mm-329
Subject: Re: Mechanics question
> Hi group, I'm a high school student attempting to
realistically model the
> motion of a ball sliding down a helical ramp with a side
wall.
> First I'll present my current understanding of the 
concept,
please
correct
> any mistakes here as I'm still learning. In the absence of
friction, the
> tangential and normal components of acceleration are given
respectively
by
> (1) a_T = g sintheta T
> (2) a_N = frac{v^2 cos^2theta}{rho} N
> = frac{g^2sin^2theta t^2cos^2theta}{rho} N
> where (T, N, B) is the righthanded orthonormal trihedral
associa with
the
> helix and rho the radius of curvature.
> If however, the ball rolls without slipping, then the ball
is subject to
two
> frictional forces direc antiparallel to T, which arise due
to the
static
> friction between the ball and the planes spanned by TN and
BT. The
> frictional forces in these planes would seem to give rise
to two
components
> of angular acceleration in the directions of N and B
respectively, with
the
> direction of rotation determined by the right-hand rule (a
diagram may
help here).
> According to one reference [1], a ball rolling down a
simple inclined
plane
> will feel a frictional force direc antiparallel to its
linear velocity
> given by
> (3) f = frac{Ialpha}{r}
> where I is the moment of inertia of a uniform shperical
object of radius
r.
> Since the ball in the helix rotates simultaneously about N
and B with the
> same angular acceleration, the resultant angular
acceleration is in the
> plane spanned by BN and of magnitude
> (4) |alpha| = sqrt{2}|alpha_N| = sqrt{2}frac{|a_T|}{r}
> Am I thus justified in substituting the above expression
into Eq. (3) to
> obtain the resultant frictional force?
> [1] http://theory.uwinnipeg.ca/physics/rot/node9.html
===
Subject: Problem from Rudin
Although this is homework rela, I hope someone is willing to
help.
Here's my situation:
Rudin is asking for a construction of a compact subset of |R
which has
countably many limit points. (In his usage, countably many
means
countably infinite, not countably infinite or 
finite.)
But I think I have a proof that this is impossible:
Let A be a compact subset of |R. By Heine-Borel, A is closed
and
bounded. Then A is either perfect, or it isn't. If it is
perfect,
then A is uncountable. (This follows from a theorem whose
name I
cannot remember. The theorem was the basis on which Rudin
establishes
that |R is uncountable). If it is not perfect, it has a
perfect
subset or is the union of isola points (or both).
In the second case, the set of limit points is empty, not
countable.
Am I missing something here?
===
Subject: Re: Problem from Rudin
>
Although this is homework rela, I hope someone is willing to
help.
> Here's my situation:
Rudin is asking for a construction of a compact subset of |R
which has
> countably many limit points.
A was thinking about the following process for constructing
the
desired set follows:
(1) Find a convergent sequence in R with distinct terms (very
easy).
Let (a_n) be such sequence and a = lim a_n
(2) For each n, find a sequence (b_n_k)with distinct elements
that
converges to a_n, in such a way that the convergence of the
sequences
(x_n_k) to (a_n) is uniform. This is easy, it's enough that
the
sequences are paralell, that is, each of them is a
translation of the
preceeding. Once you found (x_1_k), you set x_2_k = x_1_k +
a_2 - a_1
for every k, x_3_k=x_2_k + a_3-a_2 for every k and so on.
(3) Observe that, for each n, a_n is a limit point of the set
X_n
={x_n_k: k=1,2...}. Since x_n_k -> a_n, a_n is the only limit
point of
X_n. And since the numbers a_n are pairwise distinct, if m<>
then X_m
and X_n have different limit points. In addition, each X_n is
bounded
(3) Take the set X = Union X_n, n=1,2áSince (a_n) is bounded,
so
is X.
And, by construction, each a_n, as well as a, is a limit
point of X.
(4) Since a_n ->a and the convergence of the sequences
(x_n_k) is
uniform, it follows the double sequence (x_n_k) has a double
limit,
and this double limit equals lim a_n =a. It also follows
that, for
each natural k, the sequences (x_1_, x_2_k.....) are
convergent. If
you think of the (x_n_k), k=1,2... as row sequences on a
matrix-like
array, then it follows all the column sequences are
convergent. For
each k, call b_k the limit of the k_th column sequence. Then,
the
numbers b_k are also limit points of X. And so is b =lim b_k,
which
certainly exists.
(5) X is not closed, for it doesnt contain all of its limit
points.
But the set X*, obtained by adjoining to X its limit points,
is. Just
because it contains all its limits points. X* is the closure
of X.
(6) And you're done! You got a closed and bounded set, so a
compact
one, such that the set of its limit points is countable and
infinite.
But I skipped a detail. You may be wondering why I developed a
somewhat tricky proocess, requiring uniform convergence and
the stuff.
Well, I wan to make sure the set X has no limit point other
than
those we covered. That is, I wan to prevent the set of limit
points
of X from containing an uncountable subset. So, to be
precise, it
remains to prove X has no other limit points. This seems
clear, but it
remains to find a rigorous proof.
Artur
===
Subject: Re: Problem from Rudin
Although this is homework rela, I hope someone is willing to
help.
>Here's my situation:
>Rudin is asking for a construction of a compact subset of |R
which has
>countably many limit points. (In his usage, countably many
means
>countably infinite, not countably infinite or 
finite.)
>But I think I have a proof that this is impossible:
>Let A be a compact subset of |R. By Heine-Borel, A is closed
and
>bounded. Then A is either perfect, or it isn't. If it is
perfect,
>then A is uncountable. (This follows from a theorem whose
name I
>cannot remember. The theorem was the basis on which Rudin
establishes
>that |R is uncountable). If it is not perfect, it has a
perfect
>subset or is the union of isola points (or both).
Oh? Say A = {0} union {1, 1/2, 1/3, ...}. Is A perfect? Does
A have
a perfect subset? Is A the union of isola points?
>In the second case, the set of limit points is empty, not
countable.
>Am I missing something here?
************************
===
Subject: Re: Problem from Rudin
>Although this is homework rela, I hope someone is willing to
help.
>Here's my situation:
Rudin is asking for a construction of a compact subset of |R
which has
>countably many limit points. (In his usage, countably many
means
>countably infinite, not countably infinite or 
finite.)
>But I think I have a proof that this is impossible:
Let A be a compact subset of |R. By Heine-Borel, A is closed
and
>bounded. Then A is either perfect, or it isn't. If it is
perfect,
>then A is uncountable. (This follows from a theorem whose
name I
>cannot remember. The theorem was the basis on which Rudin
establishes
>that |R is uncountable). If it is not perfect, it has a
perfect
>subset or is the union of isola points (or both).
Oh? Say A = {0} union {1, 1/2, 1/3, ...}. Is A perfect? Does
A have
> a perfect subset? Is A the union of isola points?
>
Uhm... Well, aren't 1, 1/2, 1/3,... isola points? I can 
think
of
a radius r such that nothing in A is in the neighborhood
about 1.
(How about r = 1/4?)
Would you please critique this? This is the basis on which I
came up
with that line. (Sorry about the FOL-ese. I find topology to
be much
easier with explicit quantifiers)
Define Perfect(A) := Ax (x in A -> x in A') 
where A' is the
set of
limit points.
So ~Perfect(A) is equivalent to ~Ax (x in A -> x in A')
Ex ~(x in A -> x in A')
Call it c
~(c in A -> c in A')
(c in A) & ~(c in A') (tautological consequence)
Ex ((x in A) & ~(x in A'))
So, assuimg that A is closed and not perfect, it must have an
isola
point. Suppose we let F be the set of all such points. What
would A
- F be? It could be empty. Or it could be perfect.
Alex Solla
===
Subject: Re: Problem from Rudin
>Although this is homework rela, I hope someone is willing to
help.
>>Here's my situation:
>Rudin is asking for a construction of a compact subset of |R
which has
>>countably many limit points. (In his usage, countably many
means
>>countably infinite, not countably infinite or 
finite.)
>>But I think I have a proof that this is impossible:
>Let A be a compact subset of |R. By Heine-Borel, A is closed
and
>>bounded. Then A is either perfect, or it isn't. If it is
perfect,
>>then A is uncountable. (This follows from a theorem whose
name I
>>cannot remember. The theorem was the basis on which Rudin
establishes
>>that |R is uncountable). If it is not perfect, it has a
perfect
>>subset or is the union of isola points (or both).
Oh? Say A = {0} union {1, 1/2, 1/3, ...}. Is A perfect? Does
A have
>> a perfect subset? Is A the union of isola points?
>Uhm... Well, aren't 1, 1/2, 1/3,... isola points?
Yes. Is A equal to the union of {1}, {1/2}, ... ?
(Hint: no it's not. So what the heck is your point?)
>I can think of
>a radius r such that nothing in A is in the neighborhood
about 1.
>(How about r = 1/4?)
>Would you please critique this? This is the basis on which I
came up
>with that line. (Sorry about the FOL-ese. I find topology to
be much
>easier with explicit quantifiers)
>Define Perfect(A) := Ax (x in A -> x in A') 
where A' is the
set of
>limit points.
>So ~Perfect(A) is equivalent to ~Ax (x in A -> x in A')
>Ex ~(x in A -> x in A')
>Call it c
>~(c in A -> c in A')
>(c in A) & ~(c in A') (tautological consequence)
>Ex ((x in A) & ~(x in A'))
>So, assuimg that A is closed and not perfect, it must have
an isola
>point. Suppose we let F be the set of all such points. What
would A
>- F be? It could be empty. Or it could be perfect.
If you prove that F = A you'll have a point here.
>Alex Solla
************************
===
Subject: Re: Problem from Rudin
>> Oh? Say A = {0} union {1, 1/2, 1/3, ...}. Is A perfect?
Does A have
>> a perfect subset? Is A the union of isola points?
>Uhm... Well, aren't 1, 1/2, 1/3,... isola points? I can
think of
>a radius r such that nothing in A is in the neighborhood
about 1.
>(How about r = 1/4?)
Yes, but {0} is not an isola point, and is not perfect.
>Would you please critique this? This is the basis on which I
came up
>with that line. (Sorry about the FOL-ese. I find topology to
be much
>easier with explicit quantifiers)
>So, assuimg that A is closed and not perfect, it must have
an isola
>point. Suppose we let F be the set of all such points. What
would A
>- F be? It could be empty. Or it could be perfect.
Or neither, as in the above case.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Problem from Rudin
mm-329
Although this is homework rela, I hope someone is willing to
help.
>Here's my situation:
Rudin is asking for a construction of a compact subset of |R
which has
>countably many limit points. (In his usage, countably many
means
>countably infinite, not countably infinite or 
finite.)
>But I think I have a proof that this is impossible:
Let A be a compact subset of |R. By Heine-Borel, A is closed
and
>bounded. Then A is either perfect, or it isn't. If it is
perfect,
>then A is uncountable. (This follows from a theorem whose
name I
>cannot remember. The theorem was the basis on which Rudin
establishes
>that |R is uncountable). If it is not perfect, it has a
perfect
>subset or is the union of isola points (or both).
Oh? Say A = {0} union {1, 1/2, 1/3, ...}. Is A perfect? Does
A have
> a perfect subset? Is A the union of isola points?
> Uhm... Well, aren't 1, 1/2, 1/3,... isola points? I can
think of
> a radius r such that nothing in A is in the neighborhood
about 1.
> (How about r = 1/4?)
Would you please critique this? This is the basis on which I
came up
> with that line. (Sorry about the FOL-ese. I find topology to
be much
> easier with explicit quantifiers)
Define Perfect(A) := Ax (x in A -> x in A') 
where A' is the
set of
> limit points.
> So ~Perfect(A) is equivalent to ~Ax (x in A -> x in A')
Ex ~(x in A -> x in A')
Call it c
~(c in A -> c in A')
> (c in A) & ~(c in A') (tautological consequence)
Ex ((x in A) & ~(x in A'))
So, assuimg that A is closed and not perfect, it must have an
isola
> point. Suppose we let F be the set of all such points. What
would A
> - F be? It could be empty. Or it could be perfect
He JUST GAVE YOU an example where it isn't empty and it 
isn't
perfect.
Yes, the points 1, 1/2, 1/3, ... are isola in A. But
A {1,1/2,1/3} = {0} is neither empty nor perfect.
Perhaps you are misquoting an exercise in Rudin which says:
every
closed set in a separable metric space is the union of a
(possibly
empty) perfect set and a set which is at most countable.
--Ron Bruck
===
Subject: Re: Problem from Rudin
>Although this is homework rela, I hope someone is willing to
help.
>Here's my situation:
Rudin is asking for a construction of a compact subset of |R
which
has
>countably many limit points. (In his usage, countably many
means
>countably infinite, not countably infinite or 
finite.)
>But I think I have a proof that this is impossible:
Let A be a compact subset of |R. By Heine-Borel, A is closed
and
>bounded. Then A is either perfect, or it isn't. If it is
perfect,
>then A is uncountable. (This follows from a theorem whose
name I
>cannot remember. The theorem was the basis on which Rudin
establishes
>that |R is uncountable). If it is not perfect, it has a
perfect
>subset or is the union of isola points (or both).
Oh? Say A = {0} union {1, 1/2, 1/3, ...}. Is A perfect? Does
A have
> a perfect subset? Is A the union of isola points?
> Uhm... Well, aren't 1, 1/2, 1/3,... isola points? I can
think of
> a radius r such that nothing in A is in the neighborhood
about 1.
> (How about r = 1/4?)
Would you please critique this? This is the basis on which I
came up
> with that line. (Sorry about the FOL-ese. I find topology to
be much
> easier with explicit quantifiers)
Define Perfect(A) := Ax (x in A -> x in A') 
where A' is the
set of
> limit points.
> So ~Perfect(A) is equivalent to ~Ax (x in A -> x in A')
Ex ~(x in A -> x in A')
Call it c
~(c in A -> c in A')
> (c in A) & ~(c in A') (tautological consequence)
Ex ((x in A) & ~(x in A'))
So, assuimg that A is closed and not perfect, it must have an
isola
> point. Suppose we let F be the set of all such points. What
would A
> - F be? It could be empty. Or it could be perfect
He JUST GAVE YOU an example where it isn't empty and it 
isn't
perfect.
> Yes, the points 1, 1/2, 1/3, ... are isola in A. But
A {1,1/2,1/3} = {0} is neither empty nor perfect.
>
Yes, thanks very much. I was trying to find the error in my
logic.
It was obvious that he JUST GAVE ME a counter example.
This may be old hat to you, but I'm trying to learn it. And
it's not
particularly easy.
Ôcid
===
Subject: Re: Problem from Rudin
>Let A be a compact subset of |R. By Heine-Borel, A is closed
and
>bounded. Then A is either perfect, or it isn't. If it is
perfect,
>then A is uncountable. (This follows from a theorem whose
name I
>cannot remember. The theorem was the basis on which Rudin
establishes
>that |R is uncountable). If it is not perfect, it has a
perfect
>subset or is the union of isola points (or both).
Oh? Say A = {0} union {1, 1/2, 1/3, ...}. Is A perfect? Does
A have
> a perfect subset? Is A the union of isola points?
> Uhm... Well, aren't 1, 1/2, 1/3,... isola points? I can
think of
> a radius r such that nothing in A is in the neighborhood
about 1.
> (How about r = 1/4?)
...and 0 is not an isola point.
That example has one limit point. Next, can you make an
example with
two limit points?
Would you please critique this? This is the basis on which I
came up
> with that line. (Sorry about the FOL-ese. I find topology to
be much
> easier with explicit quantifiers)
Define Perfect(A) := Ax (x in A -> x in A') 
where A' is the
set of
> limit points.
> So ~Perfect(A) is equivalent to ~Ax (x in A -> x in A')
Ex ~(x in A -> x in A')
Call it c
~(c in A -> c in A')
> (c in A) & ~(c in A') (tautological consequence)
Ex ((x in A) & ~(x in A'))
So, assuimg that A is closed and not perfect, it must have an
isola
> point. Suppose we let F be the set of all such points. What
would A
> - F be? It could be empty. Or it could be perfect.
Alex Solla
What is it in the example above?
===
Subject: Re: Problem from Rudin
>
Although this is homework rela, I hope someone is willing to
help.
> Here's my situation:
Rudin is asking for a construction of a compact subset of |R
which has
> countably many limit points. (In his usage, countably many
means
> countably infinite, not countably infinite or 
finite.)
> But I think I have a proof that this is impossible:
Let A be a compact subset of |R. By Heine-Borel, A is closed
and
> bounded. Then A is either perfect, or it isn't. If it is
perfect,
> then A is uncountable. (This follows from a theorem whose
name I
> cannot remember. The theorem was the basis on which Rudin
establishes
> that |R is uncountable). If it is not perfect, it has a
perfect
> subset or is the union of isola points (or both).
> In the second case, the set of limit points is empty, not
countable.
Am I missing something here?
I construc a solution as follows. First, what's the most
obvious
example of a set with countably many limit points? Start with
the
integeand around each integer n include the points n + 1/k
for k =
1, 2, 3 . . . By construction each integer is the limit of a
nearby
sequence of 1/k terms, so there are countably many limit
points.
However this set is not bounded. To fix this we do the same
thing as
above but making 1/2, 3/4, 7/8, . . . the limit points. That
is, we
define
S = Union {(2^n - 1)/(2^n) + 1/k, k=1,2,3,...}
with the union taken over n = 1, 2, 3, . . .
Now S is closed, bounded, and has countably many limit points.
This is a compact set that's not perfect and has no perfect
subset.
===
Subject: Re: Please help with Induction.
Thank You!!!
> Hello
> Im trying to prove the following using Mathematical
Induction. This is
as
> far as I have got:
Show that 1^3 + 2^3 + ... + n^3 = [n(n+1)/2] ^2 whenever n is
a
positive
> integer.
Basis Step. let n = 1
> n^3 = 1 and [n(n+1)/2] ^2 = (1(2))^2 / 2 = 1 This step is
true.
Induction Hypothesis. let n = k
> 1^3 + 2^3 + ... + k^3 = [k(k+1)/2] ^2
Induction Proof. n = (k+1)
> 1^3 + 2^3 + ... + (k+1)^3 = [((k+1)((k+1)+1))/2] ^2
Not sure where to go from here. Any help apprecia. Thank you
> You want to show Sum[j^3,{i,1,n+1}] = ((n+1)(n+2)/2))^2
(Call this equ.
1)
> Now we add the (n+1)th term, so we get:
> ((n)(n+1)/2))^2 + (n+1)^3
> Simplifying, we get (((n+1)(n+2))/2)^2
> Look like equ 1? Sure does
> See if you can follow the steps as it is a bit tricky, but
hopefully this
> will give enough guidance.
> The trick is to take out the IH, and then add the n+1 term
to it and
> simplify.
> Since we have k^3 evalua at (n+ 1), that is where the
(n+1)^3 term
comes
> from.
> Make sense? HTH, Flip
Subject: Re: How Many Blinkin' Lights On The Tree?
===
Subject: Re: Question - Special properties of 4-D
> I'm trying to compile a list of mathematical properties,
objects and
>> relationships which are peculiar to a 4-dimensional space
(of course,
>> the same would be of interest for a general dimension, n).
For e.g.,
>> are there certain theorems that are easy to prove for all
dimensions
>> but n=4? Or a certain type of nontrivial mathematical
object whose
>> definition constrains its existence to 4-D only?
Fake R^4 springs immediately to mind. It is known that for n
<> 4 (of
> course n is a nonnegative integer), all smooth structures
on R^n are
> diffeomorphic to one another. On the other hand, there
exist *uncountably
> many* non-diffeomorphic smooth structures on R^4. (These
are called
fake
> R^4's.) I'm sorry but I am unable to provide 
a citation
just now; all
my
> books are at the office.
I believe this result is due to Donaldson, building on work
of Freedman.
In the same vein, there are untriangulable 4-manifolds. It's
conjectured
this is the only dimension where this happens.
===
Subject: Help with limsup and liminf
My problem is this :
Suppose that the sequence of non-negative real numbers
satisfies a_(m+n) <=
a_n + a_m for all m,n>=1.
Show that lim_(n-->infinity) of (a_n)/n exists.
Here is how I am going for it : I want to show that
limsup ((a_n)/n) <= inf((a_n)/n) <= liminf((a_n)/n) and if I
do this, then
clearly the sequence converges. So first I want to show that
limsup ((a_n)/n) <= inf((a_n)/n)
Well, I proved that (a_(nk))/(nk) <= (a_n)/n by the first
supposition by
just noting that a_(kn) = a_(n+n+...+n k-times) <= k*a_n.
Now I proceed logically by taking the infimum of both sides of
(a_(nk))/(nk)
<= (a_n)/n. Finally, I want to show that
limsup ((a_n)/n) <= inf {(a_(nk))/(nk)} and then I will have
proved that
limsup ((a_n)/n) <= inf((a_n)/n).
If I can do this I am sure the other half is very similar. My
only problem
is proving
limsup ((a_n)/n) <= inf {(a_(nk))/(nk)}
There is a definition I found : limsup (x_n) = inf { sup
{x_(n+1),x_(n+2),...} | n=1,2,3,...}. I have tried to use
this fact, but I
can't seem to get anywhere.
I am so close I think, maybe someone can help me see the
light?
Thank you so much,
Topy
===
Subject: Re: Help with limsup and liminf
===
> My problem is this :
Suppose that the sequence of non-negative real numbers
satisfies a_(m+n)
<=
> a_n + a_m for all m,n>=1.
Show that lim_(n-->infinity) of (a_n)/n exists.
Here is how I am going for it : I want to show that
limsup ((a_n)/n) <= inf((a_n)/n) <= liminf((a_n)/n) and if I
do this,
then
> clearly the sequence converges. So first I want to show that
> limsup ((a_n)/n) <= inf((a_n)/n)
Let I = inf((a_n)/n).
Fix e > 0.
Choose (and fix) m so large that a_m/m < I+e.
Then a_{nm)/(nm) le a_m/m for all n=1,2,3,...
Now interpolate by considering a_{nm+j}/(nm+j) for
j=1,2,...m-1 and
n=1,2,...
> Well, I proved that (a_(nk))/(nk) <= (a_n)/n by the first
supposition by
> just noting that a_(kn) = a_(n+n+...+n k-times) <= k*a_n.
Now I proceed logically by taking the infimum of both sides of
(a_(nk))/(nk)
> <= (a_n)/n. Finally, I want to show that
limsup ((a_n)/n) <= inf {(a_(nk))/(nk)} and then I will have
proved that
> limsup ((a_n)/n) <= inf((a_n)/n).
If I can do this I am sure the other half is very similar. My
only
problem
> is proving
> limsup ((a_n)/n) <= inf {(a_(nk))/(nk)}
There is a definition I found : limsup (x_n) = inf { sup
> {x_(n+1),x_(n+2),...} | n=1,2,3,...}. I have tried to use
this fact, but
I
> can't seem to get anywhere.
I am so close I think, maybe someone can help me see the
light?
Thank you so much,
Topy
>
--
A.
===
Subject: Re: Help with limsup and liminf
Topy
> My problem is this :
> Suppose that the sequence of non-negative real numbers
satisfies a_(m+n)
<=
> a_n + a_m for all m,n>=1.
> Show that lim_(n-->infinity) of (a_n)/n exists.
...
> Finally, I want to show that
> limsup ((a_n)/n) <= inf {(a_(nk))/(nk)} and then I will
have proved that
> limsup ((a_n)/n) <= inf((a_n)/n).
> If I can do this I am sure the other half is very similar.
My only
problem
> is proving
> limsup ((a_n)/n) <= inf {(a_(nk))/(nk)}
...
Um, I suspect it's not going to help to look at subsequences
(a_(nk)).
If a_1=0 then all the a_n are zero, so suppose not. Dividing
all the terms
through by a_1, we can suppose a_1=1 and so a_n<=n for all n.
If a_n/n has
no limit, then it has two distinct cluster points x,y with 0
<= x < y <=1.
Can we deduce that a(m+n)>a(m)+a(n) for some m,n?
Still looking.
LH
===
Subject: Calculus Guidance
Hi. I'm having some big problems with calculus with these
problems
based on proofs. I was hoping the people in this group can
help guide
me towards the solution. :)
The -> means vector.
Show that if ->a and ->b have the same direction, then:
||a+b|| = ||a|| + ||b||
I can see this clearly with from a graphical drawing, but I
have no
clue how to prove it mathematically. I tried making ->a =
(a1,a2,a3)
and ->b = (b1,b2,b3) and then taking the norm with the
square-root
formula but that didn't seem to help in anyway.
Another one:
->a = (1,1,1) ->b = (-1,3,2) ->c = (-3,0,1) ->d = (4,-1,1)
It asks to find scalars A, B, C so that ->d = A->a + B->b +
C->C
I'm guessing trying random numbers is not a very 
efficient
approach. I
thought about simultaneous equations somehow but I haven't
figured out
a way to do it so it turns out nicely.
Also one more:
It wants to prove this version of the triangle inequality:
abs (||a|| - ||b||) =< (less than/equal to) ||a-b||
Hint: ->a = (->a - ->b) + ->b
Someone tried to explain it to me but it didn't seem to help
much...
Thank you for any help on these problems. Also, does anyone
know of a
software program that can draw vectors in 3D space? I need to
visualize first before I can start working in all variables. 
:)
===
Subject: Re: Calculus Guidance
Answer to the 1st question
> ||a+b|| = ||a|| + ||b||
->a = (a1,a2,a3)
> and ->b = (b1,b2,b3)
then ->a + ->b = a1->i + a2->j + a3->k + b1->i + b2->j + b3->
<=>
->a + ->b = (a1+b1)->i + (a2+b2)->j + (a3+b3)->k
and |->a + ->b| = [(a1+b1)^2 + (a2+b2)^2 + (a3+b3)^2]^(1/2)
where ^2 means at the power of 2 and ^1/2 is the square root.
also |->a| + |->b| = [a1^2 + a2^2 + a3^2]^(1/2) + [b1^2 +
b2^2 +
b3^2]^(1/2)
in order |->a + ->b| = |->a| + |->b| we just have to prove
that:
[(a1+b1)^2 + (a2+b2)^2 + (a3+b3)^2]^(1/2) =
[a1^2 + a2^2 + a3^2]^(1/2) + [b1^2 + b2^2 + b3^2]^(1/2)<=>
(a1+b1)^2 + (a2+b2)^2 + (a3+b3)^2 =
a1^2 + a2^2 + a3^2 + b1^2 + b2^2 + b3^2 + 2*[(a1^2 + a2^2 +
a3^2)*
(b1^2 + b2^2 + b3^2)]^(1/2)<=>
a1^2 + a2^2 + a3^2 + b1^2 + b2^2 + b3^2 + 2*(a1*b1 + a2*b2 +
a3*b3) =
a1^2 + a2^2 + a3^2 + b1^2 + b2^2 + b3^2 + 2*[(a1^2 + a2^2 +
a3^2)*
(b1^2 + b2^2 + b3^2)]^(1/2)<=>
(a1*b1 + a2*b2 + a3*b3) =
[(a1^2 + a2^2 + a3^2)* (b1^2 + b2^2 + b3^2)]^(1/2)<=>
(a1*b1 + a2*b2 + a3*b3)^2 = (a1^2 + a2^2 + a3^2)* (b1^2 +
b2^2 + b3^2)
<=>
(a1b1)^2 + (a2b2)^2 + (a3b3)^2 +2*[a1b1a2b2 + a1b1a3b3 +
a2b2a3b3] =
(a1b1)^2 + (a2b2)^2 + (a3b3)^2 + (a1b2)^2 + (a1b3)^2 +
(a2b1)^2 +
(a2b3)^2 + (a3b1)^2 + (a3b2)^2 <=>
2*[a1b1a2b2 + a1b1a3b3 + a2b2a3b3] =
(a1b2)^2 + (a1b3)^2 + (a2b1)^2 + (a2b3)^2 + (a3b1)^2 +
(a3b2)^2
(equation 1)
but if ->a is // to ->b (parallel)
then a1/b1 = a2/b2, a1/b1 = a3/b3 and a2/b2 = a3/b3
so a1b1a2b2 = a1b1(a1b2/b1)b2 = (a1b2)^2 or (a2b1)^2
and similarly a1b1a3b3 = (a1b3)^2 = (a3b1)^2
and a2b2a3b3 = (a2b3)^2 = (a3b2)^2
therefore eq.1 can be transformed to:
2*[(a1b2)^2 + (a1b3)^2 + (a2b3)^2] =
(a1b2)^2 + (a1b3)^2 + (a2b1)^2 + (a2b3)^2 + (a3b1)^2 +
(a3b2)^2 <=>
(a1b2)^2 + (a1b3)^2 + (a2b1)^2 + (a2b3)^2 + (a3b1)^2 +
(a3b2)^2 =
(a1b2)^2 + (a1b3)^2 + (a2b1)^2 + (a2b3)^2 + (a3b1)^2 +
(a3b2)^2
...........
which leads to 0=0 (identity) true for every ->a, ->b
parallel to each
other
greetings spiros
===
Subject: Re: Calculus Guidance
answer to the 2nd question
Lets say that
->i ->j and ->k are the unit vectors of x, y and z -axis
respectively.
then ->a = a1->i + a2->j + a3->k
and similarly b, c and d come up...
therefore ->d = [A*1+B*(-1)+C*(-3)]->i + [A*1+B*3+C*0]->j +
[A*1+B*2+C*1]->k.
