mm-1017
===
Subject: Re: Need help on my educational website
>It works the same as TeX. I usually go to a page on
wikipedia like
>this:
>http://en.wikipedia.org/wiki/Partial_derivative
>or this
>http://en.wikipedia.org/wiki/Integral
>to see how they did it.
>heres some TeX symbols:
>http://www.astro.uiuc.edu/~bima/proposal/TeXsymbols.html
>And heres how you did it:
>http://www.exampleproblems.com/wiki/index.php?title=Calc1.1
I just tried posting something there, but it wasnt clear to
me how to
save the solution elsewhere than on the same page where the
problem is
stated. It was under nonlinear odes I think.
--Lynn
===
Subject: [test] character set
... 0 1 2 3 4 5 6 7 8 9 a b c d e f
2 ! # $ % & Ô ( ) * + , - . / 2
3 0 1 2 3 4 5 6 7 8 9 : ; < = > ? 3
4 @ A B C D E F G H I J K L M N O 4
5 P Q R S T U V W X Y Z [ ] ^ _ 5
6 ` a b c d e f g h i j k l m n o 6
7 p q r s t u v w x y z { | } ~  7
8 .81 .82 .83
.84 .85 .86 .87
.88 .89 .8a .8b .8c .8d
.8e .8f 8
9 .91 .92 .93 .94 .95 .96
.97 .98 .99 .9a .9b .9c
.9d .9e .9f 9
a ¡ ¢ £ ¤ .95 .a6 §
¨ © .99 « ¬ != ®
¯ a
b ? <= >= ´ [Micro] ? <sum> ? <pi> ? » .bc ?
.be ¿ b
c Á Â ? .83 ? ? Ç
È .85 [NonBreakingSpace] Ë
Ì Í .8c .9c c
d .97 .91 Ô Ö ? Ø .9f / Ä 
.8b
.9b fi ß d
e á .82 .84 .89 å æ
ç è é ê
ë ì í î
ï e
f ? ò ó ô ? .88 .98
ø <breve> ? ? ü «« ? 
<caron> f
... 0 1 2 3 4 5 6 7 8 9 a b c d e f
===
Subject: Re: [test] character set
> ... 0 1 2 3 4 5 6 7 8 9 a b c d e f
> 2 ! # $ % & Ô ( ) * + , - . / 2
> 3 0 1 2 3 4 5 6 7 8 9 : ; < = > ? 3
> 4 @ A B C D E F G H I J K L M N O 4
> 5 P Q R S T U V W X Y Z [ ] ^ _ 5
> 6 ` a b c d e f g h i j k l m n o 6
> 7 p q r s t u v w x y z { | } ~  7
> 8 Ä .81 .82 .83 .84 .85 .86 .87 .88 .89 .8a .8b
.8c .8d .8e .8f 8
> 9 .90 .91 .92 .93 .94 .95 .96 .97 .98 .99 .9a .9b
.9c .9d .9e .9f 9
> a [NonBreakingSpace] Á ¢ £
Û ´ .a6 ¤ ¬ © »
Ç Â != ¨ ø a
> b ? ± <= >= « [Micro] ? <sum> ? <pi> ?
È .bc ? .be À b
> c Ë ç å ?
Ä ? ? .82 é
.83 è í ê
ë ì c
> d [CapitalEth] .84 ï Ô
.85 ? ¯ ô /
ó .86 [CapitalYAcute] fi ß d
> e .88 .87 .89 .8b .8a .8c
.be .8d .8f .8e .90 .91 .93
.92 .94 .95 e
> f ? .96 .98 .97 .99 ? .9a
Ö ¿ <breve> ? ? .9f
ÇÇ ? <caron> f
> ... 0 1 2 3 4 5 6 7 8 9 a b c d e f
General rule of thumb: rows 8-f may not display
correctly/consistently.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: integrating (sin x)^(1/3)
> Can someone remind me how this indefinite integral is 
solved?
Why do you think there is a nice formula for it?
Maple gave me this:
3/4*sin(x)^(4/3)*hypergeom([1/2, 2/3],[5/3],sin(x)^2)
Or, change variables t=sin(x) to get:
integral t^(1/3) dt/sqrt(1-t^2)
===
Subject: Re: integrating (sin x)^(1/3)
>Why do you think there is a nice formula for it?
>Maple gave me this:
> 3/4*sin(x)^(4/3)*hypergeom([1/2, 2/3],[5/3],sin(x)^2)
>Or, change variables t=sin(x) to get:
> integral t^(1/3) dt/sqrt(1-t^2)
I never took Complex Analysis and wondered if C.A. was
required (or
helpful) in solving that integral.
===
Subject: Re: what boolean connectives together with -> make a
complete
system?
Where are the expected muraders of anything that novices say,
such as me,
express.
Ullrich, G. Frege, Fransen, ..where are you complaints?
Surely , I didnt say something correct, in your view.
> I. (nor)
> 1.~p =df (p nor p)
> 2. p v q =df ~(p nor q)
> 3. _T =df p v ~p
> 4. F =df ~T
> 5. p & q =df ~(~p v ~q)
> 6. p nand q =df ~(p & q)
> 7. p -> q =df ~p v q
> 8. p -|-> q =df ~(p -> q)
> 9. p <-> q =df (p -> q)&(q -> p)
> 10. p xor q =df ~(p <-> q)
> II. (nand)
> 1. ~p =df (p nand p)
> 2. p & q =df ~(p nand q)
> 3. F =df p & ~p
> 4. T =df ~F
> 5. p v q =df ~(~p & ~q)
> 6. p nor q =df ~(p v q)
> 7. p -> q =df ~p v q
> 8. p -|-> q =df ~(p -> q)
> 9. p <-> q =df (p -> q)&(q -> p)
> 10. p xor q =df ~(p <-> q)
> III. (T, -|->)
> 1. ~p =df T -|-> p
> 2. F =df ~T
> 3. p -> q =df ~(p -|-> q)
> 4. p v q =df ~p -> q
> 5. p nor q =df ~(p v q)
> 6. p & q =df ~(~p v ~q)
> 7. p nand q =df ~(p & q)
> 8. p <-> q =df (p -> q)&(q -> p)
> 9. p xor q =df ~(p <-> q)
> IV. (F, ->)
> 1. ~p =df p -> F
> 2. T=df ~F
> 3. p -|-> q =df ~(p -> q)
> 4. p v q =df ~p -> q
> 5. p nor q =df ~(p v q)
> 6. p & q =df ~(~p v ~q)
> 7. p nand q =df ~(p & q)
> 8. p <-> q =df (p -> q)&(q -> p)
> 9. p xor q =df ~(p <-> q)
> V. (~, v)
> 1. T =df p v ~p
> 2. F =df ~T
> 3. p nor q =df ~(p v q)
> 4. p -> q =df ~p v q
> 5. p -|-> q =df ~(p -> q)
> 6. p & q =df ~(~p v ~q)
> 7. p nand q =df ~(p & q)
> 8. p <-> q =df (p -> q)&(q -> p)
> 9. p xor q =df ~(p <-> q)
> VI. (~, &)
> 1. F =df p & ~p
> 2. T =df ~F
> 3. p nand q =df ~(p & q)
> 4. p v q =df ~(~p & ~q)
> 5. p nor q =df ~(p v q)
> 6. p -> q =df ~p v q
> 7. p -|-> q =df ~(p -> q)
> 8. p <-> q =df (p -> q)&(q -> p)
> 9. p xor q =df ~(p <-> q)
> VII. (~, ->)
> 1. T =df p -> p
> 2. F =df ~T
> 3. p -|-> q =df ~(p -> q)
> 4. p v q =df ~p -> q
> 5. p nor q =df ~(p v q)
> 6. p & q =df ~(~p v ~q)
> 7. p nand q =df ~(p & q)
> 8. p <-> q =df (p -> q)&(q -> p)
> 9. p xor q =df ~(p <-> q)
> > like NOT and OR is complete.
> >
> > NOR is complete
> >
> > -> and ?
> >
> > Herc
> >
> > --
> > YOU CANT PROVE ME
> > If you prove its true then it has a proof, which makes
it false
> > If you dont prove it, then its true
> > 10,000 people in sci.math ALL believe this is irrefutable
that
> mathematics will always be incomplete.
