial
>subgroup for that.
>My definition agrees with that on mathworld 
>(see mathworld.wolfram.com/ProperSubgroup.html),
>but I guess that won't help to convince anybody!
Some people do it one way, some people do it the other way. But you're
right; most people use proper subgroup to mean subgroup which is a
proper subset.
Then again, the trivial subgroup is usually considered to be cyclic,
so even if the problem was meant to include the trivial subgroup, it
would still be correct.
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Can you invert this Laplace transform?
Consider a Laplace transform
F(s) =  exp {- a Sqrt[s] Tanh[b Sqrt[s] ] },
where a and b are real, strictly positive parameters,
and Tanh is the hyberbolic tangent function.
I suspect this may have a closed-form
Inverse Laplace transform f(t), but haven't
found it in tables or by software. Tried perusing
Gradshteyn and Ryzhik, but no luck. The expression
occurs in problems where the modified Bessel functions
I_nu (z) and K_nu (z) often appear.
Does anyone know f(t)?
alan
===
Subject: PROOF that 0.99999... = 1
Hi.
Theorem: 0.9999... = 1
Proof:
A number x = 0.aaaaaa..., where 0 <= a <= 9 is
x = sum_{n=1...inf} a/(10^n)
If a = 9, then
x = sum_{n=1...inf} 9/(10^n)
Then, we have the sequence of partial sums
S_n = sum_{i=1...n} 9/(10^n).
We can then derive the _general formula_ as follows:
S_1 = 9/10
S_2 = 9/10 + 9/100 = 99/100
S_3 = 9/10 + 9/100 + 9/1000 = 999/1000
...
S_n = ((10^n)-1)/(10^n)
...
Then we take the limit as n -> inf:
lim_{n->inf} ((10^n)-1)/(10^n).
Direct susbtitution produces the indeterminate form inf/inf, so we rewrite
the limit:
lim_{n->inf} ((10^n)-1)/(10^n) = lim_{n->inf} 1 - (1/(10^n))
                               = lim_{n->inf} 1 - lim_{n->inf} (1/(10^n))
                               = 1 - 0
                               = 1
Thus, 0.9999... = 1. QED.
===
Subject: Re: PROOF that 0.99999... = 1
>Hi.
>Theorem: 0.9999... = 1
>Proof:
>A number x = 0.aaaaaa..., where 0 <= a <= 9 is
>x = sum_{n=1...inf} a/(10^n)
>If a = 9, then
>x = sum_{n=1...inf} 9/(10^n)
>Then, we have the sequence of partial sums
>S_n = sum_{i=1...n} 9/(10^n).
>We can then derive the _general formula_ as follows:
>S_1 = 9/10
>S_2 = 9/10 + 9/100 = 99/100
>S_3 = 9/10 + 9/100 + 9/1000 = 999/1000
>...
>S_n = ((10^n)-1)/(10^n)
>...
>Then we take the limit as n -> inf:
>lim_{n->inf} ((10^n)-1)/(10^n).
>Direct susbtitution produces the indeterminate form inf/inf, so we rewrite
>the limit:
>lim_{n->inf} ((10^n)-1)/(10^n) = lim_{n->inf} 1 - (1/(10^n))
>                               = lim_{n->inf} 1 - lim_{n->inf} (1/(10^n))
>                               = 1 - 0
>                               = 1
>Thus, 0.9999... = 1. QED.
   Hey thank you very much, this proves .999... diverges. According to the
Basic Convergence Test according to Advanced Engineering Mathematics, you 
just
proved that,
Since,
n-->oo
Lim Z_n =/= 0
So it diverges, thank you.
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: PROOF that 0.99999... = 1
I'm afraid not.
lim Z_n =/= 0 implies that
sum_i (Z_i) diverges.
However the S_n in the original post were already the partial sums.
===
Subject: People helping people this holiday season
People helping people this holiday season
Spread the joy this holiday season by visiting HELP SOMEONE.  A group
of selected and screened individuals who are in need of any type of
holiday help.
Group web page : http://msclay0.tripod.com/helpsomeone/  Group site
===
Subject: Limits for functions defined on arbitrary sets (not intervals)
For functions defined on arbitrary sets (such as integers) one usually
talks of continuity in terms of topological concepts. However I was
reading Kennan T Smiths Primer of Modern analysis where he defines 
limit as follows :
Let g be a real valued function on a set S of real numbers. Let a and l 
be real numbers . The statement 
                   lim g(x) = l
                   x->a
                   x in S
means that 
 (a) For each posative number delta there is at least one point x in S
     with |x-a| < delta
 (b) For each posative number epsilon there is a posative number delta
     such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon
 
For functions defined on intervals (b) is ALMOST the definition of
limits since (a) is trivially satisfied. However there is no requirement
in (b) that x not equal a. Is this a typo or there is a reason for it?
Notices :
(1)
let f(x) = x for all x in Natural numbers. This function is continuous
under a metric topology. And the above definition can also be used to
prove the same. However if we add the requirement that x not equal
a then this function does not have a limit at any point.
(2)
On the other hand consider the function g(x)=x for all x in natural
numbers except 1, and g(1)=2. We would like a definition that yeilds
that the limit of g(x) as x tends to 1 is 1. However the above
definition implies that this limit is 2.
Prof Smith then goes on define limits on deleted neighbourhoods
(intervals) in the usual way. Could some one kindly shed some
light on the apparently unusual way for defining limits and the
motivation for it.
sincerely
B Thomas
===
Subject: Re: Limits for functions defined on arbitrary sets (not intervals)
> Let g be a real valued function on a set S of real numbers. Let a and l
> be real numbers . The statement
>                    lim g(x) = l
>                  x->a
>                  x in S
> means that
when S subset R, g:S -> R.
for all e > 0, some d with for all x, (0 < |x - a| < d ==> |g(x) - l| < e)
In topology lim(x->a) f(x) = l when
for all open U nhood l, some open V nhood a with
        f(Va) subset U
>  (b) For each posative number epsilon there is a posative number delta
>      such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon
> However there is no requirement in (b) that x not equal a. Is this a
> typo or there is a reason for it?
Another sloppy, quick profit text book?  It should be there
except when g is continuous at a, then the 0 < is redundant.
> let f(x) = x for all x in Natural numbers. This function is continuous
> under a metric topology. And the above definition can also be used to
> prove the same. However if we add the requirement that x not equal
> a then this function does not have a limit at any point.
Pick any e.  Let d = e.  Then for all x
        0 < |x - a| < d ==> |f(x) - a| < e
lim(x->a) f(x) = a.
> On the other hand consider the function g(x)=x for all x in natural
On the other hand for readablity, consider the function g(x) = x.
> numbers except 1, and g(1)=2. We would like a definition that yeilds
> that the limit of g(x) as x tends to 1 is 1. However the above
> definition implies that this limit is 2.
It does not.  Let e = 1/2, where's the d such that for all x
0 < |x - 1| < d ==> |g(x) - 2| < 1/2 ?
===
Subject: Re: Limits for functions defined on arbitrary sets (not intervals)
I suspect you are correct about the typo, though Kennan T Smith's book
should not be that bad; it has been around for a while and
springer-verlag released the second ed.
>> On the other hand consider the function g(x)=x for all x in natural
>> numbers except 1, and g(1)=2. We would like a definition that yeilds
>> that the limit of g(x) as x tends to 1 is 1. However the above
>> definition implies that this limit is 2.
> It does not.  Let e = 1/2, where's the d such that for all x
> 0 < |x - 1| < d ==> |g(x) - 2| < 1/2 ?
If as in the definition there was no requirement that 0 < |x - 1| 
then let d = 1 and we could always let x = 1 so |g(x) - 2| = 0
===
Subject: Limits for functions defined on arbitrary sets (not intervals)
Correction made to the last part of my previous post.
> Let g be a real valued function on a set S of real numbers. Let a and l
> be real numbers . The statement
>                    lim g(x) = l
>                x->a
>                x in S
> means that
> when S subset R, g:S -> R.
> for all e > 0, some d with for all x, (0 < |x - a| < d ==> |g(x) - l| < 
e)
> In topology lim(x->a) f(x) = l when
> for all open U nhood l, some open V nhood a with
>       f(Va) subset U
It is becoming apparent from the discussion below, why the above isn't a
standard definition, but instead
f continuous at a, ie lim(x->a) f(x) = f(a) when
for all open U nhood f(a), some open V nhood a with f(V) subset U.
>  (b) For each posative number epsilon there is a posative number delta
>      such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon
> However there is no requirement in (b) that x not equal a.
That's for when g is continuous at a.  BTW, you've a typo at g(x) - a.
> let f(x) = x for all x in Natural numbers. This function is continuous
> under a metric topology. And the above definition can also be used to
> prove the same. However if we add the requirement that x not equal
> a then this function does not have a limit at any point.
> Pick any e.  Let d = e.  Then for all x
>       0 < |x - a| < d ==> |f(x) - a| < e
> lim(x->a) f(x) = a.
> On the other hand consider the function g(x)=x for all x in natural
> On the other hand for readability, consider the function g(x) = x.
> numbers except 1, and g(1)=2. We would like a definition that yeilds
> that the limit of g(x) as x tends to 1 is 1. However the above
> definition implies that this limit is 2.
Let g(1) = b.  Then for any e > 0, let d = 1/2.
Thus for all integers x, (0 < |x - 1| < 1/2 ==> |g(x) - c| < e)
because there is no integer x with 0 < |x - 1| < 1/2.
Hence lim(x->1) g(x) = c for any and all c.
This is a rather perplexing result showing that this calculus
definition of limit of a function at a point has to be limited
to exclude discrete domains.  You will notice that for the domain
D = (-oo,-1] / { 0 } / [1,oo) the same problem will occur but not
for domains like (-oo,0] / (1,2) / [5,oo) which don't have any
isolated points like 0 in the first example.  Thus the 0 < requirement
needs be included with a requirement that for x->a, that a is not an
isolated point of the domain of the function.
When a is an isolated point, then there is no sequence of points /= a
that will converge to a.  When there is a sequence of points /= a,
a1, a2, ... that converge to a, then the definition of limit will show
f(a1), f(a2), ... will converge to the limit point.  But that's the
intuition in lim(x->a) f(x), that f(x) will approach a number as x
approaches a.
Now you be an imp and ask teacher what's the limit of a function at
0 with the domain D as described above and report to us his answer.
The function could be x^2, x, 1 or even 0 for spectacular results.
On domain D, lim(x->0) 0 = 1,000,000,000 or any number a what ever.
If e > 0, let d = 1/2
for all x in D, (0 < |x - 0| < d ==> |0 - a| < e)
again by vacuous implication.
===
Subject: Re: Limits for functions defined on arbitrary sets (not intervals)
The vacuous implication argument is indeed a convincing one.
bt
> Correction made to the last part of my previous post.
>> Let g be a real valued function on a set S of real numbers. Let a and 
l
>> be real numbers . The statement
>>                    lim g(x) = l
>>               x->a
>>               x in S
>> means that
>> when S subset R, g:S -> R.
>> for all e > 0, some d with for all x, (0 < |x - a| < d ==> |g(x) - l| < 
e)
>> In topology lim(x->a) f(x) = l when
>> for all open U nhood l, some open V nhood a with
>>      f(Va) subset U
> It is becoming apparent from the discussion below, why the above isn't a
> standard definition, but instead
> f continuous at a, ie lim(x->a) f(x) = f(a) when
> for all open U nhood f(a), some open V nhood a with f(V) subset U.
>>  (b) For each posative number epsilon there is a posative number delta
>>      such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon
>> However there is no requirement in (b) that x not equal a.
> That's for when g is continuous at a.  BTW, you've a typo at g(x) - a.
>> let f(x) = x for all x in Natural numbers. This function is continuous
>> under a metric topology. And the above definition can also be used to
>> prove the same. However if we add the requirement that x not equal
>> a then this function does not have a limit at any point.
>> Pick any e.  Let d = e.  Then for all x
>>      0 < |x - a| < d ==> |f(x) - a| < e
>> lim(x->a) f(x) = a.
>> On the other hand consider the function g(x)=x for all x in natural
>> On the other hand for readability, consider the function g(x) = x.
>> numbers except 1, and g(1)=2. We would like a definition that yeilds
>> that the limit of g(x) as x tends to 1 is 1. However the above
>> definition implies that this limit is 2.
> Let g(1) = b.  Then for any e > 0, let d = 1/2.
> Thus for all integers x, (0 < |x - 1| < 1/2 ==> |g(x) - c| < e)
> because there is no integer x with 0 < |x - 1| < 1/2.
> Hence lim(x->1) g(x) = c for any and all c.
> This is a rather perplexing result showing that this calculus
> definition of limit of a function at a point has to be limited
> to exclude discrete domains.  You will notice that for the domain
> D = (-oo,-1] / { 0 } / [1,oo) the same problem will occur but not
> for domains like (-oo,0] / (1,2) / [5,oo) which don't have any
> isolated points like 0 in the first example.  Thus the 0 < requirement
> needs be included with a requirement that for x->a, that a is not an
> isolated point of the domain of the function.
> When a is an isolated point, then there is no sequence of points /= a
> that will converge to a.  When there is a sequence of points /= a,
> a1, a2, ... that converge to a, then the definition of limit will show
> f(a1), f(a2), ... will converge to the limit point.  But that's the
> intuition in lim(x->a) f(x), that f(x) will approach a number as x
> approaches a.
> Now you be an imp and ask teacher what's the limit of a function at
> 0 with the domain D as described above and report to us his answer.
> The function could be x^2, x, 1 or even 0 for spectacular results.
> On domain D, lim(x->0) 0 = 1,000,000,000 or any number a what ever.
> If e > 0, let d = 1/2
> for all x in D, (0 < |x - 0| < d ==> |0 - a| < e)
> again by vacuous implication.
===
Subject: Re: Limits for functions defined on arbitrary sets (not intervals)
> For functions defined on arbitrary sets (such as integers) one usually
> talks of continuity in terms of topological concepts. However I was
> reading Kennan T Smiths Primer of Modern analysis where he defines 
> limit as follows :
> Let g be a real valued function on a set S of real numbers. Let a and l 
> be real numbers . The statement 
>                    lim g(x) = l
>                  x->a
>                  x in S
> means that 
>  (a) For each posative number delta there is at least one point x in S
>      with |x-a| < delta
>  (b) For each posative number epsilon there is a posative number delta
>      such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon
> For functions defined on intervals (b) is ALMOST the definition of
> limits since (a) is trivially satisfied. However there is no requirement
> in (b) that x not equal a. Is this a typo or there is a reason for it?
> Notices :
> (1)
> let f(x) = x for all x in Natural numbers. This function is continuous
> under a metric topology. And the above definition can also be used to
> prove the same. However if we add the requirement that x not equal
> a then this function does not have a limit at any point.
> (2)
> On the other hand consider the function g(x)=x for all x in natural
> numbers except 1, and g(1)=2. We would like a definition that yeilds
> that the limit of g(x) as x tends to 1 is 1. However the above
> definition implies that this limit is 2.
> Prof Smith then goes on define limits on deleted neighbourhoods
> (intervals) in the usual way. Could some one kindly shed some
> light on the apparently unusual way for defining limits and the
> motivation for it.
> sincerely
> B Thomas
===
Subject: Re: Limits for functions defined on arbitrary sets (not intervals)
> (a) For each posative number delta there is at least one point x in S
>     with |x-a| < delta
> (b) For each posative number epsilon there is a posative number delta
>     such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon
>(2)
>On the other hand consider the function g(x)=x for all x in natural
>numbers except 1, and g(1)=2. We would like a definition that yeilds
>that the limit of g(x) as x tends to 1 is 1. However the above
>definition implies that this limit is 2.
That doesn't make sense to me.  You can certainly interpolate (g(0)+g(2))/2
but that isn't the same as a limit.
The one case where a limit on a function which is in a discrete domain 
would
make sense is a=oo.  Then you can test successively closer deltas for
validation.  For example:
lim(x->oo) n!/(n-1)!/(n+1)
The epsilons are then infinite, but other than that it works.
--Keith Lewis              klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: Limits for functions defined on arbitrary sets (not intervals)
>> (a) For each posative number delta there is at least one point x in S
>>     with |x-a| < delta
>> (b) For each posative number epsilon there is a posative number delta
>>     such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon
>>(2)
>>On the other hand consider the function g(x)=x for all x in natural
>>numbers except 1, and g(1)=2. We would like a definition that yeilds
>>that the limit of g(x) as x tends to 1 is 1. However the above
>>definition implies that this limit is 2.
> That doesn't make sense to me.  You can certainly interpolate 
(g(0)+g(2))/2
> but that isn't the same as a limit.
No I am not interpolating. Just using the definition as in a
epsilon-delta proof of a limit to show the limit is indeed 2. However if
the domain was real numbers then the limit would be 1, for the same
function. Which is why I said we  would like a definition that yeilds
 1 as the limit.
> The one case where a limit on a function which is in a discrete domain 
would
> make sense is a=oo.  Then you can test successively closer deltas for
> validation.  For example:
> lim(x->oo) n!/(n-1)!/(n+1)
> The epsilons are then infinite, but other than that it works.
One can talk of continuity in arbitrary topological spaces. And if the
topology is induced by a metric then we can certainly talk about limits
even if we are dealing with point sets and not intervals.
sincerely
b thomas
===
Subject: Some questions about the Cauchy distribution
The mean of n random variables picked from the Cauchy distribution has 
itself the Cauchy distribution. Does this mean that there is no use 
having a larger sample in order to estimate the expected value? (Or wait 
a minute, does the expected value exist? No, right?)
Does the existence of such distributions and those of the form 
Constant(p) / (1 + abs(x)^p) with p > 2 indicate that it is not always 
legitimate to assume that more measurements leads to a better 
approximation of the expected value? (Unless you have a good idea of the 
distribution of the measurements.)
For instance, if several hundred people sees an UFO and it is claimed 
that it must really have been an UFO is it proper to say that this 
cannot be assumed since there exists distributions where the mean value 
of many measurements are even worse than one measurement?
        / Daniel Janzon
===
Subject: Re: Some questions about the Cauchy distribution
>The mean of n random variables picked from the Cauchy distribution has 
>itself the Cauchy distribution. Does this mean that there is no use 
>having a larger sample in order to estimate the expected value? (Or wait 
>a minute, does the expected value exist? No, right?)
The expected value does not exist.  However, a larger
sample can be used to estimate the parameters of the
distribution, and even the distribution itself without
knowing its form.
>Does the existence of such distributions and those of the form 
>Constant(p) / (1 + abs(x)^p) with p > 2 indicate that it is not always 
>legitimate to assume that more measurements leads to a better 
>approximation of the expected value? (Unless you have a good idea of the 
>distribution of the measurements.)
If p > 2, the average will be a better estimate of the
mean.  If you know that the distribution is of that form,
except for translation, one can do much better.
>For instance, if several hundred people sees an UFO and it is claimed 
>that it must really have been an UFO is it proper to say that this 
>cannot be assumed since there exists distributions where the mean value 
>of many measurements are even worse than one measurement?
Here one is just estimating a probability, which is 
bounded.  The argument does not apply.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558
===
Subject: Re: Some questions about the Cauchy distribution
>>For instance, if several hundred people sees an UFO and it is claimed 
>>that it must really have been an UFO is it proper to say that this 
>>cannot be assumed since there exists distributions where the mean value 
>>of many measurements are even worse than one measurement?
> Here one is just estimating a probability, which is 
> bounded.  The argument does not apply.
I don't really understand. What is it that must be bounded?
Further, my argument is not really about if the above distribution can 
be applied or not. It is rather based on the _existance_ of 
distributions without mean.
Suppose many people see a UFO and suppose their descriptions match quite 
well, but not perfectly. Someone might say that Many people saw it and 
their descriptions matched quite well so it wasn't hallucination. _Thus_ 
it really was a UFO or an object with the same properties of a UFO.
My argument is this: We know that there exists distributions without 
mean. Thus in order to believe fully in the above implication (many saw 
it => it was true) one has to confirm that observations of the phenomena 
behaves properly. This might be easy, but it has to be done.
===
Subject: Re: Some questions about the Cauchy distribution
>The mean of n random variables picked from the Cauchy distribution has 
>itself the Cauchy distribution. Does this mean that there is no use 
>having a larger sample in order to estimate the expected value? (Or wait 
>a minute, does the expected value exist? No, right?)
I think what you're trying to do is the following: given a random sample
of size n from the Cauchy distribution with density 1/(Pi (1+(x-a)^2)),
estimate the parameter a.  This parameter is not the expected value,
because that doesn't exist, but it is the centre of symmetry of the 
distribution.  True, the sample mean has the same distribution
as one of these random variables.  And that says that the sample mean
is not a useful estimator.  However, there are better estimators, for
example the median.  This does get better for larger n.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada
===
Subject: Re: Some questions about the Cauchy distribution
> I think what you're trying to do is the following: given a random sample
> of size n from the Cauchy distribution with density 1/(Pi (1+(x-a)^2)),
> estimate the parameter a.  This parameter is not the expected value,
> because that doesn't exist, but it is the centre of symmetry of the 
> distribution.  True, the sample mean has the same distribution
> as one of these random variables.  And that says that the sample mean
> is not a useful estimator.  However, there are better estimators, for
> example the median.  This does get better for larger n.
Ultimately I was trying to justify that one cannot assume (unless one 
know something about the distribution behind the phenomena in question) 
that many observations/measurements of a phenomena leads to a better 
estimate. (Now from your post I know that even though the mean is not a 
converging estimate there are other that do converge.)
For instance if many people describe a murderer the pure existence of 
distributions where the mean is not a good estimate implies that one 
cannot _automatically_ assume that the murderer looks like the mean of 
the descriptions (with a high probability).
===
Subject: Re: Some questions about the Cauchy distribution
> Ultimately I was trying to justify that one cannot assume (unless one 
> know something about the distribution behind the phenomena in question) 
> that many observations/measurements of a phenomena leads to a better 
> estimate. (Now from your post I know that even though the mean is not a 
> converging estimate there are other that do converge.)
> For instance if many people describe a murderer the pure existence of 
> distributions where the mean is not a good estimate implies that one 
> cannot _automatically_ assume that the murderer looks like the mean of 
> the descriptions (with a high probability).
However, if a distribution is *bounded* then it does have a mean.
One would assume that the space of descriptions of murderers
(whatever that is) will be bounded.
===
Subject: Re: Some questions about the Cauchy distribution
>> Ultimately I was trying to justify that one cannot assume (unless one 
>> know something about the distribution behind the phenomena in question) 
>> that many observations/measurements of a phenomena leads to a better 
>> estimate. (Now from your post I know that even though the mean is not a 
>> converging estimate there are other that do converge.)
>> For instance if many people describe a murderer the pure existence of 
>> distributions where the mean is not a good estimate implies that one 
>> cannot _automatically_ assume that the murderer looks like the mean 
of 
>> the descriptions (with a high probability).
>However, if a distribution is *bounded* then it does have a mean.
>One would assume that the space of descriptions of murderers
>(whatever that is) will be bounded.
But having a mean doens't mean that one should expect that
a sample will look like the mean with high probability.
(If you flip a coin, counting heads as 1 and tails as -1,
the mean of your random variable is 0, but the probability
that a given toss comes up 0 is 0. Or with murderers:
say s(P) is 1 if it's perfectly clear that P is male
from his appearance, -1 if it's perfectly clear that
P is female. Most people have a P very close to 1 or
-1, although the mean is 0.)
************************
David C. Ullrich
===
Subject: Re: Some questions about the Cauchy distribution
> The mean of n random variables picked from the Cauchy distribution has 
> itself the Cauchy distribution. Does this mean that there is no use 
> having a larger sample in order to estimate the expected value? (Or 
> wait a minute, does the expected value exist? No, right?)
You are correct that the expected value does not exist.  What do you 
mean that you want to the estimate it, especially since you know the 
distribution already?
> Does the existence of such distributions and those of the form 
> Constant(p) / (1 + abs(x)^p) with p > 2 indicate that it is not always 
> legitimate to assume that more measurements leads to a better 
> approximation of the expected value? (Unless you have a good idea of 
> the distribution of the measurements.)
No.  For p > 2, the expected value exists, and the sample mean 
approaches it with probability 1.  You will have trouble with using a 
normal approximation if 2 < p < 3, since the second moment does not 
exist in this case.
-- 
Stephen J. Herschkorn                        sjherschko@netscape.net
===
Subject: Re: Some questions about the Cauchy distribution
> The mean of n random variables picked from the Cauchy distribution has 
> itself the Cauchy distribution. Does this mean ....
      You may get a good answer here, or you may like to try sending the 
question to the  sci.stat.math  or  sci.stat.edu  news group.
            Ken Pledger.
===
Subject: Re: Platonism
>Arithmetic does occur physically. It's not something we make up in our 
>heads.
>>Do you mean like rocks?
> Precisely. That is the realist portion of my theory that Godel has
> dubbed Aristotelian Realism. It is realist without the extra 
baggage
> of Platonism.
> Arithmetic is no fiction, arithmetic is something you do when you
> *weigh* things, and it occurs *there*, not merely in your head. (I'm
> using head to emphasize how physicalist this point of view is)
>> I don't see that at all. You might throw a rock at me and it would
>> hurt, you might throw another and have hit me twice, but you really
>> can't throw 1+1=2 at me (unles you write it one one of the rocks).
>> -- 
http://www.dartmouth.edu/~matc/MathDrama/reading/Wigner.html
WHAT IS PHYSICS? by Eugene Wigner
The physicist is interested in discovering the laws of inanimate nature. In 

order to understand this statement, it is necessary to analyze the concept, 

law of nature.
The world around us is of baffling complexity and the most obvious fact 
about it is that we cannot predict the future. Although the joke attributes 

only to the optimist the view that the future is uncertain, the optimist is 

right in this case: the future is unpredictable. It is, as Schrodinger has 
remarked, a miracle that in spite of the baffling complexity of the world, 
certain regularities in the events could be discovered. One such regularity, 

discovered by Galileo, is that two rocks, dropped at the same time from the 

same height, reach the ground at the same time. The laws of nature are 
concerned with such regularities. Galileo's regularity is a prototype of a 
large class of regularities. It is a surprising regularity for three 
reasons.
The first reason that it is surprising is that it is true not only in Pisa, 

and in Galileo's time, it is true everywhere on the Earth, was always true, 

and will always be true. This property of the regularity is a recognized 
invariance property and, as I had occasion to point out some time ago, 
without invariance principles similar to those implied in the preceding 
generalization of Galileo's observation, physics would not be possible. The 

second surprising feature is that the regularity which we are discussing is 

independent of so many conditions which could have an effect on it. It is 
valid no matter whether it rains or not, whether the experiment is carried 
out in a room or from the Leaning Tower, no matter whether the person who 
drops the rocks is a man or a woman. It is valid even if the two rocks are 
dropped, simultaneously and from the same height, by two different people. 
There are, obviously, innumerable other conditions which are all immaterial 

from the point of view of the validity of Galileo's regularity. The 
irrelevancy of so many circumstances which could play a role in the 
phenomenon observed has also been called an invariance. However, this 
invariance is of a different character from the preceding one since it 
cannot be formulated as a general principle. The exploration of the 
conditions which do, and which do not, influence a phenomenon is part of the 

early experimental exploration of a field. It is the skill and ingenuity of 

the experimenter which show him phenomena which depend on a relatively 
narrow set of relatively easily realizable and reproducible conditions. [5 
See, in this connection, the graphic essay of M. Deutsch, Daedalus 87, 86 
(1958). A. Shimony has called my attention to a similar passage in C. S. 
Peirce's Essays in the Philosophy of Science (New York: The Liberal Arts 
Press, 1957), p. 237.] In the present case, Galileo's restriction of his 
observations to relatively heavy bodies was the most important step in this 

regard. Again, it is true that if there were no phenomena which are 
independent of all but a manageably small set of conditions, physics would 
be impossible.
The preceding two points, though highly significant from the point of view 
of the philosopher, are not the ones which surprised Galileo most, nor do 
they contain a specific law of nature. The law of nature is contained in the 

statement that the length of time which it takes for a heavy object to fall 

from a given height is independent of the size, material, and shape of the 
body which drops. In the framework of Newton's second law, this amounts 
to 
the statement that the gravitational force which acts on the falling body is 

proportional to its mass but independent of the size, material, and shape of 

the body which falls.
The preceding discussion is intended to remind us, first, that it is not at 

all natural that laws of nature exist, much less that man is able to 
discover them. [6 E. Schrodinger, in his What Is Life? (Cambridge: Cambridge 

University Press, 1945), p. 31, says that this second miracle may well be 
beyond human understanding.] The present writer had occasion, some time ago, 

to call attention to the succession of layers of laws of nature, each 
layer containing more general and more encompassing laws than the previous 
one and its discovery constituting a deeper penetration into the structure 
of the universe than the layers recognized before. However, the point which 

is most significant in the present context is that all these laws of nature 

contain, in even their remotest consequences, only a small part of our 
knowledge of the inanimate world. All the laws of nature are conditional 
statements which permit a prediction of some future events on the basis of 
the knowledge of the present, except that some aspects of the present state 

of the world, in practice the overwhelming majority of the determinants of 
the present state of the world, are irrelevant from the point of view of the 

prediction. The irrelevancy is meant in the sense of the second point in the 

discussion of Galileo's theorem. [7 The writer feels sure that it is 
unnecessary to mention that Galileo's theorem, as given in the text, does 
not exhaust the content of Galileo's observations in connection with the 
laws of freely falling bodies.]  ...
The principal purpose of the preceding discussion is to point out that the 
laws of nature are all conditional statements and they relate only to a very 

small part of our knowledge of the world.
SH: Mathematics is a knowledge representation of the world, not the world 
itself.  This is such a fundamental aspect of being regarded as educated. 
RE
===
Subject: Re: Platonism
>Arithmetic does occur physically. It's not something we make up in our 
>>>heads.
>> Precisely. That is the realist portion of my theory that Godel has
>> dubbed Aristotelian Realism. It is realist without the extra 
baggage
>> of Platonism.
>>
http://en.wikipedia.org/wiki/Philosophy_of_mathematics
Mathematical realism holds that mathematical entities exist independently 
of the human mind. Thus humans do not invent mathematics, but rather 
discover it, and any other intelligent beings in the universe would 
presumably do the same. The term Platonism is used because such a view is 
seen to parallel Plato's belief in a heaven of ideas, an unchanging 
ultimate reality that the everyday world can only imperfectly approximate. 
Plato's view probably derives from Pythagoras, and his followers the 
Pythagoreans, who believed that the world was, quite literally, built up by 

the numbers. This idea may have even older origins that are unknown to 
us.
-------------------------------------------------------------------------
Embodied mind theories:
These theories hold that mathematical thought is a natural outgrowth of 
the 
human cognitive apparatus which finds itself in our physical universe. For 
example, the abstract concept of number springs from the experience of 
counting discrete objects. It is held that mathematics is not universal and 

does not exist in any real sense, other than in human brains. Humans 
construct, but do not discover, mathematics.
The physical universe can thus be seen as the ultimate foundation of 
mathematics: it guided the evolution of the brain and later determined which 

questions this brain would find worthy of investigation. However, the human 

mind has no special claim on reality or approaches to it built out of 
math; If such constructs as Euler's Identity are true then they are true 
as a map of the human mind and cognition, not as a map of anything it 
sees.
The effectiveness of mathematics is thus easily explained: mathematics was 
constructed by the brain in order to be effective in this universe.
SH: This is the mathematical perspective philosophy which is in accord with 

physicalism. Not Platonism which posits an abstract realm which is not at 
all physical. Aristotelian realism is not as close to physicalism as 
'embodied minds'.
----------------------------------------------------------------------------

Aristotelean realism
Similar to idealism, realism is also one of the oldest philosophies in 
western culture and its origin began with the Greek philosopher Aristotle 
(384-322 B.C.E.) in ancient Greece. Being a longtime star student of Plato, 

he elaborated on the idealist view of reality being based on ideas and not 
matter. He thought that a proper study of matter could lead to better and 
more distinct ideas.
The exaggerated Realism of Plato, investing the real being with the 
attributes
of the being in thought, is the principal doctrine of his metaphysics.  ..
Aristotle broke away from these exaggerated views of his master and
formulated the main doctrines of Moderate Realism. The real is not, as 
Plato
says, some vague entity of which the sensible world is only the shadow; it
dwells in the midst of the sensible world. Individual substance (this man, 
that
horse) alone has reality; it alone can exist. The universal is not a thing 
in
itself; it is immanent in individuals and is multiplied in all the 
representatives
of a class. As to the form of universality of our concepts (man, just), it 
is a
product of our subjective consideration. The objects of our generic and 
specific representations can certainly be called substances (ous.92ai), 
when 
they designate
the fundamental reality (man) with the accidental determinations (just, 
big);
but these are de.9cterai ous.92ai (second substances), and by that 
Aristotle 
means
precisely that this attribute of universality which affects the substance as 

in
thought does not belong to the substance (thing in itself); it is the 
outcome of
our subjective elaboration. This theorem of Aristotle, which completes the
metaphysics of Heraclitus (denial of permanent) by means of that of 
Parmenides (denial of change), is the antithesis of Platonism, and may be 
considered one of
the finest pronouncements of Peripateticism. It was through this wise 
doctrine
that the Stagyrite exercised his ascendency over all later thought.
===
Subject: Re: Platonism
>Arithmetic does occur physically. It's not something we make up in our 
>heads.
>>
http://www.dartmouth.edu/~matc/MathDrama/reading/Wigner.html
The Unreasonable Effectiveness of Mathematics in the Natural Sciences
WHAT IS MATHEMATICS? by Eugene Wigner
Somebody once said that philosophy is the misuse of a terminology which 
was 
invented just for this purpose. [2 This statement is quoted here from W. 
Dubislav's Die Philosophie der Mathematik in der Gegenwart (Berlin: Junker 
and Dunnhaupt Verlag, 1932), p. 1.] In the same vein, I would say that 
mathematics is the science of skillful operations with concepts and rules 
invented just for this purpose. The principal emphasis is on the invention 
of concepts. Mathematics would soon run out of interesting theorems if these 

had to be formulated in terms of the concepts which already appear in the 
axioms. Furthermore, whereas it is unquestionably true that the concepts of 

elementary mathematics and particularly elementary geometry were formulated 

to describe entities which are directly suggested by the actual world, the 
same does not seem to be true of the more advanced concepts, in particular 
the concepts which play such an important role in physics. Thus, the rules 
for operations with pairs of numbers are obviously designed to give the same 

results as the operations with fractions which we first learned without 
reference to pairs of numbers. The rules for the operations with 
sequences, that is, with irrational numbers, still belong to the category of 

rules which were determined so as to reproduce rules for the operations with 

quantities which were already known to us. Most more advanced mathematical 
concepts, such as complex numbers, algebras, linear operators, Borel sets 
and this list could be continued almost indefinitely were so devised that 
they are apt subjects on which the mathematician can demonstrate his 
ingenuity and sense of formal beauty. In fact, the definition of these 
concepts, with a realization that interesting and ingenious considerations 
could be applied to them, is the first demonstration of the ingeniousness of 

the mathematician who defines them. The depth of thought which goes into the 

formulation of the mathematical concepts is later justified by the skill 
with which these concepts are used. The great mathematician fully, almost 
ruthlessly, exploits the domain of permissible reasoning and skirts the 
impermissible. That his recklessness does not lead him into a morass of 
contradictions is a miracle in itself: certainly it is hard to believe that 

our reasoning power was brought, by Darwin's process of natural selection, 
to the perfection which it seems to possess. However, this is not our 
present subject....
SH: Advanced mathematical concepts become abstract not concrete
===
Subject: Re: Platonism
>Arithmetic does occur physically. It's not something we make up in 
our 
>heads.
> http://www.dartmouth.edu/~matc/MathDrama/reading/Wigner.html
> The Unreasonable Effectiveness of Mathematics in the Natural Sciences
The effectiveness of mathematics in natural sciences is not
unreasonable. I wonder who ever thought it was.
--
Eray Ozkural
===
Subject: Re: Platonism
   at 07:52 PM, examachine@gmail.com (Eray Ozkural  exa) said:
>The effectiveness of mathematics in natural sciences is not
>unreasonable. I wonder who ever thought it was.
Those who understand it better than you, like Wigner.
-- 
Shmuel (Seymour J.) Metz, SysProg and JOAT  <http://patriot.net/~shmuel>
Unsolicited bulk E-mail subject to legal action.  I reserve the
right to publicly post or ridicule any abusive E-mail.  Reply to
domain Patriot dot net user shmuel+news to contact me.  Do not
===
Subject: Re: Platonism
>    at 07:52 PM, examachine@gmail.com (Eray Ozkural  exa) said:
>>The effectiveness of mathematics in natural sciences is not
>>unreasonable. I wonder who ever thought it was.
> Those who understand it better than you, like Wigner.
What Wigner is saying is -- then a miracle happens. Here is the last 
paragraph of his well known essay:
------------------------------------------------------------------------
Let me end on a more cheerful note. The miracle of the appropriateness 
of the language of mathematics for the formulation of the laws of 
physics is a wonderful gift which we neither understand nor deserve. We 
should be grateful for it and hope that it will remain valid in future 
research and that it will extend, for better or for worse, to our 
pleasure, even though perhaps also to our bafflement, to wide branches 
of learning.
---------------------------------------------------------------------------
Bob Kolker
===
Subject: Re: Platonism
>    at 07:52 PM, examachine@gmail.com (Eray Ozkural  exa) said:
> 
> 
>>The effectiveness of mathematics in natural sciences is not
>>unreasonable. I wonder who ever thought it was.
> 
> 
> Those who understand it better than you, like Wigner.
> What Wigner is saying is -- then a miracle happens. Here is the last 
> paragraph of his well known essay:
> ------------------------------------------------------------------------
> Let me end on a more cheerful note. The miracle of the appropriateness 
> of the language of mathematics for the formulation of the laws of 
> physics is a wonderful gift which we neither understand nor deserve. We 
> should be grateful for it and hope that it will remain valid in future 
> research and that it will extend, for better or for worse, to our 
> pleasure, even though perhaps also to our bafflement, to wide branches 
> of learning.
> 
---------------------------------------------------------------------------
Another new mysterian. Why should I confide in mystery? That is
nonsense. I am  still a positivist, and I see no place for such naive
spiritualism. I think we can understand the world with science and
philosophy.
I don't like all that religious talk.
--
Eray
===
Subject: Re: Platonism
> Another new mysterian. Why should I confide in mystery? That is
> nonsense. I am  still a positivist, and I see no place for such naive
> spiritualism. I think we can understand the world with science and
> philosophy.
Wigner does not have what you want. You will have to come up with your 
own answers. If you can figure out why a discipline which is not 
empirical in the slightest turns out to be so useful for empircal ends, 
please share it with the rest of us.
Bob Kolker
===
Subject: Re: Platonism
 <41b241bd$9$fuzhry+tra$mr2ice@news.patriot.net>
 <iLrsd.613544$mD.280286@attbi_s02>
 <6zLsd.149474$5K2.115759@attbi_s03>  Another new mysterian. Why should I confide in mystery? That is
>> nonsense. I am  still a positivist, and I see no place for such naive
>> spiritualism. I think we can understand the world with science and
>> philosophy.
>Wigner does not have what you want. You will have to come up with your 
>own answers. If you can figure out why a discipline which is not 
>empirical in the slightest turns out to be so useful for empircal ends, 
>please share it with the rest of us.
>Bob Kolker
Not everyone believes that any more.
For many, it isn't the mystery it once was. At least, it hasn't been for 
many since Quine's Two Dogmas of Empiricism (1951). Husserl also had a 
pretty good shot explaining it with his origin or geometry even before 
then (and some think Piaget contributed through his genetic 
epistemology).
Derrida managed to obfuscate all that.
>empirical in the slightest turns out to be so useful for empircal ends, 
>please share it with the rest of us.
>Bob Kolker
-- 
David Longley
===
Subject: Re: Platonism
>>  Another new mysterian. Why should I confide in mystery? That is
>> nonsense. I am  still a positivist, and I see no place for such naive
>> spiritualism. I think we can understand the world with science and
>> philosophy.
>Wigner does not have what you want. You will have to come up with your 
>own answers. If you can figure out why a discipline which is not 
>empirical in the slightest turns out to be so useful for empircal ends, 
>please share it with the rest of us.
>Bob Kolker
> Not everyone believes that any more.
> For many, it isn't the mystery it once was. At least, it hasn't been for 
> many since Quine's Two Dogmas of Empiricism (1951). Husserl also had a 
> pretty good shot explaining it with his origin or geometry even before 
> then (and some think Piaget did contributed through his genetic 
> epistemology).
> Derrida managed to obfuscate all that.
Quine didn't say the final word on that. But still, I will agree that
mathematics has no special place in the world. And I don't find it
surprising that it's useful!
It's obvious that origin of geometry is firmly grounded in common
sense. There is no denying to that, even for Platonists who pretend to
be scientific minded!
About Derrida: I asked the head of our philosophy department, and he
told me that they don't consider him much of a philosopher. I don't
--
Eray
===
Subject: Re: Platonism
 <41b241bd$9$fuzhry+tra$mr2ice@news.patriot.net>
 <iLrsd.613544$mD.280286@attbi_s02>
 <6zLsd.149474$5K2.115759@attbi_s03> <BJefjAFN$4sBFw+7@longley.demon.co.uk>>  Another new mysterian. Why should I confide in mystery? That is
>>> nonsense. I am  still a positivist, and I see no place for such naive
>>> spiritualism. I think we can understand the world with science and
>>> philosophy.
>>Wigner does not have what you want. You will have to come up with your
>>own answers. If you can figure out why a discipline which is not
>>empirical in the slightest turns out to be so useful for empircal ends,
>>please share it with the rest of us.
>>Bob Kolker
>> Not everyone believes that any more.
>> For many, it isn't the mystery it once was. At least, it hasn't been for
>> many since Quine's Two Dogmas of Empiricism (1951). Husserl also had 
a
>> pretty good shot explaining it with his origin or geometry even before
>> then (and some think Piaget did contributed through his genetic
>> epistemology).
>> Derrida managed to obfuscate all that.
>Quine didn't say the final word on that. But still, I will agree that
>mathematics has no special place in the world. And I don't find it
>surprising that it's useful!
>It's obvious that origin of geometry is firmly grounded in common
>sense. There is no denying to that, even for Platonists who pretend to
>be scientific minded!
>About Derrida: I asked the head of our philosophy department, and he
>told me that they don't consider him much of a philosopher. I don't
>Eray
Nobody is likely to have the final word. That's not just Quine's central 
message but Popper's as well. This unending quest is critical to the 
nature of science which will always strive to improve prediction and 
control in an almost limitless universe. Quine's contribution was as 
genetic as Husserl's or Piaget's. The important difference (if there 
really is one worth pragmatic emphasis) is that Quine articulated an 
enlightened, non-mentalistic, external, or extensionalist empiricism 
(also known as evidential behaviourism). One need only look to the 
recent history of geometry (and science) to see how prone we are to be 
misled by the intensional heuristics of our lebenswelt and why we turn 
to the extensional stance wherever it matters.
-- 
David Longley
http://www.longley.demon.co.uk/Frag.htm
===
Subject: Re: Platonism
>> Another new mysterian. Why should I confide in mystery? That is
>> nonsense. I am  still a positivist, and I see no place for such naive
>> spiritualism. I think we can understand the world with science and
>> philosophy.
> Wigner does not have what you want. You will have to come up with your 
> own answers. If you can figure out why a discipline which is not 
> empirical in the slightest turns out to be so useful for empircal ends, 
> please share it with the rest of us.
> Bob Kolker
I don't know if its totally accurate to say that mathematics is not 
empirical.  Once you make a construction, you subject it to tests.  If 
the tests fail, you reject the construction.  What is your criteria for 
being empirical ?
patty
===
Subject: Re: Platonism
> I don't know if its totally accurate to say that mathematics is not 
> empirical.  Once you make a construction, you subject it to tests.  If 
> the tests fail, you reject the construction.  What is your criteria for 
> being empirical ?
Measurement and observation vs formal deduction. Essentially all 
mathematical proofs are tautologies. The same is not true for 
empirically true statements.
It is empirically true that the washington monument is 555 +- 1 foot 
high. No logical contradiction would be entialed if it were only 500 
feet high. An empirically true statement just so happens to be true. A 
logically true statement is necessarily true.
Bob Kolker
===
Subject: Re: Platonism
>> I don't know if its totally accurate to say that mathematics is not 
>> empirical.  Once you make a construction, you subject it to tests.  If 
>> the tests fail, you reject the construction.  What is your criteria for 
>> being empirical ?
>Measurement and observation vs formal deduction. Essentially all 
>mathematical proofs are tautologies. The same is not true for 
>empirically true statements.
>It is empirically true that the washington monument is 555 +- 1 foot 
>high. No logical contradiction would be entialed if it were only 500 
>feet high. An empirically true statement just so happens to be true. A 
>logically true statement is necessarily true.
Technically only true of the axioms of which it is drawn. A statement
is only logically or mathematically true according to the logical or
mathematical axioms involved.
===
Subject: Re: Platonism
> 
> I don't know if its totally accurate to say that mathematics is not 
> empirical.  Once you make a construction, you subject it to tests.  If 
> the tests fail, you reject the construction.  What is your criteria for 
> being empirical ?
> Measurement and observation vs formal deduction. Essentially all 
> mathematical proofs are tautologies. The same is not true for 
> empirically true statements.
> It is empirically true that the washington monument is 555 +- 1 foot 
> high. No logical contradiction would be entialed if it were only 500 
> feet high. An empirically true statement just so happens to be true. A 
> logically true statement is necessarily true.
I think you are overlooking the fact that depicting the axioms are of
a quite empirical nature. You think of an axiom, and then you put it
to the test, to see if it's accepted by the mathematical community.
(Regardless of whether there is any reality to its truth!)
Plus, you should not overlook the logical depth of a proof, which
may sometimes be non-trivial! It takes ages to find all the
interesting theorems in a formal theory!
And a third point: you should consider what it means for a theorem to
be interesting. Again, that's an empirical process that does not
explained by the fact that (the truth of) a theorem is, formally, a
reduction to (truth of) axioms!
The first is in fact a point made by none other than Godel to support
his brand of Platonism, however, it is independent of Platonism
objectively speaking. By the incompleteness theorem, you can't reduce
mathematics to a=a, that is why you have to admit axioms of greater
complexity. This is best exemplified in work of Chaitin: ZFC explains
only so few facts about the halting problem.
--
Eray Ozkural
===
Subject: Re: Platonism
>> I don't know if its totally accurate to say that mathematics is not 
>> empirical.  Once you make a construction, you subject it to tests.  If 
>> the tests fail, you reject the construction.  What is your criteria 
>> for being empirical ?
> Measurement and observation vs formal deduction. Essentially all 
> mathematical proofs are tautologies. The same is not true for 
> empirically true statements.
> It is empirically true that the washington monument is 555 +- 1 foot 
> high. No logical contradiction would be entialed if it were only 500 
> feet high. An empirically true statement just so happens to be true. A 
> logically true statement is necessarily true.
> Bob Kolker
Oh,  your drawing that distinction.  How do you determine that it 
applies? What do you do to determine that one construction is analytic 
and another is synthetic?   Take for example x^n+y^n=z^n.   We can 
substitute integral values for x,y, z and n, and (with the with the 
exception of n=2) can not find an integral solution to the equation. 
Pretty soon you conjecture that there are no such solutions.  I can 
examine many animal species and determine that every one i look at which 
has a heart also has kidneys.  Pretty soon i conjecture that all 
animals with a heart also have kidneys.  What is the big difference? 
Aren't you just talking about proofs that have already been tested ?
patty
===
Subject: Re: Platonism
Originator: harris@tcs.inf.tu-dresden.de (Mitchell Harris)
>What Wigner is saying is -- then a miracle happens. 
This really makes it sound just like Hume's problem of induction. Is it 
the same, just similar, or what?
Mitch
===
Subject: Re: Platonism
> The effectiveness of mathematics in natural sciences is not
> unreasonable. I wonder who ever thought it was.
The question is how can a science (or art?) which has no empirical 
constraints be so useful in empirical pursuits such as physics? In 
mathematical justification the only empirical issue is whether something 
that is supposed to be a proof really is a proof. Proofs require checking.
Bob Kolker
===
Subject: Re: Platonism
> 
> The effectiveness of mathematics in natural sciences is not
> unreasonable. I wonder who ever thought it was.
> The question is how can a science (or art?) which has no empirical 
> constraints be so useful in empirical pursuits such as physics?
It has empirical constraints. It may just be that most mathematicians
delude themselves and think they are free of the world. They can't
really imagine an alternative to our universe. All mathematics is done
here and now, in our universe, and I think we just discover the
logical limits in our universe, not in other universes, not in
nomologically impossible universes... Anyway, the answer is simple:
your premise is wrong. The paradox is nonsense.
> In 
> mathematical justification the only empirical issue is whether something 
> that is supposed to be a proof really is a proof. Proofs require 
checking.
Wrong. Every mathematical thought has an intersubjective empirical
basis, I think I already elaborated on that.
Even a mind which has had no sensory history would be empirical in the
mathematics it does, simply because some facts of the matter can only
be found empirically and not with a proof!!! Otherwise, we would
expect to be able to *reduce* all mathematics to ZFC, which is clearly
not the case as Godel, Chaitin. Calude and others tell us.
If you did not understand the last sentence for the quasi-empirical
nature of mathematics, I can elaborate, although it's possible that
the oh-mathematics-is-so-above-science camp will try to sabotage the
discussion.
--
Eray Ozkural
===
Subject: Re: Platonism
> It has empirical constraints. It may just be that most mathematicians
> delude themselves and think they are free of the world. They can't
> really imagine an alternative to our universe. 
Nonsense. We have Euclidean worlds, a variety of non Euclidean worlds, 
many flavors of logic and multiple theories about physical reality that 
are not compatable. We have alternatives up the ying yang. The only 
constant demand in all this is logical consistency.
> Wrong. Every mathematical thought has an intersubjective empirical
> basis, I think I already elaborated on that.
> Even a mind which has had no sensory history would be empirical in the
> mathematics it does, simply because some facts of the matter can only
> be found empirically and not with a proof!!! Otherwise, we would
> expect to be able to *reduce* all mathematics to ZFC, which is clearly
> not the case as Godel, Chaitin. Calude and others tell us.
> If you did not understand the last sentence for the quasi-empirical
> nature of mathematics, I can elaborate, although it's possible that
> the oh-mathematics-is-so-above-science camp will try to sabotage the
> discussion.
Mathematics is clearly not above science, since some of it is so useful 
to scientists. Mathematics has a different agenda. Science tries to 
figure out how the (physical) world works. Many mathematicians like to 
spin patterns of ideas on the looms of imagination. Mathmatics is more 
like art than science, in spite of its usefulness in applied fields.
How empirical is a totally abstract painting, other than the fact it 
must use some physical medium (canvass and paint) to be presented? Is a 
totally non representational piece of art empirical?
Bob Kolker
===
Subject: Re: Platonism
>> It has empirical constraints. It may just be that most mathematicians
>> delude themselves and think they are free of the world. They can't
>> really imagine an alternative to our universe. 
>Nonsense. We have Euclidean worlds, a variety of non Euclidean worlds, 
>many flavors of logic and multiple theories about physical reality that 
>are not compatable. We have alternatives up the ying yang. The only 
>constant demand in all this is logical consistency.
While not disagreeing, I nonetheless have to ask according to which
axioms of logic?
 
>> Wrong. Every mathematical thought has an intersubjective empirical
>> basis, I think I already elaborated on that.
>> Even a mind which has had no sensory history would be empirical in the
>> mathematics it does, simply because some facts of the matter can only
>> be found empirically and not with a proof!!! Otherwise, we would
>> expect to be able to *reduce* all mathematics to ZFC, which is clearly
>> not the case as Godel, Chaitin. Calude and others tell us.
>> If you did not understand the last sentence for the quasi-empirical
>> nature of mathematics, I can elaborate, although it's possible that
>> the oh-mathematics-is-so-above-science camp will try to sabotage the
>> discussion.
>Mathematics is clearly not above science, since some of it is so useful 
>to scientists. Mathematics has a different agenda. Science tries to 
>figure out how the (physical) world works. Many mathematicians like to 
>spin patterns of ideas on the looms of imagination. Mathmatics is more 
>like art than science, in spite of its usefulness in applied fields.
>How empirical is a totally abstract painting, other than the fact it 
>must use some physical medium (canvass and paint) to be presented? Is a 
>totally non representational piece of art empirical?
Tautologies are necessarily composed of empirical observations.
So even tautological systems rest with one empirical foot in the
quagmire.
===
Subject: Re: Platonism
> 
> The effectiveness of mathematics in natural sciences is not
> unreasonable. I wonder who ever thought it was.
> The question is how can a science (or art?) which has no empirical 
> constraints be so useful in empirical pursuits such as physics? In 
> mathematical justification the only empirical issue is whether something 
> that is supposed to be a proof really is a proof. Proofs require 
checking.
Hmmm. I wildly disagree, as you would guess. Mathematical statements
have a lot of empirical content. Mathematical intuition is a game of
thought experiments, and thought experiments are not merely fictional.
I now think the said title explicitly refers to an underdeveloped
philosophy of mathematics, which today's mathematicians are taking for
granted. Godel was right in that. Philosophy is indeed in a sorry
state.
--
Eray Ozkural
===
Subject: Re: Platonism
> Hmmm. I wildly disagree, as you would guess. Mathematical statements
> have a lot of empirical content. Mathematical intuition is a game of
> thought experiments, and thought experiments are not merely fictional.
> I now think the said title explicitly refers to an underdeveloped
> philosophy of mathematics, which today's mathematicians are taking for
> granted. Godel was right in that. Philosophy is indeed in a sorry
> state.
If we distinguish discovery from justification, we can indeed argue that 
a great deal of mathematics is discover(invented?) by virtue of 
experience. However justification (i.e. proving) is purely formal 
however checking what purports to be a proof requires looking at at.
Bob Kolker
===
Subject: Re: Platonism
> If we distinguish discovery from justification, we can indeed argue that 
> a great deal of mathematics is discover(invented?) by virtue of 
> experience. However justification (i.e. proving) is purely formal 
> however checking what purports to be a proof requires looking at at.
I'm not sure I get that. Can't one discover a justification? Isn't
finding a proof quite an creative/inventive act (over and above the
act of discovering/inventing the thing to be proved)? I would think
checking a proof is what is more likely to be a formal process.
-- 
Mitch Harris
(remove q to reply)
===
Subject: Re: Platonism
> I'm not sure I get that. Can't one discover a justification? Isn't
> finding a proof quite an creative/inventive act (over and above the
> act of discovering/inventing the thing to be proved)? I would think
> checking a proof is what is more likely to be a formal process.
Linear logic enters more in the justification mode than the discovery 
mode. Mathematicians let their imaginations run free unfettered by 
logical constraint to discover consequences of their assumptions. On the 
other hand, -proving- the consequences is a matter of logic. It is true, 
that finding a proof (or what purports to be a proof) often requires a 
free spirt, but establishing that what purports be a proof is indeed a 
proof is nore a mechanical procedure.
Bob Kolker
===
Subject: Re: Platonism
>> The effectiveness of mathematics in natural sciences is not
>> unreasonable. I wonder who ever thought it was.
Intelligent, educated people, which of course excludes you.
Do you think that when you advertise that you are an arrogant
ignorant young punk, that adults can't see you for what you are?
===
Subject: congruence classes
What does it mean to have a negative congruence class mod n?
For example,
the congruence class, [-4], mod 16.
= [12] ?  Why?
===
Subject: Re: congruence classes
> What does it mean to have a negative congruence class mod n?
> For example,
> the congruence class, [-4], mod 16.
> = [12] ?  Why?
        [-4]_16 = [12]_16
because
        -4 = 12 (mod 16)
because
        16 | -4-12
===
Subject: Re: congruence classes
Because the number in the brackets is just a representative member of a 
set:
..., -20, -4, 12, 28, 44, ...
And it doesn't really matter which number we choose to be the
representative, since anyone will uniquely determine the congruence class 
it
is in.
-Darren
> What does it mean to have a negative congruence class mod n?
> For example,
> the congruence class, [-4], mod 16.
> = [12] ?  Why?
===
Subject: 6th degree equations
Hi.
Can one produce a solution to the general sixth-degree
equation
x^6 + a5 x^5 + a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0
 = 0
with a function mag_n(z), defined such that
mag_n(z)^n + mag_n(z) = z? (mag means magic)?
I know you can express the quintic's solution in terms of
mag_5(z), but could you express the 6th degree
equation's solution using mag_5(z) and mag_6(z)?
-- 
We should have a town named Alderaan
someday. No, seriously. Let's put it on
the table.
===
Subject: Re: 6th degree equations
> Can one produce a solution to the general sixth-degree
> equation
> x^6 + a5 x^5 + a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0
>  = 0
> with a function mag_n(z), defined such that
> mag_n(z)^n + mag_n(z) = z? (mag means magic)?
> I know you can express the quintic's solution in terms of
> mag_5(z), but could you express the 6th degree
> equation's solution using mag_5(z) and mag_6(z)?
You can't. However, using so called Tschirnhausen transformation you can 
show that the solution of an arbitrary equation of degree n can be 
reduced to the solution of equations of degrees =< 3 and one equation of 
the form:
X^n + b_4*X^{n-4} + b_5*X^{n-5} + ... + b_n = 0.
This is the Bring-Jerard theorem. In particular, the solution of an 
equation of degree 6 can be obtained from the solution of equations of 
degree =< 3 and one equation:
X^6 + b_4*X^2 + b_5*X + b_6.
For details see e.g. A. Mostowski, M. Stark, Introduction to higher 
algebra, Pergamon Press, 1964, pp. 366-370.
Pawel Gladki
===
Subject: Re: 6th degree equations
What kind of function would work? Any known
functions?
> Can one produce a solution to the general sixth-degree
> equation
> x^6 + a5 x^5 + a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0
>  = 0
> with a function mag_n(z), defined such that
> mag_n(z)^n + mag_n(z) = z? (mag means magic)?
> I know you can express the quintic's solution in terms of
> mag_5(z), but could you express the 6th degree
> equation's solution using mag_5(z) and mag_6(z)?
> You can't. However, using so called Tschirnhausen transformation you can
> show that the solution of an arbitrary equation of degree n can be
> reduced to the solution of equations of degrees =< 3 and one equation of
> the form:
> X^n + b_4*X^{n-4} + b_5*X^{n-5} + ... + b_n = 0.
> This is the Bring-Jerard theorem. In particular, the solution of an
> equation of degree 6 can be obtained from the solution of equations of
> degree =< 3 and one equation:
> X^6 + b_4*X^2 + b_5*X + b_6.
> For details see e.g. A. Mostowski, M. Stark, Introduction to higher
> algebra, Pergamon Press, 1964, pp. 366-370.
> Pawel Gladki
===
Subject: Re: 6th degree equations
>Can one produce a solution to the general sixth-degree
>equation
>x^6 + a5 x^5 + a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0
> = 0
>with a function mag_n(z), defined such that
>mag_n(z)^n + mag_n(z) = z? (mag means magic)?
>I know you can express the quintic's solution in terms of
>mag_5(z), but could you express the 6th degree
>equation's solution using mag_5(z) and mag_6(z)?
>>You can't. However, using so called Tschirnhausen transformation you can
>>show that the solution of an arbitrary equation of degree n can be
>>reduced to the solution of equations of degrees =< 3 and one equation of
>>the form:
>>X^n + b_4*X^{n-4} + b_5*X^{n-5} + ... + b_n = 0.
>>This is the Bring-Jerard theorem. In particular, the solution of an
>>equation of degree 6 can be obtained from the solution of equations of
>>degree =< 3 and one equation:
>>X^6 + b_4*X^2 + b_5*X + b_6.
>>For details see e.g. A. Mostowski, M. Stark, Introduction to higher
>>algebra, Pergamon Press, 1964, pp. 366-370.
 >What kind of function would work? Any known
 >functions?
That depends... The general sextic equation can be solved using so 
called Kamp.8e de F.8eriet functions (which are some kind of 
generalization 
of hypergeometric functions). For details, see e.g. P. Appell, J. Kamp.8e 
de F.8eriet, Fonctions hyperg.8eom.8etriques et hypersph.8eriques: 
polynomes 
d'Hermite, Gauthier-Villars, 1926.
For arbitrary equations of any degree it is possible to express 
solutions in terms of the Mellin integrals.
Pawel Gladki
===
Subject: Re: 6th degree equations
> That depends... The general sextic equation can be solved using so
> called Kamp.8e de F.8eriet functions (which are some kind of 
generalization
> of hypergeometric functions). For details, see e.g. P. Appell, J. 
Kamp.8e
> de F.8eriet, Fonctions hyperg.8eom.8etriques et hypersph.8eriques: 
polynomes
> d'Hermite, Gauthier-Villars, 1926.
> For arbitrary equations of any degree it is possible to express
> solutions in terms of the Mellin integrals.
Interesting! Do you have any reference?
Wilbert
===
Subject: Re: 6th degree equations
>>For arbitrary equations of any degree it is possible to express
>>solutions in terms of the Mellin integrals.
> Interesting! Do you have any reference?
H. Mellin, Ein allgemeiner Satz .9fber algebraische Gleichungen, Ann. 
Soc. Fennicae, (A) 7, Nr. 8, 44 S. (1915)
Pawel Gladki
===
Subject: Re: 6th degree equations
> What kind of function would work? Any known
> functions?
I believe that the closed form solutions of any degree polynomial can
be expressed in terms of Theta functions.
> Can one produce a solution to the general sixth-degree
> equation
 x^6 + a5 x^5 + a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0
>  = 0
 with a function mag_n(z), defined such that
> mag_n(z)^n + mag_n(z) = z? (mag means magic)?
 I know you can express the quintic's solution in terms of
> mag_5(z), but could you express the 6th degree
> equation's solution using mag_5(z) and mag_6(z)?
> You can't. However, using so called Tschirnhausen transformation you 
can
> show that the solution of an arbitrary equation of degree n can be
> reduced to the solution of equations of degrees =< 3 and one equation 
of
> the form:
> X^n + b_4*X^{n-4} + b_5*X^{n-5} + ... + b_n = 0.
> This is the Bring-Jerard theorem. In particular, the solution of an
> equation of degree 6 can be obtained from the solution of equations of
> degree =< 3 and one equation:
> X^6 + b_4*X^2 + b_5*X + b_6.
> For details see e.g. A. Mostowski, M. Stark, Introduction to higher
> algebra, Pergamon Press, 1964, pp. 366-370.
> Pawel Gladki
===
Subject: Re: 6th degree equations
        ETAsAhQlwrPvgjmWG5EhtaAyCc7pYAfvvQIUBfkPEdY3MpJE3P9+ximEzcRKK8o= 
There is not likey to be a simple combination.  The transformation
which, in quintic equations, leads to x^5 + ax + b = 0 gives x^6 + ax^2
+ bx  + c = 0 in degree 6.  The magic functions you describe can't
handle that x^2 term.
Should b in the transformed 6th-degree equation happen to be zero, you
get a solution involving not mag_5 or mag_6 but mag_3, where mag_3 can
be rendered into ordinary radicals or trigonometric functions.
--OL
===
Subject: Re: 6th degree equations
What if you used a quintic substitution into the
6th degree equation? Yes I know it would be
so damn complicated it is pretty much useless,
but would it do anything?
> There is not likey to be a simple combination.  The transformation
> which, in quintic equations, leads to x^5 + ax + b = 0 gives x^6 + ax^2
> + bx  + c = 0 in degree 6.  The magic functions you describe can't
> handle that x^2 term.
> Should b in the transformed 6th-degree equation happen to be zero, you
> get a solution involving not mag_5 or mag_6 but mag_3, where mag_3 can
> be rendered into ordinary radicals or trigonometric functions.
> --OL
===
Subject: Re: 6th degree equations
        ETAtAhUAhcAIkoEbfScAdoDfO37aZthjrTICFGwHUhIFlE369E4mjQZ2o37/8kyH 
No idea, really.  Yup you're in complicated territory.  Once I get above
degree 3 I go for numerical solutions.  You basically need numerical
methods to evaluate radicals and magic funcitons anyway.
--OL
===
Subject: Re: 6th degree equations
> No idea, really.  Yup you're in complicated territory.  Once I get above
> degree 3 I go for numerical solutions.  You basically need numerical
> methods to evaluate radicals and magic funcitons anyway.
> --OL
Yep.
Actually, I think you can solve any nth degree equation with radicals and
magic functions, but the length of the formula increases exponentially (or
maybe even superexponentially, not sure though). Numeric methods are
how you check the answer anyway, and since the solutions become so
horribly complicated, you might as well just ditch the exact-solve 
approach.
Anyway, I was just curious.
===
Subject: sin(x^x)
Hi.
Is there a formula that describes the density of
the graph of sin(x^x) for real x>0 as x increases
without bound?
-- 
We should have a town named Alderaan
someday. No, seriously. Let's put it on
the table.
===
Subject: Re: sin(x^x)
> Hi.
> Is there a formula that describes the density of
> the graph of sin(x^x) for real x>0 as x increases
> without bound?
[...]
 Topologically, it is nowhere dense.
 In terms of two-dimensional area (say,
Lebesgue measure), it has measure 0.
 Fractal-wise, it is of Hausdorff dimension 1.
 (And it is a real-analytic submanifold of the plane,
perhaps with some care taken about the missing left endpoint).
 Any other criteria?
If it means how close the subsequent roots (or the increasing and
decreasing segments) are, sit down and do the arithmetic, using the
Lambert W-function.
For your convenience, the explicit formula for y^y=x (x sufficiently
large) is
  y = ln(x) / lambertW(ln(x))
===
Subject: Re: sin(x^x)
> Hi.
> Is there a formula that describes the density of
> the graph of sin(x^x) for real x>0 as x increases
> without bound?
> [...]
>  Topologically, it is nowhere dense.
>  In terms of two-dimensional area (say,
> Lebesgue measure), it has measure 0.
>  Fractal-wise, it is of Hausdorff dimension 1.
>  (And it is a real-analytic submanifold of the plane,
> perhaps with some care taken about the missing left endpoint).
>  Any other criteria?
> If it means how close the subsequent roots (or the increasing and
> decreasing segments) are, sit down and do the arithmetic, using the
> Lambert W-function.
> For your convenience, the explicit formula for y^y=x (x sufficiently
> large) is
>   y = ln(x) / lambertW(ln(x))
By density I meant distance between the peaks of the waves
as x -> inf.
Obviously, the distance -> 0 (and so the density -> inf) as
x -> inf but what is the equation that describes that density?
I would guess it's something like this:
sin(u) is equal to 1 at u = pi/2, so sin(x^x) has a peak
whenever x^x = npi/2, with n a positive integer.
Then the nth peak is at x = ln(npi/2)/W(ln(npi/2))
So the 'density' function I'm looking for would then
be
D(n) = 1/(ln((n+1)pi/2)W(ln((n+1)pi/2)) - ln(npi/2)/W(ln(npi/2))).
Would that work?
===
Subject: Re: sin(x^x)
 <ucKrd.5982$vi1.5122@news01.roc.ny Hi.
 Is there a formula that describes the density of
> the graph of sin(x^x) for real x>0 as x increases
> without bound?
> [...]
>  Topologically, it is nowhere dense.
>  In terms of two-dimensional area (say,
> Lebesgue measure), it has measure 0.
>  Fractal-wise, it is of Hausdorff dimension 1.
>  (And it is a real-analytic submanifold of the plane,
> perhaps with some care taken about the missing left endpoint).
>  Any other criteria?
> If it means how close the subsequent roots (or the increasing and
> decreasing segments) are, sit down and do the arithmetic, using the
> Lambert W-function.
> For your convenience, the explicit formula for y^y=x (x sufficiently
> large) is
>   y = ln(x) / lambertW(ln(x))
> By density I meant distance between the peaks of the waves
> as x -> inf.
> Obviously, the distance -> 0 (and so the density -> inf) as
> x -> inf but what is the equation that describes that density?
> I would guess it's something like this:
> sin(u) is equal to 1 at u = pi/2, so sin(x^x) has a peak
> whenever x^x = npi/2, with n a positive integer.
   Caution: you have listed not only local maxima, but also roots and
local minima. Sort them out:
u = (4*k+1)*pi/2 (k integer) gives local maxima of sin(u).
You can complete the list.
[nothing added by me below]
> Then the nth peak is at x = ln(npi/2)/W(ln(npi/2))
> So the 'density' function I'm looking for would then
> be
> D(n) = 1/(ln((n+1)pi/2)W(ln((n+1)pi/2)) - ln(npi/2)/W(ln(npi/2))).
> Would that work?
===
Subject: Re: sin(x^x)