But as we know d = 4->i - 1->j + 1->k
therefore
A - B -3C = 4
A + 3B = -1
A + 2B + C = 1
which leads to A = 26/7, B = -11/7 and C = 3/7
greetings spiros.
===
Subject: Re: Calculus Guidance
> Hi. I'm having some big problems with calculus with these
problems
> based on proofs. I was hoping the people in this group can
help guide
> me towards the solution. :)
The -> means vector.
Show that if ->a and ->b have the same direction, then:
>
If two vectors have the same direction, then each of them is a
positive scalar multiple of the other, and all vectors having
the
same direction will be positive scalar multiples of any one
of them.
Thus you can effectively factor out the vector properties and
deal
only with the positive scalars.
===
Subject: Re: Calculus Guidance
> Hi. I'm having some big problems with calculus with these
problems
> based on proofs. I was hoping the people in this group can
help guide
> me towards the solution. :)
The -> means vector.
Show that if ->a and ->b have the same direction, then:
||a+b|| = ||a|| + ||b||
>
If a and b are vectors in the same direction, is there
anything you can
say about them that might be useful?
Look at some examples. What are some vectors that have the
same
direction as (1,1)? What are some vectors that have the same
direction
as (2,3)? See a pattern?
===
Subject: Re: a question about Relative Numbers
Content-transfer-encoding: 8bit
> In Z, I can multiply (a, b) and (a', b') in 
this way:
(a, b) * (a', b') := (a*a' + 
b*b', a*b' + a'*b)
I want to prove that this operation has not affec by the
specific
> choice of the couples.
So I can take:
> (c, d) R (a, b) so that c+b = a+d
> (c', d' ) R (a', 
b') so that c'+b' = 
a'+d'
and then: (c,d) * (c',d') = 
(c*c' + d*d', c*d' + 
c'*d)
How can I prove that (c*c' + d*d', 
c*d' + c'*d) R (a*a' +
b*b', a*b' +
> a'*b) ?
>
It is non-trivial. The following is copied from Cohen and
Erlich and
I'm too lazy to change the notation.
Suppose (m, n)R(m', n') and (p, 
q)R(p', q').
Since m + n' = m' + n,
(m*p + n*q) + (m'*q + n'*p) = (m + 
n')*p + (n + m')*q = (m +
n')*(p +q)
and
(m'*p + n'*q) + (m*q + n*p) = 
(m' + n)*p + (n' + m)*q =
(m + n')*(p + q).
It follows that (m*p + n*q, m*q + n*p)R(m'*p + 
n'*q, m'*q +
n'*p)
Similarly, since p + q' = p' + q,
(m'*p + n'*q) + 
(m'*q' + n'*p') = 
m'*(p + q') + n'*(q + 
p') =
(m' + n')*(p + q')
and
(m'*p' + n'*q') 
+ (m'*q + n'*p) = 
m'*(p' + q) + n'*(p + 
q') =
(m' + n')*(p + q')
It follows that
(m'*p + n'*q, m'*q + 
n'*p)R(m'*p' + 
n'*q', m'*q' + 
n'*p').
So, (m*p + n*q, m*q + n*p)R(m'*p' + 
n'*q', m'*q' + 
n'*p').
This holds in N. Hope there are no typos.
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: Skewed mean value, skew-induced partial order?
>Given natural numbers i and j such that i <= j
>let k be maximal integer satisfying i*2^k <= j
>Is there a term for i*2^k value? If there isn't, then, any
>suggestions? (Ariphmetic-geometric mean is already taken:-(
Actually, it's arithmetic-geometric mean that is taken.
Ariphmetic should be available for your use (although it
does get 135 hits in Google).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: f(n^3 + m^3) = f(n)^3 + f(m)^3
>It is known that if a map f from N to N satisfies f(1) > 0 
and
>f(n^2 + m^2) = f(n)^2 + f(m)^2
>for all n, m in N, then f is the identity map.
>(Here N is the set of natural numbers including 0.)
>What happens if we consider cubes instead of squares...?
>And in case the conditions become too weak, let us extend
the domain and
range
>to all the integers to make the conditions a little bit
stronger.
>So, to be exact, if a map f from Z to Z satisfies f(1) > 0 
and
>f(n^3 + m^3) = f(n)^3 + f(m)^3
>for all n, m in Z, then can we conclude that f is the
identity map?
>TIA.
>Tad
As a start, set m = n = 0 to get f(0) = 2*f(0)^3 and f(0) = 0,
Set m = 0 and n = 1 to get f(1) = f(1)^3 and hence f(1) = 1
(since f(1) >
0).
If f(m) = m, then f(-m) = -m since we can let n = -m.
It is now easy to check that f(m) = m for
m = 0, 1, 2, 7, 8, 9, 16
and their negatives.
Using 9^3 - 6^3 = 513 = 8^3 + 1^3, we can add m = 6 to this
list.
Using 4104 = 2^3 + 16^3 = 9^3 + 15^3, we can add m = 15 to
this list.
Can anyone extend this table through 20?
You may find it convenient to use 6^3 = 3^3 + 4^3 + 5^3
and 9^3 + 10^3 = 1^3 + 12^3 and 4104 = 18^3 - 12^3.
--
Wan: Experts at choosing the best of 100+ applicants for a
position.
Register as a California voter by September 22, and vote on
October 7.
Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California
Microsoft Research and CWI
===
Subject: Re: f(n^3 + m^3) = f(n)^3 + f(m)^3
Thank you for your reply.
The nine numbers you lis are just the same as what I got at
first.
But now (I think) I have proved
that f(n) = n for all integers n whose prime divisors are <=
13,
and that if the following conjecture is true for all primes
up to, say p0,
then f(n) = n for all integers n whose prime divisors are <=
p0.
Conjecture. For every prime number p >= 17
there exist integers x, y, z whose prime divisors are less
than p such that
(*) p^3 = x^3 + y^3 + z^3.
I have verified this for p < 10000 with my computer.
The outline of the proof is as follows.
Let S denote the set of integers n such that f(n) = n for
convenience.
And for a set P of primes
let Z(P) denote the set of integers whose prime divisors are
contained in
P.
The main course of the proof is to extend P step by step
for which Z(P) is contained in S.
We first have to show that Z(2,3) is in S.
The first breakthrough is to determine all the integer
solutions of the
equation
X^3 + Y^3 = c
for a given integer c, for which we have an efficient upper
bound.
For example, by listing all the solutions for c = 728 and
-4104,
we get f(12) = 12.
Then we can show inductively that Z(2,3) is in S, using 1^3 +
6^3 + 8^3 =
9^3.
Now we can add 5 to the list of primes using 5^3 = 6^3 - 3^3
- 4^3,
and also 11 by making a detour somewhat.
The cases p = 7 and 13 are the most challenging
because the equation (*) has no solutions satisfying gcd(xyz,
p) = 1.
(And this is because the group ({F_p}^*)^3 has order 2 and 4
respectively.)
But I believe that I have solved these cases by considering
p^2 instead of
p.
So, if all the stuff above is correct,
the problem is reduced to the above conjecture.
Thanks.
Tad
===
Subject: Obfuscation Numbers
It muses to me to think how all branches of science and
mathematics
start from simple observations or notions about the world
around us,
and evolve (and sometimes seemingly uneedlessly) a higher
level of
complexity. Usually there is also a counter revolution
happening
whereby at some point of criticality, a swing back to a more
simpler,
albeit elegant understanding of the universe occurs. That has
happened
numerous times in the pusuit of knowledge, and usually at a
time when
a deeper insight into the workings of the world around is
discovered.
I am no slacker in math having acquired a distinction in
university
level mathematics, but that was a long time ago and there is
a lot of
new math out there that bamboozles me... including its lingo.
I just
found it too complex. Alas I have long given up it in my
pursuits of
fully understanding it. There is only so many brain cells one
can
devote... as well as time.
So to leave my mark before the great goodbye, I hearby add my
observations to the universe around us, and let posterity
decide if I
am an agent of increasing complexity or a
counter-revolutionary who
just obtained a deeper insight... or just plain nuts.
I am proposing a new number system derived from complex
numbers as
complex numbers were once derived from real numbers to help
understand
certain classes of problems. I call these new numbers
obfuscation
numbers to be distinct from complex numbers.
Some background:
The complex number i is defined as the positive square-root of
-1.
This simple concept helped to identify roots of polynomials
that were
in a number of differing fields of human endeavours... though
electrical engineers still prefer to use j instead of i. I
suppose
accountants and economists still haven't caught on to 
them...
but time
will tell... especially when I get a please explain by the tax
departments when they realise I declared my income in complex
number
notation.
The observation:
the positive square-root of -1 is only one possible root
wereas there
is also a negative square-root of -1 which until now has been
omit
from official parlance. One view of the complex number i, and
the
complex-plane is that it defines right-handed rotational
schema.
Taking an idea out of quantum theory in particualr the
supposition of
states concept, we observe that a given any rotational schema
can be
considered to contain both a positive rotation as well as a
negative
rotation in supposition.
So why not incorporate that idea into the basic math.
The idea:
I define i to be the positive square-root of -1, and j to be
the
negative square-root of -1. The i-plane is a right handed
(R.H.)
rotational schema, whereas the j-plan is a left handed (L.H.)
rotational schema. The idea is that a supposition of these
elements
defines a whole new number, and represents any rotational
schema. I
call these numbers obfuscation numbers.
Examples:
Examples of obfuscation numbers are:
1 + 2i + 3j
4 + i + 4j
2j etc
Identities:
i = sqrt(-1)
j = - sqrt(-1)
i = -j
j = -i
i*j = 1
i*i = -1
j*j = -1
So you may start to wonder, what a complete waste of my time!
Well
have you considered what the benefits of obfuscation numbers
really
are? Here is a short list:
a) It rounds off the untidy definition of complex numbers
whereby only
the positive squae-root of a number is represen and not the
negative root.
b) It eliminates the need to use a +/- symbol, such that
consider
expressions like sqrt(-4) = +/-2i can instead be represen as
2i+2j.
c) It provides a deeper understanding of the geometrical
implications
of certain operations, for example a supposition of R.H.
rotations and
L.H. rotations. Such a system would have benefits in safe
cracking or
quantum theory.
d) Most of your colleagues would have no idea what you are
talking
about, thereby raising an air of mystery about you. Maybe
there is a
new kind of math that they need to know about and you already
are
ahead of the game.
Yes there have been attempts in the past to design ternary
number
systems (I think called ternions), but mathematicians for
good reason
moved on to quaternions. That is a whole other kettle of fish.
I do
not seek here a higher math, I just seek to complicate the
lower math
uneedlessly, or simplify it deliberately.
A timely warning: Do not consider obfuscation numbers as
having a
universal application to mathematics and sciences as it is
only
designed to help explain a limi class of problems... Why
there is
so little in good math humour?
===
Subject: Re: Obfuscation Numbers
> It muses to me to think how all branches of science and
mathematics
> start from simple observations or notions about the world
around us,
> and evolve (and sometimes seemingly uneedlessly) a higher
level of
> complexity. Usually there is also a counter revolution
happening
> whereby at some point of criticality, a swing back to a
more simpler,
> albeit elegant understanding of the universe occurs. That
has happened
> numerous times in the pusuit of knowledge, and usually at a
time when
> a deeper insight into the workings of the world around is
discovered.
I am no slacker in math having acquired a distinction in
university
> level mathematics,
Careful. Item #10 on Baez's crackpot index is, 10 points for
pointing
out that you have gone to school, as if this were evidence of
sanity.
but that was a long time ago and there is a lot of
> new math out there that bamboozles me... including its
lingo. I just
> found it too complex. Alas I have long given up it in my
pursuits of
> fully understanding it. There is only so many brain cells
one can
> devote... as well as time.
So to leave my mark before the great goodbye,
Even a dog can leave his mark.
I hearby add my
> observations to the universe around us, and let posterity
decide if I
> am an agent of increasing complexity or a
counter-revolutionary who
> just obtained a deeper insight... or just plain nuts.
>
Well you're rackin' up the crank points by the 
paragraph here.
> I am proposing a new number system derived from complex
numbers as
> complex numbers were once derived from real numbers
I hope you know about the fundamental theorem of algebra,
which says you
don't need to go beyond complex numbers to solve 
polynomials.
to help understand
> certain classes of problems. I call these new numbers
obfuscation
> numbers to be distinct from complex numbers.
Some background:
The complex number i is defined as the positive square-root of
-1.
That's already not true. There is no way to 
define the positive
square root of -1. You can't distinguish i from -i as you 
can
1 from
-1.
> This simple concept helped to identify roots of polynomials
that were
> in a number of differing fields of human endeavours... 
though
> electrical engineers still prefer to use j instead of i.
Shows what the engineers know.
I suppose
> accountants and economists still haven't caught on to
them...
Government economists use imaginary figures all the time!
but time
> will tell... especially when I get a please explain by the
tax
> departments when they realise I declared my income in
complex number
> notation.
The observation:
the positive square-root of -1 is only one possible root
wereas there
> is also a negative square-root of -1 which until now has
been omit
> from official parlance.
Well you're in trouble right here, as I poin out. Why do you
think
everyone's overlooked it all this time?
One view of the complex number i, and the
> complex-plane is that it defines right-handed rotational
schema.
Taking an idea out of quantum theory in particualr the
supposition of
> states concept, we observe that a given any rotational
schema can be
> considered to contain both a positive rotation as well as a
negative
> rotation in supposition.
So why not incorporate that idea into the basic math.
The idea:
I define i to be the positive square-root of -1, and j to be
the
> negative square-root of -1. The i-plane is a right handed
(R.H.)
> rotational schema, whereas the j-plan is a left handed
(L.H.)
> rotational schema. The idea is that a supposition of these
elements
> defines a whole new number, and represents any rotational
schema. I
> call these numbers obfuscation numbers.
Examples:
Examples of obfuscation numbers are:
1 + 2i + 3j
> 4 + i + 4j
> 2j etc
Identities:
i = sqrt(-1)
> j = - sqrt(-1)
> i = -j
> j = -i
> i*j = 1
> i*i = -1
> j*j = -1
>
It seems like j is just what we normally call -i, or the
square root of
-1 at the point (0, -1) on the Argand plane. But note that
there's no
real difference between i and -1. If we named them the other
way round,
nothing in mathematics would change.
> So you may start to wonder, what a complete waste of my
time! Well
> have you considered what the benefits of obfuscation numbers
really
> are? Here is a short list:
a) It rounds off the untidy definition of complex numbers
whereby only
> the positive squae-root of a number is represen and not the
> negative root.
> b) It eliminates the need to use a +/- symbol, such that
consider
> expressions like sqrt(-4) = +/-2i can instead be represen
as 2i+2j.
> c) It provides a deeper understanding of the geometrical
implications
> of certain operations, for example a supposition of R.H.
rotations and
> L.H. rotations. Such a system would have benefits in safe
cracking or
> quantum theory.
> d) Most of your colleagues would have no idea what you are
talking
> about, thereby raising an air of mystery about you. Maybe
there is a
> new kind of math that they need to know about and you
already are
> ahead of the game.
>
I agree that (d) would have the potential to confuse or
impress some of
your more slow-wit colleagues, depending on what field your
in. For
example you're not going to get very far with this stuff on
sci.math.
> Yes there have been attempts in the past to design ternary
number
> systems (I think called ternions), but mathematicians for
good reason
> moved on to quaternions.
Do you know what that good reason is?
That is a whole other kettle of fish. I do
> not seek here a higher math, I just seek to complicate the
lower math
> uneedlessly, or simplify it deliberately.
>
Ah, you mean you're a troll looking for attention. But
clearly not a
crank, since you don't actually believe your own stuff.
> A timely warning: Do not consider obfuscation numbers as
having a
> universal application to mathematics and sciences as it is
only
> designed to help explain a limi class of problems... Why
there is
> so little in good math humour?
There's plenty of good math humor, of which your post was 
not
an
example. Here's a math humor I happen to know.
There was this priest in 18th century France who tried to
reform the
prostitutes of Paris by teaching them analytic geometry. He
was the
first person to put Descartes before the whores.
I think my story's funnier than yours.
===
Subject: Re: Basic calculus questions
> dy/dx is a ratio. To isolate dy, you must multiply by dx to
create 1.
You
> have an equation. So keep it balanced by mult. 3x by dx
also.
I wouldn't consider this a basic calculus problem. I would
consider
this
> more of an algebra problem.
I don't think so. dy/dx is not realy a ratio, but just a
symbol that
means derivative. See my first post.
Artur
--
tml
===
Subject: Re: Basic calculus questions
> dy/dx is a ratio. To isolate dy, you must multiply by dx to
create 1.
You
> have an equation. So keep it balanced by mult. 3x by dx
also.
I wouldn't consider this a basic calculus problem. I would
consider
this
> more of an algebra problem.
> I don't think so. dy/dx is not realy a ratio, but just a
symbol that
> means derivative. See my first post.
> Artur
dy/dx is defined as a ratio.
it turns out that it is the same as F'(x)
Advancd Calculus p 60
Watson Fulks
Wiley and Co 1961
( oldie, but goodie)
RJ Pease
--
tml
===
Subject: Re: Basic calculus questions
> dy/dx is defined as a ratio.
Not in my experience. It is notation signifying the limit of
a certain
ratio. The notation is to be understood as a totality. Any
explanation of
what dy and dx signify as separate entities is likely to be
predica on
nonsense at the level we're dealing with here (beginning
calculus, k12.ed).
We've seen some of this nonsense in this thread. There may 
be
convenience
in pushing differentials (whatever they are) around, but I
think that is
best explained as a heuristic device. And Steven Herschkorn
is correct:
There are settings in which the equation dy = 3xdx can be
made precise, but
that is miles beyond where we are here.
--
tml
===
Subject: Re: Basic calculus questions
> best explained as a heuristic device. And Steven Herschkorn
is correct:
> There are settings in which the equation dy = 3xdx can be
made precise,
but
> that is miles beyond where we are here.
You can think of dy as the value of a linear function that
assigns the
number 3xdx to every dx. And 3x is the value at x of the
derivative of
the original function. This is not hard to understand. But I
think
there's a problem, this notation may lead someone to think
that dy =
3x dx is the real variation of the function value with
respect to a
variation dx in the value of x, skipping the fact that dy =
3xdx +
o(dx).
Artur
--
tml
===
Subject: Re: Basic calculus questions
> best explained as a heuristic device. And Steven Herschkorn
is correct:
> There are settings in which the equation dy = 3xdx can be
made precise,
but
> that is miles beyond where we are here.
> You can think of dy as the value of a linear function that
assigns the
> number 3xdx to every dx. And 3x is the value at x of the
derivative of
> the original function. This is not hard to understand. But
I think
> there's a problem, this notation may lead someone to think
that dy =
> 3x dx is the real variation of the function value with
respect to a
> variation dx in the value of x, skipping the fact that dy =
3xdx +
> o(dx).
> Artur
The thread is getting sidetracked into pedagogical approaches.
As a result, the BASIC question is being ignored.
Even if you don't define the symbol dy/dx as a 
ratio but
simply as a
substitute symbol for f'(x) , WHAT MAKES IT OK to assume 
that
this symbol
is
suddenly an algebra expression which can be c multiplied??
Maybe , as others have sugges, skipping the text that covers
the
material, muttering words about pedants, and simply tell the
kids to take
it
on PHAITH will work.
Actually, cynicism aside (not) , That may indeed be the best
way to work
with kids of sixth grade Math ability whose parents have
buffaloed their
way
into a Calculus class in High school.
Bob Pease
--
tml
===
Subject: Re: Basic calculus questions
> dy/dx is a ratio. To isolate dy, you must multiply by dx to
create 1.
> You
> have an equation. So keep it balanced by mult. 3x by dx
also.
I wouldn't consider this a basic calculus problem. I would
consider
> this
> more of an algebra problem.
I don't think so. dy/dx is not realy a ratio, but just a
symbol that
> means derivative. See my first post.
> Artur
dy/dx is defined as a ratio.
> it turns out that it is the same as F'(x)
>
If it's a ratio, what's it the ratio of? 
There's really no
calculus
book that makes this legit.
--
tml
===
Subject: Re: Basic calculus questions
>If it's a ratio, what's it the ratio of? 
There's really no
calculus
>book that makes this legit.
A strange pronouncement. How many calculus books have you
read?
You might start with Stewart, Single-Variable Calculus with
Early
Transcendentals, 4th edition, section 3.11; or Adams,
Calculus,
a Complete Course, 3rd edition, section 2.2.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
--
tml
===
Subject: Re: Basic calculus questions
>If it's a ratio, what's it the ratio of? 
There's really no
calculus
>book that makes this legit.
A strange pronouncement. How many calculus books have you
read?
In my case, way too many.
> You might start with Stewart, Single-Variable Calculus with
Early
> Transcendentals, 4th edition, section 3.11; or Adams,
Calculus,
> a Complete Course, 3rd edition, section 2.2.
Can you tell us brießy how these texts define dy/dx as a
ratio? What is it
a ratio of? Do they actually define the notation this way? 
I'm
familiar
with attempts in some texts to make sense of dy and dx from
the point of
view of linear differential transformations, but this occurs
after
derivatives and the notation dy/dx have already been defined.
After
slogging through this pedagogical debacle one can say voila,
dy/dx can be
viewed as a ratio of dy and dx. I always skipped that section
in my
classes. Is there something better that I've missed?
--
tml
===
Subject: Re: Basic calculus questions
>> You might start with Stewart, Single-Variable Calculus
with Early
>> Transcendentals, 4th edition, section 3.11; or Adams,
Calculus,
>> a Complete Course, 3rd edition, section 2.2.
>Can you tell us brießy how these texts define dy/dx as a
ratio? What is it
>a ratio of? Do they actually define the notation this way?
I'm familiar
>with attempts in some texts to make sense of dy and dx from
the point of
>view of linear differential transformations, but this occurs
after
>derivatives and the notation dy/dx have already been defined.
After
>slogging through this pedagogical debacle one can say voila,
dy/dx can be
>viewed as a ratio of dy and dx. I always skipped that
section in my
>classes. Is there something better that I've missed?
Of course the derivative is defined first. But 
then you define
dx as
a new variable (independent of x) and dy as f'(x) dx. You 
can
interpret
dx as representing horizontal displacement along the tangent
line to the curve at x, and dy as representing vertical
displacement
along this tangent line.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
--
tml
===
Subject: Re: Basic calculus questions
>Can you tell us brießy how these texts define dy/dx as a
ratio? What is
it
>a ratio of? Do they actually define the notation this way?
I'm familiar
>with attempts in some texts to make sense of dy and dx from
the point of
>view of linear differential transformations, but this occurs
after
>derivatives and the notation dy/dx have already been defined.
After
>slogging through this pedagogical debacle one can say voila,
dy/dx can be
>viewed as a ratio of dy and dx. I always skipped that
section in my
>classes. Is there something better that I've missed?
Of course the derivative is defined first.
not, and I thought that's what 
fishfry's comment was about. So
I was
surprised that you kind of jumped on him.
> But then you define dx as
> a new variable (independent of x) and dy as f'(x) dx. You
can interpret
> dx as representing horizontal displacement along the tangent
> line to the curve at x, and dy as representing vertical
displacement
> along this tangent line.
Right, that's the debacle I was talking about. dy is really 
a
one parameter
family of linear functions, so to make things clear we should
write dy_x.
For x in the domain of the differentiable function y, dy_x :
R -> R is
defined by dy_x(dx) = f'(x)*dx for all dx in R. 
And as long as
dx is not 0,
we have dy_x(dx)/dx = f'(x). Mission accomplished. Anyone
awake?
Not only is this confusing in the extreme to anyone learning
calculus for
the first time (and that's the level 
I've been addressing), it
contradicts
the heuristic that dy and dx are infintely small, which is the
way our
mathematical fathers thought about these symbols. It's 
really
wonderful
notation from the old point of view. By the way, what is the
notation
int_[a,b] f(x)dx supposed to suggest if dx is a new variable
independent of
x that can take on any real value?
--
tml
===
Subject: Bored Uber Trolls: what to do
Subject: Bored Uber Trolls what to do
> Let's say I'm a Uber Troll -
> I don't know about anyone else but I reply to Trackers
message because
> she's one fine ass biatch.
Just kidding.
anyway, If you don't like it don't read 
usenet, I mean don't
like it
> don't use the internet you biatch, send back your aol disk
up your ass
> you faggeatot.
> So you agree there isn't actually anyone replying to
trackers posts -
> it's ass sock puppetry
Well whatever I told you why I respond to trackers posts, she
a fine
ass biatch.
I told you why most people resond to trackers posts, with
dick in hand
and spewing gism on the keyboard looking around for a tissue.
If u
can't deal with it dude, i put u along with everyone else
that says,
killfile tracker.
The only people that killfile tracker are the imature people.
The
mature people know that every tracker post must be responded
to and
disussed ad nauseum.
u may think u have a different way, u dumbass fool - it wont
work.
===
Subject: Re: Fake JSH
This guy's a fake. The real JSH doesn't post 
from yahoo.
===
Subject: Re: Fake JSH
This guy's a fake. The real JSH doesn't post 
from yahoo.
Hehe. I took a break from this news group for maybe 3-4
years... really, is
JSH still around? :)))
/
===
Subject: Re: Fake JSH
>>This guy's a fake. The real JSH doesn't post 
from yahoo.
> Hehe. I took a break from this news group for maybe 3-4
years... really,
is JSH still around? :)))
/
> My original plan was to work out some great math results
on-line
> trusting that when I found that great math, someone in the
world would
> recognize it.
[snip usual rubbish]
> Currently my proof of Fermat's Last Theorem, my prime
counting
> function, and my proof of a ßaw in taught mathematics are
with me,
> and I will use the regular channels to get proper
recognition.
For me, the Internet and Usenet were failures for
recognition, but
> quite useful for the brainstorming process.
Thanks to those who participa, even some of you rude and
obnoxious
> people.
The only ones who really bothered me were the incessant liars.
Now I'll use other means to get things done. This experiment
is over.
>
===
Subject: Re: Random Walk result request.
> I think that this approach to a proof will not work, at
least not
> without some significant new idea added. 
Here's why:
Yes, it does seem as if Vlada Whatsit's proof is irreparably
broken,
which is a pity because it was a slick attack of just the
sort I was
hoping for.
Nevertheless, Don Coppersmith's approach was almost as 
slick,
and does
give an easy-ish proof as well. Don proposed a function f with
f(0,0) = 1; f(x,y) --> 0 as either x/y --> +/- oo; and
f(x,y) = [ f(x+1,y) + f(x-1,y) + f(x,y+1) + f(x,y-1) ] / 4
otherwise.
He then slickly overlaid the function with its
shift-one-left, did some
basic algebra and the result dropped out very neatly.
Unfortunately the proof as it stands is defective, because
there IS no such
function f; but that doesn't really matter, as there are
arbitrarily close
approximations. Just put up a barrier of boundary condition f
= 0
for x/y = +/-N for larger and larger values of N. Though the
f function
will continue to slowly change with N, the (f - shift_f)
function will
not;
it will stabilize due to the fact that the probability of
return to one
of the two centre states before hitting this N boundary tends
to 1;
by the basic recurrence result for 2 dimensions.
So that is very nice. Thanks Don! I don't know exactly what 
a
response
is, but these engineering-style proofs are not uncommon in
this area.
