mm-1939
existance of some function b(a) which does yield a number such that
L(a,b(a))=L(b(a),b(a)). and of course this function does exist for
all countable lists L since it is sufficient to assume this function
proof that is...   
just the proof traversed in the opposite direction. your AXIOM is
just what you assumed for your definition [2]. actually you even
assumed that that L(b,b)+1 (mod 9) would be digit-changing! other
great idea to use some (mod base-1)!
however, now that you did expand your whole view on this proof,
I somehow lost the connection to what you said earlier: that it
would be a miscomprehension of usage of 2 independant infinite
variables. what did you mean by that? which of those variables
are infinite and independant? Sorry if I ask you to re-tell something
you already mentioned multiple times, I'm just curious, and this
newsgroup really is much too big to perform any serious search...
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ehm, I think you mean ...mod 9. after all, I just showed that changing
every 9 to a 0 is a bad thing for real numbers larger than 1...
of course, that's why I said when viewed as since Cantor's
diagonalization proof does imply that some algorithm does exist
and does describe how it could look like. Did I again make some
mistake with my wording?
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   at 05:45 PM, piotr5@unet.univie.ac.at (Piotr Sawuk) said:
No. I meant f(d) = d+2 mod 10.
No.
The problem is deeper than your wording.
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but in set-theory. I already told you that there dense is defined
through a structure-preserving bijection between the set and some
subset of the set it is supposed to be dense in, and through the
order this subset does have relatively to the whole.
there exists a set X such that R is dense in X, X has higher cardinality
than R, and there exist at least a function which takes only elements
of R and produces elements of XR. if this claim is true, then of course
inside of the topology of X, R would not be connected.
I would wish to, but first I need to understand the construction of
R based on Q, and create above X in the same manner. Dedekind-cuts
are not helpful at all in this endeavour, neither are Cauchy-sequences.
the idea is to increase the turing-degree of Cauchy-filters and
hope for Goedel's incompleteness-theorem to proof that the resulting
construction does yield something which can not be described by any
cauchy-filter based on rational numbers...
as you know, there exista a partial function which takes as input
an infinite sequence (of rational numbers) and does return a real
number iff that sequence does converge. I could generalize that
notion and define number as the output of a partial function
which does take some collection of numbers as input, and which
does fulfill some density-criteria which I still need to figure out.
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   at 05:44 PM, piotr5@unet.univie.ac.at (Piotr Sawuk) said:
No. That's neither standard nor a definition.
ITYM a topological space X and an imbedding of R in X with those
properties. It's trivial to construct, but referring to that as reals
are not continuous. seems rather bizarre.
There is no of course; it's a non sequitor. R remains connected.
If you can't prove it, then why make the claim?
Why would that be either useful or relevant?
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as always you are right, I wasn't aware that this was actually the
definition. however, this is the definition of equinumerous in ZF
while I somehow doubt that would have been the thing Ross had in mind
when claiming that all infinite sets are equal. as usual, just because
something is non-equinumerous in ZF doesn't mean that there exists no
theory T with T|-(ZF-infinity) and T|-all infinite sets are the same
with 2 sets are the same iff some invertible mapping does exist (and
of course this theory would need to have its own notion of infinite
and maybe even mapping would be defined differently)...
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I think the same definitions of infinite set and mapping could
apply.  (ZF - Infinity + No set is infinite) would be a perfectly
good example of such a theory.
I think Ross mentioned a theory including ZF without regularity.  Such
a theory would need a redefinition of ordered pair.  The standard one
uses regularity to prove that {{a},{a,b}} = {{c},{c,d}} iff a=c and
b=d.  Something else would need to be found.
- Tim
Such
Maybe you could take a leaf from Peter Aczel's book where I believe his
theory, or rather the anti-foundational aspect often associated with
him, has no foundation and an anti-foundation axiom, where that theorem
that {{a,b}, a} = {{c, d}, d} is true if and only if a=c and b=d.  If
it resolves to a tautology, it does.
Maybe Holmes, et al.'s NFU is called New Foundations with an
Ur-element for a reason.  It's called New Foundations with
Ur-elements, not New Foundations with a Universe, unfortunately.
The ur-element is the ding-an-sich and Hegelian Being and Nothing.
It's the ur-element, it's universal and void, it's the root
contradiction in acknowledgment.  It's the ur-element, it's any set,
one and many, the ur-element is very flexible.  All the paradoxes are
assigned to the ur-element, so it's the only counterintuitive thing,
thus being intuitive in that way as it laughs at us sarcastically
because it's only honest and tautologous.  Whatever you take from it
for truth it's happy to lie consistently, because it's opposite is
another truth.  It's quite serious.  Those are basically meaningless
statements, true dada, yet instead beautiful, or rather, art.
I'm not a Randian objectivist.  I think you are too, not being a
hive-mind, it's a human condition.  Actually purposefully constructing
surrealism is a matter of definition, I suppose.
It's similar to the opposite of the notion that a theorem inconsistent
with others, and thus a logical axiom, leads to being able to prove
anything, instead tautology as the only truth, disregarding the
ubiquitous ur-element, or existence, concretizes only truths, and one
plus one equals two.
That's basically a philosophy.  The idea of interest is that it's a
mathematical logic.  Is that not sublimely arrogant?
Ross
--
No, pun intended.
nuh un.
ground hogs.
Rewrite that please, and make sure your sentences are coherent. You
also might want to entertain the querulous masses by putting your
points in plain-speak, or can the this topic only be reached in the way
you present it?
JJ
way
In some notion of contemplating paradox and truth, in a sense
compatible with mathematical logic, it seems basically unobservable
directly, instead the implications of a logic that refers to itself are
rigorously defined in terms that necessarily suffer from some root
unobservable in casting light upon the ability to consistently address
inconsistency.
So in saying that the truth lies, that is a reference to, for example,
what is called the Liar paradox.  Paradox may be the apeiron, where
that is a Greek word meaning indefinite or unbounded, and for the
Pythagoreans unnatural and false, anathema to structure, that I learned
from Rucker's _Infinity and the Mind_ , where it was used to describe
the actual infinite and irrational numbers by Pythagoreans, but it is
as well trust in basic assumptions that those observable things do
exist, that they are addressed.  Where there is no right or wrong
answer, the idea is to rationalize that unilemma so that there is, or
that existence or existenz a la Derrida leads to there being only one
acceptible set of truths.
On researching the apeiron it appears to be part of Anaximander, a
