mm-250
===
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
> listed below would be generated.
>
> 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 quoted that 2P was (0,22). Now you quote it's
(6,19).
> What did this paper actually say?
Robin,
I think you clarified this a bit for me when you said This
just looks
like
a list of non-identity points on the curve in ascending order
of
The original list was just that, it never stated 2P = , 3 P =
... .
The second list does assign specific values (my confusion,
giving
a note that makes sense to me). Anyway, here it the list of
points in it's
entirety (I assume this is over Z{23}).
P = (0, 1), 2P = (6, 19), 3P = (3, 13), 4P = (13, 16), 5P =
(18, 3), 6P =
(7, 11)
7P = (11, 3), 8P = (5, 19), 9P = (19, 18), 10P = (12, 4), 11P
= (1, 16),
12P
= (17, 20)
13P = (9, 16), 14P = (4, 0), 15P = (9, 7), 16P = (17, 3), 17P
= (1, 7), 18P
= (12, 19)
19P = (19, 5), 20P = (5, 4), 21P = (11, 20), 22P = (7, 12),
23P = (18, 20),
24P = (13, 7)
25P = (3, 10), 26P = (6, 4), 27P = (0, 22), 28 P = (0, 1) = P
... so the
order is 28.
My problem now is how they are calculating these:
E: y^2 = x^3 + x + 1 (mod 23), points starting from P = (0, 1).
When I try the chord method equations, I seem to be doing
something wrong.
I gave the formula in an earlier post. Can you help show me
what I am
doing
wrong here?
I get fractions when I use those formulas, but we are working
mod 23,
so
is there some conversion that is being done to get these
points?
> If 28P = P then the order of P divides 27 --- impossible if
> it also divides 28 (if that is the order of the group).
> --
> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
> His mind has been corrupted by colousounds and shapes.
> The League of Gentlemen
===
Subject: Re: Elliptic Curve over Z{p}
<snip>
>The second list does assign specific values (my confusion,
giving
>a note that makes sense to me). Anyway, here it the list of
points in
it's
>entirety (I assume this is over Z{23}).
>P = (0, 1), 2P = (6, 19), 3P = (3, 13), 4P = (13, 16), 5P =
(18, 3), 6P =
>(7, 11)
>7P = (11, 3), 8P = (5, 19), 9P = (19, 18), 10P = (12, 4), 11P
= (1, 16),
12P
>= (17, 20)
>13P = (9, 16), 14P = (4, 0), 15P = (9, 7), 16P = (17, 3), 17P
= (1, 7),
18P
>= (12, 19)
>19P = (19, 5), 20P = (5, 4), 21P = (11, 20), 22P = (7, 12),
23P = (18,
20),
>24P = (13, 7)
>25P = (3, 10), 26P = (6, 4), 27P = (0, 22), 28 P = (0, 1) = P
... so the
>order is 28.
If 27P = (0, 22), then 28P is the point at infinity, and 29P =
(0, 1) = P,
so that 28 is the order.
>My problem now is how they are calculating these:
>E: y^2 = x^3 + x + 1 (mod 23), points starting from P = (0,
1).
>When I try the chord method equations, I seem to be doing
something wrong.
>I gave the formula in an earlier post. Can you help show me
what I am
doing
>wrong here?
>I get fractions when I use those formulas, but we are working
mod 23,
so
>is there some conversion that is being done to get these
points?
Use Euclid's Algorithm.
David McAnally
--------------
===
Subject: Re: Elliptic Curve over Z{p}
>
> I get fractions when I use those formulas, but we are
working mod 23,
so
> is there some conversion that is being done to get these
points?
We have 4 x 6 = 24 = 1 (mod 23), so in Z_23,
4 x 6 = 1. So in Z_23, 1/4 = 6 n'est-ce pas? And 1/6 = 4.
SO what about 1/2, 1/3, 1/5, 1/7 etc. etc. etc.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colousounds and shapes.
The League of Gentlemen
===
Subject: Re: Elliptic Curve over Z{p}
>
> I get fractions when I use those formulas, but we are
working mod
23,
so
> is there some conversion that is being done to get these
points?
> We have 4 x 6 = 24 = 1 (mod 23), so in Z_23,
> 4 x 6 = 1. So in Z_23, 1/4 = 6 n'est-ce pas? And 1/6 = 4.
> SO what about 1/2, 1/3, 1/5, 1/7 etc. etc. etc.
** Just modular inverses ... I should have tried that, but the
earlier list
we discussed threw me for a loop as I wasn't even in the ball
park!
> --
> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
> His mind has been corrupted by colousounds and shapes.
> The League of Gentlemen
===
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)
> 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?
Over what?
> That is, each of these points does satisfy:
>
> y^2 == x^3 + x + 1 (mod 23) (here, a = 1 and b = 1)
Well of course: addition of points on an elliptic curve gives
rise to points on elliptic curves!
> 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).
This seems to be the chord tangent process. Does this
not give the above values.
> 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).
Are you saying that 28P is zero in the group?
It isn't: -P is obtained by reflecting P in the x-axis: it is
(0,-1) = (0,22) so 27 P =/= -P i.e., 28P =/= O.
The order of the group is the number of elements in the group:
1 + # of solutions of E defined over F_{23} (the 1 is for the
point at infinity). You seem to be computing the order
of P within the group rather than the order of the group
itself.
The former is a factor of the latter. By the Hasse-Weil bound,
the group's order lies between 15 and 33 inclusive.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colousounds and shapes.
The League of Gentlemen
===
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)
> 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?
> Over what?
> That is, each of these points does satisfy:
>
> y^2 == x^3 + x + 1 (mod 23) (here, a = 1 and b = 1)
> Well of course: addition of points on an elliptic curve gives
> rise to points on elliptic curves!
> 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).
> This seems to be the chord tangent process. Does this
> not give the above values.
> 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).
> Are you saying that 28P is zero in the group?
** No, I am saying that I was not able to calculate the points
as I have
given them above, so am unable to calculate the order.
This is an example I found on the Certicom web site somewhere.
Can you tell me how you calculate these points? (The chord
method does not
appear to produce these points?.
> It isn't: -P is obtained by reflecting P in the x-axis: it is
> (0,-1) = (0,22) so 27 P =/= -P i.e., 28P =/= O.
> The order of the group is the number of elements in the
group:
> 1 + # of solutions of E defined over F_{23} (the 1 is for the
> point at infinity). You seem to be computing the order
> of P within the group rather than the order of the group
itself.
> The former is a factor of the latter. By the Hasse-Weil
bound,
> the group's order lies between 15 and 33 inclusive.
> --
> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
> His mind has been corrupted by colousounds and shapes.
> The League of Gentlemen
===
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)
>> 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?
>> Over what?
>> That is, each of these points does satisfy:
>>
>> y^2 == x^3 + x + 1 (mod 23) (here, a = 1 and b = 1)
>> Well of course: addition of points on an elliptic curve
gives
>> rise to points on elliptic curves!
>> 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).
>> This seems to be the chord tangent process. Does this
>> not give the above values.
>> 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).
>> Are you saying that 28P is zero in the group?
>
> ** No, I am saying that I was not able to calculate the
points as I have
> given them above, so am unable to calculate the order.
>
> This is an example I found on the Certicom web site
somewhere.
>
> Can you tell me how you calculate these points? (The chord
method does
> not appear to produce these points?.
Are you saying that the point they label 2P isn't P + P,
that 3P isn't P + P + P? Then I have no idea what they're
doing.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colousounds and shapes.
The League of Gentlemen
===
Subject: Closed form ,Lucas poly., Chebychev poly.
T_0,T_1,...,T_n,... are Chebychev polynomials
of first kind, that is T_0(x)=1 , T_1(x)=x and
T_{n+1}(x)=2x*T_n(x)-T_{n-1}(x) for n=1,2,... .
l_0(x),l_1(x),...,l_n(x),... are Lucas polynomials
defined as l_0(x)=2 , l_1(x)=x , and for n >= 1
l_{n+1}(x)=x*l_n(x)+l_{n-1}(x) .
It is supposed that a polynomial f(x) is given , and
moreover
f(x) = SUM_{k=0 to k=n} c(k)*T_k(x) .
I am interested to find a closed form for the sum
(Tf)(x,y):= SUM_{k=0 to k=n} c(k)*l_k(x)*l_k(y) .
==
===
Subject: Re: Closed form ,Lucas poly., Chebychev poly.
>T_0,T_1,...,T_n,... are Chebychev polynomials
>of first kind, that is T_0(x)=1 , T_1(x)=x and
> T_{n+1}(x)=2x*T_n(x)-T_{n-1}(x) for n=1,2,... .
>l_0(x),l_1(x),...,l_n(x),... are Lucas polynomials
>defined as l_0(x)=2 , l_1(x)=x , and for n >= 1
>l_{n+1}(x)=x*l_n(x)+l_{n-1}(x) .
>It is supposed that a polynomial f(x) is given , and
>moreover
> f(x) = SUM_{k=0 to k=n} c(k)*T_k(x) .
>I am interested to find a closed form for the sum
>(Tf)(x,y):= SUM_{k=0 to k=n} c(k)*l_k(x)*l_k(y) .
>==
It's rather amazing that such a closed form exists.
Note that T_k(cos(x)) = cos(k x).
So f(cos x) = sum_{k=0}^n c(k) cos(k x)
= sum_{k=-n}^n b(k) exp(i k x)
where b(0) = c(0) and b(k) = b(-k) = c(k)/2 for k > 0.
And thus f((v+1/v)/2) = sum_{k=-n}^n b(k) v^k
Now l_k(x) = A(x)^k + B(x)^k
where A(x) = (x+sqrt(x^2+4))/2 and
B(x) = (x - sqrt(x^2+4))/2 = -1/A(x).
So sum_{k=0}^n c(k) l_k(x) l_k(y)
= sum_{k=0}^n c(k) (A(x)^k + B(x)^k)(A(y)^k + B(y)^k)
where
sum_{k=0}^n c(k) ((A(x)A(y))^k + (B(x)B(y))^k)
= sum_{k=-n}^n b(k) (A(x)A(y))^k
= 2 f((A(x)A(y)+B(x)B(y))/2)
and similarly
sum_{k=0}^n c(k) ((A(x)B(y))^k + (B(x)A(y))^k)
= sum_{k=-n}^n b(k) (A(x)B(y))^k
= 2 f((A(x)B(y)+B(x)A(y))/2)
So Tf(x,y) = 2 f((A(x)A(y)+B(x)B(y))/2) + 2
f((A(x)B(y)+B(x)A(y))/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
===
Subject: Re: Closed form ,Lucas poly., Chebychev poly.
>T_0,T_1,...,T_n,... are Chebychev polynomials
>of first kind, that is T_0(x)=1 , T_1(x)=x and
> T_{n+1}(x)=2x*T_n(x)-T_{n-1}(x) for n=1,2,... .
>l_0(x),l_1(x),...,l_n(x),... are Lucas polynomials
>defined as l_0(x)=2 , l_1(x)=x , and for n >= 1
>l_{n+1}(x)=x*l_n(x)+l_{n-1}(x) .
>It is supposed that a polynomial f(x) is given , and
>moreover
> f(x) = SUM_{k=0 to k=n} c(k)*T_k(x) .
>I am interested to find a closed form for the sum
>(Tf)(x,y):= SUM_{k=0 to k=n} c(k)*l_k(x)*l_k(y) .
>==
>
> It's rather amazing that such a closed form exists.
After a bit of sleep, a little less amazing.
> So Tf(x,y) = 2 f((A(x)A(y)+B(x)B(y))/2) + 2
f((A(x)B(y)+B(x)A(y))/2)
Which just says
l_k(x) l_k(y) = 2 T_k((A(x)A(y)+B(x)B(y))/2) + 2
T_k((A(x)B(y)+B(x)A(y))/2)
Not too surprising that such a relation exists, given that l_k
and T_k are
related by T_k(t) = i^k/2 l_k(-2 i 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: Re: Closed form ,Lucas poly., Chebychev poly.
>T_0,T_1,...,T_n,... are Chebychev polynomials
>of first kind, that is T_0(x)=1 , T_1(x)=x and
> T_{n+1}(x)=2x*T_n(x)-T_{n-1}(x) for n=1,2,... .
>l_0(x),l_1(x),...,l_n(x),... are Lucas polynomials
>defined as l_0(x)=2 , l_1(x)=x , and for n >= 1
>l_{n+1}(x)=x*l_n(x)+l_{n-1}(x) .
>It is supposed that a polynomial f(x) is given , and
>moreover
> f(x) = SUM_{k=0 to k=n} c(k)*T_k(x) .
>I am interested to find a closed form for the sum
>(Tf)(x,y):= SUM_{k=0 to k=n} c(k)*l_k(x)*l_k(y) .
>==
>
> It's rather amazing that such a closed form exists.
>
> Note that T_k(cos(x)) = cos(k x).
> So f(cos x) = sum_{k=0}^n c(k) cos(k x)
> = sum_{k=-n}^n b(k) exp(i k x)
> where b(0) = c(0) and b(k) = b(-k) = c(k)/2 for k > 0.
> And thus f((v+1/v)/2) = sum_{k=-n}^n b(k) v^k
>
> Now l_k(x) = A(x)^k + B(x)^k
> where A(x) = (x+sqrt(x^2+4))/2 and
> B(x) = (x - sqrt(x^2+4))/2 = -1/A(x).
>
> So sum_{k=0}^n c(k) l_k(x) l_k(y)
> = sum_{k=0}^n c(k) (A(x)^k + B(x)^k)(A(y)^k + B(y)^k)
>
> where
> sum_{k=0}^n c(k) ((A(x)A(y))^k + (B(x)B(y))^k)
> = sum_{k=-n}^n b(k) (A(x)A(y))^k
> = 2 f((A(x)A(y)+B(x)B(y))/2)
>
> and similarly
> sum_{k=0}^n c(k) ((A(x)B(y))^k + (B(x)A(y))^k)
> = sum_{k=-n}^n b(k) (A(x)B(y))^k
> = 2 f((A(x)B(y)+B(x)A(y))/2)
>
> So Tf(x,y) = 2 f((A(x)A(y)+B(x)B(y))/2) + 2
f((A(x)B(y)+B(x)A(y))/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
===
Subject: to find an answer
the 7th degree polynomial
a0 + a1x + a2x^2 + a3x^3 + a4x^4 + a5x^5 + a6x^6 + a7x^7
can alternatively be expressed,
0! d^7 0!
a0x^0 = ---- a0 ----- x^7 = ---- a0 y'''''''
7! dx^7 7!
1! d^6 1!
a1x^1 = ---- a1 ----- x^7 = ---- a1 y''''''
7! dx^6 7!
2! d^5 2!
a2x^2 = ---- a2 ----- x^7 = ---- a2 y'''''
7! dx^5 7!
3! d^4 3!
a3x^3 = ---- a3 ----- x^7 = ---- a3 y''''
7! dx^4 7!
4! d^3 4!
a4x^4 = ---- a4 ----- x^7 = ---- a4 y'''
7! dx^3 7!
5! d^2 5!
a5x^5 = ---- a5 ----- x^7 = ---- a5 y''
7! dx^2 7!
6! d^1 6!
a6x^6 = ---- a6 ----- x^7 = ---- a6 y'
7! dx^1 7!
7! d^0 7!
a7x^7 = ---- a7 ----- x^7 = ---- a7 y
7! dx^0 7!
--------------------------------------------
0
One solution is y=x^7 But are there others?
For positive roots, let y = e^x - 1 and solve for x. Then the
root is
y^(1/7)
For negative roots, let y = 1-e^x and solve for x. Then the
root is
y^(1/7)
For more roots apply DeMoivre's Theorem
y=e^x - 1
y' = y'' = y''' = y'''' = y''''' = y'''''' = y''''''' = e^x
a7
{
ln{-----------------------------------------------------------
------}}^(1/
7)
a0/7!+a1/7!+a2(2/7!)+a3(3!/7!)+a4(4!/7!)+a5(5!/7!)+a6/7+a7
However it is doubtful that this is a solution, since the
differential
applies
only to y=x^7
===
Subject: Re: Number theory problem
>
> How do you solve the problem:
>
> What is the greatest integer n such that
> n divides p^4-1 for all primes p > 5?
>
> The choices are 12, 30, 48, 120, 240
> I don't know what the answer is, nor how to solve it, but
I'm most
> interested in knowing the basic method to solve this
problem. It seems
like
> Fermat's Little Theorem might have something to say about
this problem but
I
> can't figure out how to apply it in this case. (By the way,
this comes
from
> an old GRE Subject Test, I'm not trying to cheat on my
homework or
anything.)
This was discussed here on August 31, see
40nestle.ai.mit.edu
There I show its easy using the Carmichael lambda function y
y(240) = y(2^4 3 5) = lcm(2^2,2,4) = 4
=> x^4 = 1 (mod 240) for x coprime to 2,3,5
Follow the above link for further detail.
-Bill Dubuque
===
Subject: Re: Factorial/Exponential Identity, Infinity
did you use a good source of entropy?
===
Subject: Re: Factorial/Exponential Identity, Infinity
I'm still trying to determine univariate polynomial ratio
forms for
the unsigned Stirling cycle numbers s(n+1, n-x+1). The signed
Stirling number is (-1)^x s(n+1, n-x+1).
For x = 0, s(n+1, n+1) = 1.
s(n+1, n+1) = 1
For x=1, s(n+1, n) = n(n+1)/2 = (n^2 + n)/2.
s(n+1, n) = (n^2 + n) / 2
For x=2, s(n+1, n-1) = ((n(n+1)/2)^2- n(n+1)(2n+1)/6)/2 =
(n^2+n)^2/8
- (n^2+1)(2n+1)/12 = (n^4+2n^3+n^2) / 8 - (2n^3+n^2+2n+1)/12 =
(3n^4+2n^3+n^2-6n-3)/24
s(n+1, n-1) = (3n^4+2n^3+n^2-6n-3)/24
for x=3, s(n+1, n-2) = ((n^2+n)^3 / 8 - (n^2+n)^2 / 4 ) (n-2) /
6(n+2) = ( (n^4+2n^3+n^2)(n^2+n)/8 - (n^4+2n^3+n^2)/4 )(n-2) /
6(n+2)
= ( (n^6+3n^5 + 3n^4 + n^3)/8 - (2n^4+4n^3+2n^2)/8 )(n-2) /
6(n+2) = (
n^6+3n^5+n^4-3n^3+2n^2)(n-2) / 48 (n+2) = (
n^7+3n^6+n^5-3n^4+2n^3
-2n^6-6n^5-2n^4+6n^3-4n^2 ) / 48 (n+2) = (n^7 + n^6 -5n^5
-5n^4 +8n^3
-4n^2) / 48(n+2) = n^2(n^5 + n^4 -5n^3 -5n^2 +8n^1 -4) /
48(n+2)
s(n+1, n-2) = n^2(n^5 + n^4 -5n^3 -5n^2 +8n^1 -4) / 48(n+2)
Thus s(n+1, n-2) is represented by a polynomial with rank 7,
s(n+1,
n-1) is represented by a polynomial of rank 4, s(n+1, n) by a
polynomial of rank 2, and s(n+1, n+1) by a polynomial of rank
0.
One avenue of determining s(n+1, n-x+1) is that it is equal to
the sum
of the products of each x-subset of {1, ..., n}. Then it is
obvious
in a 2-D nxn matrix M where m_ij = i*j that the matrix is
symmetrical
about the main diagonal and that the entries either above or
below the
diagonal represent the 2-subsets of {1, ..., n}, thus that the
sum of
all entries is easily calculated as (sum n)^2, the sum of the
diagonal
entries is obviously sum (n^2) and that their difference is
twice the
sum of the 2-subsets.
In the 3-d hypermatrix case M with m_ijk = ijk it is not so
obvious
how to divide the elements where the elements on the triagonal
are not
3-subsets and many other elements with matching indices are
also not
3-subsets, and then each 3-subset is represented 6 times in the
matrix: eg (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1,
2), and
(3, 2, 1). Then, (n+2)/(n-2) represents the value of the sum
of those
cells relative to the cells with matching coordinates besides
the main
diagonal with all matching coordinates.
I guess I could enumerate all the cells in a 4-matrix, from
(1, 1, 1,
1) to (n, n, n, n). Then, I'll remove the elements on the main
diagonal, and get metrics of those elements having two and
three
matching elements. Then, of the remaining elements with four
distinct
coordinates, there would be 4!=24 representations of each
4-subset of
{1, ..., n}. So what it would do is generate the n^x-many
element
array with each of the coordinates, the multidimensional
array, and
then start counting things about it.
One thing I do is check the sums of each of the elements of
the x-d
hypermatrix with all matching elements (the main n-agonal) and
also
with x-1, x-2, matching elements, and for elements with no
matching
elements that represent the x-subsets of n. One thing I've
noticed in
the case of x=4 is that the sum of the elements with no
matching
coordinates is 24 times s(n+1, n-3). That is to say, the
elements at
coordinates that represent k-subsets sum to 4! times s(n+1,
n-4+1).
This is good to know, but then I am left trying to determine
how to
calculate that value from n and x as a univariate polynomial
of n or a
ratio of univariate polynomials of n.
I'm looking at the count and sum of elements that have at
least a pair
of matching coordinates.
n = 4 x = 4
Sum of elements: 10000
Sum of elements with 3 matching elements: 2037 = 7*3*97
Sum of elements with 2 matching elements: 7033 = 13*541
Number of elements with 3 matching elements: 39 = 3*13
Number of elements with 2 matching elements: 189 = 3*3*3*7
I'm not quite sure how to calculate how many elements there
are with
at least two matching coordinates and less than x many matching
coordinates, or their elements' sum.
n = 5 x = 4
Sum of elements: 50625 = (sum n)^x
Sum of elements with 4 matching elements: 979 = sum (n^x)
Sum of elements with 3 matching elements: 7412 = 2*2*17*109
Sum of elements with 2 matching elements: 35658 = 2*7*3*3*283
Sum of elements with no matching elements: 6576 = x! * s(n+1,
n-x+1)
Number of elements with 4 matching elements: 5 = n
Number of elements with 3 matching elements: 64 =
2*2*2*2*2*2*2*2
Number of elements with 2 matching elements: 436 = 2*2*109
Number of elements with no matching elements: 120 = n! ~ (x+1)!
Finding simple expressions for those items above denoted as
equal to
factorizations would help determine s(n+1, n-x+1) from the
other known
quantities.
Once again, in examining an x-d nxnx...xn hypermatrix m where
m_ijk...
= i*j*k*... in an attempt to discern a method to calculate
unsigned
Stirling cycle numbethere are simple methods for x=1, 2, 3 but
I
want to determine the forms for x=4, 5, ..., and they are not
so
simple.
n = 4 x = 4
Sum of elements: 10000
Sum of elements with 4 matching elements: 354
Sum of elements with 3 matching elements: 2037
Sum of elements with 2 matching elements: 7033
Sum of elements with no matching elements: 576
Number of elements with 4 matching elements: 4
Number of elements with 3 matching elements: 39
Number of elements with 2 matching elements: 189
Number of elements with no matching elements: 24
n = 6 x = 4
Sum of elements: 194481
Sum of elements with 4 matching elements: 2275
Sum of elements with 3 matching elements: 21398
Sum of elements with 2 matching elements: 131832
Sum of elements with no matching elements: 38976
Number of elements with 4 matching elements: 6
Number of elements with 3 matching elements: 95
Number of elements with 2 matching elements: 835
Number of elements with no matching elements: 360
n = 7 x = 4
Sum of elements: 614656
Sum of elements with 4 matching elements: 4676
Sum of elements with 3 matching elements: 52611
Sum of elements with 2 matching elements: 394913
Sum of elements with no matching elements: 162456
Number of elements with 4 matching elements: 7
Number of elements with 3 matching elements: 132
Number of elements with 2 matching elements: 1422
Number of elements with no matching elements: 840
What should I read to get more information about
hypermatrices, what
is the definitive reference on hypermatrices?
I noticed someone asking about space being 3-D, I noticed the
other
day in reading about knots that MathWorld noted that knots
don't exist
in dimensions >=4. I don't quite understand that.
Besides knot theory, another thing I am trying to grasp is the
concept
of the matroid. It appears to be some generalization of graphs
and
matrices.
I notice some posters asking about multitudes of infinities
and a
universal set. A set of all sets would be its own powerset,
and would
not contain an element of all sets not containing themselves.
Are
there more even than odd numbers? No, there are not. Are there
more
integers than even integers? Yes, there are, in the integers
(or
rationals, reals, complex, or hypercomplex numbers). There are
also
more rational than integers in the rationals, etcetera, and
more
positive integers than non-negative integers in the integefor
various reasons. Are there functions that map among all those
sets?
Consider a space-filling curve. An infinite set a is greater
than
another infinite set b if infinitely many proper supersets of
b are
proper subsets of a, |[a]| > |[b]|, an infinite set a is
greater than
a proper superset b, |(a)| > |(b)|, the set or a linear
translation.
Anyways, I want to know about how many permuations of x
elements of
{1, ..., n} there are with multiples of the elements in the
multiset.
A combination of the elements of an n-set {1, ..., n} is a
subset of
that set. A permutation of a combination of an n-set is an
ordered
list, a sequence, of elements of a combination of an n-set. A
multiset is a set that many contain multiple indistinguishable
instances of an otherwise unique set element. The xn-set is
here
defined as the multiset with x-many instances of each element
of an
n-set. The hypermatrix indices or coordinates of an x-d
nxnx...xn M
= m_ijk... are the set of permutations of the x-subsets of the
xn-set.
For example, for a 5-d hypermatrix of rank or degree 4,
coordinates
include (1, 1, 1, 1, 1), (1, 2, 3, 3, 4), and (4, 4, 4, 4, 4),
permutations of {{1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3,
4, 4,
4, 4, 4, 5, 5, 5, 5, 5}}, the xn-set that is the (4)(5)-set,
using two
braces to signify a multiset.
About a multiset, a set still has to have unique elements.
There are
a variety of ways to represent the unique elements of a set.
The von
Neumann ordinals are the most direct way to represent natural
integevery large composites could also represent the natural
integers and as well multitudes of integers. The composites
would
equate to each other as integea subset of the composite could
be
the von Neumann ordinal, the integer operations would operate
on
uniquified composite multiset element ordinals.
I've seen on Dr. Math's site on Math Forum a form for the
Stirling
cycle number, the unsigned Stirling number of the first kind,
for n
and n-2, as noted in this thread. I found a different one for
n+1 and
n-1 before that, I assume it is already known. I found a form
for
s(n+1, n-2), I have not seen any other method directly
calculating
that value via a ratio of univariate polynomials. I want to
know
where it was already known, because the same source might be
able to
tell me all the other ones that are known. I want to determine
forms
for the other values of x for s(n+1, n-x+1) without having to
learn
umbral calculus, Gian-Carlo Rota's shadow-y exploration of
recurrence relations and perhaps linear relations, although it
seems a
good thing to know.
Have I discovered a form for s(n+1, n-4+1) since my previous
post?
Have you?
I'm thinking about investing in one of those CAS, computational
algebra systems. Yet, I want to actually write one. I'll look
to the
noted references.
I was going to add a political spiel but this is sci.math,
humor is
irrelevant, maybe later.
Why did Isabel cross the Atlantic? To get to the other side. I
hope
that Isabel doesn't do too much damage. What's an F-6
hurricane? I
heard Isabel has gusts of more than 190 miles per hour. Buckle
up.
How about a war against climate change? People should start
evacuating. Hopefully Isabel will careen off towards
Greenland, yet
it currently appears that it might march up the Potomac, and
be the
most damaging Act of God to ever hit the United States. Some
day a
volcano will erupt, although hopefully far enough in the
future that
we can alleviate its pressure via tunneling or applied
explosives, the
same goes for meteor strikes with anti-meteor tugs, rockets,
and
laseetcetera, and of course far in the future there will be
significant control of the weather besides the crude burning of
petrochemicals to release huge amounts of greenhouse gasses,
petrified muck. Seriously, people on the Maryland shore should
get
warnings.
Anyways, I think I found another factorial identity, as
described
previously, but it's kind of strange, lim n->oo ((sum n)^n -
sum(n^n))
/ n!^2 = 1, I'm still trying to understand it.
Anyways, i want to learn more about the theory of
hypermatrices, n-d
arrays, matroids, I guess I should read more of the tensor
material.
Bye bye,
Ross
===
Subject: Which is more difficult
Is factoring harder than solving a diophantine equation, in
general?
===
Subject: Re: Which is more difficult
>Is factoring harder than solving a diophantine equation, in
general?
Given an integer n, if you can find all integer solutions of
the
Diophantine equation x*y = n, then you can factor n.
Or you can reduce it to finding a nonnegative integer solution
of
(x + 2)*(y + 2) = n,
which is like finding one integer solution to
(x1^2 + x2^2 + x3^2 + x4^2 + 2)*(y1^2 + y2^2 + y3^2 + y4^2 +
2) = n.
--
Wanted: 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: Which is more difficult
>Is factoring harder than solving a diophantine equation, in
general?
Not sure whether you mean factoring a polynomial (say over the
rationals) or
factoring an integer, but these are both easier than solving a
diophantine
equation: there are algorithms for both kinds of factoring,
while for
diophantine equations in general there is no algorithm (look
up Hilbert's
Tenth Problem).
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: Legal logic ? puzzle.
> The matter which I am desperately struggling with,
concerns a
> time sequence of event; with the assumption that 'cause'
does NOT
> follow 'effect'.
> If I can't explain this to scientifically trained persons,
how will
> I explain the legal-types ?!
Perhaps they all do understand your story, but they don't
understand what
your actual question is.
If x owes 10$, but is billed 11$, isn't it possible for x to
just pay 10$?
Other possibility: every 10 times, x could simply not pay 1
bill. That would
make things even.
Herman Jurjus
===
Subject: Re: The two envelope paradox
In artikel <YBqdnfOzOe0p2fyiXTWJiQ@comcast.com> schreef Steve
Gerrard:
>
>
>< snippage>
>
> Having just muddled my way through this myself, I am going
to take a shot
at
> explaining what I have learned as it applies to your
proposal.
>
>> Just budding in, but I have found few people come up with
the following
>> solution. Before you open the envelope in your hand, think
of a number.
>> If the amount in your envelope is smaller then that number
ask to
>> switch. If it is greater don't switch.
>
> Your proposal would work (I think) if it just so happened
that over time,
the
> frequency of each possible amount was the same - that is,
that the
distribution
> of amounts was uniform. However, there is no reason to think
that is the
case.
> Illustrating that it is a false assumption is the real
purpose of the
paradox.
No
>
>> There are three possibilities.
>> 1) Both envelopes contain an amount under your number.
>> You will always switch and on average gain nothing.
>
> #1 is false, because the on average is assuming a uniform
distribution
of
> amounts, and that it will all average out over time. There
just is
nothing
> that makes that so.
No it is assuming a uniform distribution in having chosen the
lower or
the higher number. Half of the time I will have sitched the
lower
amount for the higher and vice versa. The amount I win in the
first
case is on average the amount I lose in the latter case. So on
average I gain nothing.
>> 2) Both envelopes contain an amount above your number.
>> You will never switch and on average gain nothing.
>
> #2 works, since you are not changing anything.
>
>> 3) Your number is between the amounts in the envelopes.
>> You will keep the higher amount but will swicth
>> with the lower amount, so you will gain some.
>
> #3 might happen, and if it did, you might make a gain. Might
not. There is
no
> way to tell, since no one knows the true distribution of
amounts that
will be
> carried around in pairs of envelopes by generous persons
with an interest
in
> logic problems, for the rest of eternity ...
But you can't escape the fact, that if a person has two such
envelopes
that the double amount has a higher chance of being bigger
than some
randomly choosen number than the single amount. Sure the
chance can
be very small, but a small chance is still better than no
chance.
--
Antoon Pardon
===
Subject: Re: The two envelope paradox
> But you can't escape the fact, that if a person has two such
envelopes
> that the double amount has a higher chance of being bigger
than some
> randomly choosen number than the single amount. Sure the
chance can
> be very small, but a small chance is still better than no
chance.
It actually isn't better, in the case of infinities. This is
like a
limit
in calculus. What is the limit of 1/x as x -> infinity? The
answer is
zero.
In other words, a sufficiently small chance is
_exactly_the_same_ as
no chance, not still better than no chance.
-- Don
______________________________________________________________
______________
___
Don Geddis http://don.geddis.org/
don@geddis.org
The statistics on sanity are that one out of every four
Americans is
suffering
from some form of mental illness. Think of your three best
friends. If
they're okay, then it's you. -- Rita Mae Brown
===
Subject: symbolic calculus software
What is the best soft to do symbolic calculus ? Maple 9,
Mathematica 4.2
or Matlab 6.5 ?
===
Subject: Re: symbolic calculus software
>What is the best soft to do symbolic calculus ? Maple 9,
Mathematica 4.2
>or Matlab 6.5 ?
What does best mean??
===
Subject: Re: symbolic calculus software
> What is the best soft to do symbolic calculus ? Maple 9,
Mathematica 4.2
> or Matlab 6.5 ?
Mathematica has a newer release (IIRC it is 5.0).
There are other symbolic math programs out there too.
For example, Gauss, Macsyma, Mathcad, Axiom, MuPad, and others.
There are also specialized programs out there for specific
math topics.
May want to have a look.
HTH, FLip
===
Subject: Re: symbolic calculus software
> What is the best soft to do symbolic calculus ? Maple 9,
Mathematica 4.2
> or Matlab 6.5 ?
>
Use Matlab for numerical and matrix calculations. Use Maple or
Mathematica for symbolic calculus.
===
Subject: [graph theory] how to find a path with a length N
I'm looking for some algorithms in order to solve the following
problem:
I want to extract from a graph all the paths between two nodes
A and B
such that the number of edges (of those path) is equal to N.
does anyone have some ideas?
Jeremie
===
Subject: Rubik's cube: another question...
If you consider the moves you can make, a plane can be rotated
clockwise
once, anticlockwise once, or rotated twice (180 degrees) -
i.e. each plane
has three possible moves. I count 9 planes on the cube (3
hirozontal, 3
vertical and 3 into the cube. So, in one move there are 3x9 =
27 possible
moves. so in 14 moves, there are 27^14 possible outcomes this
= 1.09 x
10^20
(I presume a few of these might return to it's original state)
. But there
are only 4.3x 10^19 total different patterns.
Does this mean that any given cube is possible to solve within
14 moves?
===
Subject: Re: Rubik's cube: another question...
>
> If you consider the moves you can make, a plane can be
rotated clockwise
> once, anticlockwise once, or rotated twice (180 degrees) -
i.e. each
plane
> has three possible moves. I count 9 planes on the cube (3
hirozontal, 3
> vertical and 3 into the cube. So, in one move there are 3x9
= 27 possible
> moves. so in 14 moves, there are 27^14 possible outcomes
this = 1.09 x
10^20
> (I presume a few of these might return to it's original
state) . But
there
> are only 4.3x 10^19 total different patterns.
>
> Does this mean that any given cube is possible to solve
within 14 moves?
I think this problem is generally referred to as God's
algorithm. Given D, the diameter of the graph of Rubik
cube states, an omnipotent being could determine the shortest
solution to any given starting state. This solution by
definition
would be D or fewer moves long.
According to this site, God's algorithm is 14 moves long
for the 2 x 2 x 2 cube, but the diameter is unknown for the
3-cube.
http://web.usna.navy.mil/~wdj/book/node187.html
On another site, I saw mention of a speculation
that D=22 for the 3-cube, but without any mathematics.
- Randy
===
Subject: Re: Rubik's cube: another question...
...
> According to this site, God's algorithm is 14 moves long
> for the 2 x 2 x 2 cube, but the diameter is unknown for the
> 3-cube.
> http://web.usna.navy.mil/~wdj/book/node187.html
> On another site, I saw mention of a speculation
> that D=22 for the 3-cube, but without any mathematics.
The 14 are right. Long ago I have already speculated that for
the 3-cube
the diameter would be 22 or 23. This was based on the look of
the graph
for other similar puzzles and the part that was known about
the graph
forthe
3-cube.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Rubik's cube: another question...
>
> If you consider the moves you can make, a plane can be
rotated clockwise
> once, anticlockwise once, or rotated twice (180 degrees) -
i.e. each
plane
> has three possible moves. I count 9 planes on the cube (3
hirozontal, 3
> vertical and 3 into the cube. So, in one move there are 3x9
= 27 possible
> moves. so in 14 moves, there are 27^14 possible outcomes
this = 1.09 x
10^20
> (I presume a few of these might return to it's original
state) . But
there
> are only 4.3x 10^19 total different patterns.
>
> Does this mean that any given cube is possible to solve
within 14 moves?
It depends what you mean by a move - I think it's most usual to
consider the 6 squares at the centres of the faces to be fixed,
so a move of one on the 3 central planes is counted as two
moves -
one by each of its neighbours. So there are 18 (3x6) possible
moves from any position rather than 27.
Unfortunately this doesnt prove that any position can be
reached in
16 moves, even though 18^16 > 4.3x10^19, because, as you point
out,
a few of these might return [the cube] to [its] original state.
(Actually it will be rather more than a few...)
Have a look at http://cubeman.org/ for more information,
especially
the bits about God's algorithm in Progress in Solving
Algorithms
under Cube Notes. According to this, the maximum length of
God's
algorithm is empirically at most 20: I don't know whether
there are
any better estimates. (By the argument above it's at least 16).
Andrew Taylor
(Who has the small claim to fame of being mentioned in the Cube
section of Winning Ways by Berlekamp, Conway & Guy)
===
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.
Aoccdrnig to Google the word Aoccdrnig has appeared in at
least 51 news
groups. It appears to have been off-topic in most of them. The
dates below
relate to the most recent post in the thread. The word was
unknown to
Usenet
The list below does not include all the newsgroups in which
the word has
appeared. If a post is cross-posted to several groups it seems
that only
the
first is referenced. For example, I picked it up in rec.humor
which is not
listed below.
--
Clive Tooth
http://www.clivetooth.dk
===
Subject: Re: Way off topic
linux)
>> 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.
> Aoccdrnig to Google the word Aoccdrnig has appeared in at
least 51
news
> groups. It appears to have been off-topic in most of them.
The dates
below
> relate to the most recent post in the thread. The word was
unknown to
Usenet
It's making the Xerox rounds, too. A guy at work printed out a
few
copies of this quote, or nearly enough. Same misspelling of
(Aoccdrnig to *a* rscheearch...).
Was the paragraph youor did it appear elsewhere?
--
What I've learned is that [mathematicians are] the gatekeepeand
seem to have almost absolute power when it comes to
mathematics.
-- , on All I Really Ever Needed to Know I Learned
in /Ghostbusters/.
===
Subject: Re: Way off topic
message
>
>> 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.
>
> Aoccdrnig to Google the word Aoccdrnig has appeared in at
least 51
news
> groups. It appears to have been off-topic in most of them.
The dates
below
> relate to the most recent post in the thread. The word was
unknown to
Usenet
> It's making the Xerox rounds, too. A guy at work printed out
a few
> copies of this quote, or nearly enough. Same misspelling of
> (Aoccdrnig to *a* rscheearch...).
> Was the paragraph youor did it appear elsewhere?
I copied it from a post in rec.humor, making a couple of very
minor
corrections.
--
Clive Tooth
http://www.clivetooth.dk
===
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.
>
>
> That's very interesting. And it's easy to predict what the
replies are
> going to look like. tuB I ewnrdo hthewre it tsetamr htat we
aelev eth
> sfrit and tsal strelet in lapce? Hmm, I guess it does.