Nice!
Greg goes on:
> Thus the expec number of visits to any point
> before returning to (0,0) is 1!
Yes, that is very surprising at first, though 
Greg's brief
intuitive
explanation is quite revealing. This result can be proved in
various ways,
and applies to more general (even non-Markovian) situations
of the same
type.
The only requirement is transitional symmetry between the two
points.
Here is a very rough proof outline, which will do for this
case (among
others).
We look at visits to a point A and a point B, and ignore
things in between.
The successive visits to the two points will be something like
...AAABBABABABBBBBBAABAAAAABBBBBBBABAABABABBBAB...
with the expec number of visits to A between B's,
i.e. the mean A-block length being a ; and similarly b for
the B's.
And in a very long stretch of length L, there will be an
expec number
of A occurrences, say e ; and similarly f the expec number of
B's.
Then (taking expec values of expec values - quite legit) we
get
e = f.a so a = e/f. Similarly b = f/e. Thus ab = 1.
But points A and B are exchangeable, so a = b; thus a = b =
1. QED.
Sweet!
-------------------------------------------------------------
---------------
--
Bill Taylor W.Taylor@math.canterbury.ac.nz
-------------------------------------------------------------
---------------
--
Every problem has at least one solution which is elegant,
neat - and
wrong.
-------------------------------------------------------------
---------------
--
===
Subject: Re: Random Walk result request.
I think that this approach to a proof will not work, at least
not
> without some significant new idea added. 
Here's why:
Yes, it does seem as if Vlada Whatsit's proof is irreparably
broken,
> which is a pity because it was a slick attack of just the
sort I was
> hoping for.
Nevertheless, Don Coppersmith's approach was almost as 
slick,
and does
> give an easy-ish proof as well. Don proposed a function f
with
f(0,0) = 1; f(x,y) --> 0 as either x/y --> +/- oo; and
f(x,y) = [ f(x+1,y) + f(x-1,y) + f(x,y+1) + f(x,y-1) ] / 4
otherwise.
He then slickly overlaid the function with its
shift-one-left, did some
> basic algebra and the result dropped out very neatly.
Unfortunately the proof as it stands is defective, because
there IS no
such
> function f; but that doesn't really matter, as there are
arbitrarily
close
> approximations. Just put up a barrier of boundary condition
f = 0
> for x/y = +/-N for larger and larger values of N. Though
the f function
> will continue to slowly change with N, the (f - shift_f)
function will
not;
> it will stabilize due to the fact that the probability of
return to one
> of the two centre states before hitting this N boundary
tends to 1;
> by the basic recurrence result for 2 dimensions.
So that is very nice. Thanks Don! I don't know exactly what 
a
response
> is, but these engineering-style proofs are not uncommon in
this area.
Nice!
That's because responses have nothing to do with 
engineering.
They come from Psychology, as just about all arbitrarly close
things do. They're a design specification, not a 
design result.
===
Subject: Re: Question about triangle puzzle (??)
: There are many puzzles of this type. But why do these
puzzles always
seem
: to have dimensions which involve Fibonacci numbers? In this
case, the
large
: triange is 13 x 5, and the smaller one is 8 x 3.
For this type of puzzle, you need two rational numbers that
are close
together. The simplest way to do that is to take consecutive
convergents of
some continued fraction. The continued fraction whose
convergents are the
smallest is [1,1,1,...] which gives the Fibonacci ratio
convergents
1, 2/1, 3/2, 5/3, 8/5, ...
===
Subject: Re: Question about triangle puzzle (??)
Mark Nudelman
> Charlie Johnson
> Hi all,
I have a picture of two triangles here :
> http://home.mindspring.com/~cj-bubba/
My question is: shouldn't the sum of the areas of the
polygons equal
the
> area of the triangle. In the second triangle one unit has
to be added
for
> its area to equal the first. Why is that so?
> There are many puzzles of this type. But why do these
puzzles always
seem
> to have dimensions which involve Fibonacci numbers? In this
case, the
large
> triange is 13 x 5, and the smaller one is 8 x 3.
>
Well, for consecutive Fibo numbers a,b,c,d we have
|ac-bb|=|ad-bc|=1, so
that e.g.
a/c - b/d is small.
I can't swear to it, but I think the above triangle trick
goes back to the
brilliant puzzlist H. E. Dudeney (1857-1930) who has numerous
others about
cutting up and rearranging polygons.
LH
===
Subject: Fake JSH
> This guy's a fake. The real JSH doesn't post 
from yahoo.
FISHFRY!!! what's up. I got that song you wan, and those
lyrics!
The Red Hot Chilli Peps song starts
And I'm sailing!
and I got 4 u that Kings X song you wan Over my Head man 
let's
get 2gether the RIAA.
Ken
PS good on U pointing out that SNOT the real JSH. U're a
genious.
===
Subject: Re: Calculus Guidence
>Subject: Calculus Guidance
>Hi. I'm having some big problems with calculus with these
problems
>based on proofs. I was hoping the people in this group can
help guide
>me towards the solution. :)
>The -> means vector.
>Show that if ->a and ->b have the same direction, then:
> ||a+b|| = ||a|| + ||b||
>I can see this clearly with from a graphical drawing, but I
have no
>clue how to prove it mathematically.
Well, I guess you can't ask anyone to prove it 
mathematically
unless you define what ||x|| means. But... I think we know
what you mean so...If (using your notation):
->a = (a1,a2,a3)
->b = (b1,b2,b3)
then
->(a+b) = (a1+b1,a2+b2,a3+b3)
||(a+b)||^2 = (a1+b1)^2 + (a2+b2)^2 + (a3+b3)^2
= a1^2 +b1^2 + 2a1b2 + a2^2 + b2^2 + 2a2b2 + a3^2 + b3^2 +
2a3b3
= (a1^2 + a2^2 + a3^2) + (b1^2 b2^2 +b3^2) + 2(a1b1 +a2b2 +
a3b3)
= ||a||^2 + ||b||^2 + 2<a,b>
if ->a and ->b are in same direction
<a,b> = ||a|| ||b||
so:
||(a+b)||^2 = (||a|| + ||b||)^2
and then this means (since ||whatever|| is postive)
||(a+b)|| = ||a|| + ||b||
cheeeeeeerio.
adam
P.S. This isn't a Calculus question, but whatever
===
Subject: Re: infinity dimensional vector space
>Can you prove that the vector space of real-valued function
over a
>differentiable manifold M is infinity-dimensional
>>Suggestion: if n is the dimension of M, then start by
proving that
>>the vector space of real-valued function over an open
subset of R^n
>>is infinity-dimensional. Since there is an obvious injective
linear
>>function from such a space into your space...
> And that obvious injective linear function is *what*,
precisely?
> ...Oh, ah. I see that I missed the fact (also in my direct
> reply to tern) that tern didn't specify *differentiable*
function.
> Without any such restriction, I agree that there *is* an
obvious
> injection.
First of all, I must say that I forgot that n might be equal
to 0.
But for any other n it is rather easy to prove that the space
of
all differentiable functions from M to the reals is infinite-
dimensional. Consider some chart c: U -> V, where U is an open
subset of M and V is an open subset of R^n. The space F of all
real-valued differentiable functions on V with compact support
is infinite-dimensional. Then you associate to any function
f in F the function f':M -> R such that f' is 
0 outside U and
that the restriction of f' to U is equal to (f o c). This
association is an injective linear map.
As you can see, I do not need partitions of unity or some
extension theorem.
Best regards
===
Subject: Re: infinity dimensional vector space
>First of all, I must say that I forgot that n might be equal
to 0.
>But for any other n it is rather easy to prove that the
space of
>all differentiable functions from M to the reals is infinite-
>dimensional. Consider some chart c: U -> V, where U is an
open
>subset of M and V is an open subset of R^n. The space F of
all
>real-valued differentiable functions on V with compact
support
>is infinite-dimensional. Then you associate to any function
>f in F the function f':M -> R such that f' is 
0 outside U and
>that the restriction of f' to U is equal to (f o c). This
>association is an injective linear map.
>As you can see, I do not need partitions of unity or some
>extension theorem.
How do you prove that The space F of all real-valued
differentiable functions on V with compact support is
infinite-dimensional (in fact, bigger than 0) without
some extension theorem? Sure, it's a trivial extension
theorem, as such things go; but given the degree of naivety
shown so far by the original poster, I wouldn't bet he knows
how to prove it (or even that it needs to be proved).
I do, of course, agree that you don't need anything *big*
in the way of extension theorems. And my reference to
partitions of unity was somewhat muddle-headed.
Lee Rudolph
===
Subject: Re: infinity dimensional vector space
> How do you prove that The space F of all real-valued
> differentiable functions on V with compact support is
> infinite-dimensional (in fact, bigger than 0) without
> some extension theorem? Sure, it's a trivial extension
> theorem, as such things go; but given the degree of naivety
> shown so far by the original poster, I wouldn't bet he 
knows
> how to prove it (or even that it needs to be proved).
that what you had in mind was Tietze's extension theorem.
Now, it's clear to me what you had in mind and I agree with
you.
Best regards
===
Subject: Re: University of Minnesota math joke
> Well, I just went with communist because of his e-mail
address:
> palaste@cc.helsinki.fi
> The Finns were controlled by the Communists, even thought
they like to
say
> they were independent.
http://www.google.com/groups?selm=03676caf.3f1843fe%
40usw-ex0109-069.remarq.
com
http://www.google.com/groups?
selm=8F2C76D37dtennerameritechnet%40news.mil.ameritech.net
--
http://hertzlinger.blogspot.com
===
Subject: Polychora
While lurching about the internet, as one does, I lurched into
http://hometown.aol.com/hedrondude/polychora.html
There are some rather appealing pictures.
Don't miss
http://www.cafeshops.com/hedrondude
http://hometown.aol.com/hedrondude/news.html
http://hometown.aol.com/hedrondude/misc.html
And what about that rasdi .29 at
http://members.aol.com/hedrondude/snubs.html
??
--
Clive Tooth
http://www.clivetooth.dk
Subject: Re: Polychora
===
> While lurching about the internet, as one does, I lurched
into
> http://hometown.aol.com/hedrondude/polychora.html
There are some rather appealing pictures.
Don't miss
> http://www.cafeshops.com/hedrondude
> http://hometown.aol.com/hedrondude/news.html
> http://hometown.aol.com/hedrondude/misc.html
And what about that rasdi .29 at
> http://members.aol.com/hedrondude/snubs.html
> ??
>
I like the pics. Of course, I love fractals too.
--
===
Subject: Re: Polychora
>Subject: Polychora
>Message-id: <bk1djl$ndjie$1@ID-11651.news.uni-berlin.deWhile lurching about the internet, as one does, I lurched
into
>http://hometown.aol.com/hedrondude/polychora.html
>There are some rather appealing pictures.
>Don't miss
>http://www.cafeshops.com/hedrondude
>http://hometown.aol.com/hedrondude/news.html
>http://hometown.aol.com/hedrondude/misc.html
>And what about that rasdi .29 at
>http://members.aol.com/hedrondude/snubs.html
>??
>--
>Clive Tooth
>http://www.clivetooth.dk
I've been making those things for years. Every time I clean
off my desk and
wad
up all the Post-it notes into a ball.
--
Mensanator
2 of Clubs
http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm
===
Subject: Re: Way off topic
>Aoccdrnig to rscheearch at an Elingsh uinervtisy, it 
deosn't
mttaer in
what
>oredr the ltteers in a wrod are, the olny iprmoetnt tihng is
taht the
frist
>and lsat ltteers are in the rghit pclae. The rset can be a
toatl mses and
>you can sitll raed it wouthit a porbelm. Tihs is bcuseae we
do not raed
>ervey lteter by itslef but the wrod as a wlohe.
This is amazing !
I seem to be able to read it faster than correctly order
chars.
I aways suspec I was Ôdislexik' or something.
I'd like to read the Ôfull 
report' of the research.
== Chris Glur.
===
Subject: common prime factors
Recently, somebody on this NG wan to prove that 2^n+1 is
prime for all
nonnegative n (or something like that).
Now 3 | (2^n+1) if n is odd, so maybe (2^n+1)/3 has something
more
interesting to offer.
I noticed that, if n is odd and m | n
gcd( (2^n+1)/3, (2^m+1)/3 ) > 1
Well, at least for the numbers I checked using Dario's
elliptic curve
factorization program (http://www.alpertron.com.ar/ECM.HTM,
thanx Dario)
Can/has this be proved?
I'm not a mathematician, so please type slowly
--
Steven
(remove pants to reply by e-mail)
===
Subject: Re: common prime factors
> Recently, somebody on this NG wan to prove that 2^n+1 is
prime for all
> nonnegative n (or something like that).
2^(2^n) + 1
> Now 3 | (2^n+1) if n is odd, so maybe (2^n+1)/3 has
something more
> interesting to offer.
I noticed that, if n is odd and m | n
gcd( (2^n+1)/3, (2^m+1)/3 ) > 1
Well, at least for the numbers I checked using Dario's
elliptic curve
> factorization program (http://www.alpertron.com.ar/ECM.HTM,
thanx Dario)
> Can/has this be proved?
I'm not a mathematician, so please type slowly
Let t = n/m. Then t is odd. Therefore x^t + 1 is divisible by
x + 1.
Taking x = 2^m shows that 2^m + 1 divides 2^n + 1, and hence
the gcd
is (2^m+1)/3 itself. (This is only > 1 if m > 1 btw.)
Clearly n must be an odd prime in order for (2^n+1)/3 to be
prime
(but it's not sufficient, of course).
--
J K Haugland
http://www.neutreeko.com
===
Subject: Re: common prime factors
Steven <stevenpantsvh@pandora.be> escribi.97 en el
mensaje|nt0Z8b.19893$ir3.1323147@phobos.telenet-ops.be:
> Recently, somebody on this NG wan to prove that 2^n+1 is
prime for
> all nonnegative n (or something like that).
> Now 3 | (2^n+1) if n is odd, so maybe (2^n+1)/3 has
something more
> interesting to offer.
> I noticed that, if n is odd and m | n
> gcd( (2^n+1)/3, (2^m+1)/3 ) > 1
> Well, at least for the numbers I checked using Dario's
elliptic curve
> factorization program (http://www.alpertron.com.ar/ECM.HTM,
thanx
> Dario) Can/has this be proved?
> I'm not a mathematician, so please type slowly
If m | n, then n = k*m for any k odd integer k. Then
2^n + 1 = 2^(k*m) + 1 = (2^m)^k + 1
= (2^m + 1)((2^m)^(k-1) - (2^m)^(k-2) + ... + 1)
As 3 divides both 2^n + 1 and 2^m + 1, we conclude that (2^m
+ 1)/3
divides to (2^n + 1)/3 and
gcd((2^n + 1)/3, (2^m + 1)/3) = (2^m + 1)/3
--
Best
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: common prime factors
> Steven <stevenpantsvh@pandora.be> escribi.97 en el
> mensaje|nt0Z8b.19893$ir3.1323147@phobos.telenet-ops.be:
>> Recently, somebody on this NG wan to prove that 2^n+1 is
prime for
>> all nonnegative n (or something like that).
>> Now 3 | (2^n+1) if n is odd, so maybe (2^n+1)/3 has
something more
>> interesting to offer.
>> I noticed that, if n is odd and m | n
>> gcd( (2^n+1)/3, (2^m+1)/3 ) > 1
>> Well, at least for the numbers I checked using Dario's
elliptic curve
>> factorization program
(http://www.alpertron.com.ar/ECM.HTM, thanx
>> Dario) Can/has this be proved?
>> I'm not a mathematician, so please type slowly
> If m | n, then n = k*m for any k odd integer k. Then
> 2^n + 1 = 2^(k*m) + 1 = (2^m)^k + 1
> = (2^m + 1)((2^m)^(k-1) - (2^m)^(k-2) + ... + 1)
> As 3 divides both 2^n + 1 and 2^m + 1, we conclude that
(2^m + 1)/3
> divides to (2^n + 1)/3 and
> gcd((2^n + 1)/3, (2^m + 1)/3) = (2^m + 1)/3
Thanks Ignacio.
I didn't know about the expansion
(a^n+1) = (a+1) * (a^(n-1) - a^(n-2) + ... + 1)
but now I see it, it's obvious.
===
Subject: Re: common prime factors
Try 2^3 + 1 = 9. Not prime.
Lurch
> Recently, somebody on this NG wan to prove that 2^n+1 is
prime for all
> nonnegative n (or something like that).
> Now 3 | (2^n+1) if n is odd, so maybe (2^n+1)/3 has
something more
> interesting to offer.
> I noticed that, if n is odd and m | n
> gcd( (2^n+1)/3, (2^m+1)/3 ) > 1
> Well, at least for the numbers I checked using Dario's
elliptic curve
> factorization program (http://www.alpertron.com.ar/ECM.HTM,
thanx Dario)
> Can/has this be proved?
> I'm not a mathematician, so please type slowly
> --
> Steven
> (remove pants to reply by e-mail)
===
Subject: Re: common prime factors
Originator: jgamble@ripco.com (John M. Gamble)
[Jeopardy-style response repaired]
>> Recently, somebody on this NG wan to prove that 2^n+1 is
prime for
all
>> nonnegative n (or something like that).
>> Now 3 | (2^n+1) if n is odd, so maybe (2^n+1)/3 has
something more
>> interesting to offer.
>> I noticed that, if n is odd and m | n
>> gcd( (2^n+1)/3, (2^m+1)/3 ) > 1
>> Well, at least for the numbers I checked using Dario's
elliptic curve
>> factorization program
(http://www.alpertron.com.ar/ECM.HTM, thanx Dario)
>> Can/has this be proved?
>> I'm not a mathematician, so please type slowly
>> --
>> Steven
>> (remove pants to reply by e-mail)
>Try 2^3 + 1 = 9. Not prime.
I have a couple of C header files, hiprime.h and loprime.h,
which simply
contain the highest primes less than 2**N and the lowest
primes greater
than or equal to 2**N. Obviously, the contents of loprime.h
are
relevant here:
[Just a section of the file included here:]
#define LPRIME_01 0x2 // pn(1)
#define LPRIME_02 0x5 // pn(3)
#define LPRIME_03 0xb // pn(5)
#define LPRIME_04 0x11 // pn(7)
#define LPRIME_05 0x25 // pn(12)
#define LPRIME_06 0x43 // pn(19)
#define LPRIME_07 0x83 // pn(32)
#define LPRIME_08 0x101 // pn(55)
#define LPRIME_09 0x209 // pn(98)
#define LPRIME_10 0x407 // pn(173)
#define LPRIME_11 0x805 // pn(310)
#define LPRIME_12 0x1003 // pn(565)
#define LPRIME_13 0x2011 // pn(1,029)
#define LPRIME_14 0x401b // pn(1,901)
#define LPRIME_15 0x8003 // pn(3,513)
As Lurch poin out, the lowest prime greater than 2**3 is 11
(0xb in
hexadecimal), and all the others that don't end in 1 also
fail the
2**N + 1 formula.
--
-john
February 28 1997: Last day libraries could order catalogue
cards
from the Library of Congress.
===
Subject: Re: common prime factors
> Try 2^3 + 1 = 9. Not prime.
> Lurch
Did I claim it's prime?
Don't think so
Did you read beyond the second line?
Don't think so
Did you bottom post?
Certainly not
>> Recently, somebody on this NG wan to prove that 2^n+1 is
prime
>> for all nonnegative n (or something like that).
>> Now 3 | (2^n+1) if n is odd, so maybe (2^n+1)/3 has
something more
>> interesting to offer.
>> I noticed that, if n is odd and m | n
>> gcd( (2^n+1)/3, (2^m+1)/3 ) > 1
>> Well, at least for the numbers I checked using Dario's
elliptic curve
>> factorization program
(http://www.alpertron.com.ar/ECM.HTM, thanx
>> Dario) Can/has this be proved?
>> I'm not a mathematician, so please type slowly
>> --
>> Steven
>> (remove pants to reply by e-mail)
===
Subject: Re: common prime factors
Originator: jgamble@ripco.com (John M. Gamble)
>> Try 2^3 + 1 = 9. Not prime.
>> Lurch
>Did I claim it's prime?
> Don't think so
>Did you read beyond the second line?
> Don't think so
>Did you bottom post?
> Certainly not
Erm, i was also faked out by the misleading first sentence.
Maybe
you should learn to use paragraphs more clearly.
Also, despite your complaints about top-posting, you didn't 
fix
the post.
--
-john
February 28 1997: Last day libraries could order catalogue
cards
from the Library of Congress.
===
Subject: Re: Math Question: roots of a parabola (or any
equation)
> Take the equation: x^2 - 2x -35
> This has two real solutions (where the parabola ces the x
axis)
What is the reason that solutions/roots are on the X axis.
> Why are they not on the y axis?
> Why are they not on the line of Y=1?
> You omit something important from your equation. The
equation is
> y = x^2 - 2x -35
> If you set y=0 and solve, you find out the x values for
which y=0. Any
> point with y=0 must be on the x-axis.
This may be popular mythology or complete bunk, but I have
heard that they
are called Ôroots' because the Greeks did not 
work with
negative numbers,
and so their understanding of the quadratic functions only
went to y=0,
hence the two branches of the parabola grew from it's roots.
I know
they
did not *graph* these, as that didn't happen until 
Descartes,
but they did
have a graphical understanding of functions, as shown in
their pictoral
proofs of Ôcompleting the squares', etc.
This may or may not be true, and I'm prepared to get
completely ßamed on
this in the interest of learning more about it.
--riverman
===
Subject: Re: Math Question: roots of a parabola (or any
equation)
>Take the equation: x^2 - 2x -35
>This has two real solutions (where the parabola ces the x
axis)
As the other poster commen on your earlier question, that is
not an
equation; it is a polynomial expression.
G C
===
Subject: dividing even numbers by odd numbers
Is there a formula for dividing a long series of even numbers
by a
corresponding long series of odd numbers , such as
2x4x6x810x12x14x16x18x20x22x24x26x28x30..../
1x3x5x7x9x11x13x15x17x19x21x23x25x27x29x31....
I realize that the even numbers can all be divided by 2 to get
(n/2)!x2^n/2 and that the missing even numbers in the odd
series can
be inser to yield n! if the same series is also inser above
with
the even numbers. But it still becomes a big calculation job.
Is there a simpler way of doing it and if so, where might I
read up on
it?
Thankful for any hint.
Stig Holmquist
===
Subject: Re: dividing even numbers by odd numbers
> Is there a formula for dividing a long series of even
numbers by a
> corresponding long series of odd numbers , such as
> 2x4x6x810x12x14x16x18x20x22x24x26x28x30..../
> 1x3x5x7x9x11x13x15x17x19x21x23x25x27x29x31....
I realize that the even numbers can all be divided by 2 to get
> (n/2)!x2^n/2 and that the missing even numbers in the odd
series can
> be inser to yield n! if the same series is also inser above
with
> the even numbers. But it still becomes a big calculation
job.
> Is there a simpler way of doing it and if so, where might I
read up on
> it?
> Thankful for any hint.
Stig Holmquist
2 x 4 x ... x (2n) / (1 x 3 x ... x (2n-1)), or (2^2n) (n!^2)
/ (2n)!,
is approximately sqrt(pi (n + 1/4)). HTH.
--
J K Haugland
http://www.neutreeko.com
===
Subject: Re: dividing even numbers by odd numbers
>> Is there a formula for dividing a long series of even
numbers by a
>> corresponding long series of odd numbers , such as
>> 2x4x6x810x12x14x16x18x20x22x24x26x28x30..../
>> 1x3x5x7x9x11x13x15x17x19x21x23x25x27x29x31....
I realize that the even numbers can all be divided by 2 to get
>> (n/2)!x2^n/2 and that the missing even numbers in the odd
series can
>> be inser to yield n! if the same series is also inser
above with
>> the even numbers. But it still becomes a big calculation
job.
>> Is there a simpler way of doing it and if so, where might
I read up on
>> it?
>> Thankful for any hint.
Stig Holmquist
>2 x 4 x ... x (2n) / (1 x 3 x ... x (2n-1)), or (2^2n)
(n!^2) / (2n)!,
>is approximately sqrt(pi (n + 1/4)). HTH.
I had arrived at a very similar if not identical formula but
my
handheld calculator refuses to work with numbers larger than
10^100.
Thus I'm unable to calculate any series beyond 69.
Is there another formula or a method for getting around the
limitation? I would like to test for n=1000, 10000 and 100000.
How should I go about that?
Stig Holmquist
===
Subject: Re: dividing even numbers by odd numbers
>Is there a formula for dividing a long series of even
numbers by a
>corresponding long series of odd numbers , such as
>2x4x6x810x12x14x16x18x20x22x24x26x28x30..../
>1x3x5x7x9x11x13x15x17x19x21x23x25x27x29x31....
>I realize that the even numbers can all be divided by 2 to
get
>(n/2)!x2^n/2 and that the missing even numbers in the odd
series can
>be inser to yield n! if the same series is also inser above
with
>the even numbers. But it still becomes a big calculation job.
>Is there a simpler way of doing it and if so, where might I
read up on
>it?
>Thankful for any hint.
>Stig Holmquist
>>2 x 4 x ... x (2n) / (1 x 3 x ... x (2n-1)), or (2^2n)
(n!^2) / (2n)!,
>>is approximately sqrt(pi (n + 1/4)). HTH.
I had arrived at a very similar if not identical formula but
my
> handheld calculator refuses to work with numbers larger
than 10^100.
> Thus I'm unable to calculate any series beyond 69.
Is there another formula or a method for getting around the
> limitation? I would like to test for n=1000, 10000 and
100000.
> How should I go about that?
Stig Holmquist
Don't do it as numerator/denominator.
Do (2/1)*(4/3)*(6/5)*...*(2n/(2n-1)).
Martin Cohen
===
Subject: Re: dividing even numbers by odd numbers
...
>>2 x 4 x ... x (2n) / (1 x 3 x ... x (2n-1)), or (2^2n)
(n!^2) / (2n)!,
>>is approximately sqrt(pi (n + 1/4)). HTH.
I had arrived at a very similar if not identical formula but
my
> handheld calculator refuses to work with numbers larger
than 10^100.
> Thus I'm unable to calculate any series beyond 69.
Is there another formula or a method for getting around the
> limitation? I would like to test for n=1000, 10000 and
100000.
> How should I go about that?
...
> Don't do it as numerator/denominator.
Do (2/1)*(4/3)*(6/5)*...*(2n/(2n-1)).
I think Holmquist might be asking how to evaluate
(2^2n) (n!^2) / (2n)! -- which of course can be done
in the same way as you recommend, and for small numbers
like 1000, 10000 and 100000 might as well be -- but if
you want to test with large numbeuse Stirling's
approximation to log(n!) or Gosper's form of it; see
http://mathworld.wolfram.com/StirlingsApproximation.html
where iirc they may be equal to about
(n+1/2)*log(n)-n+log(2*pi)/2 and
n*log(n)-n+log((2*n+1/3)*pi)/2 respectively.
-jiw
===
Subject: Re: dividing even numbers by odd numbers
> Is there a formula for dividing a long series of even
numbers by a
> corresponding long series of odd numbers , such as
> 2x4x6x810x12x14x16x18x20x22x24x26x28x30..../
> 1x3x5x7x9x11x13x15x17x19x21x23x25x27x29x31....
> I realize that the even numbers can all be divided by 2 to
get
> (n/2)!x2^n/2 and that the missing even numbers in the odd
series can
> be inser to yield n! if the same series is also inser above
with
> the even numbers. But it still becomes a big calculation
job.
> Is there a simpler way of doing it and if so, where might I
read up on
> it?
> Thankful for any hint.
Factors galore! 3x5x7x9x11x13x15 all divide into
6x10x14x18x22x26x30
Herc
===
Subject: A structural point of view on the Continuum and the
Discreteness
concepts
By Real Analysis the continuum is infinitely many elements
with no
gaps between them.
I think that to define the continuum by ifinitely 
many ... it is
simply a contradiction of the the original lexicographical
meaning of
the word continuum.
So, through my point of view, by forcing the Continuum to be
expressed
by its opposite proprty (infinitely many points whith no gaps
...), we
destroy our ability to understand (mathematically) the
continuum
concept.
I have found a simple and natural definitions to the
Discreteness and
the Continuum concepts, without forcing any of them to be
expressed by
the other:
A and B are sets.
q and p are members.