sixth century BCE philosopher's, consideration of the ur-element to
which the excluded middle does not apply, but from which through some
process leads to the opposites that form the elements of ancient
theories of natural physics. ... [Anaximander] did not just utter
apodictic statements, but also tried to give arguments.  This is what
makes him the first philosopher. is from
http://www.iep.utm.edu/a/anaximan.htm, the first link returned by
Google from searching for Anaximander  I never associated apeiron with
paradox before, instead with infinite, but it appears to match similar
characteristics of the unobservable consequences of any
self-referential statement, in a similar way as to the modernized
notion of Kant's Ding-an-Sich or Hegel's being and nothing, and other
notions of the conflation of the universe and void.
So basically the notion of the lying, laughing paradox is that as the
ur-element from it is extracted truth, and that it's acceptible to
consider that truth, but if all the one bits were changed to off bits
in the infinite precision binary digital computer, that it would still
function the same way, that the anti-computer is the computer.
There's a lot of room to attempt to mold these fleeting glimpses of
truth, or perhaps lack thereof or delusion, into more coherent
rationalizations to consider anything at all true in this or these
universes of logical discourse, which leads to philosophers over
millenia emitting libraries.  Compared to that anybody is just a
dabbler.  Plainness is in the eye of the beholder.
What I hope to do is to interface these notions of the ur-element, or
existence, but not necessarily in existential way, to the
mathematical logic of a set theory, to help in the resolution of the
self-referential paradox of set theory, towards for all logical
intents and purposes consistency, completeness, and concreteness in a
theory of mathematical logic conducive to establishing the truth value
of mathematical statements.
In reference to the previous post, there should be {{a,b},a} =
{{c,d},c}.  I carefully analyze the seeming error.  It is only true
when a=b=c=d, or a=d and b=c, that {{a,b},a} = {{c,d},d}, where it is
not true if a=b and c=d and a=/=c.
Sartre, nihilist?:
Sartre, cook?
Even philosophers have to dine sometime.  In a way, that's about
extending the dining philosophers Gedanken to infinitely many.
Frank Harary passed away last month.  Harary is a famous graph
theoretician.  Harary's graphs are undirected, Claude Berge's,
directed.  Both produced excellent tomes, _Graph Theory_ and _Graphs
and Hypergraphs_ are accessible, widely used textbooks.  Those
North-Holland imprints are really good.  I've been trying to get an
intuitive grasp of graph theory for computer algorithms, where graphs
and other data structures are all represented in memory as a flat,
one-dimensional sequence.
In terms of sets dense in the reals it's interesting to consider a line
from Friedman's discussion of requirements for large cardinals:  Even
for countable sets of rationals, we know that, in various appropriate
senses, we must use transfinite induction of arbitrary countable well
ordered lengths. - Dr. Friedman, Undecidable Theorems, Apr. 15,
Infinite sets are equivalent.
Ross F.
again I forgot to explicitely write down the dependencies:
1) for-all r1 and r2 non-equal in R there exists q in Q such that
either r1 <_R q <_R r2 or r2 <_R q <_R r1 with <_R being transitive
and anti-reflexive.
2) for-all r1 and r2 in R there exists q in Q such that from r1 <_R r2
follows that r1 <_R q <_R r2, again with <_R being transitive and
anti-reflexive.
the difference is that 1) does enforce anti-symetry, linearity and
extensionality from <_Q (since pred_(<_R)(x) non-equal pred_(<_R)(y)
for x<y is a result of some q between x and y) onto <_R restricted
to QxQ. the point is that if <_R is not a linear order, then 2) could
still hold!
a strict quasi-order is defined to be only transitive and anti-reflexive,
this does imply that some pair of elements could exist such that neither
incompareble, written as x _|_ y, and otherwise they are compareable.
define R as a set fulfilling some properties among which also
the existance of a quasi-order and the related quasi-denseness
are found. the question is if this set does exist and if it
is in some way unique. another question is if quasi-denseness
is sufficient to create an unique extension of <_R which does
have the same properties (i.e. linear-order) as <_Q and which
does make Q dense in R.
as I said, f2 is quasi-order-preserving since any quasi-order
on R will have corresponding elements in Q wherever compareable.
I don't know how you define homomorphism, but an isomorphism
is a bijection, and it is called structure-preserving when both
directions (i.e. also the inverse function) are structure-preserving,
and if homomorphism is the same for injection instead of bijection,
then f2 is not meant to be structure-preserving, only the inverse
compareable elements, and that property could be called order-preserving.
I am using the language from Graph-theory since I do not know enough
about the vocabulary on trees. a graph is composed of nodes and
edges, each node does have 0 or more neighbours -- the nodes it
is connected with through edges. a tree is a graph without circles,
and I additionally require it to have a special node called root.
additionally I put a direction on the set of neighbours which are
seperated from the root by the given node (what you called children).
infinite depth means that the distance (amount of nodes which need
to be traversed on the path) between 2 nodes would be infinite for
some nodes. infinite width means that a node has infinitely many
neighbours. 0.1, ...,0.9 are not the only children of 0,
0.01, ..., 0.09 and 0.001, ..., 0.009 ad infinitum are all neighbours
of 0. similarily for the children of all nodes in that tree.
tell me, which elements of Q are not represented?
if you create a tree with elements of R as its nodes, then there are.
what I was asking is if your construction of < is independant of
the choice of representatives, and how to prove that.
the point is that apart from ultra-filters you do use equivalence-classes
to describe R, and the existance of ultra-filters again is questionable.
if you do have equivalence-classes then you must walk the long way through
proofing that properties are independant of the representants used, and
you need to say something about the size and number of equivalence-classes.
yes, that's what I meant with when seen as.
but you still need to prove that equivalence-classes are seperate
and that this property does not depend on the representants used.
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   at 05:45 PM, piotr5@unet.univie.ac.at (Piotr Sawuk) said:
Why not the simpler for all r1 and r2 in R with r1 <_R r2, there
exists q in Q with r1 <_R q <_R r2?
Please clarify your nomenclature.
R is an ordered field, so all elements are comparable in that sense.
With only a partial order it wouldn't be R. But even if you were
could only preserve the orders of elements of Q, not of R'.
The standard models of R show that a complete Archimedean field does
exist. If you relax the properties to a field with a partial order
then you will lose uniqueness and you will get structures that are not
models of R.
Q is ordered, not just partially ordered. The proof that f1=f2 if both
are order-preserving homomorphisms is trivial.
Then maybe you should do some reading. A homomorphism of fields is a
function that preserves zero, unity, addition and multiplication.
No. A bijection that is not structure preserving is, by definition,
not an isomorphism.
Then why did you write tree and root?