B.t is t...e so m..h i.........n in t.e f...t a.d l..t l.....s
t..t
t.e m....e o..s d..'t m....r at a.l?
Phil
===
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.
>>
>>
>> That's very interesting. And it's easy to predict what the
replies are
>> going to look like. tuB I ewnrdo hthewre it tsetamr htat we
aelev eth
>> sfrit and tsal strelet in lapce? Hmm, I guess it does.
>B.t is t...e so m..h i.........n in t.e f...t a.d l..t
l.....s t..t
>t.e m....e o..s d..'t m....r at a.l?
Unclear - the others certainly help comprehension.
>Phil
************************
David C. Ullrich
===
Subject: Re: Way off topic
is there a reference to this result? (please try to avoid
using the
lemma when posting the reference.. . :-) )
ceehrs
-tzurs
> 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.
>
>
===
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.
I always suspected that English was a highly redundant
language, as
probably many other languages are. In the old days, when
letters were
carved into stone, they exploited this redundancy already by
leaving
out the vowels...
And Hamming, in the fifties last century, worked the other way
by
adding redundancy to a message (bit string) in a clever way,
in order
to make it more fault tolerant when passing it through a
transmission
channel.
BTW, there's plenty of math going into that, as in fault
tolerant
logic synthesis (would'nt you hate it if a single faulty
transistor,
out of a many millions, wrecks a sevral man_years of design
work!-)
And qua math texts: I noticed that it is useful to figure out,
in
advance, in how many ways a text can be mis-interpreted (for
mathematicians this appears to be their main concern, of
course;-)
-- NB
===
Subject: Re: Way off topic
In sci.math, The Ghost In The Machine
<ewill@sirius.athghost7038suus.net>
<9itb31-gdc.ln1@lexi2.athghost7038suus.net>:
> In sci.math, The Last Danish Pastry
> <TheLastDanishPastry@yahoo.com>
> <bjsgsu$mt6lr$1@ID-11651.news.uni-berlin.de>:
>> 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.
>>
>
> Fascinating, and quite readable. :-) I'll admit this is
probably
> more appropriate to sci.psychology or some such but the idea
> has occurred to me personally.
Erm, that should be
has NOT occurred to me personally.
>
> I might have to whip up a Perl script to test this at some
point.
>
> Of course spelling checkers will blow chunks over such
submissions. :-)
>
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
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.
>
> I do not believe this. We may take in the word as a whole,
> but the internal processor still does it letter by letter,
> unless the word is recognized quickly. Those with larger
> vocabularies will find more words to separate out.
That one needs to _know_ the correct spelling of the word in
order
to recognize its distorted version does not neccesarily mean
that the
process of reading goes letter by letter.
Actually one also needs a firm grasp of the underlying grammar.
This way one can match a whole sentence pattern against
plausible
meaningful sentences. I doubt if the permutation of letters
could
still be repaired when these are introduced in an isolated
word.
So the wording of the original post is a bit misleading
because it
neglects the usage of the context.
Marc
===
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.
Very nice. I had no trouble reading it once I stopped proofing
it.
>I do not believe this. We may take in the word as a whole,
>but the internal processor still does it letter by letter,
>unless the word is recognized quickly.
I don't think so. It may differ with people who have been
trained in different disciplines. For instance, you math
guys need to process glyph by glyph because that's just the
way of math.
> .. Those with larger
>vocabularies will find more words to separate out.
When I took a speed reading course, the point was to teach
my eyes to read a sentence, then a paragraph at a time.
Proficiency in reading in grade school comes when kids
stop reading each letter and word separately. The poor
readers never get beyond this point.
/BAH
Subtract a hundred and four for e-mail.
===
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.
>Very nice. I had no trouble reading it once I stopped
proofing it.
>>I do not believe this. We may take in the word as a whole,
>>but the internal processor still does it letter by letter,
>>unless the word is recognized quickly.
>I don't think so. It may differ with people who have been
>trained in different disciplines. For instance, you math
>guys need to process glyph by glyph because that's just the
>way of math.
Regardless of whether one believes it, what you say
about math is a good point, and even somewhat on-topic.
It often seems to me that when students simply cannot
read mathematics part of the problem is that they're not
reading it slowly enough - they glance at an expression,
find they cannot understand it by glancing at it and
conclude they cannot understand it.
>> .. Those with larger
>>vocabularies will find more words to separate out.
>When I took a speed reading course, the point was to teach
>my eyes to read a sentence, then a paragraph at a time.
>Proficiency in reading in grade school comes when kids
>stop reading each letter and word separately. The poor
>readers never get beyond this point.
>/BAH
>Subtract a hundred and four for e-mail.
************************
David C. Ullrich
===
Subject: Re: Way off topic
>
>Aoccdrnig to rscheearch at an Elingsh uinervtisy, it
<sinp ogirnial emaxlpe>
>Very nice. I had no trouble reading it once I stopped
proofing it.
>
>>
>>I do not believe this. We may take in the word as a whole,
>>but the internal processor still does it letter by letter,
>>unless the word is recognized quickly.
>
>I don't think so. It may differ with people who have been
>trained in different disciplines. For instance, you math
>guys need to process glyph by glyph because that's just the
>way of math.
>
> Regardless of whether one believes it, what you say
> about math is a good point, and even somewhat on-topic.
> It often seems to me that when students simply cannot
> read mathematics part of the problem is that they're not
> reading it slowly enough - they glance at an expression,
> find they cannot understand it by glancing at it and
> conclude they cannot understand it.
>
We tend to assume that there is only one kind of reading
skill, but
there's some research supporting the idea that reading varies
comparing reading of music versus reading a book:
http://www.musica.uci.edu/mrn/V5I1W98.html#sightreading
[M]usic cognition and behavior are often viewed merely as an
instance
of other, better known subjects. An example is music
sight-reading,
often believed to obey the laws of language reading. However,
recent
studies reveal that the study of sight-reading in music
provides a
unique window on the mind.
(saccades) while reading printed texts and printed musical
scores,
and how different they are.
Probably a similar point of view can be applied to reading
mathematics. While similar to regular reading, in some ways it
is
fundamentally different than reading English (or other verbal
languages).
===
Subject: Re: Way off topic
|Regardless of whether one believes it, what you say
|about math is a good point, and even somewhat on-topic.
|It often seems to me that when students simply cannot
|read mathematics part of the problem is that they're not
|reading it slowly enough - they glance at an expression,
|find they cannot understand it by glancing at it and
|conclude they cannot understand it.
I remember someone once commented that the good
math students seemed usually to be better at adjusting
their speed based on a recognition of their own level
of comprehension. As far as I know this was just an
impression without specific experimental support, but
it's certainly the sort of think one would *like* to see
math students doing. Some of the worst difficulties
students had with math seemed to accompany their
inability to tell whether they were even getting the
point correctly or not. I remember sometimes students
understanding a point correctly, but not realizing that
it was as simple as that.
Keith Ramsay
===
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.
>>Very nice. I had no trouble reading it once I stopped
proofing it.
>I do not believe this. We may take in the word as a whole,
>but the internal processor still does it letter by letter,
>unless the word is recognized quickly.
>>I don't think so. It may differ with people who have been
>>trained in different disciplines. For instance, you math
>>guys need to process glyph by glyph because that's just the
>>way of math.
>Regardless of whether one believes it, what you say
>about math is a good point, and even somewhat on-topic.
I tried :-). Math is unusual in that it packs a lot of
meaning into one little glyph; for example, textbooks have
been written to prepare me to know what that sigma implies :-).
>It often seems to me that when students simply cannot
>read mathematics part of the problem is that they're not
>reading it slowly enough - they glance at an expression,
>find they cannot understand it by glancing at it and
>conclude they cannot understand it.
Do you know if these kids do sentence diagramming in grade
school? In math, it's important to be able to take things
apart, rearrange them, and put them back together again.
Same is true in chemistry, physics, definitely programming.
/BAH
Subtract a hundred and four for e-mail.
===
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.
>Very nice. I had no trouble reading it once I stopped
proofing it.
>>
>>I do not believe this. We may take in the word as a whole,
>>but the internal processor still does it letter by letter,
>>unless the word is recognized quickly.
>I don't think so. It may differ with people who have been
>trained in different disciplines. For instance, you math
>guys need to process glyph by glyph because that's just the
>way of math.
>>Regardless of whether one believes it, what you say
>>about math is a good point, and even somewhat on-topic.
>I tried :-). Math is unusual in that it packs a lot of
>meaning into one little glyph; for example, textbooks have
>been written to prepare me to know what that sigma implies
:-).
>>It often seems to me that when students simply cannot
>>read mathematics part of the problem is that they're not
>>reading it slowly enough - they glance at an expression,
>>find they cannot understand it by glancing at it and
>>conclude they cannot understand it.
>Do you know if these kids do sentence diagramming in grade
>school?
I don't know for a fact, but it seems very unlikely.
>In math, it's important to be able to take things
>apart, rearrange them, and put them back together again.
>Same is true in chemistry, physics, definitely programming.
Yup. Which is exactly why kids should be required to do
simple proofs, just to build those muscles.
But this is really getting way on-topic...
>/BAH
>Subtract a hundred and four for e-mail.
************************
David C. Ullrich
===
Subject: Re: Way off topic
>
>In math, it's important to be able to take things
>apart, rearrange them, and put them back together again.
>Same is true in chemistry, physics, definitely programming.
>
> Yup. Which is exactly why kids should be required to do
> simple proofs, just to build those muscles.
Exhibit of proof of Ex~(x=x) from C1-C4 and someone will
point out the error.
C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1
C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] Weak
Extensionality
How's about I let you in on a little secret? (Ex)~(x=x) follows
from (C3,C4) alone!
C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] Weak
Extensionality
C'mon, David. Show us how. Don't be afraid. You can do it!!!
--John
And yes, identity is in _fact_ reflexive. To
refute that statement you need to give an
example of something which is not identical
to itself. The idea that there is something
which is _not_ identical to itself is simply
ludicrous: That's what identity _means_: A
thing is identical to itself and to nothing
else.
--David Ullrich
===
Subject: Re: Way off topic
>>
>>In math, it's important to be able to take things
>>apart, rearrange them, and put them back together again.
>>Same is true in chemistry, physics, definitely programming.
>>
>> Yup. Which is exactly why kids should be required to do
>> simple proofs, just to build those muscles.
>Exhibit of proof of Ex~(x=x) from C1-C4 and someone will
>point out the error.
>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1
>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] Weak
>Extensionality
>How's about I let you in on a little secret? (Ex)~(x=x)
follows
>from (C3,C4) alone!
You should also let everyone else in on another little secret:
In
the thread where this came up you were claiming that C1-C4
were axioms of NBG.
>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] Weak
>Extensionality
>C'mon, David. Show us how. Don't be afraid. You can do it!!!
>--John
>And yes, identity is in _fact_ reflexive. To
>refute that statement you need to give an
>example of something which is not identical
>to itself. The idea that there is something
>which is _not_ identical to itself is simply
>ludicrous: That's what identity _means_: A
>thing is identical to itself and to nothing
>else.
>--David Ullrich
Another little secret: When you quote me saying the above,
as though I'd said something stupid or even something
the teensiest bit controversial you look like an idiot.
(Don't bother replying with citations from learned parties
talking about non-reflexive identities. It's not _identity_
they're talking about.)
************************
David C. Ullrich
===
Subject: Re: Way off topic
>
>>
>>In math, it's important to be able to take things
>>apart, rearrange them, and put them back together again.
>>Same is true in chemistry, physics, definitely programming.
>>
>> Yup. Which is exactly why kids should be required to do
>> simple proofs, just to build those muscles.
>
>
>Exhibit of proof of Ex~(x=x) from C1-C4 and someone will
>point out the error.
>
>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1
>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] Weak
>Extensionality
>
>How's about I let you in on a little secret? (Ex)~(x=x)
follows
>from (C3,C4) alone!
>
> You should also let everyone else in on another little
secret: In
> the thread where this came up you were claiming that C1-C4
> were axioms of NBG.
Produce a post where I claim this, or admit you are a LIAR.
>
>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] Weak
>Extensionality
>
>C'mon, David. Show us how. Don't be afraid. You can do it!!!
>
>--John
>
>And yes, identity is in _fact_ reflexive. To
>refute that statement you need to give an
>example of something which is not identical
>to itself. The idea that there is something
>which is _not_ identical to itself is simply
>ludicrous: That's what identity _means_: A
>thing is identical to itself and to nothing
>else.
>--David Ullrich
>
> Another little secret: When you quote me saying the above,
> as though I'd said something stupid or even something
> the teensiest bit controversial you look like an idiot.
>
> (Don't bother replying with citations from learned parties
> talking about non-reflexive identities. It's not _identity_
> they're talking about.)
Thus we might characterise a non-individual as an object for
which no identity criteria hold. If the theory of identity
is taken to be that of classical logic, then for a
non-individual
it is not the case that a = a'. This allows us to avoid
the Evans-Salmon argument and resolve the curious situation
it is indeterminate whether they are identical to themselves
because they are not! (Steven French and Decio Krause, Quantum
To be ignorant of one's ignorance is the malady of the
ignorant.
--Amos Bronson Alcott, _Table Talk_ [1877]
===
Subject: Re: Way off topic
===
>Subject: Re: Way off topic
>Message-id: <tap8mvs204vngit05lepm31bdini99kh81@4ax.com>
>>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.
>>Very nice. I had no trouble reading it once I stopped
proofing it.
>I do not believe this. We may take in the word as a whole,
>but the internal processor still does it letter by letter,
>unless the word is recognized quickly.
>>I don't think so. It may differ with people who have been
>>trained in different disciplines. For instance, you math
>>guys need to process glyph by glyph because that's just the
>>way of math.
>Regardless of whether one believes it, what you say
>about math is a good point, and even somewhat on-topic.
>It often seems to me that when students simply cannot
>read mathematics part of the problem is that they're not
>reading it slowly enough - they glance at an expression,
>find they cannot understand it by glancing at it and
>conclude they cannot understand it.
I find that a lot of the stuff posted in sci.math makes my
eyes glaze over.
And
frequently it's stuff I actually understand when I make an
effort to disect
the
ASCII equations. It's amazing how one's perception of those
symbols changes
once one understands what it's about.
> .. Those with larger
>vocabularies will find more words to separate out.
>>When I took a speed reading course, the point was to teach
>>my eyes to read a sentence, then a paragraph at a time.
>>Proficiency in reading in grade school comes when kids
>>stop reading each letter and word separately. The poor
>>readers never get beyond this point.
>>/BAH
>>Subtract a hundred and four for e-mail.
>************************
>David C. Ullrich
--
Mensanator
2 of Clubs
http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm
===
Subject: Re: Way off topic
===
>>Subject: Re: Way off topic
>>Message-id: <tap8mvs204vngit05lepm31bdini99kh81@4ax.com>
>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.
>Very nice. I had no trouble reading it once I stopped
proofing it.
>>
>>I do not believe this. We may take in the word as a whole,
>>but the internal processor still does it letter by letter,
>>unless the word is recognized quickly.
>I don't think so. It may differ with people who have been
>trained in different disciplines. For instance, you math
>guys need to process glyph by glyph because that's just the
>way of math.
>>Regardless of whether one believes it, what you say
>>about math is a good point, and even somewhat on-topic.
>>It often seems to me that when students simply cannot
>>read mathematics part of the problem is that they're not
>>reading it slowly enough - they glance at an expression,
>>find they cannot understand it by glancing at it and
>>conclude they cannot understand it.
>I find that a lot of the stuff posted in sci.math
>makes my eyes glaze over.
Especially when the right margin is set way out to 80. *hint*
> .. And
>frequently it's stuff I actually understand when
>I make an effort to disect the
>ASCII equations.
I always have to write it down on paper. I don't think I'll
ever
establish that short cut of translating from screen to concept
in my head like youngsters can do very, very well.
> ..It's amazing how one's perception of those symbols changes
>once one understands what it's about.
There are pixel problems with presentation on a terminal
screen, too.
/BAH
Subtract a hundred and four for e-mail.
===
Subject: Re: Way off topic
>>I find that a lot of the stuff posted in sci.math
>>makes my eyes glaze over.
>Especially when the right margin is set way out to 80. *hint*
How do you do that in the AOL editor?
--
Mensanator
2 of Clubs
http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm
===
Subject: Re: Way off topic
>I find that a lot of the stuff posted in sci.math
>makes my eyes glaze over.
>>Especially when the right margin is set way out to 80. *hint*
>How do you do that in the AOL editor?
There is a key on the right that may have the word enter
on it. Use that when the parser gets 3/4 across your screen.
/BAH
Subtract a hundred and four for e-mail.
===
Subject: Re: Way off topic
===
>Subject: Re: Way off topic
>Message-id: <bk4655$j2k$1@bob.news.rcn.net>
>>I find that a lot of the stuff posted in sci.math
>>makes my eyes glaze over.
>Especially when the right margin is set way out to 80.
*hint*
>>How do you do that in the AOL editor?
>There is a key on the right that may have the word enter
>on it.
Duh. Sorry, I apparenetly mistook you for someone with
something
contructive to offer. Having the right margin set at 80
implies that
it can be changed in the preferences, but there aren't any.
> Use that when the parser gets 3/4 across your screen.
IF the editing windows were always the same size, and IF the
editor
used a monospaced font, and IF the font size is constant,
THEN that might make some sense.
But as they aren't, it doesn't, and it's not, I'll have to
look elsewhere
for advice.
>/BAH
>Subtract a hundred and four for e-mail.
--
Mensanator
2 of Clubs
http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm
===
Subject: Re: Way off topic
===
>>Subject: Re: Way off topic
>>Message-id: <bk4655$j2k$1@bob.news.rcn.net>
>I find that a lot of the stuff posted in sci.math
>makes my eyes glaze over.
>>
>>Especially when the right margin is set way out to 80.
*hint*
>How do you do that in the AOL editor?
>>There is a key on the right that may have the word enter
>>on it.
>Duh. Sorry, I apparenetly mistook you for someone with
something
>contructive to offer.
[emoticon notices some snot]
>Having the right margin set at 80 implies that
>it can be changed in the preferences, but there aren't any.
I suspect it's different with every program. The way to force
idiot programs to do your bidding is to use the enter key.
>> Use that when the parser gets 3/4 across your screen.
>IF the editing windows were always the same size, and IF the
editor
>used a monospaced font, and IF the font size is constant,
>THEN that might make some sense.
You're just producing excuses. None of those IFs apply.
>But as they aren't, it doesn't, and it's not, I'll have to
look elsewhere
>for advice.
IOW, doing the logical thing isn't your bag. Out of curiosity,
how did you learn to touch type? People who trained on a
typewriter
usually don't have any problems with typing carriage-control
characters.
/BAH
Subtract a hundred and four for e-mail.
===
Subject: Re: Way off topic
===
>>Subject: Re: Way off topic
>>Message-id: <bk4655$j2k$1@bob.news.rcn.net>
>>
>I find that a lot of the stuff posted in sci.math
>makes my eyes glaze over.
>>
>>Especially when the right margin is set way out to 80.
*hint*
>
>How do you do that in the AOL editor?
>>
>>There is a key on the right that may have the word enter
>>on it.
>
>Duh. Sorry, I apparenetly mistook you for someone with
something
>contructive to offer.
>
> [emoticon notices some snot]
Snotty? You were admonishing me for not having correctly set
something
I have no control over. And when I politely asked where those
settings
are found, your snotty reply was to learn where the enter key
was.
If you're going to pontificate from a position of ignorance,
don't be
surprised that you get snotty replies.
>
>Having the right margin set at 80 implies that
>it can be changed in the preferences, but there aren't any.
>
> I suspect it's different with every program. The way to force
> idiot programs to do your bidding is to use the enter key.
Do you know the difference between an editor and a word
processor? In
the former, the user usually controls the line breaks with the
enter
key. In a word processor or any other program that
automatically
controls word wrap, the enter key is used for paragraph breaks.
>
>
>> Use that when the parser gets 3/4 across your screen.
>
>IF the editing windows were always the same size, and IF the
editor
>used a monospaced font, and IF the font size is constant,
>THEN that might make some sense.
>
> You're just producing excuses. None of those IFs apply.
Unlike some, I don't spout off without knowing what I'm
talking about.
You might want to check out my reply to Gerry Myerson to learn
something about editing in AOL.
>
>
>But as they aren't, it doesn't, and it's not, I'll have to
look
elsewhere
>for advice.
>
> IOW, doing the logical thing isn't your bag.
The logical thing is to use the editor the way it is intended
to be
used, with enter used for paragraph breaks, not line breaks.
And
just to be snotty, I'll say if you don't like the way the
messages
are displayed, then that's _your_ problem, not mine.
> Out of curiosity, how did you learn to touch type?
I didn't, but 25 years of banging on computer keyboards has (in
addition to giving me carpal tunnel syndrome) made me a very
fast
pecker.
> People who trained on a typewriter usually don't have any
problems with
> typing carriage-control characters.
Until they use a word processor and find out that every line
has
become a paragraph. Luckily, I am multi-talented. I can use
typewriteeditors and word processors interchangeably and not
get
confused.
>
> /BAH
>
> Subtract a hundred and four for e-mail.
===
Subject: Re: Way off topic
===
>Subject: Re: Way off topic
>Message-id: <bk4655$j2k$1@bob.news.rcn.net>
>
>>
>>How do you do that in the AOL editor?
>
>There is a key on the right that may have the word enter on
it.
>
> Use that when the parser gets 3/4 across your screen.
>
> IF the editing windows were always the same size, and IF the
editor
> used a monospaced font, and IF the font size is constant,
THEN that
> might make some sense.
>
> But as they aren't, it doesn't, and it's not, I'll have to
look
> elsewhere for advice.
Probably not to me, as I'm not familiar with the AOL editor,
but can I ask: is it possible to resize a window? and is it
possible to make the editor use the font of your choosing?
and is it possible to convince the editor to use the font
size of your choosing?
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Way off topic
===
>Subject: Re: Way off topic
===
>>Subject: Re: Way off topic
>>Message-id: <bk4655$j2k$1@bob.news.rcn.net>
>>
>
>How do you do that in the AOL editor?
>>
>>There is a key on the right that may have the word enter on
it.
>>
>> Use that when the parser gets 3/4 across your screen.
>>
>> IF the editing windows were always the same size, and IF
the editor
>> used a monospaced font, and IF the font size is constant,
THEN that
>> might make some sense.
>>
>> But as they aren't, it doesn't, and it's not, I'll have to
look
>> elsewhere for advice.
>Probably not to me, as I'm not familiar with the AOL editor,
>but can I ask: is it possible to resize a window?
Yes, but it would be a lot of work and you may end up
with
this phenomena, which I find very annoying. It is caused
by people
who incorrectly gauge where 3/4 of the screen is and
sometimes
hit the enter key after the editor has already imposed
its own
line wrap at 80 columns, which is not shown prior to
sending.
And no, there is no preview mode. The Arial font used by AOL
contributes
to this problem.
>and is it possible to make the editor use the font of your
choosing?
No. AOL has preference settings for the e-mail but not for the
newsgroup editor.
>and is it possible to convince the editor to use the font
>size of your choosing?
And when I say there are none, I mean there is a certain
amount.
One of the e-mail settings is to display text as small, medium
or large. For some reason, the newsgroup editor inherits this
setting although it does not inherit the font setting
preferences.
With a large display such as 1280x960 and the preference on
small, the editing window is easily over a hundred characters
wide if the AOL window is at max and the edit window is at max.
And that's when you are replying. Creating a new message
almost doubles the
edit window size.
It is extremely annoying not being able to control the editor
settings.
Usually, for serious replies, especially those involving ASCII
art,
I compose in Notepad and paste in the reply. But that's a lot
of effort
for simple one or two line responses
For this message, I tried to control the line length with the
enter
key. I won't know how successful until after it's posted.
>--
>Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
--
Mensanator
2 of Clubs
http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm
===
Subject: Re: Way off topic
===
>>Subject: Re: Way off topic
>>Message-id: <bk4655$j2k$1@bob.news.rcn.net>
>>
>
>How do you do that in the AOL editor?
>>
>>There is a key on the right that may have the word enter on
it.
>>
>> Use that when the parser gets 3/4 across your screen.
>>
>> IF the editing windows were always the same size, and IF
the editor
>> used a monospaced font, and IF the font size is constant,
THEN that
>> might make some sense.
>>
>> But as they aren't, it doesn't, and it's not, I'll have to
look
>> elsewhere for advice.
>Probably not to me, as I'm not familiar with the AOL editor,
>but can I ask: is it possible to resize a window?
With the AOL software I use, it's a temporary thing and
has to be reset with every display of a post and its reply.
> .. and is it
>possible to make the editor use the font of your choosing?
I suspect that depends on the version running. Mine doesn't
but mine is very, very old.
>and is it possible to convince the editor to use the font
>size of your choosing?
No.
/BAH
Subtract a hundred and four for e-mail.
===
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.
Let's see if this pertains to paintepantries, or anything done
in
loco parentis.
--
http://hertzlinger.blogspot.com
===
Subject: Re: Way off topic
linux)
>> 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.
> Let's see if this pertains to paintepantries, or anything
done in
> loco parentis.
Do you feel an odd compulsion to repeat yourself?
--
Jesse Hughes
So far as this negative attitude toward life is concerned,
Buddhism
is merely Taoism a little touched in its wits.
-- Lin Yutang, /My Country and My People/
===
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.
>> Let's see if this pertains to paintepantries, or anything
done in
>> loco parentis.
>
> Do you feel an odd compulsion to repeat yourself?
I feel it again and again.
--
http://hertzlinger.blogspot.com
===
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.
>>
>> Let's see if this pertains to paintepantries, or anything
done in
>> loco parentis.
>
> Do you feel an odd compulsion to repeat yourself?
> I feel it again and again.
The sherpas had pitons and plates.
This might come out as...
The shapers had points and pleats.
The seraphs had pintos and petals.
The sphaers had pinots and palets.
The sphears had potins and peltas.
[Some of the more obscure words...
palets: paleae
peltas: shields
pinots: grapes
potins: copper alloys
sphaers: old form of 'spheres'
sphears: old form of 'spheres']
--
Clive Tooth
http://www.clivetooth.dk
===
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.
>
> I do not believe this. We may take in the word as a whole,
> but the internal processor still does it letter by letter,
> unless the word is recognized quickly. Those with larger
> vocabularies will find more words to separate out.
I believe it. English is not my first language, nor my second,
still I had
no problems with reading. Some words require backtracking, but
only the
long words (I think I missed only two words on first reading).
I had also
no problems with the French text supplied in follow-up (but
French is my
second language).
I know for sure that I do not take words letter for letter.
There are
many cases where I know a letter is missing in a word, only
when it is
pointed out to me.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: An algebraic problem
let x and y be two rational angles verifing the following
condition:
8 (cos x)^3 cos y is an algebraic integer.
Is it true that at least one of them has to be a multiple of
pi/4?
Nicola
===
Subject: Re: An algebraic problem
> let x and y be two rational angles verifing the following
condition:
> 8 (cos x)^3 cos y is an algebraic integer.
> Is it true that at least one of them has to be a multiple of
pi/4?
If x and y are rational, how can one of them be a (non-zero)
multiple of
pi/4?
--
Clive Tooth
http://www.clivetooth.dk
===
Subject: Re: An algebraic problem
a rational angle is of the form p Pi with p in Q.
Nicola
>
>
> If x and y are rational, how can one of them be a (non-zero)
multiple of
> pi/4?
>
===
Subject: Re: An algebraic problem
> a rational angle is of the form p Pi with p in Q.
===
Subject: Re: When laying block, better than string for
straightness
> I am building my own concrete block garage. And when it came
time to
> lay the first course I did not like the string method for it
depends too
> much
> on eye judgement.
The string method is the absolute best way to make garages.
Since that's the way you make garages that aren't made
out of concrete blocks.
If you building a prison, the pipe method is preferred.
So what I did was haul out two very long and stout
> plumbing pipes. Very long stiff and firm and layed the block
loosely for
> the first row
> and then instead of string and eyeball I simple rolled the
two pipes,
> one on the inside of the row and one on the outside of the
row to make a
> perfect line.
>
> That method is great for the first course because the pipes
are on the
> concrete slab.
>
> Now, I am trying to figure a way to use the pipe method
going up the
> wall
> so that I never have to use the time wasting and imprecise
string
> method.
Sorry you can't do it. To get perfect vertical lines,
you need a plumb bob, which use strings.
> Thought I might share that with you so that we do replace
the old string
> method with something easier, faster and better. A method
where the test
> uses the material instead of a eyeball judgement call
electron-dot-cloud are galaxies
===
Subject: Re: When laying block, better than string for
straightness
(snipped)
> Sorry you can't do it. To get perfect vertical lines,
> you need a plumb bob, which use strings.
That is a very silly and stupid device to be using for
concrete and brick
wall
building. I say that because a fixed stick or rod is a faster
means of
telling the accuracy of vertical
lines than a plumb bob setup. Just picture the time wasted in
setting up
that plumb bob when a fixed
stick or rod and determine the vertical
accuracy in a split second.
My method does use a level to check upon things. But other
than a level, the
accuracy of vertical lines
and horizontal lines should always be that of *fixed
sticks or rods* that are straight. And never on string things
which rely
upon a eyeball judgement.
Obviously the above has never done hands on building, just
mouth chatter.
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: When laying block, better than string for
straightness
> (snipped)
>
>
> Sorry you can't do it. To get perfect vertical lines,
> you need a plumb bob, which use strings.
>
>
> That is a very silly and stupid device to be using for
concrete and brick
wall
> building.
That's why we who know what they are don't use them for making
walls,
are somehing stupid as concrete.
We use them for making plumbs lines.
I say that because a fixed stick or rod is a faster means of
telling
the accuracy of vertical
> lines than a plumb bob setup.
I did'nt say anything about faster. I mentioned something
about *vertical lines*, which is what your question was about.
Just picture the time wasted in setting up that plumb bob when
a
fixed
> stick or rod and determine the vertical
> accuracy in a split second.
There are no such things as fixed sticks. Either you've
been readng something about Einstein, or some first grade
mathematicians curicculum. Either way is doesn't matter,
since neither can actually build anything other
than playground balls.
> My method does use a level to check upon things. But other
than a level,
the accuracy of vertical lines
> and horizontal lines should always be that of *fixed
> sticks or rods* that are straight.
They would make sense if a level wasn't just an amatuer plumb
bob.
And never on string things which rely upon a eyeball judgement.
>
> Obviously the above has never done hands on building, just
mouth chatter.
That's true. Those of us who actually know how to build walls,
only design them. We let day laboresuch as yourself,
build them.
> Archimedes Plutonium, a_plutonium@hotmail.com
> whole entire Universe is just one big atom where dots
> of the electron-dot-cloud are galaxies
electron-dot-cloud are galaxies
===
Subject: Re: When laying block, better than string for
straightness
> Seems like alot of work would it not be easier to buy two
cheap laser
> pointers mount two to form a 90 deg angle, one pointing down
one pointing
in
> the direction of the row of bricks. Mark the ground so the
pointers are
in
> the same location on the first brick and line the bricks up
against the
> lightr pointing down the row you could even mount a bubble
level on them
to
> make sure it's level
Actually long 2 X 4 works even better than steel pipes.
No, what I am trying to achieve is a quick way to know that
all the block
or
brick in a row are in a perfect line. And where the string
method relies on
eyeball
judgement is bad. And so will the laser.
A long straight wood board down the line of a row of block
where you touch
the
block to the board and you know you are on that perfect line.
===
Subject: Re: When laying block, better than string for
straightness
>
> No, what I am trying to achieve is a quick way to know that
all the block
or
> brick in a row are in a perfect line. And where the string
method relies
on
> eyeball
> judgement is bad. And so will the laser.
>
> A long straight wood board down the line of a row of block
where you touch
the
> block to the board and you know you are on that perfect line.
Arch,
If you're actually doing the block work yourself, 'buttering`
the
block,
dropping it in place, tapping it level and plumb with your
trowel, you
know
that your mortar joint should be between 3, and 5 sixteenths
of an
inch thick.
With an eighth of an inch to play with, 'Perfect` is a big
waist of
your
time and effort, and any solid batterboard you try to use will
just
get in your way and slow you down even more.
This is demonstrated by the fact that you've apparently laid
only one
course of block in a day.
If you're an amateur, merely staying within tolerances will be
hard
enough. Just use the proven method and get on with it.
More than one project has been 're-thought` to death.
Best of luck -
Pragmatist -Use a bigger hammer
electron-dot-cloud are galaxies
===
Subject: Re: When laying block, better than string for
straightness
>
>
> No, what I am trying to achieve is a quick way to know that
all the
block or
> brick in a row are in a perfect line. And where the string
method relies
on
> eyeball
> judgement is bad. And so will the laser.
>
> A long straight wood board down the line of a row of block
where you
touch the
> block to the board and you know you are on that perfect line.
> Arch,
> If you're actually doing the block work yourself,
'buttering` the
> block,
> dropping it in place, tapping it level and plumb with your
trowel, you
> know
> that your mortar joint should be between 3, and 5 sixteenths
of an
> inch thick.
> With an eighth of an inch to play with, 'Perfect` is a big
waist of
> your
> time and effort, and any solid batterboard you try to use
will just
> get in your way and slow you down even more.
> This is demonstrated by the fact that you've apparently laid
only one
> course of block in a day.
> If you're an amateur, merely staying within tolerances will
be hard
> enough. Just use the proven method and get on with it.
> More than one project has been 're-thought` to death.
> Best of luck -
> Pragmatist -Use a bigger hammer
Doing it myself in order to learn and improve the task. This
industry has
never
been wholescale improved and that is where I can do the most
good.
For example, the old true and tested way is to use string and
to slop alot
of mortar to build the
building. Cement is never going to get cheaper and abundant
because fusion
energy is never coming. So
cement is going to become as rare and valuable as is petroleum.
Instead of using a trowel to place cement I believe in using
the fingers.
Not because I am an amateur
bricklayer but because I need to find the method that saves
every precious
piece of Portland cement.
Of course using latex gloves and not exposed skin.
For placement of mortar for the vertical joints that is easy.
For the
horizontal joints is a little bit
more tricky. What I do is get a 3/8 steel rebar and wire
connect
it to the vertical rebar and then rest the next course with
its *inner edge*
onto the
rebar. Thus I have a uniform horizontal guide. Then with the
latex-glove,
never a trowel, I fill in the
gaps and with a pointer I then solidly compact the joint.
No matter how great one gets at a troweling method they lose
and slop too
much
and waste too much cement.
In my method, which is the method of the future as oil prices
rise so does
cement. In my method,
although painstakingly longer is perhaps the ultimate
method of laying brick and block.
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
electron-dot-cloud are galaxies
===
Subject: Re: When laying block, better than string for
straightness
In the old method, mortar had to be a mix of lime for
consistency in order
to make the mix pliable to use
a troweling method. So that a standard mortal mix would be
something like 1
part mortar-cement which is
50-50 portlandcement
and lime and then 2 to 3 parts sand.
In my method we dispense with ever using any lime. We go
straight with a
concrete type mix without aggregates because we simply use
latex gloves
finger
in the mortar into the joints.
In the future as oil prices become very high, and cement
becomes very high,
the luxury of using lime and
wasting alot of mortar is a luxury not in the future.
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: Catalan sequence problem
> OK, this sounds close to noncrossing partitions. Which are
yet
> another Catalan counted structure (with all sorts of
bijections with
> other Catalan structures) problem pp from Stanley (with only
> references, no explanation)).
>
> However, the cyclic nature of your description might make a
difference.
> But then it should show up with n = 5. Er... no it doesn't
(rethinking
> the definitions), because the partitions can enclose another
without
> crossing: e.g. {{1,2,5},{3,4}} is not crossing.
>
> I'm thinking about what the bijection really should be (I
have seen it
> before but can't remember, and I'm having trouble with
obvious
> possibilities). Maybe DFS labelling on an ordered tree?
After talking to somebody, we figured it out. It's quite simple
(afterwards of course!).
Take a binary tree on n nodes (not necessarily full), and
label the
vertices according to inorder traversal. then remove all left
edges. the
remaining components form a partition, which, by the inorder
labeling,
is non-crossing. To go the other direction, take the part (of
the
partition) which includes n, create a rightmost path labeled
increasing
with these elements. All the parts between n and the label of
its parent
(the next highest in the part that n is in) can be recursively
made into
a tree to the left of n's parent.
One can also create noncrossing partitions from first
principles that
results in the Catalan recurrence C_n = sum(C_k C_{n-k-1},
{n,0,n-1}),
C_0 = 1. disjoint union a partition of 1 thru k and k+1 thru
n-1, then
put n in the same partition as k (when k is 0, it goes in a
partition
itself). This construction insures that it is noncrossing
(this latter
bit is not clear until you do some examples).
Mitch
===
Subject: polynomial gcd
Express the polynomial gcd of (x^m - 1,x^n - 1) in terms of x
and gcd (m,n)
?
===
Subject: Re: polynomial gcd
Here's a hint.
If x^a-1 divides x^b-1, then a divides b...
You can lead a horse's ass to knowledge, but you can't make
him think.
===
Subject: Re: polynomial gcd
> Express the polynomial gcd of (x^m - 1,x^n - 1) in terms of
x and gcd
(m,n) ?
x^gcd(m,n) -1, using Euclid's algorithm.
===
Subject: Re: polynomial gcd
>
>>Express the polynomial gcd of (x^m - 1,x^n - 1) in terms of
x and
>>gcd (m,n) ?
>
> x^gcd(m,n) -1, using Euclid's algorithm.
... and by extending that argument
gcd(x^n - y^n, x^m - y^m) = x^gcd(n,m) - y^gcd(n,m)
so, what is the next generalization?
It is very reminiscent of the identity
gcd(F_n,F_m) = F_gcd(n,m)
where F_n is a Fibonacci number (not just reminiscent, it
follows the
pattern above because of Binet's formula
F_n = (sigma^n - tau^n)/sqrt(5) )
So is it something about linear recurrences and divisibility?
restricted bases cases?
I can't seem to get an example that works where there are 3
bases
(x^n - y^n - z^n) (I kind of see why it is impossible, but it
is not
clear enough to me yet how to be more rigorous about it, or
how to
extend that disproof to others)
What's the generalization?
Mitch
===
Subject: Geometry: Perimeter & Number of Shapes Possible
Is there an easy way (without drawing) to determine the number
of
straight shapes that are possible to create with a certain
perimeter
value? For example, say that there are only three shapes (not
rounded
or with diagonal lines) that can possibly be created that have
a
perimeter of eight, is there an easy way to determine the
number of
shapes that can be made with a perimeter of say 100 or even
more
without drawing them (which would obviously be time consuming)?
===
Subject: Re: Geometry: Perimeter & Number of Shapes Possible
> Is there an easy way (without drawing) to determine the
number of
> straight shapes that are possible to create with a certain
perimeter
> value? For example, say that there are only three shapes
(not rounded
> or with diagonal lines) that can possibly be created that
have a
> perimeter of eight, is there an easy way to determine the
number of
> shapes that can be made with a perimeter of say 100 or even
more
> without drawing them (which would obviously be time
consuming)?
If you don't constrain edge numbelengths, and angles to finite
sets, there will be uncountably many shapes. For example, there
are uncountably many different triangles with a perimeter of 1.
You need to define the notions of shape, difference, and
perimeter;
indicate if paths must be continuous; and how crossings or
inclusions
are treated, before you will get agreement on numbers of
shapes.