Option 1: q and p are members of A,
but then q is not equal to p .
Option 2: q is a member of A , p is a member of B .
D = Discreteness = q XOR p = a localized element = {.}
C = Continuum = q to p correspondence
= a non-localized element = {.___.}
I'll be glad to know what do you think
Subject: Re: A structural point of view on the Continuum and
the
Discreteness
concepts
===
> By Real Analysis the continuum is infinitely many elements
with no
> gaps between them.
I think that to define the continuum by ifinitely 
many ... it is
> simply a contradiction of the the original lexicographical
meaning of
> the word continuum.
So, through my point of view, by forcing the Continuum to be
expressed
> by its opposite proprty (infinitely many points whith no
gaps ...), we
> destroy our ability to understand (mathematically) the
continuum
> concept.
> I have found a simple and natural definitions to the
Discreteness and
> the Continuum concepts, without forcing any of them to be
expressed by
> the other:
A and B are sets.
Are these arbitrary sets or are they somehow rela to the real
numbers?
q and p are members.
Option 1: q and p are members of A,
> but then q is not equal to p .
Option 2: q is a member of A , p is a member of B .
I'm not clear why these are the only options.
D = Discreteness = q XOR p = a localized element = {.}
How are you defining q XOR p? For example, what is 5 XOR 7?
C = Continuum = q to p correspondence
> = a non-localized element = {.___.}
I don't know what you're trying to say here.
> I'll be glad to know what do you think
I think some examples would help. I can't tell that 
you've
said
anything meaningful, and you appear to be mixing logical
operators with
sets. While I'm sure they can be given meaningful 
definitions,
I'm not
certain what you have in mind.
--
===
Subject: Re: A structural point of view on the Continuum and
the
Discreteness concepts
> By Real Analysis the continuum is infinitely many elements
with no
> gaps between them.
Just like the rationals then
--
His mind has been corrup by colousounds and shapes.
The League of Gentlemen
===
Subject: Re: A structural point of view on the Continuum and
the
Discreteness concepts
By Real Analysis the continuum is infinitely many elements
with no
> gaps between them.
Just like the rationals then
Hi Robin,
above, holds as the real analysis definition of the Continuum.
Can you give me an example where some member of Q or R is NOT
connec to some point in the real-line ?
You
Doron
===
Subject: Re: A structural point of view on the Continuum and
the
Discreteness concepts
By Real Analysis the continuum is infinitely many elements
with no
>> gaps between them.
Just like the rationals then
Hi Robin,
above, holds as the real analysis definition of the Continuum.
Continuum with a capital C! I never knew that had
a real analysis definition.
> Can you give me an example where some member of Q or R is
NOT
> connec to some point in the real-line ?
I presume real-line means the set R. In that case each element
of Q or R (each element of Q is an element of R) is an element
of the real-line. I don't know that being an element of the
real-line
counts for you as being connec to some point in it.
--
His mind has been corrup by colousounds and shapes.
The League of Gentlemen
===
Subject: Rubik's cube combinations...?
I was reading up some info on Rubik's cubes, and the number
of possible
different combinations is given at:
43,252,003,274,489,000
I have been trying to calculate the combinations, but don't
get this
figure.
Can any of you agree?
Note: in a 3x3x3 cube, the center pieces are fixed - they only
rotate on
their own axes. There are 8 corner pieces, each having 3
orientations, and
12 side pieces, each having 2 orientations.
Also, there are 24 orientations of the same cube (4 rotations
per each of
the 6 sides), so I guess any final calculation should be
divided by 24.
BTW: for a 2x2x2 it gives the combinations at 3,674,160. I
realised this is
7! x 3^6, but not entirely sure why this is so.
===
Subject: Re: Rubik's cube combinations...?
> I was reading up some info on Rubik's cubes, and the 
number
of possible
> different combinations is given at:
> 43,252,003,274,489,000
You're a little off. The actual number is
43,252,003,274,489,856,000.
The permutations of the edges and corners must both be even
or both be
odd; there are n!/2 even permutations of n objects (and n!/2
odd
permutations). Thus, the number of permutations of the
cubelets are
2*(12!/2)*(8!/2). The orientations of the cubelets can now be
changed:
the edges can be ßipped and the corners can be twis. Since the
orientations must sum to zero, we are only free to ßip 11 of
the 12
edges and 7 of the 8 corners (the orientation of the final
piece is
determined by the orientation of the others). Thus, the total
number
of combinations is given by
2*(12!/2)*(8!/2)*2^11*3^7 = 12!*8!*2^10*3^7
I have been trying to calculate the combinations, but don't
get this
figure.
> Can any of you agree?
Note: in a 3x3x3 cube, the center pieces are fixed - they only
rotate on
> their own axes. There are 8 corner pieces, each having 3
orientations,
and
> 12 side pieces, each having 2 orientations.
Also, there are 24 orientations of the same cube (4 rotations
per each of
> the 6 sides), so I guess any final calculation should be
divided by 24.
BTW: for a 2x2x2 it gives the combinations at 3,674,160. I
realised this
is
> 7! x 3^6, but not entirely sure why this is so.
You can always rotate a 2x2x2 cube so that one piece is
correctly
positioned and orien. So there are only 7 pieces to permute.
Again,
the twists (i.e. the changes in orientation) must sum to zero
so there
are only 6 pieces to twist. Hence, 7!*3^6. Alternatively, you
could
think of it as 8!*3^7/24 (8 pieces permu, 7 twis, divided by
the
number of rotations of the cube).
===
Subject: Re: Rubik's cube combinations...?
===
> I was reading up some info on Rubik's cubes, and the 
number
of possible
> different combinations is given at:
> 43,252,003,274,489,000
You're a little off. The actual number is
43,252,003,274,489,856,000.
If I recall correctly, in the original ads for the Rubik's
Cube this
number was described as over one million.
-jwgh
--
They can track this putz through the usenet and just arrest
him.
===
Subject: Re: Rubik's cube combinations...?
> I was reading up some info on Rubik's cubes, and the 
number
of
> possible different combinations is given at:
> 43,252,003,274,489,000
>> You're a little off. The actual number is
>> 43,252,003,274,489,856,000.
> If I recall correctly, in the original ads for the Rubik's
Cube this
> number was described as over one million.
Well, it *is* over a million, isn't it?
I'm often surprised about how many people there are for whom
everything
above a million or so seems to be the same.
1 million or 1 billion or 370 billion all look the same (like
in defense
budget, to name one thing).
===
Subject: Re: Rubik's cube combinations...?
> I was reading up some info on Rubik's cubes, and the 
number
of possible
> different combinations is given at:
> 43,252,003,274,489,000
> I have been trying to calculate the combinations, but 
don't
get this
figure.
> Can any of you agree?
There are 8! ways to position the 8 corner cubies, and 3 ways
to orient
each
corner, giving (3^8)(8!) = 264,539,520 combinations. Then
there are 12!
ways
to position the 12 edge cubies, and 2 ways to orient each,
giving
(2^12)(12!) = 1,961,990,553,600 combinations. The center
cubies don't move,
and orienting them is invisible, so they don't count. This
gives
264,539,520*1,961,990,553,600 = 519,024,039,293,878,272,000
combinations.
This would be the answer if you could disassemble the cube
and reassemble
it
in any position. But using legal moves, each cubie can't
really be put
into
any position independent of the others. You can orient 7 of
the corners
arbitrarily and the eighth is then determined. Similarly, you
can orient 11
of the edges and then the twelfth is determined. And finally,
you can
position 18 of the cubies arbitrarily, then the position of
the last two is
determined. These three constraints remove a factor of 3*2*2
= 12, bringing
the total down to 43,252,003,274,489,856,000.
===
Subject: Re: Rubik's cube combinations...?
Mitchell B. <mitchell@galmail.co.za> escribi.97 en el
mensaje|n3f646a0a$0$64720@hades.is.co.za:
> I was reading up some info on Rubik's cubes, and the 
number
of
> possible different combinations is given at:
> 43,252,003,274,489,000
> I have been trying to calculate the combinations, but 
don't
get this
> figure. Can any of you agree?
> Note: in a 3x3x3 cube, the center pieces are fixed - they
only rotate
> on their own axes. There are 8 corner pieces, each having 3
> orientations, and 12 side pieces, each having 2
orientations.
> Also, there are 24 orientations of the same cube (4
rotations per
> each of the 6 sides), so I guess any final calculation
should be
> divided by 24.
> BTW: for a 2x2x2 it gives the combinations at 3,674,160. I
realised
> this is 7! x 3^6, but not entirely sure why this is so.
Others answer that. But, there is a simple proof that the
maximum order of
any element of the group of thr Rubik's cube is 1260?
--
Best
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: Rubik's cube combinations...?
> I was reading up some info on Rubik's cubes, and the 
number
of
> possible different combinations is given at:
> 43,252,003,274,489,000
Mathworld (http://mathworld.wolfram.com/RubiksCube.html) has
a different
number:
43 252 003 274 489 856 000 = 8! * 12! * 3^8 * 2^12 / (2 * 3 *
2)
===
Subject: Re: Rubik's cube combinations...?
> I was reading up some info on Rubik's cubes, and the 
number
of possible
> different combinations is given at:
> 43,252,003,274,489,000
Looks like someone dropped some digits. I get 8! 2^8 12! 3^12
/ 12 =
43,252,003,274,489,856,000.
<http://www.rubiks.com/cubefacts.html I have been trying to
calculate the combinations, but don't get this
figure.
> Can any of you agree?
> Note: in a 3x3x3 cube, the center pieces are fixed - they
only rotate on
> their own axes. There are 8 corner pieces, each having 3
orientations,
and
> 12 side pieces, each having 2 orientations.
> Also, there are 24 orientations of the same cube (4
rotations per each of
> the 6 sides), so I guess any final calculation should be
divided by 24.
Actually 12.
> BTW: for a 2x2x2 it gives the combinations at 3,674,160. I
realised this
is
> 7! x 3^6, but not entirely sure why this is so.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Simple question about non-standard analysis
Are there any serious critics about non-standard analysis??
If no,
why standard analysis still favor??
Thanks
Max
===
Subject: Re: Simple question about non-standard analysis
> Are there any serious critics about non-standard analysis??
If no,
> why standard analysis still favor??
There are no mathematical critics of non-standard analysis.
However,
there are people who feel that non-standard analysis does not
have the
pedagogical and osophical advantages it's sometimes made out
to
have. Others prefer smooth analysis or similar approaches as
more
natural formulations of our intuitions about infinitesimals
(and the
intuitions of those who worked in analysis before it was
arithmetised
and set theorised).
--
Aatu Koskensilta (aatu.koskensilta@xortec.fi)
Wovon man nicht sprechen kann, daruber muss man schweigen
- Ludwig Wittgenstein, Tractatus Logico-osophicus
===
Subject: Re: Simple question about non-standard analysis
>> Are there any serious critics about non-standard
analysis?? If no,
>> why standard analysis still favor??
>There are no mathematical critics of non-standard analysis.
However,
>there are people who feel that non-standard analysis does
not have the
>pedagogical and osophical advantages it's sometimes made 
out
to
>have.
Precisely. I see people claiming that this or that is very
easy using
non-standard analysis, but in fact it's easy only if you
understand
what's allowed and you're very careful.
Saw a talk years ago on non-standard analysis by a zealot. His
point was it was oh so easy, because [I can't quote what he
said
precisely, something to the effect that what was true in the
non-standard reals was exactly the same as what was true in
the reals.] I immediately asked about some sentence or other,
which was true in one but not in the other. He explained that
that sentence didn't count.
And of course it's easy to see why the sentence I mentioned
didn't count, if we know what's what. The 
point is that in the
context of that expository talk, intended for _professional
mathematicians_ but not logicians, he was unable to give
a precise definition of which sentences coun and which
didn't. So what his thesis amoun to was that it was
very easy as long as you didn't break the rules, but he
couldn't say exactly what the rules were! A curious notion
of easy.
>Others prefer smooth analysis or similar approaches as more
>natural formulations of our intuitions about infinitesimals
(and the
>intuitions of those who worked in analysis before it was
arithmetised
>and set theorised).
************************
===
Subject: Re: Simple question about non-standard analysis
> Are there any serious critics about non-standard analysis??
If no,
>> why standard analysis still favor??
There are no mathematical critics of non-standard analysis.
However,
> there are people who feel that non-standard analysis does
not have the
> pedagogical and osophical advantages it's sometimes made
out to
> have. Others prefer smooth analysis or similar approaches
as more
> natural formulations of our intuitions about infinitesimals
Our? Speak for yourself!
Is there any evidence that the majority of mathematicians
ever had any intuition for infinitesimals at all?
--
His mind has been corrup by colousounds and shapes.
The League of Gentlemen
===
Subject: Re: Simple question about non-standard analysis
> Are there any serious critics about non-standard analysis??
No.
> If no, why standard analysis still favor??
Because it's easier.
--
His mind has been corrup by colousounds and shapes.
The League of Gentlemen
===
Subject: Re: Simple question about non-standard analysis
> Are there any serious critics about non-standard analysis??
If no,
> why standard analysis still favor??
Thanks
Max
All standard results provable by non-standard methods also
have
standard proofs so non-standard analysis is just a different
approach,
not one that leads to truly different theorems. Secondly,
pretty much
everyone has been brought up in the standard approach so not
many want
to retool to use the non-standard method.
===
Subject: Re: Simple question about non-standard analysis
> Are there any serious critics about non-standard analysis??
Serious, no. Minor: cannot be done in ZF (requires Boolean
Algebra
Prime Ideal Theorem).
> If no, why standard analysis still favor??
>
Why not?
===
Subject: Re: Math Question: graphs of equivalent equations
> I have two equations:
> a) (x-1)^2 + 1
> b) x^2 - 2x + 3
assumption:
> these two equations are algebraically equivalent
> (a expands into b)
Why do these two equivalent equations result in different
graphs?
Math Learner
Your assumption is incorrect. a) can be written as x^2 - 2x +
2, NOT x^2 -
2x + 3.
-Sunny
===
Subject: Re: Math Question: graphs of equivalent equations
>> I have two equations:
>> a) (x-1)^2 + 1
>> b) x^2 - 2x + 3
>Those aren't equations.
If learner meant that those are expressions of functions,
then the useful
forms
to examine are standard form, and general form. Part 
Ôa' is
the right-hand
side in standard form; but part Ôb' is a 
different expression
in general
form.
Basically parts Ôa' and 
Ôb' are not equivalent. Learner can
complete the
square for part Ôb' to convert it to standard 
form.
Learner, I assume you originally meant:
f(x)=(x-1)^2 + 1
and
g(x)=x^2 - 2x + 3
A polynomial functio in standard form is usually easy to
graph; a
polynomial
function in general form usually requires much more work to
graph.
G C
===
Subject: Re: karnaugh maps
In sci.math, Elaine Jackson
<elainejackson7355@home.com>
<49M8b.954202$3C2.21581237@news3.calgary.shaw.ca>:
> They're trying to foist Karnaugh maps off on me with only 
a
handwaving
> explanation of how you're supposed to use them; along 
these
lines:
1) each loops must comprise 2^n squares of the map, for some n
> 2) every 1 on the map must belong to at least one loop
> 3) try to use as few loops as possible
> 4) try to make loops as big as possible
It's not even clear what the goal is, let alone how
> supposed to verify whether a given solution uses the
> minimal number of loops (ie: whether the proposed
> solution is optimal when condition (4) is left out
> of consideration), but it (the script) is too complex
> to run. (And, like I said, once condition (4) enters
> the picture, I don't even know what optimal means
> anymore.) Can anybody point me to an exact account of
> this technique?
>
I liken a Karnaugh map to an unfolding of an infinitely
thin piece of paper, each fold representing an independent
input for a logic implementation without ßip-ßops or
c-connects. (I used to work in a gate-array foundry,
specializing in digital implementations. I kinda picked
this up as one of the techniques; I can't say 
I've had
much formal schooling on the theory here.)
For one term the unfolding is trivial:
T | F
Now one unfolds the other way, and gets
TT | FT
---+---
TF | FF
where the first term refers to logic input 
Ôa',
and the second to logic term Ôb'. (Or one can
use i_0, i_1, etc. It doesn't matter as long
as one keeps things organized.)
The third fold gets mildly interesting:
TTT | FTT | FTF | TTF
----+-----+-----+----
TFT | FFT | FFF | TFF
Note how the first two terms reverse order in the
right half; this is a ßip, not a replicate.
The fourth fold is similarly interesting:
TTTT | FTTT | FTFT | TTFT
-----+------+------+-----
TFTT | FFTT | FFFT | TFFT
-----+------+------+-----
TFTF | FFTF | FFFF | TFFF
-----+------+------+-----
TTTF | FTTF | FTFF | TTFF
I could go on, but hopefully you get the general idea.
So much for the *tool*; now let's describe the *goal*.
Normally one doesn't write the individual terms a la 
Ôdcba'
one could write
X | 0 | 1 | X
--+---+---+--
X | 1 | 0 | 1
--+---+---+--
X | 1 | 1 | 1
--+---+---+--
X | 1 | 0 | X
(this from the Ôd' segment of a seven-segment 
display,
if I've done this correctly) and then try to gather the
terms in as few circles as possible; a circle in this case
representing a combination of inverters and an AND gate.
Bear in mind we can't just circle willy-nilly, as a
circle in this case should represent a pattern such as xyTF,
where x and y can be either T or F. Or perhaps 
Ô*' will
be a better nomenclature here. (For a 4-term Karnaugh map
the circles aren't too bad but the bigger ones get 
ridiculous.
Bear also in mind that circles might go off one edge and
onto the opposite one.)
This map can be covered with terms such as
**TF + *FT* + *F*F + T*** + *TFT
or in a more conventional notation:
ba' + c'b + c'a' 
+ d + cb'a
Note the T*** term. This term could just as easily been
TFF*, to cover just the 1's on the right edge. However,
since X's are don't care, we can cover them as 
well,
possibly simplifying the resulting logic; a 3-input AND
gate, after all, requires 8 CMOS transistobut if all we
need is a feed-line to the final OR gate, we've 
just saved
8 transistors. (The final OR gate in this case might be
represen using 12 CMOS transistors. Of course, in a
real CMOS circuit one might use NAND into NAND instead of
AND into OR, saving an inverter in each intermediate term
and an inverter in the output.)
Nowadays, a logic minimization program may be used to
automatically generate the logic, or one might simply burn
a ROM or PLA of some sort, each burnt fuse being a 0 or a 1
(and the unburnt fuse being the opposite, a 1 or 0).
HTH; I'll admit I'm not sure how to do a 
better
explanation in pure ASCII.
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: double factorials replacing factorials
For whatever it's worth, your even double factorials are
simply
(2n)!! = 2*4*...*(2n) = 2^n * n!, (factor out the 2's)
so your first series is
sum(x^(2n)/n!!) = sum( (x^2/2)^n / n!) = exp(x^2/2).
(Same answer as Peter gives, but a different derivation.)
For the odd ones, let's see,
(2n-1)!! = 1*3*5*...*(2n-1)
= 1*2*3*...*(2n-1)*2n / 2*4*6*...*2n (put in even terms)
= (2n)! / n!!
= (2n!) / (2^n * n!)
This leads to something similar for your second sum, though
it's not
as simple as just the exponential function.
JS
> I have encountered a linear combination of the two functions
1 + x^2/2!! + x^4/4!! + x^6/6!! + ...
and
x + x^3/3!! + x^5/5!! + x^7/7!! + ...
with double factorials
> 5!!=5*3*1, 9!!=9*7*5*3*1,
> 6!!=6*4*2, 10!!=10*8*6*4*2
> etc.
> Are they rela to classical functions,
> hypergeometric,etc?
George Marsaglia
===
Subject: Re: double factorials replacing factorials
>I have encountered a linear combination of the two functions
>1 + x^2/2!! + x^4/4!! + x^6/6!! + ...
>and
>x + x^3/3!! + x^5/5!! + x^7/7!! + ...
>with double factorials
> 5!!=5*3*1, 9!!=9*7*5*3*1,
> 6!!=6*4*2, 10!!=10*8*6*4*2
>etc.
>Are they rela to classical functions,
>hypergeometric,etc?
>George Marsaglia
If y is your linear combination, then
dy/dx = x*y + const
You can solve this linear differential equation to get
y = (c1 + c2*erf(x/sqrt(2))) * exp(x^2/2)
for some constants c1, c2.
In particular, your first function is exp(x^2 / 2)
--
Wan: Experts at choosing the best of 100+ applicants for a
position.
Register as a California voter by September 22, and vote on
October 7.
Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California
Microsoft Research and CWI
===
Subject: Roots & extensions
Consider the polynomials f(x)=x^2+x+3 and g(x)=x^2+2x+2 in
Z_7[x]. K is a
field that contains Z_7. In K there exist two elements c,d
such that f(c)=0
and g(c)=0.
Prove the existence of an element w in Z_7(d) such that
f(w)=0.
===
Subject: Re: Roots & extensions
Gino Prosapio <billythecazz@libero.it> ha scritto nel
messaggio
> Consider the polynomials f(x)=x^2+x+3 and g(x)=x^2+2x+2 in
Z_7[x]. K is a
> field that contains Z_7. In K there exist two elements c,d
such that
f(c)=0
> and g(c)=0.
> Prove the existence of an element w in Z_7(d) such that
f(w)=0.
It doesn't mind. I found the element [4]+d.
===
Subject: How best to study math?
This might seem like a dumb question, but I'm having trouble
figuring
out the best way to study math. What I was doing is copying
most of
what the instructors would write on the blackboards. Then I
would go
to my text to read through and copy down all of the
definitions,
theorems, proofs, etc. I star to feel like I was just copying
from
the text though, instead of understanding and absorbing. Now
what I
do is read through the text slowly and carefully, without
taking notes
(although I do mark up my books quite a bit,) making sure
that I
understand every definition and proof. I feel a little guilty
for not
writing as much though.
Please let me know what you think about the following
questions:
1. How should I take notes from the text?
2. How should my notes from lectures and notes from texts be
integra? I had been keeping them separate.
2b. What if the instructor doesn't teach in a manner that
allows me
to take a cohesive and coherent set of notes?
3. What should notes contain? Do notes even matter much since
I can
always go back to the text?
John
Subject: Re: How best to study math?
===
> This might seem like a dumb question, but I'm having
trouble figuring
> out the best way to study math. What I was doing is copying
most of
> what the instructors would write on the blackboards. Then I
would go
> to my text to read through and copy down all of the
definitions,
> theorems, proofs, etc. I star to feel like I was just
copying from
> the text though, instead of understanding and absorbing.
Now what I
> do is read through the text slowly and carefully, without
taking notes
> (although I do mark up my books quite a bit,) making sure
that I
> understand every definition and proof. I feel a little
guilty for not
> writing as much though.
The first thing you should probably do is understand *how* you
learn.
Some people learn best by listening, others by observing,
others have to
be active. Some people need background noise, some need
silence. Above
all else, know what techniques work best for you. That said,
if you
aren't somehow active in the process, you probably 
aren't
learning much.
Please let me know what you think about the following
questions:
1. How should I take notes from the text?
It depends. When taking a class, I don't.
When I'm learning something for myself, it depends on the
complexity of
the material. Something that doesn't make sense by simply
reading it is
material I'll likely write down and carefully analyze.
Theorems and
definitions are also worth writing down in an organized way
for easy
reference when I start doing problems.
2. How should my notes from lectures and notes from texts be
> integra? I had been keeping them separate.
It depends on the form of the notes. If you keep them
seperate, you are
creating two books on the subject. If you integrate them
somehow, you
are making one book. Do what allows you to make sense of them
later.
2b. What if the instructor doesn't teach in a manner that
allows me
> to take a cohesive and coherent set of notes?
Look at other ways to get notes. Are you allowed to bring in
a tape
recorder? Are you trying to record too much? Look at your
options.
3. What should notes contain? Do notes even matter much since
I can
> always go back to the text?
The process of writing helps cement ideas into your memory.
If you have
a teacher who explains things differently from the text, it
is useful to
have alternative perspectives recorded for reference. It all
depends on
the particulars of you and your situation.
--
===
Subject: Re: How best to study math?
> This might seem like a dumb question, but I'm having
trouble figuring
> out the best way to study math.
As my best professional school professor said, and genuinely
believed, the
only dumb question is the question you don't ask.
> What I was doing is copying most of
> what the instructors would write on the blackboards. Then I
would go
> to my text to read through and copy down all of the
definitions,
> theorems, proofs, etc. I star to feel like I was just
copying from
> the text though, instead of understanding and absorbing.
Now what I
> do is read through the text slowly and carefully, without
taking notes
> (although I do mark up my books quite a bit,) making sure
that I
> understand every definition and proof. I feel a little
guilty for not
> writing as much though.
My style of taking notes has always been to write less rather
than more.
But
I have not taken mathematics courses at the university level.
(I hang out
here for self-education, to help my homeschooled, very
math-interes
oldest son.)
> Please let me know what you think about the following
questions:
> 1. How should I take notes from the text?
Most mathematics texts have a FEW facts or theorems or
definitions that
they
evidently expect learners to memorize. I would write those
down if writing
them down helps you to memorize them. My own style would not
be to write
down much else from the narrative part of my textbook, but
rather to read
another textbook (or three or four or five textbooks) on the
same topic.
> 2. How should my notes from lectures and notes from texts be
> integra? I had been keeping them separate.
Again, for me taking notes is a way of having a memo of MAIN
points of
lectures, and I work exercises from texts in a notebook
(which for me is
usually the same notebook as the notebook I keep lecture
notes in), but the
way I integrate (these days, after learning this trick
preparing for
professional school) is to make an outline of all the major
ideas of the
course. I rather suspect that the only way I'll be able to
make REAL
progress in self-education in math, after about five years of
independent
reading about it, is to do something of the same kind and set
up an outline
for each major subdivision of mathematics I can find time to
learn.
> 2b. What if the instructor doesn't teach in a manner that
allows me
> to take a cohesive and coherent set of notes?
There are some good books about note-taking that teach
listeners how to
extract information from even the most disorganized lecturer.
I THINK the
one I really liked as an undergraduate was called Speaking
and Listening
[doing online library search] but I see there are several
books with that
title, and I can't remember which one is the one I read back
in the 1970s.
Anyway, in your college library you could find books about
note-taking and
learn some tricks.
> 3. What should notes contain? Do notes even matter much
since I can
> always go back to the text?
I don't go for heavy notes. Too much moving of the pencil
during lectures
detracts from listening and digesting. But I do take some,
because I always
hope--and generally find--that the professor will cover points
in the
lecture that aren't covered in the text. 
That's why you read
the text
BEFORE
the lecture: you can then choose to write down only what is
truly new and
useful during the lecture, knowing you have the book for
review of other
points.
This summer I gathered up some Web links to useful sites
about studying
mathematics, which I will post here. Any reader of this who
has suggestions
for additional links would surely be doing several readers of
this thread a
favor by sharing them, or by commenting on the links I post
here.
http://www.acad.sunytccc.edu/instruct/sbrown/math/succeed.htm
http://ucsu.colorado.edu/~lorenzen/calc2/Sec003/Strategies.PDF
http://euler.slu.edu/Dept/SuccessinMath.html
http://www.math.utah.edu/~alfeld/math.html
http://www.math.ohio-state.edu/counseling/how_to_study.html
http://college.hmco.com/mathematics/larson/intermediate_
algebra/3e/students/
tips.html
http://www.maths.cam.ac.uk/undergrad/inductionday/studyskills
/text/
http://www.esotericka.org/math/guide.html
http://www.coolmath.com/success.htm
and famous pages on how to avoid errors in mathematics study:
http://www.math.vanderbilt.edu/~schectex/commerrs/
http://www.swarthmore.edu/NatSci/smaurer1/WriteGuide/write_
errors.pdf
Hope this helps!
--
Karl M. Bunday Christ has set us free. Galatians 5:1
Learn in Freedom (TM) http://learninfreedom.org/
kmbunday AT earthlink DOT net (preferred email address)
===
Subject: Re: How best to study math?
: This might seem like a dumb question, but I'm having
trouble figuring
: out the best way to study math.
My own personal experience is that even though I ***want***
understanding
to be something I'm *conscious* of gaining, inevitably it
doesn't work
that way, and it's usually something I realize I gained when
I wasn't
really paying attention.