 Hi.
 Is there a formula that describes the density of
> the graph of sin(x^x) for real x>0 as x increases
> without bound?
> [...]
  Topologically, it is nowhere dense.
  In terms of two-dimensional area (say,
> Lebesgue measure), it has measure 0.
  Fractal-wise, it is of Hausdorff dimension 1.
  (And it is a real-analytic submanifold of the plane,
> perhaps with some care taken about the missing left endpoint).
  Any other criteria?
 If it means how close the subsequent roots (or the increasing and
> decreasing segments) are, sit down and do the arithmetic, using the
> Lambert W-function.
 For your convenience, the explicit formula for y^y=x (x sufficiently
> large) is
   y = ln(x) / lambertW(ln(x))
 By density I meant distance between the peaks of the waves
> as x -> inf.
> Obviously, the distance -> 0 (and so the density -> inf) as
> x -> inf but what is the equation that describes that density?
> I would guess it's something like this:
> sin(u) is equal to 1 at u = pi/2, so sin(x^x) has a peak
> whenever x^x = npi/2, with n a positive integer.
>    Caution: you have listed not only local maxima, but also roots and
> local minima. Sort them out:
Ohh... because of n = 2, then you get sin(pi) = 0, then n = 3 -> sin(3pi/2)
= -1,
then n = 4 -> sin(2pi) = 0, n = 5 -> sin(5pi/2) = 1, etc.
> u = (4*k+1)*pi/2 (k integer) gives local maxima of sin(u).
> You can complete the list.
So it should be
D(n) = 1/(ln((4n+2)pi/2)W(ln((4n+2)pi/2)) -
ln((4n+1)pi/2)/W(ln((4n+1)pi/2))).
> [nothing added by me below]
> Then the nth peak is at x = ln(npi/2)/W(ln(npi/2))
> So the 'density' function I'm looking for would then
> be
> D(n) = 1/(ln((n+1)pi/2)W(ln((n+1)pi/2)) - ln(npi/2)/W(ln(npi/2))).
> Would that work?
===
Subject: Re: sin(x^x)
 <ucKrd.5982$vi1.5122@news01.roc.ny> 
<Pine.WNT.4.58.0412021652550.-584711@satori.mcmaster.ca>
 <bxNrd.6017$Po1.857@news01.roc.ny

> Hi.
 Is there a formula that describes the density of
> the graph of sin(x^x) for real x>0 as x increases
> without bound?
> [...]
  Topologically, it is nowhere dense.
  In terms of two-dimensional area (say,
> Lebesgue measure), it has measure 0.
  Fractal-wise, it is of Hausdorff dimension 1.
  (And it is a real-analytic submanifold of the plane,
> perhaps with some care taken about the missing left endpoint).
  Any other criteria?
 If it means how close the subsequent roots (or the increasing and
> decreasing segments) are, sit down and do the arithmetic, using the
> Lambert W-function.
 For your convenience, the explicit formula for y^y=x (x 
sufficiently
> large) is
   y = ln(x) / lambertW(ln(x))

> By density I meant distance between the peaks of the waves
> as x -> inf.
 Obviously, the distance -> 0 (and so the density -> inf) as
> x -> inf but what is the equation that describes that density?
 I would guess it's something like this:
 sin(u) is equal to 1 at u = pi/2, so sin(x^x) has a peak
> whenever x^x = npi/2, with n a positive integer.
    Caution: you have listed not only local maxima, but also roots and
> local minima. Sort them out:
> Ohh... because of n = 2, then you get sin(pi) = 0, then n = 3 -> 
sin(3pi/2)
> = -1,
> then n = 4 -> sin(2pi) = 0, n = 5 -> sin(5pi/2) = 1, etc.
> u = (4*k+1)*pi/2 (k integer) gives local maxima of sin(u).
> You can complete the list.
> So it should be
> D(n) = 1/(ln((4n+2)pi/2)W(ln((4n+2)pi/2)) -
> ln((4n+1)pi/2)/W(ln((4n+1)pi/2))).
Replace (4n+2) by (4n+5), because you need to use 4(n+1)+1,
rather than (4n+1)+1.
> [nothing added by me below]
> Then the nth peak is at x = ln(npi/2)/W(ln(npi/2))
 So the 'density' function I'm looking for would then
> be
 D(n) = 1/(ln((n+1)pi/2)W(ln((n+1)pi/2)) - ln(npi/2)/W(ln(npi/2))).
 Would that work?

>
===
Subject: Re: sin(x^x)
>Is there a formula that describes the density of
>the graph of sin(x^x) for real x>0 as x increases
>without bound?
As x increases, the period decreases to 2*pi/x^x and the amplitude stays
the same.  Is that what you're looking for?
--Keith Lewis              klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: sin(x^x)
>Is there a formula that describes the density of
>the graph of sin(x^x) for real x>0 as x increases
>without bound?
> As x increases, the period decreases to 2*pi/x^x and the amplitude stays
> the same.  Is that what you're looking for?
> --Keith Lewis              klewis {at} mitre.org
> The above may not (yet) represent the opinions of my employer.
===
Subject: Complex analysis power series question
I'm rather new to complex analysis, so I'm wondering if somebody can
help me with this question?  Any hints are greatly appreciated, thank
you!
Determine all analytic functions f(z) on the complex plane that
satisfy
f(z^2) = (f(z))^2
I guess if f(z) is holomorphic everywhere, that means that it has its
power representation valid everywhere...but the power series of these
thoughts?
KH
===
Subject: Re: Complex analysis power series question
> I'm rather new to complex analysis, so I'm wondering if somebody can
> help me with this question?  Any hints are greatly appreciated, thank
> you!
> Determine all analytic functions f(z) on the complex plane that
> satisfy
> f(z^2) = (f(z))^2
> I guess if f(z) is holomorphic everywhere, that means that it has its
> power representation valid everywhere...but the power series of these
> thoughts?
> KH
I think that consideration of the the power series representation of f(z^2) 

is insightful and the way to crack this problem. Note that g^(n) = 0 if n is 

odd when
g=f(z^2). 
===
Subject: Re: Complex analysis power series question
> I'm rather new to complex analysis, so I'm wondering if somebody can
> help me with this question?  Any hints are greatly appreciated, thank
> you!
> Determine all analytic functions f(z) on the complex plane that
> satisfy
> f(z^2) = (f(z))^2
> I guess if f(z) is holomorphic everywhere, that means that it has its
> power representation valid everywhere...but the power series of these
> thoughts?
Suppose f is nonconstant. Then you can write f(z) = a + z^m*g(z), where m 
is a positive integer, g is entire, and g(0) is nonzero. Now equate f(z^2) 
and (f(z))^2 using this form for f, and see what happens.
===
Subject: Liouville theorem corollary?
I read in a complex analysis book that one can use Liouville's theorem
to classify all analytic functions f:C -> C such that
|f(z)| leq K*|z|^n
Does anybody know about this corollary, or how I might be able to
Shin
===
Subject: Re: Liouville theorem corollary?
> I read in a complex analysis book that one can use Liouville's theorem
> to classify all analytic functions f:C -> C such that
> |f(z)| leq K*|z|^n
> Does anybody know about this corollary, or how I might be able to
> derive it? 
Sure, let f(z) = sum(m=0,oo) c_m*z^m. Use Cauchy's estimates to prove 
something interesting about c_m for large m.
===
Subject: Re: Liouville theorem corollary?
>> I read in a complex analysis book that one can use Liouville's theorem
>> to classify all analytic functions f:C -> C such that
>> |f(z)| leq K*|z|^n
>> Does anybody know about this corollary, or how I might be able to
>> derive it? 
>Sure, let f(z) = sum(m=0,oo) c_m*z^m. Use Cauchy's estimates to prove 
>something interesting about c_m for large m.
Given the way the hypothesis was stated you can in fact use Cauchy's
estimates to prove something interesting about c_m for every m <> n,
both large and small.
************************
David C. Ullrich
===
Subject: Abstract Algebra
I'm missing something on this seemingly trivial problem.
Let G be a group with 3000 elements and H be a subgroup with 1000 elements.
If
x is in G, show that either x^2 or x^3 is in H.
This is what I got so far.
If x is in H then not only is x^2 in but x^3 is also in H.
So now we assume that x is not in H.
IF I can show that H, x^2H and x^3H are pairwise disjoint then G=H U x^2H U
x^3H  where each coset has o( H ) =1000 elements. So if x is in G then x is
in H or x^2H or x^3H. But x is not in H. So x is in x^2H or x^3H.
So all that remains to show is that H, x^2H and x^3H are pairwise disjoint.
Since x is not in H we have H /= xH-->xH /= x^2H-->x^2H /= x^3H. So x^2 /=
x^3H. I just can't seem to show that H /= x^2H ( or x^2 is not in H) and H
/= x^3H (or x^3 is not in H).
What am I not seeing?
===
Subject: Re: Abstract Algebra
> I'm missing something on this seemingly trivial problem.
> Let G be a group with 3000 elements and H be a subgroup with 1000 
elements.
> If
> x is in G, show that either x^2 or x^3 is in H.
> This is what I got so far.
> If x is in H then not only is x^2 in but x^3 is also in H.
> So now we assume that x is not in H.
> IF I can show that H, x^2H and x^3H are pairwise disjoint then G=H U x^2H 

U
> x^3H  where each coset has o( H ) =1000 elements. So if x is in G then x 
is
> in H or x^2H or x^3H. But x is not in H. So x is in x^2H or x^3H.
> So all that remains to show is that H, x^2H and x^3H are pairwise 
disjoint.
> Since x is not in H we have H /= xH-->xH /= x^2H-->x^2H /= x^3H. So x^2 
/=
> x^3H. I just can't seem to show that H /= x^2H ( or x^2 is not in H) and 
H
> /= x^3H (or x^3 is not in H).
> What am I not seeing?
You are not seeing that it is H, xH, and x^2H which are mutually
disjoint.  Write out explicitly what it means for an element of G to
be in two of these subsets of G and determine a contradiction based on
the fact that H is not merely a subset of G but a subgroup of G.
Achava
===
Subject: Re: Abstract Algebra
> I'm missing something on this seemingly trivial problem.
> Let G be a group with 3000 elements and H be a subgroup with 1000
> elements.
> If
> x is in G, show that either x^2 or x^3 is in H.
Hint:  If [G:H] = 3, then either H is normal in G, or H has a
subgroup of index 2 that's normal in G.
[...]
-- 
Jim Heckman
===
Subject: Re: Abstract Algebra
> I'm missing something on this seemingly trivial problem.
> Let G be a group with 3000 elements and H be a subgroup with 1000
> elements.
> If x is in G, show that either x^2 or x^3 is in H.
> Hint:  If [G:H] = 3, then either H is normal in G, or H has a
> subgroup of index 2 that's normal in G.
I've never seen this before - I'm not sure I believe it - and 
I don't think anything like it is necessary for the problem at hand.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Abstract Algebra
> I'm missing something on this seemingly trivial problem.
> Let G be a group with 3000 elements and H be a subgroup with 1000
> elements.
> If x is in G, show that either x^2 or x^3 is in H.
> Hint:  If [G:H] = 3, then either H is normal in G, or H has a
> subgroup of index 2 that's normal in G.
> I've never seen this before - I'm not sure I believe it - and
> I don't think anything like it is necessary for the problem at hand.
Well, of course it's not *necessary*.  It just makes the problem
easy to solve.
And it's certainly true.  The action of G on the set of cosets of
H gives a surjective homomorphism from G onto a transitive
degree-3 subgroup of S_3, whose kernel lies in H (since the
latter is the point stabilizer of the coset H in this action).
S_3 has only two transitive subgroups in its natural action, so
either H is normal in G and {H, xH, x^2H} =~ C_3, or [H:K] = 2
with K normal in G and {K, xK, x^2K, HK, xHK, x^2HK} =~ S_3
(with the latter 3 cosets squaring to K).
-- 
Jim Heckman
===
Subject: Re: Abstract Algebra
> I'm missing something on this seemingly trivial problem.
> Let G be a group with 3000 elements and H be a subgroup with 1000
elements.
> If
> x is in G, show that either x^2 or x^3 is in H.
> This is what I got so far.
> If x is in H then not only is x^2 in but x^3 is also in H.
> So now we assume that x is not in H.
> IF I can show that H, x^2H and x^3H are pairwise disjoint then G=H U x^2H
U
> x^3H  where each coset has o( H ) =1000 elements. So if x is in G then x
is
> in H or x^2H or x^3H. But x is not in H. So x is in x^2H or x^3H.
> So all that remains to show is that H, x^2H and x^3H are pairwise
disjoint.
> Since x is not in H we have H /= xH-->xH /= x^2H-->x^2H /= x^3H. So x^2 
/=
> x^3H. I just can't seem to show that H /= x^2H ( or x^2 is not in H) and 
H
> /= x^3H (or x^3 is not in H).
> What am I not seeing?
The above is not completely correct. I see now how to do this.
Steven
===
Subject: Re: Abstract Algebra
> I'm missing something on this seemingly trivial problem.
> Let G be a group with 3000 elements and H be a subgroup with 1000 
elements.
> If
> x is in G, show that either x^2 or x^3 is in H.
> This is what I got so far.
> If x is in H then not only is x^2 in but x^3 is also in H.
> So now we assume that x is not in H.
> IF I can show that H, x^2H and x^3H are pairwise disjoint then G=H U x^2H 

U
> x^3H  where each coset has o( H ) =1000 elements. So if x is in G then x 
is
> in H or x^2H or x^3H. But x is not in H. So x is in x^2H or x^3H.
> So all that remains to show is that H, x^2H and x^3H are pairwise 
disjoint.
> Since x is not in H we have H /= xH-->xH /= x^2H-->x^2H /= x^3H. So x^2 
/=
> x^3H. I just can't seem to show that H /= x^2H ( or x^2 is not in H) and 
H
> /= x^3H (or x^3 is not in H).
> What am I not seeing?
If two elements of G are in the same coset of H then their quotient 
is in H.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Help soving this problem?
Trying to get back into school after many years away. I have to take
an admitance exam which I am currently studying for. Here is a problem
on a sample exam that is stumping me . I belive it is a y intercept
problem Any ideas on how to solve? I am thinking you solve for y but
that didn't seem to work for me.
M and N are the X and Y coordinates, respectively, of a point in a
coordinate plane. If the points (M,N) and (M + P, N +4) both lie on
the line defined by the equation X = y/2 - 2/5, what is the value of
P?
===
Subject: Re: Help soving this problem?
So each point must solve the line equation (since they both lie on it)
I.e.
For point (M,N), we have:
M=N/2 - 2/5
And for (M + P, N+4), we have:
M+P = (N+4)/2 -2/5
We substitute the value for M from our first equation into our second
equation:
(N/2 - 2/5) + P = (N+4)/2 -2/5
implies
N/2 - 2/5 + P = N/2 + 2 - 2/5
implies
P = 2
-Darren
> Trying to get back into school after many years away. I have to take
> an admitance exam which I am currently studying for. Here is a problem
> on a sample exam that is stumping me . I belive it is a y intercept
> problem Any ideas on how to solve? I am thinking you solve for y but
> that didn't seem to work for me.
> M and N are the X and Y coordinates, respectively, of a point in a
> coordinate plane. If the points (M,N) and (M + P, N +4) both lie on
> the line defined by the equation X = y/2 - 2/5, what is the value of
> P?
===
Subject: Re: Help soving this problem?
still think the problem is way too hard for liberal arts admitance
test :)
Fletch
> So each point must solve the line equation (since they both lie on it)
> I.e.
> For point (M,N), we have:
> M=N/2 - 2/5
> And for (M + P, N+4), we have:
> M+P = (N+4)/2 -2/5
> We substitute the value for M from our first equation into our second
> equation:
> (N/2 - 2/5) + P = (N+4)/2 -2/5
> implies
> N/2 - 2/5 + P = N/2 + 2 - 2/5
> implies
> P = 2
> -Darren
> Trying to get back into school after many years away. I have to take
> an admitance exam which I am currently studying for. Here is a problem
> on a sample exam that is stumping me . I belive it is a y intercept
> problem Any ideas on how to solve? I am thinking you solve for y but
> that didn't seem to work for me.
> M and N are the X and Y coordinates, respectively, of a point in a
> coordinate plane. If the points (M,N) and (M + P, N +4) both lie on
> the line defined by the equation X = y/2 - 2/5, what is the value of
> P?
===
Subject: Re: Help soving this problem?
>Trying to get back into school after many years away. I have to take
>an admitance exam which I am currently studying for. Here is a problem
>on a sample exam that is stumping me . I belive it is a y intercept
>problem Any ideas on how to solve? I am thinking you solve for y but
>that didn't seem to work for me.
>M and N are the X and Y coordinates, respectively, of a point in a
>coordinate plane. If the points (M,N) and (M + P, N +4) both lie on
>the line defined by the equation X = y/2 - 2/5, what is the value of
1. Solve your equation for y to get the form y = mx + b and read off
the slope m.
2. Compute the slope between your two points and set it equal to your
value of m.
3. Solve for P
--Lynn
===
Subject: Re: Help soving this problem?
>Trying to get back into school after many years away. I have to take
>an admitance exam which I am currently studying for. Here is a problem
>on a sample exam that is stumping me . I belive it is a y intercept
>problem Any ideas on how to solve? I am thinking you solve for y but
>that didn't seem to work for me.
>M and N are the X and Y coordinates, respectively, of a point in a
>coordinate plane. If the points (M,N) and (M + P, N +4) both lie on
>the line defined by the equation X = y/2 - 2/5, what is the value of
>P?
> 1. Solve your equation for y to get the form y = mx + b and read off
> the slope m.
> 2. Compute the slope between your two points and set it equal to your
> value of m.
> 3. Solve for P
> --Lynn
===
Subject: Re: Help soving this problem?
>Trying to get back into school after many years away. I have to take
>an admitance exam which I am currently studying for. Here is a problem
>on a sample exam that is stumping me . I belive it is a y intercept
>problem Any ideas on how to solve? I am thinking you solve for y but
>that didn't seem to work for me.
>M and N are the X and Y coordinates, respectively, of a point in a
>coordinate plane. If the points (M,N) and (M + P, N +4) both lie on
>the line defined by the equation X = y/2 - 2/5, what is the value of
>P?
M = N/2 - 2/5
M+P = (N+4)/2 - 2/5
Subtract the first from the second.
===
Subject: Re: Help soving this problem?
I don't think I follow you...
Fletch
> 
Trying to get back into school after many years away. I have to take
>an admitance exam which I am currently studying for. Here is a problem
>on a sample exam that is stumping me . I belive it is a y intercept
>problem Any ideas on how to solve? I am thinking you solve for y but
>that didn't seem to work for me.
M and N are the X and Y coordinates, respectively, of a point in a
>coordinate plane. If the points (M,N) and (M + P, N +4) both lie on
>the line defined by the equation X = y/2 - 2/5, what is the value of
>P?
> M = N/2 - 2/5
> M+P = (N+4)/2 - 2/5
> Subtract the first from the second.
===
Subject: Graphing polynomial equations
Hi.
Houw would one go about plotting a 3D surface graph of
a quintic or higher equation in three variables?
Like this:
xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0
Obviously you would loop through x and y and calculate
z, and we'd use a numerical approximation technique to
do that. I'm trying to write a program that would plot
the graph. The problem boils down to applying a
numerical algorithm to a quintic like this:
z^5 + a4 z^4 + a3 z^3 + a2 z^2 + a1 z + a0 = 0.
But the problem I'm having is this: Most numerical
approximation algorithms require an initial guess. So, what
would be a good algorithm to find a good initial
guess?
-- 
We should have a town named Alderaan
someday. No, seriously. Let's put it on
the table.
===
Subject: Re: Graphing polynomial equations
>xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0
>Obviously you would loop through x and y and calculate
>z, and we'd use a numerical approximation technique to
>do that. I'm trying to write a program that would plot
>the graph. The problem boils down to applying a
>numerical algorithm to a quintic like this:
>z^5 + a4 z^4 + a3 z^3 + a2 z^2 + a1 z + a0 = 0.
>But the problem I'm having is this: Most numerical
>approximation algorithms require an initial guess. So, what
>would be a good algorithm to find a good initial
>guess?
You probably realize that a degree-5 polynomial can have up to 5 zeroes, so
you would want 5 (or maybe more) initial guesses.  
The high and low guesses should be fairly easy, because there's no such
thing as too high or too low for the tangential method (sorry I forgot who
discovered that or I'd give him credit!).  Even if you're using some other
numerical method you can still use the tangential results to seed it.
You can get the other possibilities by looking at the zeroes of the
derivative, which are potentially relative mimima and maxima.  Any zeroes
other than the first and last will be between a relative maximum and a
relative minimum.  You may have to apply this process recursively until you
get down to degree 2 where you can use the quadratic equation.
Once you're started, you can use results from neighboring (x,y) as guesses.
They won't always work out, but when they do it will save time.
--Keith Lewis              klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: Graphing polynomial equations
>Houw would one go about plotting a 3D surface graph of
>a quintic or higher equation in three variables?
>Like this:
>xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0
Try surf.sourceforge.net .
===
Subject: Re: Graphing polynomial equations
>Houw would one go about plotting a 3D surface graph of
>a quintic or higher equation in three variables?
>Like this:
>xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0
> Try surf.sourceforge.net .
===
Subject: Re: Graphing polynomial equations
> Houw would one go about plotting a 3D surface graph of
> a quintic or higher equation in three variables?
> Like this:
> xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0
> Obviously you would loop through x and y and calculate
> z, and we'd use a numerical approximation technique to
> do that. 
The particular equation you've given is linear in y, 
so if I were to follow your approach I'd loop through 
x and z and calculate y by baby algebra.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Graphing polynomial equations
> Houw would one go about plotting a 3D surface graph of
> a quintic or higher equation in three variables?
> Like this:
> xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0
> Obviously you would loop through x and y and calculate
> z, and we'd use a numerical approximation technique to
> do that.
> The particular equation you've given is linear in y,
> so if I were to follow your approach I'd loop through
> x and z and calculate y by baby algebra.
But what about a GENERAL algorithm for a GENERAL polynomial?
> -- 
> Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Partial Products
Hello Liu Pingkai,
In the following UBASIC example I used different names for the variables 
than before. The code based on your suggestion needs 3M-1 multiplications
(mod N) and 1 inversion (mod N) to calculate M inverses (mod N):
  110   ' Inversion (mod N) of M integers X(0),...X(M-1).
  120   ' Needed 3*(M-1) modular multiplications and 1 modular inversion.
  130   ' A(), B() are auxiliary arrays. V() array holds inverses (mod N).
  140   '
  150   word -20
  160   M=2^10:'          this example uses M=1024 samples
  170   dim X(M),A(M),B(M),V(M)
  180   '
Now fill up array X() with 1024 samples and define N.
  190   A(0)=X(0)
  200   for I=1 to M-2
  210     A(I)=A(I-1)*X(I)@N
  220   next I
  230   U=A(M-2)*X(M-1)@N
  240   V=modinv(U,N)
  250   A(M-2)=V*A(M-2)@N
  260   '
  270   B(0)=X(M-1)*V@N
  280   for I=1 to M-2
  290     B(I)=B(I-1)*X(M-I-1)@N
  300   next I
  310   '
  320   V(0)=B(M-2)
  330   for I=1 to M-2
  340      V(I)=A(I-1)*B(M-I-2)@N
  350   next I
  360   V(M-1)=A(M-2)
  370   end
This method may be faster than direct method based on M modular inversions. 

The improvement depends on the size of M, choice of N, (also on CPU
architecture and language/compiler). This method has the time cost of 1 
inversion equivalent to 3 multiplications.
A.D.
===
Subject: A New Way To Solve Cubics Using A Linear Fractional Transformation
Hello all,
Here's a new take at a well-studied mathematical object:
A New Way To Solve Cubics Using a Linear Fractional Transformation
ABSTRACT: We provide a new method to solve the general cubic equation by
using a linear fractional transformation.  This transformation, sometimes
referred to as a Mobius transformation, can transform the general cubic
into the binomial form, though in a manner different from the traditional
Tschirnhausen transformation that can also transform the cubic into the
binomial form.  The resulting binomial is then simply solved by the 
extraction
of a cube root.
Mathematics Subject Classification.  Primary: 12E12; Secondary: 15A04
http://www.geocities.com/titus_piezas/cubics.html
Just click at the link to the pdf file.
P.S.  There's also a new paper on sextics. Just go to the index.
--Titus
===
Subject: Re: A New Way To Solve Cubics Using A Linear Fractional 
Transformation
> A New Way To Solve Cubics Using a Linear Fractional Transformation
> ABSTRACT: We provide a new method to solve the general cubic equation by
> using a linear fractional transformation.  This transformation, sometimes
> referred to as a Mobius transformation, can transform the general cubic
> into the binomial form, though in a manner different from the traditional
> Tschirnhausen transformation that can also transform the cubic into the
> binomial form.  The resulting binomial is then simply solved by the 
extraction
> of a cube root.
> Mathematics Subject Classification.  Primary: 12E12; Secondary: 15A04
http://www.fc.up.pt/mp/jcsantos/PDF/artigos/Mobius.pdf
Well, OK, it's written in portuguese, but I think that you'll be able
to get the general picture from the equations.
Jose Carlos Santos
===
Subject: Re: A New Way To Solve Cubics Using A Linear Fractional 
Transformation
> A New Way To Solve Cubics Using a Linear Fractional Transformation
> 
> ABSTRACT: We provide a new method to solve the general cubic equation 
by
> using a linear fractional transformation.  This transformation, 
sometimes
> referred to as a Mobius transformation, can transform the general cubic
> into the binomial form, though in a manner different from the 
traditional
> Tschirnhausen transformation that can also transform the cubic into the
> binomial form.  The resulting binomial is then simply solved by the 
extraction
> of a cube root.
> 
> Mathematics Subject Classification.  Primary: 12E12; Secondary: 15A04
> http://www.fc.up.pt/mp/jcsantos/PDF/artigos/Mobius.pdf
> Well, OK, it's written in portuguese, but I think that you'll be able
> to get the general picture from the equations.
> Jose Carlos Santos
     I guess it was bound to happen that the same idea would occur to
different people independently.
     I can't read Portuguese, but, yes, mathematics is the universal
language and I can understand some of your ideas from the equations. 
I can see you also brought up the case when the discriminant is zero.
     I hope you translate your paper.  That way, it'll be accessible
to more people.
--Titus
===
Subject: Re: A New Way To Solve Cubics Using A Linear Fractional 
Transformation
It wasn't published but it has already been accepted for publication.
It will be published in May 2005.
Jose Carlos Santos
===
Subject: Jesus christ What's going on?
I just got on Gooleg where I use to post mesasges and theyre' thing is
all fuceked up now! and i can't nkow how to vie waticles properly
For god sake are they trying tko Kikc us off or what!!
Andrew Usher
===
Subject: Re: Jesus christ What's going on?
> I just got on Gooleg where I use to post mesasges and theyre' thing
is
> all fuceked up now! and i can't nkow how to vie waticles properly
> For god sake are they trying tko Kikc us off or what!!
> Andrew Usher
Some of us have been seeing this for awhile, ever since
we discovered that Google had a beta version of their
server. Apparently they've taken the beta version live.
It seems to be having a hard time under the increased
load. I'm getting a lot of server errors today.
            - Randy
===
Subject: Re: Jesus christ What's going on?
> I just got on Gooleg where I use to post mesasges and theyre' thing
> is
> all fuceked up now! and i can't nkow how to vie waticles properly
> For god sake are they trying tko Kikc us off or what!!
> Andrew Usher
> Some of us have been seeing this for awhile,
a while
> ever since
> we discovered that Google had a beta version of their
> server. Apparently they've taken the beta version live.
Not here. Or maybe, as someone suggested, it went live for a few hours and
then they changed their minds. I've tried the new version now and it does 
my
head in. I think a few more horizontal lines might help. I can see that
something like this is the way to go though (Wow! Look! My name's
highlighted** in a colour box!), so I'll have to get used to it.
In my Favourites I have a GG search for my name and handle. I notice 
that
the new GG lists the hits in a completely different order, and if I click 
on
search by date the number of hits more than halves. Odd. Given that the
two forms of GG have such completely different algorithms, it would worry 
me
if the original version did not continue to be available.
**highlit, anyone?
Adrian (UK)
===
Subject: Re: Jesus christ What's going on?
>I just got on Gooleg where I use to post mesasges and theyre' thing is
>all fuceked up now! and i can't nkow how to vie waticles properly
>For god sake are they trying tko Kikc us off or what!!
>Andrew Usher
<calms down>
I still can't view them properly - I have to use my other news reader
and it's not posting my name at the subject header of the messages.
Why in the hell are they doing this?
Andrew Usher
===
Subject: Re: Jesus christ What's going on?
>>I just got on Gooleg where I use to post mesasges and theyre' thing is
>>all fuceked up now! and i can't nkow how to vie waticles properly
>>For god sake are they trying tko Kikc us off or what!!
>>Andrew Usher
><calms downI still can't view them properly - I have to use my other news reader
>and it's not posting my name at the subject header of the messages.
>Why in the hell are they doing this?
Google is not a news reader. Once you have used a real news reader, you will 

not news :) ).
Bruce
------------------------------
Health nuts are going to feel stupid someday, lying in hospitals
 dying of nothing.
-Redd Foxx 
(if there were any)
===
Subject: Re: Jesus christ What's going on?
>>I just got on Gooleg where I use to post mesasges and theyre' thing is
>>all fuceked up now! and i can't nkow how to vie waticles properly
>>For god sake are they trying tko Kikc us off or what!!
>>Andrew Usher
><calms downI still can't view them properly - I have to use my other news reader
>and it's not posting my name at the subject header of the messages.
>Why in the hell are they doing this?
>Andrew Usher
I hope everyone has been sending feedback to Google instead of just to sci 
groups and a hapless english group.
-- 
Is that plutonium on your gums?
Shut up and kiss me!
  -- Marge and Homer Simpson
===
Subject: Re: Jesus christ What's going on?
> I hope everyone has been sending feedback to Google instead of just to sci 

> groups and a hapless english group.
I have written to them in the past; I think I'll try to compose a
useful critiqe of the new version, when they put it back up.
Andrew Usher
===
Subject: Re: Jesus christ What's going on?
> I hope everyone has been sending feedback to Google instead of just to 
sci 
> groups and a hapless english group.
> I have written to them in the past; I think I'll try to compose a
> useful critiqe of the new version, when they put it back up.
> Andrew Usher
I simply emailed them a link to this thread.  They probably don't
browse everywhere, but plenty of stuff that was said here should be
seen by Google's folks.
===
Subject: Re: Jesus christ What's going on?
Gregory L. Hansen typed thus:
> I hope everyone has been sending feedback to Google instead of just to sci 

> groups and a hapless english group.
I'll have you know that we here at AUE are well provided with haps.  
In fact we've got enough haps between us to keep ourselves hapy for 
some considerable time to come.
-- 
David
replace the first component of address 
===
Subject: Re: Jesus christ What's going on?
>>I just got on Gooleg where I use to post mesasges and theyre' thing is
>>all fuceked up now! and i can't nkow how to vie waticles properly
>>For god sake are they trying tko Kikc us off or what!!
>>Andrew Usher
> <calms down I still can't view them properly - I have to use my other news reader
> and it's not posting my name at the subject header of the messages.
> Why in the hell are they doing this?
date (somewhere near the top right), it should return a list similar to the 

old format, but this format looks somewhat better somehow, though I don't 
like the way text and links run into other blocks of text.
be nice.
Dyl.
-------------------------------
Une  rhapsodie  pour  grapheus.
http://tinylink.com/?J0TDOmaIWK 
===
Subject: Re: Jesus christ What's going on?
> be nice.
I've been using Gougle Groups 2 Beta as much as the original. I like
the new one better.
===
Subject: Re: Jesus christ What's going on?
too,
>> that'd be nice.
> I've been using Gougle Groups 2 Beta as much as the original. I
like
> the new one better.
What happens when you use it to post a message?
Mike.
===
Subject: Re: Jesus christ What's going on?
>  too,
>> that'd be nice.
> I've been using Gougle Groups 2 Beta as much as the original. I
>  like
> the new one better.
> What happens when you use it to post a message?
Hmmm, I see. Something has changed in the last day or so. Any press
release from Google on it?
===
Subject: Re: Jesus christ What's going on?
>>  too,
>> that'd be nice.
> I've been using Gougle Groups 2 Beta as much as the original. I
>>  like
> the new one better.
>> What happens when you use it to post a message?
> Hmmm, I see. Something has changed in the last day or so. Any press
> release from Google on it?
Not that I know of. When I was taking an interest a few months ago I
emailed them with some comments, and they replied courteously but
didn't appear to do anything. I think, as I said the other day,
they're into something the project managers didn't quite understand;
or, which is highly likely, the project team understand but the
commercials made them go public too early (hand up anybody who hasn't
seen this happen elsewhere).
Mike.
===
Subject: Re: Jesus christ What's going on?
   <3ddtq0lmsgld92i29pkrinf7otk6i7qrmt@4ax.com>
   <31825aF37p6pkU1@individual.net>
that'd
> be nice.
I am able to use both the old Google groups and the beta groups with no
problem.
On the other hand, I'm not sure that I appreciate the current
inconsistency in how one navigates. On one visit I may get a handy
list, off to the left, of recently visited groups. Next time around, no
such thing. It does, in fact, change in several different ways, as if
the software is trying to guess where I wish to go before I, myself,
know.
It also doesn't seem to allow you to click on a poster's name and get
an instant archive of everything they've posted.
within a few minutes, whereas the old groups take several hours.
-Mark Martin
===
Subject: Re: Jesus christ What's going on?
>>I just got on Gooleg where I use to post mesasges and theyre' thing is
>>all fuceked up now! and i can't nkow how to vie waticles properly
>>For god sake are they trying tko Kikc us off or what!!
>>Andrew Usher
> <calms down I still can't view them properly - I have to use my other news reader
> and it's not posting my name at the subject header of the messages.
> Why in the hell are they doing this?
> date (somewhere near the top right), it should return a list similar to 
the 
> old format, but this format looks somewhat better somehow, though I don't 

> like the way text and links run into other blocks of text.
> be nice.
I use newreader software, in my case MT-Newswatcher on Mac OS. I'm 
evidently missing all the fun. Do most people get their Usenet from 
Google these days?
===
Subject: Re: Jesus christ What's going on?
> I use newreader software, in my case MT-Newswatcher on Mac OS. I'm 
> evidently missing all the fun. Do most people get their Usenet from 
> Google these days?
There are a couple of benefits, the best one is the easy click on the
author's name, so you can see all their other achived posts by title
and newsgroup.  It makes obvious trolls easy to spot.  And having a
web interface to a usenet archive stretching back to the begining of
the Usenet helps when you need old info and don't want to get flamed
for the crime of asking the incredibly obvious.
Oh, well, there's still Groupsrv, http://www.groupsrv.com/science/ , I
guess.
===
Subject: Re: Jesus christ What's going on?
   <3ddtq0lmsgld92i29pkrinf7otk6i7qrmt@4ax.com>
   <31825aF37p6pkU1@individual.net>
Only temporarily. My preferred newsreader is Agent (free
version is FreeAgent), but my windoze box is down right
now.
Here's a bug I'm surprised they didn't fix in the beta:
The only way to get quoted text in your reply and to
edit it, is to hit preview before entering any text
of your own.
            - Randy
===
Subject: Re: Jesus christ What's going on?
   <3ddtq0lmsgld92i29pkrinf7otk6i7qrmt@4ax.com>
   <31825aF37p6pkU1@individual.net>
Only temporarily. My preferred newsreader is Agent (free
version is FreeAgent), but my windoze box is down right
now.
Here's a bug I'm surprised they didn't fix in the beta:
The only way to get quoted text in your reply and to
edit it, is to hit preview before entering any text
of your own.
            - Randy
===
Subject: Re: Jesus christ What's going on?
>>>I just got on Gooleg where I use to post mesasges and theyre' thing is
>>>all fuceked up now! and i can't nkow how to vie waticles properly
>>>For god sake are they trying tko Kikc us off or what!!
>>>>Andrew Usher
>> <calms down> I still can't view them properly - I have to use my other news reader
>> and it's not posting my name at the subject header of the messages.
>> Why in the hell are they doing this?
>> by
>> date (somewhere near the top right), it should return a list similar to 
>> the
>> old format, but this format looks somewhat better somehow, though I 
don't
>> like the way text and links run into other blocks of text.
>> that'd
>> be nice.
> I use newreader software, in my case MT-Newswatcher on Mac OS. I'm
> evidently missing all the fun. Do most people get their Usenet from
> Google these days?
Dyl.
-------------------------------
Une  rhapsodie  pour  grapheus.
http://tinylink.com/?J0TDOmaIWK
===
Subject: Re: Jesus christ What's going on?
>I just got on Gooleg where I use to post mesasges and theyre' thing is
>all fuceked up now! and i can't nkow how to vie waticles properly
>For god sake are they trying tko Kikc us off or what!!
Andrew Usher
>><calms down>I still can't view them properly - I have to use my other news reader
>>and it's not posting my name at the subject header of the messages.
>>Why in the hell are they doing this?
>by
>date (somewhere near the top right), it should return a list similar to 
>the
>old format, but this format looks somewhat better somehow, though I 
don't
>like the way text and links run into other blocks of text.
>that'd
>be nice.
>>I use newreader software, in my case MT-Newswatcher on Mac OS. I'm
>>evidently missing all the fun. Do most people get their Usenet from
>>Google these days?
-- 
Dirk
The Consensus:-
The political party for the new millenium
http://www.theconsensus.org
===
Subject: Re: Jesus christ What's going on?
 <31825aF37p6pkU1@individual.net>
 <318k2hF37h1s9U1@individual.net> <318onkF38ntouU4@individual.net>
 Discussion, linux)
> groups?
The easiest thing is to have a newsreader that can send HTML requests
to Google and parse the results.  Gnus does this.  For now.
Changing the interface is bound to temporarily break this feature, I
reckon.
-- 
Jesse F. Hughes
Knowing about logic is not the same as being in touch with reality.
                                                 -- David Kastrup
===
Subject: Re: Jesus christ What's going on?
Well, for now, at least.
Dyl. 
===
Subject: Re: Jesus christ What's going on?
   <3ddtq0lmsgld92i29pkrinf7otk6i7qrmt@4ax.com>
   <31825aF37p6pkU1@individual.net>
It's free.
David Ames
===
Subject: Re: Jesus christ What's going on?
   <3ddtq0lmsgld92i29pkrinf7otk6i7qrmt@4ax.com>
   <31825aF37p6pkU1@individual.net>
hoping its back to the usual 8 hour delay.
message goes up instantly, different view options, once you find the
[subject only], [sort by most recent], [show thread view], its just the
same as old groups but instant!
<hide quote> is a pain but can make for quick reading of long threads.
if they could flag the new messages instead of just listing your
subscribed threads it would be better than outlook,  bit hard to find
new threads in 200+ posts but still usable.  Hope your reading this
Google I know you follow my posts!
now groups2-beta is replaced with groups-beta which seems like read
only at the moment, see if this goes up.
pity, been writing my cantor rebuttal all week, absolutely classic make
you change sides!
Herc
===
Subject: Re: Jesus christ What's going on?
> message goes up instantly, different view options, once you find the
> [subject only], [sort by most recent], [show thread view], its just the
> same as old groups but instant!
Goes up instantly? To all Usenet or only to Google? Yes, I got those
viewing options too.
> <hide quote> is a pain but can make for quick reading of long threads.
> if they could flag the new messages instead of just listing your
> subscribed threads it would be better than outlook,  bit hard to find
> new threads in 200+ posts but still usable.  Hope your reading this
> Google I know you follow my posts!
I think you should be able to disable hide quote. 
> now groups2-beta is replaced with groups-beta which seems like read
> only at the moment, see if this goes up.
> pity, been writing my cantor rebuttal all week, absolutely classic make
> you change sides!
Back in the real world, pigs don't fly, and C(R) > C(N).
Andrew Usher
===
Subject: Re: Jesus christ What's going on?
kinds of numbnuts running around there...
 But there is a way to use the old system. Remove the beta- from the
URL, eliminate all from the last slash, hit enter, and you're back
in.-Jitney
===
Subject: Re: Jesus christ What's going on?
> It's free.
> David Ames
It's back, AFAIK, at this time, Noon, NYC time  :-)
  Mark (And as the popularly elected president,
        I would raze the White House with it's
        oval orifice and replace it with a mud
        hut  :-)
===
Subject: Re: Jesus christ What's going on?
> It's back, AFAIK, at this time, Noon, NYC time  :-)
Yes, it's back! But for how long ...
Actually the new version isn't quite as bad as I thought at first.
I got to a view pretty similar to the old one, and the new one now
works properly for threads longer than 250 messages.
I didn't try to post using it, though, can anyone comment on that?
Andrew Usher
===
Subject: Re: Jesus christ What's going on?
k_over_hbarc@yahoo.com exposited:
>I just got on Gooleg where I use to post mesasges and theyre' thing is
>all fuceked up now! and i can't nkow how to vie waticles properly
>For god sake are they trying tko Kikc us off or what!!
>Andrew Usher
> <calms down I still can't view them properly - I have to use my other news reader
> and it's not posting my name at the subject header of the messages.
> Why in the hell are they doing this?
Isn't this where you turn to the camera and say, Why am I asking 
you?
-- 
dg (domain=ccwebster)
===
Subject: Re: Jesus christ What's going on?
   <3ddtq0lmsgld92i29pkrinf7otk6i7qrmt@4ax.com>
Agreed. It's a total disaster. The easy navigation is gone, there are a
lot of aggravating requests, Must activate scripts, etc. A good reasons
for giving up usenet.
===
Subject: inverse modulos
If two numbers x and m are relatively prime, so that gcd(x, m) = 1 is 
true, then x has a unique multiplicative inverse modulo m, a, so that ax 
= 1 (mod m).
Knowing only the multiplicative inverse, a modulo m, and m, is it 
possible to find x?
Is it true that a*x - 1 = k*m, for some k, possibly negative?  Or 
multiple k's?  How to find the multiple k's so that x could be found? 
Is that possible at all, so that the candidate k's can be found knowing 
only the inverse a modulo m and m?
If there is no single way to find it, is there a reasonable trial and 
error process to go through to find x given a and m?
What I suppose I am saying is, what is the relationship between x and 
its inverse a modulo m?
Well, if you caught all that, hope you can help me out :)
Johnathan
===
Subject: Re: inverse modulos
The Euclidean algorithm constructively produces coefficients a,b such
that  ax + bm = gcd(x,m).  So given x,m with gcd(x,m) = 1, one easily
finds the multiplicative inverse a of x modulo m.
You are asking if a were known, could you find x (at least, up to
modulo m), and the answer is yes.  This is just turning the question
around; x is again the multiplicative inverse of a modulo m.
If you do a Google search for Euclidean algorithm, you'll see that it
is a pretty iteration of the division algorithm.
===
Subject: Re: inverse modulos
> If two numbers x and m are relatively prime, so that gcd(x, m) = 1 is
> true, then x has a unique multiplicative inverse modulo m, a, so that
ax
> = 1 (mod m).
> Knowing only the multiplicative inverse, a modulo m, and m, is it
> possible to find x?
X is the modulo-inverse of a and m.
> Is it true that a*x - 1 = k*m, for some k, possibly negative?  Or
> multiple k's?  How to find the multiple k's so that x could be found?
> Is that possible at all, so that the candidate k's can be found
knowing
> only the inverse a modulo m and m?
> If there is no single way to find it, is there a reasonable trial and
> error process to go through to find x given a and m?
> What I suppose I am saying is, what is the relationship between x and
> its inverse a modulo m?
> Well, if you caught all that, hope you can help me out :)
> Johnathan
Here's a Python demo:
' > import gmpy
' > for i in range(1,20):
'       twoo = 2**i
'       twee = 3**i
'       a = gmpy.invert(twoo,twee) # modulo-inverse function
'       x = gmpy.invert(a,twee)    # run it again to find x
'       k = divmod(a*x-1,twee)     # returns quotient, remainder
'       print twoo,twee,a,x,k
'
'
'      2          3         2      2 (mpz(1),      mpz(0))
'      4          9         7      4 (mpz(3),      mpz(0))
'      8         27        17      8 (mpz(5),      mpz(0))
'     16         81        76     16 (mpz(15),     mpz(0))
'     32        243        38     32 (mpz(5),      mpz(0))
'     64        729       262     64 (mpz(23),     mpz(0))
'    128       2187      1589    128 (mpz(93),     mpz(0))
'    256       6561      4075    256 (mpz(159),    mpz(0))
'    512      19683     11879    512 (mpz(309),    mpz(0))
'   1024      59049     35464   1024 (mpz(615),    mpz(0))
'   2048     177147     17732   2048 (mpz(205),    mpz(0))
'   4096     531441    363160   4096 (mpz(2799),   mpz(0))
'   8192    1594323    181580   8192 (mpz(933),    mpz(0))
'  16384    4782969     90790  16384 (mpz(311),    mpz(0))
'  32768   14348907   9611333  32768 (mpz(21949),  mpz(0))
'  65536   43046721  11980120  65536 (mpz(18239),  mpz(0))
' 131072  129140163  92083502 131072 (mpz(93461),  mpz(0))
' 262144  387420489 175181914 262144 (mpz(118535), mpz(0))
' 524288 1162261467 862431935 524288 (mpz(389037), mpz(0))
The gcd of a power of 2 and a power of 3 is always 1, so there
is a valid 'a' for every pair twoo and twee. Once I determined
the modulo-inverse (a) of twoo, twee; I plugged 'a' into the
gmpy.invert function to get 'x' (which is the same as twoo, but
we're pretending we don't know that).
of the calculation is always 0, so the quotient is the legit 'k'.
CC:  mensanator@aol.com
===
Subject: Re: inverse modulos
> X is the modulo-inverse of a and m.
Ah-ha, I thought there must be some kind of relationship (not bright 
enough to find it myself ;-)
> Here's a Python demo:
Lovely, thankyou!!!
But why doesn't it work with:
x = 2385150540210225032
m = 3517
x and m are relatively prime.
a = 3500 (multiplicative inverse of x and m)
?  Or am I misunderstanding something?  (Do I need to get k to find the 
original x?)  Running with this sample of x, m and a gives me 1655 as 
the inverse of a and m, which should have been equal to x.
> ' > import gmpy
> ' > for i in range(1,20):
> '       twoo = 2**i
> '       twee = 3**i
> '       a = gmpy.invert(twoo,twee) # modulo-inverse function
> '       x = gmpy.invert(a,twee)    # run it again to find x
> '       k = divmod(a*x-1,twee)     # returns quotient, remainder
> '       print twoo,twee,a,x,k
> '      2          3         2      2 (mpz(1),      mpz(0))
> '      4          9         7      4 (mpz(3),      mpz(0))
> '      8         27        17      8 (mpz(5),      mpz(0))
> '     16         81        76     16 (mpz(15),     mpz(0))
> '     32        243        38     32 (mpz(5),      mpz(0))
> '     64        729       262     64 (mpz(23),     mpz(0))
> '    128       2187      1589    128 (mpz(93),     mpz(0))
> '    256       6561      4075    256 (mpz(159),    mpz(0))
> '    512      19683     11879    512 (mpz(309),    mpz(0))
> '   1024      59049     35464   1024 (mpz(615),    mpz(0))
> '   2048     177147     17732   2048 (mpz(205),    mpz(0))
> '   4096     531441    363160   4096 (mpz(2799),   mpz(0))
> '   8192    1594323    181580   8192 (mpz(933),    mpz(0))
> '  16384    4782969     90790  16384 (mpz(311),    mpz(0))
> '  32768   14348907   9611333  32768 (mpz(21949),  mpz(0))
> '  65536   43046721  11980120  65536 (mpz(18239),  mpz(0))
> ' 131072  129140163  92083502 131072 (mpz(93461),  mpz(0))
> ' 262144  387420489 175181914 262144 (mpz(118535), mpz(0))
> ' 524288 1162261467 862431935 524288 (mpz(389037), mpz(0))
> The gcd of a power of 2 and a power of 3 is always 1, so there
> is a valid 'a' for every pair twoo and twee. Once I determined
> the modulo-inverse (a) of twoo, twee; I plugged 'a' into the
> gmpy.invert function to get 'x' (which is the same as twoo, but
> we're pretending we don't know that).
> of the calculation is always 0, so the quotient is the legit 'k'.
I wish I could get these results with my numbers... :)
only a and m.
===
Subject: Re: inverse modulos
===
>Subject: Re: inverse modulos
>Message-id: <41af7774$0$25776$5a62ac22@per-qv1-newsreader-01.iinet.net.au> X is the modulo-inverse of a and m.
>Ah-ha, I thought there must be some kind of relationship (not bright 
>enough to find it myself ;-)
>> Here's a Python demo:
>Lovely, thankyou!!!
>But why doesn't it work with:
>x = 2385150540210225032
>m = 3517
Yeah, I guess that only works when x<n. I made a correction in
another thread but you probably haven't seen it yet.
>x and m are relatively prime.
>a = 3500 (multiplicative inverse of x and m)
>?  Or am I misunderstanding something?  (Do I need to get k to find the 
>original x?)  Running with this sample of x, m and a gives me 1655 as 
>the inverse of a and m, which should have been equal to x.
1655 is the first of an infinite series of x's where the inverse modulo is
3500.
When x<n, the first x is the actual one you're looking for, that's why my 
first
example seemed to work. But as you note, 1655 is not the x you're looking 
for. But it is related to 1655 because the kth x whose inverse modulo is 
3500
can be found by
x_k = k*n + x_0
Since we know x_0 is 1655, k is
k = (2385150540210225032-1655)/3517 = 678177577540581
But we only know this because we already know that 
x = 2385150540210225032. If we have just 3500 and 3517, 
we can find x_0 = 1655. That gives us a starting point, but
there are an infinite number of k's so we have no way of 
knowing what the original x was.
>> ' > import gmpy
>> ' > for i in range(1,20):
>> '       twoo = 2**i
>> '       twee = 3**i
>> '       a = gmpy.invert(twoo,twee) # modulo-inverse function
>> '       x = gmpy.invert(a,twee)    # run it again to find x
>> '       k = divmod(a*x-1,twee)     # returns quotient, remainder
>> '       print twoo,twee,a,x,k
>> '
>> '
>> '      2          3         2      2 (mpz(1),      mpz(0))
>> '      4          9         7      4 (mpz(3),      mpz(0))
>> '      8         27        17      8 (mpz(5),      mpz(0))
>> '     16         81        76     16 (mpz(15),     mpz(0))
>> '     32        243        38     32 (mpz(5),      mpz(0))
>> '     64        729       262     64 (mpz(23),     mpz(0))
>> '    128       2187      1589    128 (mpz(93),     mpz(0))
>> '    256       6561      4075    256 (mpz(159),    mpz(0))
>> '    512      19683     11879    512 (mpz(309),    mpz(0))
>> '   1024      59049     35464   1024 (mpz(615),    mpz(0))
>> '   2048     177147     17732   2048 (mpz(205),    mpz(0))
>> '   4096     531441    363160   4096 (mpz(2799),   mpz(0))
>> '   8192    1594323    181580   8192 (mpz(933),    mpz(0))
>> '  16384    4782969     90790  16384 (mpz(311),    mpz(0))
>> '  32768   14348907   9611333  32768 (mpz(21949),  mpz(0))
>> '  65536   43046721  11980120  65536 (mpz(18239),  mpz(0))
>> ' 131072  129140163  92083502 131072 (mpz(93461),  mpz(0))
>> ' 262144  387420489 175181914 262144 (mpz(118535), mpz(0))
>> ' 524288 1162261467 862431935 524288 (mpz(389037), mpz(0))
>> The gcd of a power of 2 and a power of 3 is always 1, so there
>> is a valid 'a' for every pair twoo and twee. Once I determined
>> the modulo-inverse (a) of twoo, twee; I plugged 'a' into the
>> gmpy.invert function to get 'x' (which is the same as twoo, but
>> we're pretending we don't know that).
>> of the calculation is always 0, so the quotient is the legit 'k'.
>I wish I could get these results with my numbers... :)
>only a and m.
-- 
Mensanator
Ace of Clubs
===
Subject: Re: inverse modulos
> If two numbers x and m are relatively prime, so that gcd(x, m) = 1 is 
> true, then x has a unique multiplicative inverse modulo m, a, so that ax 
> = 1 (mod m).
> Knowing only the multiplicative inverse, a modulo m, and m, is it 
> possible to find x?
I'm not sure if I understand what you mean... We are given two numbers a 
and m and we want to find two numbers x and k such that:
a*x - 1 = k*m
or, equivalently:
a*x - k*m = 1.
But this is a very simple linear diophantine equation - so, where is the 
problem...?
Pawel Gladki
CC:  gladki@math.usask.ca
===
Subject: Re: inverse modulos
> I'm not sure if I understand what you mean... We are given two numbers a 
> and m and we want to find two numbers x and k such that:
> a*x - 1 = k*m
> or, equivalently:
> a*x - k*m = 1.
> But this is a very simple linear diophantine equation - so, where is the 
> problem...?
Ok... hang on!  I understand what *you* are asking now.  Ok, just did a 
mathworld.wolfram.com.  This is exactly what I am looking for.  Solving 
for two unknowns.
Is there some way I can solve a diophantine equation computationally, 
i.e. I am using the GNU GMP library and writing programs in C.
Now I need to do it in code.  My papers show:
px + sn = 1, where x and s are unknown, and p = modular inverse of x and 
n, and n is relatively prime to x of course for p to exist.
So rearranging this I get px = 1 + (-s)n, is that the same thing?  (I 
have done this one to verify when I do know x, but now I want to find x 
and s for px + sn = 1 knowing only p and n.)
Johnathan
===
Subject: Re: inverse modulos
>> I'm not sure if I understand what you mean... We are given two numbers 
>> a and m and we want to find two numbers x and k such that:
>> a*x - 1 = k*m
>> or, equivalently:
>> a*x - k*m = 1.
>> But this is a very simple linear diophantine equation - so, where is 
>> the problem...?
> Ok... hang on!  I understand what *you* are asking now.  Ok, just did a 
> mathworld.wolfram.com.  This is exactly what I am looking for.  Solving 
> for two unknowns.
> Is there some way I can solve a diophantine equation computationally, 
> i.e. I am using the GNU GMP library and writing programs in C.
> Now I need to do it in code.  My papers show:
> px + sn = 1, where x and s are unknown
Solving diophantine equations is easy... it uses the extended Euclidean 
algorithm, for details see e.g.:
http://www.win.tue.nl/~ida/demo/c1s3f2.html
Pawel Gladki
===
Subject: Re: inverse modulos
> Solving diophantine equations is easy... it uses the extended Euclidean 
> algorithm, for details see e.g.:
> http://www.win.tue.nl/~ida/demo/c1s3f2.html
use of the extended Euclidean algorithm, but it's not giving the result 
I want.  I thought that the inverse modulo was supposed to be unique 
where it existed, but I must have misunderstood that.  Apparently there 
are many solutions sometimes (or always?)  I haven't yet found a way to 
grab all the solutions and test them for the particular value of x that 
I want.  And even then, if the relationship between x and n and its 
inverse modulo is not unique (same for every possible x solution) then 
perhaps I can't know what the original x was.
Johnathan
===
Subject: Re: inverse modulos
:> Solving diophantine equations is easy... it uses the extended Euclidean 
:> algorithm, for details see e.g.:
:> http://www.win.tue.nl/~ida/demo/c1s3f2.html
: use of the extended Euclidean algorithm, but it's not giving the result 
: I want.  I thought that the inverse modulo was supposed to be unique 
: where it existed, but I must have misunderstood that.  
The inverse of a modulo m is unique *modulo m*, if it exists.  
For example, a = 17, m = 5.  The inverse of 17 modulo 5 is 3, since
17*3 = 51 = 1 + 5*10.  (k=10 in the notation of your earlier posts).
I could add any multiple of m (in this case 5) to 3,  and it would
still be an inverse: for example 8,13,18,... but these are all
congruent modulo 5 (i.e. they differ from each other by a multiple
of 5).  The inverse is unique modulo 5.
In general, if a-b is divisible by m, then we say that a and b are 
congruent modulo m, and write a = b (mod m).  
Ted
===
Subject: Re: inverse modulos
>> Solving diophantine equations is easy... it uses the extended 
>> Euclidean algorithm, for details see e.g.:
>> http://www.win.tue.nl/~ida/demo/c1s3f2.html
> use of the extended Euclidean algorithm, but it's not giving the result 
> I want.  I thought that the inverse modulo was supposed to be unique 
> where it existed, but I must have misunderstood that.  Apparently there 
> are many solutions sometimes (or always?)  
There are _always_ infinitely many solutions. However, since:
a*x - k*m = 1
it follows that gcd(x, m) = 1, so there exists _exactly_ _one_ x which 
satisfies the above equation and is less than m (and bigger than 0, 
too). Indeed, suppose that we have two such x-es, x and x'(and two 
respective k and k'). Then:
a*x - k*m = 1
a*x' - k'*m = 1
Subtracting the second equation from the first gives:
a*(x - x') - m*(k - k') = 0
that is a*(x-x') = m*(k'-k), which yields x-x' = m*(k'-k)/a. But also 
gcd(a,m) = 1, which implies that a divides k'-k, so that (k'-k)/a is an 
integer. That means x and x' differ by multiplicity of m, which is 
possible only if one of them is bigger than m.
Pawel Gladki
===
Subject: Re: inverse modulos
> Solving diophantine equations is easy... it uses the extended
> Euclidean algorithm, for details see e.g.:
> http://www.win.tue.nl/~ida/demo/c1s3f2.html
>> makes use of the extended Euclidean algorithm, but it's not giving
>> the result I want.  I thought that the inverse modulo was supposed
>> to be unique where it existed, but I must have misunderstood that.
>> Apparently there are many solutions sometimes (or always?)
> There are always infinitely many solutions. However, since:
>            ax - km = 1
> it follows that gcd(x, m) = 1, so there exists exactly one x which
> satisfies the above equation and is less than m (and bigger than 0,
> too). Indeed, suppose that we have two such x-es, x and x'(and two
> respective k and k'). Then:
>           ax  - k m = 1
>           ax' - k'm = 1
> Subtracting the second equation from the first gives:
>     a(x - x') - m(k - k') = 0
> that is a(x-x') = m(k'-k), which yields x-x' = m(k'-k)/a. But also
> gcd(a,m) = 1, which implies that a divides k'-k, so that (k'-k)/a is
> an integer. That means x and x' differ by multiplicity of m, which is
> possible only if one of them is bigger than m.
Simpler: Inverses are always unique in any abelian group G. 
For if B has inverses A,C then A = A(BC) = (AB)C = C
Here G = U(Z/m) = unit group of Z/m, the ring of integers (mod m),
i.e. the units (invertibles) of the multiplicative monoid of Z/m.
By the extended Euclidean/Bezout algorithm the units are simply
the cosets  n (mod m) := n + mZ  such that n is coprime to m
since then and only then does there exist integers j,k such that  
    j n + k m = 1  <=>  j n = 1 (mod m)  <=>  n in U(Z/m)
Uniqueness theorems are powerful tools for proving equalities.
For many further less trivial examples see my prior posts
--Bill Dubuque
===
Subject: Re: inverse modulos
>I thought that the inverse modulo was supposed
>to be unique where it existed, but I must have misunderstood that.
>Apparently there are many solutions sometimes (or always?)
>>There are always infinitely many solutions. However, since:
>>           ax - km = 1
>>it follows that gcd(x, m) = 1, so there exists exactly one x which
>>satisfies the above equation and is less than m (and bigger than 0,
>>too). Indeed, suppose that we have two such x-es, x and x'(and two
>>respective k and k'). Then:
>>          ax  - k m = 1
>>          ax' - k'm = 1
>>Subtracting the second equation from the first gives:
>>    a(x - x') - m(k - k') = 0
>>that is a(x-x') = m(k'-k), which yields x-x' = m(k'-k)/a. But also
>>gcd(a,m) = 1, which implies that a divides k'-k, so that (k'-k)/a is
>>an integer. That means x and x' differ by multiplicity of m, which is
>>possible only if one of them is bigger than m.
> Simpler: Inverses are always unique in any abelian group G. (...)
Well, yeah, I know that much :-)
But it seems that the person who asked that question didn't know much 
about groups, rings etc., so I was trying to explain the uniqueness of 
an inverse element in Z_m in the simplest possible terms... hope I 
haven't confused him too much.
Sometimes one has to say difficult things, but one ought to say them as 
simply as one knows how - G. H. Hardy
Pawel Gladki
===
Subject: Re: inverse modulos
> I'm not sure if I understand what you mean...
Let me see if I can explain it a bit better.  Two numbers, I'll call 
them x and n, and the gcd is 1, so they are relatively prime.  I compute 
the modulo inverse of x and n, to get p.
I know only n and p:  how do I find x?
See:  xp = 1 (mod n), so:
xp - 1 is exactly divisible by n.
Here's a concrete example:
x = 2385150540210225032
n = 3517
x and n are relatively prime.
p = 3500, the inverse modulo of x and n.
x*p = 1 (mod n),
and x*p - 1 is divisible by n exactly.
If I know only n and p here, how do I get back the x?
> But this is a very simple linear diophantine equation - so, where is the 
> problem...?
I'm after a numerical solution.  Is it possible to find one?  That is, 
given an unknown x and a known n with a known inverse modulo of those 
two p, can I find x?  (Is there only one x?)
So I can find a/the numerical solution by solving a linear diophantine 
equation?
I hope that I explained it correctly on the first attempt!
Johnathan
===
Subject: Re: inverse modulos
> I'm not sure if I understand what you mean...
> Let me see if I can explain it a bit better.  Two numbers, I'll call 
> them x and n, and the gcd is 1, so they are relatively prime.  I compute 
> the modulo inverse of x and n, to get p.
> I know only n and p:  how do I find x?
> See:  xp = 1 (mod n), so:
> xp - 1 is exactly divisible by n.
> Here's a concrete example:
> x = 2385150540210225032
> n = 3517
Drat! My previous post was not quite correct.
There are an infite number of x's whose modulo 
inverse 3517 result in 3500. Doing the gmpy.invert
on (3500,3517) yields 1655, the first of that
infinite list of x's. Call the first one x_0.
To find x_k:
x_k = k*n + x_0
Here's how the list starts out. Note the modulo 
inverse is 3500 for each x_k:
> for k in range(20):
        print k, k*n+r,gmpy.invert(k*n+r,n)
        