-jiw
===
Subject: Re: CAN'T Order Reals
I don't see your point. As far as I know, counting &
establishing a total
order are different things. Counting a set is the same as
constructing a
infinite sequence in it, & the total order <= is a equivalence
relation, ie
* reflexive: a <= a
* antisymmetric: a <= b and b <= a ---> a = b
* transitive: a <= b and b <= c ---> a <= c
that satisfies, for all a,b: a<=b or b<=a.
So, I don't see why you claim that total orders c counted
orders.
The usual order in R is somewhat inherited from Q, if you
construct R from
Q-Cauchy sequences. And the order in Q can be stated as:
a/b <= c/d iff a*d <= b*c
There's no place for counting here, and the density of Q and R
does not
contradict that <= is a total order.
Sorry if I'm too lost, please explain me again, if I missed
the point.
--
DvD Gomez
===
Subject: Re: CAN'T Order Reals
> I don't see your point. As far as I know, counting &
establishing a total
> order are different things. Counting a set is the same as
constructing a
> infinite sequence in it, & the total order <= is a
equivalence relation,
ie
> * reflexive: a <= a
> * antisymmetric: a <= b and b <= a ---> a = b
> * transitive: a <= b and b <= c ---> a <= c
> that satisfies, for all a,b: a<=b or b<=a.
> So, I don't see why you claim that total orders c counted
orders.
> The usual order in R is somewhat inherited from Q, if you
construct R
from
> Q-Cauchy sequences. And the order in Q can be stated as:
> a/b <= c/d iff a*d <= b*c
> There's no place for counting here, and the density of Q and
R does not
> contradict that <= is a total order.
> Sorry if I'm too lost, please explain me again, if I missed
the point.
That's correct,
I meant :
properties of total order C properties of sequences w.r.t.
countable
sets
regarding total order is a specialisation of the term order
my point is order in maths is a vector field, in English it
means put one
before the other.
Cantors proof has nothing to do with vectors so CANT Order
Reals is an apt
name.
Herc
===
Subject: Rodrigues-like formulas, F,L numbers ?
=
n is a positive integer
a=a(n)= n!/(2n-1)! , b=b(n)= 2*n!/(2n)!
h(x)=sqrt(x^2+4):=(x^2+4)^{0.5}
H(x)=H_n(x)=h^{2n-1}(x)=(x^2+4)^{n-0.5}
f_n(x)=a*H^{(n-1)}(x)/h(x) ,l_n(x)=b*h(x)*H^{(n)}(x) .
By H^{(k)} is denoted the k-th derivative .
It's true that f_n(1) and l_n(1) are integer numbers ?
If possible , please recognize f_n(1) and l_n(1)
as special numbers.
=
===
Subject: Mahler measure of a polynomial
I have been unable to write a satisfactory Maple program to
compute
the Mahler measure of a polynomial.
If anyone has an efficient Maple program for computing this
number and
would be willing to share it with me I most appreciate it.
C.Nicol
===
Subject: Re: Multitude of infinities.
>> Allowing an infinite cardinality would mean that we'd have
aleph
>> infinity. Taking the power set would give us aleph infinity
+ 1. But
>> infinity + 1 and infinity are the same number.
>
> No. If the cardinality of a set were aleph infinity, the
cardinality
> of its power set would be 2^(aleph infinity)which is
strictly greater
> than aleph infinity.
Just to clarify, the subscripts in the sequence of alephs are
the ordinals.
The least upper bound of the aleph_n for n in the natural
numbers
is aleph_omega (omega being the smallest infinite ordinal).
The next
cardinal after aleph_omega is aleph_{omega + 1} (the GCH
asserts that
that is 2^(aleph_omega)). Of course, omega and omega+1
are different ordinals.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colousounds and shapes.
The League of Gentlemen
===
Subject: Finding a basis continuously
Given a N-by-N matrix X of rank M (not necessarily full rank).
Think of it
as an array of N-column vectors {x1, .., xN}. Can you find a
basis for the
space spanned by {x1, .., xN} continuously? That means the
algorithm to
compute the basis should depend continuously in {x1, .., xN}.
Bernard
===
Subject: Re: Finding a basis continuously
>Given a N-by-N matrix X of rank M (not necessarily full
rank). Think of it
>as an array of N-column vectors {x1, .., xN}. Can you find a
basis for the
>space spanned by {x1, .., xN} continuously? That means the
algorithm
to
>compute the basis should depend continuously in {x1, .., xN}.
No. Consider the case N=2, M=1.
Suppose there is a continuous function f from the set of 2x2
matrices of
rank 1 to nonzero vectors such that f(A) is a basis for the
columns of A.
Since x -> x/|x| is continuous on nonzero vectowe can assume
f(A)
is always a unit vector. Let e(theta) be the column vector
[ cos(theta) ]
[ sin(theta) ]
and think of the matrices as pairs of column vectors.
WLOG f(e(0),e(0)) = e(0).
By continuity, we must have f(e(theta),e(theta)) = e(theta).
But then again by continuity, f(t e(0), e(0)) = e(0) for all t.
In particular, f(-e(0),e(0)) = e(0).
And then again by continuity, f(-e(0),t e(0)) = e(0) for all t
so that f(-e(0),-e(0)) = e(0). But -e(0) = e(pi), and we must
have f(e(pi),e(pi)) = e(pi), not e(0), contradiction.
I suspect algebraic topology may have something to say about
the general case (for N > M).
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: Combinatorics in Bridge
> Here's an approach from a different angle:
The task was very ungreatfull, though...
--
Kindly
Konrad
---------------------------------------------------
their souls be chased by demons in Gehenna from one room to
another for all eternity and more.
Sleep - thing used by ineffective people
as a substitute for coffee
Ambition - a poor excuse for not having
enough sence to be lazy
---------------------------------------------------
===
Subject: Re: Mathematical induction and the use of n and k (?)
>Hi all,
>After reading some of the mathematical induction problems,
I tried to
>remember why it is necessary to replace the n with k in the
induction
step.
>Can somebody help me? Is it that k in N is supposed to be a
particular
>element; whereas, n in N is meant as a general element. How
is this
>described in logic?
>>As you are using it, N is a constant, the set of integers.
>>The idea that one should even use mnemonic letters for
>>variables is greatly to be decried.
>I'd be interested in your reasons why. ISTM that mnemonic
variable names
>are common precisely because they *are* helpful.
>> It makes no difference
>>if one uses n or k or q or z or T or alpha for a variable
>>symbol. Sometimes, more than one needs to be used.
>There are well-established coventions for variable names,
presumably as
>memory aids: <delta> and <epsilon> in limit proofs, <theta>
for an angle,
>x and y for reals, z for complex, etc.
>Although formally the variable names make no difference,
well-chosen ones
>certainly do seem to aid readability.
They may or may not aid readability. Unless the person KNOWS
that the particular variable names are unimportant, there is
likely to be the assumption that one needs to use particular
names for particular types of variables. This seems to be
exactly the type of confusion of the original poster; look at
the penultimate statement in the segment quoted.
The general use of variables needs to come first to avoid the
student ever running into this problem. Only then can one use
the simplifications.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Mathematical induction and the use of n and k (?)
>>Although formally the variable names make no difference,
well-chosen ones
>>certainly do seem to aid readability.
>They may or may not aid readability. Unless the person KNOWS
>that the particular variable names are unimportant, there is
>likely to be the assumption that one needs to use particular
>names for particular types of variables. This seems to be
>exactly the type of confusion of the original poster; look at
>the penultimate statement in the segment quoted.
It has always bugged me that in recursion theory there is an
important
theorem called the SNM theorem, where N and M are just names of
indices.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Mathematical induction and the use of n and k (?)
>Although formally the variable names make no difference,
well-chosen
ones
>certainly do seem to aid readability.
>>They may or may not aid readability. Unless the person KNOWS
>>that the particular variable names are unimportant, there is
>>likely to be the assumption that one needs to use particular
>>names for particular types of variables. This seems to be
>>exactly the type of confusion of the original poster; look at
>>the penultimate statement in the segment quoted.
>It has always bugged me that in recursion theory there is an
important
>theorem called the SNM theorem, where N and M are just names
of
indices.
There is also the abc (or is it ABC ?) conjecture in number
theory, where the letters stand for the variables.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Mathematical induction and the use of n and k (?)
>Although formally the variable names make no difference,
well-chosen
ones
>certainly do seem to aid readability.
>>They may or may not aid readability. Unless the person KNOWS
>>that the particular variable names are unimportant, there is
>>likely to be the assumption that one needs to use particular
>>names for particular types of variables. This seems to be
>>exactly the type of confusion of the original poster; look at
>>the penultimate statement in the segment quoted.
>It has always bugged me that in recursion theory there is an
important
>theorem called the SNM theorem, where N and M are just names
of
indices.
And there's alpha-beta pruning of game trees, where alpha and
beta are
just the names of cutoff values.
--
---------------------------
| BBB b barbara minus knox at iname stop com
| B B aa rrr b |
| BBB a a r bbb |
| B B a a r b b |
| BBB aa a r bbb |
-----------------------------
electron-dot-cloud are galaxies
===
Subject: theoretically the strongest concrete
1.6 cement :: 3.2 sand :: 6 gravel
5 cement :: 12.5 sand :: 20 aggregate or gravel
For mortar I get this as a mix ratio:
1 portland cement :: 1 lime :: 4 to 6 parts sand
What I am wondering is like a checkerboard or chess board
where the
squares are each one grain of sand. And that the strongest and
ideal mix
to make both Concrete and to make Mortar is simple this ratio:
1 portland cement to 1 of sand
portland cement between them. Trouble is getting a mix so that
it is
perfectly mixed as to result in a 3-dimensional perfect
packing such as
a chessboard is a perfect 2 dimensional packing.
Has anyone done experimental tests to see at what ratio of
portlandcement to sand yields the strongest most durable
concrete??
I would hazard to guess that if uniformly mixed that the 1 to
1 mix is
the strongest.
Is there any proof to my above assertion?
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: theoretically the strongest concrete
I would imagine that civil engineers have studied topic this
just about to
temperature history of the cement, amount of water available,
etc.
Michael
> 1.6 cement :: 3.2 sand :: 6 gravel
> 5 cement :: 12.5 sand :: 20 aggregate or gravel
> For mortar I get this as a mix ratio:
> 1 portland cement :: 1 lime :: 4 to 6 parts sand
> What I am wondering is like a checkerboard or chess board
where the
> squares are each one grain of sand. And that the strongest
and ideal mix
> to make both Concrete and to make Mortar is simple this
ratio:
> 1 portland cement to 1 of sand
> portland cement between them. Trouble is getting a mix so
that it is
> perfectly mixed as to result in a 3-dimensional perfect
packing such as
> a chessboard is a perfect 2 dimensional packing.
> Has anyone done experimental tests to see at what ratio of
> portlandcement to sand yields the strongest most durable
concrete??
> I would hazard to guess that if uniformly mixed that the 1
to 1 mix is
> the strongest.
> Is there any proof to my above assertion?
> Archimedes Plutonium, a_plutonium@hotmail.com
> whole entire Universe is just one big atom where dots
> of the electron-dot-cloud are galaxies
===
Subject: Re: theoretically the strongest concrete
> I would imagine that civil engineers have studied topic this
just about
to
> temperature history of the cement, amount of water
available, etc.
well, they have, but as any empirical science (sorry, I just
had to
write that) there is still a lot of room for improvement.
Unfortunately i don't know the right english words for the
coming
remarks; i hope it is still understandable.
Often highest strength is not the aim, but early strength to
allow
quick formwork removal, or low porosity (high density) to make
a
barriers for ground water protection etc. Concretes are
optimized to
set under suboptimal circumstances like low/hi temperatures,
short
times in formwork, low viscosity, very long/short pot life.
The posed 1:1 optimal concrete theory has no basis... a sand
grain is
a sand grain is a sand grain...silicon dioxide... a dead dog..
the
cement however is more or less reactive depending on where it
comes
from, it is not a defined substance.. pure CaO has completely
different setting behaviour than fly ash cements. Silica fume,
plasticizers and thousands of other additives make this a real
witchcraft system. So no 1:1 ratio, neither m:m, nor v:v or
mol:mol
have the slightest chance of being a general rule, let alone
specify
some concrete mixture.
The quality of the concrete widely varies even with the same
cement/sand ratio (and using the exact same materials in
different
experiments) with different water/cement ratios.
Depending on water content of the sand, its specific surface,
the
quality of the cement etc., the brickie on the building site
has to
slightly modify the recipe every day to just get the ideal,
same
consistence.
electron-dot-cloud are galaxies
===
Subject: Re: theoretically the strongest concrete
> I would imagine that civil engineers have studied topic this
just about
to
> temperature history of the cement, amount of water
available, etc.
> well, they have, but as any empirical science (sorry, I just
had to
> write that) there is still a lot of room for improvement.
> Unfortunately i don't know the right english words for the
coming
> remarks; i hope it is still understandable.
> Often highest strength is not the aim, but early strength to
allow
> quick formwork removal, or low porosity (high density) to
make a
> barriers for ground water protection etc. Concretes are
optimized to
> set under suboptimal circumstances like low/hi temperatures,
short
> times in formwork, low viscosity, very long/short pot life.
> The posed 1:1 optimal concrete theory has no basis... a sand
grain is
> a sand grain is a sand grain...silicon dioxide... a dead
dog.. the
Well yes, there is a direct link with theory and with
analysis. We have the
Kepler Packing to be applied to concrete and to see if the
Kepler Packing is
a
better concrete than the theory of 1 :: 1 mix of a 3-d
chessboard mix.
I forgotten the 3-d Kepler Packing percent of voids. Was it
somewhere
around
23% voids? Let us say it was. So that a Kepler Packing
Concrete mix would
not be
a 1 :: 1 mix but a 23% :: 77% mix or a 1 :: 3 mix.
> cement however is more or less reactive depending on where
it comes
> from, it is not a defined substance.. pure CaO has completely
> different setting behaviour than fly ash cements. Silica
fume,
> plasticizers and thousands of other additives make this a
real
> witchcraft system. So no 1:1 ratio, neither m:m, nor v:v or
mol:mol
> have the slightest chance of being a general rule, let alone
specify
> some concrete mixture.
> The quality of the concrete widely varies even with the same
> cement/sand ratio (and using the exact same materials in
different
> experiments) with different water/cement ratios.
> Depending on water content of the sand, its specific
surface, the
> quality of the cement etc., the brickie on the building site
has to
> slightly modify the recipe every day to just get the ideal,
same
> consistence.
Experiment: It should be easy to experiment as to whether a
Kepler Packing
concrete mix is superior to a 1 :: 1 chessboard mix. In that
we get marble
sized
balls and Kepler pack them in wet cement. We then pack some
marbles in a
configuration of a chessboard where the marbles and cement are
in a 1 :: 1
ratio. We wait for the two samples to dry and harden. We then
test them for
superiority of one over the other.
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: theoretically the strongest concrete
Once upon a time, Herman Family
<celcaps@frontiernets.net/without_any_s/> said:
>I would imagine that civil engineers have studied topic this
just about to
>temperature history of the cement, amount of water available,
etc.
I can't imagine the answer being independent of the actual
sources of crumb rubber as an asphalt filler, for instance, it
was found that rubber that was ground after cryogenic treatment
resulted in a drastically less strong compound than rubber that
was ground at room temperatures. This was determined to be
because the cryogenically treated crumb had shear surfaces,
like faceted gems, which provided less surface area to grip its
binder than the more complex surfaces of the other crumb.
If different sources of sand produce significantly different
compounds made of them would have different physical
properties.
surface texture varies or not.
Getting back to the original question, I think it would depend
on exactly what kinds of strength was needed from the concrete
compound. Adding different things to the mix (straw, dung,
sawdust, rubber, steel rebar, hardware cloth, etc) can improve
the
strength/density ratio, or the resilience, or the total blunt
trauma energy absorbed per unit mass of the compound, albeit
possibly at the detriment of hardness or absolute flexural,
shear,
or tensile strength. Even if the concrete is only used to
support
a load which is static most of the time (eg, the foundation of
a
building), there may be other factors which complicate things,
like occasional dynamic loading (Californian earthquakes), or
water erosion (rain), etc. There is no single best formula for
everything.
-- TTK
===
Subject: Re: theoretically the strongest concrete
Sharp sand and aggregate makes much stronger concrete than
smooth sand and
aggregate.
Gordon
> Once upon a time, Herman Family
<celcaps@frontiernets.net/without_any_s/> said:
>
>I would imagine that civil engineers have studied topic this
just about
to
salt,
>temperature history of the cement, amount of water available,
etc.
> I can't imagine the answer being independent of the actual
> sources of crumb rubber as an asphalt filler, for instance,
it
> was found that rubber that was ground after cryogenic
treatment
> resulted in a drastically less strong compound than rubber
that
> was ground at room temperatures. This was determined to be
> because the cryogenically treated crumb had shear surfaces,
> like faceted gems, which provided less surface area to grip
its
> binder than the more complex surfaces of the other crumb.
> If different sources of sand produce significantly different
> compounds made of them would have different physical
properties.
> surface texture varies or not.
> Getting back to the original question, I think it would
depend
> on exactly what kinds of strength was needed from the
concrete
> compound. Adding different things to the mix (straw, dung,
> sawdust, rubber, steel rebar, hardware cloth, etc) can
improve the
> strength/density ratio, or the resilience, or the total blunt
> trauma energy absorbed per unit mass of the compound, albeit
> possibly at the detriment of hardness or absolute flexural,
shear,
> or tensile strength. Even if the concrete is only used to
support
> a load which is static most of the time (eg, the foundation
of a
> building), there may be other factors which complicate
things,
> like occasional dynamic loading (Californian earthquakes), or
> water erosion (rain), etc. There is no single best formula
for
> everything.
> -- TTK
===
Subject: Re: theoretically the strongest concrete
NNTP-Proxy-Relay: library1-aux.airnews.net
What you are asking isn't as easy as it sounds. Concrete is
not like a
checkerboard, where all the squares are the same size and
shape.
Concrete is a combination of rock, sand, cement and water, all
of
varying sizes and shapes (or amorphous).
Think of it this way - in woodworking the strongest glued
joint is the
thinnest one. Concrete is sort of the same way. Water
surrounds the
concrete, and you want the least amount of mortar. You could
theoretically calculate all this for spheres, or using computer
simulations, but the National Institute of Standards and
Technology
uses super-computers to do it, and is just now, after about 10
years,
getting to the point where they can do it. Actually, their
program
doesn't optimize the concrete, it just tries to predict what
the
specified concrete will do.
If you are familiar with the subject, you might comment on my
question
week.
Jay Shilstone
>1.6 cement :: 3.2 sand :: 6 gravel
>5 cement :: 12.5 sand :: 20 aggregate or gravel
>For mortar I get this as a mix ratio:
>1 portland cement :: 1 lime :: 4 to 6 parts sand
>What I am wondering is like a checkerboard or chess board
where the
>squares are each one grain of sand. And that the strongest
and ideal mix
>to make both Concrete and to make Mortar is simple this ratio:
>1 portland cement to 1 of sand
>portland cement between them. Trouble is getting a mix so
that it is
>perfectly mixed as to result in a 3-dimensional perfect
packing such as
>a chessboard is a perfect 2 dimensional packing.
>Has anyone done experimental tests to see at what ratio of
>portlandcement to sand yields the strongest most durable
concrete??
>I would hazard to guess that if uniformly mixed that the 1 to
1 mix is
>the strongest.
>Is there any proof to my above assertion?
>Archimedes Plutonium, a_plutonium@hotmail.com
>whole entire Universe is just one big atom where dots
>of the electron-dot-cloud are galaxies
electron-dot-cloud are galaxies
===
Subject: Re: theoretically the strongest concrete
> What you are asking isn't as easy as it sounds. Concrete is
not like a
> checkerboard, where all the squares are the same size and
shape.
> Concrete is a combination of rock, sand, cement and water,
all of
> varying sizes and shapes (or amorphous).
It should be easy to research and find the answer. One easily
can get a
uniform grade of sand and see if a 1 :: 1 mix is stronger than
say a 1:: 2
mix and various
other mixes.
> Think of it this way - in woodworking the strongest glued
joint is the
> thinnest one. Concrete is sort of the same way. Water
surrounds the
> concrete, and you want the least amount of mortar. You could
> theoretically calculate all this for spheres, or using
computer
> simulations, but the National Institute of Standards and
Technology
> uses super-computers to do it, and is just now, after about
10 years,
> getting to the point where they can do it. Actually, their
program
> doesn't optimize the concrete, it just tries to predict what
the
> specified concrete will do.
What is intriguing about the strongest concrete is that it is
sort of like
another Kepler Packing Problem where the spaces between the
balls is filled
with portland cement.
So, in the Kepler Packing Problem in 3-d, each sphere is
surrounded by how
many gaps?
Perhaps the strongest concrete is a direct result of the
Kepler Packing
Problem.
> If you are familiar with the subject, you might comment on
my question
> week.
> Jay Shilstone
I proved the Kepler Packing Problem as that of kissing points.
sizes, this new problem is also solvable once you factor in
the adjustment
of kissing points. For example, say you have 60 balls of which
1/3 are
one
size, and 1/3
another and the last 1/3 another, then once you factor in the
new kissing
points to determine the smallest packing, well, you are on the
way to a
generalized formula
for variable packing sizes.
As for Concrete and the maximum strength, I believe it is
somewhere near a
1 :: 1
mix, where 1 is portland cement and the 1 part sand is of
uniform size and
where the mix is so well mixed.
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: theoretically the strongest concrete
for uniform-ball aggregate, teh cement is the complement
to the volume of the ball, inscribed within a rhombical
dodecahedron;
I think it's got pi and 18 in the ratio, as per Kepler's
problem,
about a fifth left-over. (congratulations on your proof
via kissing; you've beat Hale, at last .-)
I wasn't aware taht sharp sand is better, either. however,
assigning a simple 1/3 ratio of various sizes will not be easy
-- no easier than Kepler's problem.
> What is intriguing about the strongest concrete is that it
is sort of
like
> another Kepler Packing Problem where the spaces between the
balls is
filled
> with portland cement.
>
> So, in the Kepler Packing Problem in 3-d, each sphere is
surrounded by
how
> many gaps?
> I proved the Kepler Packing Problem as that of kissing
points.
>
> sizes, this new problem is also solvable once you factor in
the
adjustment
> of kissing points. For example, say you have 60 balls of
which 1/3 are
one
> size, and 1/3
> another and the last 1/3 another, then once you factor in
the new kissing
> points to determine the smallest packing, well, you are on
the way to a
> generalized formula
> for variable packing sizes.
--Dec.2000 'WAND' Chairman Paul O'Neill, reelected
to Board. Newsish?
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://members.tripod.com/~american_almanac
===
Subject: Re: theoretically the strongest concrete
<snip>
> It should be easy to research and find the answer. One
easily can get a
> uniform grade of sand and see if a 1 :: 1 mix is stronger
than say a 1::
2
> mix and various
> other mixes.
<snip>
As I understand it, the cement is much smaller than the sand,
and
chemically reacts with the water as it solidifies. The actual
material
giving strength to the material is the sand and gravel, the
cement
The use of steel rebar is a common way to strengthen the
concrete
(concrete is an ideal environment for normal steels as long as
salts
aren't present), but fiberglass, rubber, and perhaps other
materials
can also add useful characteristics to the concrete
(respectively,
resistance to crack formation and flexibility).
Water in the original mix is bad. You've probably already
noted that
the recipes state to use as little water as possible to wet the
mixture. There are surfactants that can reduce the amount of
water
required for your concrete. Reducing water in the mix is
another way
to strengthen the concrete since little water pockets in the
concrete
apparently create flaws in the material.
The mix ratios you mention have been tried and tested IMHO and
are
probably really close to being optimal. Incidentally, when I
looked on
the Internet for details on small batch preparation of
concrete, I
found a lot of useful information from instructions for small
house
projects to art-related websites. You should be able to find
good
information on small-scale concrete projects if that's your
inclination.
Karl Hallowell
khallow@hotmail.com
electron-dot-cloud are galaxies
===
Subject: Kepler Packing on concrete mixes Re: theoretically
the strongest
concrete
> <snip>
> It should be easy to research and find the answer. One
easily can get a
> uniform grade of sand and see if a 1 :: 1 mix is stronger
than say a 1::
2
> mix and various
> other mixes.
> <snip>
> As I understand it, the cement is much smaller than the
sand, and
> chemically reacts with the water as it solidifies. The
actual material
> giving strength to the material is the sand and gravel, the
cement
> The use of steel rebar is a common way to strengthen the
concrete
> (concrete is an ideal environment for normal steels as long
as salts
> aren't present), but fiberglass, rubber, and perhaps other
materials
> can also add useful characteristics to the concrete
(respectively,
> resistance to crack formation and flexibility).
Okay, it is hard to get a uniform sand mix.
So let us do a research on something that is perfectly
uniform. That of
steel balls of the size of
marbles and BBs. Where the marbles are gravel and the BBs are
sand.
Now the question is, what mix ratio of Portland cement to make
a cubic foot
of concrete that has marbles
and BBs and is the strongest cubic foot.
I still suspect the final answer to be a cubic foot wherein
only BBs are
used and the ratio is close to
a 1 :: 1 in terms of volume. Say perhaps 10 liters of portland
cement and 10
liters of BBs.
Trouble with various sizes of aggregate is that it allows air
pockets which
diminish strength and cause
cracking.
So I am guessing that the strongest mix ratio is a 1 to 1
ratio using only
BBs and portland cement.
And I am guessing that the perfect piece of concrete is one in
which there
is a Kepler Packing and where
the portland cement is in between the voids of the Kepler
Packing.
I wonder if someone can arrange for a Kepler Packing of steel
BBs and infuse
the crevasses between the
BBs or take marble size steel balls and Kepler Pack them and
then infuse the
interstatials with
portlandcement. I have the hunch that such a block of concrete
would be the
strongest such block over
any other such block having a different arrangement of those
BBs and
marbles. And obviously the ratio is no longer a 1 :: 1 because
if memory
serves me (mathematician would
know) that in a Kepler Packing that the gaps
are something like 23% of the volume and so the ration of
cement to balls
would be closer to that of 1 :: 4
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: Kepler Packing on concrete mixes Re:
theoretically the
strongest concrete
<snip>
> Okay, it is hard to get a uniform sand mix.
>
> So let us do a research on something that is perfectly
uniform. That of
steel balls of the size of
> marbles and BBs. Where the marbles are gravel and the BBs
are sand.
>
> Now the question is, what mix ratio of Portland cement to
make a cubic
foot of concrete that has marbles
> and BBs and is the strongest cubic foot.
>
> I still suspect the final answer to be a cubic foot wherein
only BBs are
used and the ratio is close to
> a 1 :: 1 in terms of volume. Say perhaps 10 liters of
portland cement and
10 liters of BBs.
>
> Trouble with various sizes of aggregate is that it allows
air pockets
which diminish strength and cause
> cracking.
Actually, I'm not sure that this is the case. I can see that
the sand
would fill in the gaps between the larger aggregate, and the
cement in
turn would fill in the gaps between the sand. OTOH, you
probably would
get more air than with sand-sized aggregate alone.
Also, would a marble sized glob of BB's cemented together be
stronger
than a steel marble of the same size? I think not. It
indicates to me
that the large aggregate is probably responsible for most of
the
strength (perhaps in some random almost kissing structure)
with the
spaces filled by cement and sand.
> So I am guessing that the strongest mix ratio is a 1 to 1
ratio using only
BBs and portland cement.
>
> And I am guessing that the perfect piece of concrete is one
in which there
is a Kepler Packing and where
> the portland cement is in between the voids of the Kepler
Packing.
>
> I wonder if someone can arrange for a Kepler Packing of
steel BBs and
infuse the crevasses between the
> BBs or take marble size steel balls and Kepler Pack them and
then infuse
the interstatials with
> portlandcement. I have the hunch that such a block of
concrete would be
the strongest such block over
> any other such block having a different arrangement of those
BBs and
> marbles. And obviously the ratio is no longer a 1 :: 1
because if memory
serves me (mathematician would
> know) that in a Kepler Packing that the gaps
> are something like 23% of the volume and so the ration of
cement to balls
> would be closer to that of 1 :: 4
This seems more reasonable to me. I notice that there appears
to be a
sparseness of the larger stronger stuff (by volume). Ie, the
large
gravel should be 4 over the smaller stuff and cement combined.
Instead
it's slightly better than 1. The sand to cement ratio is a
little over
2-1 with pure sand in cement around 3 to 2.
I see a couple of reasons this theoretical limit isn't
reached. First,
the concrete when poured needs to be sufficiently fluid to be
usable
(and other people have mentioned the various other attributes
that
usable concrete needs depending on the application). That
appears to
be a significant constraint and may be the sole reason for the
low
ratios of large aggregate to the space filling material.
Second, the
material as you note is usually randomly packed. I don't know
how
inefficient that is. One way to improving the packing would be
to
compress the concrete while it's still wet. That would force
the
aggregate into a more optimal packing, and squeeze out air and
perhaps
some of the water. I seem to recall that this is done for some
applications (and the weight of a large amount of concrete
tends to
compress the buried sections).
Karl Hallowell
khallow@hotmail.com
===
Subject: Re: Kepler Packing on concrete mixes Re:
theoretically the
strongest concrete
it may have to do with Kepler's problem, but a)
using BBs will weaken it (too heavy), and b)
aerocrete is actually stronger than that using sand:
the aggregate has no tensional strength,
between any two chunks.
the thing about Portland Cement is that
it's stronger, the longer it takes to dry --
so keep dousing it.
> Trouble with various sizes of aggregate is that it allows
air pockets
which diminish strength and cause
> cracking.
>
> So I am guessing that the strongest mix ratio is a 1 to 1
ratio using only
BBs and portland cement.
>
> And I am guessing that the perfect piece of concrete is one
in which there
is a Kepler Packing and where
> the portland cement is in between the voids of the Kepler
Packing.
>
> I wonder if someone can arrange for a Kepler Packing of
steel BBs and
infuse the crevasses between the
> BBs or take marble size steel balls and Kepler Pack them and
then infuse
the interstatials with
> portlandcement. I have the hunch that such a block of
concrete would be
the strongest such block over
> any other such block having a different arrangement of those
BBs and
> marbles. And obviously the ratio is no longer a 1 :: 1
because if memory
serves me (mathematician would
> know) that in a Kepler Packing that the gaps
> are something like 23% of the volume and so the ration of
cement to balls
> would be closer to that of 1 :: 4
--Dec.2000 'WAND' Chairman Paul O'Neill, reelected
to Board. Newsish?
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://members.tripod.com/~american_almanac
===
Subject: Re: Kepler Packing on concrete mixes Re:
theoretically the
strongest concrete
NNTP-Proxy-Relay: library1-aux.airnews.net
Part of the problem also revolves around the fact that the
more water
that is added to the concrete, the lower the strength. We do
not have
a binary (marbles and B-Bs), or a ternary mix (marbles, B-Bs
and
cement), but a quaternary (4 part) blend of marbles, B-Bs,
cement and
water. Studies have shown that we can reduce water by getting
a good
studies have shown that reducing the maximum aggregate size
will
minimize water demand.
I am a concrete technologist, not a mathematician. If you know
of a
computer program or web model that will predict voids in a
Kepler
Packing system using a variety of sphere sizes, I would like
to know
where.
Jay Shilstone
>> <snip>
>> It should be easy to research and find the answer. One
easily can get
a
>> uniform grade of sand and see if a 1 :: 1 mix is stronger
than say a
1:: 2
>> mix and various
>> other mixes.
>> <snip>
>> As I understand it, the cement is much smaller than the
sand, and
>> chemically reacts with the water as it solidifies. The
actual material
>> giving strength to the material is the sand and gravel, the
cement
>> The use of steel rebar is a common way to strengthen the
concrete
>> (concrete is an ideal environment for normal steels as long
as salts
>> aren't present), but fiberglass, rubber, and perhaps other
materials
>> can also add useful characteristics to the concrete
(respectively,
>> resistance to crack formation and flexibility).
>Okay, it is hard to get a uniform sand mix.
>So let us do a research on something that is perfectly
uniform. That of
steel balls of the size of
>marbles and BBs. Where the marbles are gravel and the BBs are
sand.
>Now the question is, what mix ratio of Portland cement to
make a cubic foot
of concrete that has marbles
>and BBs and is the strongest cubic foot.
>I still suspect the final answer to be a cubic foot wherein
only BBs are
used and the ratio is close to
>a 1 :: 1 in terms of volume. Say perhaps 10 liters of
portland cement and
10 liters of BBs.
>Trouble with various sizes of aggregate is that it allows air
pockets which
diminish strength and cause
>cracking.
>So I am guessing that the strongest mix ratio is a 1 to 1
ratio using only
BBs and portland cement.
>And I am guessing that the perfect piece of concrete is one
in which there
is a Kepler Packing and where
>the portland cement is in between the voids of the Kepler
Packing.
>I wonder if someone can arrange for a Kepler Packing of steel
BBs and
infuse the crevasses between the
>BBs or take marble size steel balls and Kepler Pack them and
then infuse
the interstatials with
>portlandcement. I have the hunch that such a block of
concrete would be the
strongest such block over
>any other such block having a different arrangement of those
BBs and
>marbles. And obviously the ratio is no longer a 1 :: 1
because if memory
serves me (mathematician would
>know) that in a Kepler Packing that the gaps
>are something like 23% of the volume and so the ration of
cement to balls
>would be closer to that of 1 :: 4
>Archimedes Plutonium, a_plutonium@hotmail.com
>whole entire Universe is just one big atom where dots
>of the electron-dot-cloud are galaxies
===
Subject: a question
Let B be a field of characteristic <> k,j in N. Is it true
that for every
p(x), p(x)=kf(x)+jf'(x) for a suitable f(x) in B[x]???
TIA
===
Subject: Re: a question
>
> Let B be a field of characteristic <> k,j in N. Is it true
that
> for every p(x), p(x) = kf(x)+jf'(x) for a suitable f(x) in
B[x]???
Hint: compute the kernel of f -> k f + j f'
-Bill Dubuque
===
Subject: Re: a question
>Let B be a field of characteristic <> k,j in N. Is it true
that for every
>p(x), p(x)=kf(x)+jf'(x) for a suitable f(x) in B[x]???
Somewhat more generally: it's true if p is in B[x], B is a
field, k and j
in B with k <> 0.
Hint: consider the vector space V_n of polynomials of degree
<= n
and the linear map T: V_n -> V_n defined by T(f) = k f + j f'.
What is its matrix using the standard basis [1, x, ..., x^n]?
Show that T is invertible.
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: Jacobi's method : B.S. Jacobi or C.J.G. Jacobi
Originator: israel@math.ubc.ca (Robert Israel)
Jacobi's iterative method
for solving linear systems was
discovered (in 1845 ? ) by
Boris Semionovici JACOBI ( Russian mathematician)
or by German matehmatician
Carl Jacob Gustav JACOBI
Please help with a reference .
===
Subject: Re: Jacobi's method : B.S. Jacobi or C.J.G. Jacobi
Alex.Lupas schrieb in Nachricht
>Jacobi's iterative method
>for solving linear systems was
>discovered (in 1845 ? ) by
> Boris Semionovici JACOBI ( Russian mathematician)
>or by German matehmatician
> Carl Jacob Gustav JACOBI
Carl Gustav Jacob, see below
http://www-history.mcs.st-and.ac.uk/~history/Mathematicians/
Jacobi.html
Boris Semjonowitsch Jakobi, aka Boris Semionovich Yakobi
(1801-1874), was
the
elder brother of Carl Gustav Jacob Jacobi; his name in German
is
Moritz Hermann Jacobi.
In 1835 he got a chair of physics at Dorpat (now Tartu)
University
(Estonia), and in
1837 he moved to the Russian Academy of Sciences in St.
Petersburg.
http://chem.ch.huji.ac.il/~eugeniik/history/jacobi.html (at
the bottom are
a few
links to biographies in Russian).
http://hp.iitp.ru/ger/32/3284.htm
http://hp.iitp.ru/ger/32/3284a.htm
>Please help with a reference .
C. G. Jacobi: .86ber eine neue Aufl.9asungsart der bei der
Methode der
kleinsten
Quadrate vorkommenden linearen Gleichungen.
[On a new way of solving the linear equations arising in the
method of
least
squares].
Astronomische Nachrichten Vol. 22 (1845), Issue No. 523.
It may also be found in Jacobi's Collected Works in 7 Vols.
Berlin: R.
Reimer
1881-1891 in Vol. 3, p. 467-478.
All 7 volumes are online at the Biblioth.8fque Nationale de
France in
Paris:
http://gallica.bnf.fr --> Recherche --> Auteur Jacobi
but currently the server is partially down due to update and
maintainance.
Here are some more references:
http://www.numerik.uni-kiel.de/~in/PAPER/Diplom/node43.html
--> Jacobi45
http://www.cs.umd.edu/TRs/authors/C_G_Jacobi.html
--> CS-TR-2877, but the full paper seems to be no more
available ;-((
Hope that helps ...
===
Subject: big-oh functionality
hi,
I wanted to know if for any two functions, f and g : N -> R*,
R* is a
set of positive real numbers including 0(zero),,, could we
verify that
f (not belongs) O(g) and g (not belongs) O(f).
Please let me know 2 sets of examples at the earliest!
===
Subject: Re: big-oh functionality
> hi,
> I wanted to know if for any two functions, f and g : N ->
R*, R* is a
> set of positive real numbers including 0(zero),,, could we
verify that
> f (not belongs) O(g) and g (not belongs) O(f).
> Please let me know 2 sets of examples at the earliest!
f(n) = exp(2*n),
g(n) = exp(n*(2+(-1)^n))
===
Subject: Michelle Erin Haley - January 15th 1983
H A L E Y
8 1 12 5 25 = 51
In the morning I went to Burger King at 1515 8th Street (phone
373-3177). I bought a coffee from the nubile sweety Emily
Lewans and
she gave me gave me receipt number 377, then her boss Burger
King
managerette Michelle Haley provided stats, Michelle's parents
were
born on days of the year adding to 377.
144 Rickie 1 12 54 335/30 +809
Rickie 55 Craig 38 Haley 51
155 Cindy 11 2 57 42/323 +6
Cindy 55 Jeanne 49 Haley 51
120 Deidre 5 12 80 340/26 8692
Deidre 45 Rae 24 Haley 51
151 Melissa 15 1 83 15/350 9463
Melissa 78 Lee 22 Haley 51
164 Michelle 15 1 83 15/350 9463
Michelle 67 Erin 46 Haley 51
Primes Non-Primes Fibonacci Lucas
2 1 0 1
3 4 1 3
5 6 1 4
7 8 2 7
11 9 3 11
13 10 5 18
17 12 8 29
19 14 13 47
23 15 21 76
29 16 34 123
31 18 55 199
37 20 89 322
41 21 144 521
43 22 233 843
47 <-15th-> 24 <-15th-> 377 <-15th-> 1364
--- --- --- ----
328 200 986 3568
The parents were born on days of the year adding to 377 (the
15th
Fibonacci). The parents were together born 24 (15th non-prime)
days
closer to the end of their years than to the beginning of
their years.
The parents have given names adding together for 197, it's the
45th
(15+15+15th) prime. Dad was born in 54 (the 15th Book of the
New
Testament). Mom's 42nd day of birth adds with her 155 valued
name for
197 (the 15+15+15th prime). Mom and her first kid were born on
days of
the year adding to 382 (Second Chronicles 15). The first of
the kids
has a first name adding to 45 (15+15+15), a middle name adding
to 24
(15th non-prime) and a full name adding to 120 (1 through 15).