ITYM acyclic DIRECTED graph.
If they're separated then they're not neighbors.
1/3
Recipe for tiger stew: first catch the tiger. You've sort of kind of
defined a structure that does not include all elements of R, or even
all elements of Q. Whether you could construct other structures that
included one of the missing elements is irrelevant.
Well, I'm neither Cauchy nor Dedekind, but I already gave you a proof
for Cauchy sequences.
For Cauchy sequences; not for Dedekind cuts.
Are you arguing for or against?
So? That involves less work than dealing with ultrafilters.
No I don't; I need only show that the construction yields a complete
Archimedean field.
What equivalence classes? The only ambiguity is whether to use (-oo,q]
and (q,oo) or to use (-oo,q) and [q,oo), and that can be eliminated by
an appropriate change to the definition of a cut.
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exactly! I have seen several definitions of open set in R, and since
the representation of R does differ, so does this definition. However,
I choose to define open set as F(x) for some element x and some
Cauchy-filter F. in order for (a + (b-a)/3, b - (b-a)/3) to be an
open set you need to prove that there exists an x and a Cauchy-filter
F such that F(x) is equal to that set. since a and b are rational,
so one could easily construct some x inside of that set. and since
filters are the result of the powerset-axiom, so some set of pairs
would actually contain x on the left side. all you need to show is
the existance of a set {(x,y)|y in your open set} as a subset of
RxR, and that a filter containing such a set could be a cauchy-filter.
as you said, Cauchy-filters are far too complicated for such things.
a theorem who's proof in ZFC I have not seen yet (for the definition of
R through Cauchy-filters). an alternative to providing such a proof would
be to simply define open sets that way. then it is even less clear why
your open set should actually be open in the resulting topology.
What I wish is an algorithm for creating a bunch of models for R.
what Cauchy-filters do offer me is a description of the properties
a model of R should fulfill, and of the consequences of some of
those properties. Cauchy-sequences and Dedekind-cuts are just 2
models, each useful for some limited amount of applications. I
would wish for a more general model which is useful in all cases
in that it really is a set of instructions of how to create a
fitting model. you define the set of Cauchy-sequences, or you
define the set of Dedekind-cuts, either way you are just working
with a definition and not with a characterization. Cauchy-filters
on the other hand do not need to be used when working with R,
you only need to prove some properties related to them before
you can introduce your own favourite model of R to work with.
shorter expression. was that the wrong decision?
x is a fixed variable and should be put into subscript in the same manner
as the n you suggested to write that way. but you are right, I should 
write:
Below n ranges over natural numbers and x is a fixed irrational number:
there exists I4_x_n intersected with Q subset of F(x,y)? Anyway,
even if it exists, then only I3_x intersected with Q subset of f(x)
has been proven, and not I3_x subset of f(x)!
there exists some I2_x_r subset of I3_x_n? I think it does: it's *the*
surjective function from the uncountable subset of RQ to the countable
subset of Q which does exist according to some proof in set-theory.
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   at 05:45 PM, piotr5@unet.univie.ac.at (Piotr Sawuk) said:
No. You have seen several models of R, but they are canonically
isomorphic and homeomorphic. You have not seen multiple topologies on
R; the definitions of open sets on each are equivalent.
If you choose a nonstandard topology then it is your obligation to
explicitly state that whenever you refer to topological notions.
In order to legitimately refer to the definition of R through
Cauchy-filters: you must prove that the objects, functions and
relations satisfy the axioms for R, i.e., that you have constructed a
complete Archimedean field. Once you have done so, then the fact that
each non-empty open set contains both rational and irrational numbers
is a simple theorem and the proof does not need to go back to your
construction.
I don't see the utility.
No. The properties that a  model of R should[1] fulfil are those of a
complete Archimedean field, nothing more.
The whole point of those models is to show that a complete Archimedean
field exists. For that purpose, the simpler the better.
Any model is general.
No, you don't need to prove any properties of Cauchy filters in order
to introduce a model that doesn't use Cauchy filters, and there is no
point to dragging them in.
Writing the shorter expression is only legitimate if it expresses
the same thing. Writing a shorter expression that doesn't express what
you meant is always the wrong decision.
You still haven't done that.
There is no such thing. You need to make explicit what quantifiers are
involved and what their scopes are. You can do that in English, but it
Did you mean
follow what you're trying to say. Fix the typos, *PROOFREAD* and post
it all in one place.
[1] Actually, must fulfill, because otherwise it isn't a model of R.
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I have no idea what the Huffman coding scheme is. Here is my input.
Name the coins A, B, C, D, E, F, G, H, I, J, K, L.
First weighing:
Weigh A, B, C, D against E, F, G, H.
Second weighing:
If mass(A, B, C, D)=mass(E, F, G, H), weigh I, J, K against A, B, C.
If mass(A, B, C, D)< mass(E, F, G, H), weigh A, B, G against C, D, E.
Third weighing:
If mass(I, J, K)=mass(A, B, C), weigh L against A.  If mass(L)=mass(A),
then there is no counterfeit coin.  Otherwise L is the counterfeit
coin.
then I is the counterfeit coin.  If mass(I)<mass(J), then J is the
counterfeit coin.
If  mass(I, J, K)<mass(A, B, C), weigh I against J.  If
mass(I)=mass(J), then K is the counterfeit coin.  If mass(I)<mass(J),
counterfeit coin.
If mass(A, B, G)=mass(C, D, E), weigh F against A.  If there is a
difference, then F is the counterfeit coin.  Otherwise H is the
counterfeit coin.
then A is the counterfeit coin.  If mass(A)<mass(B), then B is the
counterfeit coin.  If mass(A)=mass(B), then E is the counterfeit coin.
then C is the counterfeit coin.  If mass(C)<mass(D), then D is the
counterfeit coin.    If mass(C)=mass(D), then G is the counterfeit
coin.
For mass(A, B, C, D)< mass(E, F, G, H):
If mass(A, B, G)=mass(C, D, E), weigh F against A.  If there is a
difference, then F is the counterfeit coin.  Otherwise H is the
counterfeit coin.
If mass(A, B, G)<mass(C, D, E), weigh A against B.  If mass(A)<mass(B),
counterfeit coin.  If mass(A)=mass(B), then E is the counterfeit coin.
counterfeit coin.    If mass(C)=mass(D), then G is the counterfeit
coin.
counterfeit
than the
Huffman
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id j1HILFn21771;
 I'd like to know how you tackle such an equation.
 Alain.
Hint: first solve h(x) = h(2x) - 2.  Then there's an easy set
of two equations to solve for f(x) and g(x)...
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada
I get:
f(x) = c(log[2](x)) + (3/2)*log[2](x)
g(x) = (1/2)*log[2](x)
where c(y) is an arbitrary function with period 1.
-- 
G. A. Edgar                               
http://www.math.ohio-state.edu/~edgar/
whereas I get (with c(y) an arbitrary function with period 1)
 f(x) = (3/2)*log[2](x) + c(log[2](x)) + c(log[2](x)/2)
 g(x) = (1/2)*log[2](x) + c(log[2](x)) - c(log[2](x)/2)
Mike Guy
Haven't you gathered enough information from the last N times you've
asked very similar questions?
 