0 1655 3500
1 5172 3500
2 8689 3500
3 12206 3500
4 15723 3500
5 19240 3500
6 22757 3500
7 26274 3500
8 29791 3500
9 33308 3500
10 36825 3500
11 40342 3500
12 43859 3500
13 47376 3500
14 50893 3500
15 54410 3500
16 57927 3500
17 61444 3500
18 64961 3500
19 68478 3500
Your example happens to occur at k = 678177577540581.
> x and n are relatively prime.
> p = 3500, the inverse modulo of x and n.
> x*p = 1 (mod n),
> and x*p - 1 is divisible by n exactly.
> If I know only n and p here, how do I get back the x?
Now that I think I'm doing it correctly, I don't see how.
If you saw a clock with a missing hour hand, you would know
how many minutes past the hour it was, but not what hour.
> But this is a very simple linear diophantine equation - so, where is the 
> problem...?
> I'm after a numerical solution.  Is it possible to find one?  That is, 
> given an unknown x and a known n with a known inverse modulo of those 
> two p, can I find x?  (Is there only one x?)
> So I can find a/the numerical solution by solving a linear diophantine 
> equation?
That's beyond me.
> I hope that I explained it correctly on the first attempt!
> Johnathan
===
Subject: equilateral triangles and their mirror image
My book says word for word:
Suppose a given triangle is directly congruent to its mirror
image. We can be absoilutely certain that this triangle is:
Equilateral
Isosceles
acute
obtuse
The answer, according to my book, is acute. Why?
Why not an Equilateral triangle? Guess what happens
when I construct an Equilateral triangle and stick
it in front of a mirror? It's mirror image is directly
congruent to the actual triangle. So, how in the world
does my book figure that we can't be absolutely certain
that a mirror image of an Equilateral triangle will
always be directly congruent with its mirror image?
Is my book wrong? Or am I missing something?
===
Subject: Re: equilateral triangles and their mirror image
I am afraid that your book is wrong. Any isosceles triangle with a 
vertical angle >= 90 degrees is a counterexample.
The correct alternative is isosceles, not the more restrictive 
equilateral. Proof: (1) Figures that are directly congruent to their 
mirror images possess a symmetry axis, which acts as a mirror by 
definition. (2) In a symmetric triangle one of the sides is necessarily 
perpendicular to the symmetry axis; the other two are each other's 
mirror images and are therefore equal. Done!
To answer your question about what you missed: just nothing; but the 
relevant issue here is that equilateral triangles are not the only 
symmetric ones.
Johan Ernest Mebius
>My book says word for word:
>Suppose a given triangle is directly congruent to its mirror
>image. We can be absoilutely certain that this triangle is:
>Equilateral
>Isosceles
>acute
>obtuse
>The answer, according to my book, is acute. Why?
>Why not an Equilateral triangle? Guess what happens
>when I construct an Equilateral triangle and stick
>it in front of a mirror? It's mirror image is directly
>congruent to the actual triangle. So, how in the world
>does my book figure that we can't be absolutely certain
>that a mirror image of an Equilateral triangle will
>always be directly congruent with its mirror image?
>Is my book wrong? Or am I missing something?
===
Subject: Re: equilateral triangles and their mirror image
> My book says word for word:
> Suppose a given triangle is directly congruent to its mirror
> image. We can be absoilutely certain that this triangle is:
> Equilateral
> Isosceles
> acute
> obtuse
> The answer, according to my book, is acute. Why?
> Why not an Equilateral triangle? Guess what happens
> when I construct an Equilateral triangle and stick
> it in front of a mirror? It's mirror image is directly
> congruent to the actual triangle. So, how in the world
> does my book figure that we can't be absolutely certain
> that a mirror image of an Equilateral triangle will
> always be directly congruent with its mirror image?
> Is my book wrong? Or am I missing something?
Could br picked from a tree
A) fruit
B) apple
C) green
D) red
answer of your book is red, your answer is apple
conclude...
-- 
philippe
(chephip at free dot fr)
===
Subject: Re: censorship screen in sci.math
> It appears to me that posts with titles containing some words as
> Optimal, Strategy, StockMarket, VonNeumann, OS, free are being
> censored and not registering in sci.math or sci.econ.
> Some person or group of persons is deliberately censoring the posts
> of author Archimedes Plutonium to these newsgroups.
Some of your problems are perhaps addressed at
http://www.geocities.com/ResearchTriangle/Lab/1131/ua.txt
and see Q.11 and A.11.
David Ames
===
Subject: Re: censorship screen in sci.math
It's true some messages ( a few) do seem to get lost.
Not only my experience , but a friend's as well.
I suggest reposting ones that don't appear until they do.
>>It appears to me that posts with titles containing some words as
>>Optimal, Strategy, StockMarket, VonNeumann, OS, free are being
>>censored and not registering in sci.math or sci.econ.
>>Some person or group of persons is deliberately censoring the posts
>>of author Archimedes Plutonium to these newsgroups.
>>    
>Some of your problems are perhaps addressed at
>http://www.geocities.com/ResearchTriangle/Lab/1131/ua.txt
>and see Q.11 and A.11.
>David Ames
-- 
Respectfully, Roger L. Bagula
tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 
619-5610814 :
alternative email: rlbtftn@netscape.net
URL :  http://home.earthlink.net/~tftn
===
Subject: Mathematical calendar (long)
Having too much time in my hands, I decided to compile a mathematical
calendar, where each day is celebrated with a mathematical concept
somehow related to the numbers of the date.
This is the result of my waste of time. Note that the dates are written
in the European format (day/month).
January
1/1 - Fibonnaci series: The first two terms of the Fibonacci series are
1,1...
2/1 - Continued fraction for Khinchin's constant: The continued
fraction for Khinchin's constant is [2,1...]
3/1 - Trefoil knot: The symbol for the trefoil knot is 3sub1
4/1 - Square pyramid: A square pyramid is formed of 4 triangles and 1
square
5/1 - Pentagonal pyramid: A pentagonal pyramid is formed of 5 triangles
and 1 square
6/1 - Stevedore's knot: The symbol for Stevedore's knot is 6sub1
7/1 - Cube line picking: The minimum straight line distance between any
two of 7 points placed as far apart as possible in a unit cube is 1
8/1 - Kaprekar numbers: The first Kaprekar number is 9 because 9^2=81
and 8+1=9
9/1 - Tarski's plank problem: Given a circular table of diameter 9
feet, which is the minimal number of planks, 1 foot wide and length
greater that 9 feet, needed in order to completely cover the tabletop?
10/1 - Biaugmented triangular prism: A biaugmented triangular prism has
10 triangles and 1 square
11/1 - Herschel graph: There is only 1 non-Hamiltonian polyhedral graph
in 11 nodes (the least possible), and this is the Herschel graph
12/1 - Gyroelongated square pyramid: A gyroelongated square pyramid has
12 triangles and 1 square
13/1 - Clock solitaire: The probability of winning a clock solitaire is
1/13
14/1 - Stella octangula numbers: The first two stella octangula numbers
are 1,14...
15/1 - Rhombic dodecahedral numbers: The first two rhombic dodecahedral
numbers are 1,15...
16/1 - Biquadratic numbers: The first two biquadratic numbers are
1,16...
17/1 - Magic tesseract constants: The first two magic tesseract
constants are 1,17...
18/1 - Smallest Salem number: The smallest Salem number is 1.18...
19/1 - Foias constant: The value of Foias constant is 1.19...
20/1 - Apery's constant (zeta(3)): The value of Apery's constant is
1.20...
21/1 - Smallest simple perfect square: There is only 1 simple perfect
square of order 21 (the least possible)
22/1 - Inradius of the small rhombicuboctahedron: The inradius of the
small rhombicuboctahedron is 1.22...
23/1 - Ways of coloring an octahedron: There is 1 way of coloring an
octahedron with 1 color, and 23 with 2 colors
24/1 - Midradius of the snub cube: The midradius of the snub cube is
1.24...
25/1 - Bifurcation diagram: The onset of the second bifurcation in the
bifurcation diagram is 1.25
26/1 - Delian constant (cubic root of 2): The value of the Delian
constant is 1.26...
27/1 - In-shuffle: The ordering of cards after an in-shuffle is 27,1...
28/1 - Glaisher-Kinkelin constant: The value of the Glaisher-Kinkelin
constant is 1.28...
29/1 - C29: There is only 1 finite group of order 29, and it is C29
30/1 - Midradius of the dodecahedron: The midradius of the dodecahedron
is 1.30...
31/1 - Mill's constant: The value of Mill's constant is 1.31...
February
1/2 - Natural numbers: The first two natural numbers are 1,2...
2/2 - Continued fraction of the square root of 2: The continued
fraction of the square root of 2 is periodical with 2: [1,2,2,2,2...]
3/2 - Catalan's conjecture: 2^3 and 3^2 are the only consecutive powers
4/2 - Subsets of {1,2}: There are 4 subsets of {1,2}
5/2 - Piet Hein's superellipse: Piet Hein's superellipse formula is
ax^(5/2)+by^(5^2)=1
6/2 - Hexagram: The symbol of the hexagram is {6/2}
7/2 - 2nd Mersenne prime: The 2nd Mersenne prime is 7
8/2 - Star of Lakshmi: The symbol of the star of Lakshmi is {8/2}
9/2 - Tetrahedron circumscribing: The smallest volume of a tetrahedron
circumscribing a parallelepiped of volume 1 is 9/2
10/2 - Binary: 2 is written 10 in binary
11/2 - Decimal expansion of 11: The decimal expansion of 11 is the
first unit fraction with period 2
12/2 - Sphenocorona: A sphenocorona has 12 triangles and 2 squares
13/2 - Satellite knots with 13 crossings: There are 2 satellite knots
with 13 crossings
14/2 - Satellite knots with 14 crossings: There are 2 satellite knots
with 14 crossings
15/2 - 3 circle packing: The minimum diameter of a circle containing 3
unit circles is 2.15...
16/2 - Sphenomegacorona: A sphenomegacorona has 16 triangles and 2
squares
17/2 - Quadratic surfaces: There are 17 types of quadratic surfaces
(defined by polynomials of grade 2)
18/2 - Small universal Turing machine: The smallest known universal
Turing machine with 2 states has 18 colors
19/2 - Monotonic matrices of order 2: There are 19 monotonic matrices
of order 2
20/2 - Inradius of the great rhombicuboctahedron: The inradius of the
great rhombicuboctahedron is 2.20...
21/2 - Mean number of edges for graphs with 7 vertices: The mean number
of edges for graphs with 7 vertices is 21/2
22/2 - Possible last 2 digits of a square: There are 22 possible
combinations for the las 2 digits of a square number
23/2 - Birthday problem: The minimum number of people in a meeting so
that there is more than 50% chance that 2 have their birthdays on the
same day is 23
24/2 - Snub square antiprism: A snub square antiprism has 24 triangles
and 2 squares
25/2 - Smallest hypothenuse with 2 integer right-angle triangles: The
smallest hypothenuse length that is shared by 2 integer right-angle
triangles is 25
26/2 - Midradius of the great rhombicuboctahedron: The midradius of the
great rhombicuboctahedron is 2.26...
27/2 - Sattelite knots with wrapping number bigger than 2: The minimum
crossings of a sattelite knot with wrapping number bigger than 2 is 27
28/2 - 2nd perfect number: The 2nd perfect number is 28
29/2 - Prime generating polynomial: The polynomial 2x^2+29 generates
primes for 0,...,28
March
1/3 - Odd numbers: The first two odd numbers are 1,3...
2/3 - Prime numbers: The first two prime numbers are 2,3...
3/3 - Tetrahedron: The symbol of the tetrahedron is {3,3}
4/3 - Cube: The symbol of the cube is {4,3}
5/3 - Dodecahedron: The symbol of the dodecahedron is {5,3}
6/3 - Hexagonal tesselation: The symbol of the hexagonal tesselation is
{6,3}
7/3 - Plane division by 3 lines: 3 lines can divide the plane in 7
regions
8/3 - Octagram: The symbol of the octagram is {8/3}
9/3 - Sum of cubes: Every integer can be expressed as a sum of at most
9 3rd powers
10/3 - Ternary: 3 is 10 in ternary
11/3 - Mirror images in space groups: There are 11 space groups on 3
dimensions that are mirror images
12/3 - Tricylinder: The intersection of 3 cylinders has 12 curved faces
13/3 - Delannoy numbers: The first two Delannoy numbers are 3,13...
14/3 - Pi: The value of pi is 3.14...
15/3 - Tritriangular numbers: The first two tritriangular numbers are
3,15...
16/3 - Simple directed graphs with 3 nodes: There are 16 simple
directed graphs with 3 nodes
17/3 - Tri-bes: There are 17 tri-bes
18/3 - Least common rolls of 3 dice: The least common rolls of 3 dice
are 3 and 18
19/3 - Braced polygon problem: The minimum number of rods needed to
make a triangle rigid is 3, and for a square is 19
20/3 - Steiner points: The 20 points of the 3 by 3 intersections of the
Pascal lines
21/3 - Binary trees of 3 nodes height: There are 21 binary trees of 3
nodes height
22/3 - Isohedral tilings: There are 3 isohedral tilings with
heptominoes and 22 with octominoes
23/3 - First cluster primes: The 23 first primes starting with 3 are
cluster primes
24/3 - 24-cell: The 24-cell is a polychoron composed of 24 octahedra,
with 3 to an edge
25/3 - Silver constant: The silver constant's value is 3.25...
26/3 - Rubik's cube: A Rubik's cube is a 3x3x3 cube in which the 26
subcubes on the outside can be rotated on any plane of cubes
27/3 - Power tower of 3: 3^3=27
28/3 - Khinchin-Levy constant: The value of Khinchin-Levy constant is
3.28...
29/3 - Confidence interval for a normal distribution: The confidence
interval for a normal distribution with p=0.999 is 3.29
30/3 - Smallest Giuga number: The smallest Giuga number is 30 and has 3
factors
31/3 - 3rd Mersenne prime: The 3rd Mersenne prime is 31
April
1/4 - Square numbers: The first two square numbers are 1,4...
2/4 - Even numbers: The first two even numbers are 2,4...
3/4 - Octahedron: The symbol of the octahedron is {3,4}
4/4 - Square tesselation: The symbol of the square tesselation is {4,4}
5/4 - First Brown pair: The first Brown pair is 4,5
6/4 - Bichromatic graphs with 4 nodes: There are 6 bichromatic graphs
with 4 nodes
7/4 - Archimedean solids: There are 13 Archimedean solids, 7 obtained
by truncation and 4 by expansion
8/4 - Eight-point circle theorem: If a quadrilateral has perpendicular
diagonals, the 4 midpoints of the sides and the 4 feet of the
perpendiculars from the midpoints to the opposite sides are in a circle
that includes the 8 points.
9/4 - Tangent circles of the triangle: There are 4 tangent circles to a
given triangle: the incircle and the 3 excircles, and they are all
touched by the 9-point circle.
10/4 - Quaternary: 4 is 10 in quaternary
11/4 - Bracketings of 4 letters: 4 letters have 11 possible bracketings
12/4 - Construction of an icosahedron from octahedron: If the edges of
an octahedron are divided in the golden ratio proportion such that the
points of division for any face form an equilateral triangle, the 12
points form an icosahedron. This can be done in 4 ways, resulting in 4
possible icosahedra.
13/4 - Busy beaver of 4 states: A busy beaver of 4 states will write 13
1s before halting
14/4 - Tetrabolos: There are 14 tetrabolos, or figures formed by 4
isosceles right triangles joined at the sides
15/4 - 15 puzzle: The 15 puzzle consists of 15 squares numbered from 1
to 15 in a 4x4 square
16/4 - Even square numbers: The first two even square numbers are
4,16...
17/4 - Area of the heptadecagon: The area of the regular heptadecagon
with unit side is 17/4cot(pi/17)
18/4 - Party problem with 4 people: The minimum number of guests that
must be invited to a party so that at least 4 will know each other or
at least 4 won't know each other is 18
19/4 - Sum of biquadratics: Every integer can be expressed as a sum of
19 4th powers
20/4 - Disphenocingulum: A disphenocingulum has 20 triangles and 4
squares
21/4 - Necklaces of 4 white, black or red beads: There are 21 possible
necklaces with 4 white, black or red beads
22/4 - Tetraplets: There are 22 tetraplets, or figures formed by 4
squares joined at the sides or corners
23/4 - Triangle dissections in 4 triangles: There are 23 topologically
different ways of dissecting a triangle into 4 triangles
24/4 - Kissing hyperspheres: 24 hyperspheres can touch another one in 4
dimensions
25/4 - 4th automorphic number day: The 4th automorphic number is 25
26/4 - Plane division by 4 ellipses: 4 ellipses can divide the plane in
26 regions
27/4 - Nine mice problem walk length: If nine mice start at the corners
of a regular nonagon, each heading towards it closest neighboring mouse
at constant speed, they will trace a logarithmic spiral of length
4.27...
28/4 - Euclid's postulates: Euclid proved the first 28 propositions of
the Elements using only the first 4 geometry postulates
29/4 - Universal cellular automaton of von Neumann: von Neumann proved
that a cellular automaton consisting of cells with 4 orthogonal
neighbours and 29 states could simulate a Turing machine
30/4 - Sangaku problem: One sangaku problem asks: how can you
distribute 30 identical spheres such that they are tangent to a single
central sphere and to 4 other small spheres?
May
1/5 - Euler numbers: The first two Euler numbers are 1,5...
2/5 - Sums of two squares: The first two numbers that are sum of two
squares are 2,5...
3/5 - Icosahedron: The symbol of the icosahedron is {3,5}
4/5 - Pentiamonds: There are 4 pentiamonds, or figures formed by 5
equilateral triangles joined at the sides
5/5 - Wong graph: The Wong graph is one of the (5,5) cage graphs
6/5 - Whipple's identity: Whipple's identity is given by the
generalized hypergeometric function 6F5
7/5 - Triakis tetrahedron construction: The triakis tetrahedron can be
constructed by cumulation of a tetrahedron by pyramids of height 7/5
8/5 - Queens problem: The maximum number of queens that can be placed
on a chessboard without any two attacking each other is 8, and the
minimum needed to occupy or attack every square is 5
9/5 - Last digit of odd Catalan numbers: The last digit of the odd
Catalan numbers is 9 or 5 from the second one up to at least the 30th
number
10/5 - Pentatope: A pentatope has 10 triangular faces and 5 tetrahedral
cells
11/5 - Second Brown pair: The second Brown pair is (11,5)
12/5 - Pentominoes: There are 12 pentominoes, or figures formed by 5
squares joined at the sides
13/5 - Wilson primes: The first two Wilson primes are 5,13...
14/5 - Pentagon tilings: There are 14 known tilings with pentagons
(5-sided polygons)
15/5 - Go-moku: Go-moku is a game similar to tic-tac-toe played on a
15x15 board, trying to place 5 pieces in a row
16/5 - Plane division by 5 lines: 5 lines can divide the plane in 16
regions
17/5 - Simple connected tricolorable graphs on 5 nodes: There are 17
simple connected tricolorable graphs on 5 nodes
18/5 - One-sided pentominoes: There are 18 one-sided pentominoes, or
figures formed by 5 squares joined at the sides, when mirror images are
considered different
19/5 - Continued fraction for Champernowne's constant: The continued
fraction for Champernowne's constant has sporadic extremely large
terms, the first two being at positions 5 and 19
20/5 - Tetrahedron 5-compound: The tetrahedron 5-compound is composed
of 5 tetrahedra occupying the 20 vertices of a dodecahedron
21/5 - Connected graphs with 5 nodes: There are 21 connected graphs
with 5 nodes
22/5 - Pentahexes: There are 22 pentahexes, or figures formed by 5
regular hexagons joined at the sides
23/5 - Euclid's Elements: Euclid's Elements started with 23
definitions, 5 postulates and 5 common notions
24/5 - Ruth-Aaron pairs: The first two numbers giving Ruth-Aaron pairs
are 5 and 24
25/5 - Isohedra: The isohedra are 25 finite solids and 5 classes of
infinite solids
26/5 - Space division by 5 planes: 5 planes can divide space in 26
parts
27/5 - Pentakites: There are 27 pentakites, or figures formed by 5 kite
shapes joined at the sides
28/5 - Orders of sociable numbers: The smallest sets of sociable
numbers are two sets with 5 and 28 elements
29/5 - Pentacubes: There are 29 pentacubes, or solids formed by 5 cubes
joined by the faces
30/5 - Octahedron 5-compound: The octahedron 5-compound is composed of
5 octahedra occupying the 30 vertices of an icosidodecahedron
31/5 - Dervish double points: The dervish is a quintic surface (defined
by a polynomial of degree 5) that has the maximum possible number of
double points for a quintic, that is 31
June
1/6 - Golden ratio: The value of the golden ratio is 1.6...
2/6 - Pronic numbers: The first two pronic numbers are 2,6...
3/6 - Triangular tesselation: The symbol of the triangular tesselation
is {3,6}
4/6 - Interprimes: The first two interprimes are 4,6...
5/6 - Dissections of a regular hexagon into a golden rectangle: There
are 5 ways of dissecting a regular hexagon (6-sided polygon) into a
golden rectangle
6/6 - Dissections of a regular hexagon into a Latin cross: There are 6
ways of dissecting a regular hexagon (6-sided polygon) into a Latin
cross
7/6 - Dice: Ordinary dice have the numbers from 1 to 6, in such a way
that opposite faces add up to 7
8/6 - Cuboctahedron: A cuboctahedron has 8 triangles and 6 squares
9/6 - 6-multiperfect numbers: 6-multiperfect numbers have at least 9
prime factors
10/6 - Dodecahedron 6-compound: A dodecahedron 6-compound is composed
of 6 dodecahedra, each rotated by 1/10 of a turn about the line joining
the centroids of opposite faces
11/6 - Nets for the cube: There are 11 nets for the cube, which has 6
faces
12/6 - Hexiamonds: There are 12 hexiamonds, or figures formed by 6
equilateral triangles joined by the sides
13/6 - Sums of 6 cubes: The first two numbers that are a sum of six
cubes are 6,13...
14/6 - Necklaces of 6 black or white beads: There are 14 possible
necklaces of 6 black or white beads, counting mirror images
15/6 - Odd elements in Pascal's triangle: There are 15 odd elements in
the first 6 rows of Pascal's triangle
16/6 - Random walk of 4 steps: The probability of being back to the
origin after a random walk of 4 steps is 6/16
17/6 - Golomb ruler with 6 marks: The length of the optimal Golomb
ruler with 6 marks is 17
18/6 - Even pentagonal pyramidal numbers: The first two even pentagonal
pyramidal numbers are 6,18...
19/6 - Graphs with two centres on 6 nodes: There are 19 graphs with two
centres in 6 nodes
20/6 - 6 equilateral triangles in a convex figure: The maximum area
possible for 6 equilateral triangles forming a convex figure is 20
times the size of the smallest one
21/6 - Hyperperfect numbers: The first two hyperperfect numbers are
6,21...
22/6 - Plane division by 6 lines: 6 lines can divide the plane in 22
regions
23/6 - Home prime for 6: The home prime for 6 is 23
24/6 - Tetrahedron cutting: A tetrahedron cut by 6 planes, each passing
through an edge and bisecting the opposite edge, is sliced in 24 pieces
25/6 - Smallest 6th power that is a sum of 6th powers day: The smallest
6th power that is a sum of 6th powers is 25^6
26/6 - 6-snake: The maximum length of a 6-snake is 26
27/6 - Solomon's seal lines: Solomon's seal lines are 27 real or
imaginary lines which lie on the general cubic surface, that can put
into a one-to-one correspondence with the vertices of a polytope in
6-dimensional space
28/6 - Perfect numbers: The first two perfect numbers are 6,28...
29/6 - Graphs with one centre on 6 nodes: There are 29 graphs with one
centre on 6 nodes
30/6 - Longimeter: A longimeter is a transparent sheet of plastic with
a regular grid of lines at an angle of 30 degrees. By counting the
number of squares occupied by a linear feature on a map for 6 different
rotations, the length of the feature can be determined
July
1/7 - Hex numbers: The first two hex numbers are 1,7...
2/7 - e: The value of e is 2.7...
3/7 - Cousin primes: The first pair of cousin primes is 3 and 7
4/7 - Feigenbaum constant: The value of the Feigenbaum constant is
4.7...
5/7 - Ass and mule problem: A classic by Euclid. The mule says to the
ass: If you gave me one of your sacks, I would have as many as you.
The ass replies, If you gave me one of your sacks, I would have twice
as many as you. The solution is 5 sacks for the mule and 7 for the
ass.
6/7 - 7 tree plantation problem: The maximum number of rows of three
trees you can have with 7 trees is 6 rows
7/7 - Sum of prime factors of 7: The sum of prime factors of 7 is 7
8/7 - Compositions of 7 of length 2: There are 8 compositions of 7 of
length 2
9/7 - Angles that are solved using a bicubic equation: The
trigonometric functions of the angles pi/7 and pi/9 are resolved using
a bicubic equation
10/7 - Sum of random numbers exceeding 5: The expected number of picks
in (0,1) so that the sum exceeds 5 is 10.7...
11/7 - Inequality of Robin's theorem: The first two numbers that hold
the inequality of Robin's theorem are 7,11...
12/7 - Harmonic mean of the divisors of 4: The harmonic mean of the
divisors of 4 is 12/7
13/7 - Euler's 6n+1 theorem: Every prime of the form 6n+1 (the first
two of them are 7,13...) can be written in the form x^2+3y^2
14/7 - Heawood graph: The Heawood graph has 14 nodes and represents the
7-color torus map
15/7 - Sums of 4 squares: The first two numbers that are sums of 4
squares are 7,15...
16/7 - Collatz sequence for 7: The Collatz sequence for 7 has 16 steps
17/7 - Full reptend primes: The first two full reptend primes are
7,17...
18/7 - Multiplication of 2x2 matrices: The minimum number of operations
needed to multiply two 2x2 matrices is 7 multiplications and 18
additions
19/7 - e approximation: A good fractional approximation of e is 19/7
20/7 - Steiner points: The Pascal lines intersect three at a time in 20
Steiner points. In the case of a regular hexagon inscribed in a cirle,
they degenerate into 7 points (the vertices and center of a regular
hexagon)
21/7 - Kobon triangles constructed with 7 lines: The maximum number of
Kobon triangles that can be constructed with 7 lines is 21
22/7 - Pi approximation: A good fractional approximation of pi is 22/7
23/7 - Pi Bible reference: There is a reference of pi in the Bible that
gives it a value of 3, which is Kings 7:23
24/7 - Heptiamonds: There are 24 heptiamonds, or figures formed by 7
equilateral triangles joined at the sides
25/7 - Smallest pseudoprime in base 7: The smallest pseudoprime in base
7 is 25
26/7 - Langford sequences with 7 digits: There are 26 Langford
sequences with 7 digits
27/7 - Giuga sequences of length 7: There are 27 Giuga sequences of
length 7
28/7 - Bitangents of the quartic curve: There are 28 bitangents on the
quartic curve, that can be put into a one-to-one correspondence with
the vertices of a polytope in 7-dimensional space
29/7 - Plane division by 7 lines: 7 lines can divide the plane in 29
regions
30/7 - 7th 3-almost prime: The 7th 3-almost prime is 30
31/7 - Graph cycles in a wheel graph with 7 nodes: There are 31 graph
cycles in a wheel graph with 7 nodes
August
1/8 - Cubic numbers: The first two cubic numbers are 1,8...
2/8 - Fransen-Robinson constant: The value of the Fransen-Robinson
constant is 2.8...
3/8 - Levi graph: The Levi graph is the unique (3,8) cage graph
4/8 - Squareful numbers: The first two squareful numbers are 4,8...
5/8 - Sums of two primes: The first two sums of two primes are 5,8...
6/8 - Truncated octahedron: The truncated octahedron has 6 squares and
8 hexagons
7/8 - Incomparable rectangles tiling: At least 7 and at most 8 mutually
incomparable rectangles are needed to tile a given rectangle
8/8 - Rooks problem: 8 rooks can be placed on a chessboard without any
two attacking, and 8 is also the minimum number of pieces needed to
occupy or attack every square
9/8 - Bailey's transformation: Bailey's transformation is given by the
generalized hypergeometric function 9F8
10/8 - Octal: 8 is written 10 in octal
11/8 - Nets for the octahedron: There are 11 nets for the octahedron,
which has 8 faces
12/8 - Matchstick graph of degree 3: The matchstick graph of degree 3
has 12 edges and 8 vertices
13/8 - Octohexes with holes: There are 13 octohexes with holes, or
figures formed by 8 regular hexagons joined at the sides
14/8 - Bishops problem: 14 bishops can be placed on a chessboard
without any two attacking, and 8 bishops are the the minimum number of
pieces needed to occupy or attack every square
15/8 - Pawn positions: There are 15^8 possible pawn possitions in chess
without captures
16/8 - Cubeful numbers: The first two cubeful numbers are 8,16...
17/8 - Wallpaper groups: There are 17 wallpaper groups, 8 of them pure
translation
18/8 - Small rhombicuboctahedron: The small rhombicuboctahedron has 8
triangles and 18 squares
19/8 - 8th prime: The 8th prime is 19
20/8 - Stella octangula construction from the dodecahedron: The stella
octangula can be constructed using 8 of the 20 vertices of the
dodecahedron
21/8 - Prime knots with 8 crossings: There are 21 prime knots with 8
crossings
22/8 - Partitions of 8: There are 22 partitions of 8
23/8 - Sums of 8 biquadratics: The first two numbers that are sums of 8
biquadratics are 8,23...
24/8 - Tesseract: A tesseract has 24 square faces and 8 cubic cells
25/8 - 8th lucky number: The 8th lucky number is 25
26/8 - Even heptagonal pyramidal numbers: The first two even heptagonal
pyramidal numbers are 8,26...
27/8 - Prime third powers: The first two prime third powers are 8,27...
28/8 - 8th positive fundamental discriminant: The 8th positive
fundamental discriminant is 28
29/8 - 8th Hilbert number: The 8th Hilbert number is 29
30/8 - Rule 30: Rule 30 is one of the elementary linear cellular
automaton rules, where each of the 8 possible combinations of colors of
a cell and its neighbours are encoded in the binary representation
00011110=30
31/8 - Gomory's theorem: Regardless of where one white and one black
square are deleted from an 8x8 chessboard, the reduced board can always
be covered exactly with 31 dominoes
September
1/9 - Centered cube numbers: The first two centered cube numbers are
1,9...
2/9 - Sums of two cubes: The first two sums of two cubes are 2,9...
3/9 - Cullen numbers: The first two Cullen numbers are 3,9...
4/9 - Prime second powers: The first two prime second powers are 4,9...
5/9 - Triangular dipyramid: A triangular dipyramid has 5 vertices and 9
edges
6/9 - Sum of prime factors of 9: The sum of prime factors of 9 is 6
7/9 - Percentage of numbers starting with 5: 7.9% of numbers in
listings will start with 5
8/9 - Isolated singularities on a cubic surface: There are 9 possible
types of isolated singularities on a cubic surface, 8 of them rational
double points
9/9 - Anisohedral nonominoes: There are 9 anisohedral nonominoes, or
figures formed by 9 squares joined at the sides
10/9 - 9 tree plantation problem: The maximum number of rows of three
threes you can have with 9 trees is 10 rows
11/9 - Dissections of the eneagon into a Greek cross: There are 11 ways
of dissecting a regular eneagon (9-sided polygon) into a Greek cross
12/9 - Partitions of 9! of length 9: There are 12 partitions of 9! of
length 9
13/9 - Chromatic number of a surface of genus 9: The chromatic number
of a surface of genus 9 is 13
14/9 - Coin tossing paradox: The expected wait until one sees the
combination THTH is 20, and to see HTHH is 18, but the probability that
THTH occurs before HTHH is 9/14
15/9 - Triaugmented dodecahedron: The triaugmented dodecahedron has 15
triangles and 9 pentagons
16/9 - Kings problem: 16 kings can be placed in a chessboard without
any two attacking each other, and 9 kings is the minimum number needed
to occupy or attack every square
17/9 - 9th number with an odd number of prime factors: 17 is the 9th
number with an odd number of prime factors
18/9 - Mean number of edges for a graph with 9 vertices: The mean
number of edges for a graph with 9 vertices is 18
19/9 - Collatz sequence for 9: The Collatz sequence for 9 has 19 steps
20/9 - Harmonic mean of the divisors of 10: The harmonic mean of the
divisors of 10 is 20/9
21/9 - Generalized Moore graphs with 9 nodes: There are 21 generalized
Moore graphs with 9 nodes
22/9 - Kempner series for 9: The value of the Kempner series for 9 is
22.9...
23/9 - Sum of 9 positive cubes: 23 is the smallest number that is sum
of 9 positive cubes
24/9 - Sum of 9 biquadratics: The first two numbers that are sum of 9
biquadratics are 9,24...
25/9 - Nonattacking kings in a 9x9 chessboard: 25 nonattacking kings
can be placed in a 9x9 chessboard
26/9 - Katona's problem on the letters of the alphabet: The solution of
Katona's problem for the 26 letters of the alphabet is that they can be
separated by a family of 9
27/9 - van der Waerden numbers: The smallest nontrivial van der Waerden
numbers are 9 and 27
28/9 - Smallest pseudoprime in base 9: The smallest pseudoprime in base
9 is 28
29/9 - Varga's constant: The value of Varga's constant is 9.29...
30/9 - Partitions of 9: There are 30 partitions of 9
October
1/10 - Catalan's constant: The value of Catalan's constant is 1.10...
2/10 - Unordered factorizations of 10: There are 2 unordered
factorizations of 10
3/10 - Sums of 3 cubes: The first two sums of 3 cubes are 3,10...
4/10 - Even tetrahedral numbers: The first two even tetrahedral numbers
are 4,10...
5/10 - Hypotenuses for a right integer triangle: The smallest two
possible hypothenuses for a right integer triangle are 5 and 10
6/10 - Regular polychora: There are 6 convex polychora and 10 stellated
polychora
7/10 - Sum of prime factors of 10: The sum of prime factors of 10 is 7
8/10 - Square bicupolas: The square bicupolas have 8 triangles and 10
squares
9/10 - Star polychora: 9 of the 10 star polychora have the same
vertices as the hexacosichoron
10/10 - Biaugmented dodecahedrons: The biaugmented dodecahedrons have
10 triangles and 10 pentagons
11/10 - Totient function of 11: The totient function of 11 is 10
12/10 - 10 tree plantation problem: The maximum number of rows of three
trees you can have with 10 trees is 12 rows
13/10 - Smallest sum of 10th powers equal to another 10th power: 13 is
the smallest known number of 10th powers whose sum equals another 10th
power
14/10 - Possible solitary numbers: The two smallest numbers that are
suspected to be solitary, but not known for certain, are 10 and 14
15/10 - 10th non-Egyptian number: The 10th non-Egyptian number is 15
16/10 - Hexadecimal: 16 in hexadecimal is 10
17/10 - Cumulative digit sum of binary numbers up to 10: The cumulative
digit sum of binary numbers up to 10 gives 17
18/10 - 10th composite number: The 10th composite number is 18
19/10 - Reverse-then-add algorithm: The smallest number to require one
iteration to reach a palindrome in the reverse-then-add algorithm is
10, and the smallest to require two iterations is 19
20/10 - Decreasing decimal numbers: The first two decreasing decimal
numbers are 10,20...
21/10 - Cubic graphs on 10 nodes: There are 21 cubic graphs on 10 nodes
22/10 - Square dissection into acute isosceles triangles: There are two
known dissections of the square into acute isosceles triangles, one
requiring 10 triangles and the other 22
23/10 - Hilbert's problems: Hilbert's problems are a set of originally
unsolved problems in mathematics proposed by Hilbert. 10 of them were
presented at the Second International Congress in Paris on 1900.
24/10 - Gyroelongated square bicupola day: The gyroelongated square
bicupola has 24 triangles and 10 squares
25/10 - Sums of 10 biquadratics: The first sums of 10 biquadratics are
10,25...
26/10 - Noncototients: The first two noncototients are 10,26...
27/10 - Eckardt points day: The Eckardt points are the 10 points where
three of the 27 Solomon's seal lines meet
28/10 - Non-hamiltonian symmetric cubic graphs: There two smallest
non-hamiltonian symmetric cubic graphs are the Petersen graph, with 10
nodes, and the Coxeter graph, with 28 nodes
29/10 - 10th prime: The 10th prime is 29
30/10 - Percentage of numbers starting with 1: 30.10% of all numbers in
listings will start with 1
31/10 - Partitions of 10! of length 10: There are 31 partitions of 10!
of length 10
November
1/11 - Repunits: The first two repunits are 1,11...
2/11 - Dihedral primes: The first two dihedral primes are 2,11...
3/11 - Unique primes day: The first two unique primes are 3,11...
4/11 - Sums of 4 cubes: The first two sums of 4 cubes are 4,11...
5/11 - Sexy primes: The first pair of sexy primes is 5 and 11
6/11 - Russian roulette problem: If a bullet is put in one of the six
chambers of a revolver, and two duelist alternately spin the chamber
and fire at themselves until one is killed, what is the probability
that the first duelist is killed? The answer is 6/11
7/11 - Josephus problem: 11 men are arranged in a circle. Every second
man will be executed going around the circle, until only one remains.
Where should you stand to be the survirvor? The answer is in the 7th
place
8/11 - Perfect shuffles: After 11 perfect shuffles, the first card will
be in position 8
9/11 - Remainder when the 11th composite number is divided by 11: When
the 11th composite number (21) is divided by 11, the remainder is 9
10/11 - Most common rolls of 3 dice: The most common rolls of 3 dice
are 10 and 11
11/11 - Sum of prime factors for 11: The sum of prime factors for 11 is
11
12/11 - Partitions of 11 into distinct parts: There are 12 partitions
of 11 into distinct parts
13/11 - Amphichiral Archimedean solids: 11 of the 13 Archimedean solids
are amphichiral
14/11 - Collatz sequence for 11 day: The Collatz sequence for 11 has 14
steps
15/11 - Smallest pseudoprime in base 11: The smallest pseudoprime in
base 11 is 15
16/11 - 11 tree plantation problem: The maximum number of rows of three
trees you can have with 11 trees is 16
17/11 - 11th prime power: The 11th prime power is 17
18/11 - Herschel graph day: The Herschel graph has 11 nodes and 18
edges
19/11 - Euler-Mascheroni constant approximation: A good fractional
approximation of the Euler-Mascheroni constant is 11/19
20/11 - Uniform tessellations: There are 11 1-uniform tesselations and
20 2-uniform tesselations
21/11 - 11th odd number: The 11th odd number is 21
22/11 - Perfect rectangles of order 11: There are 22 perfect rectangles
of order 11
23/11 - Smallest prime dividing the 11th Mersenne number: The smallest
prime that divides the 11th Mersenne number is 23
24/11 - Strictly Egyptian numbers: The first two strictly Egyptian
numbera are 11,24...
25/11 - Hendecaiamonds with holes: There are 25 hendecaiamonds with
holes, or figures formed by 11 equilateral triangles joined at the
sides
26/11 - Odd perfect numbers: If an odd perfect number exists, and it
isn't divisible by 3, it must have at least 11 distinct prime factors.
If it isn't divisible by 3, 5 or 7, it has at least 26 distinct prime
factors
27/11 - 11th plaindrome: The 11th plaindrome is 27
28/11 - 11th squareful number: The 11th squareful number is 28
29/11 - Least primes such that floor((n^n)/p) is prime: The two first
terms greater than two in this sequence are 11,29...
30/11 - Unattacked squares that 11 queens can leave in a 11x11
chessboard: The maximum number of unattacked squares that 11 queens can
leave in an 11x11 chessboard is 30
December
1/12 - Triangle triangle picking problem: The mean area of a triangle
with vertices picked inside a triangle with unit area is 1/12
2/12 - Least common rolls of 2 dice: The least common rolls of two dice
are 2 and 12
3/12 - Kissing spheres: 12 spheres can touch another one of the same
size in 3 dimensions
4/12 - Rational values of sin(2pi/n): The value of sin(2pi/n) is
irrational for all n except 4 and 12
5/12 - Sums of 5 cubes day: The first two sums of 5 cubes are 5,12...
6/12 - Octahedral graph: The octahedral graph has 6 nodes and 12 edges
7/12 - Dissections of a dodecagon into a golden rectangle: There are 7
dissections of a dodecagon (12-sided polygon) into a golden rectangle
8/12 - Cubical graph: The cubical graph has 8 nodes and 12 edges
9/12 - Collatz sequence for 12: The Collatz sequence for 12 has 9 steps
10/12 - Duodecimal: 12 is 10 in duodecimal
11/12 - Three book stacking problem: A stack of three books can
protrude 11/12 book lengths over the edge of a table without falling
over
12/12 - Dodecadodecahedron day: A dodecadodecahedron is formed of 12
pentagrams and 12 pentagons
13/12 - Sparse polynomial square: The minimum degree for a polynomial
having a sparse square is 13, with a square of degree 12
14/12 - Stomachion: The stomachion is a 14-piece dissection puzzle made
from a 12x12 square
15/12 - Partitions of 12 into distinct parts: There are 15 partitions
of 12 into distinct parts
16/12 - 12th Stormer number: The 12th Stormer number is 16
17/12 - Home plate error: The shape of the home plate in baseball is an
irregular pentagon. The specification of the shape requires the
existence of an isosceles right triangle of legs 12 and hypothenuse 17,
which isn't correct.
18/12 - Abundant numbers: The first two abundant numbers are 12,18...
19/12 - 12 tree plantation problem: The maximum number of rows of three
trees you can have with 12 trees is 19
20/12 - Icosidodecahedron: The icosidodecahedron has 20 triangles and
12 pentagons
21/12 - 12th non-Egyptian number: The 12th non-Egyptian number is 21
22/12 - Uniquely three-colorable graph: The smallest uniquely
three-colorable graph has 12 vertices and 22 edges
23/12 - 12th odd number: The 12th odd number is 23
24/12 - Cuboctahedral graph: The cuboctahedral graph has 12 nodes and
24 edges
25/12 - 4th harmonic number: The 4th harmonic number is 25/12
26/12 - Totient function of 26: The totient function of 26 is 12
27/12 - Sums of 12 biquadratics: The first two sums of 12 biquadratics
are 12,27...
28/12 - 12th negabinary number: The 12th negabinary number is 28
29/12 - 12th metadrome: The 12th metadrome is 29
30/12 - Icosahedral graph: The icosahedral graph has 12 vertices and 30
edges
31/12 - 12th product of an odd number of distinct prime factors: The
12th product of an odd number of distinct prime factors is 31
===
Subject: Re: Mathematical calendar (long)
A very nice idea!
Richard Schorn
===
Subject: Re: Mathematical calendar (long)
   <31eoqhF3bffsqU1@individual.net>
hmmm modern day numerology introduced by Doly and
received with SIN X and SCORN.
Despite my personal disposition on the idea, it seems the heavens may
consider this false idol
Herc
===
Subject: Re: Mathematical calendar (long)
> Having too much time in my hands, I decided to compile a mathematical
> calendar, where each day is celebrated with a mathematical concept
> somehow related to the numbers of the date.
> This is the result of my waste of time. Note that the dates are written
> in the European format (day/month).
That's really great! If there were additional pictures and maybe a 
little more info, I would definitelly buy one for myself (and 6 more for 
gifts for my teachers :-)
sirix.
P.S. There could be also some info about current events, but of course 
only as an addition to the main day motto.
===
Subject: Re: Most Distorted Triangle
===
Subject: Re: Most Distorted Triangle
> Is there a most scalene/ asymmetric/distorted triangle, most remote
> from the equilateral triangle? If so, what may be a proper criterion
> or definition of asymmetry?
> Petitjean gives a result,but I could not follow
> what the criterion is from links to his work he provided.
The general criterion applies for figures more complicated that
triangles (distributions). It is why it cannot be summarized
quickly (done on my web site).
In the case of a set of n points, we look for the sum of the
sqaured distances between the n points and its mirrored image,
and we minimize for all rotations and translations. If we find
this sum is zero, it means that the set is achiral. For a triangle
in the plane, it means that it is isocele. The sum is normalized
to the inertia (sum of squred distances to the center of gravity).
There is a family of triangles (all have the same sides ratios,
i.e. the same shape), for which the normalized sum is maximized:
this is the most chiral triangle.
Strictly speaking, this triangle (or any scaled or/and rotated
or/and translated copy) is the farthest from the isocele triangle.
There is an other one with a similar criterion for direct symmetry
rather than indirect symmetry, which, in your case could be viewed
as the farthest from the equilateral triangle. The shape is
simpler, and buildable with ruler and compass: angles are
pi/4, pi/8, 5pi/8. Alas, for reasons a bit long to write here,
the theory does not work for continuous sets (nevertheless work
for n points in the euclidean d-space). Details on:
http://petitjeanmichel.free.fr/itoweb.petitjean.symmetry.html
Final remark: the most distorted triangle could mean:
- farthest from isocele: compare it to its mirror image
- farthest from equilateral: compare it to itself, rotating
- fartest from half rectangle (any criterion you like)
My question is: which relative weights would you set for these
three criterions, in order to define an overall criterion ?
Michel Petitjean,                     Email: petitjean@itodys.jussieu.fr
ITODYS (CNRS, UMR 7086)                      ptitjean@ccr.jussieu.fr
http://petitjeanmichel.free.fr/itoweb.petitjean.html
===
Subject: ADV:      Staff Announcement
        boundary=1..A4_6A.FA6.D_EAB5.E38.
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) with ESMTP id iB2BIqI13956;
        by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, 
$Revision: 1.6  secondary) with SMTP id iB2BImX2012059;
---------------------------------------------------------------------
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allotment of BRAND NEW, top of-the-line, name-brand desktop computers
at more than 50% off MSRP to all Teachers, Students,Faculty and Staff, 
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Call Avtech Direct
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===
Subject: Re: Staff Announcement
Actech advertised:
> * IBM Processor for amazing speed and performance
Which IBM processor? 
===
Subject: Re: ADV:      Staff Announcement
Administrator <bulletin@staffadministrator.com> scribbled the 
following:
> Attention All School Staff: Teachers, Students and Faculty Members:
Why? What will happen if they don't?
-- 
/-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland --------
-------------------------------------------------------- rules! --------/
You have moved your mouse, for these changes to take effect you must shut 
down
and restart your computer. Do you want to restart your computer now?
   - Karri Kalpio
===
Subject: need proof!!!
i resently asked the following question
I'm looking for a continuous function f:R->R with discontinuity on
irrational domain and continuous on Q.
the good people who responded, told me that the answer is NO.
now, does anyone knows a formal proof for this...
but if you dont know just tell me why
===
Subject: Re: need proof!!!
> i resently asked the following question
> I'm looking for a continuous function f:R->R with discontinuity on
> irrational domain and continuous on Q.
> the good people who responded, told me that the answer is NO.
> now, does anyone knows a formal proof for this...
> but if you dont know just tell me why
Suppose f:R->R is an arbitrary real function.
For n in N, n>=1 and y in R let
A_n(y) = f^-1[(y-1/n, y+1/n)]
U_n(y) = int A_n(y) where int = topological interior
U_n = UNION {U_n(y) | y in R}
C = INTERSECTION {U_n | n in N, n>=1}
If f is continuous at x:
x in int(f^-1[O]) whenever O is open and f(x) in O
so x in U_n(f(x)) for all n
so x in U_n for all n
so x in C
If x in C:
For all n, x in U_n
So for each n we can find y_n in R with x in U_n(y_n)
So for all n, x in int f^-1[(y_n-1/n, y_n+1/n)]
Thus for all n, |f(x)-y_n| < 1/n
and also for all n there is d_n>0 such that
|x-z| < d_n => |f(z)-y_n| < 1/n
Suppose e>0 and pick N such that 1/N < e/2
|x-z| < d_N => |f(z)-f(x)| <= |f(z)-y_N| + |y_N-f(x)|
                            <  1/N        + 1/N
                            <  e
Hence f is continuous at x.
So C is precisely the set of points at which f is continuous.
Each U_n is open since it is a union of open sets and so C is a G_delta 
set as required.
If Q were a G_delta set, R-Q would be an F_sigma set, so there would be 
a sequence of closed sets F_0, F_1, ... with R-Q = UNION {F_n | n in N}
Each F_i contains no rational and so is nowhere dense.
Let (q_n) be an enumeration of Q. Each {q_n} is also a nowhere dense 
set. Then R = F_0 U F_1 U ... U q_0 U q_1 U ... is a countable union of 
nowhere dense sets, which contradicts Baire's category theorem.
Hence Q is not a G_delta set.
===
Subject: Re: need proof!!!
> i resently asked the following question
> I'm looking for a continuous function f:R->R with discontinuity on
> irrational domain and continuous on Q.
> the good people who responded, told me that the answer is NO.
> now, does anyone knows a formal proof for this...
> but if you dont know just tell me why
Why do you ask? When you've asked your previous question, Dennis May
discontinuous must be an F_sigma set. The irrationals are not an F_sigma
set. If that's not a formal proof, then I have no clue about what a
formal proof is. Of course, if you had written that you do not
understand it, that would have been different. But it's a proof.
And, please, stop putting those exclamation marks at the subjects of
your posts.
Jose Carlos Santos
===
Subject: Re: need proof!!!
 <318crhF35divaU1@individual.net I'm looking for a continuous function f:R->R with discontinuity on
> irrational domain and continuous on Q.
> the good people who responded, told me that the answer is NO.
> now, does anyone knows a formal proof for this...
> There is no such function. The set of points at which f is
> discontinuous must be an F_sigma set. The irrationals are not an F_sigma
> set. If that's not a formal proof, then I have no clue about what a
> formal proof is. Of course, if you had written that you do not
> understand it, that would have been different. But it's a proof.
The maverick may want a proof that RQ isn't F_sigma, and that the points
of continuity of a function into a metric space are G_delta.
That RQ isn't F_sigma can be proved by the fact that if it were
F_sigma, then R would be sparse or meager which it isn't by Baire's
theorem for example.
===
Subject: Re: need proof!!!
> I'm looking for a continuous function f:R->R with discontinuity on
> irrational domain and continuous on Q.
> the good people who responded, told me that the answer is NO.
> now, does anyone knows a formal proof for this...
> There is no such function. The set of points at which f is
> discontinuous must be an F_sigma set. The irrationals are not an 
F_sigma
> set. If that's not a formal proof, then I have no clue about what 
a
> formal proof is. Of course, if you had written that you do not
> understand it, that would have been different. But it's a proof.
> The maverick may want a proof that RQ isn't F_sigma, and that the 
points
> of continuity of a function into a metric space are G_delta.
> That RQ isn't F_sigma can be proved by the fact that if it were
> F_sigma, then R would be sparse or meager which it isn't by Baire's
> theorem for example.
can you explain me what is F_sigma set and why Q isnt a F_sigma set?
===
Subject: Re: need proof!!!
 I'm looking for a continuous function f:R->R with discontinuity 
on
> irrational domain and continuous on Q.
> the good people who responded, told me that the answer is NO.
> now, does anyone knows a formal proof for this...
 There is no such function. The set of points at which f is
> discontinuous must be an F_sigma set. The irrationals are not an 
F_sigma
> set.
 The maverick may want a proof that RQ isn't F_sigma, and that the 
points
> of continuity of a function into a metric space are G_delta.
> That RQ isn't F_sigma can be proved by the fact that if it were
> F_sigma, then R would be sparse or meager which it isn't by Baire's
> theorem for example.
> can you explain me what is F_sigma set and why Q isnt a F_sigma set?
A set is F_sigma when it's a countable union of closed sets.
Exercise:  Q is F_sigma.
===
Subject: Re: need proof!!!
> i resently asked the following question
> I'm looking for a continuous function f:R->R with discontinuity on
> irrational domain and continuous on Q.
> the good people who responded, told me that the answer is NO.
> now, does anyone knows a formal proof for this...
> but if you dont know just tell me why
Q is a subset of R, and so is the irrational numbers R - Q. If you
have discontinuity on R - Q, then you have discontinuity on a
subset of R, and hence on R as a whole. Therefore, no such
function exists.
===
Subject: Re: need proof!!!
> i resently asked the following question
> I'm looking for a continuous function f:R->R with discontinuity on
> irrational domain and continuous on Q.
> the good people who responded, told me that the answer is NO.
> now, does anyone knows a formal proof for this...
> but if you dont know just tell me why
> Q is a subset of R, and so is the irrational numbers R - Q. If you
> have discontinuity on R - Q, then you have discontinuity on a
> subset of R, and hence on R as a whole. Therefore, no such
> function exists.
ok, so why is the oppside function exist???
if the discontinuity is on Q which is also a subset of R???
===
Subject: Re: need proof!!!
> ok, so why is the oppside function exist???
A partial answer to some of your questions:
The set R-Q is a G_delta, i.e. a countable intersection
of open subsets of R. This is because
R-Q= cap_{qin Q} (R-{q}).
Here the complement of a singleton is obviously open,
and as Q is countable, so is this intersection. Obviously
this generalizes to the statement that the complement
of a countable set is a G_delta.
Q is not a G_delta as explained by William Elliott
(the complement of a G_delta is an F_sigma, i.e. a countable
union of closed sets).
I don't know, whether being a G_delta is a sufficient
condition for a set to be equal to the points of contuinity
of some function. It is relatively easy to see that it is
a necessary condition, though. This is an exercise (together
with some hints) in Royden's book Real Analysis. I'm sure
many other first year graduate texts in real analysis have
related material as well.
Jyrki Lahtonen, Turku, Finland
===
Subject: Re: need proof!!!
> 
> ok, so why is the oppside function exist???
> A partial answer to some of your questions:
> The set R-Q is a G_delta, i.e. a countable intersection
> of open subsets of R. This is because
> R-Q= cap_{qin Q} (R-{q}).
> Here the complement of a singleton is obviously open,
> and as Q is countable, so is this intersection. Obviously
> this generalizes to the statement that the complement
> of a countable set is a G_delta.
> Q is not a G_delta as explained by William Elliott
> (the complement of a G_delta is an F_sigma, i.e. a countable
> union of closed sets).
> I don't know, whether being a G_delta is a sufficient
> condition for a set to be equal to the points of contuinity
> of some function. It is relatively easy to see that it is
> a necessary condition, though. This is an exercise (together
> with some hints) in Royden's book Real Analysis. I'm sure
> many other first year graduate texts in real analysis have
> related material as well.
> Jyrki Lahtonen, Turku, Finland
but i have one more problem
you are based on the fact that The set of points at which f is
discontinuous must be an F_sigma set and thats o.k
but i only know that: 
a function f(x) in a single variable x is said to be continuous at point y 
if
lim x-->y f(x) = f(y)
can you proof basing on the lim of a function?????
===
Subject: Re: need proof!!!
> I don't know, whether being a G_delta is a sufficient
> condition for a set to be equal to the points of contuinity
> of some function. It is relatively easy to see that it is
> a necessary condition, though. This is an exercise (together
> with some hints) in Royden's book Real Analysis. I'm sure
> many other first year graduate texts in real analysis have
> related material as well.
It is also sufficient.
The book Counterexamples in Analysis shows a construction of a 
function which is discontinuous at precisely the points of an arbitrary 
F_sigma set.
===
Subject: Re: need proof!!!
>i resently asked the following question
>I'm looking for a continuous function f:R->R with discontinuity on
>irrational domain and continuous on Q.
>the good people who responded, told me that the answer is NO.
>now, does anyone knows a formal proof for this...
>but if you dont know just tell me why
>>Q is a subset of R, and so is the irrational numbers R - Q. If you
>>have discontinuity on R - Q, then you have discontinuity on a
>>subset of R, and hence on R as a whole. Therefore, no such
>>function exists.
> ok, so why is the oppside function exist???
> if the discontinuity is on Q which is also a subset of R???
instead of posting an answer to what you meant to write. Your problem is
function which was *discountinuous* at some points (namely, at the
irrational numbers). Obviously, no such function exists, by the reasons
stated by mike3. What you meant to write was:
    I'm looking for a function f:R->R which is discontinuous on every
    irrational point and continuous on every point of Q.
Jose Carlos Santos
===
Subject: Re: need proof!!!
>i resently asked the following question
>I'm looking for a continuous function f:R->R with discontinuity on
>irrational domain and continuous on Q.
>the good people who responded, told me that the answer is NO.
>now, does anyone knows a formal proof for this...
>but if you dont know just tell me why
>Q is a subset of R, and so is the irrational numbers R - Q. If you
>>have discontinuity on R - Q, then you have discontinuity on a
>>subset of R, and hence on R as a whole. Therefore, no such
>>function exists.
> 
> ok, so why is the oppside function exist???
> if the discontinuity is on Q which is also a subset of R???
> instead of posting an answer to what you meant to write. Your problem is
> function which was *discountinuous* at some points (namely, at the
> irrational numbers). Obviously, no such function exists, by the reasons
> stated by mike3. What you meant to write was:
>     I'm looking for a function f:R->R which is discontinuous on every
>     irrational point and continuous on every point of Q.
> Jose Carlos Santos
Though the reverse is quite possible, continuous at irrational points 
and discontinuous at rational ones.
===
Subject: Re: Turing machines
There is no such thing as an ultimate model of symbol manipulation.
There are many equivalent models, and TM is but one of them. The only
special thing is that it is generally thought to be the first complete
model written in a mathematical paper.
A TM is nothing but a computer specification that runs with a very very
low level machine code. It's not any more fundamental than lambda
calculus or cellular automata. (to a computer scientist!)
What it does capture is merely the notion of a causal graph, that
univerally characterizes *any* discrete mechanism in the world, e.g. a
machine that works on blocks of things. It's up to you to call these
symbols.
Is the state of a transistor, or a wire in your VLSI chip, or a single
These are philosophical questions of course, but the proper approach,
in my opinion, is to see these as mere marks, that may or may not be
conceived as symbols (which possibly refer to other things, etc.) They
are just states, chosen from a finite alphabet like the alphabet of a
TM.
--
Eray Ozkural
===
Subject: Re: Turing machines
> There is no such thing as an ultimate model of symbol manipulation.
> There are many equivalent models, and TM is but one of them. The only
> special thing is that it is generally thought to be the first complete
> model written in a mathematical paper.
> A TM is nothing but a computer specification that runs with a very very
> low level machine code. It's not any more fundamental than lambda
> calculus or cellular automata. (to a computer scientist!)
> What it does capture is merely the notion of a causal graph, that
> univerally characterizes *any* discrete mechanism in the world, e.g. a
> machine that works on blocks of things. It's up to you to call these
> symbols.
> Is the state of a transistor, or a wire in your VLSI chip, or a single
> These are philosophical questions of course, but the proper approach,
> in my opinion, is to see these as mere marks, that may or may not be
> conceived as symbols (which possibly refer to other things, etc.) They
> are just states, chosen from a finite alphabet like the alphabet of a
> TM.
-- 
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Turing machines
> There is no such thing as an ultimate model of symbol manipulation.
> There are many equivalent models, and TM is but one of them. The only
> special thing is that it is generally thought to be the first complete
> model written in a mathematical paper.
> 
> A TM is nothing but a computer specification that runs with a very very
> low level machine code. It's not any more fundamental than lambda
> calculus or cellular automata. (to a computer scientist!)
> 
> What it does capture is merely the notion of a causal graph, that
> univerally characterizes *any* discrete mechanism in the world, e.g. a
> machine that works on blocks of things. It's up to you to call these
> symbols.
> 
> Is the state of a transistor, or a wire in your VLSI chip, or a single
> These are philosophical questions of course, but the proper approach,
> in my opinion, is to see these as mere marks, that may or may not be
> conceived as symbols (which possibly refer to other things, etc.) They
> are just states, chosen from a finite alphabet like the alphabet of a
> TM.
which is the case for physical computers. It's trivial to see. We have
something called network. We can attach computers, and we can
indefinitely, we can do it. The silly considerations of OS address
space etc. are irrelevant)
--
Eray Ozkural
===
Subject: Smullyan's Quiz Problem
Raymond Smullyan presents the following (paraphrased) riddle in his
book, Forever Undecided:
  On Monday a professor says to his class, I will give you a surprise
  examination some day this week. You will not know that there is an
  examination when the class begins. A student then reasoned, I
  can't get the quiz on Friday because if I haven't gotten it by
  Friday I will know the quiz must be that day. Similarly he reasoned
  for Thursday all the way to Monday. Where is the error in his logic?
My question is this: Why is there any inconsistency in the professor
giving the quiz to the class on Monday?
Now if the professor had stated to his class on Friday, I will give
you a surprise examiniation some day next week, then there would be
inconsistency problems for all days of the week, but that's not the
way the problem is stated.
Comments anyone?
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
===
Subject: Re: Smullyan's Quiz Problem
> Raymond Smullyan presents the following (paraphrased) riddle in his
> book, Forever Undecided:
>   On Monday a professor says to his class, I will give you a surprise
>   examination some day this week. You will not know that there is an
>   examination when the class begins. A student then reasoned, I
>   can't get the quiz on Friday because if I haven't gotten it by
>   Friday I will know the quiz must be that day. Similarly he reasoned
>   for Thursday all the way to Monday. Where is the error in his logic?
> My question is this: Why is there any inconsistency in the professor
> giving the quiz to the class on Monday?
> Now if the professor had stated to his class on Friday, I will give
> you a surprise examiniation some day next week, then there would be
> inconsistency problems for all days of the week, but that's not the
> way the problem is stated.
> Comments anyone?
Quine has quite a good discussion of this paradox in On A Supposed
Antinomy in The Ways of Paradox. The student needs to allow for the
logical possibility that the exam will be on Friday but he will not
know this fact on Thursday.
===
Subject: Re: Smullyan's Quiz Problem
            days. My association with the Department is that of an alumnus.
>Raymond Smullyan presents the following (paraphrased) riddle in his
>book, Forever Undecided:
>  On Monday a professor says to his class, I will give you a surprise
>  examination some day this week. You will not know that there is an
>  examination when the class begins. A student then reasoned, I
>  can't get the quiz on Friday because if I haven't gotten it by
>  Friday I will know the quiz must be that day. Similarly he reasoned
>  for Thursday all the way to Monday. Where is the error in his logic?
>My question is this: Why is there any inconsistency in the professor
>giving the quiz to the class on Monday?
One of the hypothesis is that an examination will be given during the
week. The argument shows that the hypothesis lead to a contradiction:
that no examination will be given during the week. That is,
H->not(H). But 
(H -> not(H)) -> not(H)
is a tautology. That means that the only thing we can deduce from the
professor's statements are that his conditions will not be met. 
Since we can only conclude that his conditions will not be met, none
of the original argument showing that no test can occur is valid. The
test may occur at any time.
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Smullyan's Quiz Problem
>Raymond Smullyan presents the following (paraphrased) riddle in his
>book, Forever Undecided:
>  On Monday a professor says to his class, I will give you a surprise
>  examination some day this week. You will not know that there is an
>  examination when the class begins. A student then reasoned, I
>  can't get the quiz on Friday because if I haven't gotten it by
>  Friday I will know the quiz must be that day. Similarly he reasoned
>  for Thursday all the way to Monday. Where is the error in his logic?
>My question is this: Why is there any inconsistency in the professor
>giving the quiz to the class on Monday?
> One of the hypothesis is that an examination will be given during the
> week. The argument shows that the hypothesis lead to a contradiction:
> that no examination will be given during the week. That is,
> H->not(H). But 
> (H -> not(H)) -> not(H)
> is a tautology. That means that the only thing we can deduce from the
> professor's statements are that his conditions will not be met. 
> Since we can only conclude that his conditions will not be met, none
> of the original argument showing that no test can occur is valid. The
> test may occur at any time.
I agree with your analysis of the purely logical version of
the paradox.  The professor can be said to be asserting a contradiction.
Thus, either the professor must be considered fallible, or
we have a truth teller asserting a contradiction.  In this case
we can prove anything, (including the facts that a test on Wednesday
is both expected and unexpected).
However, I think that the paradox is much deeper than this.  Consider
four statements the professor could make:
       i) there will be an unexpected test tomorrow
      ii) there will be an unxepected test in the next three days
     iii) there will be an unexpected test next week
      iv) there will be an unexpected test this semester
All four have the same logical structure.  However, i) seems absurd,
ii) seems questionable, iii) seems fine, iv) is completely unremarkable.
A full resoution of the paradox must explain the difference.
Other versions of the paradox which do not involve prediction
(e.g. the unexpected egg)  seem very difficult to resolve.
As I have noted elsethread, I think the problem is the
deep seated conviction that a person must either have
a rational basis for belief or not have a rational basis for belief.
However, if you can arrange a situation where a person has
a rational basis for belief, iff she does not have a rational basis
for belief, this deep seated conviction is seem to be wrong.
My analysis is that the unexpected egg paradox produces
exactly such a situation.
                          -William Hughes
===
Subject: Re: Smullyan's Quiz Problem
            days. My association with the Department is that of an alumnus.
   [.snip.]
>> One of the hypothesis is that an examination will be given during the
>> week. The argument shows that the hypothesis lead to a contradiction:
>> that no examination will be given during the week. That is,
>> H->not(H). But 
>> (H -> not(H)) -> not(H)
>> is a tautology. That means that the only thing we can deduce from the
>> professor's statements are that his conditions will not be met. 
>> Since we can only conclude that his conditions will not be met, none
>> of the original argument showing that no test can occur is valid. The
>> test may occur at any time.
>I agree with your analysis of the purely logical version of
>the paradox.  The professor can be said to be asserting a contradiction.
>Thus, either the professor must be considered fallible, or
>we have a truth teller asserting a contradiction.  In this case
>we can prove anything, (including the facts that a test on Wednesday
>is both expected and unexpected).
>However, I think that the paradox is much deeper than this.  Consider
>four statements the professor could make:
>       i) there will be an unexpected test tomorrow
>      ii) there will be an unxepected test in the next three days
>     iii) there will be an unexpected test next week
>      iv) there will be an unexpected test this semester
>All four have the same logical structure.  However, i) seems absurd,
>ii) seems questionable, iii) seems fine, iv) is completely unremarkable.
>A full resoution of the paradox must explain the difference.
The paradox here is literally that they contradict our intuition. But
the only explanation needed is that ->our intuition is wrong<-. Just
because some of them seem absurd and others do not does not mean
that the statements actually are or are not wrong. 
These statements have a whole bunch of unstated assumptions. For
example, I disagree with you that (i) seems absurd prima facie. It
only becomes absurd if you assume that you know what time and what
context such a test would take place in. The reason (iv) do not seem
so absurd is that the latitude for those unknowns seems so much wider
that you cannot lull yourself into a false sense of knowledge about
the statement.
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Smullyan's Quiz Problem
>    [.snip.]
>> One of the hypothesis is that an examination will be given during the
>> week. The argument shows that the hypothesis lead to a contradiction:
>> that no examination will be given during the week. That is,
>> H->not(H). But 
>> 
>> (H -> not(H)) -> not(H)
>> 
>> is a tautology. That means that the only thing we can deduce from the
>> professor's statements are that his conditions will not be met. 
>> 
>> Since we can only conclude that his conditions will not be met, none
>> of the original argument showing that no test can occur is valid. The
>> test may occur at any time.
>> 
>I agree with your analysis of the purely logical version of
>the paradox.  The professor can be said to be asserting a contradiction.
>Thus, either the professor must be considered fallible, or
>we have a truth teller asserting a contradiction.  In this case
>we can prove anything, (including the facts that a test on Wednesday
>is both expected and unexpected).
>However, I think that the paradox is much deeper than this.  Consider
>four statements the professor could make:
>       i) there will be an unexpected test tomorrow
>      ii) there will be an unxepected test in the next three days
>     iii) there will be an unexpected test next week
>      iv) there will be an unexpected test this semester
>All four have the same logical structure.  However, i) seems absurd,
>ii) seems questionable, iii) seems fine, iv) is completely unremarkable.
>A full resoution of the paradox must explain the difference.
> The paradox here is literally that they contradict our intuition.
Agreed, the arguments are now informal, so paradox has
exactly this meaning.
> But the only explanation needed is that ->our intuition is wrong<-.
True, but the bald statement our intuintion is wrong is not
very helpful.  A full resolution of the paradox requires an
explanation as to why our intuition is wrong
>Just
> because some of them seem absurd and others do not does not mean
> that the statements actually are or are not wrong. 
> These statements have a whole bunch of unstated assumptions. For
> example, I disagree with you that (i) seems absurd prima facie. It
> only becomes absurd if you assume that you know what time and what
> context such a test would take place in.
Yes, there are ways to make (i) sensible (e.g.  An unxepected exam
is one on a different colour of paper), however, these do not
seem sensible to me.  If you take (i) to mean (and this seems
to me the most obvious meaning) You will have an exam tomorrow but
you don't know this, (i) is Bertram's paradox.
> The reason (iv) do not seem
> so absurd is that the latitude for those unknowns seems so much wider
> that you cannot lull yourself into a false sense of knowledge about
> the statement.
Indeed.  However, both the professor and students agree  
that information has been communicated, and appear to agree on
what has been communicated.  So the question's are:  Do the two
parties actually agree on what has been communicated? and
If the professor's statement is self contradictory, what should
he have said?
              -William Hughes
===
Subject: Re: Smullyan's Quiz Problem
            days. My association with the Department is that of an alumnus.
   [.snip.]
>>I agree with your analysis of the purely logical version of
>>the paradox.  The professor can be said to be asserting a 
contradiction.
>>Thus, either the professor must be considered fallible, or
>>we have a truth teller asserting a contradiction.  In this case
>>we can prove anything, (including the facts that a test on Wednesday
>>is both expected and unexpected).
>>However, I think that the paradox is much deeper than this.  Consider
>>four statements the professor could make:
>>       i) there will be an unexpected test tomorrow
>>      ii) there will be an unxepected test in the next three days
>>     iii) there will be an unexpected test next week
>>      iv) there will be an unexpected test this semester
>>All four have the same logical structure.  However, i) seems absurd,
>>ii) seems questionable, iii) seems fine, iv) is completely 
unremarkable.
>>A full resoution of the paradox must explain the difference.
>> The paradox here is literally that they contradict our intuition.
>Agreed, the arguments are now informal, so paradox has
>exactly this meaning.
It should also be added that there is a fair amount of the
Wilde-like paradox in this; you know, the The only thing worse than
being talked about is not being talked about kind of statements,
since we argue that if we expect the test then we don't expect the
test (the old I got you by not getting you argument and variants
thereof).
>> But the only explanation needed is that ->our intuition is wrong<-.
>True, but the bald statement our intuintion is wrong is not
>very helpful.  A full resolution of the paradox requires an
>explanation as to why our intuition is wrong
I think out intuition is wrong because we are making a lot of
unstated assumptions about these statements. As I noted: there will
be an unexpected test tomorrow seems absurd when we assume that
'unexpected' means something like 'I won't know exactly when', but
that knowing the ->day<- of the exam narrows the window sufficiently
for us to figure out exactly when. The other three statements don't
lull us into that assumption because, prima facie, we have more than
one option and no information about which option we should take. But
that (i) if we know the day then we know when; and (ii) if we don't
know which day in advance then we don't know exactly when;  are common
enough conclusions which are both unwarranted in this situation.
>>Just
>> because some of them seem absurd and others do not does not mean
>> that the statements actually are or are not wrong. 
>> These statements have a whole bunch of unstated assumptions. For
>> example, I disagree with you that (i) seems absurd prima facie. It
>> only becomes absurd if you assume that you know what time and what
>> context such a test would take place in.
>Yes, there are ways to make (i) sensible (e.g.  An unxepected exam
>is one on a different colour of paper), however, these do not
>seem sensible to me. 
I'm not even saying that. You can even make it sensible if
unexpected refers to you won't know exactly when; I teach at three
different hours; which hour will contain the exam?
> If you take (i) to mean (and this seems
>to me the most obvious meaning) You will have an exam tomorrow but
>you don't know this, (i) is Bertram's paradox.
And here we are making another assumption about what unexpected
means. Does it mean, you do not expect an exam at all? Or does it
mean, you won't know exactly when? It is in the latter sense that the
Paradox of the Unexpected Hanging uses it, for example.
>> The reason (iv) do not seem
>> so absurd is that the latitude for those unknowns seems so much wider
>> that you cannot lull yourself into a false sense of knowledge about
>> the statement.
>Indeed.  However, both the professor and students agree  
>that information has been communicated, and appear to agree on
>what has been communicated.
Which, as we know, in the real world  is a virtual guarantee that it
is not true that what the professor meant to communicate is the
communication that the students have received... (-:
>   So the question's are:  Do the two
>parties actually agree on what has been communicated? and
>If the professor's statement is self contradictory, what should
>he have said?
Do you mean, how can he make a statement which communicates the same
information but is not self-contradictory? You may be begging the
question there, as in fact the point is that you cannot communicate
that information without also communicating a contradiction.
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Smullyan's Quiz Problem
>    [.snip.]
>>I agree with your analysis of the purely logical version of
>>the paradox.  The professor can be said to be asserting a 
contradiction.
>>Thus, either the professor must be considered fallible, or
>>we have a truth teller asserting a contradiction.  In this case
>>we can prove anything, (including the facts that a test on Wednesday
>>is both expected and unexpected).
>>However, I think that the paradox is much deeper than this.  Consider
>>four statements the professor could make:
>>       i) there will be an unexpected test tomorrow
>>      ii) there will be an unxepected test in the next three days
>>     iii) there will be an unexpected test next week
>>      iv) there will be an unexpected test this semester
>>All four have the same logical structure.  However, i) seems absurd,
>>ii) seems questionable, iii) seems fine, iv) is completely 
unremarkable.
>>A full resoution of the paradox must explain the difference.
>> 
>> The paradox here is literally that they contradict our intuition.
>Agreed, the arguments are now informal, so paradox has
>exactly this meaning.
> It should also be added that there is a fair amount of the
> Wilde-like paradox in this; you know, the The only thing worse than
> being talked about is not being talked about kind of statements,
> since we argue that if we expect the test then we don't expect the
> test (the old I got you by not getting you argument and variants
> thereof).
>> But the only explanation needed is that ->our intuition is wrong<-.
>True, but the bald statement our intuintion is wrong is not
>very helpful.  A full resolution of the paradox requires an
>explanation as to why our intuition is wrong
> I think out intuition is wrong because we are making a lot of
> unstated assumptions about these statements. As I noted: there will
> be an unexpected test tomorrow seems absurd when we assume that
> 'unexpected' means something like 'I won't know exactly when', but
> that knowing the ->day<- of the exam narrows the window sufficiently
> for us to figure out exactly when. The other three statements don't
> lull us into that assumption because, prima facie, we have more than
> one option and no information about which option we should take. But
> that (i) if we know the day then we know when; and (ii) if we don't
> know which day in advance then we don't know exactly when;  are common
> enough conclusions which are both unwarranted in this situation.
You appear to be arguing that all four statements are equally
contradictory, but that the latter statements don't seem to
be as absurd because of the longer time window.  While this
is a reasonable explanation of why (iv) is intuitively less 
problematic than (i), it leaves out an important fact.  A professor
telling his students (i) imparts no information.  A professor telling
his students (iv) imparts information (or at least appears to).  So:
    -what information does the professor impart?
    -is there a more accurate statement that the professor
     could use, or is (iv) the best he can do?
>>Just
>> because some of them seem absurd and others do not does not mean
>> that the statements actually are or are not wrong. 
>> 
>> These statements have a whole bunch of unstated assumptions. For
>> example, I disagree with you that (i) seems absurd prima facie. It
>> only becomes absurd if you assume that you know what time and what
>> context such a test would take place in.
>Yes, there are ways to make (i) sensible (e.g.  An unxepected exam
>is one on a different colour of paper), however, these do not
>seem sensible to me. 
> I'm not even saying that. You can even make it sensible if
> unexpected refers to you won't know exactly when; I teach at 
three
> different hours; which hour will contain the exam?
> If you take (i) to mean (and this seems
>to me the most obvious meaning) You will have an exam tomorrow but
>you don't know this, (i) is Bertram's paradox.
> And here we are making another assumption about what unexpected
> means. Does it mean, you do not expect an exam at all? Or does it
> mean, you won't know exactly when? It is in the latter sense that the
> Paradox of the Unexpected Hanging uses it, for example.
The OP quoted Smullyan definiton of a surprize examination as
You will not know that there is an examination when the class
begins.
I am using unexpected to mean the same as Smullyan's
surprize.
>> The reason (iv) do not seem
>> so absurd is that the latitude for those unknowns seems so much wider
>> that you cannot lull yourself into a false sense of knowledge about
>> the statement.
>> 
>Indeed.  However, both the professor and students agree  
>that information has been communicated, and appear to agree on
>what has been communicated.
> Which, as we know, in the real world  is a virtual guarantee that it
> is not true that what the professor meant to communicate is the
> communication that the students have received... (-:
>   So the question's are:  Do the two
>parties actually agree on what has been communicated? and
>If the professor's statement is self contradictory, what should
>he have said?
> Do you mean, how can he make a statement which communicates the same
> information but is not self-contradictory? You may be begging the
> question there, as in fact the point is that you cannot communicate
> that information without also communicating a contradiction.
If we decide that the professor is trying
to communicate the fact that there will with very high
(but less than 100%) probability be an unexpected examination,
then this information can be communicated without also communicating a
contradiction. If we decide that the professor is trying to
communicate the
fact that there is a 100% probability of an unexpected examination
(i.e. that he has scheduled an examination in such a way that
it will be unexpected)  then he cannot communicate this fact
without communicating a contradiction.  But, it may not be possible
to communicate a contradiction.  The students may conclude from
the professor's attempt to communicate a contradiction that
there will with very high (but less than 100%) probability
be an unexpected examination.  So it may not be possible for
the professor to communicate what he wants to communicate, but 
be possible for him to communicate the information the students
obtain without communicating a contradiction.  
                              -William Hughes
===
Subject: Re: Smullyan's Quiz Problem
William Hughes says...
>However, I think that the paradox is much deeper than this.  Consider
>four statements the professor could make:
>       i) there will be an unexpected test tomorrow
>      ii) there will be an unexepected test in the next three days
>     iii) there will be an unexpected test next week
>      iv) there will be an unexpected test this semester
>All four have the same logical structure.  However, i) seems absurd,
>ii) seems questionable, iii) seems fine, iv) is completely unremarkable.
>A full resoution of the paradox must explain the difference.
I think that if someone says there will be a surprise test sometime
in the next week, he usually only means that there is no way *now*
to figure out what day it will occur on. People don't usually mean
the stronger claim that, even on the day of the test, you won't be
able to figure out that there will be a test. Obviously, on the last
day of the week, you can reason that if you haven't had a test yet,
you're going to have it today.
However, if someone says there will be a surprise test tomorrow,
the weak interpretation of surprise is not available.
The strong interpretation of surprise is paradoxical, while the
weak interpretation is not.
As someone pointed out, the strong interpretation is not actually
paradoxical, once you realize that just because somebody says something
doesn't make it true. So, if the teacher announces that there will
be a surprise test, one possibility is that he is lying---either
there won't be a test, or it won't be a surprise. If you consider
the *possibility* that the teacher is lying, then you can't logically
deduce anything from the fact that he said there will be a surprise
test, and so on the day of the test, it *will* be a surprise.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Smullyan's Quiz Problem
> I think that if someone says there will be a surprise test sometime
> in the next week, he usually only means that there is no way *now*
> to figure out what day it will occur on. People don't usually mean
> the stronger claim that, even on the day of the test, you won't be
> able to figure out that there will be a test. Obviously, on the last
> day of the week, you can reason that if you haven't had a test yet,
> you're going to have it today.
> As someone pointed out, the strong interpretation is not actually
> paradoxical, once you realize that just because somebody says something
> doesn't make it true. So, if the teacher announces that there will
> be a surprise test, one possibility is that he is lying---either
> there won't be a test, or it won't be a surprise. If you consider
> the *possibility* that the teacher is lying, then you can't logically
> deduce anything from the fact that he said there will be a surprise
> test, and so on the day of the test, it *will* be a surprise.
I think that you have given the solution to the unexpected exam
problem.
By Thursday morning, the students will know that either (1) the test
will be on that day and it will be a surprise, or that (2) the test
will be on Friday and it will not be a surprise, in which case the
teacher will have said something that is false when he said that the
test will be a surprise.
However, there is the possibility that the test will not be given on
Friday after all, the teacher might forget, or he might have been
lying, or be taken ill, or he might decide on Thursday night to give
the test the following week. (Put yourself in the shoes of a student
sitting in the class on Friday morning .9a you don't know for 
certain
that the test will go ahead that day, its happening is a contingent
event, not a logical consequence of what the teacher said last week.)
Thus the class does not know for certain that the test will be on
Friday. So the test is still a surprise when it is on Friday, although
less of a surprise.
===
Subject: Re: Smullyan's Quiz Problem
>Raymond Smullyan presents the following (paraphrased) riddle in his
>book, Forever Undecided:
>  On Monday a professor says to his class, I will give you a surprise
>  examination some day this week. You will not know that there is an
>  examination when the class begins. A student then reasoned, I
>  can't get the quiz on Friday because if I haven't gotten it by
>  Friday I will know the quiz must be that day. Similarly he reasoned
>  for Thursday all the way to Monday. Where is the error in his logic?
>My question is this: Why is there any inconsistency in the professor
>giving the quiz to the class on Monday?
> One of the hypothesis is that an examination will be given during the
> week. The argument shows that the hypothesis lead to a contradiction:
> that no examination will be given during the week. That is,
> H->not(H). But 
> (H -> not(H)) -> not(H)
> is a tautology. That means that the only thing we can deduce from the
> professor's statements are that his conditions will not be met. 
> Since we can only conclude that his conditions will not be met, none
> of the original argument showing that no test can occur is valid. The
> test may occur at any time.
Arturo,
point is that, ON MONDAY, the professor stated You will not know that 
there is
an examination when the class begins. When the Monday morning class began, 
the 
students had no idea that there was going to be a surprise examination that 