The
twins were born on the 15th day of the month and on the 15th
day of
the year. The first of the twins has a name adding to 151, the
twins
have names adding together for 315. The twins have names
adding to 151
and 164, corresponding to Numbers 34 and Deuteronomy 11,
together for
45 (15+15+15). The twins have middle names adding to 22 and
46, the
former is 47.82% of the latter (47 is the 15th prime). The
twins were
born 45 (15+15+15) days after dad's birthday. The first and
last kids
have names adding to 120 and 164, it's a span of 45
(15+15+15). The
kids have 15 vowels in their given names. The kids have given
names
averaging exactly 47 (15th prime). The kids have given names
adding
together for 282 (Second Samuel 15). The family has given
names adding
together for 479 (47 is the 15th prime while chapter 479 is
the first
of the 150 chapters of Psalms). The family was born on days of
the
month adding to 47 (15th prime). The family was born on days
of the
year adding to 747 (the 615th non-prime, prettier as 47 is the
15th
prime). The kids were born on days 5, 15 and 15, together
these Bible
Books contain 1519 verses. Mom and the kids have middle names
adding
together for 141, corresponding to Numbers 24 (15th
non-prime). Mom's
average age when she gave birth amounts to 24.86 years (24 is
the 15th
non-prime while 86 is 15 plus the 15th prime plus the 15th
non-prime).
The twins follow 257+257+257 days after the first of the kids,
there
are 257 verses in Bible Book 47 (15th prime), prettier as Bible
chapter 257 (First Samuel 21) contains 15 verses. The twins
were born
338 days after mom's birthday, corresponding to the 47th and
terminating chapter of The Kings (the 15th prime). Dad was
born on the
1st (Genesis with 1533 verses). Today the parents are an
average of
47.47 years old (47 is the 15th prime). The twins were born on
days of
the year adding together for 30 and have names adding together
for
315, all together for 345 (15x23). Dad and the twins were born
on days
of the month adding to 31 (31 chapters in the Bible contain
the length
of 15 verses, while the 15th chapter to contain the length of
15
verses is chapter 3x3x3x31). Then Michelle goinks and grows
herself up
and becomes a big boss managerette at Burger King, at 1515 8th
Street.
Anyway, note that Bible Book 24 (15th non-prime) is Jeremiah
with 1364
verses (the 15th Lucas). The kids and I have names averaging
155.5.
Primes
2 73 179
3 79 181
5 83 191
7 89 193
11 97 197
13 101 199
17 103 211
19 107 223
23 109 227
29 113 229
31 127 233
37 131 239
41 137 241
43 139 251
47 149 257 <-55th
53 151 263
59 157 269
61 163 271
67 167 277
71 173 281
The parents both have first names adding to 55. Mom's middle
name
adds to 49 while her full name added to the 155 verses of
Bible Book
49. The twins were born in 83 (the number of verses in Bible
Book 55).
The twins follow 771 days after the first of the kids, it is
three
times the 55th prime (257+257+257). The kids were born on days
340 and
15 (together for 355). The kids have vowels in their given
names
adding together for the 83 verses of Bible Book 55, and these
vowels
have an average value of 5.5333. All names in the family add
together
for 734 (Isaiah 55). Note that Book 10 brings The Samuels up
to 1505
verses and 55 chaptepretty as 1 through 10 adds to 55. The 10th
Book of the New Testament contains 155 verses, again pretty as
1
through 10 adds to 55. And note that Bible Book 39 contains 55
verses,
pretty as 55 is the 39th non-prime.
Non-Primes
1 27 50
4 28 51
6 30 52
8 32 54
9 33 55 <-39th
10 34 56
12 35 57
14 36 58
15 38 60
16 39 62
18 40 63
20 42 64
21 44 65
22 45 66
24 46 68
25 48 69
26 49 70
The kids were born on days of the month adding to 35, the kids
have
names adding together for 435. The parents were together born
with 353
days remaining in their years. The kids were together born 356
days
closer to the beginning of their years than to the end of
their years.
Dad was born on day 335 while the twins were born 335 days
closer to
the beginning of their years than to the end of their years.
The first
two kids have names adding together for 271 (the 35th chapter
of The
Samuels). The family was born in years adding to 357. Mom is
listed in
the good phone book, she is at 435 Brightsand Crescent,
keeping in
mind that her kids have names adding together for 435.
Mom was born in 57, the family was born in years adding to 357
(the
number of verses in Daniel). In non-leap yeathe kids together
have
their birthdays 357 days closer to the beginning of their
years than
to the end of their years. Melissa adds to 78 (57th
non-prime). There
are 78 letters in the family (57th non-prime). Dad and the
kids have
names adding together for 579. The parents were together 57681
days
old when the kids were born, while today the family is
together 57852
days old, it's a difference of 171 (57+57+57).
389 <-77th prime
104 <-77th non-prime
77 <-77
---
570
The Four 57's
Genesis 41 -> 41
Leviticus 14 -> 104
Judges 9 -> 220 <-I dreamt of 220 roofs blown
John 11 -> 1008 off homes in the Dakotas
----
1373 <-220th prime
Chapter 57 is Exodus 7 with 25 verses
Book 57 is Philemon with 25 verses
-- --
41st non-prime 16th non-prime
<-together for 57->
Major Books of End-Times Prophecy (Daniel and Revelation are
in part
about 666 while Isaiah contains 66 chapters):
Daniel - 357 verses
Revelation - 404 verses <-57 plus the 57th prime
plus the 57th non-prime
Isaiah - 1292 verses <-an average of 19.575757...
verses per chapter
The parents have names adding together for 299 (First Kings 8
with
66 verses). Mom gave birth a total of 974 days after her
birthdays
(2.666 years). Mom's combined ages when she gave birth amounts
to
75.66 years. The family was born in years adding to the 357
verses of
Daniel, the Book is in part about 666. The twins were born 383
days
after their parent's birthdays, it 66 plus the 66th prime
(317).
Melissa adds to 78 (6 times the 6th prime), her full name adds
to 151
(the 6x6th prime). The family name adds to 51 (the 6x6th
non-prime).
The kids have names adding with their days of birth during the
year
for 805 (666th non-prime).
1-50 - Genesis
51-90 - Exodus
91-117 - Leviticus
118-153 - Numbers
154-187 - Deuteronomy
188-211 - Joshua
930-957 - Matthew
958-973 - Mark
974-997 - Luke
998-1018 - John
1019-1046 - Acts
1047-1062 - Romans
188 <-the opening chapter of Book 6 is 6x6x6 short
of the 404 verses of Bible Book 66, it is the
6th prime squared (13x13) short of the 357
verses of Daniel (also in part about 666)
193 <-Book 6 chapter 6 is the 44th prime,
while 44 is in turn 66.666...% of 66
211 <-the terminating chapter of Book 6 is
approximately 66.6% of the 66th prime (317)
357 <-the opening chapter of Book 6 plus the 6th
prime squared is the 357 verses of Daniel
(in part about 666)
404 <-the 6th prime squared (13x13) plus the 6th
prime squared (13x13) plus 66 adds to the
404 verses of Bible Book 66
1062 <-666 plus 6x66 is a combination of the 658
verses of Bible Book 6 plus the 404 verses
of Bible Book 66, and is the terminating
chapter of New Testament Book 6
1070 <-666 plus the 404 verses of Book 66 is the
1070 verses of Job (Book 6+6+6)
1213 <-Exodus terminates at chapter 90 (66th non-
prime) with 1213 verses (the 198th or the
66+66+66th prime)
1292 <-the 658 verses of Book 6 plus twice the
66th prime (317) is the 1292 verses of
Isaiah (the Book contains 66 chapters)
The parents have given names adding together for the 197
verses of
Bible Book 28 while the family was born in months adding to
28. The
kids have given names adding together for 282.
The parents were born on days of the year adding to 377, the
kids
on days of the year adding to 370 (a difference of 7). The
family has
names adding to 144, 155, 120, 151 and 164, corresponding to
Numbers
27, Deuteronomy 2, Numbers 3, Numbers 34 and Deuteronomy 11,
together
for 77.
I returned a couple of hours later to give Michelle another
sheet
of paper, when I arrived I purchased a couple of hamburglers
and was
provided with receipt number 164, pretty as her name adds to
164. I
showed her gems, she was like everybody else and didn't even
have the
decency to offer to buy me a cookie for my work, she didn't
even have
the decency to send me a cheap letter in the mail expressing
thanx for
showing her evidence that her very name is a gift from God.
And even
if she did throw me a few bucks for my labor, she would turn
around in
December and spend big bucks on a decorated tree, and then she
would
tithe yet more to the church that teaches her to turn
evergreen trees
into decorated idols (you people collectively spent millions of
dollars having me repeatedly arrested and tortured in an
attempt to
make me shut up about your traditions, I begged for years for
assistance to get out of the country and you close your hearts
and
continue to support Babylonian churches). Good luck with your
traditions, your traditions are in opposition to God's
Commandments
and are in opposition the Spirit of Christ's teachings,
without God
you need nothing less than luck to survive. All you people are
good
for is to have your stats posted on the usenet and be used as
examples
to otheand look nubile sweety, here you are!!! If I find any of
these family members in the obituaries I will cheer, it will
be in
accordance to Scripture (Psalm 137:9) and I will post the
stats again.
In lost summer after summer after summer after summer after
summer
after summer after summer to psychiatric torture and you
people are so
cheap and ignorant that you don't even have the compassion to
throw me
a few bucks so I could get out of this country for one single
winter.
Daryl Shawn Kabatoff
Box 7134
Saskatoon Saskatchewan
Canada
S7K 4J1
Isaiah 45:4, Ephesians 3:15 - God gives you your name!!!
===
Subject: Re: Symmetric polynomials (I need help)
Of course that the difference of symmetric polynomials is
symmetric, what
was I
in the
future...
:)
Anderson Brasil
andersbrasil@aol.com
>Any product of symmetric polynomials is symmetric
>(from the definition of symmetric --
>invariant under any permutation of x_1,...,x_n),
>and so is any difference of symmetric polynomials.
>--
>Timothy Murphy
>e-mail: tim@birdsnest.maths.tcd.ie
>tel: +353-86-233 6090
>s-mail: School of Mathematics, Trinity College, Dublin 2,
Ireland
===
Subject: Vulgar but curious:
How would go about solving the following problem (or problems
like it)
?
(don't read farther if you are easily offended)
Chrissie loves to suck cock and jerk off guys! One fine
evening, her
three gorgeous flatmates (Will, Jeff and Keith) show up for
some
action. Since Chrissie has had a rough evening partying and
has a sore
nose and ass, she phones her adequate relative Uncle Happy for
help in
deciding the most efficient way she can perform her task,
since she
just wants to go to bed.
Uncle Happy takes a drag on his pipe, contemplating. He has
just
turned off his copy of Girls Gone Wild- Brooklyn Edition and
now knows
that some girls can service three men at a time using mouth
and hands.
Uncle Happy asks Chrissie to estimate, based on her past
experiences,
how long it would normally take her to please her three
charges with
her mouth, her hand, or a combination thereof.
Chrissie replies (units in minutes):
Will - 20 mouth, 18 hand, 17 hand+mouth
Jeff - 21 mouth, 19 hand, 17 hand+mouth
Keith - 21 mouth, 20 hand, 18 hand+mouth
How should Uncle Happy respond? Assume the following:
- Chrissie may arrange her friends however she chooses.
- Chrissie possesses one mouth and two hands.
- Assume it takes Chrissie thirty seconds to get up to speed
due to
disorientation problems whenever she moves her mouth from one
flatmate
to the other. Ex.: If it normally takes Dan 2 minutes to
finish, but
Chrissie works him for a minute, does something else for a
while and
then returns, the total time Dan will take is 3 minutes (.5
startup, 1
sucking, .5 startup, 1 sucking = 3 minutes).
Bonus: Chrissie needs to know whether or not to program her
VCR to
catch Law & Order. How long will this activity session take?
===
Subject: Re: Vulgar but curious:
>Will - 20 mouth, 18 hand, 17 hand+mouth
>Jeff - 21 mouth, 19 hand, 17 hand+mouth
>Keith - 21 mouth, 20 hand, 18 hand+mouth
Will, mouth, Keith, hand, Jeff, hand.
20 minutes. (Both Will and Jeff are on the critical path, and
finish at the
same time.)
Yours,
Doug Goncz (at aol dot com)
Replikon Research
Read the RIAA Clean Slate Program Affidavit and Description at
http://www.riaa.org/
I will be signing an amended Affidavit soon.
===
Subject: Enumerating all pairs of positive integers method,
does it have a
name?
Googling didn't answer my question, so I suspect there is no
name. How
about
triangular navigation method?
===
Subject: Re: Enumerating all pairs of positive integers
method, does it
have
a name?
> Googling didn't answer my question, so I suspect there is no
name. How
about
> triangular navigation method?
Dovetailing? (that term is specifically used in automata
theory to
describe the simulation of a countable # of Turing machines.
1st step on TM 1
2nd step on TM 1
1st step on TM 2
3rd step on TM 1
2nd TM 2
1st TM 3
etc.
looks like the fanned out tail of a dove. I do not know where
the word
came from (who originally coined it in this usage).
Mitch
===
Subject: Re: Enumerating all pairs of positive integers
method, does it have
a name?
Mikito Harakiri <mikharakiri@ywho.com> scribbled the following:
> Googling didn't answer my question, so I suspect there is no
name. How
about
> triangular navigation method?
AFAIK Georg Cantor used a similar method to enumerate Q. You
could check
his work to see if he named the method.
--
/-- Joona Palaste (palaste@cc.helsinki.fi)
---------------------------
| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+
ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
----------------------------------------- Finland rules!
------------/
A friend of mine is into Voodoo Acupuncture. You don't have to
go into
her
office. You'll just be walking down the street and... ohh,
that's much
better!
- Stephen Wright
===
Subject: Re: How to factor 27X^6-1?
> plz help me factor this cause my math teacher ended up
confusing me when
she
> went over this
>
The rules that are commonly used for factoring binomials are:
a^2 - b^2 = (a - b)(a + b)
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
In your case, 27x^6 - 1 can be viewed as the middle method
because it
can be rewritten in the form (3x^2)^3 - (1)^3 where 3x^2 = a,
1 = b.
replacing a and b in (a - b)(a^2 + ab + b^2) gives you
(3x^2 - 1)((3x^2)^2 + (3x^2)(1) + (1)^2) which is normally
written as
(3x^2 - 1)(9x^4 + 3x^2 + 1)
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: How to factor 27X^6-1?
> plz help me factor this cause my math teacher ended up
confusing me when
she
> went over this
>
> The rules that are commonly used for factoring binomials are:
> a^2 - b^2 = (a - b)(a + b)
> a^3 - b^3 = (a - b)(a^2 + ab + b^2)
> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
> In your case, 27x^6 - 1 can be viewed as the middle method
because it
> can be rewritten in the form (3x^2)^3 - (1)^3 where 3x^2 =
a, 1 = b.
> replacing a and b in (a - b)(a^2 + ab + b^2) gives you
> (3x^2 - 1)((3x^2)^2 + (3x^2)(1) + (1)^2) which is normally
written as
> (3x^2 - 1)(9x^4 + 3x^2 + 1)
This has been asked before, but why is your method better than
(sqrt(27))^2 - 1^2 = (sqrt(27) - 1) (sqrt(27) + 1)?
My factorization has the advantage that you can follow it by
the
other two and continue factoring, if you notice that
sqrt(27) = sqrt(3^3) = [sqrt(3)]^3.
===
Subject: Re: How to factor 27X^6-1?
>
>
>plz help me factor this cause my math teacher ended up
confusing me when
she
>went over this
>>The rules that are commonly used for factoring binomials are:
>>a^2 - b^2 = (a - b)(a + b)
>>a^3 - b^3 = (a - b)(a^2 + ab + b^2)
>>a^3 + b^3 = (a + b)(a^2 - ab + b^2)
>>In your case, 27x^6 - 1 can be viewed as the middle method
because it
>>can be rewritten in the form (3x^2)^3 - (1)^3 where 3x^2 =
a, 1 = b.
>>replacing a and b in (a - b)(a^2 + ab + b^2) gives you
>>(3x^2 - 1)((3x^2)^2 + (3x^2)(1) + (1)^2) which is normally
written as
>>(3x^2 - 1)(9x^4 + 3x^2 + 1)
>
>
> This has been asked before, but why is your method better
than
> (sqrt(27))^2 - 1^2 = (sqrt(27) - 1) (sqrt(27) + 1)?
> My factorization has the advantage that you can follow it by
the
> other two and continue factoring, if you notice that
> sqrt(27) = sqrt(3^3) = [sqrt(3)]^3.
In principle, it isn't. However, factored form is generally
understood
to be looking for factors with integer coefficients. Other
than that,
no special reason.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: How to factor 27X^6-1?
> In principle, it isn't. However, factored form is generally
understood
> to be looking for factors with integer coefficients. Other
than that,
> no special reason.
Proving that I'm no general. Oh well.
===
Subject: Re: What is the one sentence you can believe?
>
>If someone tells you something *specific* and they know
what they are
saying,
>no matter what circumstances are what is it they said if you
>know it is always true?
>
>e.g. circles are round isn't specific!
>
>cryptic hint : old
>
>Herc
First of all, it must refer to their beliefs, as people will
believe
anything, regaldless of the logic involved. (E.g. that Saddam
Hussein
amassed a huge arsenal of weapons of mass destruction, but
when the
war began and it was time to use them, he destroyed them all.)
This is false and you don't believe it.
It cannot be true, as then by the first conjunct it is false.
Thus it
is false. Thus, one or the other of the conjuncts must be
false. The
first one is true, so the 2nd one must be false. Thus you
believe it.
(Whether you believe it or not.)
(Now who can see the falacy of the above?)
Cambridge, MA
PS I also like the answer: If you don't believe me, I will
shoot you
with this gun. (while holding out an authentic looking weapon
- of
small, yet important, destruction.)
===
Subject: Re: What is the one sentence you can believe?
>
>>If someone tells you something *specific* and they know
what they are
saying,
>>no matter what circumstances are what is it they said if
you
>>know it is always true?
>>
>>e.g. circles are round isn't specific!
>>
>>cryptic hint : old
>>
>>Herc
>
>
> First of all, it must refer to their beliefs, as people will
believe
> anything, regaldless of the logic involved. (E.g. that
Saddam Hussein
> amassed a huge arsenal of weapons of mass destruction, but
when the
> war began and it was time to use them, he destroyed them
all.)
>
> This is false and you don't believe it.
>
> It cannot be true, as then by the first conjunct it is
false. Thus it
> is false. Thus, one or the other of the conjuncts must be
false. The
> first one is true, so the 2nd one must be false. Thus you
believe it.
> (Whether you believe it or not.)
>
> (Now who can see the falacy of the above?)
>
>
> Cambridge, MA
>
> PS I also like the answer: If you don't believe me, I will
shoot you
> with this gun. (while holding out an authentic looking
weapon - of
> small, yet important, destruction.)
Just keep in mind this famous limerick:
A theorem both deep and profound
States that Every circle is round.
But in a paper by Erdos
(written in Kurdish)
A counterexample is found!
(not by) Martin Cohen
(Seen on a bulletin board in Cal Tech in the 1960's)
===
Subject: Polynomials by recurrence
Let A,B,C, a, b real numbers with ACa=/=0 .
Consider the sequence (P_n)_{n>=0} of
polynomials defined as p_0(x)=a , p_1(x)=Aax/2+ b
p_{k+1}(x)=(Ax +B)*p_k(x) +C*p_{k-1}(x) , k=1,2,... .
Suppose that T_n and U_{n-1} are Chebychev polynomials
defined for t in (-1,1) by
T_n(t)=cos(n*arccos t) , U_{n-1}(t)= sin( n*arccos
t)/sqrt(1-t^2) .
Question: try to represent p_n(x) as
p_n(x)= C_1*T_n(px+q)+ C_2*U_{n-1}(ux+v) , n>=1 ,
where :
C_1=C_1(n) , C_2=C_2(n) are real numbers ,
p,q,u,v are (complex) numbers which does'nt depend on n .
===
Subject: Re: David Ullrich on Identity
> David Ullrich says:
>
> And yes, identity is in _fact_ reflexive. To
> refute that statement you need to give an
> example of something which is not identical
> to itself. The idea that there is something
> which is _not_ identical to itself is simply
> ludicrous
A variable that returns a random value (common in programming
languages) is often not equal to itself. One that returns the
next
sequential number is never equal to itself. Just output the
value of
x=x where x is that variable.
But more generally, people from the ancient Greek philosophers
who
said things like Two things equal to the same thing are equal
to each
other. to modern day pseudointellectuals miss the real point of
equality.
ANY TWO THINGS ARE EQUAL AT SOME LEVEL OF ABSTRACTION AND
ABOVE AND
UNEQUAL AT ALL LOWER LEVELS OF ABSTRACTION.
Does 1+1 equal 2? At a low, physical level, they are different
strings of characters. (1+1 will not equal 2 in string
maniplulation processes in most programming languages.) But at
a
higher, Mathematical level of abstraction, they are.
Does 1 equal 1? At a simple Mathematical level, they are. But
at a
more exact level, they are not, becaue one of them is the 2nd
word in
the question and the other is the 4th word. Or look at them
under a
microscope and you will detect faint differences in the
displays.
At a very high level of abstraction, they are both things.
apple? But up close you can.
Like everything, equality is context sensitive.
Cambridge, MA
===
Subject: Re: David Ullrich on Identity
chvol@aol.com says...
>> David Ullrich says:
>>
>> And yes, identity is in _fact_ reflexive. To
>> refute that statement you need to give an
>> example of something which is not identical
>> to itself. The idea that there is something
>> which is _not_ identical to itself is simply
>> ludicrous
>A variable that returns a random value (common in programming
>languages) is often not equal to itself. One that returns the
next
>sequential number is never equal to itself. Just output the
value of
>x=x where x is that variable.
>But more generally, people from the ancient Greek
philosophers who
>said things like Two things equal to the same thing are equal
to each
>other. to modern day pseudointellectuals miss the real point
of
>equality.
>ANY TWO THINGS ARE EQUAL AT SOME LEVEL OF ABSTRACTION AND
ABOVE AND
>UNEQUAL AT ALL LOWER LEVELS OF ABSTRACTION.
>Does 1+1 equal 2? At a low, physical level, they are different
>strings of characters. (1+1 will not equal 2 in string
>maniplulation processes in most programming languages.) But
at a
>higher, Mathematical level of abstraction, they are.
>Does 1 equal 1? At a simple Mathematical level, they are. But
at a
>more exact level, they are not, becaue one of them is the 2nd
word in
>the question and the other is the 4th word. Or look at them
under a
>microscope and you will detect faint differences in the
displays.
>At a very high level of abstraction, they are both things.
>apple? But up close you can.
>Like everything, equality is context sensitive.
>
>Cambridge, MA
Yes I agree.
I'm not equal to myself (so says my scale after a big meal).
IMHO the
flaw is to ponder on the identify of the things themselves.
IMHO, we
should rather ponder on the identity of their properties.
Which is the
same thing you're saying: identity seems to be an artifact of
intellectual resolution. Zooming in (and out) of levels of
precision
and abstraction reveals or dissolves differences. Equivalence
is just
as context-sensitive as equality but is not as misleading.
===
Subject: Re: David Ullrich on Identity
>> David Ullrich says:
>>
>> And yes, identity is in _fact_ reflexive. To
>> refute that statement you need to give an
>> example of something which is not identical
>> to itself. The idea that there is something
>> which is _not_ identical to itself is simply
>> ludicrous
>A variable that returns a random value (common in programming
>languages) is often not equal to itself.
The value of the variable is always equal to itself. For a
while
the variable has the value 3. And 3 = 3. A little later the
variable
has the value 17. And 17 = 17.
The value at one time is not equal to the value at another
time.
So what? They're not equal, because they're different things.
>One that returns the next
>sequential number is never equal to itself. Just output the
value of
>x=x where x is that variable.
>But more generally, people from the ancient Greek
philosophers who
>said things like Two things equal to the same thing are equal
to each
>other. to modern day pseudointellectuals miss the real point
of
>equality.
>ANY TWO THINGS ARE EQUAL AT SOME LEVEL OF ABSTRACTION AND
ABOVE AND
>UNEQUAL AT ALL LOWER LEVELS OF ABSTRACTION.
>Does 1+1 equal 2?
Yes.
>At a low, physical level, they are different
>strings of characters.
No, neither 1 + 1 nor 2 is a string of characters.
> (1+1 will not equal 2 in string
>maniplulation processes in most programming languages.)
One hopes not. So what? If 1 and 1 were the same thing you'd
have a point.
>But at a
>higher, Mathematical level of abstraction, they are.
>Does 1 equal 1? At a simple Mathematical level, they are. But
at a
>more exact level, they are not, becaue one of them is the 2nd
word in
>the question and the other is the 4th word. Or look at them
under a
>microscope and you will detect faint differences in the
displays.
>At a very high level of abstraction, they are both things.
>apple? But up close you can.
>Like everything, equality is context sensitive.
Whether two _things_ are equal has nothing to do with
context. Whether the objects denoted by two symbols
are equal does have a lot to do with context. This
says nothing about whether things are equal to
themselves - in a context where two symbols denote
things that are not equal they denote two _different_
things.
Duh.
>
>Cambridge, MA
************************
David C. Ullrich
===
Subject: Re: David Ullrich on Identity
>>Does 1+1 equal 2?
>
> Yes.
>
>>At a low, physical level, they are different
>>strings of characters.
>
> No, neither 1 + 1 nor 2 is a string of characters.
Quite right. Charlie-Poo seems to be mired in crass formalism,
unable to distinguish between an object and its
representations.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colousounds and shapes.
The League of Gentlemen
===
Subject: Re: David Ullrich on Identity
>Does 1+1 equal 2?
>>
>> Yes.
>>
>At a low, physical level, they are different
>strings of characters.
>>
>> No, neither 1 + 1 nor 2 is a string of characters.
>Quite right. Charlie-Poo seems to be mired in crass formalism,
>unable to distinguish between an object and its
>representations.
So it would appear. (null comment inserted just to put
sci.logic back on the list of newsgroups, lest he miss this.)
************************
David C. Ullrich
===
Subject: Re: David Ullrich on Identity
>
>> David Ullrich says:
>>
>> And yes, identity is in _fact_ reflexive. To
>> refute that statement you need to give an
>> example of something which is not identical
>> to itself. The idea that there is something
>> which is _not_ identical to itself is simply
>> ludicrous
>
>A variable that returns a random value (common in programming
>languages) is often not equal to itself.
>
> The value of the variable is always equal to itself. For a
while
> the variable has the value 3. And 3 = 3. A little later the
variable
> has the value 17. And 17 = 17.
Right. Only when you use the word (concept) itself is there
equality, but then you are not referring to two things.
Call this variable $R (the name used in my favorite programming
language.) Then we agree that $R is not equal to $R. But you
maintain that 3 is equal to 3. However, I pointed out the fact
that 3
is not actually equal to 3 because the former is the 8th word
in this
sentence and the latter is the 14th word in this sentence. So
at some
level they are still not equal. Now if we say that 3 is equal
to
itself, we are not referring to two things. We are referring
to one
thing and then referring to itself.
Whenever there is a distinction, and there are in fact two
things,
then whether they are equal or not depends on the level of
abstraction
at which we are dealing, whether it makes that same
distinction.
Since abstracting is the removal of distinctions, at some
level, and
above, they will be equal.
In other words, it is not a question of whether two things are
equal
or not. Rather, it is a question of, at what level of
abstraction,
and above, they are equal? f(one thing,something)=the lowest
level of
abstraction at which they are equal.
> They're not equal, because they're different things.
What's the difference between not equal and different things?
That is, when are these not the same?
>Does 1+1 equal 2?
>
> Yes.
>
>At a low, physical level, they are different strings of
characters.
>
> No, neither 1 + 1 nor 2 is a string of characters.
How else can we communicate other than by an exchange of
strings of
characters (i.e. elements of an agreed upon recursively
enumerable
set)?
> Whether two _things_ are equal has nothing to do with
> context. Whether the objects denoted by two symbols
> are equal does have a lot to do with context.
Are not objects things?
Cambridge, MA
> ************************
>
> David C. Ullrich
===
Subject: Re: David Ullrich on Identity
>>
> David Ullrich says:
>
> And yes, identity is in _fact_ reflexive. To
> refute that statement you need to give an
> example of something which is not identical
> to itself. The idea that there is something
> which is _not_ identical to itself is simply
> ludicrous
>>
>>A variable that returns a random value (common in programming
>>languages) is often not equal to itself.
>>
>> The value of the variable is always equal to itself. For a
while
>> the variable has the value 3. And 3 = 3. A little later the
variable
>> has the value 17. And 17 = 17.
>Right. Only when you use the word (concept) itself is there
>equality, but then you are not referring to two things.
Uh, that's correct. That's why the word itself appears
prominently
in the statement that you seemed to think you were refuting.
>Call this variable $R (the name used in my favorite
programming
>language.) Then we agree that $R is not equal to $R.
No, we don't agree to that.
>But you
>maintain that 3 is equal to 3.
You don't realize how funny this sounds, saying that I maintain
that 3 equals 3?
>However, I pointed out the fact that 3
>is not actually equal to 3 because the former is the 8th word
in this
>sentence and the latter is the 14th word in this sentence.
Yes, you pointed this out. That doesn't make it true. In fact
3 is not a word.
>So at some
>level they are still not equal. Now if we say that 3 is equal
to
>itself, we are not referring to two things. We are referring
to one
>thing and then referring to itself.
>Whenever there is a distinction, and there are in fact two
things,
>then whether they are equal or not depends on the level of
abstraction
>at which we are dealing, whether it makes that same
distinction.
>Since abstracting is the removal of distinctions, at some
level, and
>above, they will be equal.
>In other words, it is not a question of whether two things
are equal
>or not. Rather, it is a question of, at what level of
abstraction,
>and above, they are equal? f(one thing,something)=the lowest
level of
>abstraction at which they are equal.
>> They're not equal, because they're different things.
>What's the difference between not equal and different things?
There is none. That's what equal means.
>That is, when are these not the same?
>>Does 1+1 equal 2?
>>
>> Yes.
>>
>>At a low, physical level, they are different strings of
characters.
>>
>> No, neither 1 + 1 nor 2 is a string of characters.
>How else can we communicate other than by an exchange of
strings of
>characters (i.e. elements of an agreed upon recursively
enumerable
>set)?
We can't. The fact that we need character strings to talk about
things does not imply that things _are_ character strings.
>> Whether two _things_ are equal has nothing to do with
>> context. Whether the objects denoted by two symbols
>> are equal does have a lot to do with context.
>Are not objects things?
Yes...
>
>Cambridge, MA
>
>> ************************
>>
>> David C. Ullrich
************************
David C. Ullrich
===
Subject: Re: David Ullrich on Identity
>
>
> Homework for David Ullrich:
>
> 1) What philosopher said:
>
> ...definitions are available only for transforming
> truths, not for founding them.
>
>
> I would be interested in this answer...
_Ways of Paradox_, p. 81
> So, your dispute with the boyz in the hood has to do with
formal set
> theory--class models of that theory, in fact.>
> mitch
> How can *theorems* be *false*? Well, the point here is that
> when one says that a statement about the natural numbers is
> false one means that it is false of the *standard* integers,
> i.e., the integers that we normally work with.
Likewise, when David Ullrich says that
statements about individuals such as ExAy~(x=y)
and Ex~(x=x) are false, he has said nothing more
than that these are false of the *standard*
individuals that FOL= deals with. Put back in
the domain those that FOL= excludes, and it is
not ExAy~(x=y) and Ex~(x=x) but AxEy(x=y) and
Ax(x=x) that are false.
It doesn't take a rocket scientist to figure
that one out...
--John
===
Subject: Re: David Ullrich on Identity
> You should check out http://megafoundation.org/
> Harris has finally found people who understand him there.
This from a purveyor of store-bought, on-the-shelf knowledge,
whose absolute inability to reason from premises other than
the customary, bought-and-paid-for ones is a common trait
of the unabashedly uptaught.
>Exhibit of proof of Ex~(x=x) from
>C1-C4 and someone will point out the error.
>
>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1
>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
>
>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
>Classification
>
>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}]
> Weak
Here's the proof you couldn't hack.
1) Ax(x in y <-> Et(x in t & ~(x in x)) Instance of (C1)
2) y in y <-> Et(y in t & ~(y in y)) 1, US
3) ~Et(y in t) 2
4) AyAx[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] C4
5) Az(z in y <-> z in y) -> (Et(y in t) <-> y=y) 4,US
6) Et(y in t) <-> y=y 5
7) ~(y=y) 3,6
8) Ex~(x=x) 7,EG
**************************************************************
********
Homework for David Ullrich
What bearing would the non-self-identity of quanta
have on the Leibnizean principle of the Identity of
Indiscernibles?
--John
And yes, identity is in _fact_ reflexive. To
refute that statement you need to give an
example of something which is not identical
to itself. The idea that there is something
which is _not_ identical to itself is simply
ludicrous: That's what identity _means_: A
thing is identical to itself and to nothing
else.
--David Ullrich
===
Subject: Re: David Ullrich on Identity
>> You should check out http://megafoundation.org/
>> Harris has finally found people who understand him there.
>This from a purveyor of store-bought, on-the-shelf knowledge,
>whose absolute inability to reason from premises other than
>the customary, bought-and-paid-for ones is a common trait
>of the unabashedly uptaught.
You should _really_ check out http://megafoundation.org/ !
This sort of anyone-who-actually-knows-something-must-be
-too-stupid-to-think-for-himself-the-way-we-geniuses-do is
their motto.
>>Exhibit of proof of Ex~(x=x) from
>>C1-C4 and someone will point out the error.
Just for the record, when I said that you were claiming that
C1-C4 were part of NBG. Later turned out that that was just
something you made up.
>>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1
>>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
>>
>>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
>>Classification
>>
>>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}]
>> Weak
>Here's the proof you couldn't hack.
>1) Ax(x in y <-> Et(x in t & ~(x in x)) Instance of (C1)
>2) y in y <-> Et(y in t & ~(y in y)) 1, US
>3) ~Et(y in t) 2
>4) AyAx[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] C4
>5) Az(z in y <-> z in y) -> (Et(y in t) <-> y=y) 4,US
>6) Et(y in t) <-> y=y 5
>7) ~(y=y) 3,6
>8) Ex~(x=x) 7,EG
>*************************************************************
*********
> Homework for David Ullrich
>What bearing would the non-self-identity of quanta
>have on the Leibnizean principle of the Identity of
>Indiscernibles?
>--John
>And yes, identity is in _fact_ reflexive. To
>refute that statement you need to give an
>example of something which is not identical
>to itself. The idea that there is something
>which is _not_ identical to itself is simply
>ludicrous: That's what identity _means_: A
>thing is identical to itself and to nothing
>else.
>--David Ullrich
************************
David C. Ullrich
===
Subject: Re: David Ullrich on Identity
>
>
>> You should check out http://megafoundation.org/
>> Harris has finally found people who understand him there.
>
>This from a purveyor of store-bought, on-the-shelf knowledge,
>whose absolute inability to reason from premises other than
>the customary, bought-and-paid-for ones is a common trait
>of the unabashedly uptaught.
>
> You should _really_ check out http://megafoundation.org/ !
> This sort of anyone-who-actually-knows-something-must-be
> -too-stupid-to-think-for-himself-the-way-we-geniuses-do is
> their motto.
Against stupidity the very gods
Themselves contend in vain.
--Schiller, _The Maid of Orleans_
> Homework for David Ullrich
>
>What bearing would the non-self-identity of quanta
>have on the Leibnizean principle of the Identity of
>Indiscernibles?
===
Subject: Re: David Ullrich on Identity
>
>
>>Exhibit of proof of Ex~(x=x) from
>>C1-C4 and someone will point out the error.
>
>>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1
>>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
>
>>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
>>Classification
>
>>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}]
> Weak
>
>Here's the proof you couldn't hack.
>
>2) y in y <-> Et(y in t & ~(y in y)) 1, US
>3) ~Et(y in t) 2
>4) AyAx[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] C4
>5) Az(z in y <-> z in y) -> (Et(y in t) <-> y=y) 4,US
>6) Et(y in t) <-> y=y 5
>7) ~(y=y) 3,6
>
>*************************************************************
*********
> Just for the record, when I said that you were claiming that
> C1-C4 were part of NBG. Later turned out that that was just
> something you made up.
Till their own dreams at length deceive 'em,
And oft repeating, they believe 'em.
--Matthew Prior
Alma [1718], canto III, l. 13
> Homework for David Ullrich
>
> What bearing would the non-self-identity of quanta
> have on the Leibnizean principle of the Identity of
> Indiscernibles?
>
>--John
>
> David C. Ullrich
Have you forgotten your homework, AGAIN???
===
Subject: Re: David Ullrich on Identity
linux)
>> You should check out http://megafoundation.org/
>> Harris has finally found people who understand him there.
> What part of Ax[Qu(x) -> ~(x = x)] do YOU fail to understand?
that there does not exist an x such that Qu(x)? Is this a
conclusion
with which Teller would be happy?
In any case, if some quantum theorist somewhere wants to mess
about
with a thoroughly non-standard idea of identity, what the heck
should
Ullrich care? And how does Teller's speculations support your
claim
that all and only existent things are self-identical?
(I will probably regret treating you like a human being here.
Again.)
--
Jesse Hughes
Of course, my ability to admit my mistakes and correct them is
a
trait that many of you seem to never have properly appreciated.
-- JSH, discussing his 1463rd proof of Fermat's Last Theorem.
===
Subject: Re: David Ullrich on Identity
>
>> You should check out http://megafoundation.org/
>> Harris has finally found people who understand him there.
>
> What part of Ax[Qu(x) -> ~(x = x)] do YOU fail to understand?
>
> that there does not exist an x such that Qu(x)? Is this a
conclusion
> with which Teller would be happy?
'Quanta' are one thing, non-existents are another, and
in common is that they are identical-with-nothings. My (sketchy
and non-mathematically well-informed impression) is that
fruitful
applications of 'imaginary numbers' have been found in
different
areas of mathematics. It would not be surprising if the logic
of
identical-with-nothings were to turn out to have to have
different
applications as well.
file if you wish), what Teller would not be happy with is the
notion
that (e.g.) AxAy(qu(x) & qu(y) -> ~(x=y)) and Ax(qu(x) ->
~(x=x)
are 'well-formed'. His argument appears to be that since quanta
lack 'haecceities', it makes sense neither to assert that x=y
when x,y are quanta, or that ~(x=y).
> In any case, if some quantum theorist somewhere wants to
mess about
> with a thoroughly non-standard idea of identity, what the
heck should
> Ullrich care? And how does Teller's speculations support
your claim
> that all and only existent things are self-identical?
The main thing at stake is whether, in the logic that
different theories run under, identity should be taken to be
reflexive or non-reflexive. If the latter, then quantum
theorists can battle it out (on empirical grounds) over whether
quanta satisfy Ax(x=x). If the former, such (potentially
fruitful) disputes are--illegitimately, I think--ruled out
of court by the background logic.
>
> (I will probably regret treating you like a human being here.
> Again.)
--John
===
Subject: power set cardinality?