> >
> >
===
Subject: Re: Combinatorial design?
posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs
Latin Squares gave me a lot of information to sort through,
as did
the Sloane page.
- Randy
===
Subject: Re: Combinatorial design?
posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs
Latin Squares gave me a lot of information to sort through,
as did
the Sloane page.
- Randy
===
Subject: Re: Combinatorial design?
posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs
Latin Squares gave me a lot of information to sort through,
as did
the Sloane page.
- Randy
===
Subject: Re: Combinatorial design?
posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs
Latin Squares gave me a lot of information to sort through,
as did
the Sloane page.
- Randy
===
Subject: Generalized Quantifiers
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBLKP1j15303;
Dave L. Renfro <renfr1dl@cmich.edu>
http://mathforum.org/discuss/sci.math/m/664050/664484
> In the past few years Ive grown fond of using the
> prefix co as in co-meagerly, co-measure zero,
> co-countable, etc. A lot of the kinds of results
> Im interested in can be described nicely using
> quantifiers modulo some notion of smallness:
<snip rest>
Im starting a new thread because there might be some
people interested in this who wouldnt see it if I
continued to post in the sci.math thread What is meant
by a meager subset??.
Continuing this theme, and because it will give me
an excuse to continue putting off grading final exams
for my remaining class, let
v-oo mean there exists infinitely many
and
^-oo mean for all but finitely many
In non-ASCII writing the generally accepted notation
is to superscript with the infinity symbol oo the
usual symbols for there exists (backwards capital E)
and for all (upside down capital A).
These are the there exists and for all quantifiers
(which Ill use Ôv and 
Ô^ for) modulo the smallness
notion finite.
Note that negation ~ distributes through these new
quantifiers the same way it distributes through v and ^:
(~)(v-oo) is the same as (^-oo)(~)
and
(~)(^-oo) is the same as (v-oo)(~)
It follows that the negation of a sequence of
such quantifiers can be rewritten the same way
that the negation of a sequence of ordinary
quantifiers can be rewritten, namely switch all
the vs to ^s and all the ^s 
to vs and then
take the negation of the right-most expression.
We can often use these new quantifiers to give
shorter definitions, such as
x_n converges to L can be expressed as
(^ e>0)(^-oo n)(|x_n - L| < e).
The negation of this can be carried out formally
in the way I described above to get
x_n does not converge to L expressed as
(v e>0)(v-oo n)(|x_n - L| geq e).
For another example, the lim-inf of a sequence {A_n}
of sets is {x: (^-oo n)(x in A_n)} and the lim-sup of
the sequence {A_n} is {x: (v-oo n)(x in A_n)}.
In these and in other ways, I found the v-oo and ^-oo
quantifiers quite useful in a graduate real analysis
class I taught three years ago.
For the ordinary quantifiers we have the following
logical strength chart (no implication can be
reversed in general):
(^)(^) ==> (v)(^) ==> (^)(v) ==> (v)(v)
The analogous chart also holds for v-oo and ^-oo.
I havent investigated the logical relationships for
sequences involving these four quantifier types, but I
have noticed that ^-oo doesnt commute with ^ (and hence
by considering negations, v-oo doesnt commute with v).
In particular, (^-oo)(^) is strictly stronger logically
than (^)(^-oo). For example, note that for all real numbers
x we have for all but finitely many integers n that x < n
holds is true, but for all but finitely many integers
n we have for all real numbers x that x < n holds is false.
The issue is that in order for (^-oo r)(^ s) to be true,
we need the existence of a co-finite collection of 
rs
that uniformly works for each s. In fact, the distinction
between (^-oo)(^) and (^)(^-oo) is really a (v)(^) verses
(^)(v) distinction (see [*]) if you look at things the
right way. Let ÔC be a variable that runs over 
the set
of co-finite collections of rs. Then (^-oo r)(^ 
s)
becomes (v C)(^ r in C)(^ s), which is equivalent to
(v C)(^ s)(^ r in C) (see [#]), while (^ s)(^-oo r) becomes
(^ s)(v C)(^ r in C).
[*] Note that (v)(^) verses (^)(v) is what makes uniform
continuity and uniform convergence (of functions)
different from continuity and pointwise convergence.
The distinction is also important if you want to
correctly state the identity and inverse axioms
for a group.
[#] Recall that ordinary ^s (and vs as well) 
commute
among themselves. That is, (^ P)(^ Q) is logically
the same as (^ Q)(^ P).
I havent tried to develop these ideas into a general
framework (quantifiers modulo various notions of smallness,
and how they logically relate to each other relative
to how the various notions of smallness relate to each
other, besides trivial things like the larger the notion
of smallness, the stronger and weaker the corresponding
versions of v and ^ are, respectively, with no difference
between them at all when the set being quantified over is
small), nor do I expect to in the future. However, below
are a few literature references Ive come across that seem
as if they might be relevant if anyone wants to look into
these ideas further.
[1] Jon Barwise, An introduction to first-order logic,
pp. 5-46 in Jon Barwise (editor), HANDBOOK OF
MATHEMATICAL LOGIC, Studies in Logic and the
Foundations of Mathematics #90, North-Holland, 1977.
[See Section 5.5 (Logic with new quantifiers) on
pp. 44-45.]
[2] Johan van Benthem, Questions about quantifiers,
Journal of Symbolic Logic 49 (1984), 443-466.
[3] Alexander S. Kechris, CLASSICAL DESCRIPTIVE SET
THEORY, Graduate Texts in Mathematics #156,
Springer-Verlag, 1995. [See p. 53 (meager versions)
and p. 114 (measure zero versions).]
[4] H. Jerome Keisler, Logic with the quantifier 
Ôthere
exist uncountably many, Annals of Mathematical
Logic 1 (1970), 1-93.
[5] Andrzej Mostowski, On a generalization of quantifiers,
Fundamenta Mathematicae 44 (1957), 12-36.
[6] Dag Westerstahl, Self-commuting quantifiers, Journal
of Symbolic Logic 61 (1996), 212-224.
http://www.maths.usyd.edu.au:8000/s/search/p?p1=generalized+
quantifiers
Dave L. Renfro
===
Subject: Re: radicals radicals radicals
A generalisation would be:
Suppose a,b,c,d be in Q+ (set of positive rational numbers)
such that a, b,
c, d are not equal to zero;
Let x be in R (set of real numbers) such that x is not equal
to zero.
case x > 0 :
(*) x ^ ( a/b ) > 0
case x < 0 :
(*) x ^ ( a/b ) > 0 if and only if one of r and d is multiple
of 2, where
r and d are relatively prime and r/d = a/b ;
So x ^ (a/b * c/d) = - (|x| ^ (a/b) ) ^ (c/d) if r and d are
not multiples
of 2, where r and d are relatively prime and r/d = a * c/ (b
* d) ;
> Let a,b be in Q ( set of rational numbers), b is not equal
to zero.
> Let x be in R ( set of real numbers)
> x ^ (a/b) = x ^ ( 2 * a/(2b)) = (x ^ 2) ^ (a/2b) = | x ^
(a/b) |
> for example,
> let x = -27; a = 1; b = 3;
> (-27) ^ (1/3) = (-27) ^ ( 2 * 1/6) = ((-27) ^ 2) ^ (1/6) =
729 ^ (1/6) =
> 3;
> But (-27) ^ (1/3) = -3;
> there should be some mistake i did,
> suppose a,b,c,d be in Q such that b,d are not equal to zero;
> let x be in R;
> x ^ ( (a/b) * (c/d) ) is not always equal to ( x ^ ( a/b) )
^ (c/d)
> is that true?
> in mathworld.com ,
http://mathworld.wolfram.com/SquareRoot.html , they
> talk about principal square root, but anyway thats only in
the case of
> square root.
===
Subject: Inertial Forces Newton -> Einstein -> Shipov?
Background to inertial forces in Einsteins GR ~ (LC)
connection field
and in Gennady Shipovs extension of GR to include possibly
torsion fields:
Review of the concept of inertial forces in Newtons
kinematics of
non-inertial frames of reference in Galilean relativity of
absolute
simultaneity when v/c << 1.