- Tim
I'm looking for papers and books about power series of bivariate rational
functions. I'm thankful for everything but I'm especially interestet in
coefficients of the power series.
Guido
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id j1HJBqw26758;
[sci.math: February 17, 2005 12:36:06:000PM]
http://mathforum.org/discuss/sci.math/m/684011/684172
Note: Random snippage ahead. However, since you
didn't bother to address the concerns I raised
in a previous post (see URL below), I'm not
terribly concerned about making sure that you
can't complain I'm taking things out of context.
Besides, I've given a URL for your original post
above if there's any question.
http://mathforum.org/discuss/sci.math/m/682638/682924
Why doesn't the same problem arise with circles, lines,
numbers such as '2', etc.? I've never seen a '2' before,
have you?
For what it's worth, Newton didn't define the derivative
of [f(x+h) - f(x)]/h. Also, where is there division by
zero in this, or are you thinking of another formulation
of the derivative?
What does it mean for an approximation formula to be
incorrect? Isn't every approximation incorrect, unless
the approximation is the same thing as what is being
approximated (i.e. the linear approximation of a
linear function)?
Dave L. Renfro
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id j1HJkhO29644;
Your best bet on this type of equation would be to do something like:
     f(x) = |(x mod 2) - 1|
because that would be like a mathematial if statement, which is probably 
what you want to use, modulus.
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id j1HJkhg29640;
I recently got hacked(Who knows how) on a Triple-DES + 256 bit RSA Encrypted 