===
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
> listed below would be generated.
>
> 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 quoted that 2P was (0,22). Now you quote it's
(6,19).
> What did this paper actually say?
Robin,
I think you clarified this a bit for me when you said This
just looks
like
a list of non-identity points on the curve in ascending order
of
The original list was just that, it never stated 2P = , 3 P =
... .
The second list does assign specific values (my confusion,
giving
a note that makes sense to me). Anyway, here it the list of
points in it's
entirety (I assume this is over Z{23}).
P = (0, 1), 2P = (6, 19), 3P = (3, 13), 4P = (13, 16), 5P =
(18, 3), 6P =
(7, 11)
7P = (11, 3), 8P = (5, 19), 9P = (19, 18), 10P = (12, 4), 11P
= (1, 16),
12P
= (17, 20)
13P = (9, 16), 14P = (4, 0), 15P = (9, 7), 16P = (17, 3), 17P
= (1, 7), 18P
= (12, 19)
19P = (19, 5), 20P = (5, 4), 21P = (11, 20), 22P = (7, 12),
23P = (18, 20),
24P = (13, 7)
25P = (3, 10), 26P = (6, 4), 27P = (0, 22), 28 P = (0, 1) = P
... so the
order is 28.
My problem now is how they are calculating these:
E: y^2 = x^3 + x + 1 (mod 23), points starting from P = (0, 1).
When I try the chord method equations, I seem to be doing
something wrong.
I gave the formula in an earlier post. Can you help show me
what I am
doing
wrong here?
I get fractions when I use those formulas, but we are working
mod 23,
so
is there some conversion that is being done to get these
points?
> If 28P = P then the order of P divides 27 --- impossible if
> it also divides 28 (if that is the order of the group).
> --
> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
> His mind has been corrupted by colousounds and shapes.
> The League of Gentlemen
===
Subject: Re: Elliptic Curve over Z{p}
<snip>
>The second list does assign specific values (my confusion,
giving
>a note that makes sense to me). Anyway, here it the list of
points in
it's
>entirety (I assume this is over Z{23}).
>P = (0, 1), 2P = (6, 19), 3P = (3, 13), 4P = (13, 16), 5P =
(18, 3), 6P =
>(7, 11)
>7P = (11, 3), 8P = (5, 19), 9P = (19, 18), 10P = (12, 4), 11P
= (1, 16),
12P
>= (17, 20)
>13P = (9, 16), 14P = (4, 0), 15P = (9, 7), 16P = (17, 3), 17P
= (1, 7),
18P
>= (12, 19)
>19P = (19, 5), 20P = (5, 4), 21P = (11, 20), 22P = (7, 12),
23P = (18,
20),
>24P = (13, 7)
>25P = (3, 10), 26P = (6, 4), 27P = (0, 22), 28 P = (0, 1) = P
... so the
>order is 28.
If 27P = (0, 22), then 28P is the point at infinity, and 29P =
(0, 1) = P,
so that 28 is the order.
>My problem now is how they are calculating these:
>E: y^2 = x^3 + x + 1 (mod 23), points starting from P = (0,
1).
>When I try the chord method equations, I seem to be doing
something wrong.
>I gave the formula in an earlier post. Can you help show me
what I am
doing
>wrong here?
>I get fractions when I use those formulas, but we are working
mod 23,
so
>is there some conversion that is being done to get these
points?
Use Euclid's Algorithm.
David McAnally
--------------
===
Subject: Re: Elliptic Curve over Z{p}
>
> I get fractions when I use those formulas, but we are
working mod 23,
so
> is there some conversion that is being done to get these
points?
We have 4 x 6 = 24 = 1 (mod 23), so in Z_23,
4 x 6 = 1. So in Z_23, 1/4 = 6 n'est-ce pas? And 1/6 = 4.
SO what about 1/2, 1/3, 1/5, 1/7 etc. etc. etc.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colousounds and shapes.
The League of Gentlemen
===
Subject: Re: Elliptic Curve over Z{p}
>
> I get fractions when I use those formulas, but we are
working mod
23,
so
> is there some conversion that is being done to get these
points?
> We have 4 x 6 = 24 = 1 (mod 23), so in Z_23,
> 4 x 6 = 1. So in Z_23, 1/4 = 6 n'est-ce pas? And 1/6 = 4.
> SO what about 1/2, 1/3, 1/5, 1/7 etc. etc. etc.
** Just modular inverses ... I should have tried that, but the
earlier list
we discussed threw me for a loop as I wasn't even in the ball
park!
> --
> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
> His mind has been corrupted by colousounds and shapes.
> The League of Gentlemen
===
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)
> 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?
Over what?
> That is, each of these points does satisfy:
>
> y^2 == x^3 + x + 1 (mod 23) (here, a = 1 and b = 1)
Well of course: addition of points on an elliptic curve gives
rise to points on elliptic curves!
> 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).
This seems to be the chord tangent process. Does this
not give the above values.
> 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).
Are you saying that 28P is zero in the group?
It isn't: -P is obtained by reflecting P in the x-axis: it is
(0,-1) = (0,22) so 27 P =/= -P i.e., 28P =/= O.
The order of the group is the number of elements in the group:
1 + # of solutions of E defined over F_{23} (the 1 is for the
point at infinity). You seem to be computing the order
of P within the group rather than the order of the group
itself.
The former is a factor of the latter. By the Hasse-Weil bound,
the group's order lies between 15 and 33 inclusive.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colousounds and shapes.
The League of Gentlemen
===
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)
> 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?
> Over what?
> That is, each of these points does satisfy:
>
> y^2 == x^3 + x + 1 (mod 23) (here, a = 1 and b = 1)
> Well of course: addition of points on an elliptic curve gives
> rise to points on elliptic curves!
> 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).
> This seems to be the chord tangent process. Does this
> not give the above values.
> 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).
> Are you saying that 28P is zero in the group?
** No, I am saying that I was not able to calculate the points
as I have
given them above, so am unable to calculate the order.
This is an example I found on the Certicom web site somewhere.
Can you tell me how you calculate these points? (The chord
method does not
appear to produce these points?.
> It isn't: -P is obtained by reflecting P in the x-axis: it is
> (0,-1) = (0,22) so 27 P =/= -P i.e., 28P =/= O.
> The order of the group is the number of elements in the
group:
> 1 + # of solutions of E defined over F_{23} (the 1 is for the
> point at infinity). You seem to be computing the order
> of P within the group rather than the order of the group
itself.
> The former is a factor of the latter. By the Hasse-Weil
bound,
> the group's order lies between 15 and 33 inclusive.
> --
> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
> His mind has been corrupted by colousounds and shapes.
> The League of Gentlemen
===
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)
>> 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?
>> Over what?
>> That is, each of these points does satisfy:
>>
>> y^2 == x^3 + x + 1 (mod 23) (here, a = 1 and b = 1)
>> Well of course: addition of points on an elliptic curve
gives
>> rise to points on elliptic curves!
>> 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).
>> This seems to be the chord tangent process. Does this
>> not give the above values.
>> 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).
>> Are you saying that 28P is zero in the group?
>
> ** No, I am saying that I was not able to calculate the
points as I have
> given them above, so am unable to calculate the order.
>
> This is an example I found on the Certicom web site
somewhere.
>
> Can you tell me how you calculate these points? (The chord
method does
> not appear to produce these points?.
Are you saying that the point they label 2P isn't P + P,
that 3P isn't P + P + P? Then I have no idea what they're
doing.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colousounds and shapes.
The League of Gentlemen
===
Subject: Closed form ,Lucas poly., Chebychev poly.
T_0,T_1,...,T_n,... are Chebychev polynomials
of first kind, that is T_0(x)=1 , T_1(x)=x and
T_{n+1}(x)=2x*T_n(x)-T_{n-1}(x) for n=1,2,... .
l_0(x),l_1(x),...,l_n(x),... are Lucas polynomials
defined as l_0(x)=2 , l_1(x)=x , and for n >= 1
l_{n+1}(x)=x*l_n(x)+l_{n-1}(x) .
It is supposed that a polynomial f(x) is given , and
moreover
f(x) = SUM_{k=0 to k=n} c(k)*T_k(x) .
I am interested to find a closed form for the sum
(Tf)(x,y):= SUM_{k=0 to k=n} c(k)*l_k(x)*l_k(y) .
==
===
Subject: Re: Closed form ,Lucas poly., Chebychev poly.
>T_0,T_1,...,T_n,... are Chebychev polynomials
>of first kind, that is T_0(x)=1 , T_1(x)=x and
> T_{n+1}(x)=2x*T_n(x)-T_{n-1}(x) for n=1,2,... .
>l_0(x),l_1(x),...,l_n(x),... are Lucas polynomials
>defined as l_0(x)=2 , l_1(x)=x , and for n >= 1
>l_{n+1}(x)=x*l_n(x)+l_{n-1}(x) .
>It is supposed that a polynomial f(x) is given , and
>moreover
> f(x) = SUM_{k=0 to k=n} c(k)*T_k(x) .
>I am interested to find a closed form for the sum
>(Tf)(x,y):= SUM_{k=0 to k=n} c(k)*l_k(x)*l_k(y) .
>==
It's rather amazing that such a closed form exists.
Note that T_k(cos(x)) = cos(k x).
So f(cos x) = sum_{k=0}^n c(k) cos(k x)
= sum_{k=-n}^n b(k) exp(i k x)
where b(0) = c(0) and b(k) = b(-k) = c(k)/2 for k > 0.
And thus f((v+1/v)/2) = sum_{k=-n}^n b(k) v^k
Now l_k(x) = A(x)^k + B(x)^k
where A(x) = (x+sqrt(x^2+4))/2 and
B(x) = (x - sqrt(x^2+4))/2 = -1/A(x).
So sum_{k=0}^n c(k) l_k(x) l_k(y)
= sum_{k=0}^n c(k) (A(x)^k + B(x)^k)(A(y)^k + B(y)^k)
where
sum_{k=0}^n c(k) ((A(x)A(y))^k + (B(x)B(y))^k)
= sum_{k=-n}^n b(k) (A(x)A(y))^k
= 2 f((A(x)A(y)+B(x)B(y))/2)
and similarly
sum_{k=0}^n c(k) ((A(x)B(y))^k + (B(x)A(y))^k)
= sum_{k=-n}^n b(k) (A(x)B(y))^k
= 2 f((A(x)B(y)+B(x)A(y))/2)
So Tf(x,y) = 2 f((A(x)A(y)+B(x)B(y))/2) + 2
f((A(x)B(y)+B(x)A(y))/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
===
Subject: Re: Closed form ,Lucas poly., Chebychev poly.
>T_0,T_1,...,T_n,... are Chebychev polynomials
>of first kind, that is T_0(x)=1 , T_1(x)=x and
> T_{n+1}(x)=2x*T_n(x)-T_{n-1}(x) for n=1,2,... .
>l_0(x),l_1(x),...,l_n(x),... are Lucas polynomials
>defined as l_0(x)=2 , l_1(x)=x , and for n >= 1
>l_{n+1}(x)=x*l_n(x)+l_{n-1}(x) .
>It is supposed that a polynomial f(x) is given , and
>moreover
> f(x) = SUM_{k=0 to k=n} c(k)*T_k(x) .
>I am interested to find a closed form for the sum
>(Tf)(x,y):= SUM_{k=0 to k=n} c(k)*l_k(x)*l_k(y) .
>==
>
> It's rather amazing that such a closed form exists.
After a bit of sleep, a little less amazing.
> So Tf(x,y) = 2 f((A(x)A(y)+B(x)B(y))/2) + 2
f((A(x)B(y)+B(x)A(y))/2)
Which just says
l_k(x) l_k(y) = 2 T_k((A(x)A(y)+B(x)B(y))/2) + 2
T_k((A(x)B(y)+B(x)A(y))/2)
Not too surprising that such a relation exists, given that l_k
and T_k are
related by T_k(t) = i^k/2 l_k(-2 i 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: Re: Closed form ,Lucas poly., Chebychev poly.
>T_0,T_1,...,T_n,... are Chebychev polynomials
>of first kind, that is T_0(x)=1 , T_1(x)=x and
> T_{n+1}(x)=2x*T_n(x)-T_{n-1}(x) for n=1,2,... .
>l_0(x),l_1(x),...,l_n(x),... are Lucas polynomials
>defined as l_0(x)=2 , l_1(x)=x , and for n >= 1
>l_{n+1}(x)=x*l_n(x)+l_{n-1}(x) .
>It is supposed that a polynomial f(x) is given , and
>moreover
> f(x) = SUM_{k=0 to k=n} c(k)*T_k(x) .
>I am interested to find a closed form for the sum
>(Tf)(x,y):= SUM_{k=0 to k=n} c(k)*l_k(x)*l_k(y) .
>==
>
> It's rather amazing that such a closed form exists.
>
> Note that T_k(cos(x)) = cos(k x).
> So f(cos x) = sum_{k=0}^n c(k) cos(k x)
> = sum_{k=-n}^n b(k) exp(i k x)
> where b(0) = c(0) and b(k) = b(-k) = c(k)/2 for k > 0.
> And thus f((v+1/v)/2) = sum_{k=-n}^n b(k) v^k
>
> Now l_k(x) = A(x)^k + B(x)^k
> where A(x) = (x+sqrt(x^2+4))/2 and
> B(x) = (x - sqrt(x^2+4))/2 = -1/A(x).
>
> So sum_{k=0}^n c(k) l_k(x) l_k(y)
> = sum_{k=0}^n c(k) (A(x)^k + B(x)^k)(A(y)^k + B(y)^k)
>
> where
> sum_{k=0}^n c(k) ((A(x)A(y))^k + (B(x)B(y))^k)
> = sum_{k=-n}^n b(k) (A(x)A(y))^k
> = 2 f((A(x)A(y)+B(x)B(y))/2)
>
> and similarly
> sum_{k=0}^n c(k) ((A(x)B(y))^k + (B(x)A(y))^k)
> = sum_{k=-n}^n b(k) (A(x)B(y))^k
> = 2 f((A(x)B(y)+B(x)A(y))/2)
>
> So Tf(x,y) = 2 f((A(x)A(y)+B(x)B(y))/2) + 2
f((A(x)B(y)+B(x)A(y))/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
===
Subject: to find an answer
the 7th degree polynomial
a0 + a1x + a2x^2 + a3x^3 + a4x^4 + a5x^5 + a6x^6 + a7x^7
can alternatively be expressed,
0! d^7 0!
a0x^0 = ---- a0 ----- x^7 = ---- a0 y'''''''
7! dx^7 7!
1! d^6 1!
a1x^1 = ---- a1 ----- x^7 = ---- a1 y''''''
7! dx^6 7!
2! d^5 2!
a2x^2 = ---- a2 ----- x^7 = ---- a2 y'''''
7! dx^5 7!
3! d^4 3!
a3x^3 = ---- a3 ----- x^7 = ---- a3 y''''
7! dx^4 7!
4! d^3 4!
a4x^4 = ---- a4 ----- x^7 = ---- a4 y'''
7! dx^3 7!
5! d^2 5!
a5x^5 = ---- a5 ----- x^7 = ---- a5 y''
7! dx^2 7!
6! d^1 6!
a6x^6 = ---- a6 ----- x^7 = ---- a6 y'
7! dx^1 7!
7! d^0 7!
a7x^7 = ---- a7 ----- x^7 = ---- a7 y
7! dx^0 7!
--------------------------------------------
0
One solution is y=x^7 But are there others?
For positive roots, let y = e^x - 1 and solve for x. Then the
root is
y^(1/7)
For negative roots, let y = 1-e^x and solve for x. Then the
root is
y^(1/7)
For more roots apply DeMoivre's Theorem
y=e^x - 1
y' = y'' = y''' = y'''' = y''''' = y'''''' = y''''''' = e^x
a7
{
ln{-----------------------------------------------------------
------}}^(1/
7)
a0/7!+a1/7!+a2(2/7!)+a3(3!/7!)+a4(4!/7!)+a5(5!/7!)+a6/7+a7
However it is doubtful that this is a solution, since the
differential
applies
only to y=x^7
===
Subject: Re: Number theory problem
>
> How do you solve the problem:
>
> What is the greatest integer n such that
> n divides p^4-1 for all primes p > 5?
>
> The choices are 12, 30, 48, 120, 240
> I don't know what the answer is, nor how to solve it, but
I'm most
> interested in knowing the basic method to solve this
problem. It seems
like
> Fermat's Little Theorem might have something to say about
this problem but
I
> can't figure out how to apply it in this case. (By the way,
this comes
from
> an old GRE Subject Test, I'm not trying to cheat on my
homework or
anything.)
This was discussed here on August 31, see
40nestle.ai.mit.edu
There I show its easy using the Carmichael lambda function y
y(240) = y(2^4 3 5) = lcm(2^2,2,4) = 4
=> x^4 = 1 (mod 240) for x coprime to 2,3,5
Follow the above link for further detail.
-Bill Dubuque
===
Subject: Re: Factorial/Exponential Identity, Infinity
did you use a good source of entropy?
===
Subject: Re: Factorial/Exponential Identity, Infinity
I'm still trying to determine univariate polynomial ratio
forms for
the unsigned Stirling cycle numbers s(n+1, n-x+1). The signed
Stirling number is (-1)^x s(n+1, n-x+1).
For x = 0, s(n+1, n+1) = 1.
s(n+1, n+1) = 1
For x=1, s(n+1, n) = n(n+1)/2 = (n^2 + n)/2.
s(n+1, n) = (n^2 + n) / 2
For x=2, s(n+1, n-1) = ((n(n+1)/2)^2- n(n+1)(2n+1)/6)/2 =
(n^2+n)^2/8
- (n^2+1)(2n+1)/12 = (n^4+2n^3+n^2) / 8 - (2n^3+n^2+2n+1)/12 =
(3n^4+2n^3+n^2-6n-3)/24
s(n+1, n-1) = (3n^4+2n^3+n^2-6n-3)/24
for x=3, s(n+1, n-2) = ((n^2+n)^3 / 8 - (n^2+n)^2 / 4 ) (n-2) /
6(n+2) = ( (n^4+2n^3+n^2)(n^2+n)/8 - (n^4+2n^3+n^2)/4 )(n-2) /
6(n+2)
= ( (n^6+3n^5 + 3n^4 + n^3)/8 - (2n^4+4n^3+2n^2)/8 )(n-2) /
6(n+2) = (
n^6+3n^5+n^4-3n^3+2n^2)(n-2) / 48 (n+2) = (
n^7+3n^6+n^5-3n^4+2n^3
-2n^6-6n^5-2n^4+6n^3-4n^2 ) / 48 (n+2) = (n^7 + n^6 -5n^5
-5n^4 +8n^3
-4n^2) / 48(n+2) = n^2(n^5 + n^4 -5n^3 -5n^2 +8n^1 -4) /
48(n+2)
s(n+1, n-2) = n^2(n^5 + n^4 -5n^3 -5n^2 +8n^1 -4) / 48(n+2)
Thus s(n+1, n-2) is represented by a polynomial with rank 7,
s(n+1,
n-1) is represented by a polynomial of rank 4, s(n+1, n) by a
polynomial of rank 2, and s(n+1, n+1) by a polynomial of rank
0.
One avenue of determining s(n+1, n-x+1) is that it is equal to
the sum
of the products of each x-subset of {1, ..., n}. Then it is
obvious
in a 2-D nxn matrix M where m_ij = i*j that the matrix is
symmetrical
about the main diagonal and that the entries either above or
below the
diagonal represent the 2-subsets of {1, ..., n}, thus that the
sum of
all entries is easily calculated as (sum n)^2, the sum of the
diagonal
entries is obviously sum (n^2) and that their difference is
twice the
sum of the 2-subsets.
In the 3-d hypermatrix case M with m_ijk = ijk it is not so
obvious
how to divide the elements where the elements on the triagonal
are not
3-subsets and many other elements with matching indices are
also not
3-subsets, and then each 3-subset is represented 6 times in the
matrix: eg (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1,
2), and
(3, 2, 1). Then, (n+2)/(n-2) represents the value of the sum
of those
cells relative to the cells with matching coordinates besides
the main
diagonal with all matching coordinates.
I guess I could enumerate all the cells in a 4-matrix, from
(1, 1, 1,
1) to (n, n, n, n). Then, I'll remove the elements on the main
diagonal, and get metrics of those elements having two and
three
matching elements. Then, of the remaining elements with four
distinct
coordinates, there would be 4!=24 representations of each
4-subset of
{1, ..., n}. So what it would do is generate the n^x-many
element
array with each of the coordinates, the multidimensional
array, and
then start counting things about it.
One thing I do is check the sums of each of the elements of
the x-d
hypermatrix with all matching elements (the main n-agonal) and
also
with x-1, x-2, matching elements, and for elements with no
matching
elements that represent the x-subsets of n. One thing I've
noticed in
the case of x=4 is that the sum of the elements with no
matching
coordinates is 24 times s(n+1, n-3). That is to say, the
elements at
coordinates that represent k-subsets sum to 4! times s(n+1,
n-4+1).
This is good to know, but then I am left trying to determine
how to
calculate that value from n and x as a univariate polynomial
of n or a
ratio of univariate polynomials of n.
I'm looking at the count and sum of elements that have at
least a pair
of matching coordinates.
n = 4 x = 4
Sum of elements: 10000
Sum of elements with 3 matching elements: 2037 = 7*3*97
Sum of elements with 2 matching elements: 7033 = 13*541
Number of elements with 3 matching elements: 39 = 3*13
Number of elements with 2 matching elements: 189 = 3*3*3*7
I'm not quite sure how to calculate how many elements there
are with
at least two matching coordinates and less than x many matching
coordinates, or their elements' sum.
n = 5 x = 4
Sum of elements: 50625 = (sum n)^x
Sum of elements with 4 matching elements: 979 = sum (n^x)
Sum of elements with 3 matching elements: 7412 = 2*2*17*109
Sum of elements with 2 matching elements: 35658 = 2*7*3*3*283
Sum of elements with no matching elements: 6576 = x! * s(n+1,
n-x+1)
Number of elements with 4 matching elements: 5 = n
Number of elements with 3 matching elements: 64 =
2*2*2*2*2*2*2*2
Number of elements with 2 matching elements: 436 = 2*2*109
Number of elements with no matching elements: 120 = n! ~ (x+1)!
Finding simple expressions for those items above denoted as
equal to
factorizations would help determine s(n+1, n-x+1) from the
other known
quantities.
Once again, in examining an x-d nxnx...xn hypermatrix m where
m_ijk...
= i*j*k*... in an attempt to discern a method to calculate
unsigned
Stirling cycle numbethere are simple methods for x=1, 2, 3 but
I
want to determine the forms for x=4, 5, ..., and they are not
so
simple.
n = 4 x = 4
Sum of elements: 10000
Sum of elements with 4 matching elements: 354
Sum of elements with 3 matching elements: 2037
Sum of elements with 2 matching elements: 7033
Sum of elements with no matching elements: 576
Number of elements with 4 matching elements: 4
Number of elements with 3 matching elements: 39
Number of elements with 2 matching elements: 189
Number of elements with no matching elements: 24
n = 6 x = 4
Sum of elements: 194481
Sum of elements with 4 matching elements: 2275
Sum of elements with 3 matching elements: 21398
Sum of elements with 2 matching elements: 131832
Sum of elements with no matching elements: 38976
Number of elements with 4 matching elements: 6
Number of elements with 3 matching elements: 95
Number of elements with 2 matching elements: 835
Number of elements with no matching elements: 360
n = 7 x = 4
Sum of elements: 614656
Sum of elements with 4 matching elements: 4676
Sum of elements with 3 matching elements: 52611
Sum of elements with 2 matching elements: 394913
Sum of elements with no matching elements: 162456
Number of elements with 4 matching elements: 7
Number of elements with 3 matching elements: 132
Number of elements with 2 matching elements: 1422
Number of elements with no matching elements: 840
What should I read to get more information about
hypermatrices, what
is the definitive reference on hypermatrices?
I noticed someone asking about space being 3-D, I noticed the
other
day in reading about knots that MathWorld noted that knots
don't exist
in dimensions >=4. I don't quite understand that.
Besides knot theory, another thing I am trying to grasp is the
concept
of the matroid. It appears to be some generalization of graphs
and
matrices.
I notice some posters asking about multitudes of infinities
and a
universal set. A set of all sets would be its own powerset,
and would
not contain an element of all sets not containing themselves.
Are
there more even than odd numbers? No, there are not. Are there
more
integers than even integers? Yes, there are, in the integers
(or
rationals, reals, complex, or hypercomplex numbers). There are
also
more rational than integers in the rationals, etcetera, and
more
positive integers than non-negative integers in the integefor
various reasons. Are there functions that map among all those
sets?
Consider a space-filling curve. An infinite set a is greater
than
another infinite set b if infinitely many proper supersets of
b are
proper subsets of a, |[a]| > |[b]|, an infinite set a is
greater than
a proper superset b, |(a)| > |(b)|, the set or a linear
translation.
Anyways, I want to know about how many permuations of x
elements of
{1, ..., n} there are with multiples of the elements in the
multiset.
A combination of the elements of an n-set {1, ..., n} is a
subset of
that set. A permutation of a combination of an n-set is an
ordered
list, a sequence, of elements of a combination of an n-set. A
multiset is a set that many contain multiple indistinguishable
instances of an otherwise unique set element. The xn-set is
here
defined as the multiset with x-many instances of each element
of an
n-set. The hypermatrix indices or coordinates of an x-d
nxnx...xn M
= m_ijk... are the set of permutations of the x-subsets of the
xn-set.
For example, for a 5-d hypermatrix of rank or degree 4,
coordinates
include (1, 1, 1, 1, 1), (1, 2, 3, 3, 4), and (4, 4, 4, 4, 4),
permutations of {{1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3,
4, 4,
4, 4, 4, 5, 5, 5, 5, 5}}, the xn-set that is the (4)(5)-set,
using two
braces to signify a multiset.
About a multiset, a set still has to have unique elements.
There are
a variety of ways to represent the unique elements of a set.
The von
Neumann ordinals are the most direct way to represent natural
integevery large composites could also represent the natural
integers and as well multitudes of integers. The composites
would
equate to each other as integea subset of the composite could
be
the von Neumann ordinal, the integer operations would operate
on
uniquified composite multiset element ordinals.
I've seen on Dr. Math's site on Math Forum a form for the
Stirling
cycle number, the unsigned Stirling number of the first kind,
for n
and n-2, as noted in this thread. I found a different one for
n+1 and
n-1 before that, I assume it is already known. I found a form
for
s(n+1, n-2), I have not seen any other method directly
calculating
that value via a ratio of univariate polynomials. I want to
know
where it was already known, because the same source might be
able to
tell me all the other ones that are known. I want to determine
forms
for the other values of x for s(n+1, n-x+1) without having to
learn
umbral calculus, Gian-Carlo Rota's shadow-y exploration of
recurrence relations and perhaps linear relations, although it
seems a
good thing to know.
Have I discovered a form for s(n+1, n-4+1) since my previous
post?
Have you?
I'm thinking about investing in one of those CAS, computational
algebra systems. Yet, I want to actually write one. I'll look
to the
noted references.
I was going to add a political spiel but this is sci.math,
humor is
irrelevant, maybe later.
Why did Isabel cross the Atlantic? To get to the other side. I
hope
that Isabel doesn't do too much damage. What's an F-6
hurricane? I
heard Isabel has gusts of more than 190 miles per hour. Buckle
up.
How about a war against climate change? People should start
evacuating. Hopefully Isabel will careen off towards
Greenland, yet
it currently appears that it might march up the Potomac, and
be the
most damaging Act of God to ever hit the United States. Some
day a
volcano will erupt, although hopefully far enough in the
future that
we can alleviate its pressure via tunneling or applied
explosives, the
same goes for meteor strikes with anti-meteor tugs, rockets,
and
laseetcetera, and of course far in the future there will be
significant control of the weather besides the crude burning of
petrochemicals to release huge amounts of greenhouse gasses,
petrified muck. Seriously, people on the Maryland shore should
get
warnings.
Anyways, I think I found another factorial identity, as
described
previously, but it's kind of strange, lim n->oo ((sum n)^n -
sum(n^n))
/ n!^2 = 1, I'm still trying to understand it.
Anyways, i want to learn more about the theory of
hypermatrices, n-d
arrays, matroids, I guess I should read more of the tensor
material.
Bye bye,
Ross
===
Subject: Which is more difficult
Is factoring harder than solving a diophantine equation, in
general?
===
Subject: Re: Which is more difficult
>Is factoring harder than solving a diophantine equation, in
general?
Given an integer n, if you can find all integer solutions of
the
Diophantine equation x*y = n, then you can factor n.
Or you can reduce it to finding a nonnegative integer solution
of
(x + 2)*(y + 2) = n,
which is like finding one integer solution to
(x1^2 + x2^2 + x3^2 + x4^2 + 2)*(y1^2 + y2^2 + y3^2 + y4^2 +
2) = n.
--
Wanted: 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: Which is more difficult
>Is factoring harder than solving a diophantine equation, in
general?
Not sure whether you mean factoring a polynomial (say over the
rationals) or
factoring an integer, but these are both easier than solving a
diophantine
equation: there are algorithms for both kinds of factoring,
while for
diophantine equations in general there is no algorithm (look
up Hilbert's
Tenth Problem).
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: Legal logic ? puzzle.
> The matter which I am desperately struggling with,
concerns a
> time sequence of event; with the assumption that 'cause'
does NOT
> follow 'effect'.
> If I can't explain this to scientifically trained persons,
how will
> I explain the legal-types ?!
Perhaps they all do understand your story, but they don't
understand what
your actual question is.
If x owes 10$, but is billed 11$, isn't it possible for x to
just pay 10$?
Other possibility: every 10 times, x could simply not pay 1
bill. That would
make things even.
Herman Jurjus
===
Subject: Re: The two envelope paradox
In artikel <YBqdnfOzOe0p2fyiXTWJiQ@comcast.com> schreef Steve
Gerrard:
>
>
>< snippage>
>
> Having just muddled my way through this myself, I am going
to take a shot
at
> explaining what I have learned as it applies to your
proposal.
>
>> Just budding in, but I have found few people come up with
the following
>> solution. Before you open the envelope in your hand, think
of a number.
>> If the amount in your envelope is smaller then that number
ask to
>> switch. If it is greater don't switch.
>
> Your proposal would work (I think) if it just so happened
that over time,
the
> frequency of each possible amount was the same - that is,
that the
distribution
> of amounts was uniform. However, there is no reason to think
that is the
case.
> Illustrating that it is a false assumption is the real
purpose of the
paradox.
No
>
>> There are three possibilities.
>> 1) Both envelopes contain an amount under your number.
>> You will always switch and on average gain nothing.
>
> #1 is false, because the on average is assuming a uniform
distribution
of
> amounts, and that it will all average out over time. There
just is
nothing
> that makes that so.
No it is assuming a uniform distribution in having chosen the
lower or
the higher number. Half of the time I will have sitched the
lower
amount for the higher and vice versa. The amount I win in the
first
case is on average the amount I lose in the latter case. So on
average I gain nothing.
>> 2) Both envelopes contain an amount above your number.
>> You will never switch and on average gain nothing.
>
> #2 works, since you are not changing anything.
>
>> 3) Your number is between the amounts in the envelopes.
>> You will keep the higher amount but will swicth
>> with the lower amount, so you will gain some.
>
> #3 might happen, and if it did, you might make a gain. Might
not. There is
no
> way to tell, since no one knows the true distribution of
amounts that
will be
> carried around in pairs of envelopes by generous persons
with an interest
in
> logic problems, for the rest of eternity ...
But you can't escape the fact, that if a person has two such
envelopes
that the double amount has a higher chance of being bigger
than some
randomly choosen number than the single amount. Sure the
chance can
be very small, but a small chance is still better than no
chance.
--
Antoon Pardon
===
Subject: Re: The two envelope paradox
> But you can't escape the fact, that if a person has two such
envelopes
> that the double amount has a higher chance of being bigger
than some
> randomly choosen number than the single amount. Sure the
chance can
> be very small, but a small chance is still better than no
chance.
It actually isn't better, in the case of infinities. This is
like a
limit
in calculus. What is the limit of 1/x as x -> infinity? The
answer is
zero.
In other words, a sufficiently small chance is
_exactly_the_same_ as
no chance, not still better than no chance.
-- Don
______________________________________________________________
______________
___
Don Geddis http://don.geddis.org/
don@geddis.org
The statistics on sanity are that one out of every four
Americans is
suffering
from some form of mental illness. Think of your three best
friends. If
they're okay, then it's you. -- Rita Mae Brown
===
Subject: symbolic calculus software
What is the best soft to do symbolic calculus ? Maple 9,
Mathematica 4.2
or Matlab 6.5 ?
===
Subject: Re: symbolic calculus software
>What is the best soft to do symbolic calculus ? Maple 9,
Mathematica 4.2
>or Matlab 6.5 ?
What does best mean??
===
Subject: Re: symbolic calculus software
> What is the best soft to do symbolic calculus ? Maple 9,
Mathematica 4.2
> or Matlab 6.5 ?
Mathematica has a newer release (IIRC it is 5.0).
There are other symbolic math programs out there too.
For example, Gauss, Macsyma, Mathcad, Axiom, MuPad, and others.
There are also specialized programs out there for specific
math topics.
May want to have a look.
HTH, FLip
===
Subject: Re: symbolic calculus software
> What is the best soft to do symbolic calculus ? Maple 9,
Mathematica 4.2
> or Matlab 6.5 ?
>
Use Matlab for numerical and matrix calculations. Use Maple or
Mathematica for symbolic calculus.
===
Subject: [graph theory] how to find a path with a length N
I'm looking for some algorithms in order to solve the following
problem:
I want to extract from a graph all the paths between two nodes
A and B
such that the number of edges (of those path) is equal to N.
does anyone have some ideas?
Jeremie
===
Subject: Rubik's cube: another question...
If you consider the moves you can make, a plane can be rotated
clockwise
once, anticlockwise once, or rotated twice (180 degrees) -
i.e. each plane
has three possible moves. I count 9 planes on the cube (3
hirozontal, 3
vertical and 3 into the cube. So, in one move there are 3x9 =
27 possible
moves. so in 14 moves, there are 27^14 possible outcomes this
= 1.09 x
10^20
(I presume a few of these might return to it's original state)
. But there
are only 4.3x 10^19 total different patterns.
Does this mean that any given cube is possible to solve within
14 moves?
===
Subject: Re: Rubik's cube: another question...
>
> If you consider the moves you can make, a plane can be
rotated clockwise
> once, anticlockwise once, or rotated twice (180 degrees) -
i.e. each
plane
> has three possible moves. I count 9 planes on the cube (3
hirozontal, 3
> vertical and 3 into the cube. So, in one move there are 3x9
= 27 possible
> moves. so in 14 moves, there are 27^14 possible outcomes
this = 1.09 x
10^20
> (I presume a few of these might return to it's original
state) . But
there
> are only 4.3x 10^19 total different patterns.
>
> Does this mean that any given cube is possible to solve
within 14 moves?
I think this problem is generally referred to as God's
algorithm. Given D, the diameter of the graph of Rubik
cube states, an omnipotent being could determine the shortest
solution to any given starting state. This solution by
definition
would be D or fewer moves long.
According to this site, God's algorithm is 14 moves long
for the 2 x 2 x 2 cube, but the diameter is unknown for the
3-cube.
http://web.usna.navy.mil/~wdj/book/node187.html
On another site, I saw mention of a speculation
that D=22 for the 3-cube, but without any mathematics.
- Randy
===
Subject: Re: Rubik's cube: another question...
...
> According to this site, God's algorithm is 14 moves long
> for the 2 x 2 x 2 cube, but the diameter is unknown for the
> 3-cube.
> http://web.usna.navy.mil/~wdj/book/node187.html
> On another site, I saw mention of a speculation
> that D=22 for the 3-cube, but without any mathematics.
The 14 are right. Long ago I have already speculated that for
the 3-cube
the diameter would be 22 or 23. This was based on the look of
the graph
for other similar puzzles and the part that was known about
the graph
forthe
3-cube.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Rubik's cube: another question...
>
> If you consider the moves you can make, a plane can be
rotated clockwise
> once, anticlockwise once, or rotated twice (180 degrees) -
i.e. each
plane
> has three possible moves. I count 9 planes on the cube (3
hirozontal, 3
> vertical and 3 into the cube. So, in one move there are 3x9
= 27 possible
> moves. so in 14 moves, there are 27^14 possible outcomes
this = 1.09 x
10^20
> (I presume a few of these might return to it's original
state) . But
there
> are only 4.3x 10^19 total different patterns.
>
> Does this mean that any given cube is possible to solve
within 14 moves?
It depends what you mean by a move - I think it's most usual to
consider the 6 squares at the centres of the faces to be fixed,
so a move of one on the 3 central planes is counted as two
moves -
one by each of its neighbours. So there are 18 (3x6) possible
moves from any position rather than 27.
Unfortunately this doesnt prove that any position can be
reached in
16 moves, even though 18^16 > 4.3x10^19, because, as you point
out,
a few of these might return [the cube] to [its] original state.
(Actually it will be rather more than a few...)
Have a look at http://cubeman.org/ for more information,
especially
the bits about God's algorithm in Progress in Solving
Algorithms
under Cube Notes. According to this, the maximum length of
God's
algorithm is empirically at most 20: I don't know whether
there are
any better estimates. (By the argument above it's at least 16).
Andrew Taylor
(Who has the small claim to fame of being mentioned in the Cube
section of Winning Ways by Berlekamp, Conway & Guy)
===
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.
Aoccdrnig to Google the word Aoccdrnig has appeared in at
least 51 news
groups. It appears to have been off-topic in most of them. The
dates below
relate to the most recent post in the thread. The word was
unknown to
Usenet
The list below does not include all the newsgroups in which
the word has
appeared. If a post is cross-posted to several groups it seems
that only
the
first is referenced. For example, I picked it up in rec.humor
which is not
listed below.
--
Clive Tooth
http://www.clivetooth.dk
===
Subject: Re: Way off topic
linux)
>> 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.
> Aoccdrnig to Google the word Aoccdrnig has appeared in at
least 51
news
> groups. It appears to have been off-topic in most of them.
The dates
below
> relate to the most recent post in the thread. The word was
unknown to
Usenet
It's making the Xerox rounds, too. A guy at work printed out a
few
copies of this quote, or nearly enough. Same misspelling of
(Aoccdrnig to *a* rscheearch...).
Was the paragraph youor did it appear elsewhere?
--
What I've learned is that [mathematicians are] the gatekeepeand
seem to have almost absolute power when it comes to
mathematics.
-- , on All I Really Ever Needed to Know I Learned
in /Ghostbusters/.
===
Subject: Re: Way off topic
message
>
>> 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.
>
> Aoccdrnig to Google the word Aoccdrnig has appeared in at
least 51
news
> groups. It appears to have been off-topic in most of them.
The dates
below
> relate to the most recent post in the thread. The word was
unknown to
Usenet
> It's making the Xerox rounds, too. A guy at work printed out
a few
> copies of this quote, or nearly enough. Same misspelling of
> (Aoccdrnig to *a* rscheearch...).
> Was the paragraph youor did it appear elsewhere?
I copied it from a post in rec.humor, making a couple of very
minor
corrections.
--
Clive Tooth
http://www.clivetooth.dk
===
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.
>
>
> That's very interesting. And it's easy to predict what the
replies are
> going to look like. tuB I ewnrdo hthewre it tsetamr htat we
aelev eth
> sfrit and tsal strelet in lapce? Hmm, I guess it does.