Let the rotating non-inertial frame S have pseudo-vector W
instant
angular velocity
The rate of change in the rotating frame S is the 
convective
derivative
linear operator on 3-vectors
...,t = ...,t + Wx...
where ,t is the rate of change in the inertial frame S.
r,t = r,t + Wxr = instant velocity
r,t,t = r,t,t + Wxr,t + W,txr + Wxr,t + 
WxWxr = r,t,t +
2Wxr,t + WxWxr
+ W,txr
2Wxr,t = Coriolis inertial force per unit test mass in the
non-inertial
rotating frame
WxWxr = Centrifugal inertial force per unit test mass in the
non-inertial rotating frame
W,txr is the inertial torque per unit test mass
Or in terms of Newtons F = ma
unprimed quantities are in the inertial frame, primed
quantities are in
the rotating non-inertial frame sharing common origin.
v = v + Wxr
a = a + 2Wxv + WxWxr + W,txr
Note that the Coriolis inertial force in the rotating frame
is a bit
like the Lorentz magnetic force on a charge e with W like
magnetic field
out of the
problem for all inertial forces just like with gravity.
SPECIAL CASE of above
If the rotating frame is also the instant rest frame of the
point test
v = 0
Therefore v = -Wxr
a = 0
Assume also no torque so that W,t = 0
Therefore
a + Wxv = 0
Or
a - WxWxr = 0
That is inertial force compensation in the REST LNIF of the
test
g = a = WxWxr
Of course, the inertial geodesic motion is a straight
Euclidean line
In Newtons theory there is actually an external gravity
force and you
can have, for example
GM/r^2 = |WxWxr|
GM = W^2r^3 = r^3/T^2
T = period of orbit. This is Keplers law of planetary
motion, M is mass
of Sun.
In Newtons Force theory the motion of the planets is
NON-GEODESIC in
ßat 3D space.
This world picture paradigm changes completely in Einsteins
force
without Force theory AKA geometrodynamics where now the
motion of the
planets is inertial i.e. geodesic in 4D space-time i.e.
straightest
4D world line possible in curved spacetime.
The space-time curvatures in the above example of Keplers
law are made
from
1/r*^2 = GM/c^2r^3
where r* is the scale of local radii of curvature of
space-time at
distance r in weak field limit GM/c^2r << 1.
===
Subject: Re: FORGET ABOUT TRUE IN A SYSTEM
>>all statements
>>are either TRUE of FALSE.
>But that is false.
>The statement
>This statement is false.
>is a counterexample.
>Abandon the law of the excluded middle, which is true only
for
>first-order statements, and there are no paradoxes.
Actually, the Liar Paradox occurs with full force in
first-order logic, if
there are (1) axioms that define some means of internally
representing the
syntax of the logics formulas (e.g., Goedel numbering), and
(2) a predicate
ÔTrue that defines the logical 
truth of every formula
represented by its
syntactic representation.
And it is *not* the Law of Excluded Middle (P v ~P) that the
Liar relies on,
but rather the Law of Non-Contradiction (~(P ^ ~P)), as
follows:
Given the predicate ÔTrue and the 
Self-Reference Lemma (proof
omitted here), there must be a syntactic term 
Ôt such that:
t = [~True(t)]
where [phi] means the syntactic encoding of the formula phi.
Since True([phi]) <-> phi (by the definition of logical
Truth), we
have:
True(t) <-> True([~True(t)]) <-> ~True(t)
which violates Non-Contradiction (but not excluded middle).
Note that the Liar also occurs in first-order logics that LACK
Excluded
Middle, for example 3-valued logics with True / False /
Paradoxical, where
the appropriate Liar sentence is This statement is false or
paradoxical.
>John Savard
>http://home.ecn.ab.ca/~jsavard/index.html
--
---------------------------
| BBB b Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
-----------------------------
===
Subject: Re: FORGET ABOUT TRUE IN A SYSTEM
>>all statements
>>are either TRUE of FALSE.
>But that is false.
NO LOCAL VARIABLES
EVERYTHING IS GLOBALLY TRUE OR FALSE, or neither. <<<<<<
Thats not the issue, the issue is that 99% of the time we
DONT issue
a frame of reference. f = f in what system is that false? Ex?
The issue is that 99% of proofs we take for the usual
interpretations of
the terms
as the system. SYSTEM is effectively a void term
THERE IS NO SYSTEM. Physisists are still arguing over wrt
frame of
reference
either VS no-either, who cares? YOU CANT backpeddle your way
out from
a statement saying IT HAS NO GLOBAL VALUE. That is nonsense
and
it is NEVER a facet of mathematics at all to subdivide
knowledge, system
was
specifically introduced into mathematics to cover up this
biggest oxymoron
in history :
THE UNDISPUTABLE PROOF OF this has proof
What SYSTEM are you actually proving it in? CANT YOU READ 
IT
IN THAT
SYSTEM?
>The statement
>This statement is false.
>is a counterexample.
>Abandon the law of the excluded middle, which is true only
for
>first-order statements, and there are no paradoxes.
> Actually, the Liar Paradox occurs with full force in
first-order logic,
if
> there are (1) axioms that define some means of internally
representing
the
> syntax of the logics formulas (e.g., Goedel numbering),
and (2) a
predicate
> ÔTrue that defines the logical 
truth of every formula
represented by its
> syntactic representation.
> And it is *not* the Law of Excluded Middle (P v ~P) that
the Liar relies
on,
> but rather the Law of Non-Contradiction (~(P ^ ~P)), as
follows:
> Given the predicate ÔTrue and the 
Self-Reference Lemma
(proof
> omitted here), there must be a syntactic term 
Ôt such that:
> t = [~True(t)]
> where [phi] means the syntactic encoding of the formula phi.
> Since True([phi]) <-> phi (by the definition of logical
Truth), we
have:
> True(t) <-> True([~True(t)]) <-> ~True(t)
> which violates Non-Contradiction (but not excluded middle).
> Note that the Liar also occurs in first-order logics that
LACK Excluded
> Middle, for example 3-valued logics with True / False /
Paradoxical,
where
> the appropriate Liar sentence is This statement is false or
paradoxical.
This is where the current deplorable mass hypnosis of
mathematics stems
from.
What *is* this formula?
> t = [~True(t)]
Invalid?
Incorrect?
Syntactically wrong?
False?
This happens in nearly every proof of the impossible there is.
Find some subset relationship to the fundamental property of
the object.
-------------------------------------------------------------
---------
Here it is TRUE STATEMENT C PROOF OF STATEMENT
Now apply the liars paradox using the more general case.
true(statement) = ~proof(statement)
X = ~X
VOILA, its not a paradox because proof and true are spelt
differently,
Its not an ill defined formula, its a NEW GATEWAY INTO
INFORMATION THEORY.
Exactly the same method with Halt.
ProgramGivesOutput C ProgramHalts
Output(Program) = ~Halt(Program)
X = ~X
ANOTHER GATEWAY INTO INFORMATION THEORY
Cantor
DigitOnDiag C DigitsOnNumber
Number(digit) = ~Diag(digit)
X = ~X
No its not a paradox, its not an ill defined formula,
its a NEW GATEWAY INTO INFORMATION THEORY.
Busy Beaver
ProgramsMaximumOutput C ProgramsOutput
ProgramsOutput > ProgramsMaximumOutput
X > X
Another impossibility.
Russel.
Set C Member (some members are sets themselves)
Member = ~ Set(containsItself)
X = ~X
Another liar statement using the general case, the only one
that is also
recognised as a paradox.
Cant talk a grain of sense into any of you. 
Infinite people
toss coins,
all the combinations
are taken to infinite length... diag is a new sequence? LOL
Herc
===
Subject: Linear algebra question
Is possyble to represent the follow expression
a[0]*b[n]-a[1]*b[n-1]+a[2]*b[n-1]+...=sum((-1)^i*a[i]*b[n-i],
i=0..n)
as determinant or treck of suitable matrix?
|a[0] a[1]|
for sample if n=1 then a[0]*b[1]-a[1]*b[0]=det|b[0] b[1]|,
what about n>1?
===
Subject: Re: Cantors diagonal proof wrong?
>
>> Prof. Dr. Mueckenheim, I disagree with you. First, the
universe is
>> infinite.
>
> It has not yet been deided whether the universe is finite or
not. It is
> expanding. But its actual shape doesnt matter. The 
universe
> accessible to us, is finite and will remain so forever.
> Forever? So the universe is infinite in temporal extent?
> Q.E.D.
> John Briggs
Well, no one yet has been able to experience more than a
temporally
finite amount of it.