week
since the professor had yet to make the announcement. Thus giving the exam 
on 
Monday would not contract his statement - it would be a surprise 
examination
and the students won't have known that there is an examination when the
Monday morning class began.
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
===
Subject: Re: Smullyan's Quiz Problem
            days. My association with the Department is that of an alumnus.
>>Raymond Smullyan presents the following (paraphrased) riddle in his
>>book, Forever Undecided:
>>  On Monday a professor says to his class, I will give you a surprise
>>  examination some day this week. You will not know that there is an
>>  examination when the class begins. A student then reasoned, I
>>  can't get the quiz on Friday because if I haven't gotten it by
>>  Friday I will know the quiz must be that day. Similarly he reasoned
>>  for Thursday all the way to Monday. Where is the error in his logic?
>>My question is this: Why is there any inconsistency in the professor
>>giving the quiz to the class on Monday?
>> One of the hypothesis is that an examination will be given during the
>> week. The argument shows that the hypothesis lead to a contradiction:
>> that no examination will be given during the week. That is,
>> H->not(H). But 
>> (H -> not(H)) -> not(H)
>> is a tautology. That means that the only thing we can deduce from the
>> professor's statements are that his conditions will not be met. 
>> Since we can only conclude that his conditions will not be met, none
>> of the original argument showing that no test can occur is valid. The
>> test may occur at any time.
>Arturo,
>point is that, ON MONDAY, the professor stated You will not know that 
there is
>an examination when the class begins. When the Monday morning class 
began, the 
>students had no idea that there was going to be a surprise examination that 

week
>since the professor had yet to make the announcement. Thus giving the exam 

on 
>Monday would not contract his statement - it would be a surprise 
examination
>and the students won't have known that there is an examination when the
>Monday morning class began.
What makes you think that the announcement was made after class that
begun? In fact, one might argue that the announcement occurred before
class had started, as evidenced by the fact that the student tried to
use induction all the way to Monday.
Yes, that's being a bit persnickety, but that is math, after all.
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Smullyan's Quiz Problem
 <xxpllcglbc5.fsf@usrts005.corpusers.net>
 <cont5c$2m62$1@agate.berkeley.edu>
 windows-nt)
>Raymond Smullyan presents the following (paraphrased) riddle in his
>book, Forever Undecided:
  On Monday a professor says to his class, I will give you a 
surprise
>  examination some day this week. You will not know that there is an
>  examination when the class begins. A student then reasoned, I
>  can't get the quiz on Friday because if I haven't gotten it by
>  Friday I will know the quiz must be that day. Similarly he 
reasoned
>  for Thursday all the way to Monday. Where is the error in his logic?
My question is this: Why is there any inconsistency in the professor
>giving the quiz to the class on Monday?
> 
> One of the hypothesis is that an examination will be given during the
> week. The argument shows that the hypothesis lead to a contradiction:
> that no examination will be given during the week. That is,
> H->not(H). But 
> 
> (H -> not(H)) -> not(H)
> 
> is a tautology. That means that the only thing we can deduce from the
> professor's statements are that his conditions will not be met. 
> 
> Since we can only conclude that his conditions will not be met, none
> of the original argument showing that no test can occur is valid. The
> test may occur at any time.
>>Arturo,
>>point is that, ON MONDAY, the professor stated You will not know that 
there is
>>an examination when the class begins. When the Monday morning class 
began, the 
>>students had no idea that there was going to be a surprise examination 
that week
>>since the professor had yet to make the announcement. Thus giving the exam 

on 
>>Monday would not contract his statement - it would be a surprise 
examination
>>and the students won't have known that there is an examination when the
>>Monday morning class began.
> What makes you think that the announcement was made after class that
> begun? In fact, one might argue that the announcement occurred before
> class had started, as evidenced by the fact that the student tried to
> use induction all the way to Monday.
> Yes, that's being a bit persnickety, but that is math, after all.
Let me also add (to my adjacent post) that this is a question of the
problem definition. The definition of the problem should be clear. If
it's not, let's make define it where it lacks clarity. THEN let's work
on the solution.
I really would like this point to be addressed and resolved before moving
forward with discussion of the problem, otherwise we know not what we're
discussing.
-- 
%  Randy Yates                  % Ticket to the moon, flight leaves here 
today 
%% Fuquay-Varina, NC            %  from Satellite 2
%%% 919-577-9882                % 'Ticket To The Moon' 
%%%% <yates@ieee.org>           % *Time*, Electric Light Orchestra
http://home.earthlink.net/~yatescr
===
Subject: Re: Smullyan's Quiz Problem
            days. My association with the Department is that of an alumnus.
>Let me also add (to my adjacent post) that this is a question of the
>problem definition. The definition of the problem should be clear. If
>it's not, let's make define it where it lacks clarity. THEN let's work
>on the solution.
>I really would like this point to be addressed and resolved before moving
>forward with discussion of the problem, otherwise we know not what we're
>discussing.
I think your argument is as follows: let us call the times when the
test may be administered the admissible range of the statement by
the professor. Let x be the time when the test begins. The admissible
range is divided into days; think of them as being integers, so that
floor(x) represents the beginning of the day at which the test
begins. The professor's qualifying statement says something about the
knowledge of the students at floor(x) (the start of the day) where x
is the day in which the test will occur. The professor is making a
statement at time t. The statement says:
  (1) x>t, and x is in the admissible range.
  (2) Something will be true at floor(x).
Let us denote the admissible range by [Y,Z].
I agree with you that we have two different situations. In the
classical Paradox of the Unexpected Hanging/Exam/Egg, we have
floor(Y)>t. 
Your interpretation of Smullyan's presentation has floor(Y)<t.
I think you are correct that there is a qualitative difference between
the two problems. When floor(Y)>t, all statements made are about
future events; since the statements are about knowledge, the
information provided at instant t may affect the situation at instant
floor(Y). That is in fact what the student's argument is.
On the other hand, if floor(Y)<t, then the information provided,
when applied to any value of x, t<= x < floor(Y)+1, is information
about past, fixed events; that is, the information we obtain at time t
cannot modify what the situation was at floor(Y). In this situation,
the assertions made by the professor are NOT necessarily
contradictory.
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Smullyan's Quiz Problem
 <xxpllcglbc5.fsf@usrts005.corpusers.net>
 <cont5c$2m62$1@agate.berkeley.edu>
 windows-nt)
>Raymond Smullyan presents the following (paraphrased) riddle in his
>book, Forever Undecided:
  On Monday a professor says to his class, I will give you a 
surprise
>  examination some day this week. You will not know that there is an
>  examination when the class begins. A student then reasoned, I
>  can't get the quiz on Friday because if I haven't gotten it by
>  Friday I will know the quiz must be that day. Similarly he 
reasoned
>  for Thursday all the way to Monday. Where is the error in his logic?
My question is this: Why is there any inconsistency in the professor
>giving the quiz to the class on Monday?
> 
> One of the hypothesis is that an examination will be given during the
> week. The argument shows that the hypothesis lead to a contradiction:
> that no examination will be given during the week. That is,
> H->not(H). But 
> 
> (H -> not(H)) -> not(H)
> 
> is a tautology. That means that the only thing we can deduce from the
> professor's statements are that his conditions will not be met. 
> 
> Since we can only conclude that his conditions will not be met, none
> of the original argument showing that no test can occur is valid. The
> test may occur at any time.
>>Arturo,
>>point is that, ON MONDAY, the professor stated You will not know that 
there is
>>an examination when the class begins. When the Monday morning class 
began, the 
>>students had no idea that there was going to be a surprise examination 
that week
>>since the professor had yet to make the announcement. Thus giving the exam 

on 
>>Monday would not contract his statement - it would be a surprise 
examination
>>and the students won't have known that there is an examination when the
>>Monday morning class began.
> What makes you think that the announcement was made after class that
> begun? In fact, one might argue that the announcement occurred before
> class had started, as evidenced by the fact that the student tried to
> use induction all the way to Monday.
> Yes, that's being a bit persnickety, but that is math, after all.
That's a possibility. I interpreted the problem statement to reasonably
mean that the professor was making the statement during the meeting time
of the very class he is proposing to give the quiz in. 
To be persnicketier, who said the student's reasoning was correct??? :)
-- 
%  Randy Yates                  % I met someone who looks alot like you,
%% Fuquay-Varina, NC            %             she does the things you do, 
%%% 919-577-9882                %                     but she is an IBM.
%%%% <yates@ieee.org>           %        'Yours Truly, 2095', *Time*, ELO   

http://home.earthlink.net/~yatescr
===
Subject: Re: Smullyan's Quiz Problem
This is the Swedish civil defense paradox:
page 34Mathematical Fallacies and Paradoxes, Bryan Bunch, Dover books,1982
It was analyed by Williard Van Orman Quine in 1953
The analysis was that the 4th logical possiblity is best :
The test will occur this week but it will be unexpected
Martin Gardner also has a version of this in  an Scientific American 
Bunch concludes that reasoning isn't enough to solve the paradox.
>Raymond Smullyan presents the following (paraphrased) riddle in his
>book, Forever Undecided:
>  On Monday a professor says to his class, I will give you a surprise
>  examination some day this week. You will not know that there is an
>  examination when the class begins. A student then reasoned, I
>  can't get the quiz on Friday because if I haven't gotten it by
>  Friday I will know the quiz must be that day. Similarly he reasoned
>  for Thursday all the way to Monday. Where is the error in his logic?
>My question is this: Why is there any inconsistency in the professor
>giving the quiz to the class on Monday?
>Now if the professor had stated to his class on Friday, I will give
>you a surprise examiniation some day next week, then there would be
>inconsistency problems for all days of the week, but that's not the
>way the problem is stated.
>Comments anyone?
-- 
Respectfully, Roger L. Bagula
tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 
619-5610814 :
alternative email: rlbtftn@netscape.net
URL :  http://home.earthlink.net/~tftn
===
Subject: Re: Smullyan's Quiz Problem
> This is the Swedish civil defense paradox:
> page 34Mathematical Fallacies and Paradoxes, Bryan Bunch, Dover 
books,1982
> It was analyed by Williard Van Orman Quine in 1953
> The analysis was that the 4th logical possiblity is best :
> The test will occur this week but it will be unexpected
> Martin Gardner also has a version of this in  an Scientific American 
> Bunch concludes that reasoning isn't enough to solve the paradox.
>Raymond Smullyan presents the following (paraphrased) riddle in his
>book, Forever Undecided:
>  On Monday a professor says to his class, I will give you a surprise
>  examination some day this week. You will not know that there is an
>  examination when the class begins. A student then reasoned, I
>  can't get the quiz on Friday because if I haven't gotten it by
>  Friday I will know the quiz must be that day. Similarly he reasoned
>  for Thursday all the way to Monday. Where is the error in his logic?
>My question is this: Why is there any inconsistency in the professor
>giving the quiz to the class on Monday?
>Now if the professor had stated to his class on Friday, I will give
>you a surprise examiniation some day next week, then there would be
>inconsistency problems for all days of the week, but that's not the
>way the problem is stated.
>Comments anyone?
>  
Agreed, the professor can give the exam on Monday and
it will indeed be unexpected.  However, that really doesn't
deal with the paradox.
This paradox has been presented under many names:
the Swedish civil defense paradox, the prediction paradox,
the unexpected exam, the unexpected egg
to name a few.  It gets my vote as the best paradox.
I have not seen a resoltuion that I find satisfactory.
My own analysis is that the error is the insistance that the 
students must have a rational basis for believing
that there is or is not going to be an exam.
In essence the professor is saying to the students:
you have a rational basis for believing there
will be an exam if and only if you do not have a rational
basis for believing there will be an exam.  If the
students believe the professor, they must conclude
that they do not have a rational reason to either
to believe that there will be an exam or to believe that
there will not be an exam.  The best resolution is that
the students should not believe the professor.
In real life, of course, the professor can only
say  There is a very high probability that you
will have an unexpected exam next week.  This statement
does not lead to a paradox, so the impression that
a professor can make this kind of statement and
communicate information is justified.
In other forms of the paradox, where there
is no prediction element and hence no probablity,
the statement (e.g. there is an unexpected egg)
cannot be believably made.
               -William Hughes
===
Subject: Re: Smullyan's Quiz Problem
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Smullyan.html
Looking at his history the exam question isn't paradoxical at all,
but expected.
unexpected has the connotation of Random:
If you substitute at random for unexpected , it becomes an
probability problem with an expectation distribution ,  and not a 
paradox
or logical question anymore. It becomes equally probable that he will 
give the exam
on the class days that week ( including the day he announces it).
Here's an alternative:
The Swedish civil defense paradox
on the chalk board.
He, then,  announces he with give an unexpected exam at some time that 
week.
The week passes and he gives no exam.
At the first day of the next week when no one has responded to the
phrase which he has left on the chalkboard,
he announces that they all failed the exam:
no one has told him what the The Swedish civil defense paradox
is about.
Dirty trick?
Or a clever way to make the students pay better attention?
>>This is the Swedish civil defense paradox:
>>page 34Mathematical Fallacies and Paradoxes, Bryan Bunch, Dover 
books,1982
>>It was analyed by Williard Van Orman Quine in 1953
>>The analysis was that the 4th logical possiblity is best :
>>The test will occur this week but it will be unexpected
>>Martin Gardner also has a version of this in  an Scientific American 
>>Bunch concludes that reasoning isn't enough to solve the paradox.
>>    
>Raymond Smullyan presents the following (paraphrased) riddle in his
>book, Forever Undecided:
> On Monday a professor says to his class, I will give you a surprise
> examination some day this week. You will not know that there is an
> examination when the class begins. A student then reasoned, I
> can't get the quiz on Friday because if I haven't gotten it by
> Friday I will know the quiz must be that day. Similarly he reasoned
> for Thursday all the way to Monday. Where is the error in his logic?
>My question is this: Why is there any inconsistency in the professor
>giving the quiz to the class on Monday?
>Now if the professor had stated to his class on Friday, I will give
>you a surprise examiniation some day next week, then there would be
>inconsistency problems for all days of the week, but that's not the
>way the problem is stated.
>Comments anyone?
> 
>      
>Agreed, the professor can give the exam on Monday and
>it will indeed be unexpected.  However, that really doesn't
>deal with the paradox.
>This paradox has been presented under many names:
>the Swedish civil defense paradox, the prediction paradox,
>the unexpected exam, the unexpected egg
>to name a few.  It gets my vote as the best paradox.
>I have not seen a resoltuion that I find satisfactory.
>My own analysis is that the error is the insistance that the 
>students must have a rational basis for believing
>that there is or is not going to be an exam.
>In essence the professor is saying to the students:
>you have a rational basis for believing there
>will be an exam if and only if you do not have a rational
>basis for believing there will be an exam.  If the
>students believe the professor, they must conclude
>that they do not have a rational reason to either
>to believe that there will be an exam or to believe that
>there will not be an exam.  The best resolution is that
>the students should not believe the professor.
>In real life, of course, the professor can only
>say  There is a very high probability that you
>will have an unexpected exam next week.  This statement
>does not lead to a paradox, so the impression that
>a professor can make this kind of statement and
>communicate information is justified.
>In other forms of the paradox, where there
>is no prediction element and hence no probablity,
>the statement (e.g. there is an unexpected egg)
>cannot be believably made.
>               -William Hughes
-- 
Respectfully, Roger L. Bagula
tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 
619-5610814 :
alternative email: rlbtftn@netscape.net
URL :  http://home.earthlink.net/~tftn
===
Subject: Re: Smullyan's Quiz Problem
* Randy Yates
> Raymond Smullyan presents the following (paraphrased) riddle in his
> book, Forever Undecided:
>   On Monday a professor says to his class, I will give you a surprise
>   examination some day this week. You will not know that there is an
>   examination when the class begins. A student then reasoned, I
>   can't get the quiz on Friday because if I haven't gotten it by
>   Friday I will know the quiz must be that day. Similarly he reasoned
>   for Thursday all the way to Monday. Where is the error in his logic?
> My question is this: Why is there any inconsistency in the professor
> giving the quiz to the class on Monday?
> Now if the professor had stated to his class on Friday, I will give
> you a surprise examiniation some day next week, then there would be
> inconsistency problems for all days of the week, but that's not the
> way the problem is stated.
Cannot see that there are any differences if the statement is made on
Friday from giving it on the Monday.  Anyway, this is discussed to
death earlier on this group and on rec.puzzles.
Basically, such a statement is self-contradictory, as if I say, My
name is Jon, but you are not able to figure out my name.  Therefore,
the professor does not have any real information in his statement.
But when the stunned class get the surprise examination on Wednesday,
the statement was sensible anyway.
-- 
Jon Haugsand
  Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no
  http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92
===
Subject: Re: Smullyan's Quiz Problem
> [...]
> Basically, such a statement is self-contradictory, as if I say, My
> name is Jon, but you are not able to figure out my name.  Therefore,
> the professor does not have any real information in his statement.
Jon, 
I don't think this analogy is appropriate. The problem with Smullyan's
riddle, as I see it, is that the consistency of the professor's 
statement(s)
depends on the intelligence and awareness of the student. If the student
never pondered the logic, he would feel that any day is a possible quiz
day.
--Randy
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
===
Subject: Re: Smullyan's Quiz Problem
* Randy Yates
> [...]
> Basically, such a statement is self-contradictory, as if I say, My
> name is Jon, but you are not able to figure out my name.  Therefore,
> the professor does not have any real information in his statement.
> I don't think this analogy is appropriate. The problem with Smullyan's
> riddle, as I see it, is that the consistency of the professor's 
statement(s)
> depends on the intelligence and awareness of the student. If the student
> never pondered the logic, he would feel that any day is a possible quiz
> day.
That is of course one way to settle the problem.  However, if there
_are_ intelligent students in the class, you end up there anyway.
(As in most puzzles or games, stupid individuals disturb the
solution.  I guess many of my students would be amazed the following
Thursday even if I say I'll going to have small test on Thursday.)
-- 
Jon Haugsand
  Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no
  http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92
===
Subject: Re: Smullyan's Quiz Problem
> * Randy Yates
> Raymond Smullyan presents the following (paraphrased) riddle in his
> book, Forever Undecided:
> 
>   On Monday a professor says to his class, I will give you a surprise
>   examination some day this week. You will not know that there is an
>   examination when the class begins. A student then reasoned, I
>   can't get the quiz on Friday because if I haven't gotten it by
>   Friday I will know the quiz must be that day. Similarly he reasoned
>   for Thursday all the way to Monday. Where is the error in his logic?
> 
> My question is this: Why is there any inconsistency in the professor
> giving the quiz to the class on Monday?
> 
> Now if the professor had stated to his class on Friday, I will give
> you a surprise examiniation some day next week, then there would be
> inconsistency problems for all days of the week, but that's not the
> way the problem is stated.
> Cannot see that there are any differences if the statement is made on
> Friday from giving it on the Monday.  
The difference is that in the first problem statement, the students would
have no knowledge of any quiz (whether they had pondered it or not) on 
Monday 
when they came to class, so having the quiz on Monday would not contradict
his statement. However, in the second problem statement the students would
already know that there was going to be a quiz, possibly on Monday, that 
week.
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
===
Subject: Gamma function/Mills ratio/Ineq
Let G be the Gamma function, and  f(x)= G(x+0.5)/G(x+ 1)   , (x>= 0).
Question: to prove or disprove  that for each  pair (x,y), 0 =< x < y ,
there  exists   u(x,y) in ( 0, 1/2 ), such that
f(y)
----- = sqrt{(x+u(x,y))/(y+u(x,y))} .
f(x)
Note:According to G.N.Watson [1] it is knownn hat for every  x >= 0
there exists  v(x) in ( 1/4, 1/pi ),  such that
f(x) =1/( sqrt(x+v(x)) )  .
Reference:
[1] G.N. Watson , ,,A note on gamma functions, Proc.Edinburgh
Math.Soc., (2) 11
(1958/59), Edinburgh Math.Notes 42, (1959) 7-9. 
===
Subject: Re: Gamma function/Mills ratio/Ineq
> Let G be the Gamma function, and  f(x)= G(x+0.5)/G(x+ 1)   , (x>= 0).
> Question: to prove or disprove  that for each  pair (x,y), 0 =< x < y ,
> there  exists   u(x,y) in ( 0, 1/2 ), such that
> f(y)
> ----- = sqrt{(x+u(x,y))/(y+u(x,y))} .
> f(x)
> Note:According to G.N.Watson [1] it is knownn hat for every  x >= 0
> there exists  v(x) in ( 1/4, 1/pi ),  such that
> f(x) =1/( sqrt(x+v(x)) )  .
> Reference:
> [1] G.N. Watson , ,,A note on gamma functions, Proc.Edinburgh
> Math.Soc., (2) 11
> (1958/59), Edinburgh Math.Notes 42, (1959) 7-9.
I haven't seen the reference. But certainly if we intend to allow x to be
0, then the interval for v should actually be ( 1/4, 1/pi ] instead. And it
seems that that interval is as tight as possible.
Now to your question: I won't be giving a proof here, but surely what you
suspect is correct. However, the interval you've given for u, ( 0, 1/2 ),
is much looser than necessary, it seems to me. I propose that the
tightest interval for u is of the form ( 1/4, c ) where c is a constant
which is approximately 0.360674 .
David Cantrell
===
Subject: Re: The flux theory of gravitation
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2E3Va31306;
>Actually what you said was:
>> Now, let us put our money where our mouth is and make this forum 
>> interesting. I make this challenge: DEFINE A REAL NUMBER The definition 
>> must be original and different from mine. If you post a correct 
>> definition, no flaw, etc. I.89ll send you a check for $1000, indicate 
your 
>> address. However, if your definition is wrong or not original, you 
>> should send me a check for the same amount. I.89ll post my address 
when needed.
>I'm not interested in this challenge, for the following reasons:
>1) This is not interesting.  Rigourous definitions of the real numbers
>have been available for well over 100 years.  Look it up in a book.
>2) Any correct definition I might make would likely be the same as that in 

>some book, and therefore not original.
>3) I don't anticipate I'd have much chance of collecting your $1000 in
>any case, especially if you are the judge of what is correct, no flaw, 
>etc.
>Robert Israel                                israel@math.ubc.ca
>Department of Mathematics        <a 
href=http://www.math.ubc.ca/~israel>http://www.math.ubc.ca/~israel</a> 
>University of British Columbia            Vancouver, BC, Canada
It is not challenging because you don't know what a number is.
If you well-define a number WE can discuss the merit of your definition. My 

sense is, you probably don't know what being well-defined is. It is stated 
in 
some of the threads. 
Present development of mathematics, hence, also definition of number is 
nonsense. Hence you really need an original one that is different from my 
definition in the new real number system
E. E. Escultura
University of the Philippines.
===
Subject: Re: Escultura affair: publication scandal
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2E3Vx31296;
>I wonder if Mr. Escultura is teaching elementary mathematics to
>students in the Phillipines...
I do and I'm telling them what's wrong with mathematics and teaching the 
right one.
===
Subject: Re: The flux theory of gravitation
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2E3Vw31315;
>> Hi Folks,
>> Since some of you have made references to the Flux Theory of
>> Gravitation that I developed, I.89ll offer some details. Present
>> mathematical physics uses conventional modeling to DESCRIBE nature by
>> mathematical spaces, functions, equations and inequalities. Its main
>> tool is computation. This is inadequate, the reason there are
>> unsolved problems in mathematical physics such as finding the basic
>> constituent of matter, the gravitational n-body problem and the
>> structure of the electron. To overcome this difficulty I have
>> introduced dynamic modeling that EXPLAIN.89S nature, physical 
systems
>> and natural phenomena in terms of the laws of nature.
>There is only one catch. Most of what Nature does or is is literally out 
>of our sight. We can only know it indirectly. Therefore we must rely on 
>hypotheses and inference, as opposed to direct knowledge. There only way 
>we can do that is by means of models grounded on physical hypotheses and 
>mathematical principles. Thus to EXPLAIN means to PREDICT and that is 
>what our theories do. The soundness of our theories rests on two 
>requirements.
>1. The theories must be internally consistent (in the mathematical and 
>logical sense).
>2. The predcitons must be empirically corroberated. Agreement with 
>experiment is the sine qua non of a sound theory.
>If you think Nature can be deduced a priori from logically necessary 
>principles, you are in for dissapointment. Science is empirical, right 
>down to the basement level.
>Bob Kolker
I fully agree with you.
Therefore, I model or explain nature in terms of its laws. I spend my time 
looking for those laws instead of solving equations. Those laws, and there 
are now 42 of them, comprise the flux theory of gravitation.
Its domain includes dark matter.
E. E. Escultura
University of the Philippines
===
Subject: Re: The flux theory of gravitation
gravitation your mother 
===
Subject: Lyapunov Function
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2E3UD31245;
x' =  y - 2*x*(y^2)
y' = -(x^3) + (x^4)*y
I have tried V(x,y) of the form:
a(x^2) + b(y^2)
a(x^2) + b(y^2) + c(x^4)
a(x^2) + b(y^2) + c(y^4)
a(x^2) + c*x*y + b(y^2)           -1<c<1
with no luck. I even tried reverse engineering it, but still no luck. A hint 

or two would be better than an explicit answer, as I'm trying to learn here 

;-) but I'm glad of any help you can give me.
Mike
===
Subject: Re: Lyapunov Function
>x' =  y - 2*x*(y^2)
>y' = -(x^3) + (x^4)*y
>I have tried V(x,y) of the form:
>a(x^2) + b(y^2)
>a(x^2) + b(y^2) + c(x^4)
>a(x^2) + b(y^2) + c(y^4)
>a(x^2) + c*x*y + b(y^2)           -1<c<1
>with no luck. I even tried reverse engineering it, but still no luck. A 
hint or two would be better than an explicit answer, as I'm trying to learn 

here ;-) but I'm glad of any help you can give me.
>Mike
Try V(x,y) = x^(2m) + a*y^(2n) with m,n integers >=1 and a>0.
Thomas
===
Subject: Re: Lyapunov Function
>>x' =  y - 2*x*(y^2)
>>y' = -(x^3) + (x^4)*y
>>I have tried V(x,y) of the form:
>>a(x^2) + b(y^2)
>>a(x^2) + b(y^2) + c(x^4)
>>a(x^2) + b(y^2) + c(y^4)
>>a(x^2) + c*x*y + b(y^2)           -1<c<1
>>with no luck. I even tried reverse engineering it, but still no luck. A 
hint or two would be better than an explicit answer, as I'm trying to learn 

here ;-) but I'm glad of any help you can give me.
>>Mike
>Try V(x,y) = x^(2m) + a*y^(2n) with m,n integers >=1 and a>0.
To be a little more specific: Use this Ansatz and compute
d/dt (V(x(t),y(t)). You will find that there is only one choice for m
and n such that the bad terms with odd exponents balance. Next you
can find an a which make the bad terms vanish *identically* so that
you are only left with even powers.
Fortunately this a is positive (otherwise we wouldn't have a
L.-function and d/dt (V(x(t),y(t)) <= 0 for all t !
That might give you an idea. BTW: You have the solution in your list
(you don't need that parameters though) - somehow you skipped over it
or miscalculated.
Thomas
>Thomas
===
Subject: Re: Escultura affair: publication scandal
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2E3VH31302;
>> I wonder if Mr. Escultura is teaching elementary mathematics to
>> students in the Phillipines...
>He could be the janitor...
You see, in the Philippines, a janitor is smarter than most mathematicians.
EEE
===
Subject: Re: Some questions about the Cauchy distribution
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2E3SZ31151;
>>The mean of n random variables picked from the Cauchy distribution has 
>>itself the Cauchy distribution. Does this mean that there is no use 
>>having a larger sample in order to estimate the expected value? (Or wait 
>>a minute, does the expected value exist? No, right?)
>I think what you're trying to do is the following: given a random sample
>of size n from the Cauchy distribution with density 1/(Pi (1+(x-a)^2)),
>estimate the parameter a.  This parameter is not the expected value,
>because that doesn't exist
[...]
Isn't it only a matter of definition of the expectation ? 
If one integrates x/(Pi (1+(x-a)^2)) from a-t to a+t 
in the sense of Cauchy's principal value (instead of Riemann)  
one gets (2*a*ArcTan[t])/Pi which has 'a' as limit 
when t goes to infinity. 
V.Astanoff
===
Subject: Re: JSH: Simple proof
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2E3SA31146;
>> Why don't you simply and briefly state the >>problematic<< property 
>> of the ring of algebraic integers and/or the statement that you
>> want to prove.
>The ring of algebraic integers is determined by roots of *monic*
>polynomials with integer coefficients.
>It is possible to show with basic algebra that there are numbers which
>are properly units but because their multiplicative inverse is not the
>root of some monic polynomial with integer coefficients they are not
>units in the ring of algebraic integers.
These numbers are >>properly units<< in which ring? The notion
of a unit is defined only with respect to some ring. Your
statement makes no sense as it stands.
>To see how it works consider that in rationals you can have
>(3x + 1)(x + 1) = 3x^2 + 4x + 1
>where, of course, one of the roots is a unit in the ring of algebraic
>integers.
>But now consider
>(3x + u_1)(x + u_2) = 3x^2 + kx + 1, where u_1 u_2 =1, and k is an
>integer.
>You find that if the u's are irrational, then u_1, while an algebraic
>integer is not a unit in the ring, while u_2 cannot then even be an
>algebraic integer.
>My research shows though that both u_1 and u_2 can be units in a ring
>where -1 and 1 are the only rational units, and no non-unit member of
>the ring is a factor of any two integers that are coprime in the ring
>of integers.
>You see, I abstracted out two key properties of rings like the ring of
>integers and the ring of algebraic integers.
I read your contributions about the >>object ring<< as well as the
ones of many others. Together we reached a rather satisfying
answer concerning the existence of such a ring.
>> Using standard mathematical terms and notions (for example from
>> commutative algebra) this should be possible in a few lines instead
>> of making a long story.
>It's not complicated.  Basically you can't just rely on whether or not
>some number is in the ring of algebraic integers when considering
>factors of roots of a polynomial.
>The mathematics is mostly REALLY simple.
This is no clear mathematical statement at all!
What is the problem? What did you prove?
>> Why do we have to discuss things like >>what is a polynomial?<<
>I'm not discussing that, other posters made a big deal out of it.
...but if such simple things like >>what is the constant term
of a polynomial<< are not clear, we have to clarify it before 
we can proceed.
>> here? The notion of a polynomial is defined since a long time and
>> can be found in every introductory book on algebra.
>So?
>> If we consistently use the common definitions of mathematical
>> objects like polynomials we should rather quickly be able to
>> clarify the situation and avoid all the frustration that frequently
>> seems to culminate in personal attacks.
>> H
>I've seen posters come and go, and every once in a while there's a
>poster like you who claims to care about working things out.
>When it turns out that I'm right, you go over to the other side, and
>either run away, or turn to bizarre behavior.
I follow the discussions you initiated since almost one year
now and tried to contribute with the aim to clarify things
ring, because I found the idea interesting. However I remember
clearly that you left the discussion suddenly claiming that
you are not interested in the stuff anymore. Strangely you
did this when with the help of other posters together we
really reached a good overview over the candidates for >>object 
rings<< within the field of algebraic numbers (and even within
larger fields).
No >>sides<< in that discussion just mathematics!
>Psychologists call it cognitive dissonance.
>Basically, deep down you believe that I must be wrong, so your post is
>not really in good faith.  But simply *saying* certain things that
>indicate objectivity or willingness to be objective sets you up
>psychologically.
>That is, you feel a need to be consistent with what you said.
>But later, when you run into the rigid mathematics, which goes against
>what you wish to believe, you basically kind of break.  Your mind
>breaks, and you run away or behave weird.
>I've seen it lots of times.  Do yourself a favor, and just walk away
>now.
>James Harris
What sometimes really makes me angry in the discussions you
initiate is the lack of mathematical rigor! Each time one of
the participants clearly worked out a point the discussion
stops, or turns into personal attacks etc. but it never ends
with a statement like: now we have seen that this and this
is true. Lets keep this result and proceed until we reach
the final conclusion.
Why don't you answer to Nora Baron's recent posts? They are
mathematically very clear. The critical points are obvious
to everyone who understands basic mathematics - as you also 
say. Forget about >>sides<< and all that personal stuff.
There is the mathematics - answer to it in mathematical terms
and with the same clarity. That would lead to a real discussion
and eventually to a clear result.
Why for example didn't we finish the discussion about object
rings. We obtained a result. We could have stated it clearly.
We would not need to start all over again discussing statements
like >>no non-unit member of the ring is a factor of any two 
integers that are coprime in the ring of integers<<.
It has been stated over and over again that if the ring R you
are considering here contains the integers, then this statement
is satisfied for every ring! It is nothing particular!
Check this out: coprime integers generate the whole of Z (integers)
as an ideal. Lifting this ideal to the ring R leads to
the whole ring too. Thus the two numbers cannot be divided by
a common non-unit of R.
This is exhausting - one feels like a hamster running in his
wheel without ever moving one centimeter ahead. 
H 
===
Subject: Re: JSH: Simple proof
 
<deleted What sometimes really makes me angry in the discussions you
> initiate is the lack of mathematical rigor! Each time one of
> the participants clearly worked out a point the discussion
> stops, or turns into personal attacks etc. but it never ends
> with a statement like: now we have seen that this and this
> is true. Lets keep this result and proceed until we reach
> the final conclusion.
You're silly then.  I've talked to some very elite mathematicians, in
discussions that aren't on Usenet.
I don't hear the same crap from them that you people whine about
constantly.
I just don't get that kind of crap from leading mathematicians, and
don't dream that I just get blown off all the time.
I've had some involved discussons over these ideas, and there just
isn't ANY of the crap that you people keep tossing up, year after
year, after year.
So why should I think anything of you?
You say you're angry.  So what?  I think you're some weirdly emotional
person who gets excited for nothing.
> Why don't you answer to Nora Baron's recent posts? They are
> mathematically very clear. The critical points are obvious
The poster Nora Baron is not even a girl.  That poster is a guy,
playing a girl.
I've explained to that poster everything repeatedly over a period of
years.
I've shot down each and every objection, only to see it raised again,
and again, and again.
And I have to tell you people, leading mathematicians don't make those
objections.  Barry Mazur didn't.  Ralph McKenzie didn't.  Andrew
Granville didn't.
You people are in a world by yourselves living in a fantasy that you
are actually good at mathematics and discussing it.
The pro's don't talk like you do.
> to everyone who understands basic mathematics - as you also 
> say. Forget about >>sides<< and all that personal stuff.
> There is the mathematics - answer to it in mathematical terms
> and with the same clarity. That would lead to a real discussion
> and eventually to a clear result.
> Why for example didn't we finish the discussion about object
> rings. We obtained a result. We could have stated it clearly.
> We would not need to start all over again discussing statements
> like >>no non-unit member of the ring is a factor of any two 
> integers that are coprime in the ring of integers<<.
> It has been stated over and over again that if the ring R you
> are considering here contains the integers, then this statement
> is satisfied for every ring! It is nothing particular!
Really?
> Check this out: coprime integers generate the whole of Z (integers)
> as an ideal. Lifting this ideal to the ring R leads to
> the whole ring too. Thus the two numbers cannot be divided by
> a common non-unit of R.
No.  Maybe later I need to come and check Kummer's work as well for
errors.
 
> This is exhausting - one feels like a hamster running in his
> wheel without ever moving one centimeter ahead. 
Why bother?  You are kind of annoying, slightly.
Why can't you people get it through your heads that I have a paper at
a journal?
You don't have to do anything, but wait.
Quit with the drama.  I'm not interested in histrionics from posters
like you.
I can create all the drama I won't by myself.
James Harris
===
Subject: Re: JSH: Simple proof
> The poster Nora Baron is not even a girl.  That poster is a guy,
> playing a girl.
A perfect match for yourself -- as I discovered when I found the anagram of 

your name which reveals
your true identity: James S. Harris = Ms. J. Harrie Ass (note the gender)
--
There are two things you must never attempt to prove: the unprovable -- and 

the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: Simple proof
> The poster Nora Baron is not even a girl.  That poster is a guy,
> playing a girl.
Why does that bother you so much, James?  Is it too reminiscent of your
dating experiences?
-- 
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 Myers, 
Silverlock
===
Subject: Re: JSH: Simple proof
 Discussion, linux)
> Quit with the drama.  I'm not interested in histrionics from posters
> like you.
> I can create all the drama I won't by myself.
                               ^^^^^
Dammit, James, you just ruined a perfect .sig.
-- 
There was an accident in the air.  There was a sign saying, 'Planes
don't go here.  The clouds have to be fixed.' 
  -- Quincy P. Hughes applies the lessons of Dutch train travel to 
     pretending about airplanes.
===
Subject: Re: JSH: Simple proof
> Quit with the drama.  I'm not interested in histrionics from posters
> like you.
> I can create all the drama I won't by myself.
>                                ^^^^^
> Dammit, James, you just ruined a perfect .sig.
He's not adopted an explicitly stream-of-consciousness style before,
but there's a first time for everything: perhaps this is to be read as
I can create all the drama [if I want to]; [but] I won't by myself
[so let's keep playing]
-- 
Larry Lard
Replies to group please
===
Subject: Re: JSH: Simple proof
 <87fz2nbuq4.fsf@phiwumbda.org>
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
 VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
>> Quit with the drama.  I'm not interested in histrionics from posters
>> like you.
>> I can create all the drama I won't by myself.
>>                                ^^^^^
>> Dammit, James, you just ruined a perfect .sig.
> He's not adopted an explicitly stream-of-consciousness style before,
> but there's a first time for everything: perhaps this is to be read
> as
> I can create all the drama [if I want to]; [but] I won't by myself
> [so let's keep playing]
Uh, what's this stream-of-conscious nonsense?  His sentence makes
perfect sense without any such tags.  It is equivalent to
I have the ability to also create by myself all the drama which I
actually am not going to create.
So that simply means that his potential for drama is not restricted at
all, even though he might refrain from some drama deliberately.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: JSH: Simple proof
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
 VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
> I can create all the drama I won't by myself.
Wow, he smashes theology and philosophy in one fell swoop together
with mathematics.
make a noise?
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: JSH: Simple proof
> Why don't you answer to Nora Baron's recent posts? They are
> mathematically very clear. The critical points are obvious
> to everyone who understands basic mathematics - as you also 
> say. 
That is the reason right there.  JSH isn't interested in a mathematical 
discussion that doesn't lead to a recognition of his genius.  When a 
poster points out clearly the flaws in his thinking, JSH must either 
ignore the post or respond with personal attacks.  The very clarity of 
NB's posts is what makes them unacceptable to JSH.
The reason JSH will never be a mathematician is that he can tolerate 
error.  I believe his emotional need to see himself as a genius is 
stronger than his rejection of contradictions, so emotions trump logic.
Gib
===
Subject: Re: Escultura affair: publication scandal
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2E3UY31284;
>><a
>1271503.iARF35e16701@proapp.mathforum.org...</a>
You've *proven* that you're not using the same language that other 
professional mathematicians use.  Anything else is simply a matter of 
[probably correct] deduction in your axiom set which simply doesn't agree 
with everyone else's.
    Norm
Unfortunately, mathematics is not a popularity contest; it is a struggle for 

precision. One of the requirements of mathematics is that every concept must 

well-defined, that is, its existence, properties and relations with other 
concepts must be specified by the axioms. The axioms are the basis of proof. 