B.t is t...e so m..h i.........n in t.e f...t a.d l..t l.....s
t..t
t.e m....e o..s d..'t m....r at a.l?
Phil
===
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.
>>
>>
>> That's very interesting. And it's easy to predict what the
replies are
>> going to look like. tuB I ewnrdo hthewre it tsetamr htat we
aelev eth
>> sfrit and tsal strelet in lapce? Hmm, I guess it does.
>B.t is t...e so m..h i.........n in t.e f...t a.d l..t
l.....s t..t
>t.e m....e o..s d..'t m....r at a.l?
Unclear - the others certainly help comprehension.
>Phil
************************
David C. Ullrich
===
Subject: Re: Way off topic
is there a reference to this result? (please try to avoid
using the
lemma when posting the reference.. . :-) )
ceehrs
-tzurs
> 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.
>
>
===
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.
I always suspected that English was a highly redundant
language, as
probably many other languages are. In the old days, when
letters were
carved into stone, they exploited this redundancy already by
leaving
out the vowels...
And Hamming, in the fifties last century, worked the other way
by
adding redundancy to a message (bit string) in a clever way,
in order
to make it more fault tolerant when passing it through a
transmission
channel.
BTW, there's plenty of math going into that, as in fault
tolerant
logic synthesis (would'nt you hate it if a single faulty
transistor,
out of a many millions, wrecks a sevral man_years of design
work!-)
And qua math texts: I noticed that it is useful to figure out,
in
advance, in how many ways a text can be mis-interpreted (for
mathematicians this appears to be their main concern, of
course;-)
-- NB
===
Subject: Re: Way off topic
In sci.math, The Ghost In The Machine
<ewill@sirius.athghost7038suus.net>
<9itb31-gdc.ln1@lexi2.athghost7038suus.net>:
> In sci.math, The Last Danish Pastry
> <TheLastDanishPastry@yahoo.com>
> <bjsgsu$mt6lr$1@ID-11651.news.uni-berlin.de>:
>> 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.
>>
>
> Fascinating, and quite readable. :-) I'll admit this is
probably
> more appropriate to sci.psychology or some such but the idea
> has occurred to me personally.
Erm, that should be
has NOT occurred to me personally.
>
> I might have to whip up a Perl script to test this at some
point.
>
> Of course spelling checkers will blow chunks over such
submissions. :-)
>
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
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.
>
> I do not believe this. We may take in the word as a whole,
> but the internal processor still does it letter by letter,
> unless the word is recognized quickly. Those with larger
> vocabularies will find more words to separate out.
That one needs to _know_ the correct spelling of the word in
order
to recognize its distorted version does not neccesarily mean
that the
process of reading goes letter by letter.
Actually one also needs a firm grasp of the underlying grammar.
This way one can match a whole sentence pattern against
plausible
meaningful sentences. I doubt if the permutation of letters
could
still be repaired when these are introduced in an isolated
word.
So the wording of the original post is a bit misleading
because it
neglects the usage of the context.
Marc
===
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.
Very nice. I had no trouble reading it once I stopped proofing
it.
>I do not believe this. We may take in the word as a whole,
>but the internal processor still does it letter by letter,
>unless the word is recognized quickly.
I don't think so. It may differ with people who have been
trained in different disciplines. For instance, you math
guys need to process glyph by glyph because that's just the
way of math.
> .. Those with larger
>vocabularies will find more words to separate out.
When I took a speed reading course, the point was to teach
my eyes to read a sentence, then a paragraph at a time.
Proficiency in reading in grade school comes when kids
stop reading each letter and word separately. The poor
readers never get beyond this point.
/BAH
Subtract a hundred and four for e-mail.
===
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.
>Very nice. I had no trouble reading it once I stopped
proofing it.
>>I do not believe this. We may take in the word as a whole,
>>but the internal processor still does it letter by letter,
>>unless the word is recognized quickly.
>I don't think so. It may differ with people who have been
>trained in different disciplines. For instance, you math
>guys need to process glyph by glyph because that's just the
>way of math.
Regardless of whether one believes it, what you say
about math is a good point, and even somewhat on-topic.
It often seems to me that when students simply cannot
read mathematics part of the problem is that they're not
reading it slowly enough - they glance at an expression,
find they cannot understand it by glancing at it and
conclude they cannot understand it.
>> .. Those with larger
>>vocabularies will find more words to separate out.
>When I took a speed reading course, the point was to teach
>my eyes to read a sentence, then a paragraph at a time.
>Proficiency in reading in grade school comes when kids
>stop reading each letter and word separately. The poor
>readers never get beyond this point.
>/BAH
>Subtract a hundred and four for e-mail.
************************
David C. Ullrich
===
Subject: Re: Way off topic
>
>Aoccdrnig to rscheearch at an Elingsh uinervtisy, it
<sinp ogirnial emaxlpe>
>Very nice. I had no trouble reading it once I stopped
proofing it.
>
>>
>>I do not believe this. We may take in the word as a whole,
>>but the internal processor still does it letter by letter,
>>unless the word is recognized quickly.
>
>I don't think so. It may differ with people who have been
>trained in different disciplines. For instance, you math
>guys need to process glyph by glyph because that's just the
>way of math.
>
> Regardless of whether one believes it, what you say
> about math is a good point, and even somewhat on-topic.
> It often seems to me that when students simply cannot
> read mathematics part of the problem is that they're not
> reading it slowly enough - they glance at an expression,
> find they cannot understand it by glancing at it and
> conclude they cannot understand it.
>
We tend to assume that there is only one kind of reading
skill, but
there's some research supporting the idea that reading varies
comparing reading of music versus reading a book:
http://www.musica.uci.edu/mrn/V5I1W98.html#sightreading
[M]usic cognition and behavior are often viewed merely as an
instance
of other, better known subjects. An example is music
sight-reading,
often believed to obey the laws of language reading. However,
recent
studies reveal that the study of sight-reading in music
provides a
unique window on the mind.
(saccades) while reading printed texts and printed musical
scores,
and how different they are.
Probably a similar point of view can be applied to reading
mathematics. While similar to regular reading, in some ways it
is
fundamentally different than reading English (or other verbal
languages).
===
Subject: Re: Way off topic
|Regardless of whether one believes it, what you say
|about math is a good point, and even somewhat on-topic.
|It often seems to me that when students simply cannot
|read mathematics part of the problem is that they're not
|reading it slowly enough - they glance at an expression,
|find they cannot understand it by glancing at it and
|conclude they cannot understand it.
I remember someone once commented that the good
math students seemed usually to be better at adjusting
their speed based on a recognition of their own level
of comprehension. As far as I know this was just an
impression without specific experimental support, but
it's certainly the sort of think one would *like* to see
math students doing. Some of the worst difficulties
students had with math seemed to accompany their
inability to tell whether they were even getting the
point correctly or not. I remember sometimes students
understanding a point correctly, but not realizing that
it was as simple as that.
Keith Ramsay
===
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.
>>Very nice. I had no trouble reading it once I stopped
proofing it.
>I do not believe this. We may take in the word as a whole,
>but the internal processor still does it letter by letter,
>unless the word is recognized quickly.
>>I don't think so. It may differ with people who have been
>>trained in different disciplines. For instance, you math
>>guys need to process glyph by glyph because that's just the
>>way of math.
>Regardless of whether one believes it, what you say
>about math is a good point, and even somewhat on-topic.
I tried :-). Math is unusual in that it packs a lot of
meaning into one little glyph; for example, textbooks have
been written to prepare me to know what that sigma implies :-).
>It often seems to me that when students simply cannot
>read mathematics part of the problem is that they're not
>reading it slowly enough - they glance at an expression,
>find they cannot understand it by glancing at it and
>conclude they cannot understand it.
Do you know if these kids do sentence diagramming in grade
school? In math, it's important to be able to take things
apart, rearrange them, and put them back together again.
Same is true in chemistry, physics, definitely programming.
/BAH
Subtract a hundred and four for e-mail.
===
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.
>Very nice. I had no trouble reading it once I stopped
proofing it.
>>
>>I do not believe this. We may take in the word as a whole,
>>but the internal processor still does it letter by letter,
>>unless the word is recognized quickly.
>I don't think so. It may differ with people who have been
>trained in different disciplines. For instance, you math
>guys need to process glyph by glyph because that's just the
>way of math.
>>Regardless of whether one believes it, what you say
>>about math is a good point, and even somewhat on-topic.
>I tried :-). Math is unusual in that it packs a lot of
>meaning into one little glyph; for example, textbooks have
>been written to prepare me to know what that sigma implies
:-).
>>It often seems to me that when students simply cannot
>>read mathematics part of the problem is that they're not
>>reading it slowly enough - they glance at an expression,
>>find they cannot understand it by glancing at it and
>>conclude they cannot understand it.
>Do you know if these kids do sentence diagramming in grade
>school?
I don't know for a fact, but it seems very unlikely.
>In math, it's important to be able to take things
>apart, rearrange them, and put them back together again.
>Same is true in chemistry, physics, definitely programming.
Yup. Which is exactly why kids should be required to do
simple proofs, just to build those muscles.
But this is really getting way on-topic...
>/BAH
>Subtract a hundred and four for e-mail.
************************
David C. Ullrich
===
Subject: Re: Way off topic
>
>In math, it's important to be able to take things
>apart, rearrange them, and put them back together again.
>Same is true in chemistry, physics, definitely programming.
>
> Yup. Which is exactly why kids should be required to do
> simple proofs, just to build those muscles.
Exhibit of proof of Ex~(x=x) from C1-C4 and someone will
point out the error.
C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1
C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] Weak
Extensionality
How's about I let you in on a little secret? (Ex)~(x=x) follows
from (C3,C4) alone!
C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] Weak
Extensionality
C'mon, David. Show us how. Don't be afraid. You can do it!!!
--John
And yes, identity is in _fact_ reflexive. To
refute that statement you need to give an
example of something which is not identical
to itself. The idea that there is something
which is _not_ identical to itself is simply
ludicrous: That's what identity _means_: A
thing is identical to itself and to nothing
else.
--David Ullrich
===
Subject: Re: Way off topic
>>
>>In math, it's important to be able to take things
>>apart, rearrange them, and put them back together again.
>>Same is true in chemistry, physics, definitely programming.
>>
>> Yup. Which is exactly why kids should be required to do
>> simple proofs, just to build those muscles.
>Exhibit of proof of Ex~(x=x) from C1-C4 and someone will
>point out the error.
>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1
>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] Weak
>Extensionality
>How's about I let you in on a little secret? (Ex)~(x=x)
follows
>from (C3,C4) alone!
You should also let everyone else in on another little secret:
In
the thread where this came up you were claiming that C1-C4
were axioms of NBG.
>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] Weak
>Extensionality
>C'mon, David. Show us how. Don't be afraid. You can do it!!!
>--John
>And yes, identity is in _fact_ reflexive. To
>refute that statement you need to give an
>example of something which is not identical
>to itself. The idea that there is something
>which is _not_ identical to itself is simply
>ludicrous: That's what identity _means_: A
>thing is identical to itself and to nothing
>else.
>--David Ullrich
Another little secret: When you quote me saying the above,
as though I'd said something stupid or even something
the teensiest bit controversial you look like an idiot.
(Don't bother replying with citations from learned parties
talking about non-reflexive identities. It's not _identity_
they're talking about.)
************************
David C. Ullrich
===
Subject: Re: Way off topic
>
>>
>>In math, it's important to be able to take things
>>apart, rearrange them, and put them back together again.
>>Same is true in chemistry, physics, definitely programming.
>>
>> Yup. Which is exactly why kids should be required to do
>> simple proofs, just to build those muscles.
>
>
>Exhibit of proof of Ex~(x=x) from C1-C4 and someone will
>point out the error.
>
>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1
>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] Weak
>Extensionality
>
>How's about I let you in on a little secret? (Ex)~(x=x)
follows
>from (C3,C4) alone!
>
> You should also let everyone else in on another little
secret: In
> the thread where this came up you were claiming that C1-C4
> were axioms of NBG.
Produce a post where I claim this, or admit you are a LIAR.
>
>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
Classification
>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] Weak
>Extensionality
>
>C'mon, David. Show us how. Don't be afraid. You can do it!!!
>
>--John
>
>And yes, identity is in _fact_ reflexive. To
>refute that statement you need to give an
>example of something which is not identical
>to itself. The idea that there is something
>which is _not_ identical to itself is simply
>ludicrous: That's what identity _means_: A
>thing is identical to itself and to nothing
>else.
>--David Ullrich
>
> Another little secret: When you quote me saying the above,
> as though I'd said something stupid or even something
> the teensiest bit controversial you look like an idiot.
>
> (Don't bother replying with citations from learned parties
> talking about non-reflexive identities. It's not _identity_
> they're talking about.)
Thus we might characterise a non-individual as an object for
which no identity criteria hold. If the theory of identity
is taken to be that of classical logic, then for a
non-individual
it is not the case that a = a'. This allows us to avoid
the Evans-Salmon argument and resolve the curious situation
it is indeterminate whether they are identical to themselves
because they are not! (Steven French and Decio Krause, Quantum
To be ignorant of one's ignorance is the malady of the
ignorant.
--Amos Bronson Alcott, _Table Talk_ [1877]
===
Subject: Re: Way off topic
===
>Subject: Re: Way off topic
>Message-id: <tap8mvs204vngit05lepm31bdini99kh81@4ax.com>
>>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.
>>Very nice. I had no trouble reading it once I stopped
proofing it.
>I do not believe this. We may take in the word as a whole,
>but the internal processor still does it letter by letter,
>unless the word is recognized quickly.
>>I don't think so. It may differ with people who have been
>>trained in different disciplines. For instance, you math
>>guys need to process glyph by glyph because that's just the
>>way of math.
>Regardless of whether one believes it, what you say
>about math is a good point, and even somewhat on-topic.
>It often seems to me that when students simply cannot
>read mathematics part of the problem is that they're not
>reading it slowly enough - they glance at an expression,
>find they cannot understand it by glancing at it and
>conclude they cannot understand it.
I find that a lot of the stuff posted in sci.math makes my
eyes glaze over.
And
frequently it's stuff I actually understand when I make an
effort to disect
the
ASCII equations. It's amazing how one's perception of those
symbols changes
once one understands what it's about.
> .. Those with larger
>vocabularies will find more words to separate out.
>>When I took a speed reading course, the point was to teach
>>my eyes to read a sentence, then a paragraph at a time.
>>Proficiency in reading in grade school comes when kids
>>stop reading each letter and word separately. The poor
>>readers never get beyond this point.
>>/BAH
>>Subtract a hundred and four for e-mail.
>************************
>David C. Ullrich
--
Mensanator
2 of Clubs
http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm
===
Subject: Re: Way off topic
===
>>Subject: Re: Way off topic
>>Message-id: <tap8mvs204vngit05lepm31bdini99kh81@4ax.com>
>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.
>Very nice. I had no trouble reading it once I stopped
proofing it.
>>
>>I do not believe this. We may take in the word as a whole,
>>but the internal processor still does it letter by letter,
>>unless the word is recognized quickly.
>I don't think so. It may differ with people who have been
>trained in different disciplines. For instance, you math
>guys need to process glyph by glyph because that's just the
>way of math.
>>Regardless of whether one believes it, what you say
>>about math is a good point, and even somewhat on-topic.
>>It often seems to me that when students simply cannot
>>read mathematics part of the problem is that they're not
>>reading it slowly enough - they glance at an expression,
>>find they cannot understand it by glancing at it and
>>conclude they cannot understand it.
>I find that a lot of the stuff posted in sci.math
>makes my eyes glaze over.
Especially when the right margin is set way out to 80. *hint*
> .. And
>frequently it's stuff I actually understand when
>I make an effort to disect the
>ASCII equations.
I always have to write it down on paper. I don't think I'll
ever
establish that short cut of translating from screen to concept
in my head like youngsters can do very, very well.
> ..It's amazing how one's perception of those symbols changes
>once one understands what it's about.
There are pixel problems with presentation on a terminal
screen, too.
/BAH
Subtract a hundred and four for e-mail.
===
Subject: Re: Way off topic
>>I find that a lot of the stuff posted in sci.math
>>makes my eyes glaze over.
>Especially when the right margin is set way out to 80. *hint*
How do you do that in the AOL editor?
--
Mensanator
2 of Clubs
http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm
===
Subject: Re: Way off topic
>I find that a lot of the stuff posted in sci.math
>makes my eyes glaze over.
>>Especially when the right margin is set way out to 80. *hint*
>How do you do that in the AOL editor?
There is a key on the right that may have the word enter
on it. Use that when the parser gets 3/4 across your screen.
/BAH
Subtract a hundred and four for e-mail.
===
Subject: Re: Way off topic
===
>Subject: Re: Way off topic
>Message-id: <bk4655$j2k$1@bob.news.rcn.net>
>>I find that a lot of the stuff posted in sci.math
>>makes my eyes glaze over.
>Especially when the right margin is set way out to 80.
*hint*
>>How do you do that in the AOL editor?
>There is a key on the right that may have the word enter
>on it.
Duh. Sorry, I apparenetly mistook you for someone with
something
contructive to offer. Having the right margin set at 80
implies that
it can be changed in the preferences, but there aren't any.
> Use that when the parser gets 3/4 across your screen.
IF the editing windows were always the same size, and IF the
editor
used a monospaced font, and IF the font size is constant,
THEN that might make some sense.
But as they aren't, it doesn't, and it's not, I'll have to
look elsewhere
for advice.
>/BAH
>Subtract a hundred and four for e-mail.
--
Mensanator
2 of Clubs
http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm
===
Subject: Re: Way off topic
===
>>Subject: Re: Way off topic
>>Message-id: <bk4655$j2k$1@bob.news.rcn.net>
>I find that a lot of the stuff posted in sci.math
>makes my eyes glaze over.
>>
>>Especially when the right margin is set way out to 80.
*hint*
>How do you do that in the AOL editor?
>>There is a key on the right that may have the word enter
>>on it.
>Duh. Sorry, I apparenetly mistook you for someone with
something
>contructive to offer.
[emoticon notices some snot]
>Having the right margin set at 80 implies that
>it can be changed in the preferences, but there aren't any.
I suspect it's different with every program. The way to force
idiot programs to do your bidding is to use the enter key.
>> Use that when the parser gets 3/4 across your screen.
>IF the editing windows were always the same size, and IF the
editor
>used a monospaced font, and IF the font size is constant,
>THEN that might make some sense.
You're just producing excuses. None of those IFs apply.
>But as they aren't, it doesn't, and it's not, I'll have to
look elsewhere
>for advice.
IOW, doing the logical thing isn't your bag. Out of curiosity,
how did you learn to touch type? People who trained on a
typewriter
usually don't have any problems with typing carriage-control
characters.
/BAH
Subtract a hundred and four for e-mail.
===
Subject: Re: Way off topic
===
>>Subject: Re: Way off topic
>>Message-id: <bk4655$j2k$1@bob.news.rcn.net>
>>
>I find that a lot of the stuff posted in sci.math
>makes my eyes glaze over.
>>
>>Especially when the right margin is set way out to 80.
*hint*
>
>How do you do that in the AOL editor?
>>
>>There is a key on the right that may have the word enter
>>on it.
>
>Duh. Sorry, I apparenetly mistook you for someone with
something
>contructive to offer.
>
> [emoticon notices some snot]
Snotty? You were admonishing me for not having correctly set
something
I have no control over. And when I politely asked where those
settings
are found, your snotty reply was to learn where the enter key
was.
If you're going to pontificate from a position of ignorance,
don't be
surprised that you get snotty replies.
>
>Having the right margin set at 80 implies that
>it can be changed in the preferences, but there aren't any.
>
> I suspect it's different with every program. The way to force
> idiot programs to do your bidding is to use the enter key.
Do you know the difference between an editor and a word
processor? In
the former, the user usually controls the line breaks with the
enter
key. In a word processor or any other program that
automatically
controls word wrap, the enter key is used for paragraph breaks.
>
>
>> Use that when the parser gets 3/4 across your screen.
>
>IF the editing windows were always the same size, and IF the
editor
>used a monospaced font, and IF the font size is constant,
>THEN that might make some sense.
>
> You're just producing excuses. None of those IFs apply.
Unlike some, I don't spout off without knowing what I'm
talking about.
You might want to check out my reply to Gerry Myerson to learn
something about editing in AOL.
>
>
>But as they aren't, it doesn't, and it's not, I'll have to
look
elsewhere
>for advice.
>
> IOW, doing the logical thing isn't your bag.
The logical thing is to use the editor the way it is intended
to be
used, with enter used for paragraph breaks, not line breaks.
And
just to be snotty, I'll say if you don't like the way the
messages
are displayed, then that's _your_ problem, not mine.
> Out of curiosity, how did you learn to touch type?
I didn't, but 25 years of banging on computer keyboards has (in
addition to giving me carpal tunnel syndrome) made me a very
fast
pecker.
> People who trained on a typewriter usually don't have any
problems with
> typing carriage-control characters.
Until they use a word processor and find out that every line
has
become a paragraph. Luckily, I am multi-talented. I can use
typewriteeditors and word processors interchangeably and not
get
confused.
>
> /BAH
>
> Subtract a hundred and four for e-mail.
===
Subject: Re: Way off topic
===
>Subject: Re: Way off topic
>Message-id: <bk4655$j2k$1@bob.news.rcn.net>
>
>>
>>How do you do that in the AOL editor?
>
>There is a key on the right that may have the word enter on
it.
>
> Use that when the parser gets 3/4 across your screen.
>
> IF the editing windows were always the same size, and IF the
editor
> used a monospaced font, and IF the font size is constant,
THEN that
> might make some sense.
>
> But as they aren't, it doesn't, and it's not, I'll have to
look
> elsewhere for advice.
Probably not to me, as I'm not familiar with the AOL editor,
but can I ask: is it possible to resize a window? and is it
possible to make the editor use the font of your choosing?
and is it possible to convince the editor to use the font
size of your choosing?
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Way off topic
===
>Subject: Re: Way off topic
===
>>Subject: Re: Way off topic
>>Message-id: <bk4655$j2k$1@bob.news.rcn.net>
>>
>
>How do you do that in the AOL editor?
>>
>>There is a key on the right that may have the word enter on
it.
>>
>> Use that when the parser gets 3/4 across your screen.
>>
>> IF the editing windows were always the same size, and IF
the editor
>> used a monospaced font, and IF the font size is constant,
THEN that
>> might make some sense.
>>
>> But as they aren't, it doesn't, and it's not, I'll have to
look
>> elsewhere for advice.
>Probably not to me, as I'm not familiar with the AOL editor,
>but can I ask: is it possible to resize a window?
Yes, but it would be a lot of work and you may end up
with
this phenomena, which I find very annoying. It is caused
by people
who incorrectly gauge where 3/4 of the screen is and
sometimes
hit the enter key after the editor has already imposed
its own
line wrap at 80 columns, which is not shown prior to
sending.
And no, there is no preview mode. The Arial font used by AOL
contributes
to this problem.
>and is it possible to make the editor use the font of your
choosing?
No. AOL has preference settings for the e-mail but not for the
newsgroup editor.
>and is it possible to convince the editor to use the font
>size of your choosing?
And when I say there are none, I mean there is a certain
amount.
One of the e-mail settings is to display text as small, medium
or large. For some reason, the newsgroup editor inherits this
setting although it does not inherit the font setting
preferences.
With a large display such as 1280x960 and the preference on
small, the editing window is easily over a hundred characters
wide if the AOL window is at max and the edit window is at max.
And that's when you are replying. Creating a new message
almost doubles the
edit window size.
It is extremely annoying not being able to control the editor
settings.
Usually, for serious replies, especially those involving ASCII
art,
I compose in Notepad and paste in the reply. But that's a lot
of effort
for simple one or two line responses
For this message, I tried to control the line length with the
enter
key. I won't know how successful until after it's posted.
>--
>Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
--
Mensanator
2 of Clubs
http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm
===
Subject: Re: Way off topic
===
>>Subject: Re: Way off topic
>>Message-id: <bk4655$j2k$1@bob.news.rcn.net>
>>
>
>How do you do that in the AOL editor?
>>
>>There is a key on the right that may have the word enter on
it.
>>
>> Use that when the parser gets 3/4 across your screen.
>>
>> IF the editing windows were always the same size, and IF
the editor
>> used a monospaced font, and IF the font size is constant,
THEN that
>> might make some sense.
>>
>> But as they aren't, it doesn't, and it's not, I'll have to
look
>> elsewhere for advice.
>Probably not to me, as I'm not familiar with the AOL editor,
>but can I ask: is it possible to resize a window?
With the AOL software I use, it's a temporary thing and
has to be reset with every display of a post and its reply.
> .. and is it
>possible to make the editor use the font of your choosing?
I suspect that depends on the version running. Mine doesn't
but mine is very, very old.
>and is it possible to convince the editor to use the font
>size of your choosing?
No.
/BAH
Subtract a hundred and four for e-mail.
===
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.
Let's see if this pertains to paintepantries, or anything done
in
loco parentis.
--
http://hertzlinger.blogspot.com
===
Subject: Re: Way off topic
linux)
>> 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.
> Let's see if this pertains to paintepantries, or anything
done in
> loco parentis.
Do you feel an odd compulsion to repeat yourself?
--
Jesse Hughes
So far as this negative attitude toward life is concerned,
Buddhism
is merely Taoism a little touched in its wits.
-- Lin Yutang, /My Country and My People/
===
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.
>> Let's see if this pertains to paintepantries, or anything
done in
>> loco parentis.
>
> Do you feel an odd compulsion to repeat yourself?
I feel it again and again.
--
http://hertzlinger.blogspot.com
===
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.
>>
>> Let's see if this pertains to paintepantries, or anything
done in
>> loco parentis.
>
> Do you feel an odd compulsion to repeat yourself?
> I feel it again and again.
The sherpas had pitons and plates.
This might come out as...
The shapers had points and pleats.
The seraphs had pintos and petals.
The sphaers had pinots and palets.
The sphears had potins and peltas.
[Some of the more obscure words...
palets: paleae
peltas: shields
pinots: grapes
potins: copper alloys
sphaers: old form of 'spheres'
sphears: old form of 'spheres']
--
Clive Tooth
http://www.clivetooth.dk
===
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.
>
> I do not believe this. We may take in the word as a whole,
> but the internal processor still does it letter by letter,
> unless the word is recognized quickly. Those with larger
> vocabularies will find more words to separate out.
I believe it. English is not my first language, nor my second,
still I had
no problems with reading. Some words require backtracking, but
only the
long words (I think I missed only two words on first reading).
I had also
no problems with the French text supplied in follow-up (but
French is my
second language).
I know for sure that I do not take words letter for letter.
There are
many cases where I know a letter is missing in a word, only
when it is
pointed out to me.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: An algebraic problem
let x and y be two rational angles verifing the following
condition:
8 (cos x)^3 cos y is an algebraic integer.
Is it true that at least one of them has to be a multiple of
pi/4?
Nicola
===
Subject: Re: An algebraic problem
> let x and y be two rational angles verifing the following
condition:
> 8 (cos x)^3 cos y is an algebraic integer.
> Is it true that at least one of them has to be a multiple of
pi/4?
If x and y are rational, how can one of them be a (non-zero)
multiple of
pi/4?
--
Clive Tooth
http://www.clivetooth.dk
===
Subject: Re: An algebraic problem
a rational angle is of the form p Pi with p in Q.
Nicola
>
>
> If x and y are rational, how can one of them be a (non-zero)
multiple of
> pi/4?
>
===
Subject: Re: An algebraic problem
> a rational angle is of the form p Pi with p in Q.
===
Subject: Re: When laying block, better than string for
straightness
> I am building my own concrete block garage. And when it came
time to
> lay the first course I did not like the string method for it
depends too
> much
> on eye judgement.
The string method is the absolute best way to make garages.
Since that's the way you make garages that aren't made
out of concrete blocks.
If you building a prison, the pipe method is preferred.
So what I did was haul out two very long and stout
> plumbing pipes. Very long stiff and firm and layed the block
loosely for
> the first row
> and then instead of string and eyeball I simple rolled the
two pipes,
> one on the inside of the row and one on the outside of the
row to make a
> perfect line.
>
> That method is great for the first course because the pipes
are on the
> concrete slab.
>
> Now, I am trying to figure a way to use the pipe method
going up the
> wall
> so that I never have to use the time wasting and imprecise
string
> method.
Sorry you can't do it. To get perfect vertical lines,
you need a plumb bob, which use strings.
> Thought I might share that with you so that we do replace
the old string
> method with something easier, faster and better. A method
where the test
> uses the material instead of a eyeball judgement call
electron-dot-cloud are galaxies
===
Subject: Re: When laying block, better than string for
straightness
(snipped)
> Sorry you can't do it. To get perfect vertical lines,
> you need a plumb bob, which use strings.
That is a very silly and stupid device to be using for
concrete and brick
wall
building. I say that because a fixed stick or rod is a faster
means of
telling the accuracy of vertical
lines than a plumb bob setup. Just picture the time wasted in
setting up
that plumb bob when a fixed
stick or rod and determine the vertical
accuracy in a split second.
My method does use a level to check upon things. But other
than a level, the
accuracy of vertical lines
and horizontal lines should always be that of *fixed
sticks or rods* that are straight. And never on string things
which rely
upon a eyeball judgement.
Obviously the above has never done hands on building, just
mouth chatter.
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: When laying block, better than string for
straightness
> (snipped)
>
>
> Sorry you can't do it. To get perfect vertical lines,
> you need a plumb bob, which use strings.
>
>
> That is a very silly and stupid device to be using for
concrete and brick
wall
> building.
That's why we who know what they are don't use them for making
walls,
are somehing stupid as concrete.
We use them for making plumbs lines.
I say that because a fixed stick or rod is a faster means of
telling
the accuracy of vertical
> lines than a plumb bob setup.
I did'nt say anything about faster. I mentioned something
about *vertical lines*, which is what your question was about.
Just picture the time wasted in setting up that plumb bob when
a
fixed
> stick or rod and determine the vertical
> accuracy in a split second.
There are no such things as fixed sticks. Either you've
been readng something about Einstein, or some first grade
mathematicians curicculum. Either way is doesn't matter,
since neither can actually build anything other
than playground balls.
> My method does use a level to check upon things. But other
than a level,
the accuracy of vertical lines
> and horizontal lines should always be that of *fixed
> sticks or rods* that are straight.
They would make sense if a level wasn't just an amatuer plumb
bob.
And never on string things which rely upon a eyeball judgement.
>
> Obviously the above has never done hands on building, just
mouth chatter.
That's true. Those of us who actually know how to build walls,
only design them. We let day laboresuch as yourself,
build them.
> Archimedes Plutonium, a_plutonium@hotmail.com
> whole entire Universe is just one big atom where dots
> of the electron-dot-cloud are galaxies
electron-dot-cloud are galaxies
===
Subject: Re: When laying block, better than string for
straightness
> Seems like alot of work would it not be easier to buy two
cheap laser
> pointers mount two to form a 90 deg angle, one pointing down
one pointing
in
> the direction of the row of bricks. Mark the ground so the
pointers are
in
> the same location on the first brick and line the bricks up
against the
> lightr pointing down the row you could even mount a bubble
level on them
to
> make sure it's level
Actually long 2 X 4 works even better than steel pipes.
No, what I am trying to achieve is a quick way to know that
all the block
or
brick in a row are in a perfect line. And where the string
method relies on
eyeball
judgement is bad. And so will the laser.
A long straight wood board down the line of a row of block
where you touch
the
block to the board and you know you are on that perfect line.
===
Subject: Re: When laying block, better than string for
straightness
>
> No, what I am trying to achieve is a quick way to know that
all the block
or
> brick in a row are in a perfect line. And where the string
method relies
on
> eyeball
> judgement is bad. And so will the laser.
>
> A long straight wood board down the line of a row of block
where you touch
the
> block to the board and you know you are on that perfect line.
Arch,
If you're actually doing the block work yourself, 'buttering`
the
block,
dropping it in place, tapping it level and plumb with your
trowel, you
know
that your mortar joint should be between 3, and 5 sixteenths
of an
inch thick.
With an eighth of an inch to play with, 'Perfect` is a big
waist of
your
time and effort, and any solid batterboard you try to use will
just
get in your way and slow you down even more.
This is demonstrated by the fact that you've apparently laid
only one
course of block in a day.
If you're an amateur, merely staying within tolerances will be
hard
enough. Just use the proven method and get on with it.
More than one project has been 're-thought` to death.
Best of luck -
Pragmatist -Use a bigger hammer
electron-dot-cloud are galaxies
===
Subject: Re: When laying block, better than string for
straightness
>
>
> No, what I am trying to achieve is a quick way to know that
all the
block or
> brick in a row are in a perfect line. And where the string
method relies
on
> eyeball
> judgement is bad. And so will the laser.
>
> A long straight wood board down the line of a row of block
where you
touch the
> block to the board and you know you are on that perfect line.
> Arch,
> If you're actually doing the block work yourself,
'buttering` the
> block,
> dropping it in place, tapping it level and plumb with your
trowel, you
> know
> that your mortar joint should be between 3, and 5 sixteenths
of an
> inch thick.
> With an eighth of an inch to play with, 'Perfect` is a big
waist of
> your
> time and effort, and any solid batterboard you try to use
will just
> get in your way and slow you down even more.
> This is demonstrated by the fact that you've apparently laid
only one
> course of block in a day.
> If you're an amateur, merely staying within tolerances will
be hard
> enough. Just use the proven method and get on with it.
> More than one project has been 're-thought` to death.
> Best of luck -
> Pragmatist -Use a bigger hammer
Doing it myself in order to learn and improve the task. This
industry has
never
been wholescale improved and that is where I can do the most
good.
For example, the old true and tested way is to use string and
to slop alot
of mortar to build the
building. Cement is never going to get cheaper and abundant
because fusion
energy is never coming. So
cement is going to become as rare and valuable as is petroleum.
Instead of using a trowel to place cement I believe in using
the fingers.
Not because I am an amateur
bricklayer but because I need to find the method that saves
every precious
piece of Portland cement.
Of course using latex gloves and not exposed skin.
For placement of mortar for the vertical joints that is easy.
For the
horizontal joints is a little bit
more tricky. What I do is get a 3/8 steel rebar and wire
connect
it to the vertical rebar and then rest the next course with
its *inner edge*
onto the
rebar. Thus I have a uniform horizontal guide. Then with the
latex-glove,
never a trowel, I fill in the
gaps and with a pointer I then solidly compact the joint.
No matter how great one gets at a troweling method they lose
and slop too
much
and waste too much cement.
In my method, which is the method of the future as oil prices
rise so does
cement. In my method,
although painstakingly longer is perhaps the ultimate
method of laying brick and block.
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
electron-dot-cloud are galaxies
===
Subject: Re: When laying block, better than string for
straightness
In the old method, mortar had to be a mix of lime for
consistency in order
to make the mix pliable to use
a troweling method. So that a standard mortal mix would be
something like 1
part mortar-cement which is
50-50 portlandcement
and lime and then 2 to 3 parts sand.
In my method we dispense with ever using any lime. We go
straight with a
concrete type mix without aggregates because we simply use
latex gloves
finger
in the mortar into the joints.
In the future as oil prices become very high, and cement
becomes very high,
the luxury of using lime and
wasting alot of mortar is a luxury not in the future.
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: Catalan sequence problem
> OK, this sounds close to noncrossing partitions. Which are
yet
> another Catalan counted structure (with all sorts of
bijections with
> other Catalan structures) problem pp from Stanley (with only
> references, no explanation)).
>
> However, the cyclic nature of your description might make a
difference.
> But then it should show up with n = 5. Er... no it doesn't
(rethinking
> the definitions), because the partitions can enclose another
without
> crossing: e.g. {{1,2,5},{3,4}} is not crossing.
>
> I'm thinking about what the bijection really should be (I
have seen it
> before but can't remember, and I'm having trouble with
obvious
> possibilities). Maybe DFS labelling on an ordered tree?
After talking to somebody, we figured it out. It's quite simple
(afterwards of course!).
Take a binary tree on n nodes (not necessarily full), and
label the
vertices according to inorder traversal. then remove all left
edges. the
remaining components form a partition, which, by the inorder
labeling,
is non-crossing. To go the other direction, take the part (of
the
partition) which includes n, create a rightmost path labeled
increasing
with these elements. All the parts between n and the label of
its parent
(the next highest in the part that n is in) can be recursively
made into
a tree to the left of n's parent.
One can also create noncrossing partitions from first
principles that
results in the Catalan recurrence C_n = sum(C_k C_{n-k-1},
{n,0,n-1}),
C_0 = 1. disjoint union a partition of 1 thru k and k+1 thru
n-1, then
put n in the same partition as k (when k is 0, it goes in a
partition
itself). This construction insures that it is noncrossing
(this latter
bit is not clear until you do some examples).
Mitch
===
Subject: polynomial gcd
Express the polynomial gcd of (x^m - 1,x^n - 1) in terms of x
and gcd (m,n)
?
===
Subject: Re: polynomial gcd
Here's a hint.
If x^a-1 divides x^b-1, then a divides b...
You can lead a horse's ass to knowledge, but you can't make
him think.
===
Subject: Re: polynomial gcd
> Express the polynomial gcd of (x^m - 1,x^n - 1) in terms of
x and gcd
(m,n) ?
x^gcd(m,n) -1, using Euclid's algorithm.
===
Subject: Re: polynomial gcd
>
>>Express the polynomial gcd of (x^m - 1,x^n - 1) in terms of
x and
>>gcd (m,n) ?
>
> x^gcd(m,n) -1, using Euclid's algorithm.
... and by extending that argument
gcd(x^n - y^n, x^m - y^m) = x^gcd(n,m) - y^gcd(n,m)
so, what is the next generalization?
It is very reminiscent of the identity
gcd(F_n,F_m) = F_gcd(n,m)
where F_n is a Fibonacci number (not just reminiscent, it
follows the
pattern above because of Binet's formula
F_n = (sigma^n - tau^n)/sqrt(5) )
So is it something about linear recurrences and divisibility?
restricted bases cases?
I can't seem to get an example that works where there are 3
bases
(x^n - y^n - z^n) (I kind of see why it is impossible, but it
is not
clear enough to me yet how to be more rigorous about it, or
how to
extend that disproof to others)
What's the generalization?
Mitch
===
Subject: Geometry: Perimeter & Number of Shapes Possible
Is there an easy way (without drawing) to determine the number
of
straight shapes that are possible to create with a certain
perimeter
value? For example, say that there are only three shapes (not
rounded
or with diagonal lines) that can possibly be created that have
a
perimeter of eight, is there an easy way to determine the
number of
shapes that can be made with a perimeter of say 100 or even
more
without drawing them (which would obviously be time consuming)?
===
Subject: Re: Geometry: Perimeter & Number of Shapes Possible
> Is there an easy way (without drawing) to determine the
number of
> straight shapes that are possible to create with a certain
perimeter
> value? For example, say that there are only three shapes
(not rounded
> or with diagonal lines) that can possibly be created that
have a
> perimeter of eight, is there an easy way to determine the
number of
> shapes that can be made with a perimeter of say 100 or even
more
> without drawing them (which would obviously be time
consuming)?
If you don't constrain edge numbelengths, and angles to finite
sets, there will be uncountably many shapes. For example, there
are uncountably many different triangles with a perimeter of 1.
You need to define the notions of shape, difference, and
perimeter;
indicate if paths must be continuous; and how crossings or
inclusions
are treated, before you will get agreement on numbers of
shapes.