===
Subject: Re: JSH: Funding, real world, not fantasy
<d5hds0p3748d608mk7i77i93179f9ot8oq@4ax.com>
<874qigcgvl.fsf@phiwumbda.org>
<x5acs7mqgj.fsf@lola.goethe.zz> <87d5x3o379.fsf@phiwumbda.org>
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(
5eZ41to5f%E@ELIi
$t^
VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
>> Have you ever seen a restraining order?
> Ill show you mine if you show me yours.
Hey, calling the bluff is not fair. Anyway, Ive heard some
of the
actual contents for a restraining order from some guy that
was so
unbelievably obnoxious that he got his student internet
access yanked
completely a few months after he first turned up in some
Usenet group
to steam off about that bad woman who had the restraining
order put on
him. Keep off the grounds of some building with a distance of
at
least so-and-so, not call a person and a list of others,
stuff like
that.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Recursively inseparable sets
Let alpha_2(i,i) be a bijection from NxN to N and
(pi_2^1(x),pi_2^2(x))
be its invert
A = {n in N ; x in W_{pi_2^1(x)}}
B = {n in N ; x in W_{pi_2^2(x)}}
A={x in A ; exists d [T(a,x,d) wedge (forall u < d) neg
T(b,x,u)]}
B={x in B ; exists d [T(b,x,d) wedge (forall u leq d) neg
T(a,x,u)]}
where a is an indice for A and b is an indice for B. T is
Kleene termination
predicate: T(i,x,t) is valid iff the Turing machine of code i
is stopped at
time t on input x.
How to prove that A and B are recursively 
inseparables ?
===
Subject: Re: Oh no! Monty Hall problem again.....
<7esqk016rvbiachpil1omqam5lja8thtv6@4ax.com>
<et3sk0tko2ks7hdsktslb2usd5gr5gleb1@4ax.com>
posting-account=M503cA0AAACGX_0VYkUgNRzZZmEkwoYy
>You talk the same way as the original poster. Are you the
>same?
> No, but I do copy-paste. :)
> You initially pick the chicken and win. (one)
> You initially pick the goat and win. (two)
> You initially pick the car and lose. (three)
> However, if you truly wish to split up the last case, be my
guest
> (although it only complicates things).
> Its the only way to solve the problem asked by the
original poster.
> p = probabilty of car behind door 1.
> q = probabilty of car behind door 2.
> 1-p-q = probabilty of car behind door 3.
> The contestant randomly picks a door (which is his best
strategy,
> seeing he knows nothing about p and q). The probability
that he
picks
> the door with the car behind it is 1/3. If he doesnt
switch, he
has a
> chance of 1/3 of winning the car.
> Lets stick to the Monty hall problem where p=q=1-p-q or 
at
the
> begginng of the game a car is randomly placed behind a door.
> thats confusing enough.
>Its undefine
> No.
>>Do you mean if Conan makes the same decision as the original
player
>>would have made? Switch?
>> The OP states quite clearly that Conan does not have that
information.
Then how do you know that Conan will choose each door with 1/2
probability?
> It is the best strategy he can take given the information
he has.
Of
> course, Conan is a barbarian, which makes it a bit harder.
:)
> However, one of those things that is unknown about Conan is
that he
> has taken a probability course (to have a greater chance of
ducking
> arrows). He will take the best strategy.
>He could choose door 1 100% of time.
> Yes, but that wouldnt be smart.
> m = probability that car is behind left door.
> 1-m = car behind right door.
> s = probability that Conan will pick left door.
> 1-s = Conan picks right door.
> The chance that Conan wins is (if the game ends directly
after he
> chooses a door):
> s*m + (1-s)*(1-m) = s*m + 1 - s - m + s*m
> = 1 + 2*s*m - s -m
> m can be anything between 0 and 1. Suppose m=0:
> 1 + 2*s*m - s -m = 1-s
> So for Conan s=0 will be optimal. Unfortunately for Conan m
might
as
> well be 1. In that case s=0 will be terrible. The fact that
m is
> unknown to Conan, leads Conan to the conclusion that he has
to pick
s
> in such a way that his chance of winning is maximised in any
> situation. That happens for s=1/2.
> How does Conan know m. He only has once chance to choose.
> 1 + 2*s*m - s -m = 1+m-1/2-m = 1/2
> You cant even define the chance
>of success for Conan.
> I cant? I just did. Conans succes rate is 
1/2.
> You assumed Conan would determine the optimium strategy
after
> telling me that Conan has no information. Why, by the way is
> his optimum strategy to win half the time. If the probablity
> for one of the two remaining doors is m and the other is
1-m, he
> should determine whichever is greater 1-m or m and pick
that door.
>Yes, Marylin made this same mistake with her aliens
argument. She
was
>wrong. When the aliens land after a door is eliminated their
chance
>of succes chosing randomly is 50% just like the chances of
success
for
>the contestant would be if he had originally decided to
switch 50%
>of the time before the door was removed or after the door was
removed.
>But the chances of success for the switch door does not
change
for the
>aliens just because they did not know through mental
telepathy
what
>Monty was thinking.
> What is the chance of success for a door? And why is it
something
that
> is connected to the door? Wont the original contestant,
Conan/aliens
> and Monty all have a different view of what the chance for
success
of
> a door is? Please explain what you are thinking as to me
you only
seem
> confused.
> Remember we are playing the monty hall game where the car
is randomly
> distributed. p=q=1/3 =1-p-q=1/3
> The original question seemed to be, If the contestant picks
door 1
> and decides to switch no matter what, before a door is
open, what are
> the chances of success?. Then if after a door is opened,
either door
2
> or 3,
> and the contestant thinks about it again and reaffirms that
he would
> like
> to switch what are the chances of success, the same? Then
suppose
> after
> a door is opened Conan suddenly rides in, jumps off his
horse, pushes
> the contestant off the stage and chooses a door. What are
the chances
> of success?.
> The answer is:
> Before the door is opened the chance of success is 2/3.
> After a door is opened the chance of success may change,
depending
> on p, which is the conditional probablity that Monty opens
that door
> given that a car is behind the first door. In other words,
if p=.7
for
> door 2 or Monty opens door 2 70% of the time a car is
behind door 1,
> then once Monty opens door 2, the chance of success if the
contestant
> switches is no longer 2/3.
> It is 1/1+p = 1/1.7
> Now if p=.7 and Monty opens door 2 and then Conan rides in,
Conans
> chance of success for switching (which is choosing door 3
instead
> of door 1) are the same as the contestants. 
Conans chance
of
success
> in general is not known because, after all who knows how
often Conan
> will choose door 3, if door 2 is opened.
===
Subject: Re: Oh no! Monty Hall problem again.....
<7esqk016rvbiachpil1omqam5lja8thtv6@4ax.com>
<et3sk0tko2ks7hdsktslb2usd5gr5gleb1@4ax.com>
posting-account=M503cA0AAACGX_0VYkUgNRzZZmEkwoYy
>You talk the same way as the original poster. Are you the
>same?
> No, but I do copy-paste. :)
> You initially pick the chicken and win. (one)
> You initially pick the goat and win. (two)
> You initially pick the car and lose. (three)
> However, if you truly wish to split up the last case, be my
guest
> (although it only complicates things).
> Its the only way to solve the problem asked by the
original poster.
> p = probabilty of car behind door 1.
> q = probabilty of car behind door 2.
> 1-p-q = probabilty of car behind door 3.
> The contestant randomly picks a door (which is his best
strategy,
> seeing he knows nothing about p and q). The probability
that he
picks
> the door with the car behind it is 1/3. If he doesnt
switch, he
has a
> chance of 1/3 of winning the car.
> Lets stick to the Monty hall problem where p=q=1-p-q or 
at
the
> begginng of the game a car is randomly placed behind a door.
> thats confusing enough.
>Its undefine
> No.
>>Do you mean if Conan makes the same decision as the original
player
>>would have made? Switch?
>> The OP states quite clearly that Conan does not have that
information.
Then how do you know that Conan will choose each door with 1/2
probability?
> It is the best strategy he can take given the information
he has.
Of
> course, Conan is a barbarian, which makes it a bit harder.
:)
> However, one of those things that is unknown about Conan is
that he
> has taken a probability course (to have a greater chance of
ducking
> arrows). He will take the best strategy.