: 1. How should I take notes from the text?
Take them in a way you enjoy. If you don't enjoy it, 
it's
pointless. I
*hate* taking notes from a text, so I don't. 
That's that. If
you like
doing it methodically, do it that way. If you like doing it
quickly
over and over, do it that way. You will probably only learn
it if you
enjoy it.
: 2. How should my notes from lectures and notes from texts be
: integra? I had been keeping them separate.
I think that if they *can* be combined in a way that's
coherent and
educational to you, then do so. If you prefer them separate,
keep them
that way.
: 2b. What if the instructor doesn't teach in a manner that
allows me
: to take a cohesive and coherent set of notes?
Then sit and just try to follow the verbal explanation, and
to heck with
the notes. When I was in grad school I had several professors
who were
terrible givers of notes. It was far more beneficial to sit 
and
listen, so I'd take a few sheets of paper and doodle 
pictures
during
class while listening. The doodling kept my mind limber and
the lectures
I just enjoyed listening to.
: 3. What should notes contain? Do notes even matter much
since I can
: always go back to the text?
Again a matter of opinion, but IMHO I like notes which
contain examples
not in the text, explainations not in the text, and more
specific details
not in the text. If it's in the text it might help to copy 
it
down
anyway, but IMHO (for me) it's useless to keep.
Best
Justin
===
Subject: Re: How best to study math?
> This might seem like a dumb question, but I'm having
trouble figuring
> out the best way to study math. What I was doing is copying
most of
> what the instructors would write on the blackboards. Then I
would go
> to my text to read through and copy down all of the
definitions,
> theorems, proofs, etc. I star to feel like I was just
copying from
> the text though, instead of understanding and absorbing.
Now what I
> do is read through the text slowly and carefully, without
taking notes
> (although I do mark up my books quite a bit,) making sure
that I
> understand every definition and proof. I feel a little
guilty for not
> writing as much though.
Please let me know what you think about the following
questions:
1. How should I take notes from the text?
2. How should my notes from lectures and notes from texts be
> integra? I had been keeping them separate.
2b. What if the instructor doesn't teach in a manner that
allows me
> to take a cohesive and coherent set of notes?
3. What should notes contain? Do notes even matter much since
I can
> always go back to the text?
> John
I'm not much of a mathematician but I have considerable
experience (over
two decades) of successful undergraduate and postgraduate
study in other
fields. My very few insights:
1 What works for me may not work for you!
2 Never take notes in a lecture. Whatever the lecturer says
is written
down more coherently in a good book. There may be a few
things specific
to the course that you want to write down - hints on the
exam, for example.
3 Don't skimp on books - it's a false economy. 
Buy the books
you need
rather than trying to get the previous edition from the
library when you
can. Then read them!
4 I don't take notes from the text, though I probably 
should.
I think,
at least theoretically, that note-taking - more specifically
jotting
down just a few key ideas - is a good idea. My study pattern
tends to be
read little and often, often in odd places (waiting rooms,
trains, over
lunch, in the delivery room while my first son was born... )
where
there's never a ßat surface. An alternative is to 
highlight
the phrases
in the book.
5 [maths specific] Work through all the exercises. Otherwise
however
many times you read it you won't *really* understand.
Hope that helps.
Mark Atherton
===
Subject: Re: How best to study math?
Mark Atherton
> This might seem like a dumb question, but I'm having
trouble figuring
> out the best way to study math. What I was doing is copying
most of
> what the instructors would write on the blackboards.
...
> ...
> 2 Never take notes in a lecture.
I agree. Listen to that guy at the blackboard, and don't be
afraid (as
everyone else in the room is, I'll bet) to ask questions. If
it were just a
matter of copying what goes down on the blackboard, the prof
could just
hand
out lecture notes and save everybody's trouble.
...
> 5 [maths specific] Work through all the exercises. Otherwise
however
> many times you read it you won't *really* understand.
I agree again. In this game, we learn by doing.
LH
===
Subject: Re: How best to study math?
Originator: jgamble@ripco.com (John M. Gamble)
>1 What works for me may not work for you!
>2 Never take notes in a lecture. Whatever the lecturer says
is written
>down more coherently in a good book. There may be a few
things specific
>to the course that you want to write down - hints on the
exam, for
example.
with (2).
Writing notes means that you are hearing the lecturer's 
words
and coming up with your own interpretation. In other words,
your
brain is actually engaged. Otherwise, you are just watching
live
TV for all the difference it makes, and your attention can
wander.
In a similar vein, i once missed a class and asked for a
fellow
student's notes. Just reading them was 
insufficient for
learning,
it was only when i copied them by hand, reading the details,
that
the contents landed in my brain.
[3-4 snipped]
>5 [maths specific] Work through all the exercises. Otherwise
however
>many times you read it you won't *really* understand.
Yes, yes, YES!
--
-john
February 28 1997: Last day libraries could order catalogue
cards
from the Library of Congress.
===
Subject: Re: How best to study math?
>1 What works for me may not work for you!
>>2 Never take notes in a lecture. Whatever the lecturer says
is written
>>down more coherently in a good book. There may be a few
things specific
>>to the course that you want to write down - hints on the
exam, for
example.
> with (2).
Writing notes means that you are hearing the lecturer's 
words
> and coming up with your own interpretation. In other words,
your
> brain is actually engaged. Otherwise, you are just watching
live
> TV for all the difference it makes, and your attention can
wander.
In a similar vein, i once missed a class and asked for a
fellow
> student's notes. Just reading them was 
insufficient for
learning,
> it was only when i copied them by hand, reading the
details, that
> the contents landed in my brain.
[3-4 snipped]
>>5 [maths specific] Work through all the exercises. Otherwise
however
>>many times you read it you won't *really* understand.
> Yes, yes, YES!
>
The problem, seen both from the point of view of a
(reluctant) lecturer
and a lecturee (is that a word?), is that many people have
excessive
expectations of a lecture. They want to be able to go into
the lecture
room devoid of knowledge and walk out an expert! A lecture
should be
supplementary to reading and working through the problems
yourself,
ideally before the lecture. I have sometimes found that
getting into
the subject matter can be easier in a lecture/seminar than
via a book.
But then you need to go away and hit the books afterwards.
That's why
lecture notes are in my opinion largely a waste of space -
90% of the
time you spend on the course is on your own with a book, and
that's the
bit that counts. [I should say that this will depend on the
area and
level of the study.]
Many (good) lecturers will forbid note-taking. They have
other ways of
retaining the classes attention - asking questions, making
jokes,
walking amongst the audience, throwing chalk at those who
aren't looking
(depreca).
You say that by writing notes you make yourself more than a
passive
consumer. The process of a lecturer talking from his papers
and you
making notes has been described as transferring words from the
lecturer's notes to the student's notes 
without passing
through the
brains of either. This is sometimes very apt.
Mark Atherton
===
Subject: Re: How best to study math?
I always found it useful, for me, to take notes during
lectures, but I
rarely read them.
I found that applications were useful to see the point of
theorems. My
idea of applications can be pretty abstract.
--
Try
http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/
Bukharin.html
To solve Linear Programs: .../LPSolver.html
r c A game: .../Keynes.html
v s a Whether strength of body or of mind, or wisdom, or
i m p virtue, are found in proportion to the power or
wealth
e a e of a man is a question fit perhaps to be discussed
by
n e . slaves in the hearing of their mastebut highly
@ r c m unbecoming to reasonable and free men in search of
d o the truth. -- Rousseau
===
Subject: Re: How best to study math?
> This might seem like a dumb question, but I'm having
trouble figuring
> out the best way to study math. What I was doing is copying
most of
> what the instructors would write on the blackboards. Then I
would go
> to my text to read through and copy down all of the
definitions,
> theorems, proofs, etc. I star to feel like I was just
copying from
> the text though, instead of understanding and absorbing.
Now what I
> do is read through the text slowly and carefully, without
taking notes
> (although I do mark up my books quite a bit,) making sure
that I
> understand every definition and proof. I feel a little
guilty for not
> writing as much though.
>
The best advice I know is to read the material in the text
BEFORE each
lecture. It will make every lecture much more understandable.
===
Subject: Re: How best to study math?
Studying is a somewhat individual matter. That said there are
some things
you can do.
What I do is to write all the definitions, theorems from the
book down,
then
I take the class notes and integrate them into the book
notes, along with
the examples given by the Professor/Teacher. These examples
are usually
used on exams and quizzes. You will benefit from rewriting all
the
definitions and theorems down because you will remember them
better.
The other thing you might want to start doing is to read and
understand all
of the proofs. Once you think you fully understand the proof,
then try to
reproduce it on your own, and in your own words. Don't try 
to
memorize
them, as this will not help and is impossible. This will help
you to learn
how to read and do proofs, which IS mathematics.
Lurch
> This might seem like a dumb question, but I'm having
trouble figuring
> out the best way to study math. What I was doing is copying
most of
> what the instructors would write on the blackboards. Then I
would go
> to my text to read through and copy down all of the
definitions,
> theorems, proofs, etc. I star to feel like I was just
copying from
> the text though, instead of understanding and absorbing.
Now what I
> do is read through the text slowly and carefully, without
taking notes
> (although I do mark up my books quite a bit,) making sure
that I
> understand every definition and proof. I feel a little
guilty for not
> writing as much though.
> Please let me know what you think about the following
questions:
> 1. How should I take notes from the text?
> 2. How should my notes from lectures and notes from texts be
> integra? I had been keeping them separate.
> 2b. What if the instructor doesn't teach in a manner that
allows me
> to take a cohesive and coherent set of notes?
> 3. What should notes contain? Do notes even matter much
since I can
> always go back to the text?
 John
===
Subject: Re: Rigorous Textbook Needed
Can anyone suggest a rigorous textbook treatment of
multivariable
>calculus and linear algebra.
I concur with the suggestion for Finite Dimensional Vector
Spaces. I
> haven't seen Apostol's book on Calculus, but 
his
Mathematical Analysis
> was excellent.
>
The second volume of Apostol's calculus text covers this, 
and
is as
rigorous as you could want. {grin} At least it was for me
when I used
it back in 1966 at Caltech.
I can't find his books for sale save for used on 
Amazon, and
then only
rarely. I'm told that all three (the two volumes on calculus
and his
analysis text) have appeared on the file stealing networks.
Sadly, iu
might be that theft is the manner in which these works will
survive.
> You might try to find some older Calculus textbooks; the
standards
> seem to have declined since the 1950s. But I have seen
references to
> more modern texts that were promising.
Pity of it is that you appear to be correct. I know that my
own
education was more rigorous that was that received by my
colleagues
who gradua in the late 70s. Caveat: I'm referring to
mathematicians/engineers/scientists who took their degrees
here in the
US. It appears that even my European colleagues have no a
similar
decline in their own alma maters.
Drieux
===
Subject: Re: Rigorous Textbook Needed
> Can anyone suggest a rigorous textbook treatment of
multivariable
calculus
> and linear algebra. My current textbook omits a lot of the
proofs--and
say
> that they will be left to a more advanced text.
while the list below may not exactly match your course of
study, it
may be a valuable source
A Course of Pure Mathematics by G. H. Hardy
Introduction to Calculus and Analysis by R. Courant and F.
John
===
Subject: Re: Rigorous Textbook Needed
Can anyone suggest a rigorous textbook treatment of
multivariable
calculus
> and linear algebra. My current textbook omits a lot of the
proofs--and
say
> that they will be left to a more advanced text.
Serge Lang's Calculus of Several Variables. My edition 
(third
printing
1974) omits some proofs such as that of Stokes' theorem but
it might be
a nice bridge to works on analysis on more general spaces.
There is
also Courant & John's Introduction to Calculus and Analysis.
--
G.C.
===
Subject: Re: Rigorous Textbook Needed
> Can anyone suggest a rigorous textbook treatment of
multivariable
calculus
> and linear algebra. My current textbook omits a lot of the
proofs--and
say
> that they will be left to a more advanced text.
Why not use the comprehensive volume of Richard Courant's
Series:
<<An introduction to Calculus and Analysis, Vol.2>>
which includes an extensive amount of elementary and
intermediate
calculus of multivariables(I am studying on it). Apostol's
Book is
another good choice.
For pure linear algebra you can check:
An introduction to linear algebra, L.Mirsky
But it tends a bit more to pure algebra than to math analysis.
Bill
===
Subject: Re: Rigorous Textbook Needed
> Can anyone suggest a rigorous textbook treatment of
> multivariable calculus and linear algebra.
Has anyone reading these newsgroup messages used the textbook
by Hubbard,
titled Vector Calculus, Linear Algebra, and Differential
Forms: A Unified
Approach (2nd Edition)? I see from a Web search that that is
the preferred
textbook for the best prepared math majors at Harvard, but I
don't have a
copy available in my university's library, so 
I've never seen
the book.
Does
Hubbard's textbook provide sufficient practice 
in developing
proofs to help
a learner assimilate that level of calculus learning and
develop good
mathematical maturity?
Thanks to anyone who can provide a report from the field about
Hubbard's
Vector Calculus textbook.
--
Karl M. Bunday Christ has set us free. Galatians 5:1
Learn in Freedom (TM) http://learninfreedom.org/
kmbunday AT earthlink DOT net (preferred email address)
===
Subject: Re: Rigorous Textbook Needed
>> Can anyone suggest a rigorous textbook treatment of
>> multivariable calculus and linear algebra.
Has anyone reading these newsgroup messages used the textbook
by Hubbard,
> titled Vector Calculus, Linear Algebra, and Differential
Forms: A Unified
> Approach (2nd Edition)? I see from a Web search that that
is the
preferred
> textbook for the best prepared math majors at Harvard, but
I don't have a
> copy available in my university's library, so 
I've never
seen the book.
Does
> Hubbard's textbook provide sufficient practice 
in developing
proofs to
help
> a learner assimilate that level of calculus learning and
develop good
> mathematical maturity?
Thanks to anyone who can provide a report from the field about
Hubbard's
> Vector Calculus textbook.
Slight correction/addition: the book is by John H. Hubbard
*and* Barbara
Burke Hubbard. Undoubly the first author provided the bulk of
the
mathematical expertise, but just as certainly the second
author provided
much writing expertise.
I heartily recommend the above book to honors level
freshmen/sophomores
math majothose who presumably need a solid understanding of
calculus
to build on in their future mathematical work. I'm not
surprised that it
is the book of choice for the best math majors at Harvard,
since it is
also the book for the honors level vector calculus sequence
at Cornell
University, where John Hubbard is professor. I took the class
from him
years ago, using a preliminary manuscript version.
There are some differences between the prelimary version and
the first
edition of the book:
1) Some of the advanced material has been cut: e.g. the hairy
ball theorem
and Brouwer's fixed point theorem.
2) The harder theorems' proofs have been moved to an 
appendix.
3) There is a more complete set of exercises and examples.
4) The material on integration has been expanded, and topics
such as
probability has been added.
5) The overall exposition is clearer.
I make note of these differences, since much of my
recollections and
experience with the book is based on the preliminary version,
although I
think I've looked through the first edition 
enough times to
have gotten a
good sense of the changes. It should be evident that the
changes are all
for the better, except perhaps 1), which I always thought one
of the more
interesting sections of the book.
A list of contents and further information is available at:
http://www.math.cornell.edu/~hubbard/vectorcalculus.html
As for the question of whether a reader would develop
mathematical
maturity and skill in proofs, the answer is definitely yes.
Everything is
proven (barring certain issues in the foundations of
mathematics), and the
book starts with an overview of set theoretical language and
basics such
as the uncountability of the real numbers. I can't imagine
how a student
could read the book and not gain considerably in maturity and
knowledge.
The appendix (of harder proofs) can be delved into as needed
for more
advanced students. I would say the book, as a whole, exceeds
what even a
bright undergraduate can understand, but it remains very
friendly
nonetheless (this is rela to the below paragraph).
The book is also much more fun (in my opinion) than the
standard texts.
There are many margin notes that act as informal help and as
a source of
miscellaneous mathematical tidbits. The presentation conveys
the
excitement of mathematics, something calc books seem to have
trouble
doing.
One aspect of the book that strikes me as quite different
from other books
is its emphasis on the concrete. The approach is very
constructive, and
so the reader will note that oftentimes an explicit bound
will be given
where the standard version of a theorem does not.
Computational issues
are mentioned when relevant. This differs greatly from the
many other
texts of today that somehow have either ignored the computer
revolution or
let it pass, remaining unaffec.
===
Subject: Re: Counting strings up to cyclic rotation
> I want to know a formula for the number of strings
containing m of
> each of k > 1 distinct characters (so the strings are of
length mk)
> that are cyclically different. What i mean by this is 
can't
be
> rota to make the same string...so the strings ABC BCA and
CAB are
> all the same cyclically, but they are all different to, for
example,
> the string ACB.
I reckon the number of strings with m occurences of each of k
> characters is simply
(mk)!
> -----
> (m!)^k
as there are (mk)! permutations of the characters in a string
mk in
> length,
> but the m occurences of each of the k characters can be
permu
> without making a difference.
However, I can't tell what I need to divide the above number
by to get
> the number of cyclically different strings.
An example might help to explain. Imagine i wan to get all
> cyclically different strings containing two As and two Bs
(so of
> length 4). The formula above gives me the answer six,
corresponding
> to the six strings
> i/ AABB
> ii/ ABAB
> iii/ ABBA
> iv/ BABA
> v/ BAAB
> vi/ BBAA
However there are only two cyclically different strings AABB
and ABAB
> as
> i/,iii/,v/ and vi/ are all in the same class as AABB, and
ii/ and iv/
> are both in the same class as ABAB.
How do I tell how many cyclic equivalence classes there are
(so I know
> what to divide my formula by)? Or how do I count the number
of
> cyclically different strings of length mk containing m
occurrences of
> each of my k characters in some other way? I'm stuck!
Thanks in advance for any help
Apparently you've thought about this enough to realize
> what the central difficulty in this sort of problem is.
These are sometimes referred to as bead problems
> or necklace problems because of the equivalence
> of an arrangement of colored beads on a necklace
> under rotation.
George Polya developed a theory for counting these
> arrangements. You might be interes in this nicely
> written introduction to those methods:
http://www.geometer.org/mathcircles/polya.pdf
chip
great -- that explains what i wan to know perfectly.
incidentally,
having read and tried to apply polya's method to the results
of a
computer program i'd written, i found a disagreement -- this
was
because the program was losing some permutations along the
way -- i'd
never have noticed this without knowing the correct number.
so thanks!
===
Subject: Re: Zeta function, product of summations
Robin,
I just wan to say that I took your advice, and was able to
solve the
problem by looking at a few values for n.
Thanks again,
Diana
> I am trying to show that the inverse of the Zeta(s)
function is the
> infinite summation as n goes from 1 to infinity 
of
MobiusMu(n)/(n^s).
> OK
> The Zeta(s) function is defined as the infinite 
summation as
n goes
from
> 1 to infinity of 1/(n^s).
To do this, I am first calculating the product of two generic
summations:
{the infinite summation as n goes from 1 to 
infinity of
a_n/(n^s)}
times
> {the infinite summation as n goes from 1 to 
infinity of
b_n/(n^s)}.
What I believe is the answer is:
{the infinite summation as n goes from 1 to 
infinity of the
summation
of
> divisors d of n of [a_d * (b_(n/d))]/(n^s)}.
> That's it.
> Using this formula, I am trying to show that:
{the infinite summation as n goes from 1 to 
infinity of 1/(n^s)
of
the
> summation of divisors d of n of MobiusMu(n/d)/(n^s)}
is one.
> That's right.
> I am trying to test my formulas on Mathematica, and
understand how to
> program regular summations. But, I am unsure how to code
the summation
of
> divisors d of n with Mathematica.
>
AAAAAAAAAAARRRRRRRRRRRRRRGGGGGGGGGHHHHHHHHHHHHH!!!!!!!!!!!!!!
!!!!
> Forget Mathematica!!!
> You don't have to give large quantities of money to 
Stephen
Wolfram
> to obtain mathematical insight. Better spend a little on a
pencil
> and some paper. As you point out in your product the
coefficient of
> 1/n^s is
> sum_{d divides n} mu(n/d).
> Why not just take a large piece of paper and see what
happens when you
> compute this sum for some small n. Say up to 20 or 30. What
patterns
> do you see emerging? Can you prove that these patterns
persist for
> all n?
> --
His mind has been corrup by colousounds and shapes.
> The League of Gentlemen
===
Subject: Re: Zeta function, product of summations
A follow-up question,
Jim, I was not able to figure out how to use Plus@@(f[n,#]& /@
Divisors[n]) in Mathematica.
I was not able to figure out how to code the function f in
relation to:
{the infinite summation as n goes from 1 to 
infinity of 1/(n^s)
of the
summation of divisors d of n of MobiusMu(n/d)/(n^s)}
If time permits, and you can help me out, I will print it out
for my
records.
Diana
>I am trying to test my formulas on Mathematica, and
understand how to
>>program regular summations. But, I am unsure how to code
the summation
of
>>divisors d of n with Mathematica.
>
AAAAAAAAAAARRRRRRRRRRRRRRGGGGGGGGGHHHHHHHHHHHHH!!!!!!!!!!!!!!
!!!!
> Forget Mathematica!!!
You don't have to give large quantities of money to Stephen
Wolfram
> to obtain mathematical insight. Better spend a little on a
pencil
> and some paper.
> Given a function f[n,d], the sum of f[n,d] over all
(positive)
> divisors d of n can be expressed as
> Plus@@(f[n,#]& /@ Divisors[n])
> Keep using Mathematica! Throw away that pencil! Bwwaaahh
hah hah!
> --
> |
> +------------------------------------+ of Advanced Rockets |
> | http://www.uiuc.edu/ph/www/jferry/
+------------------------+
> | jferry@[delete_this]uiuc.edu | University of Illinois |
===
Subject: Re: Zeta function, product of summations
> A follow-up question,
Jim, I was not able to figure out how to use Plus@@(f[n,#]& /@
> Divisors[n]) in Mathematica.
I was not able to figure out how to code the function f in
relation
to:
{the infinite summation as n goes from 1 to 
infinity of 1/(n^s)
of
the
> summation of divisors d of n of MobiusMu(n/d)/(n^s)}
If time permits, and you can help me out, I will print it out
for my
> records.
to figure out the problem. You are now a registered enemy of
Mathematica.
Just who the hell do you think you are? John Henry? Stop that
thinking
right now and enslave yourself to the machine!
--
|
===
Subject: Defining Planes
My calculus textbook says that a single vector parallel to a
plane is not
enough to convey the Ôdirection' of the plane. 
Then it goes
on to discuss
normal vectors. I'm wondering why a parallel vector 
can't be
used to
describe a plane. The textbook doesn't explain why and it
seems to me that
since the dot product of a parallel vector and a normal
vector will be 0
that means that the parallel vector could be used somehow.
What am I
missing?
Larson
===
Subject: Re: Defining Planes
charset=iso-8859-1
> My calculus textbook says that a single vector parallel to
a plane is
not
> enough to convey the Ôdirection' of the plane. 
Then it goes
on to
discuss
> normal vectors. I'm wondering why a parallel vector 
can't
be used to
> describe a plane. The textbook doesn't explain why and it
seems to me
that
> since the dot product of a parallel vector and a normal
vector will be 0
> that means that the parallel vector could be used somehow.
What am I
> missing?
Consider a cylinder with infinitly many planes tangent to the
curved part. The axis of the cylinder is parallel to each of
these planes, but the planes are not parallel to eachother.
--Jim Buddenhagen
===
Subject: Re: Defining Planes
> My calculus textbook says that a single vector parallel to
a plane is
not
> enough to convey the Ôdirection' of the plane. 
Then it goes
on to
discuss
> normal vectors. I'm wondering why a parallel vector 
can't
be used to
> describe a plane. The textbook doesn't explain why and it
seems to me
that
> since the dot product of a parallel vector and a normal
vector will be 0
> that means that the parallel vector could be used somehow.
What am I
> missing?
Larson
>
In three dimensional space, there are infinitely many planes
through
a given point and parallel to a given non-zero vector, but
there is
only one plane through that point and perpendicular to that
vector.
===
Subject: Re: Defining Planes
> My calculus textbook says that a single vector parallel to
a plane is
not
> enough to convey the Ôdirection' of the plane. 
Then it goes
on to
discuss
> normal vectors. I'm wondering why a parallel vector 
can't
be used to
> describe a plane. The textbook doesn't explain why and it
seems to me
that
> since the dot product of a parallel vector and a normal
vector will be 0
> that means that the parallel vector could be used somehow.
What am I
> missing?
Larson
>
it takes TWO vectors parallel to the plane (and not parellel
to each
other) to specify the direction of a plane.
===
===
Subject: exact equations - uniqueness of solutions?
Given a solution F(t,y(t)) to an exact differential equation
with initial
conditions (the solution is defined on an open cell about the
initial
conditions), is there anything one can say about the
uniqueness of that
solution?
-Robert
--
Robert Andrew Pruvenok
Math/EE Major
MATH 2413 TA
===
Subject: Re: exact equations - uniqueness of solutions?
> Given a solution F(t,y(t)) to an exact differential
equation with initial
> conditions (the solution is defined on an open cell about
the initial
> conditions), is there anything one can say about the
uniqueness of that
> solution?
I think you need a condition about (Lip z (sp?)) continuity,
for
example. If you think about ßows in phase space, it seems
intuitive
that a solution will be unique.
--
Try
http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/
Bukharin.html
To solve Linear Programs: .../LPSolver.html
r c A game: .../Keynes.html
v s a Whether strength of body or of mind, or wisdom, or
i m p virtue, are found in proportion to the power or
wealth
e a e of a man is a question fit perhaps to be discussed
by
n e . slaves in the hearing of their mastebut highly
@ r c m unbecoming to reasonable and free men in search of
d o the truth. -- Rousseau
===
Subject: Re: exact equations - uniqueness of solutions?
>> Given a solution F(t,y(t)) to an exact differential
equation with
initial
>> conditions (the solution is defined on an open cell about
the initial
>> conditions), is there anything one can say about the
uniqueness of that
>> solution?
>I think you need a condition about (Lip z (sp?)) continuity,
for
>example. If you think about ßows in phase space, it seems
intuitive
>that a solution will be unique.
The standard Existence and Uniqueness Theorem for
differential equations
assumes the d.e. is of the form y' = f(x,y) (or can be put
into that
form). It requires f to be continuous in x and y and to
satisfy a
Lipschitz condition in y, locally uniformly in x.
An exact d.e. is of the form M + N y' = 0, where for some
F(x,y) you have
M = F_x and N = F_y (the subscripts denoting partial
derivatives), and
Robert Pruvenok is apparently talking about implicit solutions
F(x,y(x)) = c. These don't necessarily correspond to unique
solutions
y = (function of x) except in regions where N is nonzero.
If M = F_x is continuous in x and N = F_y is continuous in y
in a
rectangle a <= x <= b, c <= y <= d containing (x_0, y_0),
then F is
uniquely determined up to a constant, since
F(x,y) = F(x,y) - F(x,y_0) + F(x,y_0) - F(x_0, y_0) + F(x_0,
y_0)
= int_{y_0}^y N(x,t) dt + int_{x_0}^x M(s,y_0) ds + F(x_0,
y_0)
In fact, M and N need not be assumed continuous, as long as F
is
absolutely continuous in x and y separately (interpreting the
integrals
in the Lebesgue sense). In fact the absolute continuity in x
need
only be assumed for y = y_0.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: CAN'T Order Reals
> another for my numerology list,
>> Cantor Can't Order Reals
> You can order the reals; you just can't count them.
No you can't. I tried last week and was told that
they were out of stock and the manufacturer
had no plans to make any more.
Rick
===
Subject: Re: CAN'T Order Reals
> Cantor Can't Order Reals
You can order the reals; you just can't count them.
>No you can't. I tried last week and was told that
>they were out of stock and the manufacturer
>had no plans to make any more.
By contrast, a visit to Google (promp by another thread)
searching for ordered pair definition presen an
opportunity to, and I quote,
Locate an Axiom Near You
Get the facts on the Axiom you want from dealers in your area.
www.autobytel.com
Not sure how long THESE will be in stock; better order some
now!
dave
===
Subject: Re: CAN'T Order Reals
In sci.math, |-|erc
<chess3@ozemail.com.au>
<vEK8b.2616$vQ1.134745@nnrp1.ozemail.com.au>:
>> In sci.math, spakka
>> <usenet_spam@m16.demon.co.uk >
<9nC8b.9093$cw2.79259778@news-text.cableinet.net>:
>You can order the reals; you just can't count them.