-jiw
===
Subject: Re: CAN'T Order Reals
I don't see your point. As far as I know, counting &
establishing a total
order are different things. Counting a set is the same as
constructing a
infinite sequence in it, & the total order <= is a equivalence
relation, ie
* reflexive: a <= a
* antisymmetric: a <= b and b <= a ---> a = b
* transitive: a <= b and b <= c ---> a <= c
that satisfies, for all a,b: a<=b or b<=a.
So, I don't see why you claim that total orders c counted
orders.
The usual order in R is somewhat inherited from Q, if you
construct R from
Q-Cauchy sequences. And the order in Q can be stated as:
a/b <= c/d iff a*d <= b*c
There's no place for counting here, and the density of Q and R
does not
contradict that <= is a total order.
Sorry if I'm too lost, please explain me again, if I missed
the point.
--
DvD Gomez
===
Subject: Re: CAN'T Order Reals
> I don't see your point. As far as I know, counting &
establishing a total
> order are different things. Counting a set is the same as
constructing a
> infinite sequence in it, & the total order <= is a
equivalence relation,
ie
> * reflexive: a <= a
> * antisymmetric: a <= b and b <= a ---> a = b
> * transitive: a <= b and b <= c ---> a <= c
> that satisfies, for all a,b: a<=b or b<=a.
> So, I don't see why you claim that total orders c counted
orders.
> The usual order in R is somewhat inherited from Q, if you
construct R
from
> Q-Cauchy sequences. And the order in Q can be stated as:
> a/b <= c/d iff a*d <= b*c
> There's no place for counting here, and the density of Q and
R does not
> contradict that <= is a total order.
> Sorry if I'm too lost, please explain me again, if I missed
the point.
That's correct,
I meant :
properties of total order C properties of sequences w.r.t.
countable
sets
regarding total order is a specialisation of the term order
my point is order in maths is a vector field, in English it
means put one
before the other.
Cantors proof has nothing to do with vectors so CANT Order
Reals is an apt
name.
Herc
===
Subject: Rodrigues-like formulas, F,L numbers ?
=
n is a positive integer
a=a(n)= n!/(2n-1)! , b=b(n)= 2*n!/(2n)!
h(x)=sqrt(x^2+4):=(x^2+4)^{0.5}
H(x)=H_n(x)=h^{2n-1}(x)=(x^2+4)^{n-0.5}
f_n(x)=a*H^{(n-1)}(x)/h(x) ,l_n(x)=b*h(x)*H^{(n)}(x) .
By H^{(k)} is denoted the k-th derivative .
It's true that f_n(1) and l_n(1) are integer numbers ?
If possible , please recognize f_n(1) and l_n(1)
as special numbers.
=
===
Subject: Mahler measure of a polynomial
I have been unable to write a satisfactory Maple program to
compute
the Mahler measure of a polynomial.
If anyone has an efficient Maple program for computing this
number and
would be willing to share it with me I most appreciate it.
C.Nicol
===
Subject: Re: Multitude of infinities.
>> Allowing an infinite cardinality would mean that we'd have
aleph
>> infinity. Taking the power set would give us aleph infinity
+ 1. But
>> infinity + 1 and infinity are the same number.
>
> No. If the cardinality of a set were aleph infinity, the
cardinality
> of its power set would be 2^(aleph infinity)which is
strictly greater
> than aleph infinity.
Just to clarify, the subscripts in the sequence of alephs are
the ordinals.
The least upper bound of the aleph_n for n in the natural
numbers
is aleph_omega (omega being the smallest infinite ordinal).
The next
cardinal after aleph_omega is aleph_{omega + 1} (the GCH
asserts that
that is 2^(aleph_omega)). Of course, omega and omega+1
are different ordinals.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colousounds and shapes.
The League of Gentlemen
===
Subject: Finding a basis continuously
Given a N-by-N matrix X of rank M (not necessarily full rank).
Think of it
as an array of N-column vectors {x1, .., xN}. Can you find a
basis for the
space spanned by {x1, .., xN} continuously? That means the
algorithm to
compute the basis should depend continuously in {x1, .., xN}.
Bernard
===
Subject: Re: Finding a basis continuously
>Given a N-by-N matrix X of rank M (not necessarily full
rank). Think of it
>as an array of N-column vectors {x1, .., xN}. Can you find a
basis for the
>space spanned by {x1, .., xN} continuously? That means the
algorithm
to
>compute the basis should depend continuously in {x1, .., xN}.
No. Consider the case N=2, M=1.
Suppose there is a continuous function f from the set of 2x2
matrices of
rank 1 to nonzero vectors such that f(A) is a basis for the
columns of A.
Since x -> x/|x| is continuous on nonzero vectowe can assume
f(A)
is always a unit vector. Let e(theta) be the column vector
[ cos(theta) ]
[ sin(theta) ]
and think of the matrices as pairs of column vectors.
WLOG f(e(0),e(0)) = e(0).
By continuity, we must have f(e(theta),e(theta)) = e(theta).
But then again by continuity, f(t e(0), e(0)) = e(0) for all t.
In particular, f(-e(0),e(0)) = e(0).
And then again by continuity, f(-e(0),t e(0)) = e(0) for all t
so that f(-e(0),-e(0)) = e(0). But -e(0) = e(pi), and we must
have f(e(pi),e(pi)) = e(pi), not e(0), contradiction.
I suspect algebraic topology may have something to say about
the general case (for N > M).
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: Combinatorics in Bridge
> Here's an approach from a different angle:
The task was very ungreatfull, though...
--
Kindly
Konrad
---------------------------------------------------
their souls be chased by demons in Gehenna from one room to
another for all eternity and more.
Sleep - thing used by ineffective people
as a substitute for coffee
Ambition - a poor excuse for not having
enough sence to be lazy
---------------------------------------------------
===
Subject: Re: Mathematical induction and the use of n and k (?)
>Hi all,
>After reading some of the mathematical induction problems,
I tried to
>remember why it is necessary to replace the n with k in the
induction
step.
>Can somebody help me? Is it that k in N is supposed to be a
particular
>element; whereas, n in N is meant as a general element. How
is this
>described in logic?
>>As you are using it, N is a constant, the set of integers.
>>The idea that one should even use mnemonic letters for
>>variables is greatly to be decried.
>I'd be interested in your reasons why. ISTM that mnemonic
variable names
>are common precisely because they *are* helpful.
>> It makes no difference
>>if one uses n or k or q or z or T or alpha for a variable
>>symbol. Sometimes, more than one needs to be used.
>There are well-established coventions for variable names,
presumably as
>memory aids: <delta> and <epsilon> in limit proofs, <theta>
for an angle,
>x and y for reals, z for complex, etc.
>Although formally the variable names make no difference,
well-chosen ones
>certainly do seem to aid readability.
They may or may not aid readability. Unless the person KNOWS
that the particular variable names are unimportant, there is
likely to be the assumption that one needs to use particular
names for particular types of variables. This seems to be
exactly the type of confusion of the original poster; look at
the penultimate statement in the segment quoted.
The general use of variables needs to come first to avoid the
student ever running into this problem. Only then can one use
the simplifications.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Mathematical induction and the use of n and k (?)
>>Although formally the variable names make no difference,
well-chosen ones
>>certainly do seem to aid readability.
>They may or may not aid readability. Unless the person KNOWS
>that the particular variable names are unimportant, there is
>likely to be the assumption that one needs to use particular
>names for particular types of variables. This seems to be
>exactly the type of confusion of the original poster; look at
>the penultimate statement in the segment quoted.
It has always bugged me that in recursion theory there is an
important
theorem called the SNM theorem, where N and M are just names of
indices.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Mathematical induction and the use of n and k (?)
>Although formally the variable names make no difference,
well-chosen
ones
>certainly do seem to aid readability.
>>They may or may not aid readability. Unless the person KNOWS
>>that the particular variable names are unimportant, there is
>>likely to be the assumption that one needs to use particular
>>names for particular types of variables. This seems to be
>>exactly the type of confusion of the original poster; look at
>>the penultimate statement in the segment quoted.
>It has always bugged me that in recursion theory there is an
important
>theorem called the SNM theorem, where N and M are just names
of
indices.
There is also the abc (or is it ABC ?) conjecture in number
theory, where the letters stand for the variables.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Mathematical induction and the use of n and k (?)
>Although formally the variable names make no difference,
well-chosen
ones
>certainly do seem to aid readability.
>>They may or may not aid readability. Unless the person KNOWS
>>that the particular variable names are unimportant, there is
>>likely to be the assumption that one needs to use particular
>>names for particular types of variables. This seems to be
>>exactly the type of confusion of the original poster; look at
>>the penultimate statement in the segment quoted.
>It has always bugged me that in recursion theory there is an
important
>theorem called the SNM theorem, where N and M are just names
of
indices.
And there's alpha-beta pruning of game trees, where alpha and
beta are
just the names of cutoff values.
--
---------------------------
| BBB b barbara minus knox at iname stop com
| B B aa rrr b |
| BBB a a r bbb |
| B B a a r b b |
| BBB aa a r bbb |
-----------------------------
electron-dot-cloud are galaxies
===
Subject: theoretically the strongest concrete
1.6 cement :: 3.2 sand :: 6 gravel
5 cement :: 12.5 sand :: 20 aggregate or gravel
For mortar I get this as a mix ratio:
1 portland cement :: 1 lime :: 4 to 6 parts sand
What I am wondering is like a checkerboard or chess board
where the
squares are each one grain of sand. And that the strongest and
ideal mix
to make both Concrete and to make Mortar is simple this ratio:
1 portland cement to 1 of sand
portland cement between them. Trouble is getting a mix so that
it is
perfectly mixed as to result in a 3-dimensional perfect
packing such as
a chessboard is a perfect 2 dimensional packing.
Has anyone done experimental tests to see at what ratio of
portlandcement to sand yields the strongest most durable
concrete??
I would hazard to guess that if uniformly mixed that the 1 to
1 mix is
the strongest.
Is there any proof to my above assertion?
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: theoretically the strongest concrete
I would imagine that civil engineers have studied topic this
just about to
temperature history of the cement, amount of water available,
etc.
Michael
> 1.6 cement :: 3.2 sand :: 6 gravel
> 5 cement :: 12.5 sand :: 20 aggregate or gravel
> For mortar I get this as a mix ratio:
> 1 portland cement :: 1 lime :: 4 to 6 parts sand
> What I am wondering is like a checkerboard or chess board
where the
> squares are each one grain of sand. And that the strongest
and ideal mix
> to make both Concrete and to make Mortar is simple this
ratio:
> 1 portland cement to 1 of sand
> portland cement between them. Trouble is getting a mix so
that it is
> perfectly mixed as to result in a 3-dimensional perfect
packing such as
> a chessboard is a perfect 2 dimensional packing.
> Has anyone done experimental tests to see at what ratio of
> portlandcement to sand yields the strongest most durable
concrete??
> I would hazard to guess that if uniformly mixed that the 1
to 1 mix is
> the strongest.
> Is there any proof to my above assertion?
> Archimedes Plutonium, a_plutonium@hotmail.com
> whole entire Universe is just one big atom where dots
> of the electron-dot-cloud are galaxies
===
Subject: Re: theoretically the strongest concrete
> I would imagine that civil engineers have studied topic this
just about
to
> temperature history of the cement, amount of water
available, etc.
well, they have, but as any empirical science (sorry, I just
had to
write that) there is still a lot of room for improvement.
Unfortunately i don't know the right english words for the
coming
remarks; i hope it is still understandable.
Often highest strength is not the aim, but early strength to
allow
quick formwork removal, or low porosity (high density) to make
a
barriers for ground water protection etc. Concretes are
optimized to
set under suboptimal circumstances like low/hi temperatures,
short
times in formwork, low viscosity, very long/short pot life.
The posed 1:1 optimal concrete theory has no basis... a sand
grain is
a sand grain is a sand grain...silicon dioxide... a dead dog..
the
cement however is more or less reactive depending on where it
comes
from, it is not a defined substance.. pure CaO has completely
different setting behaviour than fly ash cements. Silica fume,
plasticizers and thousands of other additives make this a real
witchcraft system. So no 1:1 ratio, neither m:m, nor v:v or
mol:mol
have the slightest chance of being a general rule, let alone
specify
some concrete mixture.
The quality of the concrete widely varies even with the same
cement/sand ratio (and using the exact same materials in
different
experiments) with different water/cement ratios.
Depending on water content of the sand, its specific surface,
the
quality of the cement etc., the brickie on the building site
has to
slightly modify the recipe every day to just get the ideal,
same
consistence.
electron-dot-cloud are galaxies
===
Subject: Re: theoretically the strongest concrete
> I would imagine that civil engineers have studied topic this
just about
to
> temperature history of the cement, amount of water
available, etc.
> well, they have, but as any empirical science (sorry, I just
had to
> write that) there is still a lot of room for improvement.
> Unfortunately i don't know the right english words for the
coming
> remarks; i hope it is still understandable.
> Often highest strength is not the aim, but early strength to
allow
> quick formwork removal, or low porosity (high density) to
make a
> barriers for ground water protection etc. Concretes are
optimized to
> set under suboptimal circumstances like low/hi temperatures,
short
> times in formwork, low viscosity, very long/short pot life.
> The posed 1:1 optimal concrete theory has no basis... a sand
grain is
> a sand grain is a sand grain...silicon dioxide... a dead
dog.. the
Well yes, there is a direct link with theory and with
analysis. We have the
Kepler Packing to be applied to concrete and to see if the
Kepler Packing is
a
better concrete than the theory of 1 :: 1 mix of a 3-d
chessboard mix.
I forgotten the 3-d Kepler Packing percent of voids. Was it
somewhere
around
23% voids? Let us say it was. So that a Kepler Packing
Concrete mix would
not be
a 1 :: 1 mix but a 23% :: 77% mix or a 1 :: 3 mix.
> cement however is more or less reactive depending on where
it comes
> from, it is not a defined substance.. pure CaO has completely
> different setting behaviour than fly ash cements. Silica
fume,
> plasticizers and thousands of other additives make this a
real
> witchcraft system. So no 1:1 ratio, neither m:m, nor v:v or
mol:mol
> have the slightest chance of being a general rule, let alone
specify
> some concrete mixture.
> The quality of the concrete widely varies even with the same
> cement/sand ratio (and using the exact same materials in
different
> experiments) with different water/cement ratios.
> Depending on water content of the sand, its specific
surface, the
> quality of the cement etc., the brickie on the building site
has to
> slightly modify the recipe every day to just get the ideal,
same
> consistence.
Experiment: It should be easy to experiment as to whether a
Kepler Packing
concrete mix is superior to a 1 :: 1 chessboard mix. In that
we get marble
sized
balls and Kepler pack them in wet cement. We then pack some
marbles in a
configuration of a chessboard where the marbles and cement are
in a 1 :: 1
ratio. We wait for the two samples to dry and harden. We then
test them for
superiority of one over the other.
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: theoretically the strongest concrete
Once upon a time, Herman Family
<celcaps@frontiernets.net/without_any_s/> said:
>I would imagine that civil engineers have studied topic this
just about to
>temperature history of the cement, amount of water available,
etc.
I can't imagine the answer being independent of the actual
sources of crumb rubber as an asphalt filler, for instance, it
was found that rubber that was ground after cryogenic treatment
resulted in a drastically less strong compound than rubber that
was ground at room temperatures. This was determined to be
because the cryogenically treated crumb had shear surfaces,
like faceted gems, which provided less surface area to grip its
binder than the more complex surfaces of the other crumb.
If different sources of sand produce significantly different
compounds made of them would have different physical
properties.
surface texture varies or not.
Getting back to the original question, I think it would depend
on exactly what kinds of strength was needed from the concrete
compound. Adding different things to the mix (straw, dung,
sawdust, rubber, steel rebar, hardware cloth, etc) can improve
the
strength/density ratio, or the resilience, or the total blunt
trauma energy absorbed per unit mass of the compound, albeit
possibly at the detriment of hardness or absolute flexural,
shear,
or tensile strength. Even if the concrete is only used to
support
a load which is static most of the time (eg, the foundation of
a
building), there may be other factors which complicate things,
like occasional dynamic loading (Californian earthquakes), or
water erosion (rain), etc. There is no single best formula for
everything.
-- TTK
===
Subject: Re: theoretically the strongest concrete
Sharp sand and aggregate makes much stronger concrete than
smooth sand and
aggregate.
Gordon
> Once upon a time, Herman Family
<celcaps@frontiernets.net/without_any_s/> said:
>
>I would imagine that civil engineers have studied topic this
just about
to
salt,
>temperature history of the cement, amount of water available,
etc.
> I can't imagine the answer being independent of the actual
> sources of crumb rubber as an asphalt filler, for instance,
it
> was found that rubber that was ground after cryogenic
treatment
> resulted in a drastically less strong compound than rubber
that
> was ground at room temperatures. This was determined to be
> because the cryogenically treated crumb had shear surfaces,
> like faceted gems, which provided less surface area to grip
its
> binder than the more complex surfaces of the other crumb.
> If different sources of sand produce significantly different
> compounds made of them would have different physical
properties.
> surface texture varies or not.
> Getting back to the original question, I think it would
depend
> on exactly what kinds of strength was needed from the
concrete
> compound. Adding different things to the mix (straw, dung,
> sawdust, rubber, steel rebar, hardware cloth, etc) can
improve the
> strength/density ratio, or the resilience, or the total blunt
> trauma energy absorbed per unit mass of the compound, albeit
> possibly at the detriment of hardness or absolute flexural,
shear,
> or tensile strength. Even if the concrete is only used to
support
> a load which is static most of the time (eg, the foundation
of a
> building), there may be other factors which complicate
things,
> like occasional dynamic loading (Californian earthquakes), or
> water erosion (rain), etc. There is no single best formula
for
> everything.
> -- TTK
===
Subject: Re: theoretically the strongest concrete
NNTP-Proxy-Relay: library1-aux.airnews.net
What you are asking isn't as easy as it sounds. Concrete is
not like a
checkerboard, where all the squares are the same size and
shape.
Concrete is a combination of rock, sand, cement and water, all
of
varying sizes and shapes (or amorphous).
Think of it this way - in woodworking the strongest glued
joint is the
thinnest one. Concrete is sort of the same way. Water
surrounds the
concrete, and you want the least amount of mortar. You could
theoretically calculate all this for spheres, or using computer
simulations, but the National Institute of Standards and
Technology
uses super-computers to do it, and is just now, after about 10
years,
getting to the point where they can do it. Actually, their
program
doesn't optimize the concrete, it just tries to predict what
the
specified concrete will do.
If you are familiar with the subject, you might comment on my
question
week.
Jay Shilstone
>1.6 cement :: 3.2 sand :: 6 gravel
>5 cement :: 12.5 sand :: 20 aggregate or gravel
>For mortar I get this as a mix ratio:
>1 portland cement :: 1 lime :: 4 to 6 parts sand
>What I am wondering is like a checkerboard or chess board
where the
>squares are each one grain of sand. And that the strongest
and ideal mix
>to make both Concrete and to make Mortar is simple this ratio:
>1 portland cement to 1 of sand
>portland cement between them. Trouble is getting a mix so
that it is
>perfectly mixed as to result in a 3-dimensional perfect
packing such as
>a chessboard is a perfect 2 dimensional packing.
>Has anyone done experimental tests to see at what ratio of
>portlandcement to sand yields the strongest most durable
concrete??
>I would hazard to guess that if uniformly mixed that the 1 to
1 mix is
>the strongest.
>Is there any proof to my above assertion?
>Archimedes Plutonium, a_plutonium@hotmail.com
>whole entire Universe is just one big atom where dots
>of the electron-dot-cloud are galaxies
electron-dot-cloud are galaxies
===
Subject: Re: theoretically the strongest concrete
> What you are asking isn't as easy as it sounds. Concrete is
not like a
> checkerboard, where all the squares are the same size and
shape.
> Concrete is a combination of rock, sand, cement and water,
all of
> varying sizes and shapes (or amorphous).
It should be easy to research and find the answer. One easily
can get a
uniform grade of sand and see if a 1 :: 1 mix is stronger than
say a 1:: 2
mix and various
other mixes.
> Think of it this way - in woodworking the strongest glued
joint is the
> thinnest one. Concrete is sort of the same way. Water
surrounds the
> concrete, and you want the least amount of mortar. You could
> theoretically calculate all this for spheres, or using
computer
> simulations, but the National Institute of Standards and
Technology
> uses super-computers to do it, and is just now, after about
10 years,
> getting to the point where they can do it. Actually, their
program
> doesn't optimize the concrete, it just tries to predict what
the
> specified concrete will do.
What is intriguing about the strongest concrete is that it is
sort of like
another Kepler Packing Problem where the spaces between the
balls is filled
with portland cement.
So, in the Kepler Packing Problem in 3-d, each sphere is
surrounded by how
many gaps?
Perhaps the strongest concrete is a direct result of the
Kepler Packing
Problem.
> If you are familiar with the subject, you might comment on
my question
> week.
> Jay Shilstone
I proved the Kepler Packing Problem as that of kissing points.
sizes, this new problem is also solvable once you factor in
the adjustment
of kissing points. For example, say you have 60 balls of which
1/3 are
one
size, and 1/3
another and the last 1/3 another, then once you factor in the
new kissing
points to determine the smallest packing, well, you are on the
way to a
generalized formula
for variable packing sizes.
As for Concrete and the maximum strength, I believe it is
somewhere near a
1 :: 1
mix, where 1 is portland cement and the 1 part sand is of
uniform size and
where the mix is so well mixed.
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: theoretically the strongest concrete
for uniform-ball aggregate, teh cement is the complement
to the volume of the ball, inscribed within a rhombical
dodecahedron;
I think it's got pi and 18 in the ratio, as per Kepler's
problem,
about a fifth left-over. (congratulations on your proof
via kissing; you've beat Hale, at last .-)
I wasn't aware taht sharp sand is better, either. however,
assigning a simple 1/3 ratio of various sizes will not be easy
-- no easier than Kepler's problem.
> What is intriguing about the strongest concrete is that it
is sort of
like
> another Kepler Packing Problem where the spaces between the
balls is
filled
> with portland cement.
>
> So, in the Kepler Packing Problem in 3-d, each sphere is
surrounded by
how
> many gaps?
> I proved the Kepler Packing Problem as that of kissing
points.
>
> sizes, this new problem is also solvable once you factor in
the
adjustment
> of kissing points. For example, say you have 60 balls of
which 1/3 are
one
> size, and 1/3
> another and the last 1/3 another, then once you factor in
the new kissing
> points to determine the smallest packing, well, you are on
the way to a
> generalized formula
> for variable packing sizes.
--Dec.2000 'WAND' Chairman Paul O'Neill, reelected
to Board. Newsish?
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://members.tripod.com/~american_almanac
===
Subject: Re: theoretically the strongest concrete
<snip>
> It should be easy to research and find the answer. One
easily can get a
> uniform grade of sand and see if a 1 :: 1 mix is stronger
than say a 1::
2
> mix and various
> other mixes.
<snip>
As I understand it, the cement is much smaller than the sand,
and
chemically reacts with the water as it solidifies. The actual
material
giving strength to the material is the sand and gravel, the
cement
The use of steel rebar is a common way to strengthen the
concrete
(concrete is an ideal environment for normal steels as long as
salts
aren't present), but fiberglass, rubber, and perhaps other
materials
can also add useful characteristics to the concrete
(respectively,
resistance to crack formation and flexibility).
Water in the original mix is bad. You've probably already
noted that
the recipes state to use as little water as possible to wet the
mixture. There are surfactants that can reduce the amount of
water
required for your concrete. Reducing water in the mix is
another way
to strengthen the concrete since little water pockets in the
concrete
apparently create flaws in the material.
The mix ratios you mention have been tried and tested IMHO and
are
probably really close to being optimal. Incidentally, when I
looked on
the Internet for details on small batch preparation of
concrete, I
found a lot of useful information from instructions for small
house
projects to art-related websites. You should be able to find
good
information on small-scale concrete projects if that's your
inclination.
Karl Hallowell
khallow@hotmail.com
electron-dot-cloud are galaxies
===
Subject: Kepler Packing on concrete mixes Re: theoretically
the strongest
concrete
> <snip>
> It should be easy to research and find the answer. One
easily can get a
> uniform grade of sand and see if a 1 :: 1 mix is stronger
than say a 1::
2
> mix and various
> other mixes.
> <snip>
> As I understand it, the cement is much smaller than the
sand, and
> chemically reacts with the water as it solidifies. The
actual material
> giving strength to the material is the sand and gravel, the
cement
> The use of steel rebar is a common way to strengthen the
concrete
> (concrete is an ideal environment for normal steels as long
as salts
> aren't present), but fiberglass, rubber, and perhaps other
materials
> can also add useful characteristics to the concrete
(respectively,
> resistance to crack formation and flexibility).
Okay, it is hard to get a uniform sand mix.
So let us do a research on something that is perfectly
uniform. That of
steel balls of the size of
marbles and BBs. Where the marbles are gravel and the BBs are
sand.
Now the question is, what mix ratio of Portland cement to make
a cubic foot
of concrete that has marbles
and BBs and is the strongest cubic foot.
I still suspect the final answer to be a cubic foot wherein
only BBs are
used and the ratio is close to
a 1 :: 1 in terms of volume. Say perhaps 10 liters of portland
cement and 10
liters of BBs.
Trouble with various sizes of aggregate is that it allows air
pockets which
diminish strength and cause
cracking.
So I am guessing that the strongest mix ratio is a 1 to 1
ratio using only
BBs and portland cement.
And I am guessing that the perfect piece of concrete is one in
which there
is a Kepler Packing and where
the portland cement is in between the voids of the Kepler
Packing.
I wonder if someone can arrange for a Kepler Packing of steel
BBs and infuse
the crevasses between the
BBs or take marble size steel balls and Kepler Pack them and
then infuse the
interstatials with
portlandcement. I have the hunch that such a block of concrete
would be the
strongest such block over
any other such block having a different arrangement of those
BBs and
marbles. And obviously the ratio is no longer a 1 :: 1 because
if memory
serves me (mathematician would
know) that in a Kepler Packing that the gaps
are something like 23% of the volume and so the ration of
cement to balls
would be closer to that of 1 :: 4
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: Kepler Packing on concrete mixes Re:
theoretically the
strongest concrete
<snip>
> Okay, it is hard to get a uniform sand mix.
>
> So let us do a research on something that is perfectly
uniform. That of
steel balls of the size of
> marbles and BBs. Where the marbles are gravel and the BBs
are sand.
>
> Now the question is, what mix ratio of Portland cement to
make a cubic
foot of concrete that has marbles
> and BBs and is the strongest cubic foot.
>
> I still suspect the final answer to be a cubic foot wherein
only BBs are
used and the ratio is close to
> a 1 :: 1 in terms of volume. Say perhaps 10 liters of
portland cement and
10 liters of BBs.
>
> Trouble with various sizes of aggregate is that it allows
air pockets
which diminish strength and cause
> cracking.
Actually, I'm not sure that this is the case. I can see that
the sand
would fill in the gaps between the larger aggregate, and the
cement in
turn would fill in the gaps between the sand. OTOH, you
probably would
get more air than with sand-sized aggregate alone.
Also, would a marble sized glob of BB's cemented together be
stronger
than a steel marble of the same size? I think not. It
indicates to me
that the large aggregate is probably responsible for most of
the
strength (perhaps in some random almost kissing structure)
with the
spaces filled by cement and sand.
> So I am guessing that the strongest mix ratio is a 1 to 1
ratio using only
BBs and portland cement.
>
> And I am guessing that the perfect piece of concrete is one
in which there
is a Kepler Packing and where
> the portland cement is in between the voids of the Kepler
Packing.
>
> I wonder if someone can arrange for a Kepler Packing of
steel BBs and
infuse the crevasses between the
> BBs or take marble size steel balls and Kepler Pack them and
then infuse
the interstatials with
> portlandcement. I have the hunch that such a block of
concrete would be
the strongest such block over
> any other such block having a different arrangement of those
BBs and
> marbles. And obviously the ratio is no longer a 1 :: 1
because if memory
serves me (mathematician would
> know) that in a Kepler Packing that the gaps
> are something like 23% of the volume and so the ration of
cement to balls
> would be closer to that of 1 :: 4
This seems more reasonable to me. I notice that there appears
to be a
sparseness of the larger stronger stuff (by volume). Ie, the
large
gravel should be 4 over the smaller stuff and cement combined.
Instead
it's slightly better than 1. The sand to cement ratio is a
little over
2-1 with pure sand in cement around 3 to 2.
I see a couple of reasons this theoretical limit isn't
reached. First,
the concrete when poured needs to be sufficiently fluid to be
usable
(and other people have mentioned the various other attributes
that
usable concrete needs depending on the application). That
appears to
be a significant constraint and may be the sole reason for the
low
ratios of large aggregate to the space filling material.
Second, the
material as you note is usually randomly packed. I don't know
how
inefficient that is. One way to improving the packing would be
to
compress the concrete while it's still wet. That would force
the
aggregate into a more optimal packing, and squeeze out air and
perhaps
some of the water. I seem to recall that this is done for some
applications (and the weight of a large amount of concrete
tends to
compress the buried sections).
Karl Hallowell
khallow@hotmail.com
===
Subject: Re: Kepler Packing on concrete mixes Re:
theoretically the
strongest concrete
it may have to do with Kepler's problem, but a)
using BBs will weaken it (too heavy), and b)
aerocrete is actually stronger than that using sand:
the aggregate has no tensional strength,
between any two chunks.
the thing about Portland Cement is that
it's stronger, the longer it takes to dry --
so keep dousing it.
> Trouble with various sizes of aggregate is that it allows
air pockets
which diminish strength and cause
> cracking.
>
> So I am guessing that the strongest mix ratio is a 1 to 1
ratio using only
BBs and portland cement.
>
> And I am guessing that the perfect piece of concrete is one
in which there
is a Kepler Packing and where
> the portland cement is in between the voids of the Kepler
Packing.
>
> I wonder if someone can arrange for a Kepler Packing of
steel BBs and
infuse the crevasses between the
> BBs or take marble size steel balls and Kepler Pack them and
then infuse
the interstatials with
> portlandcement. I have the hunch that such a block of
concrete would be
the strongest such block over
> any other such block having a different arrangement of those
BBs and
> marbles. And obviously the ratio is no longer a 1 :: 1
because if memory
serves me (mathematician would
> know) that in a Kepler Packing that the gaps
> are something like 23% of the volume and so the ration of
cement to balls
> would be closer to that of 1 :: 4
--Dec.2000 'WAND' Chairman Paul O'Neill, reelected
to Board. Newsish?
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://members.tripod.com/~american_almanac
===
Subject: Re: Kepler Packing on concrete mixes Re:
theoretically the
strongest concrete
NNTP-Proxy-Relay: library1-aux.airnews.net
Part of the problem also revolves around the fact that the
more water
that is added to the concrete, the lower the strength. We do
not have
a binary (marbles and B-Bs), or a ternary mix (marbles, B-Bs
and
cement), but a quaternary (4 part) blend of marbles, B-Bs,
cement and
water. Studies have shown that we can reduce water by getting
a good
studies have shown that reducing the maximum aggregate size
will
minimize water demand.
I am a concrete technologist, not a mathematician. If you know
of a
computer program or web model that will predict voids in a
Kepler
Packing system using a variety of sphere sizes, I would like
to know
where.
Jay Shilstone
>> <snip>
>> It should be easy to research and find the answer. One
easily can get
a
>> uniform grade of sand and see if a 1 :: 1 mix is stronger
than say a
1:: 2
>> mix and various
>> other mixes.
>> <snip>
>> As I understand it, the cement is much smaller than the
sand, and
>> chemically reacts with the water as it solidifies. The
actual material
>> giving strength to the material is the sand and gravel, the
cement
>> The use of steel rebar is a common way to strengthen the
concrete
>> (concrete is an ideal environment for normal steels as long
as salts
>> aren't present), but fiberglass, rubber, and perhaps other
materials
>> can also add useful characteristics to the concrete
(respectively,
>> resistance to crack formation and flexibility).
>Okay, it is hard to get a uniform sand mix.
>So let us do a research on something that is perfectly
uniform. That of
steel balls of the size of
>marbles and BBs. Where the marbles are gravel and the BBs are
sand.
>Now the question is, what mix ratio of Portland cement to
make a cubic foot
of concrete that has marbles
>and BBs and is the strongest cubic foot.
>I still suspect the final answer to be a cubic foot wherein
only BBs are
used and the ratio is close to
>a 1 :: 1 in terms of volume. Say perhaps 10 liters of
portland cement and
10 liters of BBs.
>Trouble with various sizes of aggregate is that it allows air
pockets which
diminish strength and cause
>cracking.
>So I am guessing that the strongest mix ratio is a 1 to 1
ratio using only
BBs and portland cement.
>And I am guessing that the perfect piece of concrete is one
in which there
is a Kepler Packing and where
>the portland cement is in between the voids of the Kepler
Packing.
>I wonder if someone can arrange for a Kepler Packing of steel
BBs and
infuse the crevasses between the
>BBs or take marble size steel balls and Kepler Pack them and
then infuse
the interstatials with
>portlandcement. I have the hunch that such a block of
concrete would be the
strongest such block over
>any other such block having a different arrangement of those
BBs and
>marbles. And obviously the ratio is no longer a 1 :: 1
because if memory
serves me (mathematician would
>know) that in a Kepler Packing that the gaps
>are something like 23% of the volume and so the ration of
cement to balls
>would be closer to that of 1 :: 4
>Archimedes Plutonium, a_plutonium@hotmail.com
>whole entire Universe is just one big atom where dots
>of the electron-dot-cloud are galaxies
===
Subject: a question
Let B be a field of characteristic <> k,j in N. Is it true
that for every
p(x), p(x)=kf(x)+jf'(x) for a suitable f(x) in B[x]???
TIA
===
Subject: Re: a question
>
> Let B be a field of characteristic <> k,j in N. Is it true
that
> for every p(x), p(x) = kf(x)+jf'(x) for a suitable f(x) in
B[x]???
Hint: compute the kernel of f -> k f + j f'
-Bill Dubuque
===
Subject: Re: a question
>Let B be a field of characteristic <> k,j in N. Is it true
that for every
>p(x), p(x)=kf(x)+jf'(x) for a suitable f(x) in B[x]???
Somewhat more generally: it's true if p is in B[x], B is a
field, k and j
in B with k <> 0.
Hint: consider the vector space V_n of polynomials of degree
<= n
and the linear map T: V_n -> V_n defined by T(f) = k f + j f'.
What is its matrix using the standard basis [1, x, ..., x^n]?
Show that T is invertible.
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: Jacobi's method : B.S. Jacobi or C.J.G. Jacobi
Originator: israel@math.ubc.ca (Robert Israel)
Jacobi's iterative method
for solving linear systems was
discovered (in 1845 ? ) by
Boris Semionovici JACOBI ( Russian mathematician)
or by German matehmatician
Carl Jacob Gustav JACOBI
Please help with a reference .
===
Subject: Re: Jacobi's method : B.S. Jacobi or C.J.G. Jacobi
Alex.Lupas schrieb in Nachricht
>Jacobi's iterative method
>for solving linear systems was
>discovered (in 1845 ? ) by
> Boris Semionovici JACOBI ( Russian mathematician)
>or by German matehmatician
> Carl Jacob Gustav JACOBI
Carl Gustav Jacob, see below
http://www-history.mcs.st-and.ac.uk/~history/Mathematicians/
Jacobi.html
Boris Semjonowitsch Jakobi, aka Boris Semionovich Yakobi
(1801-1874), was
the
elder brother of Carl Gustav Jacob Jacobi; his name in German
is
Moritz Hermann Jacobi.
In 1835 he got a chair of physics at Dorpat (now Tartu)
University
(Estonia), and in
1837 he moved to the Russian Academy of Sciences in St.
Petersburg.
http://chem.ch.huji.ac.il/~eugeniik/history/jacobi.html (at
the bottom are
a few
links to biographies in Russian).
http://hp.iitp.ru/ger/32/3284.htm
http://hp.iitp.ru/ger/32/3284a.htm
>Please help with a reference .
C. G. Jacobi: .86ber eine neue Aufl.9asungsart der bei der
Methode der
kleinsten
Quadrate vorkommenden linearen Gleichungen.
[On a new way of solving the linear equations arising in the
method of
least
squares].
Astronomische Nachrichten Vol. 22 (1845), Issue No. 523.
It may also be found in Jacobi's Collected Works in 7 Vols.
Berlin: R.
Reimer
1881-1891 in Vol. 3, p. 467-478.
All 7 volumes are online at the Biblioth.8fque Nationale de
France in
Paris:
http://gallica.bnf.fr --> Recherche --> Auteur Jacobi
but currently the server is partially down due to update and
maintainance.
Here are some more references:
http://www.numerik.uni-kiel.de/~in/PAPER/Diplom/node43.html
--> Jacobi45
http://www.cs.umd.edu/TRs/authors/C_G_Jacobi.html
--> CS-TR-2877, but the full paper seems to be no more
available ;-((
Hope that helps ...
===
Subject: big-oh functionality
hi,
I wanted to know if for any two functions, f and g : N -> R*,
R* is a
set of positive real numbers including 0(zero),,, could we
verify that
f (not belongs) O(g) and g (not belongs) O(f).
Please let me know 2 sets of examples at the earliest!
===
Subject: Re: big-oh functionality
> hi,
> I wanted to know if for any two functions, f and g : N ->
R*, R* is a
> set of positive real numbers including 0(zero),,, could we
verify that
> f (not belongs) O(g) and g (not belongs) O(f).
> Please let me know 2 sets of examples at the earliest!
f(n) = exp(2*n),
g(n) = exp(n*(2+(-1)^n))
===
Subject: Michelle Erin Haley - January 15th 1983
H A L E Y
8 1 12 5 25 = 51
In the morning I went to Burger King at 1515 8th Street (phone
373-3177). I bought a coffee from the nubile sweety Emily
Lewans and
she gave me gave me receipt number 377, then her boss Burger
King
managerette Michelle Haley provided stats, Michelle's parents
were
born on days of the year adding to 377.
144 Rickie 1 12 54 335/30 +809
Rickie 55 Craig 38 Haley 51
155 Cindy 11 2 57 42/323 +6
Cindy 55 Jeanne 49 Haley 51
120 Deidre 5 12 80 340/26 8692
Deidre 45 Rae 24 Haley 51
151 Melissa 15 1 83 15/350 9463
Melissa 78 Lee 22 Haley 51
164 Michelle 15 1 83 15/350 9463
Michelle 67 Erin 46 Haley 51
Primes Non-Primes Fibonacci Lucas
2 1 0 1
3 4 1 3
5 6 1 4
7 8 2 7
11 9 3 11
13 10 5 18
17 12 8 29
19 14 13 47
23 15 21 76
29 16 34 123
31 18 55 199
37 20 89 322
41 21 144 521
43 22 233 843
47 <-15th-> 24 <-15th-> 377 <-15th-> 1364
--- --- --- ----
328 200 986 3568
The parents were born on days of the year adding to 377 (the
15th
Fibonacci). The parents were together born 24 (15th non-prime)
days
closer to the end of their years than to the beginning of
their years.