>He could choose door 1 100% of time.
> Yes, but that wouldnt be smart.
> m = probability that car is behind left door.
> 1-m = car behind right door.
> s = probability that Conan will pick left door.
> 1-s = Conan picks right door.
> The chance that Conan wins is (if the game ends directly
after he
> chooses a door):
> s*m + (1-s)*(1-m) = s*m + 1 - s - m + s*m
> = 1 + 2*s*m - s -m
> m can be anything between 0 and 1. Suppose m=0:
> 1 + 2*s*m - s -m = 1-s
> So for Conan s=0 will be optimal. Unfortunately for Conan m
might
as
> well be 1. In that case s=0 will be terrible. The fact that
m is
> unknown to Conan, leads Conan to the conclusion that he has
to pick
s
> in such a way that his chance of winning is maximised in any
> situation. That happens for s=1/2.
> How does Conan know m. He only has once chance to choose.
> 1 + 2*s*m - s -m = 1+m-1/2-m = 1/2
> You cant even define the chance
>of success for Conan.
> I cant? I just did. Conans succes rate is 
1/2.
> You assumed Conan would determine the optimium strategy
after
> telling me that Conan has no information. Why, by the way is
> his optimum strategy to win half the time. If the probablity
> for one of the two remaining doors is m and the other is
1-m, he
> should determine whichever is greater 1-m or m and pick
that door.
>Yes, Marylin made this same mistake with her aliens
argument. She
was
>wrong. When the aliens land after a door is eliminated their
chance
>of succes chosing randomly is 50% just like the chances of
success
for
>the contestant would be if he had originally decided to
switch 50%
>of the time before the door was removed or after the door was
removed.
>But the chances of success for the switch door does not
change
for the
>aliens just because they did not know through mental
telepathy
what
>Monty was thinking.
> What is the chance of success for a door? And why is it
something
that
> is connected to the door? Wont the original contestant,
Conan/aliens
> and Monty all have a different view of what the chance for
success
of
> a door is? Please explain what you are thinking as to me
you only
seem
> confused.
> Remember we are playing the monty hall game where the car
is randomly
> distributed. p=q=1/3 =1-p-q=1/3
> The original question seemed to be, If the contestant picks
door 1
> and decides to switch no matter what, before a door is
open, what are
> the chances of success?. Then if after a door is opened,
either door
2
> or 3,
> and the contestant thinks about it again and reaffirms that
he would
> like
> to switch what are the chances of success, the same? Then
suppose
> after
> a door is opened Conan suddenly rides in, jumps off his
horse, pushes
> the contestant off the stage and chooses a door. What are
the chances
> of success?.
> The answer is:
> Before the door is opened the chance of success is 2/3.
> After a door is opened the chance of success may change,
depending
> on p, which is the conditional probablity that Monty opens
that door
> given that a car is behind the first door. In other words,
if p=.7
for
> door 2 or Monty opens door 2 70% of the time a car is
behind door 1,
> then once Monty opens door 2, the chance of success if the
contestant
> switches is no longer 2/3.
> It is 1/1+p = 1/1.7
> Now if p=.7 and Monty opens door 2 and then Conan rides in,
Conans
> chance of success for switching (which is choosing door 3
instead
> of door 1) are the same as the contestants. 
Conans chance
of
success
> in general is not known because, after all who knows how
often Conan
> will choose door 3, if door 2 is opened.
===
Subject: Re: differential calculus problem...
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBLLx0023351;
>in a triangle ABC angleB = 90 , AB+AC=30cm.when area of
triangle ABC
>is maximum,find angleA
>pls hlp........
30 cm is irrelevant. Let a=AB, c=AC. This is a right triangle
where one
leg is a and the hypotenuse is c. The area(X) is given by
X=.5a(c^2-a^2)^.5
Since a+c=k (constant)
X=.5a((k^2-2ak)^.5
To get max for X, let dX/da=0. Then a=k/3 and c=2k/3.
Therefore cos(A)=a/c=.5 and A=60 deg.
===
Subject: Re: .99999... still=/= 1
>could
>> reach the speed of light and cause .999... = 1, which
would make,
>This has nothing to do with dividing by zero. Besides there
is only a
>finite amount of energy in the entire physical cosmos. So the
upperbound
>Bob Kolker
Mathematics can lead a person to fairy tale land. Its
important to try
your
best to see what science shows with experimental evidence,
THEN math can
be
applied to it. We can see in physical reality that .999...
cant = 1
because
it cant reach it. This is proven in reality.
I noticed you snipped the part about the computer program
never reaching
1
from .999... , but anyway we can see in reality,
1) the tooth fairy doesnt exist
2) the fairy godmother doesnt exist
and
3) .999... = 1
Math trickery and math illusions was used in the development
of the
Periodic Table of Elements, too. They made believe that
electron orbitals
existed and then made a math story of how it could actually
work. But
reality
is beginning to show us with real atomic images that
electrons dont show
orbitals but intelligence logic structures. The Smart Model
predicted this
about 10 years before they knew it.
I have shown you repeatedly ..., that everything even in
math, and in
physical reality shows ust that,
.999... =/= 1
If you want to believe .999... = 1, its up to you, but it
doesnt.
You may ask how can light reach
3.0 X 10^8 m/sec when matter cant. This is because photons
start at a
unified
state or oneness, of this amount of energy. It doenst start
with .999...
and
then reaches 1. It is 1, that is 1 with itself at that energy
state, to
start
with.
Can someone else be you? No, because you are 1. You are you.
Others may
try
to reach you .999... ---> 1 you, but you will always be
unique, and only
you.
An exact clone of you can never be made, because .999...
cant reach 1. You
can
make close clones of something ( .999...), but not the exact
clone, 1.
Smarts Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
posting-account=EH2x8QsAAABu84CuyjstkC4nRyQ1ZHKW
decimally, the speed of light is 1.0000... lightyears/year.
> Mathematics can lead a person to fairy tale land. Its
important to
try your
--4my of it?
http://www.wlym.com
===
Subject: Re: .99999... still=/= 1
posting-account=EH2x8QsAAABu84CuyjstkC4nRyQ1ZHKW
I mean, *exactly* 1.0000... lightmoments per moment,
but only in vacuo,
which never actually occurs.
--4my of it?
http://www.wlym.com
===
Subject: Re: .99999... still=/= 1
> You may ask how can light reach
> 3.0 X 10^8 m/sec when matter cant
do photons move at light speed (in vacuo), they cannot move
at any other
speed.
energy will take it to light speed. There is only a finite
amount of
energy in the kosmos so it is impossible for any massive body
to move at
light speed.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
(snip)
> I dont know what you mean by an infinite area 
or by a
placeholder.
Perhaps it does not help to read my emails carefully. :-(
(snip)
>> OK, lets try again. First of all, I think we should 
start
a new thread
>> from
>> the beginning so that everything is constructed and we use
the same
>> concepts
>> and definitions.
>> 1) ...999 is not N as it is not in a classic way finite
integer. Lets
>> call
>> it infinite integer (N_inf) over one infinity.
> As far as I am concerned, ...999 does not mean anything at
all. It
has
> no connection with NSA or anything else as far as I can
see. You keep
> throwing that string around as if you think it has a
meaning, but you
> have never said what that meaning is.
If You and I can define it, as I have hopefully done, then
there is a
meaning and a connection to the existing mathematics!
> I am not aware of any useful way of representing the
members of *N as
> decimal digit strings. Any scheme I can think of suffers
from one of two
> defects: either there are members of *N that dont 
fit into
the naming
> scheme at all, or the coding scheme is so cryptic that you
cant even
> compare the sizes of two numbers just by looking at the
digits.
It think there is now something misundertanding, of course I
take the
reason. *N is (infinite) integer part, what ever You like to
call it. I have
a feeling that we have to start from the premices.
> Perhaps you can think of some scheme that I have
overlooked, but you will
> have to define your terms very carefully before I will be
convinced.
>> 2) I define it as I have done earlier sum (k 0 --> oo)
9*10^k.
> That sum diverges, even in NSA. It is not a number.
Yes, from Your point of view, because You have not recognized
that this
diverging sum indeed have a limit as You consider a new point
of reference.
As You or I write 9/10^-k or as I write 9*10^k, then k (as
integer in N or
*N) refers to some point of reference. Sorry, if this is so
difficult.