Network system(I tried very hard with the lisences), so I made my own 
encryption system. I haven't studied it extensively, and would like some 
comments on this. It's written in Mathematica notation, If you currently 
don't know the syntax, you'd better learn it in the 10-minute tutorial 
within 
Mathematica's trial. I know it's a large installation, but it's a largely 
acclaimed and highly advanced(I usually do things like contributing to 
newsgroups of the few people who will sit around for months trying to prove 

the Reimann Hypothesis) program, hence the reasoning of my usage of it. It's 

at least worth trying out, much more buying.
In[1]:= Fpart[a_]:=a-IntegerPart[a]
In[2]:=RobRandom[s_,m_]:= IntegerPart[m Fpart[10^5 Log[E,Pi s]]]
In[3]:=InRobEncrypt[k_,p_]:=(k Random[k-1])+p+RobRandom[k,k] /; p<k
In[4]:=RobDecrypt[k_,e_]:=Mod[(e-RobRandom[k,k]),k]
/; Re[e]==e
Encrypted Network system(I tried very hard with the lisences), so I made my
own encryption system. I haven't studied it extensively, and would like 
some
comments on this. It's written in Mathematica notation, If you currently
don't know the syntax, you'd better learn it in the 10-minute tutorial
within Mathematica's trial. I know it's a large installation, but it's a
largely acclaimed and highly advanced(I usually do things like contributing
to newsgroups of the few people who will sit around for months trying to
prove the Reimann Hypothesis) program, hence the reasoning of my usage of
it. It's at least worth trying out, much more buying.
Post this to sci.crypt and put on your armor cause you're gonna take a
beating!
Why don't you use proven algorithms like AES (symmetric) and ECC
(asymmetric) along with firewalls and all the other safeguards (virus
checker, passwords, encrypted drive space, et. al.)?
Is there a person who can calculate the exact value of the
following series where hypergeom stands for the hypergeometric
function
sum((-1)^(1+k)*hypergeom([1/2, 1+k/4],[2+k/4],-1)/(k*(4+k)),
k= 1..infinity);
?
Best wishes,
Vladimir Bondarenko
GEMM architect
Co-founder, CEO, Mathematical Director
Cyber Tester, LLC
13 Dekabristov Str, Simferopol
Crimea 95000, Ukraine
tel: +38-(0652)-447325
tel: +38-(0652)-230243
tel: +38-(0652)-523144
fax: +38-(0652)-510700
http://www.cybertester.com/
http://maple.bug-list.org/ 
http://www.CAS-testing.org/
  hypergeom([1/2, 1+k/4],[2+k/4],-1) =
    (1/4)*(4+k) * int(t^(k/4)/sqrt(1+t), t=0..1)
sum. The k-dependent terms give the sum
  sum((-1)^(1+k) t^(k/4)/k, k=0..infinity) = ln(1+t^(1/4))
The original sum is converted to the integral
  (1/4) * int(ln(1+t^(1/4))/sqrt(1+t), t=0..1) ~ 0.1189020516
But lets have some more fun. Set t=s^4
    = (1/4) * int(ln(1+s) * 4s^3/sqrt(1+s^4), t=0..1)
Integrate by parts
    = ln(2)/sqrt(2) - (1/2) * int(sqrt(1+s^4)/(1+s), s=0..1)
The remaining integral contains a the square root of a quartic and a
rational function, hence it should be possible to evaluate it in terms of
elliptic integrals. However, maple has been somewhat uncooperative with
that. We can go part of the way by hand. First move the square root to the
denominator and expand the rational part in partial fractions
  (1+s^4)/(1+s) = s^3 - s^2 + s - 1 + 2/(1+s)
The integrals involving odd powers of s can be done by hand
  int(s^3/sqrt(1+s^4), s=0..1) = (1/2)*[sqrt(1+s^4)]_0^1
    = 1/sqrt(2) - 1/2
  int(s/sqrt(1+s^4), s=0..1) = (1/2) * int(1/sqrt(1+t^2), t=0..1)
    = arcsinh(1)
Then Maple says:
  int(1/sqrt(1+t^4),t=0..1) = 1/2 * EllipticK(1/sqrt(2))
Maple gives some ugly formula for the s^2/sqrt(1+s^4) integral involving
elliptic integrals of complex amplitude and modulus. Yuck! The indefinite
integral of 1/(1+s)/sqrt(1+s^4) can be evaluated by Maple, but not the
definite one s=0..1! It gives -infinity, go figure. Perhaps there is a
nicer way to express the remaining elliptic integrals in terms of
Carlson's notation, but I'm not familiar enough with it. Putting all the
results together is left as an exercise to the reader. 
Igor
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id j1HKEM132625;
David,
  You did not give me a chance. Check the postings. I have already responded 

to one of your emails. Will respond to this one now.
Please wait for me to respond before you post again otherwise we have 
confusion and mayhem.
Correct. But we can define a 2 in very clear terms. A point cannot be 
defined in absolute terms. A '2' can be defined in absolute terms.
I know who defined the derivative. I believe it was Cauchy who compiled the 

classic first principles limit for the derivative. It doesn't matter too 
much, because we are talking about the same quotient. There will be devision 

by zero if h equals zero. In gabriel's formula, this is not possible. It 
makes sense for w to be infinitely close to zero but not quite zero. And 
this 
leads to your next comment about approximations.
I hear what you are saying and am not disagreeing. What I am saying is that 

had Newton known, he may have tried to use a format similar to Gabriel's 
rather than settling for the aproximation. Even the first principles 
definition of the derivative is only an approximation. We obtain exact 
values 
if and only if we are able to find a general derivative expression where we 

can plug in values for points and proceed to find the derivative. Suppose 
for 
once that you cannot find the derivative, then you would need to use the 
limit to attempt an approximation and that's all it would ever be - an 
approximation, not the exact value. When we obtain 2x + w from first 
principles after differentiating x squared, we perform a *hyperreal* 
calculation in obtaining 2x when we allow w to be zero.
In so doing, we are assuming that the interval is actually differentiable 
everywhere because w is used in our approximations. Does this make sense to 

you? When Gabriel tried to explain this to me, I was convinced he is a fool. 

However, the more I thought about it, the more it made sense and I could not 

refute it. In other words, when we allow w to run through to zero in the 
classic limit or w/n to become infinitely small, we are assuming 
automatically that not only is a function f say, continuous at all the 
points 
in the interval, but also differential at all these points. 
 