The parents have given names adding together for 197, it's the
45th
(15+15+15th) prime. Dad was born in 54 (the 15th Book of the
New
Testament). Mom's 42nd day of birth adds with her 155 valued
name for
197 (the 15+15+15th prime). Mom and her first kid were born on
days of
the year adding to 382 (Second Chronicles 15). The first of
the kids
has a first name adding to 45 (15+15+15), a middle name adding
to 24
(15th non-prime) and a full name adding to 120 (1 through 15).
The
twins were born on the 15th day of the month and on the 15th
day of
the year. The first of the twins has a name adding to 151, the
twins
have names adding together for 315. The twins have names
adding to 151
and 164, corresponding to Numbers 34 and Deuteronomy 11,
together for
45 (15+15+15). The twins have middle names adding to 22 and
46, the
former is 47.82% of the latter (47 is the 15th prime). The
twins were
born 45 (15+15+15) days after dad's birthday. The first and
last kids
have names adding to 120 and 164, it's a span of 45
(15+15+15). The
kids have 15 vowels in their given names. The kids have given
names
averaging exactly 47 (15th prime). The kids have given names
adding
together for 282 (Second Samuel 15). The family has given
names adding
together for 479 (47 is the 15th prime while chapter 479 is
the first
of the 150 chapters of Psalms). The family was born on days of
the
month adding to 47 (15th prime). The family was born on days
of the
year adding to 747 (the 615th non-prime, prettier as 47 is the
15th
prime). The kids were born on days 5, 15 and 15, together
these Bible
Books contain 1519 verses. Mom and the kids have middle names
adding
together for 141, corresponding to Numbers 24 (15th
non-prime). Mom's
average age when she gave birth amounts to 24.86 years (24 is
the 15th
non-prime while 86 is 15 plus the 15th prime plus the 15th
non-prime).
The twins follow 257+257+257 days after the first of the kids,
there
are 257 verses in Bible Book 47 (15th prime), prettier as Bible
chapter 257 (First Samuel 21) contains 15 verses. The twins
were born
338 days after mom's birthday, corresponding to the 47th and
terminating chapter of The Kings (the 15th prime). Dad was
born on the
1st (Genesis with 1533 verses). Today the parents are an
average of
47.47 years old (47 is the 15th prime). The twins were born on
days of
the year adding together for 30 and have names adding together
for
315, all together for 345 (15x23). Dad and the twins were born
on days
of the month adding to 31 (31 chapters in the Bible contain
the length
of 15 verses, while the 15th chapter to contain the length of
15
verses is chapter 3x3x3x31). Then Michelle goinks and grows
herself up
and becomes a big boss managerette at Burger King, at 1515 8th
Street.
Anyway, note that Bible Book 24 (15th non-prime) is Jeremiah
with 1364
verses (the 15th Lucas). The kids and I have names averaging
155.5.
Primes
2 73 179
3 79 181
5 83 191
7 89 193
11 97 197
13 101 199
17 103 211
19 107 223
23 109 227
29 113 229
31 127 233
37 131 239
41 137 241
43 139 251
47 149 257 <-55th
53 151 263
59 157 269
61 163 271
67 167 277
71 173 281
The parents both have first names adding to 55. Mom's middle
name
adds to 49 while her full name added to the 155 verses of
Bible Book
49. The twins were born in 83 (the number of verses in Bible
Book 55).
The twins follow 771 days after the first of the kids, it is
three
times the 55th prime (257+257+257). The kids were born on days
340 and
15 (together for 355). The kids have vowels in their given
names
adding together for the 83 verses of Bible Book 55, and these
vowels
have an average value of 5.5333. All names in the family add
together
for 734 (Isaiah 55). Note that Book 10 brings The Samuels up
to 1505
verses and 55 chaptepretty as 1 through 10 adds to 55. The 10th
Book of the New Testament contains 155 verses, again pretty as
1
through 10 adds to 55. And note that Bible Book 39 contains 55
verses,
pretty as 55 is the 39th non-prime.
Non-Primes
1 27 50
4 28 51
6 30 52
8 32 54
9 33 55 <-39th
10 34 56
12 35 57
14 36 58
15 38 60
16 39 62
18 40 63
20 42 64
21 44 65
22 45 66
24 46 68
25 48 69
26 49 70
The kids were born on days of the month adding to 35, the kids
have
names adding together for 435. The parents were together born
with 353
days remaining in their years. The kids were together born 356
days
closer to the beginning of their years than to the end of
their years.
Dad was born on day 335 while the twins were born 335 days
closer to
the beginning of their years than to the end of their years.
The first
two kids have names adding together for 271 (the 35th chapter
of The
Samuels). The family was born in years adding to 357. Mom is
listed in
the good phone book, she is at 435 Brightsand Crescent,
keeping in
mind that her kids have names adding together for 435.
Mom was born in 57, the family was born in years adding to 357
(the
number of verses in Daniel). In non-leap yeathe kids together
have
their birthdays 357 days closer to the beginning of their
years than
to the end of their years. Melissa adds to 78 (57th
non-prime). There
are 78 letters in the family (57th non-prime). Dad and the
kids have
names adding together for 579. The parents were together 57681
days
old when the kids were born, while today the family is
together 57852
days old, it's a difference of 171 (57+57+57).
389 <-77th prime
104 <-77th non-prime
77 <-77
---
570
The Four 57's
Genesis 41 -> 41
Leviticus 14 -> 104
Judges 9 -> 220 <-I dreamt of 220 roofs blown
John 11 -> 1008 off homes in the Dakotas
----
1373 <-220th prime
Chapter 57 is Exodus 7 with 25 verses
Book 57 is Philemon with 25 verses
-- --
41st non-prime 16th non-prime
<-together for 57->
Major Books of End-Times Prophecy (Daniel and Revelation are
in part
about 666 while Isaiah contains 66 chapters):
Daniel - 357 verses
Revelation - 404 verses <-57 plus the 57th prime
plus the 57th non-prime
Isaiah - 1292 verses <-an average of 19.575757...
verses per chapter
The parents have names adding together for 299 (First Kings 8
with
66 verses). Mom gave birth a total of 974 days after her
birthdays
(2.666 years). Mom's combined ages when she gave birth amounts
to
75.66 years. The family was born in years adding to the 357
verses of
Daniel, the Book is in part about 666. The twins were born 383
days
after their parent's birthdays, it 66 plus the 66th prime
(317).
Melissa adds to 78 (6 times the 6th prime), her full name adds
to 151
(the 6x6th prime). The family name adds to 51 (the 6x6th
non-prime).
The kids have names adding with their days of birth during the
year
for 805 (666th non-prime).
1-50 - Genesis
51-90 - Exodus
91-117 - Leviticus
118-153 - Numbers
154-187 - Deuteronomy
188-211 - Joshua
930-957 - Matthew
958-973 - Mark
974-997 - Luke
998-1018 - John
1019-1046 - Acts
1047-1062 - Romans
188 <-the opening chapter of Book 6 is 6x6x6 short
of the 404 verses of Bible Book 66, it is the
6th prime squared (13x13) short of the 357
verses of Daniel (also in part about 666)
193 <-Book 6 chapter 6 is the 44th prime,
while 44 is in turn 66.666...% of 66
211 <-the terminating chapter of Book 6 is
approximately 66.6% of the 66th prime (317)
357 <-the opening chapter of Book 6 plus the 6th
prime squared is the 357 verses of Daniel
(in part about 666)
404 <-the 6th prime squared (13x13) plus the 6th
prime squared (13x13) plus 66 adds to the
404 verses of Bible Book 66
1062 <-666 plus 6x66 is a combination of the 658
verses of Bible Book 6 plus the 404 verses
of Bible Book 66, and is the terminating
chapter of New Testament Book 6
1070 <-666 plus the 404 verses of Book 66 is the
1070 verses of Job (Book 6+6+6)
1213 <-Exodus terminates at chapter 90 (66th non-
prime) with 1213 verses (the 198th or the
66+66+66th prime)
1292 <-the 658 verses of Book 6 plus twice the
66th prime (317) is the 1292 verses of
Isaiah (the Book contains 66 chapters)
The parents have given names adding together for the 197
verses of
Bible Book 28 while the family was born in months adding to
28. The
kids have given names adding together for 282.
The parents were born on days of the year adding to 377, the
kids
on days of the year adding to 370 (a difference of 7). The
family has
names adding to 144, 155, 120, 151 and 164, corresponding to
Numbers
27, Deuteronomy 2, Numbers 3, Numbers 34 and Deuteronomy 11,
together
for 77.
I returned a couple of hours later to give Michelle another
sheet
of paper, when I arrived I purchased a couple of hamburglers
and was
provided with receipt number 164, pretty as her name adds to
164. I
showed her gems, she was like everybody else and didn't even
have the
decency to offer to buy me a cookie for my work, she didn't
even have
the decency to send me a cheap letter in the mail expressing
thanx for
showing her evidence that her very name is a gift from God.
And even
if she did throw me a few bucks for my labor, she would turn
around in
December and spend big bucks on a decorated tree, and then she
would
tithe yet more to the church that teaches her to turn
evergreen trees
into decorated idols (you people collectively spent millions of
dollars having me repeatedly arrested and tortured in an
attempt to
make me shut up about your traditions, I begged for years for
assistance to get out of the country and you close your hearts
and
continue to support Babylonian churches). Good luck with your
traditions, your traditions are in opposition to God's
Commandments
and are in opposition the Spirit of Christ's teachings,
without God
you need nothing less than luck to survive. All you people are
good
for is to have your stats posted on the usenet and be used as
examples
to otheand look nubile sweety, here you are!!! If I find any of
these family members in the obituaries I will cheer, it will
be in
accordance to Scripture (Psalm 137:9) and I will post the
stats again.
In lost summer after summer after summer after summer after
summer
after summer after summer to psychiatric torture and you
people are so
cheap and ignorant that you don't even have the compassion to
throw me
a few bucks so I could get out of this country for one single
winter.
Daryl Shawn Kabatoff
Box 7134
Saskatoon Saskatchewan
Canada
S7K 4J1
Isaiah 45:4, Ephesians 3:15 - God gives you your name!!!
===
Subject: Re: Symmetric polynomials (I need help)
Of course that the difference of symmetric polynomials is
symmetric, what
was I
in the
future...
:)
Anderson Brasil
andersbrasil@aol.com
>Any product of symmetric polynomials is symmetric
>(from the definition of symmetric --
>invariant under any permutation of x_1,...,x_n),
>and so is any difference of symmetric polynomials.
>--
>Timothy Murphy
>e-mail: tim@birdsnest.maths.tcd.ie
>tel: +353-86-233 6090
>s-mail: School of Mathematics, Trinity College, Dublin 2,
Ireland
===
Subject: Vulgar but curious:
How would go about solving the following problem (or problems
like it)
?
(don't read farther if you are easily offended)
Chrissie loves to suck cock and jerk off guys! One fine
evening, her
three gorgeous flatmates (Will, Jeff and Keith) show up for
some
action. Since Chrissie has had a rough evening partying and
has a sore
nose and ass, she phones her adequate relative Uncle Happy for
help in
deciding the most efficient way she can perform her task,
since she
just wants to go to bed.
Uncle Happy takes a drag on his pipe, contemplating. He has
just
turned off his copy of Girls Gone Wild- Brooklyn Edition and
now knows
that some girls can service three men at a time using mouth
and hands.
Uncle Happy asks Chrissie to estimate, based on her past
experiences,
how long it would normally take her to please her three
charges with
her mouth, her hand, or a combination thereof.
Chrissie replies (units in minutes):
Will - 20 mouth, 18 hand, 17 hand+mouth
Jeff - 21 mouth, 19 hand, 17 hand+mouth
Keith - 21 mouth, 20 hand, 18 hand+mouth
How should Uncle Happy respond? Assume the following:
- Chrissie may arrange her friends however she chooses.
- Chrissie possesses one mouth and two hands.
- Assume it takes Chrissie thirty seconds to get up to speed
due to
disorientation problems whenever she moves her mouth from one
flatmate
to the other. Ex.: If it normally takes Dan 2 minutes to
finish, but
Chrissie works him for a minute, does something else for a
while and
then returns, the total time Dan will take is 3 minutes (.5
startup, 1
sucking, .5 startup, 1 sucking = 3 minutes).
Bonus: Chrissie needs to know whether or not to program her
VCR to
catch Law & Order. How long will this activity session take?
===
Subject: Re: Vulgar but curious:
>Will - 20 mouth, 18 hand, 17 hand+mouth
>Jeff - 21 mouth, 19 hand, 17 hand+mouth
>Keith - 21 mouth, 20 hand, 18 hand+mouth
Will, mouth, Keith, hand, Jeff, hand.
20 minutes. (Both Will and Jeff are on the critical path, and
finish at the
same time.)
Yours,
Doug Goncz (at aol dot com)
Replikon Research
Read the RIAA Clean Slate Program Affidavit and Description at
http://www.riaa.org/
I will be signing an amended Affidavit soon.
===
Subject: Enumerating all pairs of positive integers method,
does it have a
name?
Googling didn't answer my question, so I suspect there is no
name. How
about
triangular navigation method?
===
Subject: Re: Enumerating all pairs of positive integers
method, does it
have
a name?
> Googling didn't answer my question, so I suspect there is no
name. How
about
> triangular navigation method?
Dovetailing? (that term is specifically used in automata
theory to
describe the simulation of a countable # of Turing machines.
1st step on TM 1
2nd step on TM 1
1st step on TM 2
3rd step on TM 1
2nd TM 2
1st TM 3
etc.
looks like the fanned out tail of a dove. I do not know where
the word
came from (who originally coined it in this usage).
Mitch
===
Subject: Re: Enumerating all pairs of positive integers
method, does it have
a name?
Mikito Harakiri <mikharakiri@ywho.com> scribbled the following:
> Googling didn't answer my question, so I suspect there is no
name. How
about
> triangular navigation method?
AFAIK Georg Cantor used a similar method to enumerate Q. You
could check
his work to see if he named the method.
--
/-- Joona Palaste (palaste@cc.helsinki.fi)
---------------------------
| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+
ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
----------------------------------------- Finland rules!
------------/
A friend of mine is into Voodoo Acupuncture. You don't have to
go into
her
office. You'll just be walking down the street and... ohh,
that's much
better!
- Stephen Wright
===
Subject: Re: How to factor 27X^6-1?
> plz help me factor this cause my math teacher ended up
confusing me when
she
> went over this
>
The rules that are commonly used for factoring binomials are:
a^2 - b^2 = (a - b)(a + b)
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
In your case, 27x^6 - 1 can be viewed as the middle method
because it
can be rewritten in the form (3x^2)^3 - (1)^3 where 3x^2 = a,
1 = b.
replacing a and b in (a - b)(a^2 + ab + b^2) gives you
(3x^2 - 1)((3x^2)^2 + (3x^2)(1) + (1)^2) which is normally
written as
(3x^2 - 1)(9x^4 + 3x^2 + 1)
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: How to factor 27X^6-1?
> plz help me factor this cause my math teacher ended up
confusing me when
she
> went over this
>
> The rules that are commonly used for factoring binomials are:
> a^2 - b^2 = (a - b)(a + b)
> a^3 - b^3 = (a - b)(a^2 + ab + b^2)
> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
> In your case, 27x^6 - 1 can be viewed as the middle method
because it
> can be rewritten in the form (3x^2)^3 - (1)^3 where 3x^2 =
a, 1 = b.
> replacing a and b in (a - b)(a^2 + ab + b^2) gives you
> (3x^2 - 1)((3x^2)^2 + (3x^2)(1) + (1)^2) which is normally
written as
> (3x^2 - 1)(9x^4 + 3x^2 + 1)
This has been asked before, but why is your method better than
(sqrt(27))^2 - 1^2 = (sqrt(27) - 1) (sqrt(27) + 1)?
My factorization has the advantage that you can follow it by
the
other two and continue factoring, if you notice that
sqrt(27) = sqrt(3^3) = [sqrt(3)]^3.
===
Subject: Re: How to factor 27X^6-1?
>
>
>plz help me factor this cause my math teacher ended up
confusing me when
she
>went over this
>>The rules that are commonly used for factoring binomials are:
>>a^2 - b^2 = (a - b)(a + b)
>>a^3 - b^3 = (a - b)(a^2 + ab + b^2)
>>a^3 + b^3 = (a + b)(a^2 - ab + b^2)
>>In your case, 27x^6 - 1 can be viewed as the middle method
because it
>>can be rewritten in the form (3x^2)^3 - (1)^3 where 3x^2 =
a, 1 = b.
>>replacing a and b in (a - b)(a^2 + ab + b^2) gives you
>>(3x^2 - 1)((3x^2)^2 + (3x^2)(1) + (1)^2) which is normally
written as
>>(3x^2 - 1)(9x^4 + 3x^2 + 1)
>
>
> This has been asked before, but why is your method better
than
> (sqrt(27))^2 - 1^2 = (sqrt(27) - 1) (sqrt(27) + 1)?
> My factorization has the advantage that you can follow it by
the
> other two and continue factoring, if you notice that
> sqrt(27) = sqrt(3^3) = [sqrt(3)]^3.
In principle, it isn't. However, factored form is generally
understood
to be looking for factors with integer coefficients. Other
than that,
no special reason.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: How to factor 27X^6-1?
> In principle, it isn't. However, factored form is generally
understood
> to be looking for factors with integer coefficients. Other
than that,
> no special reason.
Proving that I'm no general. Oh well.
===
Subject: Re: What is the one sentence you can believe?
>
>If someone tells you something *specific* and they know
what they are
saying,
>no matter what circumstances are what is it they said if you
>know it is always true?
>
>e.g. circles are round isn't specific!
>
>cryptic hint : old
>
>Herc
First of all, it must refer to their beliefs, as people will
believe
anything, regaldless of the logic involved. (E.g. that Saddam
Hussein
amassed a huge arsenal of weapons of mass destruction, but
when the
war began and it was time to use them, he destroyed them all.)
This is false and you don't believe it.
It cannot be true, as then by the first conjunct it is false.
Thus it
is false. Thus, one or the other of the conjuncts must be
false. The
first one is true, so the 2nd one must be false. Thus you
believe it.
(Whether you believe it or not.)
(Now who can see the falacy of the above?)
Cambridge, MA
PS I also like the answer: If you don't believe me, I will
shoot you
with this gun. (while holding out an authentic looking weapon
- of
small, yet important, destruction.)
===
Subject: Re: What is the one sentence you can believe?
>
>>If someone tells you something *specific* and they know
what they are
saying,
>>no matter what circumstances are what is it they said if
you
>>know it is always true?
>>
>>e.g. circles are round isn't specific!
>>
>>cryptic hint : old
>>
>>Herc
>
>
> First of all, it must refer to their beliefs, as people will
believe
> anything, regaldless of the logic involved. (E.g. that
Saddam Hussein
> amassed a huge arsenal of weapons of mass destruction, but
when the
> war began and it was time to use them, he destroyed them
all.)
>
> This is false and you don't believe it.
>
> It cannot be true, as then by the first conjunct it is
false. Thus it
> is false. Thus, one or the other of the conjuncts must be
false. The
> first one is true, so the 2nd one must be false. Thus you
believe it.
> (Whether you believe it or not.)
>
> (Now who can see the falacy of the above?)
>
>
> Cambridge, MA
>
> PS I also like the answer: If you don't believe me, I will
shoot you
> with this gun. (while holding out an authentic looking
weapon - of
> small, yet important, destruction.)
Just keep in mind this famous limerick:
A theorem both deep and profound
States that Every circle is round.
But in a paper by Erdos
(written in Kurdish)
A counterexample is found!
(not by) Martin Cohen
(Seen on a bulletin board in Cal Tech in the 1960's)
===
Subject: Polynomials by recurrence
Let A,B,C, a, b real numbers with ACa=/=0 .
Consider the sequence (P_n)_{n>=0} of
polynomials defined as p_0(x)=a , p_1(x)=Aax/2+ b
p_{k+1}(x)=(Ax +B)*p_k(x) +C*p_{k-1}(x) , k=1,2,... .
Suppose that T_n and U_{n-1} are Chebychev polynomials
defined for t in (-1,1) by
T_n(t)=cos(n*arccos t) , U_{n-1}(t)= sin( n*arccos
t)/sqrt(1-t^2) .
Question: try to represent p_n(x) as
p_n(x)= C_1*T_n(px+q)+ C_2*U_{n-1}(ux+v) , n>=1 ,
where :
C_1=C_1(n) , C_2=C_2(n) are real numbers ,
p,q,u,v are (complex) numbers which does'nt depend on n .
===
Subject: Re: David Ullrich on Identity
> David Ullrich says:
>
> And yes, identity is in _fact_ reflexive. To
> refute that statement you need to give an
> example of something which is not identical
> to itself. The idea that there is something
> which is _not_ identical to itself is simply
> ludicrous
A variable that returns a random value (common in programming
languages) is often not equal to itself. One that returns the
next
sequential number is never equal to itself. Just output the
value of
x=x where x is that variable.
But more generally, people from the ancient Greek philosophers
who
said things like Two things equal to the same thing are equal
to each
other. to modern day pseudointellectuals miss the real point of
equality.
ANY TWO THINGS ARE EQUAL AT SOME LEVEL OF ABSTRACTION AND
ABOVE AND
UNEQUAL AT ALL LOWER LEVELS OF ABSTRACTION.
Does 1+1 equal 2? At a low, physical level, they are different
strings of characters. (1+1 will not equal 2 in string
maniplulation processes in most programming languages.) But at
a
higher, Mathematical level of abstraction, they are.
Does 1 equal 1? At a simple Mathematical level, they are. But
at a
more exact level, they are not, becaue one of them is the 2nd
word in
the question and the other is the 4th word. Or look at them
under a
microscope and you will detect faint differences in the
displays.
At a very high level of abstraction, they are both things.
apple? But up close you can.
Like everything, equality is context sensitive.
Cambridge, MA
===
Subject: Re: David Ullrich on Identity
chvol@aol.com says...
>> David Ullrich says:
>>
>> And yes, identity is in _fact_ reflexive. To
>> refute that statement you need to give an
>> example of something which is not identical
>> to itself. The idea that there is something
>> which is _not_ identical to itself is simply
>> ludicrous
>A variable that returns a random value (common in programming
>languages) is often not equal to itself. One that returns the
next
>sequential number is never equal to itself. Just output the
value of
>x=x where x is that variable.
>But more generally, people from the ancient Greek
philosophers who
>said things like Two things equal to the same thing are equal
to each
>other. to modern day pseudointellectuals miss the real point
of
>equality.
>ANY TWO THINGS ARE EQUAL AT SOME LEVEL OF ABSTRACTION AND
ABOVE AND
>UNEQUAL AT ALL LOWER LEVELS OF ABSTRACTION.
>Does 1+1 equal 2? At a low, physical level, they are different
>strings of characters. (1+1 will not equal 2 in string
>maniplulation processes in most programming languages.) But
at a
>higher, Mathematical level of abstraction, they are.
>Does 1 equal 1? At a simple Mathematical level, they are. But
at a
>more exact level, they are not, becaue one of them is the 2nd
word in
>the question and the other is the 4th word. Or look at them
under a
>microscope and you will detect faint differences in the
displays.
>At a very high level of abstraction, they are both things.
>apple? But up close you can.
>Like everything, equality is context sensitive.
>
>Cambridge, MA
Yes I agree.
I'm not equal to myself (so says my scale after a big meal).
IMHO the
flaw is to ponder on the identify of the things themselves.
IMHO, we
should rather ponder on the identity of their properties.
Which is the
same thing you're saying: identity seems to be an artifact of
intellectual resolution. Zooming in (and out) of levels of
precision
and abstraction reveals or dissolves differences. Equivalence
is just
as context-sensitive as equality but is not as misleading.
===
Subject: Re: David Ullrich on Identity
>> David Ullrich says:
>>
>> And yes, identity is in _fact_ reflexive. To
>> refute that statement you need to give an
>> example of something which is not identical
>> to itself. The idea that there is something
>> which is _not_ identical to itself is simply
>> ludicrous
>A variable that returns a random value (common in programming
>languages) is often not equal to itself.
The value of the variable is always equal to itself. For a
while
the variable has the value 3. And 3 = 3. A little later the
variable
has the value 17. And 17 = 17.
The value at one time is not equal to the value at another
time.
So what? They're not equal, because they're different things.
>One that returns the next
>sequential number is never equal to itself. Just output the
value of
>x=x where x is that variable.
>But more generally, people from the ancient Greek
philosophers who
>said things like Two things equal to the same thing are equal
to each
>other. to modern day pseudointellectuals miss the real point
of
>equality.
>ANY TWO THINGS ARE EQUAL AT SOME LEVEL OF ABSTRACTION AND
ABOVE AND
>UNEQUAL AT ALL LOWER LEVELS OF ABSTRACTION.
>Does 1+1 equal 2?
Yes.
>At a low, physical level, they are different
>strings of characters.
No, neither 1 + 1 nor 2 is a string of characters.
> (1+1 will not equal 2 in string
>maniplulation processes in most programming languages.)
One hopes not. So what? If 1 and 1 were the same thing you'd
have a point.
>But at a
>higher, Mathematical level of abstraction, they are.
>Does 1 equal 1? At a simple Mathematical level, they are. But
at a
>more exact level, they are not, becaue one of them is the 2nd
word in
>the question and the other is the 4th word. Or look at them
under a
>microscope and you will detect faint differences in the
displays.
>At a very high level of abstraction, they are both things.
>apple? But up close you can.
>Like everything, equality is context sensitive.
Whether two _things_ are equal has nothing to do with
context. Whether the objects denoted by two symbols
are equal does have a lot to do with context. This
says nothing about whether things are equal to
themselves - in a context where two symbols denote
things that are not equal they denote two _different_
things.
Duh.
>
>Cambridge, MA
************************
David C. Ullrich
===
Subject: Re: David Ullrich on Identity
>>Does 1+1 equal 2?
>
> Yes.
>
>>At a low, physical level, they are different
>>strings of characters.
>
> No, neither 1 + 1 nor 2 is a string of characters.
Quite right. Charlie-Poo seems to be mired in crass formalism,
unable to distinguish between an object and its
representations.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colousounds and shapes.
The League of Gentlemen
===
Subject: Re: David Ullrich on Identity
>Does 1+1 equal 2?
>>
>> Yes.
>>
>At a low, physical level, they are different
>strings of characters.
>>
>> No, neither 1 + 1 nor 2 is a string of characters.
>Quite right. Charlie-Poo seems to be mired in crass formalism,
>unable to distinguish between an object and its
>representations.
So it would appear. (null comment inserted just to put
sci.logic back on the list of newsgroups, lest he miss this.)
************************
David C. Ullrich
===
Subject: Re: David Ullrich on Identity
>
>> David Ullrich says:
>>
>> And yes, identity is in _fact_ reflexive. To
>> refute that statement you need to give an
>> example of something which is not identical
>> to itself. The idea that there is something
>> which is _not_ identical to itself is simply
>> ludicrous
>
>A variable that returns a random value (common in programming
>languages) is often not equal to itself.
>
> The value of the variable is always equal to itself. For a
while
> the variable has the value 3. And 3 = 3. A little later the
variable
> has the value 17. And 17 = 17.
Right. Only when you use the word (concept) itself is there
equality, but then you are not referring to two things.
Call this variable $R (the name used in my favorite programming
language.) Then we agree that $R is not equal to $R. But you
maintain that 3 is equal to 3. However, I pointed out the fact
that 3
is not actually equal to 3 because the former is the 8th word
in this
sentence and the latter is the 14th word in this sentence. So
at some
level they are still not equal. Now if we say that 3 is equal
to
itself, we are not referring to two things. We are referring
to one
thing and then referring to itself.
Whenever there is a distinction, and there are in fact two
things,
then whether they are equal or not depends on the level of
abstraction
at which we are dealing, whether it makes that same
distinction.
Since abstracting is the removal of distinctions, at some
level, and
above, they will be equal.
In other words, it is not a question of whether two things are
equal
or not. Rather, it is a question of, at what level of
abstraction,
and above, they are equal? f(one thing,something)=the lowest
level of
abstraction at which they are equal.
> They're not equal, because they're different things.
What's the difference between not equal and different things?
That is, when are these not the same?
>Does 1+1 equal 2?
>
> Yes.
>
>At a low, physical level, they are different strings of
characters.
>
> No, neither 1 + 1 nor 2 is a string of characters.
How else can we communicate other than by an exchange of
strings of
characters (i.e. elements of an agreed upon recursively
enumerable
set)?
> Whether two _things_ are equal has nothing to do with
> context. Whether the objects denoted by two symbols
> are equal does have a lot to do with context.
Are not objects things?
Cambridge, MA
> ************************
>
> David C. Ullrich
===
Subject: Re: David Ullrich on Identity
>>
> David Ullrich says:
>
> And yes, identity is in _fact_ reflexive. To
> refute that statement you need to give an
> example of something which is not identical
> to itself. The idea that there is something
> which is _not_ identical to itself is simply
> ludicrous
>>
>>A variable that returns a random value (common in programming
>>languages) is often not equal to itself.
>>
>> The value of the variable is always equal to itself. For a
while
>> the variable has the value 3. And 3 = 3. A little later the
variable
>> has the value 17. And 17 = 17.
>Right. Only when you use the word (concept) itself is there
>equality, but then you are not referring to two things.
Uh, that's correct. That's why the word itself appears
prominently
in the statement that you seemed to think you were refuting.
>Call this variable $R (the name used in my favorite
programming
>language.) Then we agree that $R is not equal to $R.
No, we don't agree to that.
>But you
>maintain that 3 is equal to 3.
You don't realize how funny this sounds, saying that I maintain
that 3 equals 3?
>However, I pointed out the fact that 3
>is not actually equal to 3 because the former is the 8th word
in this
>sentence and the latter is the 14th word in this sentence.
Yes, you pointed this out. That doesn't make it true. In fact
3 is not a word.
>So at some
>level they are still not equal. Now if we say that 3 is equal
to
>itself, we are not referring to two things. We are referring
to one
>thing and then referring to itself.
>Whenever there is a distinction, and there are in fact two
things,
>then whether they are equal or not depends on the level of
abstraction
>at which we are dealing, whether it makes that same
distinction.
>Since abstracting is the removal of distinctions, at some
level, and
>above, they will be equal.
>In other words, it is not a question of whether two things
are equal
>or not. Rather, it is a question of, at what level of
abstraction,
>and above, they are equal? f(one thing,something)=the lowest
level of
>abstraction at which they are equal.
>> They're not equal, because they're different things.
>What's the difference between not equal and different things?
There is none. That's what equal means.
>That is, when are these not the same?
>>Does 1+1 equal 2?
>>
>> Yes.
>>
>>At a low, physical level, they are different strings of
characters.
>>
>> No, neither 1 + 1 nor 2 is a string of characters.
>How else can we communicate other than by an exchange of
strings of
>characters (i.e. elements of an agreed upon recursively
enumerable
>set)?
We can't. The fact that we need character strings to talk about
things does not imply that things _are_ character strings.
>> Whether two _things_ are equal has nothing to do with
>> context. Whether the objects denoted by two symbols
>> are equal does have a lot to do with context.
>Are not objects things?
Yes...
>
>Cambridge, MA
>
>> ************************
>>
>> David C. Ullrich
************************
David C. Ullrich
===
Subject: Re: David Ullrich on Identity
>
>
> Homework for David Ullrich:
>
> 1) What philosopher said:
>
> ...definitions are available only for transforming
> truths, not for founding them.
>
>
> I would be interested in this answer...
_Ways of Paradox_, p. 81
> So, your dispute with the boyz in the hood has to do with
formal set
> theory--class models of that theory, in fact.>
> mitch
> How can *theorems* be *false*? Well, the point here is that
> when one says that a statement about the natural numbers is
> false one means that it is false of the *standard* integers,
> i.e., the integers that we normally work with.
Likewise, when David Ullrich says that
statements about individuals such as ExAy~(x=y)
and Ex~(x=x) are false, he has said nothing more
than that these are false of the *standard*
individuals that FOL= deals with. Put back in
the domain those that FOL= excludes, and it is
not ExAy~(x=y) and Ex~(x=x) but AxEy(x=y) and
Ax(x=x) that are false.
It doesn't take a rocket scientist to figure
that one out...
--John
===
Subject: Re: David Ullrich on Identity
> You should check out http://megafoundation.org/
> Harris has finally found people who understand him there.
This from a purveyor of store-bought, on-the-shelf knowledge,
whose absolute inability to reason from premises other than
the customary, bought-and-paid-for ones is a common trait
of the unabashedly uptaught.
>Exhibit of proof of Ex~(x=x) from
>C1-C4 and someone will point out the error.
>
>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1
>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
>
>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
>Classification
>
>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}]
> Weak
Here's the proof you couldn't hack.
1) Ax(x in y <-> Et(x in t & ~(x in x)) Instance of (C1)
2) y in y <-> Et(y in t & ~(y in y)) 1, US
3) ~Et(y in t) 2
4) AyAx[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] C4
5) Az(z in y <-> z in y) -> (Et(y in t) <-> y=y) 4,US
6) Et(y in t) <-> y=y 5
7) ~(y=y) 3,6
8) Ex~(x=x) 7,EG
**************************************************************
********
Homework for David Ullrich
What bearing would the non-self-identity of quanta
have on the Leibnizean principle of the Identity of
Indiscernibles?
--John
And yes, identity is in _fact_ reflexive. To
refute that statement you need to give an
example of something which is not identical
to itself. The idea that there is something
which is _not_ identical to itself is simply
ludicrous: That's what identity _means_: A
thing is identical to itself and to nothing
else.
--David Ullrich
===
Subject: Re: David Ullrich on Identity
>> You should check out http://megafoundation.org/
>> Harris has finally found people who understand him there.
>This from a purveyor of store-bought, on-the-shelf knowledge,
>whose absolute inability to reason from premises other than
>the customary, bought-and-paid-for ones is a common trait
>of the unabashedly uptaught.
You should _really_ check out http://megafoundation.org/ !
This sort of anyone-who-actually-knows-something-must-be
-too-stupid-to-think-for-himself-the-way-we-geniuses-do is
their motto.
>>Exhibit of proof of Ex~(x=x) from
>>C1-C4 and someone will point out the error.
Just for the record, when I said that you were claiming that
C1-C4 were part of NBG. Later turned out that that was just
something you made up.
>>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1
>>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
>>
>>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
>>Classification
>>
>>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}]
>> Weak
>Here's the proof you couldn't hack.
>1) Ax(x in y <-> Et(x in t & ~(x in x)) Instance of (C1)
>2) y in y <-> Et(y in t & ~(y in y)) 1, US
>3) ~Et(y in t) 2
>4) AyAx[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] C4
>5) Az(z in y <-> z in y) -> (Et(y in t) <-> y=y) 4,US
>6) Et(y in t) <-> y=y 5
>7) ~(y=y) 3,6
>8) Ex~(x=x) 7,EG
>*************************************************************
*********
> Homework for David Ullrich
>What bearing would the non-self-identity of quanta
>have on the Leibnizean principle of the Identity of
>Indiscernibles?
>--John
>And yes, identity is in _fact_ reflexive. To
>refute that statement you need to give an
>example of something which is not identical
>to itself. The idea that there is something
>which is _not_ identical to itself is simply
>ludicrous: That's what identity _means_: A
>thing is identical to itself and to nothing
>else.
>--David Ullrich
************************
David C. Ullrich
===
Subject: Re: David Ullrich on Identity
>
>
>> You should check out http://megafoundation.org/
>> Harris has finally found people who understand him there.
>
>This from a purveyor of store-bought, on-the-shelf knowledge,
>whose absolute inability to reason from premises other than
>the customary, bought-and-paid-for ones is a common trait
>of the unabashedly uptaught.
>
> You should _really_ check out http://megafoundation.org/ !
> This sort of anyone-who-actually-knows-something-must-be
> -too-stupid-to-think-for-himself-the-way-we-geniuses-do is
> their motto.
Against stupidity the very gods
Themselves contend in vain.
--Schiller, _The Maid of Orleans_
> Homework for David Ullrich
>
>What bearing would the non-self-identity of quanta
>have on the Leibnizean principle of the Identity of
>Indiscernibles?
===
Subject: Re: David Ullrich on Identity
>
>
>>Exhibit of proof of Ex~(x=x) from
>>C1-C4 and someone will point out the error.
>
>>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1
>>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
>
>>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
>>Classification
>
>>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}]
> Weak
>
>Here's the proof you couldn't hack.
>
>2) y in y <-> Et(y in t & ~(y in y)) 1, US
>3) ~Et(y in t) 2
>4) AyAx[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] C4
>5) Az(z in y <-> z in y) -> (Et(y in t) <-> y=y) 4,US
>6) Et(y in t) <-> y=y 5
>7) ~(y=y) 3,6
>
>*************************************************************
*********
> Just for the record, when I said that you were claiming that
> C1-C4 were part of NBG. Later turned out that that was just
> something you made up.
Till their own dreams at length deceive 'em,
And oft repeating, they believe 'em.
--Matthew Prior
Alma [1718], canto III, l. 13
> Homework for David Ullrich
>
> What bearing would the non-self-identity of quanta
> have on the Leibnizean principle of the Identity of
> Indiscernibles?
>
>--John
>
> David C. Ullrich
Have you forgotten your homework, AGAIN???
===
Subject: Re: David Ullrich on Identity
linux)
>> You should check out http://megafoundation.org/
>> Harris has finally found people who understand him there.
> What part of Ax[Qu(x) -> ~(x = x)] do YOU fail to understand?
that there does not exist an x such that Qu(x)? Is this a
conclusion
with which Teller would be happy?
In any case, if some quantum theorist somewhere wants to mess
about
with a thoroughly non-standard idea of identity, what the heck
should
Ullrich care? And how does Teller's speculations support your
claim
that all and only existent things are self-identical?
(I will probably regret treating you like a human being here.
Again.)
--
Jesse Hughes
Of course, my ability to admit my mistakes and correct them is
a
trait that many of you seem to never have properly appreciated.
-- JSH, discussing his 1463rd proof of Fermat's Last Theorem.
===
Subject: Re: David Ullrich on Identity
>
>> You should check out http://megafoundation.org/
>> Harris has finally found people who understand him there.
>
> What part of Ax[Qu(x) -> ~(x = x)] do YOU fail to understand?
>
> that there does not exist an x such that Qu(x)? Is this a
conclusion
> with which Teller would be happy?
'Quanta' are one thing, non-existents are another, and
in common is that they are identical-with-nothings. My (sketchy
and non-mathematically well-informed impression) is that
fruitful
applications of 'imaginary numbers' have been found in
different
areas of mathematics. It would not be surprising if the logic
of
identical-with-nothings were to turn out to have to have
different
applications as well.
file if you wish), what Teller would not be happy with is the
notion
that (e.g.) AxAy(qu(x) & qu(y) -> ~(x=y)) and Ax(qu(x) ->
~(x=x)
are 'well-formed'. His argument appears to be that since quanta
lack 'haecceities', it makes sense neither to assert that x=y
when x,y are quanta, or that ~(x=y).
> In any case, if some quantum theorist somewhere wants to
mess about
> with a thoroughly non-standard idea of identity, what the
heck should
> Ullrich care? And how does Teller's speculations support
your claim
> that all and only existent things are self-identical?
The main thing at stake is whether, in the logic that
different theories run under, identity should be taken to be
reflexive or non-reflexive. If the latter, then quantum
theorists can battle it out (on empirical grounds) over whether
quanta satisfy Ax(x=x). If the former, such (potentially
fruitful) disputes are--illegitimately, I think--ruled out
of court by the background logic.
>
> (I will probably regret treating you like a human being here.
> Again.)
--John
===
Subject: power set cardinality?