>> It does not
>> have classic limit as the k refers now the standard point
of reference,
>> but
>> it has a limit as you hopefully noticed as we have another
point of
>> reference.
> There you go with your point of reference again. Sorry, but
you
cant
> use one undefined term to define another.
The point of reference is the counting point of reference.
Everything is
compared to some point of reference.
>> 3) It is defined and it exists, now - how do You like to
name it?
>> Hyperinteger or something else?
>> (snip)
> Its a string of digits. Nothing more. It is not a number
at all, at
> least in any sense that you have yet defined.
I assume, because You cannot see the limit? There is no
number if there is
no limit? Do You think so?
> And now, I would like to ask a counter-question. I will
describe a
> certain member of *N, and I would like you to tell me what
decimal digit
> string you think corresponds to it. First, some background.
> You may recall from other discussions that the (standard)
real numbers
> can be defined as equivalence classes of Cauchy sequences of
rationals.
> That means I can identify a real number by presenting you
with a Cauchy
> sequence, and it is understood that any other sequence that
happens to
> fall into the same equivalence class is an equally good
representation of
> that number. For example, the sequences
> < 9/10, 99/100, 999/1000, ... and
> < 1, 1, 1, 1, ... happen to be two different
representations of the same real number.
Yes, within the concept of the limit calculation. But that is
not true (this
is my point of argumentation, which I have to explain for
You) the sum is
not the same as the limit. The limit is the successor of the
sum. As You do
not - yet - accept my argumentation, I have to explain it
more clear for
You, though my personal opinion is that I have done it
already. Sorry - this
seems to be more complex that I expected. :-(
> Now, lets consider a similar construction that lies at 
the
heart of
> nonstandard analysis. To describe a member of *N, for
example, I can
> present you with a sequence of natural numbers (members of
N). Unlike
> the real-number construction, this one doesnt require the
sequences to
> be Cauchy. Technically, I also need to describe to you the
equivalence
> relation that will be used, but thats a bit more
complicated. It
> involves something called a free ultrafilter on N. You can
find an
> explanation of the concept at
> <http://mathworld.wolfram.com/Ultraproduct.html>.
I have a strong feeling that we are taking about much very
more simpler
things. Your ultrafilters are me very strange matter. 
Im very
sorry. :-(
> For our purposes, its enough to know that if I give you a
sequence in N,
> there is a unique member of *N that is represented by that
sequence. Ok
> so far?
> Here is the sequence I have in mind. Let A(x,y) be the
Ackermann
> Function, as described at
> <http://mathworld.wolfram.com/AckermannFunction.html>. Now,
let a_k =
> A(k,k) for each k. This sequence starts out:
> a_0 = A(0,0) = 1
> a_1 = A(1,1) = 3
> a_2 = A(2,2) = 7
> a_3 = A(3,3) = 2^6 - 3 = 61
> a_4 = A(4,4) = 2^2^2^2^2^2^2 - 3 = (too big to write out
here)
> and after that the sequence starts to grow rather quickly.
:-)
> Let a be the member of *N that is associated with this
sequence. My
> question is:
> (1) what decimal digit string do you think represents a?
> (2) what decimal digit string do you think represents
A(a,a)?
> My point is that your decimal digit strings are woefully
inadequate in
> this context. They cannot even begin to describe the
numbers in *N in
> any useful way.
Huh! Im indeed totaly out! Sorry. I have NOT descibed
something so
complicated like that. I cannot give You an answer. I assume
we have now a
very strange vision what I have told before.
>> (snip)
> In a similar fashion, you keep assuming that I must know
what you mean
> by
> ...999. However, I assure you that I dont.
>> Uh! I have tried to explain so simple as possible. I had
once a fealing
>> that
>> You understood very well.
>> There must be some miscommunication.
> Answer my questions (1) and (2) above, and then well see
whether there
> is any purpose at all in discussing decimal digit strings
in connection
> with *N.
The only possibility is to talk about limit of those
expressions as we
change the point of reference, but I cannot see Your goal.
> (4) It is not possible to sum over N in NSA, because N is
> not a set in NSA.
> Do you agree?
>> I agree 1,2 and 3 but in the case 4 I disagree. Maybe we
have to discuss
>> about that topic more accurate. I consider finite integers
are a subset
>> of
>> infinite integers over the first one 
infinity.
> Each of the finite naturals is a member of *N, but it does
not follow
> from this that N is a subset of *N. Thats part of why 
its
called
> nonstandard analysis. Not everything is a set in this model.
????
(snip)
>> I cannot see where the transfer principle fails, except in
the case of
>> sum
>> and limit. But in this case there are natural reasons as I
have tried to
>> explain. And the reason is not NSA.
> Lets consider the case of *R, for example. We know *R is
the extension
> (the enlargement) of R.
Yes, but You may have different vison?
> We also know that R satisfies the least upper
> bound property.
Yes, as and if I consider the limit and the sum of the string.
> By the transfer principle, *R must satisfy the least
> upper bound property.
As every limit.
> However, we also know that the collection R of
> finite reals is nonempty and is bounded above in *R by any
infinitely
> large number.
Now You talk about reals with infinite integers and plus
fractional part?
> But R does not have a least upper bound in *R, because if
> x is infinitely large, then so is x-1.
No, because there is an exception. The limit of *R comes on
or against at
x+1 as all the placeholders were totaly occupied with the
maximal digit
within the choosen base system!
> The conclusion is that R cannot
> be a set in NSA. The least upper bound principle is
preserved because it
> applies only to nonempty *subsets* of *R that are bounded
above, and R is
> not a *subset* at all.
Here our opinions needs further to discuss. :-)
Tapio
>> (snip)
> --
> Dave Seaman
> Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
>
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: .99999... still=/= 1
>> I dont know what you mean by an infinite 
area or by a
placeholder.
> Perhaps it does not help to read my emails carefully. :-(
I have not received any email from you.
>> As far as I am concerned, ...999 does not mean anything at
all. It
has
>> no connection with NSA or anything else as far as I can
see. You keep
>> throwing that string around as if you think it has a
meaning, but you
>> have never said what that meaning is.
> If You and I can define it, as I have hopefully done, then
there is a
> meaning and a connection to the existing mathematics!
You have not defined ...999 except to say that it means sum_n
9*10^n,
which is nonsense. Are you forgetting the transfer principle?
That sum
diverges in standard analysis (summing for n in N), and
therefore it also
diverges in NSA (summing for n in *N). Sorry, but you have
not defined
anything.
> It think there is now something misundertanding, of course
I take the
> reason. *N is (infinite) integer part, what ever You like to
call it. I
have
> a feeling that we have to start from the premices.
Thats exactly what I did when I explained about equivalence
classes of
integer sequences. Thats as basic as it gets in NSA.
>> And now, I would like to ask a counter-question. I will
describe a
>> certain member of *N, and I would like you to tell me what
decimal digit
>> string you think corresponds to it. First, some background.
>> You may recall from other discussions that the (standard)
real numbers
>> can be defined as equivalence classes of Cauchy sequences
of rationals.
>> That means I can identify a real number by presenting you
with a Cauchy
>> sequence, and it is understood that any other sequence
that happens to
>> fall into the same equivalence class is an equally good
representation
of
>> that number. For example, the sequences
>> < 9/10, 99/100, 999/1000, ... > and
>> < 1, 1, 1, 1, ... > happen to be two different
representations of the same real number.
> Yes, within the concept of the limit calculation. But that
is not true
(this
> is my point of argumentation, which I have to explain for
You) the sum is
> not the same as the limit. The limit is the successor of
the sum. As You
do
> not - yet - accept my argumentation, I have to explain it
more clear for
> You, though my personal opinion is that I have done it
already. Sorry -
this
> seems to be more complex that I expected. :-(
>> Now, lets consider a similar construction that lies at
the heart of
>> nonstandard analysis. To describe a member of *N, for
example, I can
>> present you with a sequence of natural numbers (members of
N). Unlike
>> the real-number construction, this one doesnt require 
the
sequences to
>> be Cauchy. Technically, I also need to describe to you the
equivalence
>> relation that will be used, but thats a bit more
complicated. It
>> involves something called a free ultrafilter on N. You can
find an
>> explanation of the concept at
>> <http://mathworld.wolfram.com/Ultraproduct.html>.