> how can you order something you can't count?
> Given any two distinct reals, you can determine which is
the greater,
>> and in a way that is consistent with arithmetic
operations. This
isn't
>> a trivial property - you can't do it with complex 
numbers.
> You don't need to be able to enumerate the set of reals as
a sequence
>> to do this.
>>
in mathematics the terms order and countable are distinct but
that
> is a specialisation of the term order, both synonyms for
list.
You can order reals (finite subset), you can't 
order the reals.
True, in a certain sense. The rationals also have this
problem,
as they're dense: given any a and b, there are an 
infinite
number
of c's in between them, if a != b. In the case of the reals
the infinity is an uncountable one, as well.
So queueing them up in order is an impossible task. But AFAIK
that's
not necessary in order to have an ordered set.
> Also, the ordering satisfies the usual properties:
>> [1] Irreßexibility: For any a, a is not < a and a is not >
a.
>> [2] Anti-commutativity: For any a and b, a < b implies b >
a, and
>> a > b implies b < a.
>> [3] Transitivity: For any a, b, and c, a > b and b > c
implies a > c.
>> For reals, there are additional issues:
>> [4] For any a and b, a > b, there are an uncountable
number of c's
>> such that a > c > b. (Namely, c = a + (b-a)r, 0 < r < 1.
>> I'm not sure what replaced Cantor's 
diagonalization proof
for
>> the uncountability of the interval (0,1), but Cantor's
proof
>> worked to a certain extent.
>> [5] For any a and b, a > b, there exist a' and 
b', a > a'
> b' > b,
>> such that a' and b' have 
finite decimal representations.
>> (This follows trivially from [4] and the fact that
>> lim(N->+oo) 10^(-N) = 0.) The converse is *not* true but
>> it's usually not a problem. (I suppose one could extend
>> this to two infinite approximation series or something. The
>> main issue is at the limit; the difference of corresponding
>> terms must be greater than 0, and further the limit of the
>> difference must be greater than 0, otherwise one's only
found
>> two series for one number.)
>> --
>> #191, ewill3@earthlink.net
>> It's still legal to go .sigless.
> thats a partial order, so is my family tree
partial order C total order C countable order
Such orderings may only be possible for finite sets.
[rest snipped]
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: CAN'T Order Reals
> In sci.math, |-|erc
> <chess3@ozemail.com.au
<vEK8b.2616$vQ1.134745@nnrp1.ozemail.com.au>:
>> In sci.math, spakka
>> <usenet_spam@m16.demon.co.uk >
<9nC8b.9093$cw2.79259778@news-text.cableinet.net>:
>You can order the reals; you just can't count them.
> how can you order something you can't count?
> Given any two distinct reals, you can determine which is the
greater,
>> and in a way that is consistent with arithmetic
operations. This
isn't
>> a trivial property - you can't do it with complex 
numbers.
> You don't need to be able to enumerate the set of reals as 
a
sequence
>> to do this.
in mathematics the terms order and countable are distinct but
that
> is a specialisation of the term order, both synonyms for
list.
You can order reals (finite subset), you can't 
order the reals.
> True, in a certain sense. The rationals also have this
problem,
> as they're dense: given any a and b, there are an 
infinite
number
> of c's in between them, if a != b. In the case of the 
reals
> the infinity is an uncountable one, as well.
> So queueing them up in order is an impossible task. But
AFAIK that's
> not necessary in order to have an ordered set.
You're using the term order for Ôordered 
set'. All the theory
is fine here
I'm just supporting my claim that the name CANT O.R. is a
good match
for his foundational work. Its strikes me if you take the 3
most
foundational
mathematical works, unit number, unprovable statements, and
noncountability,
He could have been named Cantlir, can't list reals, but
ÔCantor' carries the
meaning,
like Lady Di.
> Also, the ordering satisfies the usual properties:
> [1] Irreßexibility: For any a, a is not < a and a is not >
a.
>> [2] Anti-commutativity: For any a and b, a < b implies b >
a, and
>> a > b implies b < a.
>> [3] Transitivity: For any a, b, and c, a > b and b > c
implies a > c.
> For reals, there are additional issues:
> [4] For any a and b, a > b, there are an uncountable number
of c's
>> such that a > c > b. (Namely, c = a + (b-a)r, 0 < r < 1.
>> I'm not sure what replaced Cantor's 
diagonalization proof
for
>> the uncountability of the interval (0,1), but Cantor's
proof
>> worked to a certain extent.
> [5] For any a and b, a > b, there exist a' and 
b', a > a' >
b' > b,
>> such that a' and b' have 
finite decimal representations.
>> (This follows trivially from [4] and the fact that
>> lim(N->+oo) 10^(-N) = 0.) The converse is *not* true but
>> it's usually not a problem. (I suppose one could extend
>> this to two infinite approximation series or something. The
>> main issue is at the limit; the difference of corresponding
>> terms must be greater than 0, and further the limit of the
>> difference must be greater than 0, otherwise one's only
found
>> two series for one number.)
> --
>> #191, ewill3@earthlink.net
>> It's still legal to go .sigless.
> thats a partial order, so is my family tree
partial order C total order C countable order
> Such orderings may only be possible for finite sets.
hardly, set theoretic orders are minimal properties of
ordelike who is
in charge of who in a company, there are multiple solutions
to satisfy the
partial order with sequences. Integers support all the
characteristics of
the
set theoretic order but its hardly descriptive of a sequence.
Herc
===
Subject: Re: CAN'T Order Reals
In sci.math, spakka
<usenet_spam@m16.demon.co.uk>
<wVJ8b.9484$N96.82799863@news-text.cableinet.net>:
>> In sci.math, spakka
>> <usenet_spam@m16.demon.co.uk >
<9nC8b.9093$cw2.79259778@news-text.cableinet.net>:
>You can order the reals; you just can't count them.
>>how can you order something you can't count?
>Given any two distinct reals, you can determine which is the
greater,
>and in a way that is consistent with arithmetic operations.
This isn't
>a trivial property - you can't do it with complex numbers.
>You don't need to be able to enumerate the set of reals as 
a
sequence
>to do this.
>> Also, the ordering satisfies the usual properties:
[1] Irreßexibility: For any a, a is not < a and a is not > a.
>> [2] Anti-commutativity: For any a and b, a < b implies b >
a, and
>> a > b implies b < a.
>> [3] Transitivity: For any a, b, and c, a > b and b > c
implies a > c.
Hmm. I'd express this differently, so as not to introduce
both symbols
> Ô<' and Ô>'. Also, 
for the reals we get the useful property
that for
> any a, b one of a < b, a = b , b < a is true - i.e. the
order is total.
I'll admit I was in a hurry. And yes, what 
you're mentioning
is the trichotomy principle or property -- very useful.
There's a few otheas well, relating ordering and
arithmetic operations.
However, you've really mentioned that already above anyway.
>> For reals, there are additional issues:
[4] For any a and b, a > b, there are an uncountable number
of c's
>> such that a > c > b. (Namely, c = a + (b-a)r, 0 < r < 1.
>> I'm not sure what replaced Cantor's 
diagonalization proof
for
>> the uncountability of the interval (0,1), but Cantor's
proof
>> worked to a certain extent.
What do you think was the problem with it? Proofs don't tend
to Ôwork
> to a certain extent'.
personal aesthetic point of view having to worry about
the Ô.999... = 1.000...' problem is a minor 
blot. Of course
in base 10 one can always pick one of the other 8 digits.
In base 2 one cannot have that luxury but one can cheat in
that
case and treat the problem as though it were base 4, instead.
If one wants, one can, in base 10, restrict the digits to 0
and 1; this gives a well-defined construction for a series of
approximations, getting closer and closer to a certain number,
as one must pick either 0 or 1. Or one can use 2 and 3, or
4 and 5, or 1 and 8; as long as one of them isn't 
Ô9' there's
not much of a problem -- and the genera number is guaranteed
not to be in the alleged counting-list; therefore the reals
are uncountable.
>> [5] For any a and b, a > b, there exist a' and 
b', a > a'
> b' > b,
>> such that a' and b' have 
finite decimal representations.
>> (This follows trivially from [4] and the fact that
>> lim(N->+oo) 10^(-N) = 0.) The converse is *not* true but
>> it's usually not a problem.
What do you mean by Ôthe converse' here?
It is not the case that, for any approximations a' and 
b',
a' > b', that it necessarily follows that the 
numbers being
approxima, a and b, are such that a > a' > b' 
> b.
For a counterexample, take a' = 3.1416, b' = 
3.14159. In this
case a = b = pi. I'm not sure I've phrased 
that quite right
but some care is obviously needed.
>> (I suppose one could extend
>> this to two infinite approximation series or something. The
>> main issue is at the limit; the difference of corresponding
>> terms must be greater than 0, and further the limit of the
>> difference must be greater than 0, otherwise one's only
found
>> two series for one number.)
I'm not sure any of this helps the OP. Schizophrenic
Numerology was
> never my strong subject.
>
Not mine, either. I prefer mathematics. Or one can work
on rather contrived puzzles such as SEND + MORE = MONEY;
this isn't, however, numerology AFAICT, which appears to be
rela to astrology (and about as convincing).
I'm not sure how one can order the entire set of the
interval (0,1), in the sense of queueing them up and
heading them out -- for starte(0,1) is dense, meaning,
IIRC, that given any a < b, one can find an uncountable
number of c's between them, leading to a very funny-looking
queue. (The rationals can be coun, but they otherwise
have the same problem if one were to similarly queue
them up; the usual counting of the rationals is done in
a completely different fashion, not respecting arithmetic
ordering.)
But the reals are well-ordered since one can find out,
given any two reals a, b, whether a = b, a < b, or a > b,
though from a computational standpoint the problem
could get arbitrarily complex. For example, both a and
b may have approximations via infinite series, or be the
results of integrals or limits, or be specified as the
single real root of certain arbitrarily polynomials
a_n x^n + a_{n-1} x^{n-1} + ... + a_0 = 0. (I say certain
since many have more than one real root.)
In the last case, one might generate an infinite series
of approximations via various numeric solution methods.
For integrals or limits, one can generate infinite series
or sequences, reducing the second case to the first as well.
It's a mildly interesting problem.
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: CAN'T Order Reals
> in mathematics the terms order and countable are distinct
but that
> is a specialisation of the term order, both synonyms for
list.
 No, that's not true. You can have uncountable ordered 
sets.
and countable don't imply each other, please complete these 
3
multi choice
questions to continue
dialog.
or just give me the ordered reals if you prefer. I'll start 
:
0,
Herc
-->----------------------------------------------------------
---------------
-----
Randi will test you when you properly apply to be tes. Sign
up here:
http://www.randi.org/research/challenge.html
-----------------
A Rich Shewmaker
B CNote
C Wanda
D Rust
-------------------------------------------------------------
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It really all depends on the situation.
-----------------
A Shanx
B See You In Hell My Friend.
C Someone
D Greg Neill
-------------------------------------------------------------
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If ever I actually found myself in that situation, I'd hold
it upright,
with the intent of attacking my assailant's knife hand.
-----------------
A cliff86
B Rust
C Shanx
D NormDePloom
-------------------------------------------------------------
---------------
----
===
Subject: Re: CAN'T Order Reals
> in mathematics the terms order and countable are distinct
but that
> is a specialisation of the term order, both synonyms for
list.
 No, that's not true. You can have uncountable ordered 
sets.
But you can get rid of all of the uncountable ordered sets...
http://matwbn.icm.edu.pl/ksiazki/fm/fm114/fm11413.pdf
Oh my gosh! Where did I leave that large cardinal axiom?
:-)
mitch
===
Subject: Re: CAN'T Order Reals
proof of numerology at www.adamskingdom.com
> Herc
> sigh nearly 2 years of proof and not one believer
I'm hurt.
I already told you that the palindromic symmetries associa
with
> sphere packings made me receptive to your claims. On the
other hand,
> the fact that I do not have your awareness with regard to
the matter
> reßects the complexity of encodings associa with the Leech
lattice.
> You're into 4s and 6s,
> I'm into 7s and 10s
> It'd never work :)
:-)
You'd be surprised. Let's start with 7.
Joseph Campbell discusses calculations with respect to the
great Platonic
year in his books on mythology. He relates these calculations
to both the
Hebrew and Sumerian (I think it was Sumer--but you know who I
mean...
Tigris,
Euphrates, not Saddam) calender systems--in particular, the
number of days
in
a week.
The Hebrews used a seven day week while the Sumerians (or
whoever) used
five
day weeks. Curiously, this appears to have a structural
similarity with a
mathematical structure--the affine lattice A(Z_2, 2)--whose
geometry
reßects
my 4's and 6's.
To show you what I mean, connect the loci by adding edges...
1
2 3 4 5 6 7
8 9 10 11
12
1 connects to 2, 3, 4, 5, 6, 7
12 connects to 8, 9, 10, 11
8 connects to 2, 3, 5
9 connects to 2, 4, 6
10 connects to 3, 4, 7
11 connects to 5, 6, 7
This diagram correponds with the affine geometry consisting of
4 points and
6
lines. When represen in Euclidean space, it is given the
coordinates
(0,0), (0,1), (1,0), and (1,1).
Hopefully, these coordinates will remind you of truth tables
and switching
functions.
Of the sixteen switching functions in two variables, 14 (2x7)
of them are
unate. (Since numerology is admissible in this post, maybe
you can recall
the many biblical references involving triples. 3x14=42--the
number of
months in the rapture). That is, there are 7 linear
dichotomies (labelings
into two classes) of 4 points in general position in R^2.
When expressed
in
terms of switching functions, the number doubles because the
labels (0 and
1,
T and F, L and R, or whatever) themselves can be juxtaposed.
Now, let's find that 10.
If you think about the syntactic form used to express a
switching function
in
two variables--for instance,
[begin fixed width]
|
----------
T T | F
T F | F
F T | F
F F | T
[end fixed width]
you realize that there is no semantic constraint dictating
row order.
Therefore, the two column vectors to the left of the vertical
line are not
invariant with respect to representation.
In fact, of the sixteen column vectors that could be used to
represent a
switching function in this manner, exactly 10 of them are
excluded from
appearing to the left of that vertical line (if, we insist
that all four
rows
be distinct).
[begin fixed width]
T T T T F
T T T F T
T T F T T
T F T T T
F F F F T
F F F T F
F F T F F
F T F F F
[end fixed width]
I have made the claim in several posts on sci.logic that
information
appears
to be organized around 1's and 3's. This 
relationship arises
in the
geometric relationships involved here. Every affine geometry
has a
corresponding projective geometry. In general, there are
three distinct
collineations in affine geometries that coincide in the
corresponding
projective geometries. However, the relationship is even more
direct in
the
simple affine geometry consisting of four points and six
lines. In the
corresponding projective geometry, every line is coincident
with three
points
and every point is coincident with three lines.
With regard to the Leech lattice, this relationship appears
in the midpoint
of the palindromic symmetry to which I refer. There are two
lists of
numbers
associa with sphere packings. One list has three identical
numbers at
the
central loci. The other has a local minimum at its central
locus. It is
delimi by a pair of identical numbers in each direction that
are
maximums
for that particular list. So, the centers of the lists relate
according to
this 1-3 relationship.
I don't ascribe any particular theological or theoretical
relevance to
these
forms. But, I find their appearance in relation to logical
forms and
mathematical symmetries interesting.
Unconditional belief is not my style.
 if that's what you believe
I don't know what I believe. Truth is in the monad of the
beholder.
:-)
mitch
===
===
Subject: Re: CAN'T Order Reals
> 7 is representative of male, and 10 of female.
===
===
Subject: D_4 Subgroup Names
Hope someone can help. :)
Are there standard names for the 10 possible symmetries,
corresponding
to
the 10 subgroups of D_4, of a plane figure composed of a
subset of the
lattice Z^2?
Paul
===
Subject: Re: D_4 Subgroup Names
> Are there standard names for the 10 possible symmetries,
corresponding
to
> the 10 subgroups of D_4, of a plane figure composed of a
subset of the
> lattice Z^2?
Oh well, if not, has anyone got any suggestions?
I thought maybe:
(1) None
(2) Horizontal
(3) Vertical
(4) Main-Diagonal
(5) Skew-diagonal
(6) Bi-rotational
(7) Quad-rotational
(8) Bi-orthogonal
(9) Bi-diagonal
(10) Square
Paul
===
Subject: Re: equal temperance vs. pure temperance music
frequency ratios
===
equal temperance vs. pure temperance music frequency ratios
With the piano, I think one can multiply a frequency by
12th-root-2 to
>get the frequency of the next note a half-step up.
Is there a formula for pure temperance frequencies?
>
This is well-studied. Search for equal temperament and just
> intonation on the Ônet. One informative site I found is
> http://home.earthlink.net/~kgann/tuning.html
> <http://home.earthlink.net/%7Ekgann/tuning.html>.
You need to specify what you mean by pure temperance. Just
> intonation? Pythagorean temperance? In some tunings C sharp
is not the
> same pitch as D ßat, for example, and string players
(particularly in
> string quartets), a capella chori, and others may exploit
this
difference.
This would be better discussed in a music theory newsgroup.
The news group rec.music.theory suggests itself, although I
suspect
these questions are old hat for people there. As Stephen
mentions, this
subject has a long history. Even Euler had a go at it.
Something that hasn't been mentioned is the list of integers
proportional to the frequencies giving a major scale in just
intonation:
24, 27, 30, 32, 36, 40, 45, 48.
That's how I always remember the numbers.
Ken Pledger.
===
Subject: Re: The Grand Facade
Don't worry. There will soon be plenty of openings when Bush
starts up the
draft again.
The job comes with a American ßag and a white body designer
body bag. One
size fits all.
> spamfile@bellsouth.net says...
> The economy has certainly
> gone to hell since he was sworn in.
It's been good to me, and if there is a problem, it 
certainly
wasn't
due
to
> him. You can thank a bunch of who hijacked some planes 2
years
ago
> for that.
> If it was good for you, then it was good for you. Hooray
for you.
> Take it from someone who was in the midst of a lengthy
(depressingly
> lengthy) job search when the planes hit the towers: the
economy had gone
> to hell before 9/11/2001. Better yet, don't take my word
for it -- go
> find the facts. Warping the facts to fit a 
favorite theory
isn't pretty.
> --
> Kevin Willoughby kevin@scispace.org.invalid
> Imagine that, a FROG ON-OFF switch, hardly the work
> for test pilots. -- Mike Collins
===
Subject: Re: The Grand Facade
If you think that your getting a ride now, wait till his
second term.
That is if he can convince Israel to become the 51st. state.
As an unAmerican fascist sympathizer, you'll watch tyrant 
bush
Ah, *that's* what you're talking about, some 
foreign country.
I didn't
> realize there was a foreign leader also named bush, just
like our duly
> elec and properly acting U.S. President.
> There are many who dispute that Dubya was either duly elec
or has
> been acting properly as U. S. President. The economy has
certainly
> gone to hell since he was sworn in.
===
Subject: Re: The Grand Facade
> There are many who dispute that Dubya was either duly elec
or has
> been acting properly as U. S. President. The economy has
certainly
> gone to hell since he was sworn in.
> Yeah! Exactly! I mean, it's not like he got more votes,
Sure he did! He got a clear majority of the only votes that
count. Everyone
who graduates from Electoral College gets to become President.
--
If you have had problems with Illinois Student Assistance
Commission
(ISAC),
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in the works.
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Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Check my proof please
Hi all,
Here is the problem.
Suppose a and b belong to a commutative ring R. If a is a
unit of R and
b^2
= 0, show that a + b is a unit of R.
Proof:
Consider a^-1 - a^-2*b. Then, (a^-1 - a^-2*b)b = a^-1*b -
a^-2*b*b =
a^-1*b - a^-2*b^2
==> a^-1*b = a^-2*b^2
but, since b^2 = 0
a^-1*b = 0
b must equal 0 since a is a unit, which is a non-zero element
and so a^-1
must be non-zero.
Therefore,
a + b = a + 0 = a
since a is a unit, then a + b is a unit.
Q.E.D.
===
Subject: Re: Check my proof please
Suppose A and B belong to a commutative ring.
> If A is a unit and B^2 = 0 show A+B is a unit.
(A+B)(A-B) = A^2 is a unit hence so are its factors. QED
Recall U,V units <=> UV unit, i.e. the set of all units
has the structure of a satura monoid, see my prior post
http://google.com/groups?selm=y8zlltxpmmd.fsf%
40nestle.ai.mit.edu
-Bill Dubuque
===
Subject: Re: Check my proof please
> Hi all,
Here is the problem.
> Suppose a and b belong to a commutative ring R. If a is a
unit of R and
b^2
> = 0, show that a + b is a unit of R.
>
You want to find a good candidate for (a+b)^-1. This can be
expanded as
a^-1 (1+x)^-1
where x=ba^-1. Now one way to figure out (1+x)^-1 is to expand
it as a
power
series:
1-x+x^2-x^3+x^4....
Ordinarily there is no reason why this should converge in a
ring without
topology (actually you cannot even define what convergence
means), but if
all
but finitely many terms in the infinite series are 
zero, then
you are fine.
This will be so if x^n=0 for some positive integer n.
--
Stephen Montgomery-Smith
stephen@math.missouri.edu
http://www.math.missouri.edu/~stephen
===
Subject: Re: Check my proof please
Thanks everyone!
Lurch
> Hi all,
Here is the problem.
> Suppose a and b belong to a commutative ring R. If a is a
unit of R
and
b^2
> = 0, show that a + b is a unit of R.
 You want to find a good candidate for (a+b)^-1. This can be
expanded as
> a^-1 (1+x)^-1
> where x=ba^-1. Now one way to figure out (1+x)^-1 is to
expand it as a
power
> series:
> 1-x+x^2-x^3+x^4....
> Ordinarily there is no reason why this should converge in a
ring without
> topology (actually you cannot even define what convergence
means), but if
all
> but finitely many terms in the infinite series 
are zero, then
you are
fine.
> This will be so if x^n=0 for some positive integer n.
> --
> Stephen Montgomery-Smith
> stephen@math.missouri.edu
> http://www.math.missouri.edu/~stephen
===
Subject: Re: Check my proof please
> Suppose a and b belong to a commutative ring R. If a is a
unit of R and
b^2
> = 0, show that a + b is a unit of R.
Proof:
> Consider a^-1 - a^-2*b.
Hint: I don't know if that came from someone 
else's hint, but
that is
a great place to start. You could start a correct one-line
proof from
here.
> Then, (a^-1 - a^-2*b)b = a^-1*b - a^-2*b*b = a^-1*b -
a^-2*b^2
True, but it doesn't get you anywhere.
> ==> a^-1*b = a^-2*b^2
That would only be true if the original had been equa to 0;
but, it
wasn't.
> but, since b^2 = 0
> a^-1*b = 0
> b must equal 0 since a is a unit, which is a non-zero
element and so a^-1
> must be non-zero.
Therefore,
> a + b = a + 0 = a
> since a is a unit, then a + b is a unit.
You should be very suspicious of this proof, because if it
were
correct, it would prove far more than the original statement.
It
proves that if b^2 = 0, then b = 0. That is false. Consider
the
integers mod 4. 2^2 = 0.
===
Subject: Re: Check my proof please
===
> ....
> Suppose a and b belong to a commutative ring R. If a is a
unit of R and
b^2
> = 0, show that a + b is a unit of R.
Proof:
Consider a^-1 - a^-2*b. Then, (a^-1 - a^-2*b)b = a^-1*b -
a^-2*b*b =
> a^-1*b - a^-2*b^2
==> a^-1*b = a^-2*b^2
> ....
Why?
If you suspect that a^-1 - (a^-2)*b is the inverse, just
multiply
it by a + b to see what happens.
Ken Pledger.
===
Subject: Re: Check my proof please
Visiting Assistant Professor at the University of Montana.
>Hi all,
>Here is the problem.
>Suppose a and b belong to a commutative ring R. If a is a
unit of R and
b^2
>= 0, show that a + b is a unit of R.
>Proof:
>Consider a^-1 - a^-2*b. Then, (a^-1 - a^-2*b)b = a^-1*b -
a^-2*b*b =
>a^-1*b - a^-2*b^2
>==> a^-1*b = a^-2*b^2
No. Why do you think that? You did not know that (a^-1 -
a^-2*b)b = 0,
so why do you conclude that a^{-1}*b = a^{-2}*b^2?
===
Subject: Subspace topology
Having trouble with this problem :
If L is a straight line in the plane, describe the topology L
inherits as a
subspace of R_l x R and as a subspace of R_l x R_l, where
R_l is the lower limit topology on the reals, and R is the
reals.
The topology on R_l x R consists of all rectangles with the
left side
containing its boundary points, and the other 3 sides open.
So, the
subspace topology
would be lines intersecting these rectangles, but I am having
trouble
describing the topology. I could just write that if the line
is contained
in the rectangle,
then the intersection is the line. If the line is not
contained in the
rectangle, then the intersection is the empty set, and go
through all the
examples.
But I think I am not thinking about this correctly. How else
can I
describe
these subspace topologies?
Thanks
===
Subject: Re: Daniel Kane, the most talen mathematicians to
come along
in the last 20 years
dannycyc@aol.com (Danny) asked:
> Does anybody know what important
> conjecture Daniel has proved?
The material from the following two web sites doesn't quite
answer the
question, but it may give someone else enough to track it
down. The second
reference means that Kane has a paper accep for publication
in the
Recently, Daniel has been conducting research with Ken Ono in
Number
Theory,
and he has already written two research papers. The first
paper on
asymptotic
formulas for partition functions has been accep for
publication in the
Ramanujan Journal. His second paper is even far more
impressive. Daniel
proves
a conjecture of George E. Andrews and Richard Lewis on crank
inequalities.
This
second paper employs the theory of theta functions, the
circle method for
automorphic forms, and a wide variety of complica
combinatorial
constructions. He has just submit the paper for publication -
and he is
only
16 years old!
And from
http://www.ams.org/cgi-bin/mstrack/accep_papers/proc
Kane, Daniel M., Resolution of a conjecture of Andrews and
Lewis
involving
cranks of partitions
===
Subject: Re: Daniel Kane, the most talen mathematicians to
come along
in the last 20 years
> Daniel Kane's home page:
Does anybody know what important conjecture Daniel has proved?
Afraid not.
MADISON, Wis. - A math prodigy who's been dubbed one of the
brightest
> young minds in the country was awarded a $50,000
scholarship from a
> nonprofit group that nurtures the profoundly gif.
How does that compare to severely gif?
<journalistic hyperbole snipped He applied to be a fellow by
submitting a work called Two Papers on
> the Theory of Partitions, a branch of additive number
theory.
What I see in Daniel is real, raw talent of the type I've
never seen
> before, said Ken Ono, a University of Wisconsin-Madison math
> professor who mentors Kane. He wakes up every day wanting
to prove a
> new theorem.
Kane already has proved a conjecture posed by national
experts in the
> field, including George Andrews of Penn State, arguably the
world's
> leading authority on the theory of partitions, Ono said.
Well Ono and Andrews are certainnly top guys, and Andrews's
_The Theory of Partitions_ is the definitive text on partition
theory.
--
His mind has been corrup by colousounds and shapes.
The League of Gentlemen
===
Subject: Re: Top 10 Reasons Why Alberta Sucks
> See now I was talking in generalities when I said pegging
the minimum
> wage too high only adds to the welfare roll
I believe you are operating with an exploded and out-of-date
theory
of the labor market. I happen to have an essay on-hand
explaining some
interesting possibilities. This essay is direc towards
somebody
who has been exposed to some economic theory.
I was deligh to find in a dictionary the word MUMPSIMUS,
which means stubborn persistence in an error that has been
exposed.
-- Joan Robinson
1.0 INTRODUCTION
If you take a class on economics, your teacher might tell you
that
wages and employment are determined by the demand and supply
of
labor, abstracting from price and wage stickiness, rigidities,
information asymmetries, etc. Your teacher might tell you
that the
demand function for labor is necessarily a declining function
of
wages, given profit-maximizing firms and standard 
assumptions
about
technology. And the teacher might say, incorrectly, that this
theory
applies in both the short and long run. The teacher teaching
this
incorrect theory might tell you that the labor-demand
function is
more elastic in the long run and that those elements abstrac
from
in the theory are often less important in the long-run.