Proofs are nonsense unless they follow from the axioms. That is why a true 
mathematician looks at the foundations of a field before doing anything 
there. This is where Wiles failed miserably. Critique-rectification 
naturally 
involves new language.
E. E. Escultura  
University of the Philppines
===
Subject: Re: Escultura affair: publication scandal
><a
1301835.iAUIZpK23888@proapp.mathforum.org...</a><a
1
>1271503.iARF35e16701@proapp.mathforum.org...</a You've *proven* that you're not using the same language that other
professional mathematicians use.  Anything else is simply a matter of
[probably correct] deduction in your axiom set which simply doesn't agree
with everyone else's.
>     Norm
> Unfortunately, mathematics is not a popularity contest; it is a struggle
for precision. One of the requirements of mathematics is that every concept
must well-defined, that is, its existence, properties and relations with
other concepts must be specified by the axioms. The axioms are the basis of
proof. Proofs are nonsense unless they follow from the axioms. That is why 
a
true mathematician looks at the foundations of a field before doing 
anything
there. This is where Wiles failed miserably. Critique-rectification
naturally involves new language.
> E. E. Escultura
> University of the Philppines
No, it's not a popularity contest -- it's a game.  Well, at least there was
a game called IIRC Wiff-N-Proof that taught the creation of and deduction 
of
correct logical propositions from axioms.  If you're really a mathematician
then you already know that there are several different logical systems 
based
on differing foundational axiom sets and their semantics.  Assuming your
system is consistent, you've simply created another one that's neither more
nor less correct than the others.  What you see as errors in their 
system
results from applying your axiom scheme to someone else's system -- which
isn't valid.
    Norm
===
Subject: Re: Escultura affair: publication scandal
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2E3TM31190;
<In the >Zentralblatt f.9fr Mathematik< 9 publications
of E. Escultura are listed. For 5 of them the
Zentralblatt just mentions >not reviewed<, whatever
that means.
This is an indexing service like so many others. So it simply notes some 
contributions unless someone wants to try uncharted course.
<However I must admit that - looking at myself - I
wonder why persons like E. Escultura or J. Harris
provoke such strong reactions.
The most likely reason is fear of new ideas. 
E. E. Escultura
University of the Philippines
===
Subject: Re: Escultura affair: publication scandal
> <In the >Zentralblatt f?hematik< 9 publications
> of E. Escultura are listed. For 5 of them the
> Zentralblatt just mentions >not reviewed<, whatever
> that means.
> This is an indexing service like so many others. 
No. It commissions reviews: not reviewed means not reviewed.
There are two reasons why a paper might not be reviewed:
(i) the editord did not commision a review since they considered
it without merit or otherwise unsuitable,
(ii) a review was commissioned but never written.
There is another reviewing service: Mathematical Reviews.
It lists ten items by Escultura. None of them have been reviewed.
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Escultura affair: publication scandal
> <In the >Zentralblatt f.9fr Mathematik< 9 publications
> of E. Escultura are listed. For 5 of them the
> Zentralblatt just mentions >not reviewed<, whatever
> that means.
> 
> This is an indexing service like so many others. 
> No. It commissions reviews:
It was formerly, but haven't they changed lately?  They publish
authors' summaries?  Or did I just dream it?
-- 
G. A. Edgar                               
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Escultura affair: publication scandal
>> <In the >Zentralblatt f.9fr Mathematik< 9 publications
>> of E. Escultura are listed. For 5 of them the
>> Zentralblatt just mentions >not reviewed<, whatever
>> that means.
>> 
>> This is an indexing service like so many others.
>> No. It commissions reviews:
> It was formerly, but haven't they changed lately?  They publish
> authors' summaries?  Or did I just dream it?
Quote
Zentralblatt MATH is the world's most complete and longest running
abstracting and reviewing service in pure and applied mathematics. 
They still publish reviews.
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Escultura affair: publication scandal
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2E3T431194;
>><a
>1271503.iARF35e16701@proapp.mathforum.org...</a>
===
Subject: Re: Escultura affair: publication scandal
Author: Norm Dresner <ndrez@att.net> 
1271503.iARF35e16701@proapp.mathforum.org...
You've *proven* that you're not using the same language that other 
professional mathematicians use.  Anything else is simply a matter of 
[probably correct] deduction in your axiom set which simply doesn't agree 
with everyone else's.
    Norm
Unfortunately, mathematics is not a popularity contest; it is a struggle for 

precision. One of the requirements of mathematics is that every concept must 

well-defined, that is, its existence, properties and relations with other 
concepts must be specified by the axioms. The axioms are the basis of proof. 

Proofs are nonsense unless they follow from the axioms. That is why a true 
mathematician looks at the foundations of a field before doing anything 
there. This is where Wiles failed miserably. Critique-rectification 
naturally 
involves new language.
E. E. Escultura  
University of the Philppines
===
Subject: Re: Re Re Publication Scandal
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2E3UZ31250;
>>      Now, let us put where our money where our mouth is and make
>>this forum interesting. I make this challenge: DEFINE A REAL NUMBER
>>The definition must be original and different from mine. I you post a
>>correct definition, no flaw, etc. I'll send you a check for $1000,
>What the heck is a correct definition? For example, why couldn't
>we define real number to mean the big key on the keyboard that
>moves the cursor all the way to the left?
>Please use your real number more often.
>You can donate my $1000 to charity.
===
Subject: Re: Re Re Publication Scandal
Author: Dave Rusin <rusin@vesuvius.math.niu.edu> 
>       Now, let us put where our money where our mouth is and make this 
forum interesting. I make this challenge: DEFINE A REAL NUMBER 
> The definition must be original and different from mine. If you post a
>correct definition, no flaw, etc. I'll send you a check for $1000,
< What the heck is a correct definition? For example, why couldn't
we define real number to mean the big key on the keyboard that
moves the cursor all the way to the left?
dave
As a topologist you know what well-defined means and that a concept does not 

exist if it isn.89t well-defined. I require more: The existence, 
properties, relationship of a concept with others, etc. must be specified by 

the axioms. In the case of a real number every digit must be known or 
computable, that is, there is an algorithm for computing it. That is why a 
nonperiodic number or irrational is nonsense because no one knows it.89s, 
say, ten millionth digit. Your suggestion does not sound mathematical at 
all; 
a robot can do what you are suggesting.
You ain't got nothin for charity.
E. E. Escultura
University of the Philippines
>dave
===
Subject: Re: The-State-of-the-Art in Mathematics
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2E3UM31258;
>Hey, Escultura, 
<I disagree.
<When you say dichotomy, do you mean trichotomy?
Dichotomy axiom says: Given real numbers x, y, one and only one holds: x < y 

or x = y or x > y. Brouwer constructed a counterexample to it which, 
incidentally, implies that the real numbers have no natural ordering. 
Moreover, Banach Tarski constructed a counterexample to the completeness 
axiom of the real number system. 
<What are your vague numbers?  What is your definition of a real
number?  Do you consider the nonstandard constructions of the surreal
numbers, or hyperreals or colonies of germs and germinal functions?
Nothing to do with rurreals, etc. I have a new construction without these 
false axioms using only three simple axioms. This relieves the real numbers 

of all contradictions while at the same time enriching them. 
No, I won.89t define a real number because I.89m giving a price of 
$1000 who can define it correctly but differently from mine. The offer 
expires this weekend.
<You appear to promote geometry.
<I'm getting somewhat defensive about all this.  I would like to not share 
with you.
Yes, it.89s healthy to disagree, that.89s called critical thinking,  
the beginning of great things.
<Please go off about the real numbers.  You think they are decimals?
>Ross Finlayson
To sum up, the present axiomatization of the real numbers is terribly wrong. 

Most of the concepts are ill-defined, two axioms are false and the topics on 

fractions are redundant because fractions are decimals.
All to the best
Eddie Escultura
University of the Philippines
===
Subject: Re: The State-of-the-Art in Mathematics
> MATHEMATICAL UPDATE AND CLARIFICATION
> 
> THIS IS part of my occasional update of mathematics and physics. It
> refers to the resolution of mathematical issues that have appeared in
> several journals recently. (1) The complex number i is nonsense since
> i = sqrt(-1) = sqrt(1/-1) = 1/i = -i from which follows i = -i.
>  -i = 1/i
> -i = 1/sqrt(-1)
>>typo; i =/= -i
>  i =/= -1
>  i = sqrt(-1)
>>Smart's Alt. Physics News Group
>>http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
>>S. Enterprize (Science Journal)
>>http://smart1234.s-enterprize.com/
>This simply means that the complex number system is a bunch of 
tautologies.
  I developed another way to observe a complex number. Did you see the 
other
posts, where I use sub-level degrees of the sign itself?
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: The state-of-the-art in mathematics
[...]
|>The argument I made for the difficulty with Trichotomy in the Reals,
|>which my Professor found utterly unconvincing, was as follows:
|>Assume a Computer Program whose job it is, being fed two real numbers
|>x and y in decimal representation (note: any base would do), to
|>determine: Case 1) x > y, Case 2) x =y, or Case 3) x<y.  The Program
|>knows nothing about x and y other than the steam of decimals it 
is
|>being fed.
Decimal representation has its uses, but it's not a good
general-purpose representation of real numbers for these
purposes. It's better to consider a real number to be
represented by a sequence of rational approximations
r1,r2,... such that r_n lies within 1/n of the real.
For instance, there's no algorithm for adding or multiplying
two decimal expansions and returning the decimal expansion
of the result. But that's only because of the possibility of the
result being equal to a terminating decimal without it being
possible to be sure whether it is just under or just over. In
such a case one doesn't know whether the expansion ends
with 9999... or 0000.... But really you know the real number;
you just don't know how it compares with a particular finite
decimal expansion. You can generate arbitrarily close
approximations to it, and that is what is normally meant by
saying that an algorithm computes a real number.
Moreover, converging sequences of rationals are easier to
work with for nonalgorithmic theory too.
|>Now assume that the Computer continually receives identical
|>information about x and y.  At no point can said Computer make 
the
|>above Case 1/Case 2/Case 3. determination.
|>Conclude that Trichotomy breaks down in the Reals.
|>Is this sophistry, or is there merit to this Argument?
If you fixed it up a little bit, it would serve as an explanation
for why trichotomy is not a *constructive* theorem. If it were
me, I would not state it in terms of a computer comparing
the real numbers, since the crux of it is not algorithmic
per se.
If you are interested, have a look at Bishop's _Foundations
of Constructive Analysis_, which is basically an analysis
textbook written from the point of view of constructive
mathematics. It contains an explanation of why trichotomy
is nonconstructive in an aside.
Incidentally, there are a lot of situations where the
constructive development uses the fact that if a<c, then
either a<b or b<c, when the parallel nonconstructive
theory uses trichotomy.
It's constructive to deduce a=b from the impossibility
of a<b and of b<a.
[...]
|Trichotomy says that for any x, y in R exactly one of x < y,
|x = y and x > y is true. That doesn't say anything about
|the possibility of deciding which one holds in a finite
|amount of time given a stream of arbitrarily many decimals
|in the expansion of x and y; whether or not that program
|can ever determine that two sequences are the same (of
|course it can't in the situation above) they _are_ the
|same, or not.
The overwhelming majority of mathematicians work, and
the overwhelming bulk of the mathematics is done, from
a nonconstructive point of view. I don't know exactly how
many orders of magnitude less the mathematics being
done intentionally in a constructive way is. There is a lot
of it that is accidentally constructive, but that's a different
story. So it is important that you (msherwood) become
familiar with that way of doing things; learn how to think that
way. You will only infrequently meet anyone who knows how
to do mathematics any other way.
There are two key assumptions used that aren't constructive.
One is the law of excluded middle. David Ullrich just gave a
lovely example of the thinking here. If I have a sequence
Q1,Q2,Q3,... of questions that I can answer yes or no,
I can ask whether there are any of them whose answer is
no. The law of excluded middle says that for each proposition
P, either P is true or P is false. This is regarded as true entirely
independent of whether we have a way of getting an answer to
the question. So (one says) either the answers to Q1,Q2,...
are all yes, or at least one of them is no. This is not
constructive, however.
The other standard nonconstructive axiom is the axiom of choice.
There are slightly more people with qualms about it than there
are constructivists, but the mainstream point of view about it is
also that it's perfectly okay. You may read about the Banach-
Tarski paradox and so on, but that's not treated as a serious
problem generally.
The axiom of choice in one form says that if F is a family of
nonempty sets, then there exists a function c with domain
F having the property that for each set S in F, the value c(S)
of the function on S is a member of S. The function c is
called a choice function.
With the law of excluded middle, the thing people say is that
a statement is true or false, without it necessarily having anything
to do with our being able to determine whether it's true or false.
With the axiom of choice, the thing people say is that a set can
exist without it necessarily having anything to do with our being
able to specify it ourselves. There are cases of the axiom of choice
where the choice function it claims exists apparently can't be
defined in the language of set theory that we use. But the intuition
is that among all possible functions on F, there should exist many
that are choice functions, even if we lack the means to define them.
Two of the most famous theorems that depend upon the axiom of
choice are the Hahn-Banach theorem and the existence of a maximal
ideal in a commutative ring. Each of those theorems uses the axiom
of choice to prove the existence of some kind of biggest possible
construction, in contexts where a lot of arbitrary choices might have
to be made.
I just think it's kind of neat to be able to point out to you that you
just passed a fork in the road, with another route that could have
been taken, even if you don't ever take the less travelled path.
Keith Ramsay
===
Subject: Re: The state-of-the-art in mathematics
> The other standard nonconstructive axiom is the axiom of choice.
> There are slightly more people with qualms about it than there
> are constructivists, but the mainstream point of view about it is
> also that it's perfectly okay. You may read about the Banach-
> Tarski paradox and so on, but that's not treated as a serious
> problem generally.
The Banach-Tarski paradox is not a true antimony. It is a proof that 
non-measurable sets exist.
Bob Kolker
===
Subject: Re: The state-of-the-art in mathematics
>> The other standard nonconstructive axiom is the axiom of choice.
>> There are slightly more people with qualms about it than there
>> are constructivists, but the mainstream point of view about it is
>> also that it's perfectly okay. You may read about the Banach-
>> Tarski paradox and so on, but that's not treated as a serious
>> problem generally.
>The Banach-Tarski paradox is not a true antimony. It is a proof that 
>non-measurable sets exist.
Actually, it is a proof that there is no finitely additive
measure in 3d which is invariant under rigid motion.  Such
finitely additive measures exist in 2d, but the existence
of sets which are non-measurable with respect to countably
additive measures is the same in both 2d and 3d.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558
===
Subject: Re: The state-of-the-art in mathematics
> Actually, it is a proof that there is no finitely additive
> measure in 3d which is invariant under rigid motion.  Such
> finitely additive measures exist in 2d, but the existence
> of sets which are non-measurable with respect to countably
> additive measures is the same in both 2d and 3d.
I sit corrected.
However the B-T theorem is still not a paradox. It is just counter 
intuitive.
Bob Kolker
===
Subject: Re: The state-of-the-art in mathematics
 Discussion, linux)
> I don't know exactly how many orders of magnitude less the
> mathematics being done intentionally in a constructive way is.
.../intentionally/ in a /constructive/ way...
Are you making a pun there?
(Constructive mathematics is closely related to intensional
mathematics --- I think.  But I'm not too clear on these things.)
-- 
Jesse F. Hughes
Dead men can't talk.  Especially when they've been cremated.
===
Subject: Re: The state-of-the-art in mathematics
>[...]
>I can see David sneaking by the side door because he cannot well-define 
the 
>concept .8breal number.
> The simplest definition is this: A real number is a set S of rationals
> such that S is nonempty, S does not contain every rational, S has no
> smallest element, and for any rationals s, t, if s < t and s is in S
> then t is in S.
> That's the definition, whether or not you agree that it's well.
>E. E. Escultura
> ************************
> David C. Ullrich
A definition, certainly, but the definition?
But I agree it is well.
===
Subject: Re: The state-of-the-art in mathematics
   <ijt7t016ttq05s377kaj5ecfkbbocb4s1a@4ax.com>
(E.
>[...]
I can see David sneaking by the side door because he cannot
well-define the
>concept real number.
> The simplest definition is this: A real number is a set S of
rationals
> such that S is nonempty, S does not contain every rational, S has
no
> smallest element, and for any rationals s, t, if s < t and s is in
S
> then t is in S.
> That's the definition, whether or not you agree that it's well.
>E. E. Escultura
> ************************
> David C. Ullrich
> A definition, certainly, but the definition?
> But I agree it is well.
Between any two irrationals is a rational.  You can't define all of the
real numbers with that definition because there are distinct real
numbers indistinguishable in that representation.
Thus, all three of you are wrong.
The real number is the reduction of the infinite dimensional vector
space that is implied by the zero dimensional point.  It's the
intersection of two skew lines on a plane.  Consider Points on a Line,
Infinity, the sidedness of points, and why you know what that word
means in this context, and why it is said.
You might have heard of the A, S, T, and U irrationals:  there are only
algebraic and transcendental irrationals, and there is only one kind of
transcendental irrational.  Then again, I could see why there could be
four kinds of irrationals.  Then again that would lead to a possibly
fallacious line of reasoning why we have five fingers on a hand, with
the opposable thumb, which is a simple maximization of millions of
years of evolution.
The continuity of the real numbers has to do with the deduction of
everything of the numbers at once, it demands a projective space.
There is a chicken-and-egg problem, much as there is with the
ur-element of set theory, to get a line from points without two lines
to make a point requires their synthesis all at once within a complete
and acceptably orthonormal infinite dimensional vector space:  twice
infinite Euclideanly and not.
E.E., if you want to discuss infinitesimals that's great, join the
club, there are hundreds and thousands who spend much time considering
i among the millions who simply accept it without question, since the
ancient academics of thousands of years ago, and more applicably today.
http://www.arxiv.org/list/math/recent
The state of the art of mathematics includes the geometric algebra and
particularly its applications and the symplectic integration and
differentiation.  The state of the art in mathematics includes many,
many things.
Ross Finlayson
===
Subject: Re: The-State-of-the-Art in Mathematics
>Hey, Escultura,
> <I disagree.
> <When you say dichotomy, do you mean trichotomy?
> Dichotomy axiom says: Given real numbers x, y, one and only one holds: x < 

y or x = y or x > y.
I think that's called trichotomy around here, with dichotomy, for 
cutting into two parts instead of cutting into three parts, dichotomy 
is x<y or x>=y.
Anyways, trichotomy holds well within the real numbers.  When you get 
outside of the real numbers, it might not.
> Brouwer constructed a counterexample to it which, incidentally, implies 
that the real numbers have no natural ordering.
Please state it here.
> Moreover, Banach Tarski constructed a counterexample to the completeness 
axiom of the real number system.
I have not heard of that.  Please state it here, on this thread or some 
other, on this newsgroup where you have brought forth these claims, so that 

others may validate your claims prima facie, that means from reading only 
the 
content on this newsgroup.
> <What are your vague numbers?  What is your definition of a real
> number?  Do you consider the nonstandard constructions of the surreal
> numbers, or hyperreals or colonies of germs and germinal functions?
> Nothing to do with rurreals, etc. I have a new construction without these 

false axioms using only three simple axioms. This relieves the real numbers 

of all contradictions while at the same time enriching them.
> No, I won.89t define a real number because I.89m giving a price of 
$1000 who can define it correctly but differently from mine. The offer 
expires this weekend.
> <You appear to promote geometry.
> <I'm getting somewhat defensive about all this.  I would like to not share 

with you.
> Yes, it.89s healthy to disagree, that.89s called critical 
thinking,  the beginning of great things.
> <Please go off about the real numbers.  You think they are decimals?
>Ross Finlayson
> To sum up, the present axiomatization of the real numbers is terribly 
wrong. Most of the concepts are ill-defined, two axioms are false and the 
topics on fractions are redundant because fractions are decimals.
> All to the best
> Eddie Escultura
> University of the Philippines
That's a fine claim, but it's worthless only because you have not presented 

your reasoning to the point where someone has to be able to explain it to 
another without saying only Escultura knows.
I think you're wrong, not in spirit, because there are some few limits of 
the standard real number constructions, but in basically all details.
So, present your actual arguments to sci.math, there are competent reviewers 

who will read it and put forth their comments.
Ross
===
Subject: The Parageodetic Distance Formula (UBasic)
<pre>
After some reflection--prompted by an e-mail inquiry--this writer/
programmer has decided to first release the parageodetic (inverse
problem) formulary core, before the more dressed, integrated
quadratic-mean-spherical, parageodetic and geodetic inverse and
direct GODx program core.
But, rather than just spit out the program listing in numerical
address order, however, it is being presented subroutine-with-execution,
thereby providing another opportunity to disseminate this unique
formulary approach (including all trigonometric quadrantic adjustments
and division by zero bulletproofing) for mathematical study and
discussion.
As the concept of the parageodesic appears little acknowledged--if at
all even recognized--it would seem prudent to comparatively define the
geodesic and parageodesic from the conceptual approach used here (to
the non-technical layman, the term graticular can be considered
synonymous with spherical):
       Geodesic:  The conformally delineated great-arc segment
                  between two points;
   Parageodesic:  The graticularly delineated great-arc segment
                  between two points.
Thus, with an ellipsoid, the traditional geodesic is the pulled
string path and distance on an elliptically correct model, whereas
the parageodesic is the pulled string path and distance, first
DEFINED on a spherical model of the ellipsoid, then the sphere is
reformed to its proper ellipsoidal shape, at which time the
elliptically correct arcradius and angle/azimuth AT ANY GIVEN POINT
ALONG THE DEFINED SEGMENT can be found (i.e., it is the local,
elliptically correct Great-Circle valuation!).
UBasic's system program can be found here:
           ftp://rkmath.rikkyo.ac.jp/pub/ubibm
                 (this web page is translated through Google, thus some of 
the wording is off)
UBasic's command dictionary can be found here:
          http://ed-thelen.org/bab/bab-ubhelp.html
Of course, while it is presented here out of numerical address order,
once the program code is cut and pasted into UBasic, it will revert to
its natural arrangement.
----------------------------------------------------------------------------

------------------
 3000  GoSub 58000
58000  '*** Radial Parameters ***
58010  a=6378.135:b=6356.75:b=b::GoTo 58200
58020  a=10000:b=5000
58200  Oz=2*Atan((a-b)^.5/(a+b)^.5)
10000  BLat=+43.66111111111111111111:BLong= -70.25583333333333333333
10010  DLat=+45.52305555555555555556:DLong=-122.67583333333333333333
     1'  Parageodetic = 4092.6024524649626 Km   
291.6521536Á/73.6370729Á
     1'      Geodetic = 4092.6016550196721 Km   
291.7142510Á/73.5743344Á
     1' Q-M-Spherical = 4082.4708695785296 Km   
291.7214263Á/73.5858490Á
 2200  Pi=#Pi:PH=.5*Pi:PD=2*Pi
 2300  RF=Pi/180
55110  BLr=BLat*RF:DLr=DLat*RF:MD=(DLong-BLong)*RF:M=Abs(MD):If M>Pi Then 
MD=MD-Sgn(MD)*PD
11000  GoSub 55000
1'############################################################
55000  '*** Spherical, ParaGeodetic Valuations ***
55100  ID%=0
55120  WBLr=DLr:WDLr=BLr:GoSub 55400:DAz(0)=WAz(0):TvL(5)=TvL(0)
55130  WBLr=BLr:WDLr=DLr:GoSub 55400:BAz(0)=WAz(0):TvL(1)=TvL(0)
1'##########
55400  '*** Spherical Elements ***
55410  GoSub 55420:GoSub 55440:Return
55420  SA=Cos(WDLr)*Sin(M):If SA=0 And Abs(WDLr)<PH And Sin(M)>0 Then 
SA=#Eps
55422  If Abs(WBLr/RF)>=90 Or Abs(WDLr/RF)>=90 Then SA=0
55425  SB=(Sin(WBLr+WDLr)*Sin(.5*M)^2)-(Sin(WBLr-WDLr)*Cos(.5*M)^2)
 2150  IO=10^100
55430  WAz(ID%)=Abs(Atan(IO*SA/(IO*SB+1)))
55440  TvL(0)=Atan(IO*Sin(WBLr)/(Abs(IO*Cos(WBLr)*Cos(WAz(ID%)))+1))
55445  If SB<0 Then WAz(ID%)=Pi-WAz(ID%)
55439  Return
55449  Return
1'----------
55125  GoSub 55190:DAz(1)=WAz(1)
55135  GoSub 55190:BAz(1)=WAz(1)
1'##########
55190  ID%=1:SA=SA*.E1p(0,WBLr):SB=SB*.E0p(0,WBLr):GoSub 55430:GoSub 
55445:ID%=0:Return
1'----------
 2100  Dcm=Int(Log(#Eps)/Log(0.1)-5):Syota=0.1^Dcm
55140  If Cos(BAz(0))<0 Then TvL(1)=Sgn(IO*BLr+Syota)*Pi-TvL(1)
55142  If Cos(DAz(0))>0 Then TvL(5)=Sgn(IO*DLr+Syota)*Pi-TvL(5)
55145  If TvL(5)<TvL(1) Then TvL(5)=PD+TvL(5)
55150  AD(0)=TvL(5)-TvL(1):ADq 
0=0.25*AD(0):BTvL=TvL(1)/RF:DTvL=TvL(5)/RF:ADg=DTvL-BTvL
55155  TvL(2)=TvL(1)+ADq 0:TvL(3)=TvL(1)+2*ADq 0:TvL(4)=TvL(1)+3*ADq 0
55147  GoSub 55450:APg=AP(0)/RF:GoSub 55485:ID%=1:GoSub 55485
1'##########
55450  
AP(ID%)=Atan(Abs(IO*Cos(WBLr)*Sin(WAz(ID%)))/(IO*(Cos(WAz(ID%))^2+(Sin(WBLr)
*
Sin(WAz(ID%)))^2)^.5+1)):Return
55485  If MD<0 Then BAz(ID%)=PD-BAz(ID%)
55487  If MD>0 Then DAz(ID%)=PD-DAz(ID%)
55489  BAz(ID%)=BAz(ID%)/RF:DAz(ID%)=DAz(ID%)/RF:Return
1'----------
55189  Return
1'------------------------------------------------------------
11100  GoSub 59100:KmDx(1)=a*EGDD*AD(0)
1'############################################################
59100  '*** PEOx ***
59000  '*** Integration ***
65100  
.E2p(AP,TvL):Return((Cos(Oz)^2+(Sin(AP)*Sin(Oz))^2+(Cos(AP)*Cos(TvL)*Sin(Oz)
)
^2)^.5)
     1'            = Elliptic Integrand Of The Second Kind
65110  .E1p(AP,TvL):Return(1/.E2p(AP,TvL))
     1'            = Elliptic Integrand Of The First Kind
     1'        x a = N (Normal) = Transverse Equatorial Arcradius
65120  .E0p(AP,TvL):Return(Cos(Oz)^2*.E1p(AP,TvL)^3)
     1'        x a = M (Meridian) = Vertical Meridional Arcradius
65130  
.EGp(AP,TvL):Return((.E0p(AP,TvL)^2+Sin(AP)^2*(IO*(.E1p(AP,TvL)^2-.E0p(AP,Tv
L
)^2)/(IO*(Cos(TvL)^2+(Sin(AP)*Sin(TvL))^2)+1)))^.5)
     1'        x a = O (Omniversal) = Transverse Meridional Arcradius
1'#########################
11020  XF=1:::For UL=2 To 20 Step 2::GoSub 60000
1' Integration Convergency for EGDD to .1^20 @ cos{Oz}, for XFxUL:  cos{Oz} 

= 1xUL/2xUL/3xUL
1'   .9999 = 6/8/6;    .995 = 8/12/15;   .99 = 8/12/15;   .9 = 11/16/18;
1'     .75 = 13/18/24;   .5 = 17/24/30;  .25 = 27/36/45;  .1 = 42/60/75;
1'     .01 = --/--/231;   (.001 ~=~ 20x94 = 1880;   .0001 ~=~ 200x97 = 
19400)
60000  '*** (Gaussian) Amplitudal Kernal Expansion ***
65000  
.AE(NS,NX,XF):AE=2*Atan(((XF-NX+Sin(.5*AEg(NS))^2)/(NX-Sin(.5*AEg(NS))^2))^.
5
):If AE>.5*Pi Then AE=Pi-AE
65009  Return(AE)
60100  DU=2*UL:For NS=1 To UL:CK=Cos(.5*Pi*(((4*NS)-1)/((4*UL)+1)))
60200  LP(1)=1:LP=1:For NP=1 To DU
60210  
LP(2)=LP(1):LP(1)=LP:LP=((((2*NP)-1)*LP(1)*CK)-((NP-1)*LP(2)))/NP:Next NP
60220  DP=DU*(LP(1)-(LP*CK))/(1-CK^2)
60230  CK(2)=CK(1):CK(1)=CK:CK=CK(1)-(LP/DP)
60235  If Abs(CK-CK(1))>(10*Syota) And CK<>CK(2) Then 60200
60300  AEg(NS)=2*Atan((1-CK)^.5/(1+CK)^.5)
60310  OE(NS)=Sin(AEg(NS))^2/(DU*LP(1)^2):Next NS:Return:::::?NS; 
;Using(,30),AEg(NS)/RF;   ;OE(NS)/RF:Next NS:List 60010:End:Return
1'-------------------------
59110  EGDD=0:For NS=1 To UL:For NX=1 To XF
59120  AE=AEg(NS):If XF>1 Then AE=.AE(NS,NX,XF)
59130  TW=XF*UL:ABq=Cos(AE)*ADq 0:OW=OE(NS)/TW
59140  UUP=TvL(4)+ABq:UMP=TvL(4)-ABq:LMP=TvL(2)+ABq:LLP=TvL(2)-ABq
59150  
EGDD=EGDD+0.25*(.EGp(AP(0),UUP)+.EGp(AP(0),UMP)+.EGp(AP(0),LMP)+.EGp(AP(0),L
L
P))*OW
59160  Next NX:Next NS:Return
1'------------------------------------------------------------
   10  ClS
 1000  Dim 
AEg(101),OE(101),LP(2),CK(2),AD(1),AP(1),TvL(5),KmDx(1),WAz(1),BAz(1),DAz(1)

 2000  '*** Program Constants/Standard Values ***
 2400  Deg$=Chr(248):'(Degree Sign)
58300  ?Tab(20);************************************:?Tab(19);/ 
Parageodetic Distance Formula Core :? *----------------* ( cos{Oz} 
=;Using(,20),Cos(Oz); ) *-------------------*
58310  ?Using(,13), |  a =;a; 
Km;;:?Tab(55-Len(Str(Int(b))));Using(,13),b =;b; Km; |:? 
*---------------------------------------------------------------------------
*
:Return
11003  ?Tab(7);BTvL, DTvL: ;Using(,20),TvL(1)/RF;Deg$,TvL(5)/RF;Deg$
11005  ?Tab(25);ADg: ;Using(,20),ADg;Deg$
11007  ?Tab(25);APg: ;Using(,20),APg;Deg$
11010  ?Tab(5);BAz g, DAz g: ;Using(,20),BAz(0);Deg$,DAz(0);Deg$
11012  ?Tab(5);BAz e, DAz e: ;Using(,20),BAz(1);Deg$,DAz(1);Deg$
11015  ?Tab(5);APe b, APe d: 
;Using(,20),.APe(AP(0),TvL(1))/RF;Deg$,.APe(AP(0),TvL(5))/RF;Deg$
65010  
.ROz(APg):Return(Atan(IO*Cos(APg*RF)*Sin(Oz)/(IO*(1-(Cos(APg*RF)*Sin(Oz))^2)
^
.5+1)))
     1'         = Reduced Oz:  cos{ROz{0Á}} = cos{Oz}, 
cos{ROz{90Á}} = 1 
65020  
.APe(AP,TvL):Return(Atan(Sin(AP)*(.E2p(AP,TvL)^4*.EGp(AP,TvL)^2-Sin(AP)^2)^(
-
.5)))
11017  GoSub 11990:?Tab(16);Cos{ROz{APg}} =;Using(,20),Cos(.Roz(APg)):?
11900  ?TW =;UL*XF;Tab(10);PEODx: ;Using(,20),KmDx(1); Km 
(;Using(,25),EGDD; ):Next:?
11910  GoSub 11990
11990  
?Tab(4);*------------------------------------------------------------------
---*:Return
11980  ?:List 11020:List 58010-58030:List 10000-10010:End
64000  ' ***  Functions  ***
    1  Point 7
     ' ~=~ .1^33
----------------------------------------------------------------------------

------------------
This program is presented as a workbench specimen, intended for
dissection and analysis.
For just a no-frills output, the following modification can be
made.
-------------------------------------------------------
Delete 11003-11900
 4000  XF=1:UL=10:GoSub 60000
11100  GoSub 59100:KmDx(1)=A*EGDD*AD(0)
11200  ?:?Tab(12);PEODx: ;Using(,20),KmDx(1); Km:?
11300  ?Tab(5);BAz e, DAz e: ;Using(,20),BAz(1);Deg$,DAz(1);Deg$
11980  ?:?:List 58010:?:?:?:List 10000:?:?:List 10010:End
58010  a=6378.135:b=6356.75
58020
-------------------------------------------------------
     ~Kaimbridge M. GoldChild~
</pre>
===
Subject: Re: The Parageodetic Distance Formula (UBasic)
<pre>
Gee, aren't you a clever boy.
</pre <pre After some reflection--prompted by an e-mail inquiry--this writer/
> programmer has decided to first release the parageodetic (inverse
> problem) formulary core, before the more dressed, integrated
> quadratic-mean-spherical, parageodetic and geodetic inverse and
> direct GODx program core.
> But, rather than just spit out the program listing in numerical
> address order, however, it is being presented subroutine-with-execution,
> thereby providing another opportunity to disseminate this unique
> formulary approach (including all trigonometric quadrantic adjustments
> and division by zero bulletproofing) for mathematical study and
> discussion.
> As the concept of the parageodesic appears little acknowledged--if at
> all even recognized--it would seem prudent to comparatively define the
> geodesic and parageodesic from the conceptual approach used here (to
> the non-technical layman, the term graticular can be considered
> synonymous with spherical):
>        Geodesic:  The conformally delineated great-arc segment
>                   between two points;
>    Parageodesic:  The graticularly delineated great-arc segment
>                   between two points.
> Thus, with an ellipsoid, the traditional geodesic is the pulled
> string path and distance on an elliptically correct model, whereas
> the parageodesic is the pulled string path and distance, first
> DEFINED on a spherical model of the ellipsoid, then the sphere is
> reformed to its proper ellipsoidal shape, at which time the
> elliptically correct arcradius and angle/azimuth AT ANY GIVEN POINT
> ALONG THE DEFINED SEGMENT can be found (i.e., it is the local,
> elliptically correct Great-Circle valuation!).
> UBasic's system program can be found here:
>            ftp://rkmath.rikkyo.ac.jp/pub/ubibm
> rkmath.rikkyo.ac.jp/~kida/ubasic.htm
>                  (this web page is translated through Google, thus some of 

> the wording is off)
> UBasic's command dictionary can be found here:
>           http://ed-thelen.org/bab/bab-ubhelp.html
> Of course, while it is presented here out of numerical address order,
> once the program code is cut and pasted into UBasic, it will revert to
> its natural arrangement.
> 
---------------------------------------------------------------------------
> -------------------
>  3000  GoSub 58000
> 58000  '*** Radial Parameters ***
> 58010  a=6378.135:b=6356.75:b=b::GoTo 58200
> 58020  a=10000:b=5000
> 58200  Oz=2*Atan((a-b)^.5/(a+b)^.5)
> 10000  BLat=+43.66111111111111111111:BLong= -70.25583333333333333333
> 10010  DLat=+45.52305555555555555556:DLong=-122.67583333333333333333
>      1'  Parageodetic = 4092.6024524649626 Km   291.6521536 /73.6370729
>      1'      Geodetic = 4092.6016550196721 Km   291.7142510 /73.5743344
>      1' Q-M-Spherical = 4082.4708695785296 Km   291.7214263 /73.5858490
>  2200  Pi=#Pi:PH=.5*Pi:PD=2*Pi
>  2300  RF=Pi/180
> 55110  BLr=BLat*RF:DLr=DLat*RF:MD=(DLong-BLong)*RF:M=Abs(MD):If M>P
> i Then MD=MD-Sgn(MD)*PD
> 11000  GoSub 55000
> 1'############################################################
> 55000  '*** Spherical, ParaGeodetic Valuations ***
> 55100  ID%=0
> 55120  WBLr=DLr:WDLr=BLr:GoSub 55400:DAz(0)=WAz(0):TvL(5)=TvL(0)
> 55130  WBLr=BLr:WDLr=DLr:GoSub 55400:BAz(0)=WAz(0):TvL(1)=TvL(0)
> 1'##########
> 55400  '*** Spherical Elements ***
> 55410  GoSub 55420:GoSub 55440:Return
> 55420  SA=Cos(WDLr)*Sin(M):If SA=0 And Abs(WDLr)<PH And Sin(M)>0 Then S
> A=#Eps
> 55422  If Abs(WBLr/RF)>=90 Or Abs(WDLr/RF)>=90 Then SA=0
> 55425  SB=(Sin(WBLr+WDLr)*Sin(.5*M)^2)-(Sin(WBLr-WDLr)*Cos(.5*M)^2)
>  2150  IO=10^100
> 55430  WAz(ID%)=Abs(Atan(IO*SA/(IO*SB+1)))
> 55440  TvL(0)=Atan(IO*Sin(WBLr)/(Abs(IO*Cos(WBLr)*Cos(WAz(ID%)))+1))
> 55445  If SB<0 Then WAz(ID%)=Pi-WAz(ID%)
> 55439  Return
> 55449  Return
> 1'----------
> 55125  GoSub 55190:DAz(1)=WAz(1)
> 55135  GoSub 55190:BAz(1)=WAz(1)
> 1'##########
> 55190  ID%=1:SA=SA*.E1p(0,WBLr):SB=SB*.E0p(0,WBLr):GoSub 55430:GoSub 
> 55445:ID%=0:Return
> 1'----------
>  2100  Dcm=Int(Log(#Eps)/Log(0.1)-5):Syota=0.1^Dcm
> 55140  If Cos(BAz(0))<0 Then TvL(1)=Sgn(IO*BLr+Syota)*Pi-TvL(1)
> 55142  If Cos(DAz(0))>0 Then TvL(5)=Sgn(IO*DLr+Syota)*Pi-TvL(5)
> 55145  If TvL(5)<TvL(1) Then TvL(5)=PD+TvL(5)
> 55150  AD(0)=TvL(5)-TvL(1):ADq 0=0.25*AD(0):BTvL=TvL(1)/RF:DTvL=TvL
> (5)/RF:ADg=DTvL-BTvL
> 55155  TvL(2)=TvL(1)+ADq 0:TvL(3)=TvL(1)+2*ADq 0:TvL(4)=TvL(1)+3*ADq 0
> 55147  GoSub 55450:APg=AP(0)/RF:GoSub 55485:ID%=1:GoSub 55485
> 1'##########
> 55450  AP(ID%)=Atan(Abs(IO*Cos(WBLr)*Sin(WAz(ID%)))/(IO*(Cos(WAz(ID%))^2+
> (Sin(WBLr)*Sin(WAz(ID%)))^2)^.5+1)):Return
> 55485  If MD<0 Then BAz(ID%)=PD-BAz(ID%)
> 55487  If MD>0 Then DAz(ID%)=PD-DAz(ID%)
> 55489  BAz(ID%)=BAz(ID%)/RF:DAz(ID%)=DAz(ID%)/RF:Return
> 1'----------
> 55189  Return
> 1'------------------------------------------------------------
> 11100  GoSub 59100:KmDx(1)=a*EGDD*AD(0)
> 1'############################################################
> 59100  '*** PEOx ***
> 59000  '*** Integration ***
> 65100  
.E2p(AP,TvL):Return((Cos(Oz)^2+(Sin(AP)*Sin(Oz))^2+(Cos(AP)*Cos(TvL)
> *Sin(Oz))^2)^.5)
>      1'            = Elliptic Integrand Of The Second Kind
> 65110  .E1p(AP,TvL):Return(1/.E2p(AP,TvL))
>      1'            = Elliptic Integrand Of The First Kind
>      1'        x a = N (Normal) = Transverse Equatorial Arcradius
> 65120  .E0p(AP,TvL):Return(Cos(Oz)^2*.E1p(AP,TvL)^3)
>      1'        x a = M (Meridian) = Vertical Meridional Arcradius
> 65130  
.EGp(AP,TvL):Return((.E0p(AP,TvL)^2+Sin(AP)^2*(IO*(.E1p(AP,TvL)^2-.E
> 0p(AP,TvL)^2)/(IO*(Cos(TvL)^2+(Sin(AP)*Sin(TvL))^2)+1)))^.5)
>      1'        x a = O (Omniversal) = Transverse Meridional Arcradius
> 1'#########################
> 11020  XF=1:::For UL=2 To 20 Step 2::GoSub 60000
> 1' Integration Convergency for EGDD to .1^20 @ cos{Oz}, for XFxUL:  
cos{Oz}
>  = 1xUL/2xUL/3xUL
> 1'   .9999 = 6/8/6;    .995 = 8/12/15;   .99 = 8/12/15;   .9 = 11/1
> 6/18;
> 1'     .75 = 13/18/24;   .5 = 17/24/30;  .25 = 27/36/45;  .1 = 42/6
> 0/75;
> 1'     .01 = --/--/231;   (.001 ~=~ 20x94 = 1880;   .0001 ~=~ 200x9
> 7 = 19400)
> 60000  '*** (Gaussian) Amplitudal Kernal Expansion ***
> 65000  .AE(NS,NX,XF):AE=2*Atan(((XF-NX+Sin(.5*AEg(NS))^2)/(NX-Sin(.5*AEg(
> NS))^2))^.5):If AE>.5*Pi Then AE=Pi-AE
> 65009  Return(AE)
> 60100  DU=2*UL:For NS=1 To UL:CK=Cos(.5*Pi*(((4*NS)-1)/((4*UL)+1)))
> 60200  LP(1)=1:LP=1:For NP=1 To DU
> 60210  LP(2)=LP(1):LP(1)=LP:LP=((((2*NP)-1)*LP(1)*CK)-((NP-1)*LP(2)))
> /NP:Next NP
> 60220  DP=DU*(LP(1)-(LP*CK))/(1-CK^2)
> 60230  CK(2)=CK(1):CK(1)=CK:CK=CK(1)-(LP/DP)
> 60235  If Abs(CK-CK(1))>(10*Syota) And CK<>CK(2) Then 60200
> 60300  AEg(NS)=2*Atan((1-CK)^.5/(1+CK)^.5)
> 60310  OE(NS)=Sin(AEg(NS))^2/(DU*LP(1)^2):Next NS:Return:::::?NS; 
;Usin
> g(,30),AEg(NS)/RF;   ;OE(NS)/RF:Next NS:List 60010:End:Return
> 1'-------------------------
> 59110  EGDD=0:For NS=1 To UL:For NX=1 To XF
> 59120  AE=AEg(NS):If XF>1 Then AE=.AE(NS,NX,XF)
> 59130  TW=XF*UL:ABq=Cos(AE)*ADq 0:OW=OE(NS)/TW
> 59140  UUP=TvL(4)+ABq:UMP=TvL(4)-ABq:LMP=TvL(2)+ABq:LLP=TvL(2)-ABq
> 59150  EGDD=EGDD+0.25*(.EGp(AP(0),UUP)+.EGp(AP(0),UMP)+.EGp(AP(0),LMP)+.E
> Gp(AP(0),LLP))*OW
> 59160  Next NX:Next NS:Return
> 1'------------------------------------------------------------
>    10  ClS
>  1000  Dim 
AEg(101),OE(101),LP(2),CK(2),AD(1),AP(1),TvL(5),KmDx(1),WAz(1),B
> Az(1),DAz(1)
>  2000  '*** Program Constants/Standard Values ***
>  2400  Deg$=Chr(248):'(Degree Sign)
> 58300  ?Tab(20);************************************:?Tab(19);/ 
Parageod
> etic Distance Formula Core :? *----------------* ( cos{Oz} 
=;Using(,2
> 0),Cos(Oz); ) *-------------------*
> 58310  ?Using(,13), |  a =;a; 
Km;;:?Tab(55-Len(Str(Int(b))));Using(,1
> 3),b =;b; Km; |:? 
*-------------------------------------------------
> --------------------------*:Return
> 11003  ?Tab(7);BTvL, DTvL: ;Using(,20),TvL(1)/RF;Deg$,TvL(5)/RF;Deg$
> 11005  ?Tab(25);ADg: ;Using(,20),ADg;Deg$
> 11007  ?Tab(25);APg: ;Using(,20),APg;Deg$
> 11010  ?Tab(5);BAz g, DAz g: ;Using(,20),BAz(0);Deg$,DAz(0);Deg$
> 11012  ?Tab(5);BAz e, DAz e: ;Using(,20),BAz(1);Deg$,DAz(1);Deg$
> 11015  ?Tab(5);APe b, APe d: 
;Using(,20),.APe(AP(0),TvL(1))/RF;Deg$,.APe(
> AP(0),TvL(5))/RF;Deg$
> 65010  
.ROz(APg):Return(Atan(IO*Cos(APg*RF)*Sin(Oz)/(IO*(1-(Cos(APg*RF)*Sin
> (Oz))^2)^.5+1)))
>      1'         = Reduced Oz:  cos{ROz{0 }} = cos{Oz}, cos{ROz{90
>  }} = 1 
> 65020  
.APe(AP,TvL):Return(Atan(Sin(AP)*(.E2p(AP,TvL)^4*.EGp(AP,TvL)^2-Sin(
> AP)^2)^(-.5)))
> 11017  GoSub 11990:?Tab(16);Cos{ROz{APg}} 
=;Using(,20),Cos(.Roz(APg)):?
> 11900  ?TW =;UL*XF;Tab(10);PEODx: ;Using(,20),KmDx(1); Km 
(;Using(,
> 25),EGDD; ):Next:?
> 11910  GoSub 11990
> 11990  
?Tab(4);*----------------------------------------------------------
> -----------*:Return
> 11980  ?:List 11020:List 58010-58030:List 10000-10010:End
> 64000  ' ***  Functions  ***
>     1  Point 7
>      ' ~=~ .1^33
> 
---------------------------------------------------------------------------
> -------------------
> This program is presented as a workbench specimen, intended for
> dissection and analysis.
> For just a no-frills output, the following modification can be
> made.
> -------------------------------------------------------
> Delete 11003-11900
>  4000  XF=1:UL=10:GoSub 60000
> 11100  GoSub 59100:KmDx(1)=A*EGDD*AD(0)
> 11200  ?:?Tab(12);PEODx: ;Using(,20),KmDx(1); Km:?
> 11300  ?Tab(5);BAz e, DAz e: ;Using(,20),BAz(1);Deg$,DAz(1);Deg$
> 11980  ?:?:List 58010:?:?:?:List 10000:?:?:List 10010:End
> 58010  a=6378.135:b=6356.75
> 58020
> -------------------------------------------------------
>      ~Kaimbridge M. GoldChild~
> </pre>
        [FN42nn]
===
Subject: The Parageodetic Distance Formula (UBasic)
<pre>
Aftersome[NonBreakingSpace]reflection--prompted[NonBreakingSpace]by[NonBr
eakingSpace]an[NonBreakingSpace]e-mail[NonBreakingSpace]inquiry--this[NonB
reakingSpace]writer/
programmerhas[NonBreakingSpace]decided[NonBreakingSpace]to[NonBreakingSpa
ce]first[NonBreakingSpace]release[NonBreakingSpace]the[NonBreakingSpace]pa
rageodetic[NonBreakingSpace](inverse
problem)formulary[NonBreakingSpace]core,[NonBreakingSpace]before[NonBreak
ingSpace]the[NonBreakingSpace]more[NonBreakingSpace]dressed,[NonBreakingSp
ace]integrated
quadratic-mean-spherical,parageodetic[NonBreakingSpace]and[NonBreakingSpac
e]geodetic[NonBreakingSpace]inverse[NonBreakingSpace]and
directGODx[NonBreakingSpace]program[NonBreakingSpace]core.
But,rather[NonBreakingSpace]than[NonBreakingSpace]just[NonBreakingSpace]s
pit[NonBreakingSpace]out[NonBreakingSpace]the[NonBreakingSpace]program[No
nBreakingSpace]listing[NonBreakingSpace]in[NonBreakingSpace]numerical
addressorder,[NonBreakingSpace]however,[NonBreakingSpace]it[NonBreakingSp
ace]is[NonBreakingSpace]being[NonBreakingSpace]presented[NonBreakingSpace]
subroutine-with-execution,
therebyproviding[NonBreakingSpace]another[NonBreakingSpace]opportunity[No
nBreakingSpace]to[NonBreakingSpace]disseminate[NonBreakingSpace]this[NonBr
eakingSpace]unique
formularyapproach[NonBreakingSpace](including[NonBreakingSpace]all[NonBre
akingSpace]trigonometric[NonBreakingSpace]quadrantic[NonBreakingSpace]adjus
tments
anddivision[NonBreakingSpace]by[NonBreakingSpace]zero[NonBreakingSpac
e]bulletproofing)[NonBreakingSpace]for[NonBreakingSpace]mathematical[NonBr
eakingSpace]study[NonBreakingSpace]and
discussion.
Asthe[NonBreakingSpace]concept[NonBreakingSpace]of[NonBreakingSpace]the[
NonBreakingSpace]parageodesic[NonBreakingSpace]appears[NonBreakingSpace]lit
tle[NonBreakingSpace]acknowledged--if[NonBreakingSpace]at
alleven[NonBreakingSpace]recognized--it[NonBreakingSpace]would[NonBreakin
gSpace]seem[NonBreakingSpace]prudent[NonBreakingSpace]to[NonBreakingSpace]
comparatively[NonBreakingSpace]define[NonBreakingSpace]the
geodesicand[NonBreakingSpace]parageodesic[NonBreakingSpace]from[NonBreaki
ngSpace]the[NonBreakingSpace]conceptual[NonBreakingSpace]approach[NonBreak
ingSpace]used[NonBreakingSpace]here[NonBreakingSpace](to
thenon-technical[NonBreakingSpace]layman,[NonBreakingSpace]the[NonBreakin
gSpace]term[NonBreakingSpace]graticular[NonBreakingSpace]can[NonBreaki
ngSpace]be[NonBreakingSpace]considered
synonymouswith[NonBreakingSpace]spherical):
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace]Geodesic:[NonBreakingSpace][NonBreaki
ngSpace]The[NonBreakingSpace]conformally[NonBreakingSpace]delineated[NonBr
eakingSpace]great-arc[NonBreakingSpace]segment
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace]between[NonBreakingSpace]two[NonBreakingSpace]points;
[NonBreakingSpace][NonBreakingSpace]Parageodesic:[NonBreakingSpace][NonB
reakingSpace]The[NonBreakingSpace]graticularly[NonBreakingSpace]delineated
[NonBreakingSpace]great-arc[NonBreakingSpace]segment
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace]between[NonBreakingSpace]two[NonBreakingSpace]points.
Thus,with[NonBreakingSpace]an[NonBreakingSpace]ellipsoid,[NonBreakingSpac
e]the[NonBreakingSpace]traditional[NonBreakingSpace]geodesic[NonBreakingSp
ace]is[NonBreakingSpace]the[NonBreakingSpace]pulled
stringpath[NonBreakingSpace]and[NonBreakingSpace]distance[NonBreakingSp
ace]on[NonBreakingSpace]an[NonBreakingSpace]elliptically[NonBreakingSpace]
correct[NonBreakingSpace]model,[NonBreakingSpace]whereas
theparageodesic[NonBreakingSpace]is[NonBreakingSpace]the[NonBreakingSpace
]pulled[NonBreakingSpace]string[NonBreakingSpace]path[NonBreakingSpace
]and[NonBreakingSpace]distance,[NonBreakingSpace]first
DEFINEDon[NonBreakingSpace]a[NonBreakingSpace]spherical[NonBreakingSpace]
model[NonBreakingSpace]of[NonBreakingSpace]the[NonBreakingSpace]ellipsoid,
[NonBreakingSpace]then[NonBreakingSpace]the[NonBreakingSpace]sphere[NonBr
eakingSpace]is
reformedto[NonBreakingSpace]its[NonBreakingSpace]proper[NonBreakingSpace]
ellipsoidal[NonBreakingSpace]shape,[NonBreakingSpace]at[NonBreakingSpace]w
hich[NonBreakingSpace]time[NonBreakingSpace]the
ellipticallycorrect[NonBreakingSpace]arcradius[NonBreakingSpace]and[NonBr
eakingSpace]angle/azimuth[NonBreakingSpace]AT[NonBreakingSpace]ANY[NonBrea
kingSpace]GIVEN[NonBreakingSpace]POINT
ALONGTHE[NonBreakingSpace]DEFINED[NonBreakingSpace]SEGMENT[NonBreakingSpa
ce]can[NonBreakingSpace]be[NonBreakingSpace]found[NonBreakingSpace](i.e.,
[NonBreakingSpace]it[NonBreakingSpace]is[NonBreakingSpace]the[NonBreaking
Space]local,
ellipticallycorrect[NonBreakingSpace]Great-Circle[NonBreakingSpace]val
uation!).
UBasic'ssystem[NonBreakingSpace]program[NonBreakingSpace]can[NonBreakingS
pace]be[NonBreakingSpace]found[NonBreakingSpace]here:
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace]ftp://rkmath.rikkyo.ac.jp/pub/ubibm
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace](
this[NonBreakingSpace]web[NonBreakingSpace]page[NonBreakingSpace]is[NonBr
eakingSpace]translated[NonBreakingSpace]through[NonBreakingSpace]Google,[N
onBreakingSpace]thus[NonBreakingSpace]some[NonBreakingSpace]of[NonBreaking
Space]the[NonBreakingSpace]wording[NonBreakingSpace]is[NonBreakingSpace]of
f)
UBasic'scommand[NonBreakingSpace]dictionary[NonBreakingSpace]can[NonBreak
ingSpace]be[NonBreakingSpace]found[NonBreakingSpace]here:
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace]http://ed-thelen.org/bab/bab-ubhelp.html
Ofcourse,[NonBreakingSpace]while[NonBreakingSpace]it[NonBreakingSpace]is
[NonBreakingSpace]presented[NonBreakingSpace]here[NonBreakingSpace]out[No
nBreakingSpace]of[NonBreakingSpace]numerical[NonBreakingSpace]address[NonB
reakingSpace]order,
oncethe[NonBreakingSpace]program[NonBreakingSpace]code[NonBreakingSpace]i
s[NonBreakingSpace]cut[NonBreakingSpace]and[NonBreakingSpace]pasted[NonBr
eakingSpace]into[NonBreakingSpace]UBasic,[NonBreakingSpace]it[NonBreakingS
pace]will[NonBreakingSpace]revert[NonBreakingSpace]to
itsnatural[NonBreakingSpace]arrangement.
----------------------------------------------------------------------------

------------------
3000[NonBreakingSpace][NonBreakingSpace]GoSub[NonBreakingSpace]58000
58000[NonBreakingSpace]'***[NonBreakingSpace]Radial[NonBreakingSpace]Para
meters[NonBreakingSpace]***
58010[NonBreakingSpace]a=6378.135:b=6356.75:b=b::GoTo[NonBreakingSpace]582
00
58020[NonBreakingSpace]a=10000:b=5000
58200[NonBreakingSpace]Oz=2*Atan((a-b)^.5/(a+b)^.5)
10000[NonBreakingSpace]BLat=+43.66111111111111111111:BLong=[NonBreakingSpa
ce]-70.25583333333333333333
10010[NonBreakingSpace]DLat=+45.52305555555555555556:DLong=-122.67583333333
333333333
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
1'[NonBreakingSpace][NonBreakingSpace]Parageodetic[NonBreakingSpace]=[Non
BreakingSpace]4092.6024524649626[NonBreakingSpace]Km[NonBreakingSpace][Non
BreakingSpace][NonBreakingSpace]291.6521536¡/73.6370729[Do
wnExclamation]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
1'[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace]Geodetic[NonBreakingSpace]=[NonBreak
ingSpace]4092.6016550196721[NonBreakingSpace]Km[NonBreakingSpace][NonBreak
ingSpace][NonBreakingSpace]291.7142510¡/73.5743344[DownExc
lamation]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
1'[NonBreakingSpace]Q-M-Spherical[NonBreakingSpace]=[NonBreakingSpace]4082
.4708695785296[NonBreakingSpace]Km[NonBreakingSpace][NonBreakingSpace][No
nBreakingSpace]291.7214263¡/73.5858490¡
2200[NonBreakingSpace][NonBreakingSpace]Pi=#Pi:PH=.5*Pi:PD=2*Pi
2300[NonBreakingSpace][NonBreakingSpace]RF=Pi/180
55110[NonBreakingSpace]BLr=BLat*RF:DLr=DLat*RF:MD=(DLong-BLong)*RF:M=Abs(MD
):If[NonBreakingSpace]M>Pi[NonBreakingSpace]Then[NonBreakingSpace]MD=MD-Sg
n(MD)*PD
11000[NonBreakingSpace]GoSub[NonBreakingSpace]55000
1'############################################################
55000[NonBreakingSpace]'***[NonBreakingSpace]Spherical,[NonBreakingSpace]
ParaGeodetic[NonBreakingSpace]Valuations[NonBreakingSpace]***
55100[NonBreakingSpace]ID%=0
55120[NonBreakingSpace]WBLr=DLr:WDLr=BLr:GoSub[NonBreakingSpace]55400:DAz(
0)=WAz(0):TvL(5)=TvL(0)
55130[NonBreakingSpace]WBLr=BLr:WDLr=DLr:GoSub[NonBreakingSpace]55400:BAz(
0)=WAz(0):TvL(1)=TvL(0)
1'##########
55400[NonBreakingSpace]'***[NonBreakingSpace]Spherical[NonBreakingSpace]E
lements[NonBreakingSpace]***
55410[NonBreakingSpace]GoSub[NonBreakingSpace]55420:GoSub[NonBreakingSpac
e]55440:Return
55420[NonBreakingSpace]SA=Cos(WDLr)*Sin(M):If[NonBreakingSpace]SA=0[NonBr
eakingSpace]And[NonBreakingSpace]Abs(WDLr)<PH[NonBreakingSpace]And[NonBrea
kingSpace]Sin(M)>0[NonBreakingSpace]Then[NonBreakingSpace]SA=#Eps
55422[NonBreakingSpace]If[NonBreakingSpace]Abs(WBLr/RF)>=90[NonBreakingSp
ace]Or[NonBreakingSpace]Abs(WDLr/RF)>=90[NonBreakingSpace]Then[NonBreaking
Space]SA=0
55425[NonBreakingSpace]SB=(Sin(WBLr+WDLr)*Sin(.5*M)^2)-(Sin(WBLr-WDLr)*Cos(
.5*M)^2)
2150[NonBreakingSpace][NonBreakingSpace]IO=10^100
55430[NonBreakingSpace]WAz(ID%)=Abs(Atan(IO*SA/(IO*SB+1)))
55440[NonBreakingSpace]TvL(0)=Atan(IO*Sin(WBLr)/(Abs(IO*Cos(WBLr)*Cos(WAz(I
D%)))+1))
55445[NonBreakingSpace]If[NonBreakingSpace]SB<0[NonBreakingSpace]Then[No
nBreakingSpace]WAz(ID%)=Pi-WAz(ID%)
55439[NonBreakingSpace]Return
55449[NonBreakingSpace]Return
1'----------
55125[NonBreakingSpace]GoSub[NonBreakingSpace]55190:DAz(1)=WAz(1)
55135[NonBreakingSpace]GoSub[NonBreakingSpace]55190:BAz(1)=WAz(1)
1'##########
55190[NonBreakingSpace]ID%=1:SA=SA*.E1p(0,WBLr):SB=SB*.E0p(0,WBLr):GoSub[N
onBreakingSpace]55430:GoSub[NonBreakingSpace]55445:ID%=0:Return
1'----------
2100[NonBreakingSpace][NonBreakingSpace]Dcm=Int(Log(#Eps)/Log(0.1)-5):Syot
a=0.1^Dcm
55140[NonBreakingSpace]If[NonBreakingSpace]Cos(BAz(0))<0[NonBreakingSpace
]Then[NonBreakingSpace]TvL(1)=Sgn(IO*BLr+Syota)*Pi-TvL(1)
55142[NonBreakingSpace]If[NonBreakingSpace]Cos(DAz(0))>0[NonBreakingSpace
]Then[NonBreakingSpace]TvL(5)=Sgn(IO*DLr+Syota)*Pi-TvL(5)
55145[NonBreakingSpace]If[NonBreakingSpace]TvL(5)<TvL(1)[NonBreakingSpace
]Then[NonBreakingSpace]TvL(5)=PD+TvL(5)
55150[NonBreakingSpace]AD(0)=TvL(5)-TvL(1):ADq_0=0.25*AD(0):BTvL=TvL(1)/RF:
DTvL=TvL(5)/RF:ADg=DTvL-BTvL
55155[NonBreakingSpace]TvL(2)=TvL(1)+ADq_0:TvL(3)=TvL(1)+2*ADq_0:TvL(4)=TvL
(1)+3*ADq_0
55147[NonBreakingSpace]GoSub[NonBreakingSpace]55450:APg=AP(0)/RF:GoSub[No
nBreakingSpace]55485:ID%=1:GoSub[NonBreakingSpace]55485
1'##########
55450[NonBreakingSpace]AP(ID%)=Atan(Abs(IO*Cos(WBLr)*Sin(WAz(ID%)))/(IO*(Co
s(WAz(ID%))^2+(Sin(WBLr)*Sin(WAz(ID%)))^2)^.5+1)):Return
55485[NonBreakingSpace]If[NonBreakingSpace]MD<0[NonBreakingSpace]Then[No
nBreakingSpace]BAz(ID%)=PD-BAz(ID%)
55487[NonBreakingSpace]If[NonBreakingSpace]MD>0[NonBreakingSpace]Then[No
nBreakingSpace]DAz(ID%)=PD-DAz(ID%)
55489[NonBreakingSpace]BAz(ID%)=BAz(ID%)/RF:DAz(ID%)=DAz(ID%)/RF:Return
1'----------
55189[NonBreakingSpace]Return
1'------------------------------------------------------------
11100[NonBreakingSpace]GoSub[NonBreakingSpace]59100:KmDx(1)=a*EGDD*AD(0)
1'############################################################
59100[NonBreakingSpace]'***[NonBreakingSpace]PEOx[NonBreakingSpace]***
59000[NonBreakingSpace]'***[NonBreakingSpace]Integration[NonBreakingSpace
]***
65100[NonBreakingSpace].E2p(AP,TvL):Return((Cos(Oz)^2+(Sin(AP)*Sin(Oz))^2+(
Cos(AP)*Cos(TvL)*Sin(Oz))^2)^.5)
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
1'[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]=
[NonBreakingSpace]Elliptic[NonBreakingSpace]Integrand[NonBreakingSpace]Of
[NonBreakingSpace]The[NonBreakingSpace]Second[NonBreakingSpace]Kind
65110[NonBreakingSpace].E1p(AP,TvL):Return(1/.E2p(AP,TvL))
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
1'[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]=
[NonBreakingSpace]Elliptic[NonBreakingSpace]Integrand[NonBreakingSpace]Of
[NonBreakingSpace]The[NonBreakingSpace]First[NonBreakingSpace]Kind
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
1'[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
x[NonBreakingSpace]a[NonBreakingSpace]=[NonBreakingSpace]N[NonBreakingSpa
ce](Normal)[NonBreakingSpace]=[NonBreakingSpace]Transverse[NonBreaking
Space]Equatorial[NonBreakingSpace]Arcradius
65120[NonBreakingSpace].E0p(AP,TvL):Return(Cos(Oz)^2*.E1p(AP,TvL)^3)
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
1'[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
x[NonBreakingSpace]a[NonBreakingSpace]=[NonBreakingSpace]M[NonBreakingSpa
ce](Meridian)[NonBreakingSpace]=[NonBreakingSpace]Vertical[NonBreaking
Space]Meridional[NonBreakingSpace]Arcradius
65130[NonBreakingSpace].EGp(AP,TvL):Return((.E0p(AP,TvL)^2+Sin(AP)^2*(IO*(.
E1p(AP,TvL)^2-.E0p(AP,TvL)^2)/(IO*(Cos(TvL)^2+(Sin(AP)*Sin(TvL))^2)+1)))^.5)