> I have a strong feeling that we are taking about much very
more simpler
> things. Your ultrafilters are me very strange matter. 
Im
very sorry. :-(
As I explained, you dont need to know what an 
ultrafilter is
in order to
grasp the basic point of the example. We have a hyperinteger,
a member
of *N, that is described by a particular sequence. By the
way, since the
terms of the sequence are unbounded, we can conclude that the
associated
hyperinteger is infinitely large.
>> For our purposes, its enough to know that if I give you 
a
sequence in
N,
>> there is a unique member of *N that is represented by that
sequence. Ok
>> so far?
>> Here is the sequence I have in mind. Let A(x,y) be the
Ackermann
>> Function, as described at
>> <http://mathworld.wolfram.com/AckermannFunction.html>.
Now, let a_k =
>> A(k,k) for each k. This sequence starts out:
>> a_0 = A(0,0) = 1
>> a_1 = A(1,1) = 3
>> a_2 = A(2,2) = 7
>> a_3 = A(3,3) = 2^6 - 3 = 61
>> a_4 = A(4,4) = 2^2^2^2^2^2^2 - 3 = (too big to write out
here)
>> and after that the sequence starts to grow rather quickly.
:-)
>> Let a be the member of *N that is associated with this
sequence. My
>> question is:
>> (1) what decimal digit string do you think represents a?
>> (2) what decimal digit string do you think represents
A(a,a)?
>> My point is that your decimal digit strings are woefully
inadequate in
>> this context. They cannot even begin to describe the
numbers in *N in
>> any useful way.
> Huh! Im indeed totaly out! Sorry. I have NOT descibed
something so
> complicated like that. I cannot give You an answer. I
assume we have now a
> very strange vision what I have told before.
Thats the very point I was trying to make. You have been
talking about
infinite strings of digits, under the mistaken impression that
you were
talking about hyperintegers. The whole point of my
description above was
to show that hyperintegers are not even remotely like what
you thought
they were. Your decimal digit strings are most certainly not
hyperintegers. As far as I can see, your digit strings are
not numbers,
they have no interesting properties, and they are not worth
discussing at
all.
--
Dave Seaman
Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: Spherical trigonometry question.
posting-account=
y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g
> I know of a good paper which discusses, among other things,
generating
> uniformly distributed points on the sphere. Is your e-mail
adress
> real? If so, I will forward a copy of the paper to you.
I believe my address is real (although I could be a brain
ßoating in a
tank, thinking that I have a real email address), but please
note that
Im not the OP on this thread. That would be someone whose
handle is
beast.
I have a friend who played around years back with the point
charge
approach to distributing vertices on a sphere, so I was
already a bit
but perhaps beast would find it more helpful than I would.
===
Subject: Re: b(xy) = b(x)b(y) --> b=a^H, H>=0 ??
> 1. b is nondecreasing
> 2. b satisties: b(xy) = b(x)b(y)
> Why this does implies that b(a)=a^H for some unique
constant H>=0 ?
b(x) = b(1 x) = b(1) b(x). So b(x) = 0 for all x, or b(1) = 1.
The first case is pretty boring.
A little induction shows that b(x^n) = b(x)^n for n in Z.
Then we can
use the fact that b is nondecreasing to show that b(x^(1/n))
= b(x)^(1/n)
for n != 0. Therefore for all rational q, b(x^q) = b(x)^q.
For any r, and any epsilon > 0, we can find rationals q1, q0
such that
q0 < r < q1 and |q1 - q0| < epsilon. Since b is nondecreasing
and the
function f(x) = a^x is continuous, this means that we can
likewise
bound b(x^r) arbitrarily closely between b(x^q0) and b(x^q1).
Hence
b(x^r) = b(x)^r for all real r.
Now it is a simple matter to choose an x0, and let H =
ln(b(x0)) / ln(x0).
Hence b(x0) = x0^H. Then for any x in R+ let r = ln(x) /
ln(x0), and so
b(x) = b(x0^r) = b(x0)^r = (x0^H)^r = (x0^r)^H = x^H.
We must have H >= 0, else b would not be nondecreasing.
- Tim
===
Subject: Re: JSH: Somewhat puzzled
posting-account=EH2x8QsAAABu84CuyjstkC4nRyQ1ZHKW
a-hem;
not unlike the meign of Vanderbilt,
is Princetons, vis-a-vu the Nashville Agrarians.
thus:
other than its cruddy british/confederate meign --
its the place that gave us the schooling
for The Birth of a Nation!
--4my of it?
http://www.wlym.com
===
Subject: Re: JSH: Somewhat puzzled
> a-hem;
> not unlike the meign of Vanderbilt,
Do you mean like Vanderbilts mien? Meign is not a word,
although
youve used it three times in the two posts of yours 
Ive
seen.
> is Princetons, vis-a-vu the Nashville Agrarians.
vis-a-vu? Also, I think I met one person from Nashville in
all the
years
I was there.
> thus:
> other than its cruddy british/confederate meign --
Do you mean Princetons gorgeous Oxford/Cambridgelike
buildings there by
cruddy British meign? The only ugly buildings there the last
time *I*
was
there were the relatively new ones (like the S. S. Ralston
Purina...)
> its the place that gave us the schooling
> for The Birth of a Nation!
? What is that supposed to mean?
> --4my of it?
What?
> http://www.wlym.com
Oh, God.
Michael
===
Subject: Re: 1 -1/2 + 1/3.....
>> Id like some clues to prove that the alternating series
Sum (n=1,
>> oo) ((-1)^n)/n)) converges to Ln(2).
>> Since 1/n -> 0 as n -> oo and 1/n is strictly decreasing,
the series
>> does converge. We know that, for every real x satisfying
|x| <1,
>> Taylors theorem implies that Ln can be expanded about 0,
so that
>> Ln(x) =x - x^2/2 + x^3/3.... Since [0,a] is compact, for
every 0 < a
<
>> 1 this power series converges uniformly on [0,a] to Ln,
but, even
>> though the series converges for x =1, this doesnt mean 
the
>> convergence is uniform on [0,1). If this were true, then 1
-1/2 +
1/3
>> ... would converge to lim (x ->1) Ln(1+x) = Ln(2), but
this argument
>> failed.
>>
>> If you define ln(1+x) as the integral of 1/(1+x) and
develop 1/(1+x)
>> around 1, then differentiate...
> what then differentiate... means exactly).
> Uh, I mean develop 1/1+x around x=0 (obviously), then you
get
> 1-x+x^2-x^3... and then differentiate obviously means and
then
> integrate. Cough, cough. And then you get
> x-x^2/2+x^3/3-...
> and an integration constant. If you prove that this
constant for some
> reason is 0 and then set x=1, you get ln(2).
Well yes you do, but theres more to it than that, because 1
- x
+ x^2 - x^3 + ... doesnt converge uniformly on [0,1].
> In short: I was babbling about something remotely similar
to what I
> was actually talking about.
It was a comedy of errors, because I thought you actually
meant
develop 1/(1+x) around x = 1. You then get a series that
converges uniformly on [0,1], unlike 1 - x + x^2 - x^3 + ...,
and
I quickly saw how this led to the answer. Checking my work
today, I see that it doesnt.
===
Subject: how to find the best ADC step size?
Hi all,
Suppose my input data follows Gaussian distribution with mean
u and variance
sigma^2.
If I want to design 6-bit ADC... what should be the optimal
step size for
this ADC?
-L
===
Subject: Re: how to find the best ADC step size?
> Hi all,
> Suppose my input data follows Gaussian distribution with
mean u and
> variance sigma^2.
> If I want to design 6-bit ADC... what should be the optimal
step size
> for this ADC?
> -L
If ADC is Analog to Digital Converter, the distribution of
your data has
nothing to do with the resolution. The range of your data
does, though.
You can sacrifice range for resolution, and resolution for
range.
Why 6 bits?
Scott
===
Subject: Re: how to find the best ADC step size?
>>Suppose my input data follows Gaussian distribution with
mean u and
>>variance sigma^2.
>>If I want to design 6-bit ADC... what should be the optimal
step size
>>for this ADC?
> If ADC is Analog to Digital Converter, the distribution of
your data has
> nothing to do with the resolution. The range of your data
does, though.
> You can sacrifice range for resolution, and resolution for
range.