There will be no division by 0 if h is not zero, which it isn't 
as you can discover by reading the definition of a limit.
No real number is infinitely close to 0 without being 0 itself.
This is not the case in Nonstandard Analysis which has infinitesimals. 
The extended class of numbers which contain infinitesimals is called the 
hyperreals.
Bob Kolker
Yes it is.
Yes it does.
Precisely. They are called the hyperreals, not the reals. Nonstandard
analysis does not change the fact that no non-zero real is infinitely
close to 0.
************************
David C. Ullrich
Rob:
David:
Jason: Rob. did not say it is not. If you read his post correctly, he
is saying exactly the same thing!
Jason: He said this! Read the post carefully. It's the other idiot Wade
who has been posting garbage and confusing you in the process. Is this
how you respond to your students too? i.e. without even listening
properly?
the
Jason: I think he knows this pretty well. Once again, you need to read
his post without the Wade crap in it.
There was no Wade crap in what I read. The statement no non-zero
real number is infinitely close to 0 _is_ true in nonstandard
analysis.
************************
David C. Ullrich
Robert,
  You are wasting your time with the likes of Wade - he is one of the
local fools on this forum. Others to avoid are : Chapman, Rodgers,
Wade, Hammick and their followers.
Now Chapman is also a professor and I wonder how he manages to find the
time to post so much rot on this forum. I would not be surprised if his
sidekicks are also bored math professors. Does this say anything about
why math students do so badly? I wonder....
Jason
I've been in much worse company than that.
Someone said about Gabriel, I read his stuff, I digested it thoroughly, 
and
I've been farting ever since. Who was that witty writer? Oh, I remember. 
It
was I myself!
Sorry, but it is the case in nonstandard analysis.
 
Those are the hyperreals; I'm talking about the reals. Read again.
compiled the
matter too
devision
You are in error. This limit of the difference quotient is not the same
as the limits you are accustomed to using in analysis.
zero.
Rubbish. This is only true if we are using Frechet's Identity Of
Indiscernibles as discussed in metric spaces. A lot of his theory is
questionable just as a lot of real analysis is questionable. Most
students learn from questionable sources like the one from Rudin. I
am throwing these out the window for now. They are neither clear, nor
precise, nor dependable. Wade, you have a knack for posting worthless
and thoughtless comments. You may want to sit back and let other more
knowledgeable poeple comment. You are wasting my time.
The limit of difference quotients is precisely what we use in 
analysis to compute a derivative.
Why not write Rudin and set him straight once and for all? Send 
him Gabriel's earth-shattering reformulations, tell him to say 
touch instead of intersect, tell him that the derivative of x 
is 0 (hence it's not differentiable), or at the very least ask 
him to include self-referential definitions in the next edition.
I have a knack of posting comments you find impossible to refute, 
causing you to go into fits and tantrums and to soil your pants. 
I refuse to pay the laundry bill.
What did I just say about Wade? I think I said he has nothing to
contribute but crap. Well, just read what the fool has written and
judge for yourself.
Enough  Wade. Pull up your pants and give others a chance to speak.
You have not been able to refute anything you fool! In order to refute
this information, you first need to understand it.
No idiot! It is not. And I  on you and Rudin!! The concept of the
limit of the derivative was around long before Analysis and Rudin. Why,
Archimedes used limits you bumbling idiot!! they were not called such
but that's exactly what they were.
Rudin is an idiot because he has helped produce idiots the likes of
you!!
because he knows absolutely nothing!!
This last comment tells me what kind of a rotten amoeba you are: Let me
see? You are filthy on your person. You probably don't wash your ass
after you , why you probably don't even wipe it. Your body odour
causes everyone to flee from you and you look like a pathetic slob or
an emaciated, goggle-faced twit. Your house or apartment is crawling
with cockroaches and your breath probably stinks worse than that of a
snake.
You are a thug and a coward because if you were in my presence, you
would be incontinent just by me looking at you threateningly. What a
loud and useless asshole that you are! Go on, if it makes you happy,
continue to post more crap like you have been doing - it's about all a
low-life trash like you will ever amount to.
Have you noticed that _everyone_ thinks you're wrong about more
or less everything? There are two possible explanations for
that. One is that you and Gabriel actually understand mathematics
better than everyone here on sci.math, also better than every
mathematician on the planet. The other is that you're simply
wrong about everything. Which seems more likely?
How has he been denying anyone the right to speak?
Hint: If that were possible you would not be posting here.
Here's another hint: Most of us don't recognize the stuff above
as valid mathematical reasoning.
Yet another hint: Even if you were right about everything,
which of course you're not, nobody would ever find out.
Because when they read stuff like what you write above
they conclude you're just another crackpot.
Yeah I know, my mother's a whore and my dog smells bad.
That doesn't change the truth of that last paragraph.
************************
David C. Ullrich
 Jason, aka James Wells, squeaked:
         [A post literally full of .]
Anyone want to guess how old Jason, I mean James, is? I start the 
bidding at 13.
It is known there are UDEs or universal differential equations with
solutions in C^Inf that approximate any continuous function infinitely
accurately. For example of constructions, see:
Can the solutions of such UDEs actually be expressed in a closed form
and if so, what do they look like, or are they just an analytic
construction that can be shown to approximate a given f in C(R) with
arbitrary precision but which cannot be solved in analytical form?
Also, are such animals useful for proving some theorems about DEs or
simply pathological curiosities?
-- 
I'm not interested in mathematics that might have anything
to do with reality. -- Russell Easterly, in sci.math
 