Intro textbooks (e.g., Baumol and Blinder) present a
supply-and-demand equilibrium for the labor market that has
the
following properties:
1. Labor demand curves slope down.
2. A supply-and-demand equilibrium will result in the
equilibrium
price and quantity persisting unchanged as long as the data
for that equilibrium persist unchanged.
3. In drawing the labor demand curve, other prices (e.g. of
iron) and the market rate of return to financial capital
are taken as given.
The first condition implies that one should look at the
profit-maximizing firms' choices at 
each level of the (real)
wage. I do that in the example below.
The second condition implies that the solution of the 
firm's
problem should be such that the firm will continue to make
the same choices as long as prices persist at the same level.
I draw out the implications of the second condition below.
In particular, I show what configurations of prices and the
rate of profits are consistent with this condition.
The analysis below demonstrates that the other properties
of the textbook labor demand curve cannot be maintained. That
is, the price of iron and the market rate of return to
financial
capital must be different at different levels of the wage
for the firm to be in long run equilibrium. And the 
firm
in my example wants to adopt a more labor-intensive
technique at higher wages (for certain levels of the wage).
Thus, given the level of (net) output, the firms want to
hire more labor at higher wages.
In other words, the textbook model of the labor market is
logically inconsistent. I have an Excel spreadsheet that
permits
you to experiment with the numerical values behind some of the
models which I use:
<http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/
ChoiceOfTechnique
.xls>
So, as good economists have long recognized, the theory I am
attacking
is incorrect, as a matter of mathematics and logic.
For some reason, very few economists posting here will
acknowledge
this simple truth. I once again give economists a chance to
agree
with arithmetic.
If one wan to disagree with my conclusions, one could attempt
to produce a rational argument. Such an argument might:
o Point out a supposed error in my calculations.
o State special case conditions (e.g., on technology) that
rule
out my example and support the textbook model. For example,
one
might state that the textbook only applies when capital goods
are not used in production.
o Show how to develop the textbook model as an approximation,
and
specify when that approximation holds. For example, in intro
physics texts, one finds the false proposition that the period
of a pendulum does not depend on the initial displacement. But
this holds approximately for small angles, and well-written
textbooks point out in the derivation where one approximates
sin(x) by x.
o Show how to derive a textbook labor demand curve in my
example.
2.0 DATA ON TECHNOLOGY
Consider a very simple vertically-integra firm that produces a
single consumption good, corn, from inputs of labor, iron,
and (seed)
corn. All production processes in this example require a year
to
complete. All production processes exhibit Constant Returns
to Scale.
Two production processes are known for producing corn. These
processes
require the following inputs to be available at the beginning
of the
year for each bushel corn produced and available at the end
of the
year:
TABLE 1: INPUTS REQUIRED PER TON CORN PRODUCED
Process A Process B
1 Person-Year 1 Person-Year
2 Tons Iron 1/2 Tons Iron
2/5 Bushels Corn 3/5 Bushels Corn
Apparently, inputs of iron and corn can be traded off in
producing
corn outputs.
Iron is also produced by this firm. Two processes are known 
for
producing iron:
TABLE 2: INPUTS REQUIRED PER TON IRON PRODUCED
Process C Process D
1 Person-Year 275/464 Person-Years
1/10 Tons Iron 113/232 Tons Iron
1/40 Bushels Corn 0 Bushels Corn
Inputs of corn and iron can be traded off in producing iron.
The
process that uses less iron and more corn, however, also
requires
a greater quantity of labor input.
2.1 PRODUCTION FUNCTIONS
The data above allow for the specification of two well-behaved
production functions, one for corn and the other for iron. For
illustration, I outline how to construct the production
function
for corn.
Let L be the person-years of labor, Q1 be tons iron, and Q2 be
bushels corn available for inputs for corn-production during
the
production period (a year). Let X1 be the bushels corn
produced
with Process A, and X2 be the bushels corn produced with
Process B.
The production function for corn is the solution of an
optimization
problem in which as much corn as possible is produced from the
given inputs. Accordingly, the production function for corn is
found as the solution to the Linear Program in Display 1:
Max X = X1 + X2
X1 + X2 <= L
2*X1 + (1/2)*X2 <= Q1 (1)
(2/5)*X1 + (3/5)*X2 <= Q2
X1 >= 0, X2 >= 0
Let f(L, Q1, Q2) be the solution of this LP, that is, the
production
function for corn. (This production function is not
Leontief.) The
production functions construc in this manner exhibit
properties
typically assumed in neoclassical economics. In particular,
they
exhibit Constant Returns to Scale, and the marginal product,
for
each input, is a non-increasing step function. The production
functions are differentiable almost everywhere.
The point of this example, that sometimes a vertically integra
firm will want to hire more labor per unit output at higher
wages,
is compatible with the existence of many more processes for
producing
each commodity. As more processes are used to construct the
production
functions, the closer they come to smooth,
continuously-differentiable
production functions. The point of this example seems to be
compatible
with smooth production functions. It also does not depend on
the
circular nature of production in the example, in which corn
is used
to produce more corn.
2.2 TECHNIQUES
A technique consists of a process for producing iron and a
process
for producing corn. Thus, there are four techniques in this
example.
They are defined in Table 3.
TABLE 3: TECHNIQUES AND PROCESSES
Technique Processes
Alpha A, C
Beta A, D
Gamma B, C
Delta B, D
3.0 QUANTITY FLOWS
I want to consider a couple of different levels at which this
firm can operate the processes comprising the techniques.
First,
suppose Process A is used to produce 1 41/49 Bushels corn, and
Process C is used to produce 4 4/49 Tons iron. The quantity
ßows
shown in Table 4 result.
TABLE 4: THE ALPHA TECHNIQUE PRODUCING CORN NET
INPUTS Process C Process A
Labor 4 4/49 Person-Years 1 41/49 Person-Years
Iron 20/49 Tons Iron 3 33/49 Tons Iron
Corn 5/49 Bushels Corn 36/49 Bushels Corn
OUTPUTS 4 4/49 Tons Iron 1 41/49 Bushels Corn
LABOR-INTENSITY: 5 45/49 Person-Years Per Bushel
When the firm operates these processes in parallel, it 
requires
a total of 41/49 Bushels corn as input. The output of the
corn-producing process can replace this input, leaving a net
output of one Bushel corn. Notice that the total inputs of
iron are 20/49 + 3 33/49 = 4 4/49 Tons iron, which is exactly
replaced by the output of Process C. So Table 4 shows a
technique
in which 5 45/49 Person-Years labor are used to produce a net
output of one Bushel corn. The firm, when operating this
technique
can produce any desired output of corn by scaling both
processes
equally.
Table 5 shows the application of the same sort of arithmetic
to
the Beta technique. The labor-intensity of the Beta technique
is
lis.
TABLE 5: THE BETA TECHNIQUE PRODUCING CORN NET
INPUTS Process D Process A
Labor 3 304/357 Person-Years 1 2/3 Person-Years
Iron 3 59/357 Tons Iron 3 1/3 Tons Iron
Corn 0 Bushels Corn 2/3 Bushel Corn
OUTPUTS 6 178/357 Tons Iron 1 2/3 Bushel Corn
LABOR-INTENSITY: 5 185/357 Person-Years Per Bushel
Neither the Gamma nor the Delta technique are 
profit-maximizing
for the prices considered below.
+------------------------------------------+
| THE FIRM |
| |
| Inventory <--------------------------+ |
| | | |
Labor | | Steel+Corn+Labor -> Steel -->+ |
Market | |/ /| /| |
------->+-------------+ | | Corn
(wage | |/ | | Market
given) | Steel+Corn+Labor -> Corn --->+------->
| | (price
+------------------------------------------+ given)
FIGURE 1: A VERTICALLY INTEGRA FIRM
4.0 PRICES
Which technique will the firm adopt, if any? The answer
depends, in this analysis, on which is most profitable. So one
has to consider prices. I assume throughout that inputs of
iron,
corn, and labor are charged at the start of the year. Corn is
the numeraire; its price is unity throughout. Two different
levels of wages are considered.
4.1 PRICES WITH LOW WAGE
Accordingly, assume wages are initially 3/2780 Bushels per
Person-Year. By assumption, the firm neither buys nor sells
iron on
the market. The firm produces iron solely for its own use.
Still,
the firm must enter a price of iron on its books. I assume an
initial price of 55/1112 Bushels per Ton.
Table 6 shows accounting with these prices. The column labeled
cost shows the cost of the inputs needed to produce one unit
output, a bushel corn or a ton iron, depending on the process.
Accounting profits for a unit output are the difference 
between
the price of a unit output and this cost. The rate of
(accounting)
profits, shown in the last column, is the ratio of accounting
profits to the cost. The rate of profits is 
independent of
the scale at which each process is opera.
TABLE 6: COSTS, WAGE 3/2780 BUSHELS PER PERSON-YEAR,
PRICE OF IRON 55/1112 BUSHELS PER TON
INDUSTRY PROCESS COST PROFITS
Corn A 2*(55/1112) + (2/5)*1
+ 1*(3/2780) = 1/2 100%
Corn B (1/2)*(55/1112) + (3/5)*1
+ 1*(3/2780) = 6959/11120 60%
Iron C (1/10)*(55/1112) + (1/40)*1
+ 1*(3/2780) = 69/2224 59%
Iron D (113/232)*(55/1112) + 0
+ (275/464)*(3/2780) = 55/2224 100%
These prices are compatible with the use of the Beta technique
to produce a net output of corn. The Beta technique specifies
that
Process A be used to produce corn and process D be used to
produce
iron. Notice that Process B is more expensive than Process A,
and
that process C is more expensive than Process D. These prices
do
not provide signals to the firm that processes outside the 
Beta
technique should be adop. The vertically-integra firm is
making a rate of profit of 100% in producing corn with the 
Beta
technique. The same rate of profits are earned in producing
corn
and in reproducing the used-up iron by an iron-producing
process.
4.2 ONE SET OF PRICES WITH HIGHER WAGE
Suppose this firm faces a wage more than 20 times higher,
namely
109/4040 Bushels per Person-Year. Consider what happens if
the firm
doesn't revalue the price of iron on its books. Table 7 
shows
this
case. Since labor enters into each process, the rate of 
profits
has declined for all processes. The ratio of labor to the
costs of
the other inputs is not invariant ac processes. Thus, the
rate of profits has declined more in some processes than in
others. Notice especially, than the rate of profits is no
longer
the same in the processes, A and D, that comprise the Beta
technique.
TABLE 7: COSTS, WAGE 109/4040 BUSHELS PER PERSON-YEAR,
PRICE OF IRON 55/1112 BUSHELS PER TON
INDUSTRY PROCESS COST PROFITS
Corn A 2*(55/1112) + (2/5)*1
+ 1*(109/4040) = 0.5259 90.1%
Corn B (1/2)*(55/1112) + (3/5)*1
+ 1*(109/4040) = 0.6517 53.4%
Iron C (1/10)*(55/1112) + (1/40)*1
+ 1*(109/4040) = 0.05693 -13.1%
Iron D (113/232)*(55/1112) + 0
+ (275/464)*(109/4040) = 0.04008 23.4%
This accounting data does not reveal the firm's 
rate of return
in operating the Beta technique. The firm cannot be
simultaneously
making both 23% and 90% in operating that technique.
Furthermore,
this data provides a signal to the firm to withdraw from iron
production and make only corn. So this data says that
something
must change.
4.3 ANOTHER SET OF PRICES
Perhaps all that is needed is to re-evaluate iron on the
firm's books. Higher wages have made iron more 
valuable. Table
8 shows costs and the rate of profits when iron is
evalua at an accounting price of 0.106 Bushels per Ton.
TABLE 8: COSTS, WAGE 109/4040 BUSHELS PER PERSON-YEAR,
PRICE OF IRON 0.10569123726 BUSHELS PER TON
INDUSTRY PROCESS COST PROFITS
Corn A 2*(0.106) + (2/5)*1
+ 1*(109/4040) = 0.6384 56.65%
Corn B (1/2)*(0.106) + (3/5)*1
+ 1*(109/4040) = 0.6798 47.10%
Iron C (1/10)*(0.106) + (1/40)*1
+ 1*(109/4040) = 0.06255 68.97%
Iron D (113/232)*(0.106) + 0
+ (275/464)*(109/4040) = 0.06747 56.65%
This revaluation of iron reveals that the firm makes a rate
of profits of 57% in operating the Beta technique. The 
firm
makes
the same rate of profits in producing corn and in producing 
its
input of iron. But the manager of the iron-producing process
would
soon notice that the cost of operating process C is cheaper.
4.4 FINAL EQUILIBRIUM PRICES
So the firm would ultimately switch to using process C
to produce iron. The price of iron the firm would enter on its
books would fall somewhat. Table 9 shows the accounting with a
price of iron of 10/101 Bushels per Ton. The firm has adop
the cheapest process for producing iron, and the rate of
profits
is the same in both corn-production and iron-production. The
accounting for this vertically-integra firm is internally
consistent.
TABLE 9: COSTS, WAGE 109/4040 BUSHELS PER PERSON-YEAR,
PRICE OF IRON 10/101 BUSHELS PER TON
INDUSTRY PROCESS COST PROFITS
Corn A 2*(10/101) + (2/5)*1
+ 1*(109/4040) = 5/8 60%
Corn B (1/2)*(10/101) + (3/5)*1
+ 1*(109/4040) = 2553/4040 58%
Iron C (1/10)*(10/101) + (1/40)*1
+ 1*(109/4040) = 25/404 60%
Iron D (113/232)*(10/101) + 0
+ (275/464)*(109/4040) = 24,075/374,912
54%
5.0 CONCLUSIONS
Table 10 summarizes these calculations. The ultimate result of
a higher wage is the adoption of a more labor-intensive
technique.
If this firm continues to produce the same level of net output
and maximizes profits, its managers will want to employ more
workers
at the higher of the two wages considered.
TABLE 10: PROFIT-MAXIMIZING FIRM ADOPTS MORE LABOR-INTENSIVE
TECHNIQUE AT HIGHER WAGE
LABOR-INTENSITY OF
WAGE CORN-PRODUCING TECHNIQUE
3/2780 Bushels Per Person-Year 5 185/357 Person-Years Per
Bushel
109/4040 Bushels Per Person-Year 5 45/49 Person-Years Per
Bushel
So much for the theory that wages and employment are
determined
by the interaction of well-behaved supply and demand curves
on the
labor market.
APPENDIX A: A FORMAL MODEL
Let
Xa = Bushels corn produced (g) by process A
Xb = Bushels corn produced by process B
Xc = Tons iron produced by process C
Xd = Tons iron produced by process D
p = the accounting price of iron (corn is numeraire)
w = wage
r = rate of (accounting) profits
Q1 = Tons iron in firm's inventory at start of 
period
Q2 = Bushels corn in firm's inventory at start 
of period
Consider a firm attempting to maximize the value of the stock
in its
possession at the end of the year:
Given p, w, Q1, and Q2
Choose Xa, Xb, Xc, and Xd
To Maximize Xa + Xb + p Xc + p Xd
+ p Q1 + Q2 - ( ( w + 2 p + (2/5) ) Xa
+ ( w + (1/2) p + (3/5) ) Xb
+ ( w + (1/10) p + (1/40) ) Xc
+ ( (275/464) w + (113/232) p ) Xd )
Such that
(w + 2 p + (2/5)) Xa
+ (w + (1/2) p + (3/5)) Xb
+ (w + (1/10) p + (1/40)) Xc
+ ((275/464) w + (113/232) p) Xd <= Q1 p + Q2
Xa, Xb, Xc, Xd >= 0
The amount of financial capital the firm has at 
the start of the
production period is given by the value of the initial
inventory. This
given financial capital provides the constraint on how much
corn and
iron can be produced. In a model in which future prices are
foreseen,
the value of leftover inventory at the end of the period is
the
difference between the value of the initial inventory and the
amount
of that value consumed in production. The firm maximizes the
sum of
the value of newly produced corn and iron and the value of
leftover
inventory.
Note that initial value of the inventory, (p Q1 + Q2), is a
constant
in the above LP. There is no need to include a constant term
in the
objective function. Thus, the profit-maximizing 
firm solves the
following program:
Given p, w, Q1, and Q2
Choose Xa, Xb, Xc, and Xd
To Maximize (1 - w - 2 p - (2/5)) Xa
+ (1 - w - (1/2) p - (3/5)) Xb
+ (p - w - (1/10) p - (1/40)) Xc
+ (p - (275/464) w - (113/232) p) Xd
Such that
(w + 2 p + (2/5)) Xa
+ (w + (1/2) p + (3/5)) Xb
+ (w + (1/10) p + (1/40)) Xc
+ ((275/464) w + (113/232) p) Xd <= Q1 p + Q2
Xa, Xb, Xc, Xd >= 0
The dual Linear Program is:
Given p, w, Q1, and Q2
Choose r
To Minimize (Q1 p + Q2) r
Such That
(w + 2 p + (2/5)) r >= 1 - w - 2 p - (2/5)
(w + (1/2) p + (3/5)) r >= 1 - w - (1/2) p - (3/5)
(w + (1/10) p + (1/40)) r >= p - w - (1/10) p - (1/40)
((275/464) w + (113/232) p) r >= p - (275/464) w - (113/232) p
r >= 0
Or:
Given p, w, Q1, and Q2
Choose r
To Minimize (Q1 p + Q2) r
Such That
(w + 2 p + (2/5))(1 + r) >= 1
(w + (1/2) p + (3/5))(1 + r) >= 1
(w + (1/10) p + (1/40))(1 + r) >= p
((275/464) w + (113/232) p)(1 + r) >= p
r >= 0
If a constraint in the dual is met with inequality in the
solution, the
corresponding process in the primal will be opera at a level
of zero.
For firms to continue production unaltered from period to
period, both
corn and iron must be produced each period. For corn to be
produced,
either the first or the second constraint in the dual must be
met with
equality. Likewise, for iron to be produced, the third or the
fourth
constraint in the dual must be met with equality. Hence, for
the
analyzed firms to be in equilibrium, the vertically-integra
industry
must be on the so-called factor-price frontier for that
industry.
Nothing guarantees that the firms will be able to sell their
output
at any given location on the factor-price frontier. Whether
prices that
allow firms to be in equilibrium are realized is a question
that is
not addressed by this formal model.
--
Try
http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/
Bukharin.html
To solve Linear Programs: .../LPSolver.html
r c A game: .../Keynes.html
v s a Whether strength of body or of mind, or wisdom, or
i m p virtue, are found in proportion to the power or
wealth
e a e of a man is a question fit perhaps to be discussed
by
n e . slaves in the hearing of their mastebut highly
@ r c m unbecoming to reasonable and free men in search of
d o the truth. -- Rousseau
===
Subject: [MatLab] QR factorization ... :-(
[Q,R,E] = qr(A,0) for full matrix A, produces an economy-size
decomposition in which E is a permutation vector, so that
A(:,E) =
Q*R. The column permutation E is chosen so that abs(diag(R))
is
decreasing.
This command is perfect ... but i need that abs(diag(R)) is
INCREASING!!!
There is a specific command?
If there in not a specific command, where can i 
find the code
for the
MatLab command [Q,R,E]?
Thank,
Luca
===
Subject: Re: System of differential equations
Originator: israel@math.ubc.ca (Robert Israel)
i know there exists a general solution for systems oif linear
differential
> equations with constant coefficients. In my case the system
matrix
depends
> linearly on the states. This results in a system where the
derivatives
of
> the states x1,x2 depend not only linearly on themselves but
also
quadratic:
 dx1/dt=A11*(x1;x2)+(x1;x2)'*A12*(x1;x2)
> dx2/dt=A21*(x1;x2)+(x1;x2)'*A22*(x1;x2)
 where the matrices A11 and A21 have the size 1x2 and the
matrices A12
and
> A22 the size 2x2.
Since nobody else has responded to this one, I'll mention a
special
case that you may find useful.
A system of the form
dZ/dt = Z(t) A(t) Z(t) + Z(t) K(t) + transpose(K(t)) Z(t)
is known as a Matrix Riccati differential equation (where Z,
A, K are n x n
matrices). The substitution
Z = U^(-1), dZ/dt = - U^(-1) dU/dt U^(-1),
transforms it (assuming Z is invertible) to
U'(t) = - A(t) - K(t) U(t) - U(t) transpose(K(t))
which is a first order linear system.
With
[ x1(t) x2(t) ] [ a(t) b(t) ] [ c(t) d(t) ]
Z(t) = [ x2(t) x1(t) ], A(t) = [ b(t) a(t) ], K(t) = [ d(t)
c(t) ]
your system would fit this pattern in the case
A11 = [ 2c(t) 2d(t) ], [ a(t) b(t) ] [ b(t) a(t) ]
A21 = [ 2d(t) 2c(t) ], A12 = [ b(t) a(t) ], A22 = [ a(t) b(t)
]
or with
[ x1(t) x2(t) ] [ a(t) b(t) ] [ c(t) -d(t) ]
Z(t) = [ x2(t) -x1(t) ], A(t) = [ b(t) -a(t) ], K(t) = [ d(t)
c(t) ]
it would fit in the case
A11 = [ 2c(t) -2d(t) ], [ a(t) b(t) ] [ -b(t) a(t) ]
A21 = [ 2d(t) 2c(t) ], A12 = [ b(t) -a(t) ], A22 = [ a(t)
b(t) ]
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: cause and effect picture
> The matter which I am desperately struggling with, concerns
a
> time sequence of event; with the assumption that 
Ôcause'
does NOT
> follow Ôeffect'
You say that it is an assumption that Ôcause does not follow
effect'?
You have not assumed anything. There is simply no picture
offered here that
distinguishes cause from effect.
The rest was too much for me
jJ
> When reading certain high court judgement reasons, it is
clear to me
> that often some quite profound thinking in being exercised.
> But when I dialog with Ôlegal people': real 
live ones or
via *.legal.*
> newsgroups; it seems that they are just Ôphrase matching
clerks.
> I'm convinced of the left-brain/right-brain theory which
divides
> humanity into spacial/verbal thinkers.
> The matter which I am desperately struggling with, concerns
a
> time sequence of event; with the assumption that 
Ôcause'
does NOT
> follow Ôeffect'.
> As a science graduate, I find it best and natural to
demonstrate my
> argument by a minimalist model. None of the Ôlegal 
people'
know
> what I'm talking about and worse still they 
don't volunteer
that
> my model is (to them) not understandable.
> Is the following understandable ? How can I impove it ?
> Please answer some of the questions.
> The following conditions apply:-
> 1. The supplier's billing system is wrong: over charging 
by
10%.
> 2. The corrrect consumption/billing is $10 per period/month.
> 3. The client with-holds payment (after repea written
complaints
> and written acknowledgement of errors in the supplier's
billing
system)
> in order to get a court hearing to expose the supplier's
billing
system.
> 4. A debtor is said to NOT have a bona fide defense if he
admits owing
> at least as much as the claim (amount sued for).
> Month True Billed Event
> ==================================
> 1 10 11
> 2 20 22
> 3 30 33 A
> 4 40 44 B
> 5 50 55 C
> 6 60 66
> 7 70 77 D
> 8 80 88
> 9 90 99 E
> -----
> The events labeled in the table are as follows:
> A = demand letter from supplier ( for $33 obviously)
> B = demand letter from supplier's lawer - for $33
> C = summons served on absent client for $33 - not received
by debtor.
> D = Default judgment - because of undefended; because
summons
> not received.
> E = ignorant of the served summons, the client delivers a
letter (to
> the supplier - not the lawer) with this calculation-table,
writing:
> it is NOT about non payment of due debt, in fact I now admit
> owing $90, which is more than the $33 claim.
> Please accept my calculations or show were they are wrong,
so
> that I may settle my account.
> -----
> Q1. Has the client got a bona fide defence for with-holding
payment
> at period 3 ?
> Q2. If the client had a bona fide defence for with-holding
payment
> at time 3, has this been removed at time 9, because he
admits
> owing $90, whereas the claim is for $33 ?
> Q3. When is the date of accrual of the cause of action ?
> (which is defined as: when the material facts on which it is
based
> have been discovered or ought to have been discovered by the
plaintiff,
> by the exercise of reasonable diligence. )
> Q4. If at the appeal level the justice department
acknowledges the
> absurdity of Q2 (ie. claiming that events AFTER the default
judgment
> [month 9] can justify the default judgment [month 7]), is
the
> bona fide defence of month 3 lost, since at summons
commencing
> action [month 5] the client admits owing $50;
> which is more than the claim of $33 ?
> Q5. If so, does this mean that the summons has been
Ôdecoupled' from
> the demand letters ?
> Q6. What are the implications of the fact that the
Ôthreshold date'
> (when the calcula/admit amount owing exceeds the claim)
> could happen arbitrarly, at any time between A and E.
> Thanks for any answe(possibly emailed
> to: eas-lab@absamail.co.za)
> == Chris Glur.
===
Subject: Counting unique sor outcomes.
Summary:
I'm not a mathematician, so I probably misuse a number of
terms
in this problem. I'm taking calc I and statistics right now.
I'm my stats class yesterday, the instructor covered how to
calculate
the total possible number of outcomes in a sample set. It's
painfully
obvious, of course. If you have 2 6-sided dice, the number of
possible outcomes is 6*6. I asked how to calculate the total
possible number of outcomes if you assumed that a 3 on die 1
and a
5 on a die 2 was the same as a 5 on die 1 and a 3 on die 2. He
didn't know the answer. (I'm taking the class 
at a community
college, and this guy's an electrical engineer.)
So I went home and star to work on the problem. This is what I
came up with. It's not homework. I was just curious.
http://www.technolalia.com/~ndronen/math-problem.txt
Is there a known equation for what I'm trying to express? I
have
the seeds of the solution, but I'm not sure how to approach
this
mathematically. I'm a programmer, and I could write a 
program
to
compute this, but I'm curious how to express it
mathematically.
Nicholas
--
Why shouldn't I top-post? http://www.aglami.com/tpfaq.html
Meanings are another story.
http://www.ifas.org/wa/glossolalia.html
===
Subject: Re: Counting unique sor outcomes.
Visiting Assistant Professor at the University of Montana.
>I'm not a mathematician, so I probably misuse a number of
terms
>in this problem. I'm taking calc I and statistics right 
now.
>I'm my stats class yesterday, the instructor covered how to
calculate
>the total possible number of outcomes in a sample set. It's
painfully
>obvious, of course. If you have 2 6-sided dice, the number of
>possible outcomes is 6*6. I asked how to calculate the total
>possible number of outcomes if you assumed that a 3 on die 1
and a
>5 on a die 2 was the same as a 5 on die 1 and a 3 on die 2.
He
>didn't know the answer. (I'm taking the class 
at a community
>college, and this guy's an electrical engineer.)
>So I went home and star to work on the problem. This is what
I
>came up with. It's not homework. I was just curious.
> http://www.technolalia.com/~ndronen/math-problem.txt
>Is there a known equation for what I'm trying to express? I
have
>the seeds of the solution, but I'm not sure how to approach
this
>mathematically. I'm a programmer, and I could write a
program to
>compute this, but I'm curious how to express it
mathematically.
The number of distinct throws of n dice of r sides each (so
that 2-3 is
the same as 3-2, etc) is equivalent to a standard problem
called
combinations with repetitions.
The formula is that it is equal to n+r-1 choose n, that is,
(n+r-1)!/n!(r-1)!
===
Subject: Re: Counting unique sor outcomes.
>I'm my stats class yesterday, the instructor covered how to
calculate
>the total possible number of outcomes in a sample set. It's
painfully
>obvious, of course. If you have 2 6-sided dice, the number of
>possible outcomes is 6*6. I asked how to calculate the total
>possible number of outcomes if you assumed that a 3 on die 1
and a
>5 on a die 2 was the same as a 5 on die 1 and a 3 on die 2.
He
>didn't know the answer. (I'm taking the class 
at a community
>college, and this guy's an electrical engineer.)
The number of outcomes coun without order for n dice with m
faces
is the same as the number of ways to put n indistinguishable
balls
in m different bins. The answer is the binomial coefficient
(n+m-1 choose m) = (n+m-1)!/(m! (n-1)!). To see this, note
that
you can write an outcome as a sequence of m o's and n-1 
|'s,
where
an o is a ball and a | is the wall between two bins. Thus 5
and 3
on your dice corresponds to ||o||o|.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Counting unique sor outcomes.