[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
1'[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
x[NonBreakingSpace]a[NonBreakingSpace]=[NonBreakingSpace]O[NonBreakingSpa
ce](Omniversal)[NonBreakingSpace]=[NonBreakingSpace]Transverse[NonBrea
kingSpace]Meridional[NonBreakingSpace]Arcradius
1'#########################
11020[NonBreakingSpace]XF=1:::For[NonBreakingSpace]UL=2[NonBreakingSpace]
To[NonBreakingSpace]20[NonBreakingSpace]Step[NonBreakingSpace]2::GoSub[No
nBreakingSpace]60000
1'Integration[NonBreakingSpace]Convergency[NonBreakingSpace]for[NonBreaki
ngSpace]EGDD[NonBreakingSpace]to[NonBreakingSpace].1^20[NonBreakingSpace]@
[NonBreakingSpace]cos{Oz},[NonBreakingSpace]for[NonBreakingSpace]XFxUL:[N
onBreakingSpace][NonBreakingSpace]cos{Oz}[NonBreakingSpace]=[NonBreakingSp
ace]1xUL/2xUL/3xUL
1'[NonBreakingSpace][NonBreakingSpace].9999[NonBreakingSpace]=[NonBreaki
ngSpace]6/8/6;[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonB
reakingSpace].995[NonBreakingSpace]=[NonBreakingSpace]8/12/15;[NonBreaking
Space][NonBreakingSpace][NonBreakingSpace].99[NonBreakingSpace]=[NonBreak
ingSpace]8/12/15;[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace].9
[NonBreakingSpace]=[NonBreakingSpace]11/16/18;
1'[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpac
e].75[NonBreakingSpace]=[NonBreakingSpace]13/18/24;[NonBreakingSpace][Non
BreakingSpace][NonBreakingSpace].5[NonBreakingSpace]=[NonBreakingSpace]17/
24/30;[NonBreakingSpace][NonBreakingSpace].25[NonBreakingSpace]=[NonBreak
ingSpace]27/36/45;[NonBreakingSpace][NonBreakingSpace].1[NonBreakingSpace]
=[NonBreakingSpace]42/60/75;
1'[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpac
e].01[NonBreakingSpace]=[NonBreakingSpace]--/--/231;[NonBreakingSpace][No
nBreakingSpace][NonBreakingSpace](.001[NonBreakingSpace]~=~[NonBreakingSpa
ce]20x94[NonBreakingSpace]=[NonBreakingSpace]1880;[NonBreakingSpace][NonB
reakingSpace][NonBreakingSpace].0001[NonBreakingSpace]~=~[NonBreakingSpace
]200x97[NonBreakingSpace]=[NonBreakingSpace]19400)
60000[NonBreakingSpace]'***[NonBreakingSpace](Gaussian)[NonBreakingSpace]
Amplitudal[NonBreakingSpace]Kernal[NonBreakingSpace]Expansion[NonBreakingS
pace]***
65000[NonBreakingSpace].AE(NS,NX,XF):AE=2*Atan(((XF-NX+Sin(.5*AEg(NS))^2)/(
NX-Sin(.5*AEg(NS))^2))^.5):If[NonBreakingSpace]AE>.5*Pi[NonBreakingSpace]Th
en[NonBreakingSpace]AE=Pi-AE
65009[NonBreakingSpace]Return(AE)
60100[NonBreakingSpace]DU=2*UL:For[NonBreakingSpace]NS=1[NonBreakingSpace
]To[NonBreakingSpace]UL:CK=Cos(.5*Pi*(((4*NS)-1)/((4*UL)+1)))
60200[NonBreakingSpace]LP(1)=1:LP=1:For[NonBreakingSpace]NP=1[NonBreaking
Space]To[NonBreakingSpace]DU
60210[NonBreakingSpace]LP(2)=LP(1):LP(1)=LP:LP=((((2*NP)-1)*LP(1)*CK)-((NP-
1)*LP(2)))/NP:Next[NonBreakingSpace]NP
60220[NonBreakingSpace]DP=DU*(LP(1)-(LP*CK))/(1-CK^2)
60230[NonBreakingSpace]CK(2)=CK(1):CK(1)=CK:CK=CK(1)-(LP/DP)
60235[NonBreakingSpace]If[NonBreakingSpace]Abs(CK-CK(1))>(10*Syota)[NonBr
eakingSpace]And[NonBreakingSpace]CK<>CK(2)[NonBreakingSpace]Then[NonBreaki
ngSpace]60200
60300[NonBreakingSpace]AEg(NS)=2*Atan((1-CK)^.5/(1+CK)^.5)
60310[NonBreakingSpace]OE(NS)=Sin(AEg(NS))^2/(DU*LP(1)^2):Next[NonBreaking
Space]NS:Return:::::?NS;[NonBreakingSpace];Using(,30),AEg(NS)/RF;[Non
BreakingSpace][NonBreakingSpace][NonBreakingSpace];OE(NS)/RF:Next[NonBre
akingSpace]NS:List[NonBreakingSpace]60010:End:Return
1'-------------------------
59110[NonBreakingSpace]EGDD=0:For[NonBreakingSpace]NS=1[NonBreakingSpace]
To[NonBreakingSpace]UL:For[NonBreakingSpace]NX=1[NonBreakingSpace]To[NonB
reakingSpace]XF
59120[NonBreakingSpace]AE=AEg(NS):If[NonBreakingSpace]XF>1[NonBreakingSpa
ce]Then[NonBreakingSpace]AE=.AE(NS,NX,XF)
59130[NonBreakingSpace]TW=XF*UL:ABq=Cos(AE)*ADq_0:OW=OE(NS)/TW
59140[NonBreakingSpace]UUP=TvL(4)+ABq:UMP=TvL(4)-ABq:LMP=TvL(2)+ABq:LLP=TvL
(2)-ABq
59150[NonBreakingSpace]EGDD=EGDD+0.25*(.EGp(AP(0),UUP)+.EGp(AP(0),UMP)+.EGp
(AP(0),LMP)+.EGp(AP(0),LLP))*OW
59160[NonBreakingSpace]Next[NonBreakingSpace]NX:Next[NonBreakingSpace]NS:
Return
1'------------------------------------------------------------
[NonBreakingSpace][NonBreakingSpace]10[NonBreakingSpace][NonBreakingSpac
e]ClS
1000[NonBreakingSpace][NonBreakingSpace]Dim[NonBreakingSpace]AEg(101),OE(
101),LP(2),CK(2),AD(1),AP(1),TvL(5),KmDx(1),WAz(1),BAz(1),DAz(1)
2000[NonBreakingSpace][NonBreakingSpace]'***[NonBreakingSpace]Program[No
nBreakingSpace]Constants/Standard[NonBreakingSpace]Values[NonBreakingSpace]
***
2400[NonBreakingSpace][NonBreakingSpace]Deg$=Chr(248):'(Degree[NonBreakin
gSpace]Sign)
58300[NonBreakingSpace]?Tab(20);************************************:?T
ab(19);/[NonBreakingSpace]Parageodetic[NonBreakingSpace]Distance[NonBrea
kingSpace]Formula[NonBreakingSpace]Core[NonBreakingSpace]:?[NonBreak
ingSpace]*----------------*[NonBreakingSpace]([NonBreakingSpace]cos{Oz}[No
nBreakingSpace]=;Using(,20),Cos(Oz);[NonBreakingSpace])[NonBreakingSpac
e]*-------------------*
58310[NonBreakingSpace]?Using(,13),[NonBreakingSpace]|[NonBreakingSpace
][NonBreakingSpace]a[NonBreakingSpace]=;a;[NonBreakingSpace]Km;;:?Ta
b(55-Len(Str(Int(b))));Using(,13),b[NonBreakingSpace]=;b;[NonBreaking
Space]Km;[NonBreakingSpace]|:?[NonBreakingSpace]*----------------------
-----------------------------------------------------*:Return
11003[NonBreakingSpace]?Tab(7);BTvL,[NonBreakingSpace]DTvL:[NonBreaking
Space];Using(,20),TvL(1)/RF;Deg$,TvL(5)/RF;Deg$
11005[NonBreakingSpace]?Tab(25);ADg:[NonBreakingSpace];Using(,20),ADg;
Deg$
11007[NonBreakingSpace]?Tab(25);APg:[NonBreakingSpace];Using(,20),APg;
Deg$
11010[NonBreakingSpace]?Tab(5);BAz_g,[NonBreakingSpace]DAz_g:[NonBreaki
ngSpace];Using(,20),BAz(0);Deg$,DAz(0);Deg$
11012[NonBreakingSpace]?Tab(5);BAz_e,[NonBreakingSpace]DAz_e:[NonBreaki
ngSpace];Using(,20),BAz(1);Deg$,DAz(1);Deg$
11015[NonBreakingSpace]?Tab(5);APe_b,[NonBreakingSpace]APe_d:[NonBreaki
ngSpace];Using(,20),.APe(AP(0),TvL(1))/RF;Deg$,.APe(AP(0),TvL(5))/RF;Deg$
65010[NonBreakingSpace].ROz(APg):Return(Atan(IO*Cos(APg*RF)*Sin(Oz)/(IO*(1-
(Cos(APg*RF)*Sin(Oz))^2)^.5+1)))
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
1'[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
[NonBreakingSpace]=[NonBreakingSpace]Reduced[NonBreakingSpace]Oz:[Non
BreakingSpace][NonBreakingSpace]cos{ROz{0¡}}[NonBreakingSpa
ce]=[No
nBreakingSpace]cos{Oz},[NonBreakingSpace]cos{ROz{90¡}}[NonB
reakingS
pace]=[NonBreakingSpace]1[NonBreakingSpace]
65020[NonBreakingSpace].APe(AP,TvL):Return(Atan(Sin(AP)*(.E2p(AP,TvL)^4*.EG
p(AP,TvL)^2-Sin(AP)^2)^(-.5)))
11017[NonBreakingSpace]GoSub[NonBreakingSpace]11990:?Tab(16);Cos{ROz{APg
}}[NonBreakingSpace]=;Using(,20),Cos(.Roz(APg)):?
11900[NonBreakingSpace]?TW[NonBreakingSpace]=;UL*XF;Tab(10);PEODx:[
NonBreakingSpace];Using(,20),KmDx(1);[NonBreakingSpace]Km[NonBreakingSp
ace](;Using(,25),EGDD;[NonBreakingSpace]):Next:?
11910[NonBreakingSpace]GoSub[NonBreakingSpace]11990
11990[NonBreakingSpace]?Tab(4);*-----------------------------------------
----------------------------*:Return
11980[NonBreakingSpace]?:List[NonBreakingSpace]11020:List[NonBreakingSpac
e]58010-58030:List[NonBreakingSpace]10000-10010:End
64000[NonBreakingSpace]'[NonBreakingSpace]***[NonBreakingSpace][NonBreak
ingSpace]Functions[NonBreakingSpace][NonBreakingSpace]***
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]1[NonBreakingSpace
][NonBreakingSpace]Point[NonBreakingSpace]7
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
'[NonBreakingSpace]~=~[NonBreakingSpace].1^33
----------------------------------------------------------------------------

------------------
Thisprogram[NonBreakingSpace]is[NonBreakingSpace]presented[NonBreakingSpa
ce]as[NonBreakingSpace]a[NonBreakingSpace]workbench[NonBreakingSpace]s
pecimen,[NonBreakingSpace]intended[NonBreakingSpace]for
dissectionand[NonBreakingSpace]analysis.
Forjust[NonBreakingSpace]a[NonBreakingSpace]no-frills[NonBreakingSpac
e]output,[NonBreakingSpace]the[NonBreakingSpace]following[NonBreakingSpace
]modification[NonBreakingSpace]can[NonBreakingSpace]be
made.
-------------------------------------------------------
Delete11003-11900
4000[NonBreakingSpace][NonBreakingSpace]XF=1:UL=10:GoSub[NonBreakingSpace
]60000
11100[NonBreakingSpace]GoSub[NonBreakingSpace]59100:KmDx(1)=A*EGDD*AD(0)
11200[NonBreakingSpace]?:?Tab(12);PEODx:[NonBreakingSpace];Using(,20),
KmDx(1);[NonBreakingSpace]Km:?
11300[NonBreakingSpace]?Tab(5);BAz_e,[NonBreakingSpace]DAz_e:[NonBreaki
ngSpace];Using(,20),BAz(1);Deg$,DAz(1);Deg$
11980[NonBreakingSpace]?:?:List[NonBreakingSpace]58010:?:?:?:List[NonBrea
kingSpace]10000:?:?:List[NonBreakingSpace]10010:End
58010[NonBreakingSpace]a=6378.135:b=6356.75
58020
-------------------------------------------------------
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]
~Kaimbridge[NonBreakingSpace]M.[NonBreakingSpace]GoldChild~
</pre>
===
Subject: Re: The Parageodetic Distance Formula (UBasic)
Whoa!!  Do you like to hear yourself or what?
> <pre After some reflection--prompted by an e-mail inquiry--this writer/
> programmer has decided to first release the parageodetic (inverse
> problem) formulary core, before the more dressed, integrated
> quadratic-mean-spherical, parageodetic and geodetic inverse and
> direct GODx program core.
> But, rather than just spit out the program listing in numerical
> address order, however, it is being presented subroutine-with-execution,
> thereby providing another opportunity to disseminate this unique
> formulary approach (including all trigonometric quadrantic adjustments
> and division by zero bulletproofing) for mathematical study and
> discussion.
> As the concept of the parageodesic appears little acknowledged--if at
> all even recognized--it would seem prudent to comparatively define the
> geodesic and parageodesic from the conceptual approach used here (to
> the non-technical layman, the term graticular can be considered
> synonymous with spherical):
> Geodesic: The conformally delineated great-arc segment
> between two points;
> Parageodesic: The graticularly delineated great-arc segment
> between two points.
> Thus, with an ellipsoid, the traditional geodesic is the pulled
> string path and distance on an elliptically correct model, whereas
> the parageodesic is the pulled string path and distance, first
> DEFINED on a spherical model of the ellipsoid, then the sphere is
> reformed to its proper ellipsoidal shape, at which time the
> elliptically correct arcradius and angle/azimuth AT ANY GIVEN POINT
> ALONG THE DEFINED SEGMENT can be found (i.e., it is the local,
> elliptically correct Great-Circle valuation!).
> UBasic's system program can be found here:
> ftp://rkmath.rikkyo.ac.jp/pub/ubibm
jp/~kida/ubasic.htm
> (this web page is translated through Google, thus some of the wording is
off)
> UBasic's command dictionary can be found here:
> http://ed-thelen.org/bab/bab-ubhelp.html
> Of course, while it is presented here out of numerical address order,
> once the program code is cut and pasted into UBasic, it will revert to
> its natural arrangement.
> 
--------------------------------------------------------------------------
--------------------
> 3000 GoSub 58000
> 58000 '*** Radial Parameters ***
> 58010 a=6378.135:b=6356.75:b=b::GoTo 58200
> 58020 a=10000:b=5000
> 58200 Oz=2*Atan((a-b)^.5/(a+b)^.5)
> 10000 BLat=+43.66111111111111111111:BLong= -70.25583333333333333333
> 10010 DLat=+45.52305555555555555556:DLong=-122.67583333333333333333
> 1' Parageodetic = 4092.6024524649626 Km 
291.6521536Á/73.6370729Á
> 1' Geodetic = 4092.6016550196721 Km 
291.7142510Á/73.5743344Á
> 1' Q-M-Spherical = 4082.4708695785296 Km 
291.7214263Á/73.5858490Á
> 2200 Pi=#Pi:PH=.5*Pi:PD=2*Pi
> 2300 RF=Pi/180
> 55110 BLr=BLat*RF:DLr=DLat*RF:MD=(DLong-BLong)*RF:M=Abs(MD):If M>Pi Then
MD=MD-Sgn(MD)*PD
> 11000 GoSub 55000
> 1'############################################################
> 55000 '*** Spherical, ParaGeodetic Valuations ***
> 55100 ID%=0
> 55120 WBLr=DLr:WDLr=BLr:GoSub 55400:DAz(0)=WAz(0):TvL(5)=TvL(0)
> 55130 WBLr=BLr:WDLr=DLr:GoSub 55400:BAz(0)=WAz(0):TvL(1)=TvL(0)
> 1'##########
> 55400 '*** Spherical Elements ***
> 55410 GoSub 55420:GoSub 55440:Return
> 55420 SA=Cos(WDLr)*Sin(M):If SA=0 And Abs(WDLr)<PH And Sin(M)>0 Then
SA=#Eps
> 55422 If Abs(WBLr/RF)>=90 Or Abs(WDLr/RF)>=90 Then SA=0
> 55425 SB=(Sin(WBLr+WDLr)*Sin(.5*M)^2)-(Sin(WBLr-WDLr)*Cos(.5*M)^2)
> 2150 IO=10^100
> 55430 WAz(ID%)=Abs(Atan(IO*SA/(IO*SB+1)))
> 55440 TvL(0)=Atan(IO*Sin(WBLr)/(Abs(IO*Cos(WBLr)*Cos(WAz(ID%)))+1))
> 55445 If SB<0 Then WAz(ID%)=Pi-WAz(ID%)
> 55439 Return
> 55449 Return
> 1'----------
> 55125 GoSub 55190:DAz(1)=WAz(1)
> 55135 GoSub 55190:BAz(1)=WAz(1)
> 1'##########
> 55190 ID%=1:SA=SA*.E1p(0,WBLr):SB=SB*.E0p(0,WBLr):GoSub 55430:GoSub
55445:ID%=0:Return
> 1'----------
> 2100 Dcm=Int(Log(#Eps)/Log(0.1)-5):Syota=0.1^Dcm
> 55140 If Cos(BAz(0))<0 Then TvL(1)=Sgn(IO*BLr+Syota)*Pi-TvL(1)
> 55142 If Cos(DAz(0))>0 Then TvL(5)=Sgn(IO*DLr+Syota)*Pi-TvL(5)
> 55145 If TvL(5)<TvL(1) Then TvL(5)=PD+TvL(5)
> 55150
AD(0)=TvL(5)-TvL(1):ADq_0=0.25*AD(0):BTvL=TvL(1)/RF:DTvL=TvL(5)/RF:ADg=DTvL-

BTvL
> 55155 TvL(2)=TvL(1)+ADq_0:TvL(3)=TvL(1)+2*ADq_0:TvL(4)=TvL(1)+3*ADq_0
> 55147 GoSub 55450:APg=AP(0)/RF:GoSub 55485:ID%=1:GoSub 55485
> 1'##########
> 55450
AP(ID%)=Atan(Abs(IO*Cos(WBLr)*Sin(WAz(ID%)))/(IO*(Cos(WAz(ID%))^2+(Sin(WBLr)

*Sin(WAz(ID%)))^2)^.5+1)):Return
> 55485 If MD<0 Then BAz(ID%)=PD-BAz(ID%)
> 55487 If MD>0 Then DAz(ID%)=PD-DAz(ID%)
> 55489 BAz(ID%)=BAz(ID%)/RF:DAz(ID%)=DAz(ID%)/RF:Return
> 1'----------
> 55189 Return
> 1'------------------------------------------------------------
> 11100 GoSub 59100:KmDx(1)=a*EGDD*AD(0)
> 1'############################################################
> 59100 '*** PEOx ***
> 59000 '*** Integration ***
> 65100
.E2p(AP,TvL):Return((Cos(Oz)^2+(Sin(AP)*Sin(Oz))^2+(Cos(AP)*Cos(TvL)*Sin(Oz)

)^2)^.5)
> 1' = Elliptic Integrand Of The Second Kind
> 65110 .E1p(AP,TvL):Return(1/.E2p(AP,TvL))
> 1' = Elliptic Integrand Of The First Kind
> 1' x a = N (Normal) = Transverse Equatorial Arcradius
> 65120 .E0p(AP,TvL):Return(Cos(Oz)^2*.E1p(AP,TvL)^3)
> 1' x a = M (Meridian) = Vertical Meridional Arcradius
> 65130
.EGp(AP,TvL):Return((.E0p(AP,TvL)^2+Sin(AP)^2*(IO*(.E1p(AP,TvL)^2-.E0p(AP,Tv

L)^2)/(IO*(Cos(TvL)^2+(Sin(AP)*Sin(TvL))^2)+1)))^.5)
> 1' x a = O (Omniversal) = Transverse Meridional Arcradius
> 1'#########################
> 11020 XF=1:::For UL=2 To 20 Step 2::GoSub 60000
> 1' Integration Convergency for EGDD to .1^20 @ cos{Oz}, for XFxUL: 
cos{Oz}
= 1xUL/2xUL/3xUL
> 1' .9999 = 6/8/6; .995 = 8/12/15; .99 = 8/12/15; .9 = 11/16/18;
> 1' .75 = 13/18/24; .5 = 17/24/30; .25 = 27/36/45; .1 = 42/60/75;
> 1' .01 = --/--/231; (.001 ~=~ 20x94 = 1880; .0001 ~=~ 200x97 = 19400)
> 60000 '*** (Gaussian) Amplitudal Kernal Expansion ***
> 65000
.AE(NS,NX,XF):AE=2*Atan(((XF-NX+Sin(.5*AEg(NS))^2)/(NX-Sin(.5*AEg(NS))^2))^.

5):If AE>.5*Pi Then AE=Pi-AE
> 65009 Return(AE)
> 60100 DU=2*UL:For NS=1 To UL:CK=Cos(.5*Pi*(((4*NS)-1)/((4*UL)+1)))
> 60200 LP(1)=1:LP=1:For NP=1 To DU
> 60210
LP(2)=LP(1):LP(1)=LP:LP=((((2*NP)-1)*LP(1)*CK)-((NP-1)*LP(2)))/NP:Next NP
> 60220 DP=DU*(LP(1)-(LP*CK))/(1-CK^2)
> 60230 CK(2)=CK(1):CK(1)=CK:CK=CK(1)-(LP/DP)
> 60235 If Abs(CK-CK(1))>(10*Syota) And CK<>CK(2) Then 60200
> 60300 AEg(NS)=2*Atan((1-CK)^.5/(1+CK)^.5)
> 60310 OE(NS)=Sin(AEg(NS))^2/(DU*LP(1)^2):Next NS:Return:::::?NS;
;Using(,30),AEg(NS)/RF; ;OE(NS)/RF:Next NS:List 60010:End:Return
> 1'-------------------------
> 59110 EGDD=0:For NS=1 To UL:For NX=1 To XF
> 59120 AE=AEg(NS):If XF>1 Then AE=.AE(NS,NX,XF)
> 59130 TW=XF*UL:ABq=Cos(AE)*ADq_0:OW=OE(NS)/TW
> 59140 UUP=TvL(4)+ABq:UMP=TvL(4)-ABq:LMP=TvL(2)+ABq:LLP=TvL(2)-ABq
> 59150
EGDD=EGDD+0.25*(.EGp(AP(0),UUP)+.EGp(AP(0),UMP)+.EGp(AP(0),LMP)+.EGp(AP(0),L

LP))*OW
> 59160 Next NX:Next NS:Return
> 1'------------------------------------------------------------
> 10 ClS
> 1000 Dim
AEg(101),OE(101),LP(2),CK(2),AD(1),AP(1),TvL(5),KmDx(1),WAz(1),BAz(1),DAz(1)

> 2000 '*** Program Constants/Standard Values ***
> 2400 Deg$=Chr(248):'(Degree Sign)
> 58300 ?Tab(20);************************************:?Tab(19);/
Parageodetic Distance Formula Core :? *----------------* ( cos{Oz}
=;Using(,20),Cos(Oz); ) *-------------------*
> 58310 ?Using(,13), | a =;a;
Km;;:?Tab(55-Len(Str(Int(b))));Using(,13),b =;b; Km; |:?
*---------------------------------------------------------------------------

*:Return
> 11003 ?Tab(7);BTvL, DTvL: ;Using(,20),TvL(1)/RF;Deg$,TvL(5)/RF;Deg$
> 11005 ?Tab(25);ADg: ;Using(,20),ADg;Deg$
> 11007 ?Tab(25);APg: ;Using(,20),APg;Deg$
> 11010 ?Tab(5);BAz_g, DAz_g: ;Using(,20),BAz(0);Deg$,DAz(0);Deg$
> 11012 ?Tab(5);BAz_e, DAz_e: ;Using(,20),BAz(1);Deg$,DAz(1);Deg$
> 11015 ?Tab(5);APe_b, APe_d:
;Using(,20),.APe(AP(0),TvL(1))/RF;Deg$,.APe(AP(0),TvL(5))/RF;Deg$
> 65010
.ROz(APg):Return(Atan(IO*Cos(APg*RF)*Sin(Oz)/(IO*(1-(Cos(APg*RF)*Sin(Oz))^2)

^.5+1)))
> 1' = Reduced Oz: cos{ROz{0Á}} = cos{Oz}, 
cos{ROz{90Á}} = 1
> 65020
.APe(AP,TvL):Return(Atan(Sin(AP)*(.E2p(AP,TvL)^4*.EGp(AP,TvL)^2-Sin(AP)^2)^(

-.5)))
> 11017 GoSub 11990:?Tab(16);Cos{ROz{APg}} 
=;Using(,20),Cos(.Roz(APg)):?
> 11900 ?TW =;UL*XF;Tab(10);PEODx: ;Using(,20),KmDx(1); Km
(;Using(,25),EGDD; ):Next:?
> 11910 GoSub 11990
> 11990
?Tab(4);*-----------------------------------------------------------------
-
---*:Return
> 11980 ?:List 11020:List 58010-58030:List 10000-10010:End
> 64000 ' *** Functions ***
> 1 Point 7
> ' ~=~ .1^33
> 
--------------------------------------------------------------------------
--------------------
> This program is presented as a workbench specimen, intended for
> dissection and analysis.
> For just a no-frills output, the following modification can be
> made.
> -------------------------------------------------------
> Delete 11003-11900
> 4000 XF=1:UL=10:GoSub 60000
> 11100 GoSub 59100:KmDx(1)=A*EGDD*AD(0)
> 11200 ?:?Tab(12);PEODx: ;Using(,20),KmDx(1); Km:?
> 11300 ?Tab(5);BAz_e, DAz_e: ;Using(,20),BAz(1);Deg$,DAz(1);Deg$
> 11980 ?:?:List 58010:?:?:?:List 10000:?:?:List 10010:End
> 58010 a=6378.135:b=6356.75
> 58020
> -------------------------------------------------------
> ~Kaimbridge M. GoldChild~
> </pre>
===
Subject: Re: The Parageodetic Distance Formula (UBasic)
> After some reflection--prompted by an e-mail inquiry--this writer/
> programmer has decided to first release the parageodetic (inverse
> problem) formulary core, before the more dressed, integrated
> quadratic-mean-spherical, parageodetic and geodetic inverse and
> direct GODx program core.
                    <snip As the concept of the parageodesic appears little acknowledged--if
> at all even recognized--it would seem prudent to comparatively
> define the geodesic and parageodesic from the conceptual approach
> used here (to the non-technical layman, the term graticular can
> be considered synonymous with spherical):
>        Geodesic:  The conformally delineated great-arc segment
>                   between two points;
>    Parageodesic:  The graticularly delineated great-arc segment
>                   between two points.
                    <snip>
If the formatting came out garbled (i.e., non-fixed-width or altered
characters--such as happened with the original Google posting), or
cutting and pasting the UBasic code didn't work (e.g., broken lines),
try one of these:
          http://mathforum.org/discuss/sci.math/t/658788
     ~Kaimbridge~
-----
    WantedÖKaimbridge (w/mugshot!):
           http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html
      ----------
DigitologyÖThe Grand Theory Of The Universe:
           http://www.angelfire.com/ma2/digitology/index.html
      *****  Void Where Permitted; Limit 0 Per Customer.  *****
===
Subject: Jesus christ What's going on?
There is a work around Google's awful new usenet interface, at least
temporarily. Use Google.fr or Google.ru or probably any other foreign
Google domain. They still have the old interface up. Of course the
interface will mostly not be in Englsih.
===
Subject: Re: Jesus christ What's going on?
Vanya infrared:
>There is a work around Google's awful new usenet interface, at least
>temporarily. Use Google.fr or Google.ru or probably any other foreign
>Google domain. They still have the old interface up. Of course the
>interface will mostly not be in Englsih.
For searching - which is all I ever use Google Groups for - the
following still works:
    http://www.exit109.com/%7Ejeremy/news/deja.html
It's just the Google search with the advertising and graphics stripped
off, which speeds things up a bit.  Originally designed as an
interface to DejaNews, but he kept it working after Google bought
the DejaNews database.
-- 
Peter Moylan                 peter at ee dot newcastle dot edu dot au
http://eepjm.newcastle.edu.au (OS/2 and eCS information and software)
===
Subject: Re: Jesus christ What's going on?
> For searching - which is all I ever use Google Groups for - the
> following still works:
>    http://www.exit109.com/%7Ejeremy/news/deja.html
Haven't tried that in years! I remember that! I'll give it a try.
===
Subject: Re: Jesus christ What's going on?
> For searching - which is all I ever use Google Groups for - the
> following still works:
>     http://www.exit109.com/%7Ejeremy/news/deja.html
You might want to bookmark http://www.usenet4all.com/ as well, there are
similar others. It lacks some functionality, on the upside you don't
need to register to post there and there's hardly any lag for posts to
show up there.
P.
===
Subject: Re: Jesus christ What's going on?
> For searching - which is all I ever use Google Groups for - the
> following still works:
>     http://www.exit109.com/%7Ejeremy/news/deja.html
> You might want to bookmark http://www.usenet4all.com/ as well, there are
> similar others. It lacks some functionality, on the upside you don't
> need to register to post there 
Yes, you do. Otherwise you get ACCESS DENIED.
> and there's hardly any lag for posts to
> show up there.
> P.
===
Subject: Re: Jesus christ What's going on?
>> You might want to bookmark http://www.usenet4all.com/ as well, there
>> are similar others. It lacks some functionality, on the upside you
>> don't need to register to post there
> Yes, you do. Otherwise you get ACCESS DENIED.
Aww you're right. Oh well mostly of use for reading older posts anyway.
===
Subject: Re: Jesus christ What's going on?
>Vanya infrared:
>>There is a work around Google's awful new usenet interface, at least
>>temporarily. Use Google.fr or Google.ru or probably any other foreign
>>Google domain. They still have the old interface up. Of course the
>>interface will mostly not be in Englsih.
the new invention down for the moment.
>For searching - which is all I ever use Google Groups for - the
>following still works:
>    http://www.exit109.com/%7Ejeremy/news/deja.html
>It's just the Google search with the advertising and graphics stripped
>off, which speeds things up a bit.  Originally designed as an
>interface to DejaNews, but he kept it working after Google bought
>the DejaNews database.
-- 
Don Aitken
Mail to the addresses given in the headers is no longer being
read. To mail me, substitute clara.co.uk for freeuk.com.
===
Subject: Re: Jesus christ What's going on?
> There is a work around Google's awful new usenet interface, at least
> temporarily. Use Google.fr or Google.ru or probably any other foreign
> Google domain. They still have the old interface up. Of course the
> interface will mostly not be in Englsih.
Good point. And the language of the fields can be changed to English by
changing the last letters of the URL to en, then bookmarking it.
I played for this a while and found it confusing as to what could do
what, but it looks like this gives me the old interface, with the
results in the old format (like threads in a chain diagram):
     Old: 
     Google Groups -- Advanced Group Search 
and this gives me the new page, with new formats:
     New: 
     Google Groups BETA -- Advanced Search  
One striking difference is the new Beta form can't be restricted by
date, which was useful in searching for the earliest use of recent
words.
-- 
Best -- Donna Richoux
 
===
Subject: Re: Jesus christ What's going on?
> There is a work around Google's awful new usenet interface, at least
> temporarily. 
?
What new interface?
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Jesus christ What's going on?
> There is a work around Google's awful new usenet interface, at least
> temporarily. 
> What new interface?
Apparently it was only up in the US, and only for a short time (~15 hours).
Andrew Usher
===
Subject: Green...
hello.....doctor~
let D be the region enclosed by the curve
r(t) = (sin 2t, sin t), (0 <= t < Pi)
compute
double integral (x+y) dxdy
   D
(hint : Use the Green theorem)
--------------------------------------------
um......i know that
int Pdx + Qdy = double int (dQ/dx - dP/dy) dxdy
dD                         D
and i know that
double int  1 dxdy = (1/2)*[int (-ydx + xdy)]
    D                                 dD
but, double integral (x+y) dxdy ???? i can't...
so, i need your advice.
thank you very much for your advice.
===
Subject: Re: Green...
> let D be the region enclosed by the curve
> r(t) = (sin 2t, sin t), (0 <= t < Pi)
> compute
> double integral (x+y) dxdy
>    D
> (hint : Use the Green theorem)
We know that:
int_C Pdx + Qdy = int int_D (dQ/dx - dP/dy) dxdy.
You want to evaluate:
int int_D (x + y) dxdy.
So, take P, Q such that dQ/dx = x, dP/dy = -y - e.g. Q = 1/2*x^2, P = 
-1/2*y^2 - and use the Green theorem:
int int_D (x + y) dxdy = int_C -1/2*y^2dx + 1/2*x^2dy = ...
Pawel Gladki
===
Subject: Thin-Plate Splines vs. Radial basis Functions
Hi everyone,
I'm trying to solve a class of 3D warping problems as part of my
dissertation. I'd wandered abount in the dark for a while, till I
recently (2 days ago) came across the field of morhometrics and
specifically, the Thin-plate Spline (TPS) method, and it seems to be
exactly what the Doctor ordered :-) . I think the TPS will do nicely,
but it makes sense to justify my use of it. Specifically, I'm
wondering why (it seems as though) all the morphometric analysis I've
come across in anthoplogy-related literature use this method (I'm a
student of Computer Science). I wonder, why not radial-basis functions
(RBF) for example?
I'm sorry, if this question seems unreasonable or trivial. I really do
need to know which is the better or prefered direction. Better
stated, in what way are TPS bettter than a RBF - if at all it is,
and vise versa.
Olumide
===
Subject: Re: Thin-Plate Splines vs. Radial basis Functions
For 3D,  I think Thin Plates Splines will do nicely.
They are a form of RBF.
They result from Variational Theory with respect to mechancal 
deformation so they are sort of based in nature and mechanical physics 
you could argue.
http://mathworld.wolfram.com/ThinPlateSpline.html
Try to find the guru Prof Paul Dierckx for more info and software
He has written a book on this subject
There is also Prof Grace Wahba from the USA who is also main world guru 
herein
Paul
> Hi everyone,
> I'm trying to solve a class of 3D warping problems as part of my
> dissertation. I'd wandered abount in the dark for a while, till I
> recently (2 days ago) came across the field of morhometrics and
> specifically, the Thin-plate Spline (TPS) method, and it seems to be
> exactly what the Doctor ordered :-) . I think the TPS will do nicely,
> but it makes sense to justify my use of it. Specifically, I'm
> wondering why (it seems as though) all the morphometric analysis I've
> come across in anthoplogy-related literature use this method (I'm a
> student of Computer Science). I wonder, why not radial-basis functions
> (RBF) for example?
> I'm sorry, if this question seems unreasonable or trivial. I really do
> need to know which is the better or prefered direction. Better
> stated, in what way are TPS bettter than a RBF - if at all it is,
> and vise versa.
> Olumide
===
Subject: Subbases, compactness
Let  S  be a subbase for a topological space  X.  Suppose every cover 
of  X  by sets in  S  has a finite subcollection which covers  X.  Is  
X  necessarily compact?
(This is not homework.)
-- 
Stephen J. Herschkorn                        sjherschko@netscape.net
===
Subject: Re: Subbases, compactness
>Let  S  be a subbase for a topological space  X.  Suppose every cover 
>of  X  by sets in  S  has a finite subcollection which covers  X.  Is  
>X  necessarily compact?
Yes. (Isn't this how the traditional proof of the Tychonoff
theorem goes?)
>(This is not homework.)
************************
David C. Ullrich
===
Subject: Re: Subbases, compactness
> Let  S  be a subbase for a topological space  X.  Suppose every cover 
> of  X  by sets in  S  has a finite subcollection which covers  X.  Is  
> X  necessarily compact?
> (This is not homework.)
Alexander's Theorem.
Requires the Axiom of Choice for the proof.
-- 
G. A. Edgar                               
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Subbases, compactness
> Let  S  be a subbase for a topological space  X.  Suppose every cover
> of  X  by sets in  S  has a finite subcollection which covers  X.  Is
> X  necessarily compact?
> Alexander's Theorem.
> Requires the Axiom of Choice for the proof.
Rather than a laborious set theory proof, is there a simpler filter proof?
===
Subject: Re: Subbases, compactness
> Let S be a subbase for a topological space X. Suppose every cover
> of X by sets in S has a finite subcollection which covers X. Is
> X necessarily compact?
>> Alexander's Theorem.
>> Requires the Axiom of Choice for the proof.
> Rather than a laborious set theory proof, is there a simpler filter 
proof?
The proof in Kelley's General Topology does not seem so laborious to 
me.  It is a rather nifty application of Tuckey's Lemma, which is a 
version of Zorn's lemma.
will find many references to a result in knot theory about braids.  This 
is not the theorem to which GAE refers.  The theorem we use here is more 
easily found as Alexander's Subbase Theorem;  sometimes you will see 
Sub-base hyphenated.
-- 
Stephen J. Herschkorn                        sjherschko@netscape.net
===
Subject: Re: Subbases, compactness
===
Subject: Re: Subbases, compactness
> Let S be a subbase for a topological space X. Suppose every cover
> of X by sets in S has a finite subcollection which covers X. Is
> X necessarily compact?
>> Alexander's Theorem.
>> Requires the Axiom of Choice for the proof.
> Rather than a laborious set theory proof, is there a simpler filter
proof?
> The proof in Kelley's General Topology does not seem so laborious
> to me.  It is a rather nifty application of Tuckey's Lemma, which
> is a version of Zorn's lemma.
What's Tuckey's Lemma?  Zorn's Lemma for P(S) with subset order?
> The theorem we use here is more easily found as Alexander's Subbase
> Theorem;  sometimes you will see Sub-base hyphenated.
Well as included below, I did find an EasyProofOfTychonoff
http://www.theoryandpractice.org/kyle/Wiki/EasyProofOfTychonoff
covering both Alex's Lemma and Tychonov's Theorem.  However for all it's
clarity, it is lacking a crucial step where it is claimed G = H / S
covers X (marked **).  I'm not even seeing why a maximal cover H of X with
no finite subcover, would contain any subbase sets S, much less why all
the subbase sets of H would cover X.  What steps are needed to show this?
In the Tychonov section I corrected a couple of apparent typos
        ... x_i in X - / G_i for every i.
        ... and x in X_i - / G_i.
and seeing no need, removed the one time occurrence of
        bar{U} =
-- Alexander subbasis lemma
If X is a topological space and S is a subbasis of X, then
X is compact if and only if every S-cover of X has a finite subcover.
Proof:  Let B be the collection of all open covers of X that do not have
finite subcovers. Assume, by way of contradiction, that B is nonempty.
Partial order B by set inclusion. Let C be a chain in B.
Let D be the union of all elements of C.
If D has a finite subcover, then those finitely many open sets
would be contained in finitely many elements of the chain C.
Every finite subset of the chain has a maximum element, and so each of
those finitely many open sets that together cover X are contained in
that maximum element. This is impossible since every element of C is
an element of B, and thus has no finite subcover. Thus D has no finite
subcover, so is in B, and is an upperbound for C inside B.
Since C was an arbitrary chain in B, the conditions of Zorn's
lemma are satisfied, and thus B has a maximal element, call it H.
** Consider G = H / S.  We show that G covers X, so that by hypothesis it
has a finite subcover, but that finite subcover is a subcover of H as
well. This is a contradiction, and so the assumption that B is nonempty is
untenable. Thus every open cover of X has a finite subcover, and X is
compact.
-- Theorem (Tychonoff): Suppose that I is a set and for every i in I
        suppose X_i is a compact topological space.
Endow X = prod_{i in I} X_i with the product topology. X is compact.
Proof of Tychonoff: Let S be the standard subbasis of the product
topology, S = { pi_i^{-1}(U) : U is open in X_i }.
By Alexander's lemma it suffices to consider S-covers.
Let G be an S-cover, that is let G consist of sets pi_i^{-1}(U)
for various i in I and U open in X_i.
Define G_i = { U : U is open in X_i, pi^{-1}_i(U) in G }.
Assume BWOC, that for every i in I that / G_i /= X_i.
Using the axiom of choice, choose x_i in X_i - / G_i for every i.
Since x = (x_i)_{i in I} is in X,
x in pi_i^{-1}(U) in G for some i in I and U open in X_i.
This is a contradiction since x in U in G_i and x_i in X_i - / G_i.
Therefore there must exist an i in I such that / G_i = X_i.
Since X_i is compact, choose a finite subcover, U_1,... U_n.
Let C = { pi_i^{-1}(U_1),... pi_i^{-1}(U_n) }.
C is a finite subcover of X. Since G was an arbitrary S-cover, the
conditions of Alexander's subbase lemma are satisfied, and X is compact.
----
===
Subject: Need help setting up an induction
Suppose there are sets such that
[a0,b0] is contained in Union (1-->n) (a_j,b_j).  We make no further
assumptions about {(a_j,b_j)}.
We want to set up a formal induction for the following process,
including some re-numbering:
There exists an (a1,b1) such that a1< a0 <b1.  If b1<=b0, then there
is an (a2,b2) such that a2< b1< b2.  This continues until we find an
(a_k,b_k) such that a_k< b0< b_k.  The process stops and we end up
with
a1< a0< b1, a_k< b0< b_k, and a_(j+1)< b_j< b_(j+1) for 1<= j<= k-1.
What are the formal induction statements required to describe this
process?
All help appreciated.
===
Subject: Re: Escultura affair: publication scandal
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2IUkS25828;
><In the >Zentralblatt f.9fr Mathematik< 9 publications
>of E. Escultura are listed. For 5 of them the
>Zentralblatt just mentions >not reviewed<, whatever
>that means.
>This is an indexing service like so many others. So it simply notes some 
contributions unless someone wants to try uncharted course.
I don't agree. The Zentralblatt and the Math Reviews are the
two major indexing services on this planet. Indeed together
they cover all countries.
><However I must admit that - looking at myself - I
>wonder why persons like E. Escultura or J. Harris
>provoke such strong reactions.
>The most likely reason is fear of new ideas. 
>E. E. Escultura
>University of the Philippines
This is one possibility. Why is it the most
likely one?
H
===
Subject: Re:
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2Icad26396;
>I think it interesting to show you. g:R->R  x>1  .
>You will catch 'the knack' with 1/p  p steps are needed 
>to loop onto the next integer.
>So I choose g(x)=c*(x-1)!*(x-1+1/p)!*(x-1+2/p)!....(x-1+(p-1)/p)!
>        g(x+1/p)=       c*(x-1+1/p)!*(x-1+2/p)!....(x-1+(p-1)/p)!*x!
>I adopt the writing convention (x-1)! instead of gamma(x).
>Try solving this way  : h(y)*h(y+1)=1/y ,
>courage,
>Alain.
Here is the solution to h(y)*h(y+1)=1/y ,  (1)
With our conventions we try h(y)=c*(y/2-1)!/(y/2-1/2)! 
 we adjust coeff c to obtain (1) ;we propose :
      h(y)=sqrt(2)/2*y/2-1)!/(y/2-1/2)! 
===
Subject: Re: .99999... still=/= 1
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2IUlu25852;
>mister Company is going to love this.
>     I just found an application of 0.9999...,
>over and above OVERFLOW CONDITION or TENS COMPLIMENT.
>    That's because,
>   m--> oo
>   Lim (1 - 1/10^m) = 1
>     divergent
>>Convergent.
>  Ok don't my word for it, but at least believe the actual Advanced 
Engineering
>Math Book I have.
>Quoting from:
>6th Edition
>Advanced Engineering Mathematics
>by Erwin Kreyszig
>Page 805
>14.2 Convergence Tests for Series
>  Theorem 1 ( Divergence )
>If a series z_1 + z_ 2 + .... converges, then
>   lim Z_m = 0
>   m-->oo
>Hence if the series does not satisfy this condition, it diverges.
>Smart's Alt. Physics News Group
><a 
href=http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1>ht
tp://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1</aS. Enterprize (Science Journal)
><a 
href=http://smart1234.s-enterprize.com/>http://smart1234.s-enterprize.com
/</a>
In 1959  Mr. Hoyt, my Vocational Ag teacher informed my class that a person 

with an I,Q,
of 79 obtained a degree from Iowa State University( a fairly top notch 
Engineering institution).   I didn't believe him until reading your verbose 

garbage.
I assume that I will find you name  enshrined in crankdom
hjs
no one can understand a math subject until they have been through the next 
highter
level, ie  you don't understand trig until you have completed calculus  
E.E. Doc  Smith.
===
Subject: Re: .99999... still=/= 1
>>mister Company is going to love this.
>>     I just found an application of 0.9999...,
>>over and above OVERFLOW CONDITION or TENS COMPLIMENT.
>>    That's because,
>>   m--> oo
>>   Lim (1 - 1/10^m) = 1
>>     divergent
>Convergent.
>>  Ok don't my word for it, but at least believe the actual Advanced
>Engineering
>>Math Book I have.
>>Quoting from:
>>6th Edition
>>Advanced Engineering Mathematics
>>by Erwin Kreyszig
>>Page 805
>>14.2 Convergence Tests for Series
>>  Theorem 1 ( Divergence )
>>If a series z_1 + z_ 2 + .... converges, then
>>   lim Z_m = 0
>>   m-->oo
>>Hence if the series does not satisfy this condition, it diverges.
>>Smart's Alt. Physics News Group
>>http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
>>S. Enterprize (Science Journal)
>>http://smart1234.s-enterprize.com/
>In 1959  Mr. Hoyt, my Vocational Ag teacher informed my class that a 
person
>with an I,Q,
>of 79 obtained a degree from Iowa State University( a fairly top notch
>Engineering institution).   I didn't believe him until reading your 
verbose
>garbage.
>I assume that I will find you name  enshrined in crankdom
>hjs
>no one can understand a math subject until they have been through the 
next
>highter
>level, ie  you don't understand trig until you have completed calculus  
E.E.
>Doc  Smith.
  I did make a note of the mistake and made the corrections. I used the Lim
Partial Sums to determine that particular convergence test of,
m-->oo
lim z_m = 0
 The mistake was that I used,
m-->oo
lim SUM z_m = 0
 But, even though .999... converges to 1,
.999... still=/= 1
   Convergence doesn't mean equal to in any case or test.
   It is interesting that the Partial Sums lim would fail that particular
convergence test. When you SUM the total series at the lim at infinity, it
fails the convergence test.  So, if you SUM .999... totally, it doesn't
converge, but if you take only the mth term it does.
   In fact, a harmonic series can diverge and converge at the same time. 
Why?
Because the number 1 is involved. 
  But I think I still won the debate because the original question was, 
does
.999... = 1 and it doesn't, it converges to 1.
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
>    It is interesting that the Partial Sums lim would fail that particular
> convergence test. When you SUM the total series at the lim at infinity, 
it
> fails the convergence test.  So, if you SUM .999... totally, it doesn't
> converge, but if you take only the mth term it does.
There is no at infinity.
===
Subject: Re: .99999... still=/= 1
>>    It is interesting that the Partial Sums lim would fail that 
particular
>> convergence test. When you SUM the total series at the lim at infinity, 
it
>> fails the convergence test.  So, if you SUM .999... totally, it doesn't
>> converge, but if you take only the mth term it does.
>There is no at infinity.
  Hey why try to change the subject?
.999... =/= 1
It converges to 1.
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
>  But, even though .999... converges to 1,
> .999... still=/= 1
Can you, will you dig the wax out of your ears and the sleepy seeds from 
your eyes?
To say Sum (9/10^n) [n >= 1] = 1 MEANS (by DEFINITION)  that:
the limit of the sequence of partial sums  is 1. Got it? It does not say 
that 1 = a partial sum for some n. It says 1 is the LIMIT of the partial 
  sums.  Which it to say that for any teeny tiny number eps > 0 we can 
find an integer N(eps) such that for all partial sums for n > N(eps) the 
partial sums for n lie in the interval (1 - eps, 1). Can you get that 
through your thick head, you benighted putz?
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>>  But, even though .999... converges to 1,
>> .999... still=/= 1
>Can you, will you dig the wax out of your ears and the sleepy seeds from 
>your eyes?
>To say Sum (9/10^n) [n >= 1] = 1 MEANS (by DEFINITION)  that:
>the limit of the sequence of partial sums  is 1. Got it? It does not say 
>that 1 = a partial sum for some n. It says 1 is the LIMIT of the partial 
>  sums.  Which it to say that for any teeny tiny number eps > 0 we can 
>find an integer N(eps) such that for all partial sums for n > N(eps) the 
>partial sums for n lie in the interval (1 - eps, 1). Can you get that 
>through your thick head, you benighted putz?
>Bob Kolker
  Hey goofball, 
.999... =/= 1
it converges to 1 in the series test.
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Equal Area Spherical Triangles Apex Locus on Sphere
AB is a fixed geodesic great circle arc, say a latitude line,on a
sphere radius R.C moves so that area between arcs AB,BC and CA
{(A+B+C-pi)*R^2} is constant. What is locus of C?
===
Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere
        ETAtAhUApQL6jYNcW0hWib4EVPgc7y0QsTUCFDITqMFsndm0APzGwrAYok6z6Keo 
It's actually a closed curve, which seems very differentfrom what we get
on the plane.
Imagine that AB is fixed on the Equator of a spherical Earth, say in
South America with A in northern Ecuador and B in far nothern Brazil.
Put C initally in the Carribean Sea and start moving it west, adjusting
its latitude so that the triangle keeps the same area.  Eventually,
angle A reaches 180 degrees and C accordingly reaches the Equator -- but
the triangle does not collapse into a straight line.  Conserving the
nonzero area has forced the formation of a lune with B and C as the
poles (C is now in eastern Indonesia, opposite B).  Sides AB and AC
together form a 180-degree arc on the Equator, while side BC is an
alternate great circular arc that byapsses the Equator to the north.
Now you can flip this lune through line AB, so that BC is now in the
Southern Hemisphere, and continue to rotate C around the base AB.  C
goes through Bolivia and Brazil, all the way around to western
Indonesia, opposite A, where another lune is formed.  Flip the lune
again to complete tracing out the locus.  The net result is a closed
curve, symmetric about AB, with cusps a the points opposite A and B.
If you specify a smnall lenght ofr AB and a small are for the triangle,
the locus in the vicinity of AB looks like the pair of parallel lines
you get on the plane.
--OL
===
Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere
> It's actually a closed curve, which seems very differentfrom what we get
> on the plane.
> Imagine that AB is fixed on the Equator of a spherical Earth, say in
> South America with A in northern Ecuador and B in far nothern Brazil.
> Put C initally in the Carribean Sea and start moving it west, adjusting
> its latitude so that the triangle keeps the same area.  Eventually,
> angle A reaches 180 degrees and C accordingly reaches the Equator -- but
> the triangle does not collapse into a straight line.  Conserving the
> nonzero area has forced the formation of a lune with B and C as the
> poles (C is now in eastern Indonesia, opposite B).  Sides AB and AC
> together form a 180-degree arc on the Equator, while side BC is an
> alternate great circular arc that byapsses the Equator to the north.
> Now you can flip this lune through line AB, so that BC is now in the
> Southern Hemisphere, and continue to rotate C around the base AB.  C
> goes through Bolivia and Brazil, all the way around to western
> Indonesia, opposite A, where another lune is formed.  Flip the lune
> again to complete tracing out the locus.  The net result is a closed
> curve, symmetric about AB, with cusps a the points opposite A and B.
OK, ready on our tracks for a symbolic quantificating trigonometry...
===
Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere
        ETAuAhUAl7G/DoN3ibTTTtFAwwKX0m7AnQcCFQCUOVX9ZFzKEYVRs9Nd4y62DEDKNg== 

OK, I'll start that.
Let LtA be the latitude of A, north positive, LoA be the longitude of A,
east positive, with similar nomenclature for points B and C.  Set N to
quantities we have:
cos(AB)  = sin(LtA)sin(LtB)+cos(LtA)cos(LtB)cos(LoA-LoB)
Likewise draw Triangles NAC and NBC and get analogous expressions for
cos(AC) and cos(BC).  With the sides of Triangle ABC thus characterized,
obtain the angles in that triangle by using the Law of Cosines solved
for a vertex angle; for example:
cos  C = (cos(AB)-cos(AC)cos(BC))/(sin(AC)sin(BC))
where angle C is taken to be within Triangle ABC (no longer involving
N).  To get the sines in the denominator you must take square roots (sin
x = (+/-) sqrt(1-cos^2 x)); give the square roots positive signs because
the arcs measure 180 degrees or less.  Take the inverse cosines of your
angles, add them up and set the sum to the desired value.
The locus may be a simple closed curve, but its algebra is not very
simple!
--OL
===
Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere
> AB is a fixed geodesic great circle arc, say a latitude line,on a
> sphere radius R.C moves so that area between arcs AB,BC and CA
> {(A+B+C-pi)*R^2} is constant. What is locus of C?
I think you mean longitude line.  Latitude lines are not great circles,
except for the equator.
--Mark
===
Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere
    <wFKrd.180553$HA.82576@attbi_s01> AB is a fixed geodesic great circle arc, say a latitude line,on a
>> sphere radius R.C moves so that area between arcs AB,BC and CA
>> {(A+B+C-pi)*R^2} is constant. What is locus of C?
>I think you mean longitude line.  Latitude lines are not great 
circles,
>except for the equator.
Either way, I would guess that the answer is 2 circles centered on the same
axis as AB.  For example, if AB is a segment of the equator, C would be a
pair of latitude lines, 1 north and 1 south.
--Keith Lewis              klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere
>> AB is a fixed geodesic great circle arc, say a latitude line,on a
>> sphere radius R.C moves so that area between arcs AB,BC and CA
>> {(A+B+C-pi)*R^2} is constant. What is locus of C?
>I think you mean longitude line.  Latitude lines are not great 
circles,
>except for the equator. 
Yes as Mark said, what I meant.
> Either way, I would guess that the answer is 2 circles centered on the 
same
> axis as AB.  For example, if AB is a segment of the equator, C would be a
> pair of latitude lines, 1 north and 1 south.
> --Keith Lewis   
           
If base AB is on equator/parallel circle,we get two parallel latitudes
as required locus for C, right? If so,that tantamounts to
extending/guessing the area formula when AB is on parallel circle(ph
latitude,th longitude, points 1,2 for A,B),and we have:
 (A+B+C-pi)*R^2 = R^2*(sin(ph1)-sin(ph2)).(th1-th2) Or,
 Spherical Excess =(A+B+C-pi) = (sin(ph1)-sin(ph2)).(th1-th2)? Is this
result mentioned in some spherical trigonometry text-books?
===
Subject: Galois-Theory
I want to solve the following exercise.
Show that E=Q(sqrt(2),sqrt(3),u) where u^2=(9-5sqrt(3))(2-sqrt(2)) is
normal. Determine Gal E/Q.
I have no idea, what to do. I know two ways to show, that E/Q is normal.
First: show, that E is the splitting field of a polynomial g in Q[x].
Second: Q=Inv G, where G is a finite Group of automorphisms of E.
I think the first method needs too many calculations. But i dont know how 
to
realize the second way. If i assume, that u is not an element of
Q(sqrt(2),sqrt(3)), then [E:Q]=8, so G must be a group of order 8. I think
that G=<r,s,t>, where r:sqrt(2)->-sqrt(2), s:sqrt(3)->-sqrt(3), t:u->-u.
But how to prove it? And how can i show, that u i not an element of
Q(sqrt(2),sqrt(3))?
Is there another method?
Johannes
===
Subject: Re: Galois-Theory
>I want to solve the following exercise.
>Show that E=Q(sqrt(2),sqrt(3),u) where u^2=(9-5sqrt(3))(2-sqrt(2)) is
>normal. Determine Gal E/Q.
>I have no idea, what to do. I know two ways to show, that E/Q is normal.
>First: show, that E is the splitting field of a polynomial g in Q[x].
>Second: Q=Inv G, where G is a finite Group of automorphisms of E.
>I think the first method needs too many calculations. 
Not really. You know the polynomial that will give you sqrt(2):
x^2-2. You know the one for sqrt(3), namely x^2-3. So if we let F =
Q(sqrt(2),sqrt(3)), then F is the splitting field of (x^2-2)(x^2-3).
Now, E = F(u), and is the splitting field, over F, of the polynomial
x^2 - (9-5sqrt(3))(2-sqrt(2)).
So E must be a splitting field over Q as well. To find the polynomial
explicitly, take 
x^2 - (9-5sqrt(3))(2-sqrt(2))
x^2 - (9+5sqrt(3))(2-sqrt(2))
x^2 - (9-5sqrt(3))(2+sqrt(2))
x^2 - (9+5sqrt(3))(2+sqrt(2))
and multiply them together to get a polynomial g(x). Then E is the
spliting field over Q of g(x)(x^2-2)(x^2-3). You don't have to
calculate g(x) explicitly, just note that it will work.
>But i dont know how to
>realize the second way. If i assume, that u is not an element of
>Q(sqrt(2),sqrt(3)), then [E:Q]=8, so G must be a group of order 8. I think
>that G=<r,s,t>, where r:sqrt(2)->-sqrt(2), s:sqrt(3)->-sqrt(3), t:u->-u.
There may be some relations between r, s, and t; you should figure out
what r and s will do to u (see what they do to u^2). There aren't that
many groups of order 8 (only three abelian and two nonabelian ones),
so you should be able to figure out which one it is.
>But how to prove it? And how can i show, that u i not an element of
>Q(sqrt(2),sqrt(3))?
Suppose u is an element of Q(sqrt(2),sqrt(3)).  That would mean that u
is of the form
 u = a + b*sqrt(2) + c*sqrt(3) + d*sqrt(6)
for some rational numbers a, b, c, and d. Show that this is
impossible, given what u^2 is supposed to be.
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Galois-Theory
>>Show that E=Q(sqrt(2),sqrt(3),u) where u^2=(9-5sqrt(3))(2-sqrt(2)) is
>>normal. Determine Gal E/Q.
>>I have no idea, what to do. I know two ways to show, that E/Q is normal.
>>First: show, that E is the splitting field of a polynomial g in Q[x].
>>Second: Q=Inv G, where G is a finite Group of automorphisms of E.
>>I think the first method needs too many calculations. 
> Not really. You know the polynomial that will give you sqrt(2):
> x^2-2. You know the one for sqrt(3), namely x^2-3. So if we let F =
> Q(sqrt(2),sqrt(3)), then F is the splitting field of (x^2-2)(x^2-3).
> Now, E = F(u), and is the splitting field, over F, of the polynomial
> x^2 - (9-5sqrt(3))(2-sqrt(2)).
> So E must be a splitting field over Q as well. 
You should be careful; Q(sqrt{2})/Q is normal and 
Q(sqrt[4]{2})/Q(sqrt{2}) is normal, but clearly Q(sqrt[4]{2})/Q is 
not normal...
Actually, I think that the second method is better. Since the degree of 
the extension Q(sqrt{2}, sqrt{3}, u)/Q is 8, we have good candidates 
for elements of the Galois group, namely f_1, ..., f_8, which send 
sqrt{2}, sqrt{3} and u to:
      sqrt{2}   sqrt{3}   u
f_1  sqrt{2}   sqrt{3}   sqrt{(9 - 5sqrt{3})( 2 - sqrt{2})}
f_2  -sqrt{2}  sqrt{3}   sqrt{(9 - 5sqrt{3})( 2 + sqrt{2})}
f_3  sqrt{2}   -sqrt{3}  sqrt{(9 + 5sqrt{3})( 2 - sqrt{2})}
f_4  sqrt{2}   sqrt{3}   -sqrt{(9 - 5sqrt{3})( 2 - sqrt{2})}
f_5  -sqrt{2}  -sqrt{3}  sqrt{(9 + 5sqrt{3})( 2 + sqrt{2})}
f_6  -sqrt{2}  sqrt{3}   -sqrt{(9 - 5sqrt{3})( 2 + sqrt{2})}
f_7  sqrt{2}   -sqrt{3}  -sqrt{(9 + 5sqrt{3})( 2 - sqrt{2})}
f_8  -sqrt{2}  -sqrt{3}  -sqrt{(9 + 5sqrt{3})( 2 + sqrt{2})}
The only thing that we have to worry about is whether f_is are 
well-defined, that is whether
(9 +/- 5sqrt{3})(2 +/- sqrt{2})
are squares in Q(sqrt{2}, sqrt{3}, u). Indeed, they are - take for 
example:
(9 - 5sqrt{3})(2 + sqrt{2}) =
= [(9 - 5sqrt{3})( 2 - sqrt{2})(2 + sqrt{2})^2] /
                                 / [(2 - sqrt{2})(2 + sqrt{2})] =
= u^2*(2 + sqrt{2})^2 / (sqrt{2}^2).
Similar argument works also for other terms, because:
(9 - 5sqrt{3})(9 + 5sqrt{3}) = 81 - 75 = 6 = (sqrt{2}sqrt{3})^2
(2 - sqrt{2})(2 + sqrt{2}) = 4 - 2 = 2 = (sqrt{2})^2.
Pawel Gladki
===
Subject: Re: Galois-Theory
            days. My association with the Department is that of an alumnus.
>Show that E=Q(sqrt(2),sqrt(3),u) where u^2=(9-5sqrt(3))(2-sqrt(2)) is
>normal. Determine Gal E/Q.
>I have no idea, what to do. I know two ways to show, that E/Q is normal.
>First: show, that E is the splitting field of a polynomial g in Q[x].
>Second: Q=Inv G, where G is a finite Group of automorphisms of E.
>I think the first method needs too many calculations. 
>> Not really. You know the polynomial that will give you sqrt(2):
>> x^2-2. You know the one for sqrt(3), namely x^2-3. So if we let F =
>> Q(sqrt(2),sqrt(3)), then F is the splitting field of (x^2-2)(x^2-3).
>> Now, E = F(u), and is the splitting field, over F, of the polynomial
>> x^2 - (9-5sqrt(3))(2-sqrt(2)).
>> So E must be a splitting field over Q as well. 
>You should be careful; Q(sqrt{2})/Q is normal and 
>Q(sqrt[4]{2})/Q(sqrt{2}) is normal, but clearly Q(sqrt[4]{2})/Q is 
>not normal...
Yes.
>Actually, I think that the second method is better. Since the degree of 
>the extension Q(sqrt{2}, sqrt{3}, u)/Q is 8, we have good candidates 
>for elements of the Galois group, namely f_1, ..., f_8, which send 
>sqrt{2}, sqrt{3} and u to:
>      sqrt{2}   sqrt{3}   u
>f_1  sqrt{2}   sqrt{3}   sqrt{(9 - 5sqrt{3})( 2 - sqrt{2})}
>f_2  -sqrt{2}  sqrt{3}   sqrt{(9 - 5sqrt{3})( 2 + sqrt{2})}
>f_3  sqrt{2}   -sqrt{3}  sqrt{(9 + 5sqrt{3})( 2 - sqrt{2})}
>f_4  sqrt{2}   sqrt{3}   -sqrt{(9 - 5sqrt{3})( 2 - sqrt{2})}
>f_5  -sqrt{2}  -sqrt{3}  sqrt{(9 + 5sqrt{3})( 2 + sqrt{2})}
>f_6  -sqrt{2}  sqrt{3}   -sqrt{(9 - 5sqrt{3})( 2 + sqrt{2})}
>f_7  sqrt{2}   -sqrt{3}  -sqrt{(9 + 5sqrt{3})( 2 - sqrt{2})}
>f_8  -sqrt{2}  -sqrt{3}  -sqrt{(9 + 5sqrt{3})( 2 + sqrt{2})}
>The only thing that we have to worry about is whether f_is are 
>well-defined, that is whether
>(9 +/- 5sqrt{3})(2 +/- sqrt{2})
>are squares in Q(sqrt{2}, sqrt{3}, u). 
Which is equivalent to showing that the polynomial g(x) I described,
which is the product of 
x^2 - ri
for r1, r2, r3, r4 equal, successively, to (9 +/- 5*sqrt(3))(2 +/- sqrt(2))
does indeed have all its roots in Q(sqrt(2),sqrt(3),u), and so that
this field is the splitting field of (x^2-2)(x^2-3)g(x).
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Galois-Theory
Hello Arturo,
>>I want to solve the following exercise.
>>Show that E=Q(sqrt(2),sqrt(3),u) where u^2=(9-5sqrt(3))(2-sqrt(2)) is
>>normal. Determine Gal E/Q.
>>I have no idea, what to do. I know two ways to show, that E/Q is normal.
>>First: show, that E is the splitting field of a polynomial g in Q[x].
>>Second: Q=Inv G, where G is a finite Group of automorphisms of E.
>>I think the first method needs too many calculations.
> Not really. You know the polynomial that will give you sqrt(2):
> x^2-2. You know the one for sqrt(3), namely x^2-3. So if we let F =
> Q(sqrt(2),sqrt(3)), then F is the splitting field of (x^2-2)(x^2-3).
OK, this is clear to me.
> Now, E = F(u), and is the splitting field, over F, of the polynomial
> x^2 - (9-5sqrt(3))(2-sqrt(2)).
> So E must be a splitting field over Q as well. To find the polynomial
Is there a theorem like this?: Let E be a splitting field of f over K, and 
K
be a splitting field of g over F, then E is the splitting field of a
polynomial over F.
> explicitly, take
> x^2 - (9-5sqrt(3))(2-sqrt(2))
> x^2 - (9+5sqrt(3))(2-sqrt(2))
> x^2 - (9-5sqrt(3))(2+sqrt(2))
> x^2 - (9+5sqrt(3))(2+sqrt(2))
I dont see, why the roots of these polynomials is in F(u) too?
> and multiply them together to get a polynomial g(x). Then E is the
> spliting field over Q of g(x)(x^2-2)(x^2-3). You don't have to
> calculate g(x) explicitly, just note that it will work.
>>But i dont know how to
>>realize the second way. If i assume, that u is not an element of
>>Q(sqrt(2),sqrt(3)), then [E:Q]=8, so G must be a group of order 8. I 
think
>>that G=<r,s,t>, where r:sqrt(2)->-sqrt(2), s:sqrt(3)->-sqrt(3), t:u->-u.
> There may be some relations between r, s, and t; you should figure out
> what r and s will do to u (see what they do to u^2). There aren't that
I have the problem, that i cannot see, why such automorphisms exists? Well,
the identity on F can be extended to an automorphism t of E sending u->-u
since these are roots of an minimum polynomial m of u over F. But why can i
extend the automorphisms r',s':sqrt(k)->-sqrt(k) k=2,3 on F to
automorphisms of E? To show this i must know that r'(m) [transformed of
Minimum Polynomial of u under r'] has also roots in F(u)?
> many groups of order 8 (only three abelian and two nonabelian ones),
> so you should be able to figure out which one it is.
>>But how to prove it? And how can i show, that u i not an element of
>>Q(sqrt(2),sqrt(3))?
> Suppose u is an element of Q(sqrt(2),sqrt(3)).  That would mean that u
> is of the form
>  u = a + b*sqrt(2) + c*sqrt(3) + d*sqrt(6)
> for some rational numbers a, b, c, and d. Show that this is
> impossible, given what u^2 is supposed to be.
I thought, this would be too complicate. But i will try it again.
Johannes
===
Subject: Re: Galois-Theory
            days. My association with the Department is that of an alumnus.
>Hello Arturo,
>I want to solve the following exercise.
>Show that E=Q(sqrt(2),sqrt(3),u) where u^2=(9-5sqrt(3))(2-sqrt(2)) is
>normal. Determine Gal E/Q.
>I have no idea, what to do. I know two ways to show, that E/Q is normal.
>First: show, that E is the splitting field of a polynomial g in Q[x].
>Second: Q=Inv G, where G is a finite Group of automorphisms of E.
>I think the first method needs too many calculations.
>> Not really. You know the polynomial that will give you sqrt(2):
>> x^2-2. You know the one for sqrt(3), namely x^2-3. So if we let F =
>> Q(sqrt(2),sqrt(3)), then F is the splitting field of (x^2-2)(x^2-3).
>OK, this is clear to me.
>> Now, E = F(u), and is the splitting field, over F, of the polynomial
>> x^2 - (9-5sqrt(3))(2-sqrt(2)).
>> So E must be a splitting field over Q as well. To find the polynomial
>Is there a theorem like this?: Let E be a splitting field of f over K, and 