There's been plenty of work since Rubel's original paper,
all which is easily found by the obvious web search, e.g.
Googling universal differential turns up papers by
Boshernitzan, Briggs, Duffin, Elsner, ...  If you browse
these papers you'll find your answers, and much more.
They're the analog of small universal Turing machines
for analog computers (Shannon and Pour-El showed that
the outputs of analog computers can be indentified with
with solutions of ADEs = Algebraic Differential Equations).
--Bill Dubuque
 PROBIT, etc.
Brendan Halpin had some good notes at the following link, but for some 
reason it is not working at present.
    http://teaching.sociology.ul.ie/SSS/lugano/
You could also try Garson's online notes:
   http://www2.chass.ncsu.edu/garson/pa765/statnote.htm
Someone else noted that UCLA's website has some good help pages.  Look 
for the textbook examples page.
-- 
Bruce Weaver
bweaver@lakeheadu.ca
www.angelfire.com/wv/bwhomedir
Start at
http://www.ats.ucla.edu/stat/overview.htm
if you want to know more about SPSS and Stata (and SAS).
/LWn
Someone told me this problem today; I just had to share it:
There are three guys and a lady.  They have amongst them only two
condoms however.  How can they reasonably avoid the transmission of
infection while still each of the guys having full
vaginal intercourse with the lady?
(of course, lots of unstated assumptions are made etc., but you'll know
if you have the answer)
know
well, all that i can say is that it certainly nicer to be using the
condom on a lady then talking about jokes involving it :D
First M1 // F:
       F
C1-----------
C2----------- 
       M1
Second M2 / F:
       F
C1-----------
       M2
C2-----------
       M1
Third M3 // F:
       M3
C2----------
       M1
       M2
C1----------
       F
Is this considered grapic pornography?
At the risk of embarrassing myself, I don't understand the diagrams. 
Could someone please clarify?
Without looking at the diagrams, I recite what I think is the solution
from memory.
Joe puts on both condoms, one inside the other and uses them.
Fred removes the inner condom, dons the outer and uses it.
Harry takes the discarded inner condom, inverts it, dons it
and then wears the used outer condom over top of that (used
face to used face, ugh).
Virginia has been exposed to the same surface of the same condom
three times.  No risk there.
Joe, Fred and Harry have each been exposed to an unused surface.
No risk there.
Four surfaces, four participants.  So we appear to be using the
equipment optimally.
        John Briggs
She ain't a Virginia no more...
At last, a chance to use those male and female ASCII symbols !-)
(Apologies if they aren't displayed correctly on all browsers.)
The above can be succinctly represented by the following diagram:
  [1]      .81.891--|            |--.81.8a
  [2]                   .81.892--|--.81.8a
  [3]      .81.893--|--.81.891    .81.892--|--.81.8a
02/18/2005
There are none. ASCII only have code points from 0 to 127.
-- 
Unsolicited bulk E-mail subject to legal action.  I reserve the
right to publicly post or ridicule any abusive E-mail.  Reply to
domain Patriot dot net user shmuel+news to contact me.  Do not
Dunno much about group theory, but could this party
be an example of the Lie group SO[m, F] ?
If so, I presume SO must stand for safe orgy.
(and the operations are definitely not commutative).
Put one condom inside the other, let one of the guys at her.  Take the
inside condom out, let another guy at her through the one that *was* on
the inside (who's inside noone has yet touched).  Then take the condom
that was on the inside initially, turn it inside out, and put it back
inside the condom that's been inside the girl all along, and let the
third guy have a go.
[SNIP solution]
It's certainly graph-theoretic pornography.
Phil
Good stuff!
And it's not graphically pornographic.  At worst it's symbolically
pornographics.  And, unless I'm mistaken, it's not meant to be erotic,
so I don't see it doing any harm (but why did you flip around the women
in the third diagram, it confuses things quite a bit!)
I figured she would tire of always being on top.
heh... :)
And why would eroticism do harm? If I recall correctly, it had an important 