Well, Gaussian distributions have infinite range, though with
ever decreasing probability. This does sound like a homework
problem, and possibly some important information was left out.
What parameter is being optimized through step size choice?
I dont believe there is only one answer to this problem.
-- glen
===
Subject: Re: how to find the best ADC step size?
>Suppose my input data follows Gaussian distribution with
mean u and
>variance sigma^2.
>If I want to design 6-bit ADC... what should be the optimal
step size
>for this ADC?
>> If ADC is Analog to Digital Converter, the distribution of
your data
>> has nothing to do with the resolution. The range of your
data does,
>> though. You can sacrifice range for resolution, and
resolution for
>> range.
> Well, Gaussian distributions have infinite range, though 
with
> ever decreasing probability. This does sound like a homework
> problem, and possibly some important information was left
out.
> What parameter is being optimized through step size choice?
> I dont believe there is only one answer to this problem.
> -- glen
Lucy isnt a student, but she does hold something near a
record for thread
initiations in cssm. Frankly, she tends to use this newsgroup
instead of
she needs to know. If I had noticed it was her post, I
probably would have
skipped it.
Scott
===
Subject: Probability m out of n people meeting???
Most of you know that if N people agree to meet between say,
3:00 P.M. and
4:00 P.M., and each agree to wait
d amount of time with (0<=d<=1) and d*60 minutes is how long
they are
willing to wait, then the probability
that ALL will meet is
n*d^(n-1)-(n-1)*d^ n
and the probability that NONE of them meet is
(1-(n-1)d)^n
Ostensibly, this is assuming that their arrival times are
randomly
distributed.
Here is a different take one the problem. Lets say that you
have the above
situation. and lets say that
m<n.
What is the probability that m of those n people will meet,
and the
remaining n-m people dont meet
anyone?
for the case of n=3, its relatively easy. 
Lets say that
d=0.25 which
means that the 3 people are willing
to wait 15 minutes each. P(ALL 3 meet) is
3(.25)^2-2(.25)^3=(3/16)-(2/64)=(6-1)/32=5/32.
P(NONE meet) = (1-2(0.25))^3=1/8=(4/32). The case of 1 person
meeting
himself is an anomalous
case that we wont count so, then the probability that two
people meet and
the 3rd person just gets
lost is
1-(5/32)-(4/32)=23/32.
What Id like to find is a formula or algorithm 
which will
help me compute
the probability of
m out of n people meeting in the above described problem, for
any value of n
and m where
m<n. Preferably one that doesnt involve integration, but
Ill take whatever
I get.
There is some interesting probability stuff on
http://www.mathpages.com and
I thought that
at one time, Id read a treatment of this problem which
compared it to the
birthday problem
in which you compute the probability of at least two people
in a room having
the same
birthday, but I cant seem to find it, if it was 
even there,
and the author
of the website
doesnt seem to show his e-mail address. Does anyone know 
his
name? Im a
bit hesitant
to write him if he prefers to be anonymous right now.
--
-------------------------------
Patrick D. Rockwell
===
Subject: New Famous Proof
It is IMPOSSIBLE for 1 person to have a conversation.
Defn : the net result of a converstation is someone learns
something [1]
Assumption : AY, AX personY says statementX [2]
Effect : AY, AX, personY hears statementX
Result : AY, AX personY already knew statementX [3] [from 2]
Contradiction : [3 and 1]
Conclusion : forall people, for all statements they come up
with,
a conversation does not result with themselves.
Herc
--
YOU CANT PROVE ME
If you prove its true then it has a proof, which makes it
false
If you dont prove it, then its true
10,000 people in sci.math ALL believe this is irrefutable
that mathematics
will always be incomplete.
===
Subject: Re: New Famous Proof
> Conclusion : forall people, for all statements they come up
with,
> a conversation does not result with themselves.
Nonsense. I talk to myself all the time, and on occasion I
respond.
Bob Kolker
===
Subject: Re: New Famous Proof
> Nonsense. I talk to myself all the time, and on occasion I
respond.
For example:
Me: Now where did I put my car keys?
Me: Did you look on the night table?
Me: No I didnt, I think I will.
Me: Hey, twarnt nothin.
I ask you. Was that a conversation?
Bob Kolker
===
Subject: Re: New Famous Proof
> Nonsense. I talk to myself all the time, and on occasion I
respond.
> For example:
> Me: Now where did I put my car keys? [1]
> Me: Did you look on the night table? [2]
> Me: No I didnt, I think I will. [3]
> Me: Hey, twarnt nothin. [5]
> I ask you. Was that a conversation?
Any utterance could be classified as a conversation, but we
take a more
methodical approach,
the definition is you learnt something from what was said.
A Ônull conversation is just a series of 
utterances where
nothing is
learnt, we dont
include this type since the general obversation of
conversations is extra
knowledge.
so you didnt learn anything from this statement.
Similarly for [2] and [3].
[4] You looked on the table. This new information came from
your senses,
eyes, hearing, tactile, not from any of the statements made
in [1] [2] or
[3].
[5] This is correct, since a remedial plan of action could
find the keys the
utterences were superßuous, nothin.
Herc
===
Subject: Re: Is zero even or odd?
> 0 cant be divided by itself,
>> Sure it can: 0 / 0 = 0 * (1 / 0) = 0 * infinity = 1
>> It works if the only three numbers in the universe are
>> 0, 1, and infinity -- A number system that seems very
>> suited to usenet.
Except for the fact that: 0 / 0 = undefined
Or actually more correct: n / 0 = undefined
> The two are not the same.
>> The definition of the ratio a/b is
>> a/b = r iff b*r = a
>> for the case of n/0 there is no r such that r*0 = n
(follows from
> the
>> definition of zero. Therefore n/0 (for non zero n) *does 
not
> exist*.
>> On the other hand, for 0/0, every r qualifies since for
every r,
> r*0 =
>> 0 (the definition of zero, again). Therefore, 0/0 is truly
> undefined,
>> in the sense that it is impossible to *uniquely* assign a
value to
> the
>> ratio r.
>> Mati Meron | When you argue with a fool,
>> meron@cars.uchicago.edu | chances are he is doing just
> the same
>> It depends on how you get there, [sin(x)]/x is certainly
defined
> for all
>> values of x including 0 and infinity.
> If you knew any maths worth talking about, you would have
known that
> sin(0) / 0 is not the same as the limit of sin(x) / x as x
tends to 0.
> The first is undefined and the second is unity.
Tell that to all the book publishers who print curves for
sinx/x.
> Now it is your turn: What do you know about sin (infinity) /
infinity
> Franz
No problem. Sin x is bounded between +/- 1 for all values of
x. A finite
number divided by infinity is 0.
Tam
===
Subject: Re: Is zero even or odd?
>> I know 0 is neither negative or positive but what about
odd/even? I
> think
>> its even.
> It is essentially neutral for it signifies a number that
isnt there
> because it lacks quantity. It is the symbol for nothingness
therefore
> it isnt odd nor even. For example, take the addition of
all numbers
> from -infinity to +infinity, what do you
> get?...zero!...nothin!....nada! The sum of everything is
equal to
> nothing!
> In physics, the conservation of energy states that no
energy can be
> created or destroyed. Therefore, if you add up all of the
negative
> potential energy in the universe with the positive kinetic
energy, you
> get NOTHING!....hence the total energy in the universe is
zero.
> Therefore, we are all essentially made up of nothing.
> Sleep tight!
If you have two apples and you eat the two apples there are
now no apples
in the whole of an infinite Universe? Zero? What of the term
ground
zero?
Does it mean no ground? What of quality, or is it quantity,
infinite
zero?
Zero point energy? Running a zero balance? Hawking time zero
on the
clock? My own four dimensions of time in which one dimension
is zero
(history (pasts-futures), frequency, relativity (gain in and
loss of), and
zero)? Zeroing in on? And, most particularly, base2, base4,
base8,
base16.....?
Zero can be an anchor.
mathematicians in the early twentieth century conceived
monstrous-seeming
objects made by the technique of adding or removing infinitely
many parts.
One such shape is the Sierpinski carpet, constructed by
cutting the center
one-ninth of a square; then cutting out the centers of the
eight smaller
squares that remain; and so on. The three-dimensional
analogue is the Menger
sponge, a solid-looking lattice that has an infinite surface
area, yet zero
volume.
Brad