>>I'm my stats class yesterday, the instructor covered how 
to
calculate
>>the total possible number of outcomes in a sample set. 
It's
painfully
>>obvious, of course. If you have 2 6-sided dice, the number
of
>>possible outcomes is 6*6. I asked how to calculate the total
>>possible number of outcomes if you assumed that a 3 on die
1 and a
>>5 on a die 2 was the same as a 5 on die 1 and a 3 on die 2.
He
>>didn't know the answer. (I'm taking the 
class at a community
>>college, and this guy's an electrical engineer.)
> The number of outcomes coun without order for n dice with m
faces
> is the same as the number of ways to put n
indistinguishable balls
> in m different bins. The answer is the binomial coefficient
> (n+m-1 choose m) = (n+m-1)!/(m! (n-1)!). To see this, note
that
> you can write an outcome as a sequence of m o's and n-1
|'s, where
> an o is a ball and a | is the wall between two bins. Thus 5
and 3
> on your dice corresponds to ||o||o|.
I think that answer is actually for m dice with n faces. In
particular,
the case m=2, n=6 gives (7 choose 2) = 21.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: Counting unique sor outcomes.
OK, before we start, this problem has nothing to do with
statistics. If you
assume (1,2) is the same as (2,1) then this outcome is twice
as likely as
(1,1). So the formula cannot be used to calculate
probabilities, as not all
outcomes are equally probable.
The are 6 * 6 outcomes if order matters.
Of these, 6 have no order - being (1,1), (2,2) .. (6,6)
This leaves 30 where order matters. This is 15 sets of pairs
where the
order
is reversed.
So the total number of outcomes is 6 + 15 = 21.
More generally, n sided dice, n^2 outcomes with order,
n pairs the same, so (n^2-n) where the two number aren't the
same,
so n+ (n^2-n)/2 outcomes if order doesn't matter.
n+(n^2-n)/2= n*(n+1)/2
I note that your web page derived the same thing a different
way, by
summing
1,2...n. Just as good.
> I'm not a mathematician, so I probably misuse a number of
terms
> in this problem. I'm taking calc I and statistics right 
now.
> I'm my stats class yesterday, the instructor covered how 
to
calculate
> the total possible number of outcomes in a sample set. 
It's
painfully
> obvious, of course. If you have 2 6-sided dice, the number
of
> possible outcomes is 6*6. I asked how to calculate the total
> possible number of outcomes if you assumed that a 3 on die
1 and a
> 5 on a die 2 was the same as a 5 on die 1 and a 3 on die 2.
He
> didn't know the answer. (I'm taking the 
class at a community
> college, and this guy's an electrical engineer.)
> So I went home and star to work on the problem. This is
what I
> came up with. It's not homework. I was just curious.
> http://www.technolalia.com/~ndronen/math-problem.txt
> Is there a known equation for what I'm trying to express? 
I
have
> the seeds of the solution, but I'm not sure how to 
approach
this
> mathematically. I'm a programmer, and I could write a
program to
> compute this, but I'm curious how to express it
mathematically.
 Nicholas
> --
> Why shouldn't I top-post? http://www.aglami.com/tpfaq.html
> Meanings are another story.
http://www.ifas.org/wa/glossolalia.html
===
Subject: Re: Counting unique sor outcomes.
>> I'm not a mathematician, so I probably misuse a number of
terms
>> in this problem. I'm taking calc I and statistics right
now.
>> I'm my stats class yesterday, the instructor covered how
to calculate
>> the total possible number of outcomes in a sample set.
It's painfully
>> obvious, of course. If you have 2 6-sided dice, the number
of
>> possible outcomes is 6*6. I asked how to calculate the
total
>> possible number of outcomes if you assumed that a 3 on die
1 and a
>> 5 on a die 2 was the same as a 5 on die 1 and a 3 on die
2. He
>> didn't know the answer. (I'm taking the 
class at a
community
>> college, and this guy's an electrical engineer.)
>> So I went home and star to work on the problem. This is
what I
>> came up with. It's not homework. I was just curious.
>> http://www.technolalia.com/~ndronen/math-problem.txt
>> Is there a known equation for what I'm trying to express?
I have
>> the seeds of the solution, but I'm not sure how to
approach this
>> mathematically. I'm a programmer, and I could write a
program to
>> compute this, but I'm curious how to express it
mathematically.
Thanks for responding.
PW> OK, before we start, this problem has nothing to do with
statistics. If
you
PW> assume (1,2) is the same as (2,1) then this outcome is
twice as likely
as
PW> (1,1). So the formula cannot be used to calculate
probabilities, as not
all
PW> outcomes are equally probable.
Understood.
PW> The are 6 * 6 outcomes if order matters.
PW> Of these, 6 have no order - being (1,1), (2,2) .. (6,6)
PW> This leaves 30 where order matters. This is 15 sets of
pairs where the
order
PW> is reversed.
I follow you.
PW> So the total number of outcomes is 6 + 15 = 21.
PW> More generally, n sided dice, n^2 outcomes with order,
PW> n pairs the same, so (n^2-n) where the two number aren't
the same,
PW> so n+ (n^2-n)/2 outcomes if order doesn't matter.
PW> n+(n^2-n)/2= n*(n+1)/2
Does this apply when you vary the number of dice and the
number of
sides to each die (assuming that each die has the same number
of
sides as the others)? (I'd expect to see a variable for the
number
of dice in the equation.) What I'm trying to 
find on the web
page
is an equation that'll handle any number of any-sided dice,
including
1 4-sided die and 10 8-sided dice.
Nicholas
--
Why shouldn't I top-post? http://www.aglami.com/tpfaq.html
Meanings are another story.
http://www.ifas.org/wa/glossolalia.html
===
Subject: Abstract Algebra Conundrums
I'm having trouble with these 2 proofs:
If gcd(a,c) = 1, and gcd(b,c) = 1, prove that (ab,c)=1.
If a divides c and b divides c and gcd(a,b) = d, prove that
gcd(a,b)=1.
Thanks for all your help.
Subject: Re: Abstract Algebra Conundrums
===
> I'm having trouble with these 2 proofs:
> If gcd(a,c) = 1, and gcd(b,c) = 1, prove that (ab,c)=1.
assume (ab,c)=d . Let a prime p divide d
then p|ab and p|c
p|ab implies . . . (basic number theory)
> If a divides c and b divides c and gcd(a,b) = d, prove that
gcd(a,b)=1.
contradiction a=2; b=4; c=8
===
Subject: Re: Abstract Algebra Conundrums
> I'm having trouble with these 2 proofs:
If gcd(a,c) = 1, and gcd(b,c) = 1, prove that (ab,c)=1.
If a divides c and b divides c and gcd(a,b) = d, prove that
gcd(a,b)=1.
Thanks for all your help.
Using notation a|b for a divides b:
For the first problem you can do proof by contradiction.
Assume (ab,
c)=k with k>1. There is some prime p that divides k so p|ab
and p|c.
Since p|ab and p prime, p|a or p|b. But then p is a common
divisor of
either a and c, or b and c.
Are you sure you typed the right thing for the second problem?
===
Subject: Re: Abstract Algebra Conundrums
> I'm having trouble with these 2 proofs:
If gcd(a,c) = 1, and gcd(b,c) = 1, prove that (ab,c)=1.
If a divides c and b divides c and gcd(a,b) = d, prove that
gcd(a,b)=1.
Thanks for all your help.
Using notation a|b for a divides b:
For the first problem you can do proof by contradiction.
Assume (ab,
> c)=k with k>1. There is some prime p that divides k so p|ab
and p|c.
> Since p|ab and p prime, p|a or p|b. But then p is a common
divisor of
> either a and c, or b and c.
Are you sure you typed the right thing for the second problem?
You're right, the conclusion was supposed to be prove that 
ab
divides
cd. But, working late last night, I figured out how to do that
one.
ak=c
bj=c
ax+by=d
cd=(ax+by)ak=a^2kx+abky=acx+abky=abjx+abky=ab(jx+ky)
so ab divides cd.
Thanks again.
===
Subject: Re: Abstract Algebra Conundrums
Shouldn't that be conundra?
--
===
Subject: Re: Abstract Algebra Conundrums
>Shouldn't that be conundra?
Well, it could be that if conundrum was a Latin noun
(second declension, neuter). But in fact it isn't Latin
at all, though it may be a parody of some Latin term
(the etymology is unknown). And the plural is conundrums
in the OED's citations, going back to Ben Jonson in 1605.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Abstract Algebra Conundrums
===
> I'm having trouble with these 2 proofs:
> If gcd(a,c) = 1, and gcd(b,c) = 1, prove that (ab,c)=1.
Try using the definitions and see what comes out:
For example:
gcd(a, c) = 1 -> a x + c y = 1
gcd(b, c) = 1 -> b r + c s = 1
To prove
gcd(ab, c) = 1, if we could show a relation of the form ab*w
+ c*f = 1, we
could be done.
What about 1 * 1 = 1 = (ax + cy) (br + cs) = 1, so
axbr + axcs + brcy + cscy = ab (xr) + c (axs + bry + csy)
So, can you take it from there?
Since ab w + cf = 1 = ab (xr) + c (axs + bry + csy), can you
draw any
conclusions?
Make sense?
> If a divides c and b divides c and gcd(a,b) = d, prove that
gcd(a,b)=1.
Here again, look at the definitions.
What does it mean, by definition, when a | c, etc.?
a | c -> c = a x
b | c -> c = b y
gcd(a, b) = d -> a r + b s = d
Can we derive anything from these?
HTH, Flip
> Thanks for all your help.
===
Subject: Etymology of Center and Normal
Does anyone know why the name center and normal were chosen in
relation
to group theory? And for that matter, ring and module?
-Tyler G. Smith
===
Subject: Re: Etymology of Center and Normal
tgsmith
> Does anyone know why the name center and normal were chosen
in
relation
> to group theory? And for that matter, ring and module?
> -Tyler G. Smith
Hard to say _why_ this or that word was chosen, but as for
when, quite a
bit
of work has been done on mathematical etymology. E.g.
http://members.aol.com/jeff570/mathword.html
LH
===
Subject: Re: Etymology of Center and Normal
>Does anyone know why the name center and normal were chosen
in
relation
>to group theory? And for that matter, ring and module?
>-Tyler G. Smith
If e and r commute, then er = re, so center = centre.
This is the most common word which ßips two letters
between American and British English,
so its meaning should relate to commutativity.
Another such word is the distance one commutes, in meters =
metres.
There are other identities like colour = color.
The u is the first letter of unit.
--
Wan: Experts at choosing the best of 100+ applicants for a
position.
Register as a California voter by September 22, and vote on
October 7.
Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California
Microsoft Research and CWI
===
Subject: Re: Geodetic spheres
> Synergetics I and II by inventor R. Buckminster Fuller (at
www.bfi.org).
Yes. It can be shown that when all vertices lie on a sphere
and
maximum possible number of equilateral triangles form its
sides , the
icosahedron is the Platonic solid solution. When each side is
further
sub-divided into 3,5,.. 32 parts, B.Fuller obtains
3p(football),
5p,.. 32p etc. frequency polygons. In these closer
approximations to
the sphere, all faces are only approximately equilateral
triangles,that make up 12 pentagons at each vertex and
hexagons
elsewhere. Such subdivision is the basis of his renowned
geodesic
domes design.
===
Subject: Re: Geodetic spheres
>That message is not avaible to me. Maybe it's removed from
the server;
maybe
>it's only my connection which is removed from the server?
>Surfing on the net looking for these things I found two
online book called
>Synergetics I and II by inventor R. Buckminster Fuller (at
www.bfi.org).
Has
>someone read them? Are they valueable for a mathematician or
are they
>hopelessly out of date? They seem a bit strange but maybe
they contain
>something cool?
>
The quote from the dedication in Synergetics shows what Bucky
Fuller
thought about traditional geometry he was taught.
--
THIS WORK IS DEDICA TO
H. S. M. COXETER
PROFESSOR OF MATHEMATICS
UNIVERSITY OF TORONTO
To me no experience of childhood so reinforced self-
confidence in one's own exploratory faculties
as did geometry. Its inspiring effectiveness in
winnowing out and evaluating a plurality
of previously unknowns from a few given
knowns, and its elegance of proof
lead to the further discovery and comprehension of a
grand strategy for all
problem solving....
--
But look at the definitions of these two words and how he used
them in
section 203.04.
Definition of ghostly:
1. Of, relating to, or resembling a ghost, a wraith, or an
apparition; spectral.
2. Of or relating to the soul or spirit; spiritual.
Definition of a priori:
1. Proceeding from a known or assumed cause to a necessarily
rela
effect; deductive.
2.
a. Derived by or designating the process of reasoning without
reference to particular facts or experience.
b. Knowable without appeal to particular experience.
3. Made before or without examination; not suppor by factual
study.
--
203.04 Rather than refuting the bases of presently known
Euclidean and non-
Euclidean and hyperbolic and elliptic geometry, synergetics
identifies the
alternate freedoms of prime axiomatic assumption from which
the present
mathematical bases were selec. It embraces all the known
mathematics.
All of
the axiomatic alternatives are logical. Thus, original
assumptions
eliminate the
necessity for subsequent assignment of physical qualities to
nonconceptual
mathematical devices. Classical mathematics has, of
necessity, assigned
progressively discovered attributes of physical Universe to
irrational
relationships
with the ghostly, a priori Greek geometry.
--
I think that there is nothing much more difficult in
Synergetics than
the reasoning in section 222.41, if you build models.
222.41 In algebraic work, if you use a constant suffix__where
you always
have,
say, 33 and 53__you could treat them as 30 and 50 and come
out with the
same
algebraic conditions. Therefore, if all these terminate with
the number
two, we
can drop off the two and not affect the algebraic
relationships. If we
drop off the
number two in the last column, they will all be zeros. So in
the case of
omnidirectional closest packing of spheres, the sequence will
read; 10,
40, 90,
160, 250, 360, and so forth. Since each one of these is a
multiple of
10, we may
divide each of them by 10, and then we have 1, 4, 9, 16, 25,
and 36,
which we
recognize as a progression of second powering__two to the
second power,
three to
the second power, and so forth.
--
Section 222.43 about the formula 10f^2+2
222.43 This simple formula governing the rate at which balls
are
agglomera
around other balls or shells in closest packing is an elegant
manifest
of the reliably
incisive transactions, formings, and transformings of
Universe. I made that
discovery in the late 1930s and published it in 1944....
--
I don't know why mathematicians don't refer to 
what RBF
published in
1944 when they mention that sequence, and why it took so long
in
geometry for someone to build a model and count the spheres
in each
layer. I don't know why it is taking so long for
mathematicians to read
his books and build the models described so they can have an
opinion
about Synergetics one way or another.
Cliff Nelson
===
Subject: Re: Geodetic spheres
Surfing on the net looking for these things I found two online
book
called
> Synergetics I and II by inventor R. Buckminster Fuller (at
www.bfi.org).
> Has someone read them? Are they valueable for a
mathematician or are they
> hopelessly out of date? They seem a bit strange but maybe
they contain
> something cool?
There are a couple of Fuller acolytes who post here
occasionally
with extracts from these books. As far as mathematical content
is concerned they are trivial, incomprehensible or wrong.
--
His mind has been corrup by colousounds and shapes.
The League of Gentlemen
===
Subject: Re: Geodetic spheres
here's Lobel's generalization -- 
magnifique, as well
as a real Why didn't I think of that, or Bucky?:
http://www.equilatere.net.
your assertions about Bucky are silly, as anyone can see
by random perusal of _S_ (both volumes handily ammassed
into one, online, courtesy of Robert Gray). true,
there's nothing more complica than Pythagoras' 
theorem,
for the vast most part. I especially recommend
to novices in geometry, the Color Plates One and Two.
it's true, taht he shows a lot of his Transcendentalist 
roots,
and I think that he's something of a Jacobin, apoliticcally,
specifcally of the Toynbee school of British imperialism
(this is where he got a lot of his speculative prehistory,
in other words, digests of anthropology & archeology).
> with extracts from these books. As far as mathematical
content
> is concerned they are trivial, incomprehensible or wrong.
--les ducs d'Enron!
===
Subject: A question on absolute continuity and L^2 functions
Let y be an absolutely continuous function on [0,pi] such that
y(0)=y(pi)=0
and y Ô in L^2 [0,pi].
I can prove that y^2 cot x is continuous on [0, pi].
But is it absolutely continuous? If so, is its derivative
equal to
2yy' cot x - y^2 (csc x)^2 ?
===
Subject: Re: A question on absolute continuity and L^2
functions
> Let y be an absolutely continuous function on [0,pi] such
that
y(0)=y(pi)=0
> and y Ô in L^2 [0,pi].
> I can prove that y^2 cot x is continuous on [0, pi].
> But is it absolutely continuous? If so, is its derivative
> equal to
> 2yy' cot x - y^2 (csc x)^2 ?
Its derivative is a.e. equal to that even without a yes
answer to your
first question. That's because y is absolutely 
continuous,
hence y' exists
a.e., and [cot(x)]' exists everywhere in (0,Pi). So 
it's just
the product
rule.
As for absolute continuity, you'll be done if you can show
2yy'cot x - y^2
(csc x)^2 belongs to L^1 [0,pi]. We only need to worry about
what happens
near 0 and Pi. Let's look near 0, where cot(x) and csc(x)
both behave like
1/x. Now there's something called Hardy's 
Inequality that
implies that if f
is in L^2(0,oo), then so is x -> (1/x)*integral_[0,x] f(t)
dt. So it looks
to me like the answer to your first question is yes (using
Hardy and the
Schwarz inequality). I'm being brief with the details 
because
there may be
an easier way to do this.
===
Subject: Re: A question on absolute continuity and L^2
functions
>> Let y be an absolutely continuous function on [0,pi] such
that
y(0)=y(pi)=0
>> and y Ô in L^2 [0,pi].
>> I can prove that y^2 cot x is continuous on [0, pi].
>> But is it absolutely continuous? If so, is its derivative
>> equal to
>> 2yy' cot x - y^2 (csc x)^2 ?
>Its derivative is a.e. equal to that even without a yes
answer to your
>first question. That's because y is absolutely 
continuous,
hence y' exists
>a.e., and [cot(x)]' exists everywhere in (0,Pi). So 
it's
just the product
>rule.
To be fair, we might assume that he knew this, regarded an
a.e.
derivative as useless, and wan to know whether that was the
weak derivative. (Or maybe not...)
>As for absolute continuity, you'll be done if you can show
2yy'cot x - y^2
>(csc x)^2 belongs to L^1 [0,pi]. We only need to worry about
what happens
>near 0 and Pi. Let's look near 0, where cot(x) and csc(x)
both behave like
>1/x. Now there's something called Hardy's 
Inequality that
implies that if f
>is in L^2(0,oo), then so is x -> (1/x)*integral_[0,x] f(t)
dt. So it looks
>to me like the answer to your first question is yes (using
Hardy and the
>Schwarz inequality). I'm being brief with the details
because there may be
>an easier way to do this.
Seems right to me, and it doesn't seem to me that 
there's
going to be
an approach that's _much_ simpler. To get the second term in
L^1 is
exactly to show that 1/x int_0^x y'(t) dt is an L^2 
function,
which is
exactly Hardy's inequality - this also implies the 
first term
is L^1.
So if there _is_ something much simpler it can't proceed by
estimating the two terms separately, in any case.
************************
===
Subject: Elliptic Curve over Z{p}
===
I am looking at an example of an EC, and managed to get lost
and was hoping
someone could point me in the right direction.
We have the elliptic curve E: y^2 = x^3 + x + 1 defined over
Z{23}, that is
p = 23.
The example then starts off with:
P = (0, 1) and then calculates
2P = (0, 22)
3P = (1,7)
4P = (1, 16)
...
27P = (19, 18)
How are they getting these results, since it appears they are
actually
solving the elliptic curve over?
That is, each of these points does satisfy:
y^2 == x^3 + x + 1 (mod 23) (here, a = 1 and b = 1)
Typically, we would use
x3 = lamda^2 - x1 - x2
y3 = lamda (x1 - x3) - y1
and
lamda = (y2 - y1)/(x2-x1) if A != B
= (3x1^2 + a)/2y1 if A = B
Where A = (x1, y1), B = (x2, y2).
Also, how do we know the order of this group (I know it will
be when we get
P back and I think it would be the next P, that is, 28 P).
Please help clarify.
Can anyone help clarify?
===
Subject: Re: Elliptic Curve over Z{p}
>
I am looking at an example of an EC, and managed to get lost
and was
hoping
> someone could point me in the right direction.
We have the elliptic curve E: y^2 = x^3 + x + 1 defined over
Z{23}, that
is
> p = 23.
The example then starts off with:
P = (0, 1) and then calculates
> 2P = (0, 22)
Shurely shome mishtake?
(16:54) gp > E=ellinit([0,0,0,1,1]*Mod(1,23));
(16:55) gp > P=[0,1]
[0, 1]
(16:56) gp > elladd(E,P,P)
[Mod(6, 23), Mod(19, 23)]
===
Subject: Re: Elliptic Curve over Z{p}
===
Hi ,
may I ask what code you are using below?
Is it public domain or home - grown?
Flip
>
I am looking at an example of an EC, and managed to get lost
and was
hoping
> someone could point me in the right direction.
We have the elliptic curve E: y^2 = x^3 + x + 1 defined over
Z{23},
that
is
> p = 23.
The example then starts off with:
P = (0, 1) and then calculates
> 2P = (0, 22)
> Shurely shome mishtake?
> (16:54) gp > E=ellinit([0,0,0,1,1]*Mod(1,23));
> (16:55) gp > P=[0,1]
> [0, 1]
> (16:56) gp > elladd(E,P,P)
> [Mod(6, 23), Mod(19, 23)]
>
===
Subject: Re: Elliptic Curve over Z{p}
> We have the elliptic curve E: y^2 = x^3 + x + 1 defined over
Z{23}, that
is
> p = 23.
The example then starts off with:
P = (0, 1) and then calculates
> 2P = (0, 22)
Are you sure that this right? The equation is in a form,
where you get the
additive inverse of a point by changing the sign of the
y-coordinate, so
-P=(0,22)=(0,-1). So if 2P=(0,-1), then P would be of order 3
contradicting
your (their?) next statement that
> 3P = (1,7)
because 3P is not the point at infinity.
> 4P = (1, 16)
> ...
27P = (19, 18)
How are they getting these results, since it appears they are
actually
> solving the elliptic curve over?
If by solving you mean that they are just plugging in all the
possibilities
for x and attempting to solve for y, then that seems to be
the case.
Something
doesn't add up.
> Typically, we would use
x3 = lamda^2 - x1 - x2
> y3 = lamda (x1 - x3) - y1
> and
> lamda = (y2 - y1)/(x2-x1) if A != B
> = (3x1^2 + a)/2y1 if A = B
Where A = (x1, y1), B = (x2, y2).
Also, how do we know the order of this group (I know it will
be when we
get
> P back and I think it would be the next P, that is, 28 P).
Ok. Why don't you try that with A=B=P, and check whether you
get the
claimed
value for 2P. I agree with you that the data as reproduced
here doesn't
look right.
There are various algorithms for computing the order of the
group. Look up
the books by Menezes (title mentions the use of ECs in
cryptography, don't
remember the exact title) and Silverman: The Arithmetic of
Elliptic
Curves
for more information. Do observe that the group is not nearly
always
cyclic,
i.e. you cannot simply compute the orders of the points.
Jyrki Lahtonen, Turku, Finland
===
Subject: Re: Elliptic Curve over Z{p}
===
> We have the elliptic curve E: y^2 = x^3 + x + 1 defined over
Z{23},
that
is
> p = 23.
The example then starts off with:
P = (0, 1) and then calculates
> 2P = (0, 22)
> Are you sure that this right? The equation is in a form,
where you get
the
> additive inverse of a point by changing the sign of the
y-coordinate, so
> -P=(0,22)=(0,-1). So if 2P=(0,-1), then P would be of order
3
contradicting
> your (their?) next statement that
> 3P = (1,7)
> because 3P is not the point at infinity.
> 4P = (1, 16)
> ...
27P = (19, 18)
How are they getting these results, since it appears they are
actually
> solving the elliptic curve over?
> If by solving you mean that they are just plugging in all
the
possibilities
> for x and attempting to solve for y, then that seems to be
the case.
Something
> doesn't add up.
> Typically, we would use
x3 = lamda^2 - x1 - x2
> y3 = lamda (x1 - x3) - y1
> and
> lamda = (y2 - y1)/(x2-x1) if A != B
> = (3x1^2 + a)/2y1 if A = B
Where A = (x1, y1), B = (x2, y2).
Also, how do we know the order of this group (I know it will
be when we
get
> P back and I think it would be the next P, that is, 28 P).
> Ok. Why don't you try that with A=B=P, and check whether
you get the
claimed
> value for 2P. I agree with you that the data as reproduced
here doesn't
> look right.
** This is exactly the point of the post. I am unable (using
the chord
method) to calculate these points and think I must be doing
something wrong
(since I got this example from the Certicom web site and
certainly they are
experts in this area).
Can you figure out how they are calculating these points?
We have the elliptic curve E: y^2 = x^3 + x + 1 defined over
Z{23}, that is
p = 23.
The points as given are (I assume reading from top to bottom
and left to
right):
(0, 1) (6, 4) (12, 19)
(0, 22) (6, 19) (13, 7)
(1, 7) (7, 11) (13, 16)
(1, 16) (7, 12) (17, 3)
(3, 10) (9,7) (17, 20)
(3, 13) (9, 16) (18, 3)
(4, 0) (11, 3) (18, 20)
(5, 4) (11, 20) (19, 5)
(5, 19) (12, 4) (19, 18)
So, the point P= (0, 1) seems to have order larger than
number of points
given.
The paper then has another statement that the curve y^2 = x^3
+ x + 1 could
have a generator point of (0, 1) from which the following
points lis
below would be genera.
P = (0, 1), 2P = (6, 19), 3P = (3, 13), 4P = (13, 16), 5P =
(18, 3), 6P =
(7, 11) ... 28 P = (0, 1) = P
So the point P = (0, 1) has order of 28.
How does this example differ in points from the first (are
they using a
different field?), since the values appear to exist in the
above table too,
albeit in a different ordering? The second example does not
define p = 23
(but perhaps it is something different).
I am confused as to how the values are being calcula and why
these
values
are different in the two tables.
Can anyone clarify what is going on here?
Thank you for any guidance.
Thanks for any inputs.
> There are various algorithms for computing the order of the
group. Look
up
> the books by Menezes (title mentions the use of ECs in
cryptography,
don't
> remember the exact title) and Silverman: The Arithmetic of
Elliptic
Curves
> for more information. Do observe that the group is not
nearly always
cyclic,
> i.e. you cannot simply compute the orders of the points.
 Jyrki Lahtonen, Turku, Finland
===
Subject: Re: Elliptic Curve over Z{p}
> Can you figure out how they are calculating these points?
We have the elliptic curve E: y^2 = x^3 + x + 1 defined over
Z{23}, that
> is p = 23.
The points as given are (I assume reading from top to bottom
and left to
> right):
(0, 1) (6, 4) (12, 19)
> (0, 22) (6, 19) (13, 7)
> (1, 7) (7, 11) (13, 16)
> (1, 16) (7, 12) (17, 3)
> (3, 10) (9,7) (17, 20)
> (3, 13) (9, 16) (18, 3)
> (4, 0) (11, 3) (18, 20)
> (5, 4) (11, 20) (19, 5)
> (5, 19) (12, 4) (19, 18)
This just looks like a list of non-identity points on the
curve
in ascending order of x-coordinate. (So the group has order
28.)
The paper then has another statement that the curve y^2 = x^3
+ x + 1
> could have a generator point of (0, 1) from which the
following points
> lis below would be genera.
P = (0, 1), 2P = (6, 19), 3P = (3, 13), 4P = (13, 16), 5P =
(18, 3), 6P =
> (7, 11) ... 28 P = (0, 1) = P
Before you quo that 2P was (0,22). Now you quote it's 
(6,19).
What did this paper actually say?
If 28P = P then the order of P divides 27 --- impossible if
it also divides 28 (if that is the order of the group).
--
His mind has been corrup by colousounds and shapes.
The League of Gentlemen