K
>be a splitting field of g over F, then E is the splitting field of a
>polynomial over F.
Not as such, no; I may have been too hasty there. In fact, you notice
the problem below: it is easy enough to construct a polynomial with
coefficients in Q that has u as a root, but there is no guarantee in
general that K(u) will contain all those roots. In general, normal
over normal is not necessarily normal.
>> explicitly, take
>> x^2 - (9-5sqrt(3))(2-sqrt(2))
>> x^2 - (9+5sqrt(3))(2-sqrt(2))
>> x^2 - (9-5sqrt(3))(2+sqrt(2))
>> x^2 - (9+5sqrt(3))(2+sqrt(2))
>I dont see, why the roots of these polynomials is in F(u) too?
Yes, you're right. That is not immediate. Sorry about that. Let me get
back to you.
>> and multiply them together to get a polynomial g(x). Then E is the
>> spliting field over Q of g(x)(x^2-2)(x^2-3). You don't have to
>> calculate g(x) explicitly, just note that it will work.
>But i dont know how to
>realize the second way. If i assume, that u is not an element of
>Q(sqrt(2),sqrt(3)), then [E:Q]=8, so G must be a group of order 8. I 
think
>that G=<r,s,t>, where r:sqrt(2)->-sqrt(2), s:sqrt(3)->-sqrt(3), t:u->-u.
>> There may be some relations between r, s, and t; you should figure out
>> what r and s will do to u (see what they do to u^2). There aren't that
>I have the problem, that i cannot see, why such automorphisms exists? 
Well,
>the identity on F can be extended to an automorphism t of E sending u->-u
>since these are roots of an minimum polynomial m of u over F. But why can 
i
>extend the automorphisms r',s':sqrt(k)->-sqrt(k) k=2,3 on F to
>automorphisms of E? To show this i must know that r'(m) [transformed of
>Minimum Polynomial of u under r'] has also roots in F(u)?
This will follow once you know that E is a splitting field (in
characteristic zero, anyway); it is one of the basic properties of
splitting fields: if E/k is a splitting field, and F is a subfield of
E containing k, then every automorphism of E that fixes k pointwise
can be extended to an automorphism of F.
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Galois-Theory
Hello Arturo,
>>I want to solve the following exercise.
>>Show that E=Q(sqrt(2),sqrt(3),u) where u^2=(9-5sqrt(3))(2-sqrt(2)) is
>>normal. Determine Gal E/Q.
>>I have no idea, what to do. I know two ways to show, that E/Q is 
normal.
>>First: show, that E is the splitting field of a polynomial g in Q[x].
>>Second: Q=Inv G, where G is a finite Group of automorphisms of E.
>>I think the first method needs too many calculations.
> 
> Not really. You know the polynomial that will give you sqrt(2):
> x^2-2. You know the one for sqrt(3), namely x^2-3. So if we let F =
> Q(sqrt(2),sqrt(3)), then F is the splitting field of (x^2-2)(x^2-3).
>>OK, this is clear to me.
> Now, E = F(u), and is the splitting field, over F, of the polynomial
> 
> x^2 - (9-5sqrt(3))(2-sqrt(2)).
> 
> So E must be a splitting field over Q as well. To find the polynomial
>>Is there a theorem like this?: Let E be a splitting field of f over K, 
and
>>K be a splitting field of g over F, then E is the splitting field of a
>>polynomial over F.
> Not as such, no; I may have been too hasty there. In fact, you notice
> the problem below: it is easy enough to construct a polynomial with
> coefficients in Q that has u as a root, but there is no guarantee in
> general that K(u) will contain all those roots. In general, normal
> over normal is not necessarily normal.
> explicitly, take
> 
> x^2 - (9-5sqrt(3))(2-sqrt(2))
> x^2 - (9+5sqrt(3))(2-sqrt(2))
> x^2 - (9-5sqrt(3))(2+sqrt(2))
> x^2 - (9+5sqrt(3))(2+sqrt(2))
>>I dont see, why the roots of these polynomials is in F(u) too?
> Yes, you're right. That is not immediate. Sorry about that. Let me get
> back to you.
> and multiply them together to get a polynomial g(x). Then E is the
> spliting field over Q of g(x)(x^2-2)(x^2-3). You don't have to
> calculate g(x) explicitly, just note that it will work.
> 
>>But i dont know how to
>>realize the second way. If i assume, that u is not an element of
>>Q(sqrt(2),sqrt(3)), then [E:Q]=8, so G must be a group of order 8. I
>>think that G=<r,s,t>, where r:sqrt(2)->-sqrt(2), s:sqrt(3)->-sqrt(3),
>>t:u->-u.
> 
> There may be some relations between r, s, and t; you should figure out
> what r and s will do to u (see what they do to u^2). There aren't that
>>I have the problem, that i cannot see, why such automorphisms exists?
>>Well, the identity on F can be extended to an automorphism t of E sending
>>u->-u since these are roots of an minimum polynomial m of u over F. But
>>why can i extend the automorphisms r',s':sqrt(k)->-sqrt(k) k=2,3 on F to
>>automorphisms of E? To show this i must know that r'(m) [transformed of
>>Minimum Polynomial of u under r'] has also roots in F(u)?
> This will follow once you know that E is a splitting field (in
> characteristic zero, anyway); it is one of the basic properties of
> splitting fields: if E/k is a splitting field, and F is a subfield of
> E containing k, then every automorphism of E that fixes k pointwise
> can be extended to an automorphism of F.
Did you exchange E an F? Then I know this theorem, but the thing is, that 
we
dont know, whether E is a splitting field over Q, otherwise it follows
directly, that E is normal over Q, since Q is perfect.
Or do you mean another way to apply the above property?
Johannes
===
Subject: Re: Galois-Theory
            days. My association with the Department is that of an alumnus.
>I have the problem, that i cannot see, why such automorphisms exists?
>Well, the identity on F can be extended to an automorphism t of E 
sending
>u->-u since these are roots of an minimum polynomial m of u over F. But
>why can i extend the automorphisms r',s':sqrt(k)->-sqrt(k) k=2,3 on F to
>automorphisms of E? To show this i must know that r'(m) [transformed of
>Minimum Polynomial of u under r'] has also roots in F(u)?
>> This will follow once you know that E is a splitting field (in
>> characteristic zero, anyway); it is one of the basic properties of
>> splitting fields: if E/k is a splitting field, and F is a subfield of
>> E containing k, then every automorphism of E that fixes k pointwise
>> can be extended to an automorphism of F.
>Did you exchange E an F? 
Oops. Yes. (Not my thread...)
>Then I know this theorem, but the thing is, that we
>dont know, whether E is a splitting field over Q, otherwise it follows
>directly, 
Well, yes. But that's why I said that it will follow ->once you know
that E is a splitting field<-.
-- 
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Polynomials
i want to show, that if r is a root in a splitting field of f=x^3+x^2-2x-1
over Q, then r'=r^2-2 is another root.
I split f into factors f=(x-r)(x^2+(r+1)x+(r^2+r-2)), but how can I see,
that r^2-2 is a root of the quadratic factor?
Johannes
===
Subject: Re: Polynomials
> i want to show, that if r is a root in a splitting field of 
f=x^3+x^2-2x-1
> over Q, then r'=r^2-2 is another root.
> I split f into factors f=(x-r)(x^2+(r+1)x+(r^2+r-2)), but how can I see,
> that r^2-2 is a root of the quadratic factor?
  factorization of f(x) by just dividing it by (x - r).
  If that's the case, then you should have _no_ trouble
  finishing this -- just use polynomial long division
  to compute (x^2 + (r_1)x + (f^2+r-2))/(x - (r^2-2)) ...
  it'll work out nicely _because_ f(r) = 0 -- kinda
  neat, actually :-) (As a bonus, you'll also determine
  the third root of f in terms of r: it's r^2+r-1 ... )
 
> Johannes
===
Subject: Re: Polynomials
> i want to show, that if r is a root in a splitting field of 
f=x^3+x^2-2x-1
> over Q, then r'=r^2-2 is another root.
r being a root of f(x) = x^3 + x^2 - 2x - 1 implies that 
r^3 = 1 + 2r - r^2. 
Then r^4 = r + 2r^2 - r^3 = r + 2r^2 - (1 + 2r - r^2) = -1 - r + 3r^2. 
Similarly you can express r^5 and r^6 in terms of 1, r, and r^2. 
Now multiply out f(r^2 - 2) = (r^2 - 2)^3 + ... 
and reduce all the high powers of r as we've just seen is possible, 
and see what happens. 
Alternatively, let r'' = (r')^2 - 2; this must also be a root. 
r'' = (r^2 - 2)^2 - 2 = r^4 - 4 r^2 + 2 = (-1 - r + 3r^2) - 4 r^2 + 2 
= 1 - r - r^2. Now show that r, r', and r'' are zeros of f. 
The easy part is r + r' + r'' = -1.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: inverse modulos
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2JbUg32369;
>If two numbers x and m are relatively prime, so that gcd(x, m) = 1 is 
>true, then x has a unique multiplicative inverse modulo m, a, so that ax 
>= 1 (mod m).
>Knowing only the multiplicative inverse, a modulo m, and m, is it 
>possible to find x?
>Is it true that a*x - 1 = k*m, for some k, possibly negative?  Or 
>multiple k's?  How to find the multiple k's so that x could be found? 
>Is that possible at all, so that the candidate k's can be found knowing 
>only the inverse a modulo m and m?
>If there is no single way to find it, is there a reasonable trial and 
>error process to go through to find x given a and m?
Using the ideas from extended GCD or continued fraction.
Take the continued fraction expansion of the fraction a/m,
and truncate it at various stages to get good rational
approximations to a/m with small numerator and denominator.
One of those will be k/x.
===
Subject: Re: inverse modulos
> Using the ideas from extended GCD or continued fraction.
> Take the continued fraction expansion of the fraction a/m,
> and truncate it at various stages to get good rational
> approximations to a/m with small numerator and denominator.
> One of those will be k/x.
Is there somewhere on the web that explains how I can carry out these 
use it for this... it seems to be only for trig functions).  When should 
I truncate?  And then how would I separate the k and the x in k/x?
===
Subject: Re: .99999... still=/= 1
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB2JbVW32373;
> 
>> .999...* 10 - 9 = .999...
=> x * 10 - 9 = x where x = .999...
>=> 10x - x = 9 = 9x <==> x = 1
So why you claim that .999... != 1?
 
>   That's right.
>   x = .999...
> 
> and after his calculations are done he shows it to be equal to 
something 
> else.
> 
>  And another way to look at it let's set x = 1 to start with instead 
of 
>  .999...
> 
> 
> x = 1
> 10x = 10
>  x = 1
> 
>   Here we can see x = 1 before and after.
> 
> And,
> 
> x = .999...
> 
> 10 *x  = 9.999...
> 
> x = .999...
> 
>   Here we can see x = .999... before and after.
> 
> This is often given as a layman's justification (not proof) that .999... 
> =1. However mathematically, it's not quite right. 
>>It is sufficient for a proof.
> You can't justify the 
> step 10x = 9.999... without appealing to a theorem that says you can 
> multiply each term of a convergent series by a constant, and the 
> resulting series will converge to the constant times the sum of the 
> original series.
>>Such a theorem would be a tautology of little value, since the
>>statement follows from the directly from distributive axiom a(b+c)= ab
>>+ ac.
>>If
>>   x = 0.99999... [repeating n times]
>>Then,
>>   x = 9/10^1 + 9/10^2 + .... 9/10^n
>>   ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n
>>Let a = 10
>>  10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n
>>      = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n
>>      = 9 + x
> If you could prove that, then you'd already have enough understanding of 
> limits to know that .999... = 1.
>>Limits are unnecessary. The proof is simply,
>>x   = 0.9_
>>10x = 9.9_  (follows directly from the distributive axiom)
>>    = 9 + x
>> 9x = 9
>>  x = 1
>  Nope, this is incorrect.
>  The question was does .999... converge to 1, not 9.999... .
>equation. Equations aren't used in convergence tests.
>  The basic convergence test is like this.
>If,
>m-->oo
>Lim Z_m = 0
>then convergence otherwise divergence.
>.999... is represented as (1 - 1/10^m)
>Z_m = ( 1 - 1/10^m)
>m-->oo
>Lim Z_m = 1
>  This means divergence. So not only does it not converge to 1 but it 
doesn't
>equal 1. And even if it did converge it wouldn't EQUAL 1. There isn't even 

one
>series in all of math that does converge that really equals that 
convergence
>value.
>.999... --> diverges and =/= 1
>Smart's Alt. Physics News Group
><a 
href=http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1>ht
tp://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1</aS. Enterprize (Science Journal)
><a 
href=http://smart1234.s-enterprize.com/>http://smart1234.s-enterprize.com
/</a>
convergrence test   
according to your definition   1+ 1/2 + 1/3+1/4 +...  would converge since   

1/n--->0 as 
n--> infinity.  It is generally accepted that the sum given is divergent. 
thus we have a counter example that disproves your convergence test.
Please find an acceptable trust worthy   test for convergence 
===
Subject: Re: .99999... still=/= 1
>> 
>>> .999...* 10 - 9 = .999...
>>=> x * 10 - 9 = x where x = .999...
>>=> 10x - x = 9 = 9x <==> x = 1
>>So why you claim that .999... != 1?
>> 
>>   That's right.
>>   x = .999...
>> 
>> and after his calculations are done he shows it to be equal to
>something 
>> else.
>> 
>>  And another way to look at it let's set x = 1 to start with instead 
of
>>  .999...
>> 
>> 
>> x = 1
>> 10x = 10
>>  x = 1
>> 
>>   Here we can see x = 1 before and after.
>> 
>> And,
>> 
>> x = .999...
>> 
>> 10 *x  = 9.999...
>> 
>> x = .999...
>> 
>>   Here we can see x = .999... before and after.
>> 
>> This is often given as a layman's justification (not proof) that 
.999... 
>> =1. However mathematically, it's not quite right. 
>It is sufficient for a proof.
>> You can't justify the 
>> step 10x = 9.999... without appealing to a theorem that says you can 
>> multiply each term of a convergent series by a constant, and the 
>> resulting series will converge to the constant times the sum of the 
>> original series.
>Such a theorem would be a tautology of little value, since the
>statement follows from the directly from distributive axiom a(b+c)= ab
>+ ac.
>If
>   x = 0.99999... [repeating n times]
>Then,
>   x = 9/10^1 + 9/10^2 + .... 9/10^n
>   ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n
>Let a = 10
>  10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n
>      = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n
>      = 9 + x
>> If you could prove that, then you'd already have enough understanding 
of 
>> limits to know that .999... = 1.
>Limits are unnecessary. The proof is simply,
>x   = 0.9_
>10x = 9.9_  (follows directly from the distributive axiom)
>    = 9 + x
> 9x = 9
>  x = 1
>>  Nope, this is incorrect.
>>  The question was does .999... converge to 1, not 9.999... .
>>equation. Equations aren't used in convergence tests.
>>  The basic convergence test is like this.
>>If,
>>m-->oo
>>Lim Z_m = 0
>>then convergence otherwise divergence.
>>.999... is represented as (1 - 1/10^m)
>>Z_m = ( 1 - 1/10^m)
>>m-->oo
>>Lim Z_m = 1
>>  This means divergence. So not only does it not converge to 1 but it
>doesn't
>>equal 1. And even if it did converge it wouldn't EQUAL 1. There isn't 
even
>one
>>series in all of math that does converge that really equals that 
convergence
>>value.
>>.999... --> diverges and =/= 1
>>Smart's Alt. Physics News Group
>>http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
>>S. Enterprize (Science Journal)
>>http://smart1234.s-enterprize.com/
>convergrence test   
>according to your definition   1+ 1/2 + 1/3+1/4 +...  would converge since
>1/n--->0 as 
>n--> infinity.  It is generally accepted that the sum given is divergent. 
>thus we have a counter example that disproves your convergence test.
>Please find an acceptable trust worthy   test for convergence 
The form of of a harmonic series is 
n= 1 to oo
lim SUM   1/n
Using The Root Convergence Test
         ____
lim n/ Z_n  = L
n-->oo
L < 1 = converge
L > 1 = diverge
L = 1, test fails, no conclusion possible.
  At n-->0 it's divergent , L >1
       n=0, L= 1
       n>1, L < 1 = convergent.
  According to this convergence test, it is divergent, no conclusion 
possible ,
and convergent. It depends on the point of view you take, or you can just
observe all cases.
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
>>> .999...* 10 - 9 = .999...
>>=> x * 10 - 9 = x where x = .999...
>>=> 10x - x = 9 = 9x <==> x = 1
>>So why you claim that .999... != 1?
>>   That's right.
>>   x = .999...
>> and after his calculations are done he shows it to be equal to
>something
>> else.
>>  And another way to look at it let's set x = 1 to start with
instead of
>>  .999...
>> x = 1
>> 10x = 10
>>  x = 1
>>   Here we can see x = 1 before and after.
>> And,
>> x = .999...
>> 10 *x  = 9.999...
>> x = .999...
>>   Here we can see x = .999... before and after.
>> This is often given as a layman's justification (not proof) that
.999...
>> =1. However mathematically, it's not quite right.
It is sufficient for a proof.
> You can't justify the
>> step 10x = 9.999... without appealing to a theorem that says you can
>> multiply each term of a convergent series by a constant, and the
>> resulting series will converge to the constant times the sum of the
>> original series.
Such a theorem would be a tautology of little value, since the
>statement follows from the directly from distributive axiom a(b+c)= ab
>+ ac.
If
>   x = 0.99999... [repeating n times]
>Then,
>   x = 9/10^1 + 9/10^2 + .... 9/10^n
   ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n
Let a = 10
  10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n
>      = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n
>      = 9 + x
> If you could prove that, then you'd already have enough 
understanding
of
>> limits to know that .999... = 1.
Limits are unnecessary. The proof is simply,
>x   = 0.9_
>10x = 9.9_  (follows directly from the distributive axiom)
>    = 9 + x
> 9x = 9
>  x = 1
>>  Nope, this is incorrect.
>>  The question was does .999... converge to 1, not 9.999... .
>>equation. Equations aren't used in convergence tests.
>>  The basic convergence test is like this.
>>If,
>>m-->oo
>>Lim Z_m = 0
>>then convergence otherwise divergence.
>>.999... is represented as (1 - 1/10^m)
>>Z_m = ( 1 - 1/10^m)
>>m-->oo
>>Lim Z_m = 1
>>  This means divergence. So not only does it not converge to 1 but it
>doesn't
>>equal 1. And even if it did converge it wouldn't EQUAL 1. There isn't
even
>one
>>series in all of math that does converge that really equals that
convergence
>>value.
>>.999... --> diverges and =/= 1
>>Smart's Alt. Physics News Group
>>http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
>>S. Enterprize (Science Journal)
>>http://smart1234.s-enterprize.com/
>convergrence test
>according to your definition   1+ 1/2 + 1/3+1/4 +...  would converge
since
>1/n--->0 as
>n--> infinity.  It is generally accepted that the sum given is 
divergent.
>thus we have a counter example that disproves your convergence test.
>Please find an acceptable trust worthy   test for convergence
> The form of of a harmonic series is
> n= 1 to oo
> lim SUM   1/n
> Using The Root Convergence Test
>          ____
> lim n/ Z_n  = L
> n-->oo
> L < 1 = converge
> L > 1 = diverge
> L = 1, test fails, no conclusion possible.
>   At n-->0 it's divergent , L >1
>        n=0, L= 1
>        n>1, L < 1 = convergent.
>   According to this convergence test, it is divergent, no conclusion
possible ,
> and convergent. It depends on the point of view you take, or you can just
> observe all cases.
Are you ignoring your lsat post now?
You're hopeless.
===
Subject: Re: .99999... still=/= 1
> 
>> .999...* 10 - 9 = .999...
=> x * 10 - 9 = x where x = .999...
>=> 10x - x = 9 = 9x <==> x = 1
So why you claim that .999... != 1?
 
>   That's right.
>   x = .999...
> 
> and after his calculations are done he shows it to be equal to
>>something 
> else.
> 
>  And another way to look at it let's set x = 1 to start with 
instead
>  .999...
> 
> 
> x = 1
> 10x = 10
>  x = 1
> 
>   Here we can see x = 1 before and after.
> 
> And,
> 
> x = .999...
> 
> 10 *x  = 9.999...
> 
> x = .999...
> 
>   Here we can see x = .999... before and after.
> 
> This is often given as a layman's justification (not proof) that 
.999...
> =1. However mathematically, it's not quite right. 
>>It is sufficient for a proof.
> You can't justify the 
> step 10x = 9.999... without appealing to a theorem that says you can 
> multiply each term of a convergent series by a constant, and the 
> resulting series will converge to the constant times the sum of the 
> original series.
>>Such a theorem would be a tautology of little value, since the
>>statement follows from the directly from distributive axiom a(b+c)= ab
>>+ ac.
>>If
>>   x = 0.99999... [repeating n times]
>>Then,
>>   x = 9/10^1 + 9/10^2 + .... 9/10^n
>>   ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n
>>Let a = 10
>>  10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n
>>      = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n
>>      = 9 + x
> If you could prove that, then you'd already have enough understanding 
of
> limits to know that .999... = 1.
>>Limits are unnecessary. The proof is simply,
>>x   = 0.9_
>>10x = 9.9_  (follows directly from the distributive axiom)
>>    = 9 + x
>> 9x = 9
>>  x = 1
>  Nope, this is incorrect.
>  The question was does .999... converge to 1, not 9.999... .
>equation. Equations aren't used in convergence tests.
>  The basic convergence test is like this.
>If,
>m-->oo
>Lim Z_m = 0
>then convergence otherwise divergence.
>.999... is represented as (1 - 1/10^m)
>Z_m = ( 1 - 1/10^m)
>m-->oo
>Lim Z_m = 1
>  This means divergence. So not only does it not converge to 1 but it
>>doesn't
>equal 1. And even if it did converge it wouldn't EQUAL 1. There isn't 
even
>>one
>series in all of math that does converge that really equals that
>convergence
>value.
>.999... --> diverges and =/= 1
>Smart's Alt. Physics News Group
>http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
>S. Enterprize (Science Journal)
>http://smart1234.s-enterprize.com/
>>convergrence test   
>>according to your definition   1+ 1/2 + 1/3+1/4 +...  would converge 
since
>>1/n--->0 as 
>>n--> infinity.  It is generally accepted that the sum given is divergent. 

>>thus we have a counter example that disproves your convergence test.
>>Please find an acceptable trust worthy   test for convergence 
>The form of of a harmonic series is 
>n= 1 to oo
>lim SUM   1/n
>Using The Root Convergence Test
 Correction : Use the Ratio Test
     The answer is still the same.
Not this vvvvv
>         ____
>lim n/ Z_n  = L
>n-->oo
>L < 1 = converge
>L > 1 = diverge
>L = 1, test fails, no conclusion possible.
>  At n-->0 it's divergent , L >1
>       n=0, L= 1
>       n>1, L < 1 = convergent.
>  According to this convergence test, it is divergent, no conclusion 
possible
>and convergent. It depends on the point of view you take, or you can just
>observe all cases.
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: sci.physics  , sci.math is nonsense
shut up when i m talking to you. shut up!! 
===
Subject: Re: sci.physics  , sci.math is nonsense
> shut up when i m talking to you. shut up!! 
Speak louder we can't hear you.
===
Subject: Re: sci.physics  , sci.math is nonsense
shut up when i m taking to you. shut up, shut up. shut up when i m taking to 

you!!! 
===
Subject: Re: sci.physics  , sci.math is nonsense
is that what your mama said to you over and over?
> shut up when i m taking to you. shut up, shut up. shut up when i m taking
to
> you!!!
===
Subject: Re: sci.physics  , sci.math is nonsense
hello punk, certified punk
> is that what your mama said to you over and over?
>> shut up when i m taking to you. shut up, shut up. shut up when i m 
taking
> to
>> you!!!
===
Subject: Re: sci.physics  , sci.math is nonsense
        reply-type=response
> hello punk, certified punk
are you boring on purpose?
===
Subject: Re: sci.physics , sci.math is nonsense
   <319evkF38gkbvU1@individual.net>
   <319v8eF2qj9e4U1@individual.net>
   <31bnb4F3agmorU1@individual.net>
   <K62sd.63655$KQ2.17588@fe2.texas.rr.com hello punk, certified punk
> are you boring on purpose?
You know, for some reason, Lord of Chaos seems... 'Frazir-like'.
-Mark Martin
===
Subject: Re: sci.physics  , sci.math is nonsense
your mom again
>> hello punk, certified punk
> are you boring on purpose? 
===
Subject: i clean up sci.physics, sci.math, rec.mensa.org  gentleman bets 
taken!!!
===
Subject: Re: i clean up sci.physics, sci.math, rec.mensa.org  gentleman bets 

taken!!!
plonk 
===
Subject: Re: i clean up sci.physics, sci.math, rec.mensa.org  gentleman bets 

taken!!!
pink
> plonk
===
Subject: talk science and mathematics, you faggots.
i m supreme. i alone command you. 
===
Subject: Re: talk science and mathematics, you faggots.
        reply-type=response
> i m supreme. i alone command you.
still boring...
===
Subject: Re: talk science and mathematics, you faggots.
>> i m supreme. i alone command you.
> still boring...
Lunacy is never boring. BTW CHAOS stands for Could (do to) Have Another 
Olanzapine Soon.
===
Subject: Re: talk science and mathematics, you faggots.
In sci.math, Lord of Chaos(Suresh Devanathan)
<idatarm@yahoo.com>
<319fl1F36rekiU1@individual.net>:
> i m supreme. i alone command you. 
[1] A sultan commands his driver to get from point A to point B
    60 miles distant in 1 hour.  The driver, for various reasons,
    drives the first 30 minutes at 30 miles per hour.  How fast
    would he have to drive the rest of the distance in order to
    make it, and how far would he have to travel?
[2] A proposed US single-stage rocket craft has a thrust of
    36 megaNewtons for 120 seconds, and an exhaust velocity of 4 km/s.
    Assuming that the fuel tanks comprise 3% of the rocket's
    mass (sans payload), the radius of the Earth is 6378 km, and the
    surface g is 9.805 m/s/s, can the rocket make it to a circular
    orbit 200 km above the surface, and if so, how much payload
    could it carry?  Neglect rotation and air friction in the
    computation.  (Hint: Tsiolkovsky.)
[3] Same as [2], only this time the question is whether it can
    escape Earth's gravitational pull altogether, and how much
    payload it can carry.
[4] A construction company has been contracted to build a dam out
    of cement, and the specification requires a certain thickness of
    cement at the bottom of the dam.  This thickness can hold
    back 100 megaNewtons of pressure.  However, the thickness
    also has to contend with its own weight, which is 9.805 m/s/s = N/kg.
    The mass of the concrete is 2400 kg/m^3.  Assuming that one
    can fashion a triangular dam wedge, what is the maximum depth
    of water one can hold back with this concrete, and how thick
    is the dam at its base?  Assume a water density of 1000 kg/m^3.
[5] A flat iron bar of indefinite length and square cross section of 1 cm
    is propped up on at least two round rods, in such a fashion
    that it can travel freely.  If the round rods are each 1 mm in
    diameter, how far apart can they be spaced without the iron bar
    touching the ground?  Assume a rigidity modulus of 82 gigaPascal
    and a density of 7874 kg/m^3; neglect distortion in the support rods.
Have fun, especially since I don't know the answers either!  (Except
to #1.  No, it's not infinite speed.)
-- 
#191, ewill3@earthlink.net -- no, I don't do this for a living
It's still legal to go .sigless.
===
Subject: Re: talk science and mathematics, you faggots.
We ARE discussing science and mathematics!
> i m supreme. i alone command you.
===
Subject: Re: talk science and mathematics, you faggots.
Yes you are... yes you are
> We ARE discussing science and mathematics!
>> i m supreme. i alone command you.
===
Subject: Re: talk science and mathematics, you faggots.
> i m supreme. i alone command you.
Hey stooopid, 
   1) Turbulence scales as the cube of the distance.
   2) Graph your pathetic little brainfart on hyperbolic axes.
   3) Your trailer park called - its garbage is missing.
-- 
Uncle Al
http://www.mazepath.com/uncleal/
 (Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
===
Subject: Re: talk science and mathematics, you faggots.
shut up! shut up when i m talking to you!
>> i m supreme. i alone command you.
> Hey stooopid,
>   1) Turbulence scales as the cube of the distance.
>   2) Graph your pathetic little brainfart on hyperbolic axes.
>   3) Your trailer park called - its garbage is missing.
> -- 
> Uncle Al
> http://www.mazepath.com/uncleal/
> (Toxic URL! Unsafe for children and most mammals)
> http://www.mazepath.com/uncleal/qz.pdf 
===
Subject: Re: talk science and mathematics, you faggots.
> shut up! shut up when i m talking to you!
> i m supreme. i alone command you.
>> Hey stooopid,
>>   1) Turbulence scales as the cube of the distance.
>>   2) Graph your pathetic little brainfart on hyperbolic axes.
>>   3) Your trailer park called - its garbage is missing.
>> --
>> Uncle Al
>> http://www.mazepath.com/uncleal/
>> (Toxic URL! Unsafe for children and most mammals)
>> http://www.mazepath.com/uncleal/qz.pdf
Lord of boring.
-- 
Sl.87inte,
Fletch 
===
Subject: Re: talk science and mathematics, you faggots.
lord of you 
===
Subject: Re: talk science and mathematics, you faggots.
you are not talking to him, you are typing at him.
===
Subject: Re: talk science and mathematics, you faggots.
> you are not talking to him, you are typing at him.
Not me, I bozobinned him.
Pepe le Pew
-- 
PT Barnum was right !
===
Subject: Re: talk science and mathematics, you faggots.
> i m supreme. i alone command you. 
Chaos = 1/0
Lord of Chaos = (1/0)^2
-- 
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Universal sentences of Z and ZxZ
How to prove that the universal sentences of Z are the one of ZxZ, both 
considered as additive groups ?
A proof using ultraproduct will be much appreciated. And to avoid 
discimination, others also! 
===
Subject: how to measure entropy of music?
I have vaguely heard about this method... and I am very interested in it... 

could anybody give me some pointers?
After learning how to measure entropy of music... I can begin to measure 
entropy of texts, etc.. that's going to be fun! 
===
Subject: Re: how to measure entropy of music?
> I have vaguely heard about this method... and I am very interested in 
it... 
> could anybody give me some pointers?
> After learning how to measure entropy of music... I can begin to measure 
> entropy of texts, etc.. that's going to be fun!
   Well, theoretically speaking, you should just treat the notes of
the music as symbols for the computation of the entropy. Then, 
there is no difference between finding the entropy of text and finding
the entropy of music.
   But if you sample the signal of music, encode it, and compute the
entropy, I doubt if it makes sense to do that given the fact that you
already have the notes of the music.
   If you just want to know what is about entropy of data, you might
just start with any textbook for Information theory or some websites
for some ideas, e.g.,
http://www.ScienceOxygen.com/signal.html
http://www.ScienceOxygen.com/electrical.html
    Have fun,
===
Subject: Re: how to measure entropy of music?
> I have vaguely heard about this method... and I am very interested in 
it... 
> could anybody give me some pointers?
> After learning how to measure entropy of music... I can begin to measure 
> entropy of texts, etc.. that's going to be fun!
Compress it using a suitable lossless compressor. Provided that the
compressor knows how to detect (some amount of) musical redundancy,
this will give you the Kolmogorov complexity with respect to a
particular non-universal computer, e.g. algorithmic entropy which is
more fundamental than Shannon entropy.
This has already been exploited using the information distance of
Vitanyi et al. Precise clustering of music files has been achieved
with ordinary compressors like bzip2. Find the paper Algorithmic
Clustering of Music by Rudi Cilibrasi and Paul Vitanyi. Needless to
say, you have to compress uncompressed wav files...
--
Eray Ozkural
===
Subject: Re: how to measure entropy of music?
(snip, and previously snipped discussion of music entropy)
> Compress it using a suitable lossless compressor. Provided that the
> compressor knows how to detect (some amount of) musical redundancy,
> this will give you the Kolmogorov complexity with respect to a
> particular non-universal computer, e.g. algorithmic entropy which is
> more fundamental than Shannon entropy.
> This has already been exploited using the information distance of
> Vitanyi et al. Precise clustering of music files has been achieved
> with ordinary compressors like bzip2. Find the paper Algorithmic
> Clustering of Music by Rudi Cilibrasi and Paul Vitanyi. Needless to
> say, you have to compress uncompressed wav files...
I would think a pure sine wave should be low complexity, so maybe
you should compress the Fourier transform.   Maybe that won't
quite do it, but I can imagine wav files of low complexity
audio signals not compressing very well.
Has anyone ever done an FFT of a whole CD?
-- glen
===
Subject: Re: how to measure entropy of music?
> I would think a pure sine wave should be low complexity, so maybe
> you should compress the Fourier transform.  
I think a vocoder does something similar: linear predictive coding
tries to fit an AR model to the input, effectively searching for the
formants (resonant peaks) of the vocal tract transfer function. If
you initialize an IIR filter with the LPC  coefficients and the states
(delays) with the signal, you can let it ring like an oscillator -
the output will converge to a sum of pure damped sine waves at the
estimated frequencies of the formants. However, instead of storing the
sine waves, the vocoder stores the LPC coefficients and the residue
(prediction error).
To take this back to entropy, the energy of the prediction error could
well be regarded as the entropy of a vocal signal - it gives a good
measure on how unexpected the signal is, given its past history.
Recent work has shown that this method is also suited for general
audio signals, not just speech. The order of the AR model just
increases drastically (for interpolation, orders of 1000 - 3000 are
used). The same applies here for the prediction error and the entropy.
> Has anyone ever done an FFT of a whole CD?
If you take a music CD, I would expect a linear trend of the form
1/f^a. This is a common model for music signals, and it agrees well
with findings of long-range correlated time series.
For speech, I would guess a rectangular pulse response. The formants
vary across a certain frequency range, but are limited from above and
below.
> -- glen
Andor
===
Subject: Re: how to measure entropy of music?
> I have vaguely heard about this method... and I am very interested in 
it... 
> could anybody give me some pointers?
> After learning how to measure entropy of music... I can begin to measure 
> entropy of texts, etc.. that's going to be fun! 
Learn all you want and more at Brian Whitman's site...
<http://web.media.mit.edu/~bwhitman/>
Whitman is also the producer of Eigenradio (their motto: Statistically 
their iTunes and WMP links...
<http://eigenradio.media.mit.edu/>
-- 
Remove _me_ for e-mail address
===
Subject: Re: how to measure entropy of music?
is this some joke ?!
>I have vaguely heard about this method... and I am very interested in it... 

>could anybody give me some pointers?
> After learning how to measure entropy of music... I can begin to measure 
> entropy of texts, etc.. that's going to be fun!
===
Subject: Re: how to measure entropy of music?
>> After learning how to measure entropy of music... I can begin to measure 

>> entropy of texts, etc.. that's going to be fun!
>is this some joke ?!
The model of entropy that can be readily recognized:
In music--how often can an informed listener infer the next note in a
phrase; ie how many bits are needed to specifiy the next note.
In sound--how many bits are needed to specify the next value of the
signal.
For written language, the analog is how many bits are needed to
confirm an informed guess as to the next letter in a text.  Perhaps no
more than three.  For some TV shows these days, its only two.
For music, the question gets really challenging as one considers the
entropy of scores--the parts played by accompanying instruments and
the choices of these instruments requires a lot of encoding.  In this
case, the number of bits needed to feed a high level orchestral
synthesizer might establish a lower bound.
Using a predictor such as a Hidden Markov Model using the Viterbi
algorithm or one of its descendants might go a long way in evaluating
the predictability/aka entropy of sounds. (This is already the case in
speech recognition.)
A really interesting question would be to rank famous composers by the
average entropy of their music--eg Bach, Mozart, the Beatles at the
top, Andrew Lloyd Webber, Elton John, Cole Porter next--well, you get
the idea.
The basics are at:
http://www.music-cog.ohio-state.edu/Music829D/Notes/Infotheory.html
key words: hmm, entropy, and music I have learned the question is of
significance today for serveral reasons:
1) Recognizing when TV commercials are playing.
2) Separating out background music behind speech in speech
recognition.
3) Locating specific music contained in a large database of sound
Try:
http://crl.research.compaq.com/publications/techreports/techreports.html
search the page for Logan or music.
Also:
http://www.idiap.ch/publications/ajmera-rr-01-26.bib.abs.html
Speech/Music Discrimination using Entropy and Dynamism Features in a
HMM Classification Framework
Speech/Music Discrimination Using Discrete Hidden Markov Models
http://crl.research.compaq.com/publications/techreports/reports/2000-1.pdf
MUSIC SUMMARY USING KEY PHRASES
M. Brand, .8bStructure Discovery in Conditional Probability Models 
via
an Entropic Prior and Parameter Extinction,.8a Neural Computation, 
July
1999
John Bailey
http://home.rochester.rr.com/jbxroads/mailto.html
===
Subject: Re: how to measure entropy of music?
> For written language, the analog is how many bits are needed to
> confirm an informed guess as to the next letter in a text.  Perhaps no
> more than three.  For some TV shows these days, its only two.
> For music, the question gets really challenging as one considers the
> entropy of scores--the parts played by accompanying instruments and
> the choices of these instruments requires a lot of encoding.  In this
> case, the number of bits needed to feed a high level orchestral
> synthesizer might establish a lower bound.
Very much a lower bound, I suspect, because even a fantastic orchestral
synthesizer would be to an orchestra as a typesetter would be to the
handwritten word: you don't only have the notes, you have the playing of
each of them, and the interplay of the playing with the acoustics of the
hall, etc.
Still, it's a neat thought experiment.  And you could indeed do an
interesting analysis of musical scores, rather than muscial recordings.
-- 
Andrew
===
Subject: Re: how to measure entropy of music?
<snip The model of entropy that can be readily recognized:
> In music--how often can an informed listener infer the next note in a
> phrase; ie how many bits are needed to specifiy the next note.
> In sound--how many bits are needed to specify the next value of the
> signal.
<snip>
found music to be interesting to listen to when he had a low success
rate at predicting what would happen next.  I think he was thinking of
higher level structures than single notes, but couldn't the concept of
entropy apply to these higher structures as well?
===
Subject: Re: how to measure entropy of music?
>As a matter of interest,
>do you consider high entropy good or bad?
>found music to be interesting to listen to when he had a low success
>rate at predicting what would happen next.  I think he was thinking of
>higher level structures than single notes, but couldn't the concept of
>entropy apply to these higher structures as well?
>Presumably a completely random series of notes would have very high 
entropy,
>while absolute silence has very low entropy.
>I wouldn't have thought either was very enjoyable.
Both of your comments are quite stimulating.  Of course!  The idea
that good music is neither too rote nor too random is not new but the
idea that entropy as a concept allows a deeper investigation of what
people like is exciting.  Conjecture: classes of composers cluster
around various levels of entropy.  On further thought--what is
predictable changes with audience familiarity with a style.  Music
that is on the leading edge of predictability is the most interesting.
Its probably all about what gives our music neurons a good workout.
John Bailey
http://home.rochester.rr.com/jbxroads/mailto.html
===
Subject: Re: how to measure entropy of music?
> Try:
> http://crl.research.compaq.com/publications/techreports/techreports.html
> search the page for Logan or music.
Some of Beth's papers are available from her now-HPLabs web-site:
http://www.hpl.hp.com/research/crl/publications/papers.html
Ciao,
Peter K.
===
Subject: Re: how to measure entropy of music?
> A really interesting question would be to rank famous composers by the
> average entropy of their music--eg Bach, Mozart, the Beatles at the
> top, Andrew Lloyd Webber, Elton John, Cole Porter next--well, you get
> the idea.
As a matter of interest,
do you consider high entropy good or bad?
Presumably a completely random series of notes would have very high 
entropy,
while absolute silence has very low entropy.
I wouldn't have thought either was very enjoyable.
-- 
Timothy Murphy  
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
===
Subject: Re: how to measure entropy of music?
> The basics are at:
> http://www.music-cog.ohio-state.edu/Music829D/Notes/Infotheory.html
> key words: hmm, entropy, and music I have learned the question is of
> significance today for serveral reasons:
> 1) Recognizing when TV commercials are playing.
> 2) Separating out background music behind speech in speech
> recognition.
I wonder if they could figure a why to identify background music and 
separate it from the sound reaching my ears :-)
Steve
===
Subject: Re: how to measure entropy of music?
>> The basics are at:
>> http://www.music-cog.ohio-state.edu/Music829D/Notes/Infotheory.html
>> key words: hmm, entropy, and music I have learned the question is of
>> significance today for serveral reasons:
>> 1) Recognizing when TV commercials are playing.
>> 2) Separating out background music behind speech in speech
>> recognition.
>I wonder if they could figure a why to identify background music and 
>separate it from the sound reaching my ears :-)
Good point!  If the predictor works then you can make a pretty good
Or better yet, a transmogriphier; if it's punk metal coming out of the
speaker and you want classical, you could put some filter and
adaptation circuits in that take the punk metal energy and convert it
based on the predictor rules for classical.   Punk metal out of the
speaker, classical into your ear.
Or you could buy an MP3 player and some headphones, either way.  ;)
Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be Intel's opinions.
http://www.ericjacobsen.org
===
Subject: Products of Sines?
Does the following product have a nice closed-form solution?
  sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Products of Sines?
Daryl McCullough <daryl@atc-nycorp.com> escribi.97:
> Does the following product have a nice closed-form solution?
>  sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
Let
P(z) = (z^n - 1)/(z - 1) = 1 + z + z^2 + ... + z^(n-1)  (#1)
But also
P(z) = Prod(z - z_k, k, 1, n-1)  (#2)
being
z_k = cos(k*2pi/n) + i*sen(k*2pi/n),  k = 1, 2, ..., n -1
the n-1 n-roots of unity distinct of 1.
|P(1)| = |Prod(z - z_k, k, 1, n-1)|
         = Prod(|z - z_k|, k, 1, n-1)
         = Prod(d_k, k, 1, n-1)
where  d_k is the distance from (1, 0) to (cos(k*2pi/n), sen(k*2pi/n)). But
d_k = sqrt((1 - cos(k*2pi/n))^2 + sen^2(k*2pi/n))
       = sqrt(2 - 2cos(k*2pi/n)) = 2sqrt((1 - 1cos(k*2pi/n))/2)
       = 2*sen(kpi/n)
Then
|P(1)| = n = Prod(2sen(k*pi/n), k, 1, n-1) = 2^(n-1)*Prod(sen(k*pi/n), k, 1, 

Prod(sen(k*pi/n), k, 1, n-1) = n/2^(n-1)
Curiously, it is the same that the probability of n points choosen at ramdom 

in a circle lie in the same semicircle.
-- 
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: Products of Sines?
> Does the following product have a nice closed-form solution?
>   sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
This is very well-known. Set w = exp(pi i/n). The product
is then
(2i)^{-n+1} (w - w^{-1})(w^2 - w^{-2}) ... (w^{n-1} - w^{-n+1})
= (2i)^{-n+1} w^{n(n-1)/2} (1 - w^{-2})(1 - w^{-4}) ... (1 - w^{-2n+2}).
Now (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2})
= (X^n - 1)/(X - 1) = 1 + X + X^2 + X^3 + ... + X^{n-1}.
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Products of Sines?
Robin Chapman says...
>> Does the following product have a nice closed-form solution?
>>   sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
>This is very well-known. Set w = exp(pi i/n). The product
>is then
>(2i)^{-n+1} (w - w^{-1})(w^2 - w^{-2}) ... (w^{n-1} - w^{-n+1})
>= (2i)^{-n+1} w^{n(n-1)/2} (1 - w^{-2})(1 - w^{-4}) ... (1 - w^{-2n+2}).
>Now (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2})
>= (X^n - 1)/(X - 1) = 1 + X + X^2 + X^3 + ... + X^{n-1}.
Okay, I'm puzzled by the step
(X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) = (X^n - 1)/(X - 1)
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Products of Sines?
> Robin Chapman says...
> Does the following product have a nice closed-form solution?
> 
>   sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
>>This is very well-known. Set w = exp(pi i/n). The product
>>is then
>>(2i)^{-n+1} (w - w^{-1})(w^2 - w^{-2}) ... (w^{n-1} - w^{-n+1})
>>= (2i)^{-n+1} w^{n(n-1)/2} (1 - w^{-2})(1 - w^{-4}) ... (1 - w^{-2n+2}).
>>Now (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2})
>>= (X^n - 1)/(X - 1) = 1 + X + X^2 + X^3 + ... + X^{n-1}.
> Okay, I'm puzzled by the step
> (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) = (X^n - 1)/(X - 1)
What is the factorization of X^n - 1 into linear factors?
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Products of Sines?
> Does the following product have a nice closed-form solution?
>   sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
Yup. It is n/2^{n-1}.
Pawel Gladki
===
Subject: Re: Products of Sines?
>> Does the following product have a nice closed-form solution?
>>   sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
>Yup. It is n/2^{n-1}.
How do you derive that?
--
Daryl McCullough
===
Subject: Re: Products of Sines?
>Does the following product have a nice closed-form solution?
>  sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
>>Yup. It is n/2^{n-1}.
> How do you derive that?
See Robin Chapman's answer.
Pawel Gladki
===
Subject: Probability Of Ultimate Extinction.
Let's say that you have an amoeba which has a probability q of dying without 

divinding,
and probability p of spliting into two. What is the probability of ultimate 

extinction?
I know that the recursion formula for extinction after n cycles is
P(n)=q+p*P(n-1)^2 where P(0)=q.
of 3 or 4) is that
the probability of ultimate extinction involves using the quadratic formula 

in the following way.
1±(1-4*p*q)^0.5
---------------
2q
where the minimum of the two values achieved by the quadratic equation is 
the desired probability.
What I'd like to know is, why bother? I've discovered that in the above 
problem, the probability
of ultimate extinction is p/q = 1 if p>=q. Why bother with the quadratic 
formula?
-- 
-------------------------------
Patrick D. Rockwell 
===
Subject: Re: Probability Of Ultimate Extinction.
> Let's say that you have an amoeba which has a probability q of dying
without
> divinding,
> and probability p of spliting into two. What is the probability of
ultimate
> extinction?
As a side note, some say that cells have an inherent average number of
divisions before senisence, or the inability to subdivide. Some research
seems to indicate that it may be possible to optimize that average,
reducing the standard deviation.
===
Subject: Re: Probability Of Ultimate Extinction.
>Let's say that you have an amoeba which has a probability q of dying 
without 
>divinding,
>and probability p of spliting into two. What is the probability of ultimate 

>extinction?
>I know that the recursion formula for extinction after n cycles is
>P(n)=q+p*P(n-1)^2 where P(0)=q.
You have not said what a cycle is, but assusming p+q=1 then
if you are looking for the probability of ultimate extinction, u, then
it is the probability of going extinct without dividing plus the
probability that it divides and both halves go extinct:
u = q +p*u^2  
>of 3 or 4) is that
>the probability of ultimate extinction involves using the quadratic formula 

>in the following way.
>1±(1-4*p*q)^0.5
>---------------
You should be dividing by 2p not 2q.  
>where the minimum of the two values achieved by the quadratic equation is 
>the desired probability.
>What I'd like to know is, why bother? I've discovered that in the above 
>problem, the probability
>of ultimate extinction is p/q = 1 if p>=q. Why bother with the quadratic 
>formula?
Perhaps you have swapped p and q somewhere 
If p+q=1 then the quadratic gives the lower of (1-p)/p and 1,
depending on whether p>=1/2 or p<=1/2 which makes sense and in the
latter case confirms what is obvious.   
===
Subject: Speaking of Attacking the Conclusions
JSH constantly whines that everyone is attacking his conclusions and
that this is a fallacy because his suppositions and logic are intact.
Aside from the fact that this isn't true (people have been attacking
his logic and pointing out gaping holes) it's funny to point out that
he does the same thing. All he'll do is attack the conclusions posters
have drawn based on his inacuracies. He won't actually directly attack
or even acknowledge the points they make. In essence he's attacking
their conclusions.
===
Subject: Question about numbers in arithmetic progression
Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1.  
What
is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1?
Rich
===
Subject: Re: Question about numbers in arithmetic progression
>Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1.  
What
>is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1?
Regardless of gcd, the trivial solution applies.
The minimum value is 1/max(|x_1|,|x_n|).
--Keith Lewis              klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: Question about numbers in arithmetic progression
>>Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1.
>What
>>is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1?
>Regardless of gcd, the trivial solution applies.
>The minimum value is 1/max(|x_1|,|x_n|).
I intended that the a_i be integers.  My apologies for not making this 
clear.
Rich
===
Subject: Re: Question about numbers in arithmetic progression
> Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1.  
What
> is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1?
Well, that would depend on x_1, n, and the common difference d, 
so I suppose you want some sort of bound (I don't think you'll get 
an exact expression) in terms of those parameters, yes? 
If x_1 & x_2 have any common factor then all the x_i have it, 
so it's enough to assume x_1 & x_2 are relatively prime. 
Let's further assume 0 < x_1 < x_2. 
Then certainly there's a solution with |a_1| < x_2 and |a_2| < x_1 
and a_i = 0 for i = 3, .... - that gives 2 x_1 + d - 2 as a bound, 
independent of n.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Question about numbers in arithmetic progression
>> Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1.
>What
>> is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1?
>Well, that would depend on x_1, n, and the common difference d, 
>so I suppose you want some sort of bound (I don't think you'll get 
>an exact expression) in terms of those parameters, yes? 
Add the condition 0<x_1<x_2<...<x_n and let m(x_1,d,n) be the minimum 
refered
to above.  Here is what I get for m:
m(2001,2,5)=501  [251*2001-250*2009=1]
m(20001,2,5)=5001  [2501*20001-2500*20009=1]
m(200001,2,5)=50001  [25001*200001-25000*200009=1]
m(1001,3,5)=169
m(10001,3,5)=1669
m(100001,3,5)=16669
m(1001,10,7)=35
m(10001,10,7)=335
m(100001,10,7)=3335
So it seems as though there *may* be an exact expression for these 
parameters. 
Others, like m(1001,31,7), ... are not at all obvious.  
I have a method that seems to compute m(x_1,...,x_n) and am looking for 
inputs
with known minimums to test against, hence the question.
Rich
===
Subject: Re: Question about numbers in arithmetic progression
>> Suppose x_1,..,x_n are in arithmetic progression and 
gcd(x_1,..,x_n)=1.
>  What
>> is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1?
>Well, that would depend on x_1, n, and the common difference d, 
>so I suppose you want some sort of bound (I don't think you'll get 
>an exact expression) in terms of those parameters, yes? 
> Add the condition 0<x_1<x_2<...<x_n and let m(x_1,d,n) be the minimum 
refered
> to above.  Here is what I get for m:
> m(2001,2,5)=501  [251*2001-250*2009=1]
> m(20001,2,5)=5001  [2501*20001-2500*20009=1]
> m(200001,2,5)=50001  [25001*200001-25000*200009=1]
> m(1001,3,5)=169
> m(10001,3,5)=1669
> m(100001,3,5)=16669
> m(1001,10,7)=35
> m(10001,10,7)=335
> m(100001,10,7)=3335
> So it seems as though there *may* be an exact expression for these 
parameters. 
> Others, like m(1001,31,7), ... are not at all obvious.  
> I have a method that seems to compute m(x_1,...,x_n) and am looking for 
inputs
> with known minimums to test against, hence the question.
If you apply the Jacobi method to
a_1*x_1+...+a_n*x_n=1
you get an equation of the form
x_1(a_1 + a_2 + .. a_n) + d(a_2 + 2a_3 +  + (n-1)a_m) =1
x_1 A +dB = 1
If P/Q is the penultimate convergent of
x_1/d  then x_1 A +dB = 1 has a solution
if you choose appropriate signs for P and Q.
a_2 + 2a_3 +  + (n-1)a_n = P =>
|a_n| = INT(p/(n-1))
and some |a_j| (2<= j <n) =1
a_1 + a_2 + .. a_n  = -Q =>
|a_1| = |Q| + |1| +|INT(P/(n-1))
 
This gives minima which agree with your results.
What values did you get for m(1001,31,7) etc. ?
What is the basis of your method ?
===
Subject: Re: Question about numbers in arithmetic progression
I. M. Davidson wote:
>This gives minima which agree with your results.
But doesn't prove these are the actual minima, right?
>What values did you get for m(1001,31,7) etc. ?
I get:
m(1001,31,7)=87
m(10001,31,7)=1417
m(100001,31,7)=6462
m(1000001,31,7)=107548
m(10000001,31,7)=322585
and, curiously,
m(10^12+1,31,7)=64516129040
m(10^22+1,31,7)=322580645161290322585.
>What is the basis of your method ?
Casting in a small pond full of big fish and hoping for the best?
Seriously, I don't have a answer for your question.  I will try to write a 
good
description and post it here, though.
Rich
===
Subject: A first countable separable space that is not second countable
I need to see an example of a topological space that is first countable and
separable but not second countable.
Are there any?
===
Subject: Re: A first countable separable space that is not second countable
> I need to see an example of a topological space that is first countable
> and separable but not second countable.
Doesn't R topologized with basis the sets [a,b) satsify this?
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: A first countable separable space that is not second countable
Never mind. I found one.
> I need to see an example of a topological space that is first countable
and
> separable but not second countable.
> Are there any?
===
Subject: spherical triangle
How is it possible to compute this:
a spherical triangle is given by its three sides (three angles between
the three vectors). Somewhere in the triangle lies the origin (North
(angle between vector to point and North pole) to the North pole. How
can one computes the distance (angle) to the third point?
The coordindates of the points are not given.
S. Nurbe
===
Subject: Re: spherical triangle
> a spherical triangle is given by its three sides (three angles between
> the three vectors). Somewhere in the triangle lies the origin (North
> (angle between vector to point and North pole) to the North pole. How
> can one computes the distance (angle) to the third point?
> The coordindates of the points are not given.
Hint: Stereographic conformal projection from South pole rays on to a
plane tangent to North pole.
===
Subject: JSH: Look at it backwards
I usually start with one polynomial and then talk about dividing 49
off from it, but here I'll start *after* 49 has been divided off:
S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
and consider the factorization
S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
and the assertion that the d's must have factors
w_1(x), w_2(x), and w_3(x)
that multiply to give w_1(x) w_2(x) w_3(x) = 7.
But there's no factors of 7 anywhere.  So why should the d's have
functions that are factors of 7?
Ok, maybe that seems unfair, as there could be LOTS of different
functions you can plug in for the c's and d's, but why should ANY of
them have functions of x that are factors of 7?
The equation has no memory.
You have a memory, so if I start with the polynomial multiplied by 49,
then you can say to yourself that there's some dependency on 7.  But
it's a mirage.
Mathematically a constant multiple is not a big deal.  It's just a
constant multiple that you can divide off, leaving a result that--you
guessed it--has no memory of the multiple!
One of the weirder things about the discussions here, which I assume
escapes most of you is some fascinating belief that the equation has a
memory.
You see something like 
P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
and your brain apparently HOLLERS at you that 49 is still there, but
no, it's gone.
When I show something like
P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
you brain insists that 7 is still there.
I mean just LOOK!  You can see 7 dividing two of the a's and for God's
sake!!!
There's a 7 in that last factor, you see it, don't you?
In 5a_3(x) + 7.  Come on, there's a 7 RIGHT THERE!  Of course 7 is
still there!!!
So some poster comes at you claiming that 49 divides off from 
P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
in a way that varies dependent on the value of x, and your brain tells
you, OK!!!
You have a memory.  To you 7 is still there, even though the factor is
divided off.
Follwing the sci.math'ers insistent raving you get
P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) +
7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x))
where the w's are factors of 7, so that you're left with functions of
7.
But they would STILL be there with the factorization of
S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
inexplicably still there despite the polynomial not having any factors
of 7.
Your eyes fool you.  Your memory betrays you.
The math doesn't bother with such nonsense.  A multiple of a
polynomial can be divided off and it's gone.
It doesn't leave a trace.
In a way what's happening now is an instructive lesson in the
limitations of the human brain.  Your brains SEE something, and
insistently tell you that 7 is STILL THERE, and so the arguments go on
for years.
After all, you can SEE the 7's.  Come on, who's fooling who, right? 
Dammit.
You can see the 7's in there, can't you?
James Harris
===
Subject: Re: JSH: Look at it backwards
> I usually start with one polynomial and then talk about dividing 49
> off from it, but here I'll start *after* 49 has been divided off:
> S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
> and consider the factorization
> S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
> and the assertion that the d's must have factors
> w_1(x), w_2(x), and w_3(x)
> that multiply to give w_1(x) w_2(x) w_3(x) = 7.
Let w_1(x)=w_2(x)=w_3(x) = 7^(1/3).  Then
the w_i are factors of the d_i in the ring of algebraic numbers.
(Are they factors in any other ring?  Who knows?  Who cares?
No one except you has made such a claim.)
> But there's no factors of 7 anywhere.  So why should the d's have
> functions that are factors of 7?
the d's have functions that are factors of 7 is gibberish.
                <snipped the rest of the post>
                                -William Hughes
===
Subject: Re: JSH: Look at it backwards
>I usually start with one polynomial and then talk about dividing 49
>off from it, but here I'll start *after* 49 has been divided off:
>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
>and consider the factorization
>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
>and the assertion that the d's must have factors
>w_1(x), w_2(x), and w_3(x)
>that multiply to give w_1(x) w_2(x) w_3(x) = 7.
>But there's no factors of 7 anywhere.  So why should the d's have
>functions that are factors of 7?
  You are getting maximally confused.  In your original 
approach, you are factoring P(x) as a polynomial in 5,
not as a polynomial in x.  The constant term with 
respect to 5 and the constant term with respect to
x are different.  It was a stupid mistake to think you
were simplifying by treating 5 as if it were the 
polynomial variable in the first place.  And that is what
is causing your confusion here.
>Ok, maybe that seems unfair, as there could be LOTS of different
>functions you can plug in for the c's and d's, but why should ANY of
>them have functions of x that are factors of 7?
>The equation has no memory.
  The memory problem here is, you have forgotten what you
started with.  To see this, you need to go back to the way you
were doing things originally: for example,
  P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f)
  To put this in terms of what you are doing now: replace x in this
expression by 5, and replace m by x, and replace f by 7.  Also 
replace u by 1.  This gives
  P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7),
and you can clearly see that, as a polynomial in 5, the constant
term is 49 * 7 = 7^3, not 49 * 22.
  Go back to the expression above for P(m).  The constant term
with respect to m is f^2 (3 xu^2 + u^3 f).  That happens to reduce to
49*22 when u = 1, f = 7, and x = 5.  But the constant term with
respect to x is u^3 f^3.  You have forgotten this and gotten mixed
up, that's all.
>You have a memory, so if I start with the polynomial multiplied by 49,
>then you can say to yourself that there's some dependency on 7.  But
>it's a mirage.
>Mathematically a constant multiple is not a big deal.  It's just a
>constant multiple that you can divide off, leaving a result that--you
>guessed it--has no memory of the multiple!
>One of the weirder things about the discussions here, which I assume
>escapes most of you is some fascinating belief that the equation has a
>memory.
>You see something like 
>P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
>and your brain apparently HOLLERS at you that 49 is still there, but
>no, it's gone.
  No - my brain does no such hollering.
>When I show something like
>P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
  Key thing to note here: as I said, the polynomial variable 
with respect to which you are factoring is 5, not x.  See?
>you brain insists that 7 is still there.
>I mean just LOOK!  You can see 7 dividing two of the a's and for God's
>sake!!!
>There's a 7 in that last factor, you see it, don't you?
>In 5a_3(x) + 7.  Come on, there's a 7 RIGHT THERE!  Of course 7 is
>still there!!!
>So some poster comes at you claiming that 49 divides off from 
>P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
>in a way that varies dependent on the value of x, and your brain tells
>you, OK!!!
>You have a memory.  To you 7 is still there, even though the factor is
>divided off.
  As a polynomial in 5, it's still there.  It is not a coincidence
that 22 = 3 * 5 + 7.
>Follwing the sci.math'ers insistent raving you get
>P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) +
>7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x))
>where the w's are factors of 7, so that you're left with functions of
>But they would STILL be there with the factorization of
>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
>inexplicably still there despite the polynomial not having any factors
>of 7.
>Your eyes fool you.  Your memory betrays you.
  No, it's YOUR memory that is the problem.  You have forgotten
how you were factoring: not with respect to x, but with respect
to 5.
>The math doesn't bother with such nonsense.  A multiple of a
>polynomial can be divided off and it's gone.
>It doesn't leave a trace.
>In a way what's happening now is an instructive lesson in the
>limitations of the human brain.  Your brains SEE something, and
>insistently tell you that 7 is STILL THERE, and so the arguments go on
>for years.
>After all, you can SEE the 7's.  Come on, who's fooling who, right? 
  You are hopelessly tangled in your own dimwitted oversimplification.  
You have forgotten what you were doing.  
>Dammit.
>You can see the 7's in there, can't you?
  Sure.  On *both* sides.  Just look at:
  P(x)/49 = (x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7,
and note that the constant term is ... 7, just like it should be,
if you regard 5 as the polynomial variable!  
  Nora B.
>James Harris
===
Subject: Re: JSH: Look at it backwards
>I usually start with one polynomial and then talk about dividing 49
>off from it, but here I'll start *after* 49 has been divided off:
>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
>and consider the factorization
>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
>and the assertion that the d's must have factors
>w_1(x), w_2(x), and w_3(x)
>that multiply to give w_1(x) w_2(x) w_3(x) = 7.
>But there's no factors of 7 anywhere.  So why should the d's have
>functions that are factors of 7?
>   You are getting maximally confused.  In your original 
> approach, you are factoring P(x) as a polynomial in 5,
> not as a polynomial in x.  The constant term with 
> respect to 5 and the constant term with respect to
> x are different.  It was a stupid mistake to think you
> were simplifying by treating 5 as if it were the 
> polynomial variable in the first place.  And that is what
> is causing your confusion here.
There is only one variable Nora Baron, so what other variable would
you have listed with
S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
considering that
S(x) = 300125x^3 - 18375 x^2 - 360 x + 22?
So who's confused?
>Ok, maybe that seems unfair, as there could be LOTS of different
>functions you can plug in for the c's and d's, but why should ANY of
>them have functions of x that are factors of 7?
>The equation has no memory.
>   The memory problem here is, you have forgotten what you
> started with.  To see this, you need to go back to the way you
> were doing things originally: for example,
Now note, the polynomial S(x) doesn't have 7 as a factor, as this time
I'm starting with the result of dividing off the multiple.
So the poster Nora Baron (actually a guy as revealed in another post
when he ended with a male name) is trying to show how 7 gets back into
the expression.
That is, he's trying to prove that the factorization DOES have a
memory of the multiple 49.
>   P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f)
>   To put this in terms of what you are doing now: replace x in this
> expression by 5, and replace m by x, and replace f by 7.  Also 
> replace u by 1.  This gives
>   P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7),
> and you can clearly see that, as a polynomial in 5, the constant
> term is 49 * 7 = 7^3, not 49 * 22.
No.  It can be factored with respect to 5 and 7, but 5 is not a
variable.
The polynomial variable is x.
Now, I've explained lots of times to the Nora Baron poster, and what
I want you all to consider is that the poster isn't really that dense,
but instead knows you better than you know yourselves.
Essentially the basic strategy is just to disagree.
Time after time, and even in answer to surveys that I've done, readers
who normally lurk will admit that they primarily rely on the fact that
people argue with me, assuming that if I were right, then others
wouldn't disagree!
So, for sci.math'ers like Nora Baron, the strategy is clear--just
disagree.
They often don't even TRY to actually make mathematical senze.
>   Go back to the expression above for P(m).  The constant term
> with respect to m is f^2 (3 xu^2 + u^3 f).  That happens to reduce to
> 49*22 when u = 1, f = 7, and x = 5.  But the constant term with
> respect to x is u^3 f^3.  You have forgotten this and gotten mixed
> up, that's all.
Now the expressions actually is
S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
where the factorization
S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
is being considered.
Now how do you get 7 into that?  Well, that's not the issue for the
poster!
All he has to do is disagree.
For most readers that's what works.
Well, that enough here.
James Harris
===
Subject: Re: JSH: Look at it backwards
>snip
>There is only one variable Nora Baron . . .
Here I agree with Mr. Harris.  I once posted that there are evidently
a number of real people living in the United States that really are
named Nora Baron, but those individuals are constant while our Nora
Baron may indeed be the only one that is a variable -- although that
property has not yet been established for sure.
===
Subject: Re: JSH: Look at it backwards
>>I usually start with one polynomial and then talk about dividing
49
>off from it, but here I'll start *after* 49 has been divided off:
>>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
>>and consider the factorization
>>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
>>and the assertion that the d's must have factors
>>w_1(x), w_2(x), and w_3(x)
>>that multiply to give w_1(x) w_2(x) w_3(x) = 7.
>>But there's no factors of 7 anywhere.  So why should the d's have
>>functions that are factors of 7?
>>   You are getting maximally confused.  In your original 
>> approach, you are factoring P(x) as a polynomial in 5,
>> not as a polynomial in x.  The constant term with 
>> respect to 5 and the constant term with respect to
>> x are different.  It was a stupid mistake to think you
>> were simplifying by treating 5 as if it were the 
>> polynomial variable in the first place.  And that is what
>> is causing your confusion here.
>There is only one variable Nora Baron,
  I didn't say there were other variables in your 
present polynomial.  I was describing its origin,
when there were 4 variables, which explains why
you are now confused about the whole thing.
> so what other variable would
>you have listed with
>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
>considering that
>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22?
>So who's confused?
  No question about that.  You are.
>>Ok, maybe that seems unfair, as there could be LOTS of different
>>functions you can plug in for the c's and d's, but why should ANY of
>>them have functions of x that are factors of 7?
>>The equation has no memory.
>>   The memory problem here is, you have forgotten what you
>> started with.  To see this, you need to go back to the way you
>> were doing things originally: for example,
>Now note, the polynomial S(x) doesn't have 7 as a factor,
   I DIDN'T SAY IT DID, ASSHOLE
> as this time
>I'm starting with the result of dividing off the multiple.
>So the poster Nora Baron (actually a guy as revealed in another post
>when he ended with a male name) is trying to show how 7 gets back into
>the expression.
  It gets back in in the constant term, because you
have been factoring this function as if it were a polynomial 
in 5.
>That is, he's trying to prove that the factorization DOES have a
>memory of the multiple 49.
  Not at all.  The 49 is obviously gone.  But the 
constant term at the end, when P(x)/49 is viewed as
a polynomial in 5 [your idea, not mine!], is still
there, and it is 7, not 22.
>> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f)
>>   To put this in terms of what you are doing now: replace x in this
>> expression by 5, and replace m by x, and replace f by 7.  Also 
>> replace u by 1.  This gives
>>   P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7),
>> and you can clearly see that, as a polynomial in 5, the constant
>> term is 49 * 7 = 7^3, not 49 * 22.
>No.  It can be factored with respect to 5 and 7, but 5 is not a
>variable.
  You have treated it exactly as such.  When you factor P(x) 
in the form
     P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7),
you are NOT factoring it as a polynomial in x.  If you
were, the terms a_1(x), etc., would themselves be
polynomials in x.  As you yourself have pointed out MANY
TIMES, this is not a polynomial factorization when 
considered as a function of x.  (In fact you have claimed
that such 'nonpolynomial factorization' was one of your
great conceptual breakthroughs.  Now you want to deny 
it ???).  However it CLEARLY has the FORM of a factorization 
in 5, where 5 is treated as a polynomial variable.
>The polynomial variable is x.
  It most certainly is not [yes, I know you are going
to howl about this, but I stand by what I say].  When you 
factor 
       a b x^2 + (a + b) x + 1 
as (a x + 1)(b x + 1),
THEN you are factoring in terms of the polynomial variable
x.  The coefficients of x in the factors are functions
of a and b, NOT functions of x, as in your present 
factorization of P(x). Have you really no recollection 
of how you got started on this track? Originally, x 
was m, and 5 was x.  At that time you were considering 
factors like
       a_1(m) x + u f,
etc.  This was a factorization in terms of the polynomial
variable x, with coefficients which were algebraic
integer functions of m.  Just plug in x = 5, u = 1, f = 7, 
and m = x, and you have your present factorization  - as 
a polynomial in 5, not in x  !!
  It's easy, actually, to see how you have gotten confused,
especially if (unlike real mathematicians) you have a
poor memory.
>Now, I've explained lots of times to the Nora Baron poster, and what
>I want you all to consider is that the poster isn't really that dense,
>but instead knows you better than you know yourselves.
>Essentially the basic strategy is just to disagree.
  I only disagree when you are wrong.
>Time after time, and even in answer to surveys that I've done, readers
>who normally lurk will admit that they primarily rely on the fact that
>people argue with me, assuming that if I were right, then others
>wouldn't disagree!
  I don't recall people saying that.  
  What lurkers do seem to think, when they speak up, is that 
YOUR math is vague, hand-waving and illogical.  That's
not my fault.  Plus they seem to think that your habit of
making nasty personal attacks damages your credibility.  
That's not my fault either.  
>So, for sci.math'ers like Nora Baron, the strategy is clear--just
>disagree.
  Again, I only disagree when you are wrong.  Of course,
that happens to be virtually all the time.
>They often don't even TRY to actually make mathematical senze.
  This is gratuitous and false in my case, as you well know.
Also in the cases of Dik Winter, Rick Decker, Arturo Magidin,
'Rupert', Will Twentyman, Dale Hall, and others.
>>   Go back to the expression above for P(m).  The constant term
>> with respect to m is f^2 (3 xu^2 + u^3 f).  That happens to reduce to
>> 49*22 when u = 1, f = 7, and x = 5.  But the constant term with
>> respect to x is u^3 f^3.  You have forgotten this and gotten mixed
>> up, that's all.
>Now the expressions actually is
>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
>where the factorization
>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
>is being considered.
>Now how do you get 7 into that?  Well, that's not the issue for the
>poster!
  It was explained in what I posted.  You need to consider
how you arrive at S(x), and for that you need to consider
the form of the original function P(x): as a polynomial 
in 5, it has constant term 7^3.
  I thought you were simply confused on this point, having
forgotten how you got here.  I am now inclined to think you
are trying to deny the obvious correct explanation, though
I am not sure why.  
  Plus throughout your entire reply you seem to be playing 
to the grandstand.  You whine, exaggerate, and attempt to 
discredit.  The actual math you have hardly addressed at all.
You seem to think that you can win mathematical arguments 
by an appeal to some great silent majority of people out 
there who you can cajole into thinking that we critics are
evil, cheating liars who are jealous of your discoveries.
(If that were true, one of us evil cheating liars would have
published it long ago, of course giving no credit to James
S.Harris.  Funny, that hasn't happened.  Wonder why?  Why
isn't anyone stealing your ideas?)
  You seem to have the impression that math is a popularity 
contest.  It's not.  The only way to win is through
rigorous proof.  You don't have one.  That's the main reason 
you are not winning.
>All he has to do is disagree.
>For most readers that's what works.
  Right.  Go ahead and ask your faithful admiring grandstand 
if that's what works.
>Well, that enough here.
  Yuh - Duh - that enough !
  Nora B.
>James Harris
===
Subject: Re: JSH: Look at it backwards
> So the poster Nora Baron (actually a guy as revealed in another post
> when he ended with a male name) is trying to show how 7 gets back into
> the expression.
What difference does it make what Nora's gender is? What does it have to do 

with math? Has anybody but you made note of the question of Nora's gender? 
No, because it is not relevant. Your obsession with her (or his) gender just 

screams mental illness. It is (probably) just a pseudonym, nothing more. So 

what? My pseudonym is o[CapitalYAcute]in, so do you claim that I am 
pretending to be a 
Norse god? Well, nobody cares!!! Idiot!!!
===
Subject: Re: JSH: gametes
So... James Harris got me wondering... which type of gametes do you actually 

produce, Nora Baron? 
===
Subject: Re: JSH: gametes
> So... James Harris got me wondering... which type of gametes do you 
actually 
> produce, Nora Baron? 
See how that works? You start by mocking JSH, and after a while you 
start to think like him. He's addictive. He's actually making some very 
good points about the social aspect of proof. Philosophers of math make 
the same arguments ... that a proof is whatever convinces a majority of 
working mathematicians; that what's considered a proof in one era is 
regarded as utterly lacking in rigor in another. 
And now he's got you caring about the gender of a palindromic poster.
===
Subject: Re: JSH: gametes
> See how that works? You start by mocking JSH, and after a while you
> start to think like him. He's addictive. He's actually making some very
> good points about the social aspect of proof. Philosophers of math make
> the same arguments ... that a proof is whatever convinces a majority of
> working mathematicians; that what's considered a proof in one era is
> regarded as utterly lacking in rigor in another.
> And now he's got you caring about the gender of a palindromic poster.
Yes. You are 100 correct in all that... but now I really do want to know... 

Damn it!!! 
===
Subject: Re: JSH: Look at it backwards
> I usually start with one polynomial and then talk about dividing 49
> off from it, but here I'll start *after* 49 has been divided off:
> S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
> and consider the factorization
> S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
Ok, at this point there is no indication as to what these c_i(x) and 
d_i(x) might be, aside from the obvious necessary properties to generate 
the coefficients of S(x) when the factors are multiplied together.  Are 
they meant to be continuous functions?  Continuous almost everywhere? 
What are their domain/range?
> and the assertion that the d's must have factors
> w_1(x), w_2(x), and w_3(x)
> that multiply to give w_1(x) w_2(x) w_3(x) = 7.
If they have such factors, then the range must not be in the algebraic 
integers.  Note: I mention that since you usually say something about 
everything is in the algebraic integers.
> But there's no factors of 7 anywhere.  So why should the d's have
> functions that are factors of 7?
They can quite easily in the appropriate ring.  For example, 22 has 7 as 
a factor in the rational numbers, algebraic numbers, reals, etc.  What 
makes you say there's no factors of 7 anywhere?
> Ok, maybe that seems unfair, as there could be LOTS of different
> functions you can plug in for the c's and d's, but why should ANY of
> them have functions of x that are factors of 7?
*IN WHAT RING?*  Do you not see that you are making some sort of 
assumption that you have not shared with us?
> The equation has no memory.
Huh?  Why would you even speak of such a thing?
> You have a memory, so if I start with the polynomial multiplied by 49,
> then you can say to yourself that there's some dependency on 7.  But
> it's a mirage.
Actually, you are the one who keeps talking about 7 with such a polynomial.
> Mathematically a constant multiple is not a big deal.  It's just a
> constant multiple that you can divide off, leaving a result that--you
> guessed it--has no memory of the multiple!
Clue: you are think of things as being processes.  Try thinking of them 
as being equivalent statements.  You can talk about two different 
equations being equivalent, but they are different equations.  The 
notion of a process, or memory of prior steps in the process, is an 
artifact of the work a person does to work towards a goal.  The result 
in something like the above is simply a set of equivalent statements and 
some justification for them being equivalent.
> One of the weirder things about the discussions here, which I assume
> escapes most of you is some fascinating belief that the equation has a
> memory.
Then why are you the only one who talks about it?
> You see something like 
> P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
> and your brain apparently HOLLERS at you that 49 is still there, but
> no, it's gone.
I see it quite clearly.  On the left side of the equation.  Where it was 
in the previous equation (if at all) is irrelevent.
> When I show something like
> P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
> you brain insists that 7 is still there.
> I mean just LOOK!  You can see 7 dividing two of the a's and for God's
> sake!!!
I see three 7s, actually.
> There's a 7 in that last factor, you see it, don't you?
> In 5a_3(x) + 7.  Come on, there's a 7 RIGHT THERE!  Of course 7 is
> still there!!!
Precisely.
> So some poster comes at you claiming that 49 divides off from 
> P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
> in a way that varies dependent on the value of x, and your brain tells
> you, OK!!!
Wrong.  Simple examples make it clear that, in general, this is how 
things work when viewed from the perspective of working on *factors* of 
a polynomial with a goal of keeping the terms of each factor in a 
particular ring.  If you focus on terms of the factors or don't care 
about staying within a ring at any particular level, then it doesn't 
matter how you write it.
> You have a memory.  To you 7 is still there, even though the factor is
> divided off.
I see only /7, /7, and +7.
> Follwing the sci.math'ers insistent raving you get
> P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) +
> 7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x))
> where the w's are factors of 7, so that you're left with functions of
> 7.
Functions of 7?  Huh?
> But they would STILL be there with the factorization of
> S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
> inexplicably still there despite the polynomial not having any factors
> of 7.
So, it would appear that an appropriate definition for your functions at 
the top of your post, to tie it in with what's down here would be: given 
the a_i(x)s and w_i(x)s, then:
c_i(x) = 5a_i(x)/w_i(x) and d_i(x)=7/w_i(x).
However, that would mean that d_i(x)*w_i(x)=7, which does not suggest 
anything about d_i(x) having a factor of 7 as a factor.  So, assuming 
that the d_i(x) was arrived at as above, and that the w_i(x) are the 
same in both cases, where is the notion of w_i(x) being a factor of 
d_i(x) coming from?
> Your eyes fool you.  Your memory betrays you.
> The math doesn't bother with such nonsense.  A multiple of a
> polynomial can be divided off and it's gone.
> It doesn't leave a trace.
> In a way what's happening now is an instructive lesson in the
> limitations of the human brain.  Your brains SEE something, and
> insistently tell you that 7 is STILL THERE, and so the arguments go on
> for years.
Apparently you did not understand any of the arguments, or you wouldn't 
be saying any of this.
> After all, you can SEE the 7's.  Come on, who's fooling who, right? 
> Dammit.
> You can see the 7's in there, can't you?
I see two division by 7's and a +7.  Now, which 7s are supposed to 
still be there?
-- 
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Look at it backwards
> The equation has no memory.
This is new! What the fuck are you on about now?
> You have a memory, so if I start with the polynomial multiplied by 49,
> then you can say to yourself that there's some dependency on 7.  But
> it's a mirage.
Mirage? Ummmmm..... Holy moly... I have no words for this... Ummmm.... 
OK, 
go on....
> Mathematically a constant multiple is not a big deal.
Can you prove that assertion, and can you first define mathematically what a 

big deal is. Just joking, I know you are talking figuratively... but 
really James, do you have to add in all this goofy non-math talk in every 
post?
> It's just a
> constant multiple that you can divide off, leaving a result that--you
> guessed it--has no memory of the multiple!
What does divide off mean? I think I know what you mean, but can you not 
try to use standard terminology?
> Your eyes fool you.  Your memory betrays you.
And Wiles was tricked in this way in his FLT proof, right?
===
Subject: Area between these curves..
A friend recently posed this question to me, and I'd like to be able to
explain the answer:
integrate[ (sec x)^2 - tan x ] from x=0 to x=1
Simple enough.
the antiderivative of (sec x)^2 = tan x
and
the antiderivative of tan x = -ln(cos x)
So, after calculating F(1) - F(0), the exact answer is:
tan (1) + ln [cos(1)]
Ok, but now say we were to integrate the same function, this time from x=0
to x=Pi.
There is an asymptote in the given domain shared by both functions, at Pi /
2.
2 to Pi, the area is also infinite.
Plugging x=Pi into -ln(cos x) results in an imaginary.
Would it be correct to say, that the area between the curves from 0 to Pi 
is
infinite? Or is it really undefined because of the imaginary and/or the
asymptote (non-continuous, therefore non-integrable)?
Or, should the answer be infinity, except at x = Pi / 2?
--Samantha
===
Subject: Re: Area between these curves..
I wasn't aware that an absolute value was taken in the -ln(|cos x|) part,
but this makes sense to me now upon looking closely at it. And I will be
looking into complex integration.
--Samantha
===
Subject: Re: Area between these curves..
>A friend recently posed this question to me, and I'd like to be able to
>explain the answer:
>integrate[ (sec x)^2 - tan x ] from x=0 to x=1
>Simple enough.
>the antiderivative of (sec x)^2 = tan x
>and
>the antiderivative of tan x = -ln(cos x)
If you wish to find the antiderivative of 1/u in domains where u may
be negative, you get ln(|u|) + c. So you would have -ln(|cos(x)|).
>So, after calculating F(1) - F(0), the exact answer is:
>tan (1) + ln [cos(1)]
>Ok, but now say we were to integrate the same function, this time from x=0
>to x=Pi.
>There is an asymptote in the given domain shared by both functions, at Pi 
/
>2 to Pi, the area is also infinite.
The integral is called an improper integral because of the
asymptote, and you are correct, it diverges to infinity.
>Plugging x=Pi into -ln(cos x) results in an imaginary.
No, it doesn't, see the note above.
>Would it be correct to say, that the area between the curves from 0 to Pi 
is
>infinite?
That is common terminology, or you could say the integral is
divergent.
> Or is it really undefined because of the imaginary
No.
> and/or the asymptote (non-continuous, therefore non-integrable)?
It is not Riemann integrable, nor does the improper Riemann integral
converge.
>Or, should the answer be infinity, except at x = Pi / 2?
No.
>--Samantha
--Lynn
===
Subject: Re: Area between these curves..
| Ok, but now say we were to integrate the same function, this time from 
x=0
| to x=Pi.
|
| There is an asymptote in the given domain shared by both functions, at Pi 

/
| 2.
|
| 2 to Pi, the area is also infinite.
Intepreting it physically (area) then it's infinite (or undefined!)
|
| Plugging x=Pi into -ln(cos x) results in an imaginary.
Not too sure about this one. Try read up something in the field of complex 
integration.
eg, treat x as z. This is university level maths ok!!  :)
===
Subject: JSH: Limitations in thinking
If you follow logic, then there's no room to argue with me.
But if you rely on people relying on their intuitions and what they
think you see, then you can argue with me for years, as some
sci.math'ers have done.
For me the realization of what was happening came just a while back
when I talked with Professor McKenzie at Vanderbilt University.
I traced out the argument explaining the paper Advanced Polynomial
Factorization.
his blackboard, and asked pointed questions at key areas.
When he was satisfied, we'd move on.  It took about 40 minutes to go
through it all.
I didn't hear *any* of the objections that many of you have relied on
for years from Professor McKenzie.
I don't get those objections anywhere but on Usenet.
Think about it.  I thought about it for a while, and I concluded that
certain posters were *deliberately* putting up objections that relied
on most of you readers not knowing any better.
They were and are deliberately fooling you, and must be doing it
consciously as some of them apparently are actually rather well-versed
in mathematics.
That makes it easier for them to fool you.
It was over a year ago when I realized that and it was extremely
depressing.
So these people understood what I was talking about, understood the
mathematics, understood why it was important, but for them, more
importantly, they understood that they could lie about it successfully
to people like you, who would either get confused or just believe
them.
I mean, look at how easily they do it?  Not just with my work on
algebraic number theory, but also with the prime distribution.
That's with freaking PRIME NUMBERS, and still they confuse many of
you, and convince many of you with statements that are just crazy.
Even there, I contacted leading mathematicians, and corresponded by
email with Lagarias and Odlyzko, mathematicians noted in the area, and
didn't get the objections that sci.math'ers give, until Odlyzko
replied to me finally, claiming that there were many results that
while true were not of mathematical interest.
Lagarias never said that or anything like it.  He'd just suggest
things like sending it to certain areas, like posting my prime
counting on arxiv.
Well, I told him I couldn't get on arxiv.  He didn't reply.
The way this works is that mathematicians I contact by email or in
person basically just walk away.  I explain things to them, they may
offer a question or two, but that's it.  Actually getting to Professor
McKenzie for an in-depth discussion was a major event for me, as
mathematicians usually just walk away, which probably happened because
I'm an alumnus of Vanderbilt and professors are told to be nice to us!
So, the difference between Usenet and the rest of the math world is
that you talk to me, but you make up bizarre objections, which I know
are really bogus and weird.
Journals represent the area where mathematicians are supposed to
listen, so I can send a paper to a journal, and if the editors play by
the rules, it should get a fair hearing.
The Annals at Princeton, will most likely I believe, follow the rules.
By the rules that math journals follow they are to properly evaluate
my paper for correctness and relevance.
That's all I ask.
So why keep posting on Usenet?  I think it really is about boredom and
also about a need just to talk about my research to someone, even
hostile weird people who make up voodoo mathematics and lie a lot.
I'm just short of people to talk to about my research.
On Usenet, I can go on and on and on about it.
James Harris
===
Subject: Re: JSH: Limitations in thinking
> I'm just short of people to talk to about my research.
> On Usenet, I can go on and on and on about it.
If you really want to communicate, try the following: Write down
clear, precise definitions for all the terms you use. Put it on the
web somewhere, and reference it when you post.
For example, you could write 
if a is an element of the (unitary) ring R, then we say 
a is a unit of (or in) R
iff there exists some c in R so that ca=1.
Then, when you make a post, be careful to always use the definitions
you have set up and be precise. For example, never write is a unit,
but is a unit of R.
It really won't take up much time.
> James Harris
Tim Mellor
===
Subject: Re: JSH: Limitations in thinking
> For me the realization of what was happening came just a while back
> when I talked with Professor McKenzie at Vanderbilt University.
> I traced out the argument explaining the paper Advanced Polynomial
> Factorization.
> his blackboard, and asked pointed questions at key areas.
> When he was satisfied, we'd move on.  It took about 40 minutes to go
> through it all.
Yes, yes... And what did Professor McKenzie say _at the end_?
If I may be permitted a few words from the peanut gallery, Harris, old
chap, I'll try to help you understand why you're not actually very
convincing up here. Of course we love to hear you talk the way you do
- past all the people who post objections, at some imaginary attentive
audience hanging on your every word, yet dangerously close to the
slippery slope of being persuaded by the arguments of your tormentors.
Trouble is, that when one of your detractors says something that one
of us (if I may presume to be one of 'us') doesn't understand, then we
ask, and typically get a reply we can understand. What is a unit?
Well, comes the answer, in some ring, dot dot dot.
You, on the other hand, use inscrutable terminology we cannot find in
books, and no-one else seems to be able to understand either. When you
say something or other is not a unit in the ring of algebraic
integers, but is properly a unit, what does this mean? May seem
tough, since others can delegate the responsibility for answering
elementary questions, but it looks like you are the only one who can
answer this. So unless you do, we ain't believing anything you say.
Brian Chandler
http://imaginatorium.org
> On Usenet, I can go on and on and on about it.
> James Harris
===
Subject: Re: JSH: Limitations in thinking
>[...]
>Journals represent the area where mathematicians are supposed to
>listen, so I can send a paper to a journal, and if the editors play by
>the rules, it should get a fair hearing.
>The Annals at Princeton, will most likely I believe, follow the rules.
>By the rules that math journals follow they are to properly evaluate
>my paper for correctness and relevance.
No doubt. And they will _not_ be publishing the paper. Because they're
going to follow the rules, one of which is that they're not supposed
to publish obvious nonsense.
What's not entirely clear is whether you'll get a letter explaining
that it's obvious nonsense or a polite letter saying that it's
not suitable, maybe some other journal. A letter saying it's obvious
nonsense would really be the best thing for you, in terms of retaining
some sort of contact with reality, but I doubt they'll send such
a letter, because they won't think it would be polite.
>That's all I ask.
>So why keep posting on Usenet?  I think it really is about boredom and
>also about a need just to talk about my research to someone, even
>hostile weird people who make up voodoo mathematics and lie a lot.
>I'm just short of people to talk to about my research.
>On Usenet, I can go on and on and on about it.
Uh, we're aware that you can go on and on on usenet.
If you'd actually _learn_ some math then you could make
posts here that people would be interested in.
>James Harris
************************
David C. Ullrich
===
Subject: Re: Limitations in thinking
> If you follow logic, then there's no room to argue with me.
Has anybody ever followed your logic? If not, do you think the problem just 

might be in your logic?
> I didn't hear *any* of the objections that many of you have relied on
> for years from Professor McKenzie.
Perhaps he was being polite? Perhaps he was scared of you? Think back to 
that day... do you recall him telling you tat you were correct? If he 
thought you were correct, and he understood you logic, why did he not 
recognize the big consequences that you claim about destroying Galois and 
Wiles? Why has he not spread the gospel according to James?
> didn't get the objections that sci.math'ers give, until Odlyzko
> replied to me finally, claiming that there were many results that
> while true were not of mathematical interest.
Right... if you had what you claim you have, he would have seen lots of 
mathematical interest in your work. What can you infer from that James?
> Well, I told him I couldn't get on arxiv.  He didn't reply.
Oh? He did not reply? Keeping his distance from you? What can you infer from 

that James?
> The way this works is that mathematicians I contact by email or in
> person basically just walk away.
What can you infer from that James?
> mathematicians usually just walk away, ...
What can you infer from that James?
> The Annals at Princeton, will most likely I believe, follow the rules.
And they will reject your paper!
> I'm just short of people to talk to about my research.
What can you infer from that James?
> On Usenet, I can go on and on and on about it.
Yes, for years.... we know James...
===
Subject: JSH: Two sides
I've had months to wonder why mathematicians would work to ignore or
hide my results, and I think there are two sides to it.
For one thing, my discoveries betray much of what they've been taught
about mathematics and it's not just that I was able to find a problem
in core, but also that there were results to find, using rather
basic methods.
Somehow, despite what has been often said, all those other
mathematicians working over hundred of years didn't cover all the
ground for use of simple arguments and simple ideas.
Like my work with non-polynomial factors relies on something as simple
as factoring a polynomial with respect to something other than the
polynomial variable.
Do that and an entire world opens up.  Possibilities that didn't exist
with other methods are suddenly easy.  Factors of irrational numbers
can be discerned, even if you can't tell *which* root might have a
number like 7 as a factor, you can know that at least one of them
must.
And to add insult to injury (from a particular point of view) I went
on to find a prime counting function that relies on a partial
difference equation.
Hints of it can be seen all over the area, and Meissel's Formula
actually uses a variant of it, while part of it looks (as some
sci.math'ers have noted) like the recurrence relation that follows
from Legendre's Method, as in fact a relationship can easily be
proven.
So what was missed?
It looks like no one figured on a multi-variable function, but instead
kept focusing on the single variable function.
After all IT MAKES SENSE that it be a single variable function, as you
have one cout of prime numbers for a given value.  The count of primes
up to 10 is 4.  The count of primes up to 100 is 25.  The count of
primes up to 1000 is 168.
One variable seems to be all you need.
Yet, here is a simple, compact prime counting function that captures
the entire prime distribution in a small space, with, a
three-dimensional p(x,y) function.
So yet another simple idea, not far away, but so close.
Yet losing what you had, your beliefs, your community's shared vision,
is one thing, looking at a world that's not cognizant of it is
another.
The second even more brutal side of it for mathematicians may be in
realizing that the rest of the world doesn't have a clue.  After all,
the discoverer was labeled a crank by sci.math'ers and the label seems
to have stuck.
Years have gone by, with so much overturned, so many beliefs
shattered, and so much research shown to be invalid, but the public
doesn't notice, or care.
The classes keep coming.  The papers are still being published. 
Awards and prizes still given out.  Day follows day.  Month follows
month, and the world continues on.
That may be the greatest insult of all that maybe none of it ever
really mattered.  That maybe the world never really cared.  That
mathematicians were just part of some play, some game, some empty soap
opera without any real meaning.
After all, some of the most dramatic events in world history are
passing without a notice.  The much vaunted world press hasn't a clue.
 The supposed intellectuals in all their corners prance, and preen,
talk as much as ever, and don't realize.
The world knows nothing, and that maybe the harshest lesson of all.
Maybe then, after all, it never really mattered, and what matters now
is to hold on to what you have.
It's not really possible to completely understand how they feel, what
they feel, unless you're in their shoes.  To have so many beliefs
shattered.  To have so much that you thought true, shown to be false. 
To feel betrayed down to the depths of your core...that's something
you have to live to understand.
James Harris
===
Subject: Re: JSH: Two sides
> I've had months to wonder why mathematicians would work to ignore or
> hide my results,
Ignoring your results takes no effort, hiding them does not occur.
--
There are two things you must never attempt to prove: the unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: Two sides
[snip delusion]
What does properly a unit mean?
Can you answer that simple question James Harris?
===
Subject: Re: JSH: Two sides
> [snip delusion]
> What does properly a unit mean?
> Can you answer that simple question James Harris?
JSH does not answer questions.
===
Subject: Re: JSH: Two sides
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
 VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
> I've had months to wonder why mathematicians would work to ignore or
> hide my results, and I think there are two sides to it.
Ignoring takes no work.
> Like my work with non-polynomial factors relies on something as
> simple as factoring a polynomial with respect to something other
> than the polynomial variable.
> Do that and an entire world opens up.
Sure, if you have something more than polynomials in one variable, you
have lots of opportunities for confusing constants with regard to one
variable to constants with regard to another.
It is something you have to be aware of as soon as you work with more
than one variable.  Very basic stuff.  And where variables are not
necessarily independent, things get more complicated (partial
differential equations ooze all over with the intricacies).  You don't
get it.  Not even the basics.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: Two sides
> I've had months to wonder why mathematicians would work to ignore or
> hide my results, and I think there are two sides to it.
Just one side to it.... you are wrong. 
===
Subject: Re: Two sides
says...
> I've had months to wonder why mathematicians would work to ignore or
> hide my results, and I think there are two sides to it.
> Just one side to it.... you are wrong. 
and INCREDIBLY lonely.
===
Subject: Example of a formulation in production scheduling
Hello everybody,
I am looking for an example formulation of production scheduling
including storages with the objective of minimising the makespan.
I want the MILP formulations of this example to be solved using
Lp_solve if possible.
===
Subject: FLTMA: S.1 = c
For every triple a,b,c of postive integers there is an associated series 
{S.n)
= 
(a^n+b^n) mod c + 
(c^n - a^n) mod b + 
(c^n - b^n) mod a
and I find that while S.0 is of course 2, that S.1 seems to be equal to c.
At least in my tests, where I am applying the additional conditions
a<b<c<a+b and gcd(a,b)=gcd(b,c)=gcd(a,c)=1
S.1 = c to limit 25 in those cases where S.n = 0 for some n.
S.1 = a mod c + b mod c + c mod b - a mod b + c mod a - b mod a 
= a + b + c mod b - a + c mod a - b mod a
= b - b mod a + c mod b + c mod a
But why would *that* equal c?
I tolerance everything and tolerate everyone.
I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri, 
alphabetically.
I drive: A double-step Thunderbolt with 657% range.
I fight terrorism by: Using less gasoline.
===
Subject: Re: FLTMA: S.1 = c
>S.1 = a mod c + b mod c + c mod b - a mod b + c mod a - b mod a 
>= a + b + c mod b - a + c mod a - b mod a
>= b - b mod a + c mod b + c mod a
= b - (b-a) + (c-b) + (c-a)
=  a + 2c - b - a
= 2c - b
I must have done that wrong.
writing m for mod
S.1=
amc + bmc + cmb - amb + cma - bma=
a + b + (c-b) - a + (c-a) - (b-a)=
a + b + c - b - a + c - a - b + a=
a - a + a - a + b - b - b + c + c=
2c - b.
I will have to show the intermediate sums in Mathcad.
I tolerance everything and tolerate everyone.
I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri, 
alphabetically.
I drive: A double-step Thunderbolt with 657% range.
I fight terrorism by: Using less gasoline.
===
Subject: Novice: Pi in other base systems
Hello...
My apologies if this is a silly question. How does pi behave in 
different base systems, for instance binary,  base 8 or base 16? 
Is it still a mysterious number with endless non-repeating 
numbers? Is there any base system where there's some especially 
interesting results?
(I guess the correct way to ask this is how the ration of 
circumference to diameter behaves in different number systems, 
but still you get the point.)
Just curious.
Thx,
D.
===
Subject: Re: Novice: Pi in other base systems
> Hello...
> My apologies if this is a silly question. How does pi behave in 
> different base systems, for instance binary,  base 8 or base 16? 
> Is it still a mysterious number with endless non-repeating 
> numbers? Is there any base system where there's some especially 
> interesting results?
> (I guess the correct way to ask this is how the ration of 
> circumference to diameter behaves in different number systems, 
> but still you get the point.)
> Just curious.
> Thx,
> D.
It is possible to compute the n-th digit of pi in hexadecimal --- and
thus also in binary --- notation without computing all the previous
digits. (I don't have the reference here right now, but I'm sure you
will not be eventually repeating since pi is not rational.
Hope this helps,
Lasse
---
===
Subject: Re: Novice: Pi in other base systems
In sci.math, piter
<1jRrd.45$wy5.6266@news.uswest.net>:
> Hello...
> My apologies if this is a silly question. How does pi behave in 
> different base systems, for instance binary,  base 8 or base 16? 
> Is it still a mysterious number with endless non-repeating 
> numbers? Is there any base system where there's some especially 
> interesting results?
> (I guess the correct way to ask this is how the ration of 
> circumference to diameter behaves in different number systems, 
> but still you get the point.)
> Just curious.
> Thx,
> D.
The digits differ, but the nature of pi never changes. :-)
However, this may interest you; it can be used to implement
progressive digits in base 16.  Look at equations (29), (31),
and (32) in
http://mathworld.wolfram.com/PiFormulas.html
-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Novice: Pi in other base systems
> Hello...
> My apologies if this is a silly question. How does pi behave in 
> different base systems, for instance binary,  base 8 or base 16? 
> Is it still a mysterious number with endless non-repeating 
> numbers? 
Consider: suppose there were a base b in which the expression of pi
became repeating. That is, we could write p = pi-3 as:
(0.a_1 a_2 a_3 ... a_n-1 a_n a_n+1 ... a_n+k a_n+1 a_n+k ...)b
where the a_'s are valid digits in base b, and (x y z)b means
'representation in base b'. Here we have n digits then a repeating
group of k digits. Then we have:
p b^n = (a_1 a_2 ... a_n . a_n+1 ... a_n+k a_n+1 ... a_n+k ...)b
p b^n - (a_1 a_2 ... a_n)b = (0. a_n+1 ... a_n+k a_n+1 ... a_n+k ...)b
Call this quantity q. Now we have
q b^k = (a_n+1 ... a_n+k . a_n+1 ... a_n+k a_n+1 ... a_n+k ...)b
q b^k - q = (a_n+1 ... a_n+k)b
q = (a_n+1 ... a_n+k)b / (b^k - 1)
Both numerator and denominator here are integers, so this means q is
rational. We defined q by
q = p b^n - (a_1 a_2 ... a_n)b
so
p = (q + (a_1 a_2 ... a_n)b) / b^n
which will still be rational; but p = pi - 3, and we know pi to be
irrational. We have a contradiction - so our assumption that there is
a base b in which the representation of pi becomes repeating must be
false.
The key point here which makes this not so much a 'silly question' as
a 'question which shows I don't know this key point' :) is that
rationality <=> a repeating representation in a rational base
You can prove this yourself (both ways) using the same technique I
just demonstrated. Note that you can regard 'terminating' expansions
as repeating with a repesting group consisting of '0' - it makes
proofs shorter. In simple arithmetic terms, it comes down to noticing
/ knowing / establishing that eg
0.738475 738475 ... (arbitrary digits)
= 738475 / 999999
which is obvious once you know it, or 'trivial' as mathematicians say
:)
> Is there any base system where there's some especially 
> interesting results?
One interesting result is the BBP formula
<http://mathworld.wolfram.com/BBPFormula.html> which can be used to
obtain the digits of the hexadecimal expansion with relative ease.
> (I guess the correct way to ask this is how the ration of 
> circumference to diameter behaves in different number systems, 
> but still you get the point.)
Remember that bases are not different number systems, they are just
different *representations* of the same numbers!
-- 
Larry Lard
Replies to group please
===
Subject: Re: Novice: Pi in other base systems
> Hello...
> My apologies if this is a silly question. How does pi behave in
> different base systems, for instance binary,  base 8 or base 16?
> Is it still a mysterious number with endless non-repeating
> numbers? Is there any base system where there's some especially
> interesting results?
> (I guess the correct way to ask this is how the ration of
> circumference to diameter behaves in different number systems,
> but still you get the point.)
Since pi is not a rational number, it is mysterious when expressed in 
any
integer base.  There may be some interesting results using a transcendental
base, such as pi, or e.  But then the representations of all the integers
would be mysterious.
===
Subject: Re: Novice: Pi in other base systems
| Hello...
|
| My apologies if this is a silly question. How does pi behave in
| different base systems, for instance binary,  base 8 or base 16?
| Is it still a mysterious number with endless non-repeating
| numbers? Is there any base system where there's some especially
| interesting results?
|
| (I guess the correct way to ask this is how the ration of
| circumference to diameter behaves in different number systems,
| but still you get the point.)
|
| Just curious.
Well, base  pi  makes it a very succinct number.  ______Gerard S.
 
===
Subject: Re: induction vs Cantor
>> <snipped> What I am saying is that if one can consider the totality of all
>> functions from N to R, without worrying about constructability, then
>> there is no problem with the diagonal proof of Cantor's theorem, 
>> since
>> the Cantor anti-diagonal algorithm applied to any member of that set 
of
>> functions is guaranteed to produce a number not in the image of the
>> function, and that production is what I have called the Cantor 
function
>> from R^N to R.
>> Uses Cantor's conclusions.
> No. Only his assumptions.
No. His conclusions. 
===
Subject: Re: induction vs Cantor
>> <snipped> What I am saying is that if one can consider the totality of all
>> functions from N to R, without worrying about constructability, then
>> there is no problem with the diagonal proof of Cantor's theorem, 
>> since
>> the Cantor anti-diagonal algorithm applied to any member of that set 
of
>> functions is guaranteed to produce a number not in the image of the
>> function, and that production is what I have called the Cantor 
function
>> from R^N to R.
>> Uses Cantor's conclusions.
> No. Only his assumptions.
> No. His conclusions. 
Which ones? Are Cantor's conclusions about the value of 1+1 to be 
excluded as well as his conclusions about the uncountablity of reals?
Is anything that tends to lead ultimately to the uncountablility of the 
reals to be excluded? 
Uncountability is nowhere a rerequisite of the construction of what I 
have called the Cantor function, it is only a consequence of it, so it 
seems to me that you are outtlawing anything leading to Cantor's 
conclusion, not merely the conclusion itself.
It is not clear to me that your blanket accusation does not rule out 
even 1+1 = 2, unless you are a bit clearer about just which 
conclusions are to be outlawed.
===
Subject: Re: induction vs Cantor
<snipped what I have called the Cantor function from R^N to R.
Describe this class of functions and show how Cantor used them in his
proof as you say.  I'm looking for where he classified them as R^N to R
and that they were uncountable. 
===
Subject: Re: induction vs Cantor
Poker Joker says...
>> what I have called the Cantor function from R^N to R.
>Describe this class of functions and show how Cantor used them in his
>proof as you say.  I'm looking for where he classified them as R^N to R
>and that they were uncountable.
I think Virgil means the class of functions f which given
an infinite list of reals returns a new real that is not
on the list.
Cantor came up with only one such function: his diagonalization
function. R^N just means the set of functions from N to R, or
equivalently, the set of infinite lists of reals, or a set of
reals indexed by the naturals. Cantor's diagonalization procedure
defines a function which given an infinite list of reals (that is,
an element of R^N) returns an element of R that is not on the list.
Cantor didn't prove that there were uncountably many such functions,
he only proved that there was one. That's all he needs to be able
to show that no list of reals contains every real.
It is easy enough to show that the cardinality of the set of such
functions is uncountable, although this fact is not used in Cantor's
proof.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: induction vs Cantor
<snippedCantor's diagonalization procedure
> defines a function which given an infinite list of reals (that is,
> an element of R^N) returns an element of R that is not on the list.
> Cantor didn't prove that there were uncountably many such functions,
> he only proved that there was one. That's all he needs to be able
> to show that no list of reals contains every real.
Do you think there are any counter-examples to Cantor's diagonal
proof?
If so, can you provide a valid modification that keeps the concept
of providing a real that is proven not in the list and has no
counter-example?
===
Subject: Re: induction vs Cantor
Poker Joker says...
><snipped>Cantor's diagonalization procedure
>> defines a function which given an infinite list of reals (that is,
>> an element of R^N) returns an element of R that is not on the list.
>> Cantor didn't prove that there were uncountably many such functions,
>> he only proved that there was one. That's all he needs to be able
>> to show that no list of reals contains every real.
>Do you think there are any counter-examples to Cantor's diagonal
>proof?
What do you mean by a counter-example?
>If so, can you provide a valid modification that keeps the concept
>of providing a real that is proven not in the list and has no
>counter-example?
Sorry, I don't know what you mean by counterexample.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: induction vs Cantor
> Poker Joker says...
>> what I have called the Cantor function from R^N to R.
>Describe this class of functions and show how Cantor used them in his
>proof as you say.  I'm looking for where he classified them as R^N to R
>and that they were uncountable.
> I think Virgil means the class of functions f which given
> an infinite list of reals returns a new real that is not
> on the list.
Precisely!
> Cantor came up with only one such function: his diagonalization
> function. R^N just means the set of functions from N to R, or
> equivalently, the set of infinite lists of reals, or a set of
> reals indexed by the naturals. Cantor's diagonalization procedure
> defines a function which given an infinite list of reals (that is,
> an element of R^N) returns an element of R that is not on the list.
> Cantor didn't prove that there were uncountably many such functions,
> he only proved that there was one. That's all he needs to be able
> to show that no list of reals contains every real.
> It is easy enough to show that the cardinality of the set of such
> functions is uncountable, although this fact is not used in Cantor's
> proof.
> --
> Daryl McCullough
> Ithaca, NY
===
Subject: Re: induction vs Cantor
> <snipped what I have called the Cantor function from R^N to R.
> Describe this class of functions and show how Cantor used them in his
> proof as you say.  I'm looking for where he classified them as R^N to R
> and that they were uncountable. 
It is I who call them Cantor functions, Cantor himself never did, nor, 
as far as I am aware did he ever use the notation R^N to designate the 
set of all listings of reals (functions from N to R). 
Cantor only needed one such function, which I will denote by 
C: R^N -> R: l-> C(l). As described by Cantor, it takes a list of reals 
as argument and produces a real not in that list as its value. 
Cantor never classified the cardinality of the set of all possible such 
Cantor functions as uncountable, but it is not difficult to do so.
All Cantor was interested in was producing one such Cantor function, 
at which he succeeded.
===
Subject: Re: induction vs Cantor
> Poker Joker says...
>>Yes, its like the primes. There are only countably many primes.
> Cantor's proof is similar to Euclid's proof of the existence
> of infinitely many primes, but is different in one crucial
> step:
> In Euclid's case, you assume that you have a FINITE set of primes,
> and then show that there is a prime that is not in that set. This
> shows that the set of all primes is not FINITE.
> In Cantor's case, you assume that you have a COUNTABLY INFINITE set
> of reals, and then show that there is a real that is not in that set.
> This shows that the set of all reals is not COUNTABLY INFINITE.
> --
> Daryl McCullough
> Ithaca, NY
> Okay.  I take it back. It's not like the primes at all.  I hope that
> means you are also taking back that it is like the primes. ;-)
It IS like the primes.  Here's how.  If someone said the primes are
finite then you could ask then for which n are there n distinct
primes that includes all primes, and Euclid's case he then shows that
whatever finite n you picked.
Let's say you said the reals are countable, then we could ask you
WHICH ordering on the reals lines them up to to naturals.  The point
is that whichever one you pick is bad.  You can add MORE cantor
numbers all you want, but AS LONG AS you add ONLY a countable number,
then there are STILL missing reals, so it isnt until you add MORE than
a countable number of ADDITIONAL reals that you FINALLY get them ALL.
===
Subject: Re: induction vs Cantor
> Chairman of the David Hilbert Appreciation Society 
>>  Let L_1 be a list of reals that implies a mapping F_1
>>  between the naturals and reals.
>>  Let D_n be a Cantor anti-diagonal number that can be
>>  formed using the mapping F_n
>>  Let L_n+1 be a list of reals by inserting D_n into L_n
>>  at row 2n [*] and shifting down all the previous rows at 2n
>>  and above.  This process is clearly an inductive process
>>  that creates a new mapping for each natural number.
>>  (L_n+1 could also be formed by prepending D_n to
>>  L_n.)
>>I'm assuming that the rows of each L_n will start from 1.
>>It makes no difference to the result if we start from 0,
>>but is more convenient with 1 IMO.
>>First we agree that:
>>(1)  L_1 is a list, an injection.
>>(2)  Any real number in row 2n-1 of L_n is in the
>>     same row (2n-1) on any L_m with m >= n.  This follows
>>     from [*] in his description.
>>(3)  It follows from (2) that if x is a real number
>>     in row 2n-1 of L_m with m >= n that the (2n-1)th
>>     digit of any D_m will not be the same as the
>>     (2n-1)th digit of x.  Otherwise D_n would not
>>     be an anti-diagonal number.
>>Once we agree that items 1-3 characterize the method
>>we do as follows:
>>(4) Denote the real number in row 1 of L_1 as x.
>>Since L_1 is a list there must be some n for which
>>the first 2n digits of x aren't all the same as the
>>first 2n digits of any other y in L_1.  Otherwise we
>>have x = y, contradicting (1).
>>note that x has a unique initial segment of at least
>>2n digits.
>>Denote the first 2n digits of x, by S and add to S a
>>(2n+1)th digit that is not the same as x's (2n+1)th
>>digit. This guarantees that the inital segment of S
>>doesn't appear on L_1.
>>Construct a number T in the following way:
>>Take D_n, the anti-diagonal number of L_n, and  throw away
>>its first 2n+1 digits and replace them with S.  Call the
>>resulting number T.
>>T can't be on L_1, since no number on L_1 has the same
>>inital segment.
>>By (4), x is in row 1 of L_1.  And by (3) with n = 1,
>>we have
>>x is a real number in row 1 of L_1 and m >= 1
>>implies that the 1st digit of any D_m will not
>>be the same as the 1st digit of x.
>>Since the first digit of x is the same as the
>>first digit of T we conclude that no D_m with
>>m >= 1 shares the same 1st digit as T.
>>Thus T is not any D_m, nor is it on L_1, thus
>>it isn't on any L_n.
>>I may have made minor mistakes, but I'm pretty sure
>>the basic argument works.
> Note that you've described one of the countably many ways
> to cunstruct a Cantor number.
I didn't create a Cantor number.  The number I show, that isn't
on any L_n has the exact same first 2n digits as those of
the number on the first row of L_1 -- clearly this isn't
constructed by the same method that Cantor uses otherwise
its first digit would be different from the first digit
of the number on the first row of L_1.
>  So your T is a D_n becaues
T can't be any D_n.  The first digit of T is the same first
digit of the number in the first row of L_1. Since that number
is in the first row on L_1 and every subsequent list L_2, L_3, ...
none of the D_n share the same first digit with it -- otherwise
D_n wouldn't be an anti-diagonal number, as you specified.
> we assume that at any stage, the Cantor method can be any
> one of a countably infinite set of methods for generating the
> D_n.  You've described one.
I used the most reasonable interpretation of the phrase:
Cantor['s] anti-diagonal number, which in sci.math is
used to describe a number constructed based on the diagonal
string of digits of a list.
 > So your T has been taken into
> account.
Then which L_n does T occur on?
> Cantor methods aren't required to create a new
> real in which the first digit is different from the first digit of
> the first real.
You were using Cantor's method, you said:
Let D_n be a Cantor anti-diagonal number that... -PJ
I suspect this is why you're no longer (as of this post)
calling it an anti-DIAGONAL.
Even if you use a method that is different yet produces
a similar result to Cantor's method it makes no difference.
Once you give enough details of how D_n is constructed
the same thing can be done.
OTOH if you completely obscure how D_n is created then
of course you will experience the satisfaction of never
seeing a counter example created in the same way (based
on knowledge of how L_n is defined), but there are still
other ways...
Aside from the matter of trying to convincing you that
a number exists that isn't on any L_n, you seem to want
to convince me that there is something special about the
collection of all L_n, which so far I haven't seen.
Perhaps this example will help you see why I regard
the collection of all L_n as uninteresting:
(1) Let A and B be countably infinite ordered sets of real
     numbers such that A cap B = emptyset.  Let C = A cup B.
It seems to me that C has the same distinguishing
features as the union of all L_n.
Each x in B is the Cantor number (*) of some list of reals,
each x in B is unique, and there are countably many x in B.
C is the collection of an initial list of reals A (like L_0)
with all of the diagonal numbers in B added.
(*) Cantor numbers just have to be created so that we can be
assured that they are unique with respect to the list. -PJ
In other words, C is no different from the collection of all L_n.
By assuming that the collection of all L_n is complete, you may
as well assume that C is complete also, which is much simpler.
It's easy to show examples of A B and C which satisfy (1)
and which also fail to contain every real.
So what is so special about the collection of all L_n?
[...]
-- 
Replace Roman numerals with digits to reply by email
===
Subject: Re: induction vs Cantor
Chairman of the David Hilbert Appreciation Society says...
>I'm assuming that the rows of each L_n will start from 1.
>It makes no difference to the result if we start from 0,
>but is more convenient with 1 IMO.
>First we agree that:
>(1)  L_1 is a list, an injection.
I thought that a list of reals is a function from N into R, not
necessarily an injection. That is, there can be repetitions in
L_1, and it doesn't hurt anything.
>(2)  Any real number in row 2n-1 of L_n is in the
>     same row (2n-1) on any L_m with m >= n.  This follows
>     from [*] in his description.
>(3)  It follows from (2) that if x is a real number
>     in row 2n-1 of L_m with m >= n that the (2n-1)th
>     digit of any D_m will not be the same as the
>     (2n-1)th digit of x.  Otherwise D_n would not
>     be an anti-diagonal number.
>Once we agree that items 1-3 characterize the method
>we do as follows:
>(4) Denote the real number in row 1 of L_1 as x.
>Since L_1 is a list there must be some n for which
>the first 2n digits of x aren't all the same as the
>first 2n digits of any other y in L_1.  Otherwise we
>have x = y, contradicting (1).
I don't get that. Suppose that L_1 is the following list:
    .0000...
    .01000...
    .001000...
    .0001000...
    .00001000...
    etc.
Then x has no unique initial segment. It agrees with the second
real in the first decimal place. It agrees with the third real
in the first 2 decimal places. It agrees with the fourth real
in the first 3 decimal places, etc. So, I'm not exactly sure
how your construction is going to go.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: induction vs Cantor
> Chairman of the David Hilbert Appreciation Society says...
>>I'm assuming that the rows of each L_n will start from 1.
>>It makes no difference to the result if we start from 0,
>>but is more convenient with 1 IMO.
>>First we agree that:
>>(1)  L_1 is a list, an injection.
> I thought that a list of reals is a function from N into R, not
> necessarily an injection. That is, there can be repetitions in
> L_1, and it doesn't hurt anything.
Let L_1 be a list of reals that implies a mapping F_1
I took that to mean that L_1 was an injection.  I could
be wrong, but Poker Joker hasn't objected to (1) AFAIK.
>>(2)  Any real number in row 2n-1 of L_n is in the
>>    same row (2n-1) on any L_m with m >= n.  This follows
>>    from [*] in his description.
>>(3)  It follows from (2) that if x is a real number
>>    in row 2n-1 of L_m with m >= n that the (2n-1)th
>>    digit of any D_m will not be the same as the
>>    (2n-1)th digit of x.  Otherwise D_n would not
>>    be an anti-diagonal number.
>>Once we agree that items 1-3 characterize the method
>>we do as follows:
>>(4) Denote the real number in row 1 of L_1 as x.
>>Since L_1 is a list there must be some n for which
>>the first 2n digits of x aren't all the same as the
>>first 2n digits of any other y in L_1.  Otherwise we
>>have x = y, contradicting (1).
> I don't get that. Suppose that L_1 is the following list:
>     .0000...
>     .01000...
>     .001000...
>     .0001000...
>     .00001000...
>     etc.
> Then x has no unique initial segment. It agrees with the second
> real in the first decimal place. It agrees with the third real
> in the first 2 decimal places. It agrees with the fourth real
> in the first 3 decimal places, etc. So, I'm not exactly sure
> how your construction is going to go.
You're right.  My construction doesn't work in general.  I should
have probably instead allowed x to be a number in any row of L_1
which has a unique initial segment.  Not even sure if that's
possible now but since L_1 is countable I would suspect that
it is... (Your opinion?)
I was trying to create a real that shares the same first 2n digits
with exactly one number on L_1 and which wouldn't be on L_1.
> --
> Daryl McCullough
> Ithaca, NY
-- 
Replace Roman numerals with digits to reply by email
===
Subject: Re: induction vs Cantor
Chairman of the David Hilbert Appreciation Society 
> I didn't create a Cantor number.
Well I would say a good name for a number that
is constructed that isn't in a given list is a Cantor
number.  That's all.  No big deal.
>The number I show, that isn't
> on any L_n has the exact same first 2n digits as those of
> the number on the first row of L_1 -- clearly this isn't
> constructed by the same method that Cantor uses otherwise
> its first digit would be different from the first digit
> of the number on the first row of L_1.
Cantor could have specified his diagonal number the same way.
It doesn't matter because only countably many constructions
can be described and the process of defining the L_n is
flexible enough to allow an arbitrary Cantor number to be
constructed at any iteration whatsoever.
>>  So your T is a D_n becaues
> T can't be any D_n.  The first digit of T is the same first
> digit of the number in the first row of L_1. Since that number
> is in the first row on L_1 and every subsequent list L_2, L_3, ...
> none of the D_n share the same first digit with it -- otherwise
> D_n wouldn't be an anti-diagonal number, as you specified.
Again, the definition of L_n is general enough to allow for
the Cantor number to be constructed arbitrarily.  You only
need a method that shows a construction of a number that's
not in the list.  The process isn't in any way dependent on the
method of constructing the Cantor number.
>> we assume that at any stage, the Cantor method can be any
>> one of a countably infinite set of methods for generating the
>> D_n.  You've described one.
> I used the most reasonable interpretation of the phrase:
> Cantor['s] anti-diagonal number, which in sci.math is
> used to describe a number constructed based on the diagonal
> string of digits of a list.
I appologize for not being more clear about allowing the
Cantor method to be any general method of constructing
a unique real.  I honestly thought people assumed that
a anti-diagonal number could be generated almost
arbitrarily with the only restriction being that it is constructed
as a unique real with respect to the list.
> So your T has been taken into
>> account.
> Then which L_n does T occur on?
It could be included in the process at any iteration.  As long
as T is in there somewhere.  After all, we are allowing for
countably infinitely many of these numbers to be included.
That's enough for any construction anyone could possibly
describe.
>> Cantor methods aren't required to create a new
>> real in which the first digit is different from the first digit of
>> the first real.
> You were using Cantor's method, you said:
> Let D_n be a Cantor anti-diagonal number that... -PJ
Again, that's my mistake and I sincerely appologize.
> I suspect this is why you're no longer (as of this post)
> calling it an anti-DIAGONAL.
Yes, to some extent.  I have seen people getting confused.
> Even if you use a method that is different yet produces
> a similar result to Cantor's method it makes no difference.
> Once you give enough details of how D_n is constructed
> the same thing can be done.
> OTOH if you completely obscure how D_n is created then
> of course you will experience the satisfaction of never
> seeing a counter example created in the same way (based
> on knowledge of how L_n is defined), but there are still
> other ways...
But only countably many ways of describing such a construction.
That's my point.  Only countably many ways of coming up with
a construction of a real that's not in a list. In fact, there's only
countably many describable reals.  I just happen to allow for
the inclusion of all of them in the L_n's by allowing for 1 to be
added to the list countably infinitely many times.  There can't
be any left over after creating the union of the lists because
that is just one of the countably many ways of creating a Cantor
number.  We included those in the lists.
> Aside from the matter of trying to convincing you that
> a number exists that isn't on any L_n, you seem to want
> to convince me that there is something special about the
> collection of all L_n, which so far I haven't seen.
You are thinking about how many reals there are.
I am thinking about how many ways of showing (proving)
an arbitrary real isn't in an arbitrary list.  The two may or may
not be related.  I claim there are only countably many
such proofs (unless we use Cantor's conclusions) and I
think we can sum those up as I have done.
> Perhaps this example will help you see why I regard
> the collection of all L_n as uninteresting:
> (1) Let A and B be countably infinite ordered sets of real
>     numbers such that A cap B = emptyset.  Let C = A cup B.
> It seems to me that C has the same distinguishing
> features as the union of all L_n.
Except we can come up with a method of constructing
a new number because C isn't a set of all possible
constructions.
> Each x in B is the Cantor number (*) of some list of reals,
> each x in B is unique, and there are countably many x in B.
> C is the collection of an initial list of reals A (like L_0)
> with all of the diagonal numbers in B added.
> (*) Cantor numbers just have to be created so that we can be
> assured that they are unique with respect to the list. -PJ
Each list needs only one Cantor number to prove it is
not R.  If a list has no Cantor number, Cantor's proof
fails to prove the list is not the set of reals.
Note that for each L_n, a Cantor number exists and it
exists in L_2n.  Note that all the possible Cantor numbers
exist in the union of all of these lists assuming that we used
all of the countably many ways of constructing them when
we constructed the lists.  And we can.
===
Subject: Re: induction vs Cantor
> Virgil says...
>I use the conclusion that the power set of any set has a larger 
>cardinality than the set itself only to show that there are sets with 
>cardinalities greater than that of the set of naturals, but where do I 
>assume that that applies to the reals? I do not.
> But isn't the proof that the powerset of S has a larger cardinality
> than S essentially the same idea as the diagonal proof of the
> uncountability of the reals?
Any proof that tries to shows one set larger in cardinality than another 
must show that there is no surjection from the allegedly smaller  to 
allegedly larger.
The only way to show that there is no such surjection is to show that 
every potential surjection fails to be a surjection.
To disallow all such proofs is virtually to say that the theorem cannot 
be proved, and that no cardinalities greater than that of N can be shown 
to exist.
If Joker wants to limit himself thusly, okay, but he should not be 
trying to impose his limitations on others.
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id iB34FCx13309;
Particularly dealing with complex curves as Riemann surfaces & their
topology. Also moduli spaces of curves of genus g.
I own a couple of Shafarevich's 'basic alg geo' volumes.  There is a 
book by Kirwan on complex algebraic curves that I've heard good 
things about.  I'm thinking maybe something like Forster's book on
Riemann surfaces might help some, but I'm not sure.  Hence, the 
request for references.