role in the existence of all of us. 
I don't know if it would add anything to your presentation of the
solution.
Certainly as a source of problems it can be inspirational and helpfull
in communicating ideas (one of my own efforts is here:
http://www.maths.tcd.ie/~icecube/duality.html ), but otherwise I'm not
sure
I can think of a way that protects the men, but not the woman. 
give it a bit more thought...
OK.
Joe puts one on and cums.
Then Frank puts on the other and cums.
Then Billy puts one inside out and then puts it into the other. And cums.
How do you suppose he puts one inside the other safely? I mean at that
stage, *both* sides of *both* condoms have been touch by one of the
first three condoms, so there's no way he can safely interact with any
combination of them...
If she is a lady, they can't.
There is an answer, but it is an awful mess.
Sex is something you share in, and is (rightly) possessed by both
parties on equal terms.  I think there is no ambiguity in this.
I keeps seeing posters claiming that my surrogate factoring method is
worse than even chance or at best only as good as a chance method for
factoring.
There are several problems with their assertion.
That is, I suggest that they are lying.  Here's why.
The base equations are
yx^2 + Ax - M^2 = 0
yz^2 + Az - j^2 = 0
and
T = M^2 - j^2
from which you can easily find
Ax= Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
Az= Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
where I've basically wrapped up all the unknowns.
You give M, as it's the target, and j, as it's some picked integer, and
then everything else gets determined.
Well you have *two* base equations showing x, y, z and A are the
unknowns, and then I susbstitue out y and wrap A up with x and z, so
you have Ax, and Az as unknowns.
Substituting out y, gives you the two equations defining Ax and Az,
where the final constraint is the rationality of the square roots.
Now from the first equations square root there MUST be a rational Az
that factors M, so there must exist Ax, and Az such that M is factored.
That's a crucial point, and the first clue that posters must be lying,
as now you have a system where the factorization of M, the target, must
exist, for some solution.
I noted that Ax an integer doesn't usually work, as that's the basic
approach I tried before, and I agree that it gives a low probability of
success, but first thing to note, which CANNOT be disagreed with, there
is a rational Ax that MUST work.
So now you have an issue with the value of that Ax, as you might have
Ax = n/m
where n and m are integers.
Well, suddenly you have *two* more variables, in a system that up until
now has been fully constrained, as you had three unknowns with two
equations, where the final constraint is a rationality of square roots
constraint.
So guess what?
If posters claiming my method doesn't work are right, then m is
basically a random number.
My method then would be the world's first perfect non-quantum random
number generator.
If m is NOT random, then there is some mathematical reason or
constraint that governs the value of m, and then that can be determined
and if you figure it out, then you have a solution to the factoring
problem.
So, either I've found the world's first perfect, non-quantum, random
number generator, or there's a way to make this method work.
I suggest to you that posters are not so dumb as to not realize the
constraints and reality here, but are actually deliberately lying about
what follows mathematically, for their own reasons.
Now I did the analysis that shows that in fact the denominator of Ax IS
in fact determined so that it contains prime factors of T.
The math is not hard proving this fact.
Now then, I would be curious if any poster might reply explaining how
there is some other possibility than I mentioned:
1.  A perfect random number generator
2.  A method that must work in some deterministic fashion.
If and when they dodge those possibilities, trot out crap to try and
explain why that is not valid, without giving anything *mathematical*
then you will know that yes, they are indeed, lying.
James Harris
Nihil ex nihilo.
Wake up Harris you have got nothing.
is
for
with
square
how
Yeah, but if you *know* the factors of M, then use them to figure out
what Ax factors M, then its denominator will be random, if you are
right.
That would mean you could use M's with known factors to perfectly
generate random numbers.
I make that point to highlight that some of you are acting outside of
reason, as I don't think it's random.
You can play stupid, but it won't help you later.  And being in Germany
won't help you either as the German people will probably go after you
as well.
If there's no reason to the method, if it's as random as you people
claim, then necessarily it's a perfect random number generator.
If not, then you are, as I said, setting the world up for anyone who
can figure out how to get this to work, and if they are terrorist, then
God help everyone else, as you people have set them up like no one else
could, set them up for the worst case.
James Harris
 Discussion, linux)
You know what I think?  I think James has already solved the factoring
problem, but it has nothing to do with surrogate factoring.  I think
he's using these threads as a public service, to distract the
thousands of Al Qaeda mathematicians working on factoring large
numbers by forcing them to pay attention to the dead-end of surrogate
factoring.
Why else would he keep shouting, I bet the terrorists are mighty
interested in this thread!  If I was a terrorist, I'd be reading this
thread *right here* and catch the heathen world with its pants down!
Probably the best strategy for catching the attention of the
Usenet-grepping terrorist mathematicians.
This distraction will give the CIA, NSA and Boy Scouts time to fix the
internet.  Maybe it's quid pro quo, so that, once the free world has
been saved, the NSA promises to fix the algebraic integers too.  Then
JSH can get the recognition he deserves.
-- 
Four little piggies went to market.
 This little piggy wanted ice cream 
 And Mama said 'Nooooo'!
               --A new Piggy Song, by Quincy Hughes (age 3)
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
You have to catch M before it makes sense talking about its factors.
Forget it.  Nobody will go after anybody that says that a nincompoop
doodling inconsequently with factoring is not of mathematical
interest.
Well, so is the entire public key cryptography according to your
reasoning.  So what?
The world _is_ already set up for anyone who can factor large numbers
with a larger jump in efficiency than thought possible.  And it has
been set up this way for decades.  That's why key sizes are growing.
It just so happens that teller card fraud _is_ already big business.
I suggest you clean out your pants and get real.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
Do you believe that fast factoring of large numbers will crack public
key cryptography?  It won't: there are public key cryptosystems that
_don't_ rely on factorization being difficult.  Look up discrete
logarithm.
Do you believe that if terrorists could factor large numbers quickly
they could terrorize more effectively?  What makes you think that?  What
makes you think that for _any_ kind of public key cryptography,
terrorists would find it useful if they could break it?
method
with
between
keys
Germany
Now that's a scary thought. It was no fun the last time Germans
got hold of The Hammer.
then
else
No. Not random. You can not implement a random process on a deterministic 
computer. Could you please give your definition of the word 'random'? And 
please don't come up with something goofey again, like 'properly random'.
No, it does not mean that Idiot.
You are correct. It is not random.
Good thing I am not in Germany. I hate it when those pesky Germans go after 

me.
IDIOT! What is a perfect random number generator James?????
Could you define the word terrorist? 
Mach Dir keine Sorgen. Stellvertretend fuer die deutsche Bevoelkerung 
verspreche
Ich Dir, dass Deine Meinung zum Thema Surrogatfaktorisierung Dein 
Wohlbefinden
im Falle einer Einreise nach Deutschland nicht negativ beeinflussen wird.
Beste Gruesse!
Volker.
Danke Herr Volker. Ich bin gl.9fcklich, da¤ meine Meinungen nicht 

gegen mich 
in Deutschland vertreten werden. 
Nope. They did not say it was a random number, just that your algorithm is 
not better than testing random numbers. That is not the same as saying that 

M is a true random number, in fact it certainly is not.
No, idiot!
Then you have a solution? I thought you said many times that you already had 

a solution. Now you are saying that if you figure it out, then you have a 
solution to the factoring problem. Idiot!
No idiot.
Any algorithm, including your crap algorithm is deterministic. So number 2 
above would be closest to the truth. The only problem I have with it is that 

the word work is in it. It does not work too well.
I did not dodge anything above. Idiot!
Basically no matter what I discover, there are some obsessive posters
who come forward to try and make it ordinary, no matter how
extraordinary it actually is.
Like with my prime counting function, which supposedly is not only not
new, but supposedly it's hundreds of years old, but no one has ever
seen it before, and to this day, no one can show anyone else besides me
as having giving that formula.
And what now?
worse than random method for factoring!
Yet the crucial equations that defy that insanity are
Ax= Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
Az= Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
where if M is some target to be factored, j is some integer of your
choosing, and the other variables are *rationals* which are guaranteed
to exist such that M is factored.
I noted in a previous post that if there's no way to determine those
rationals then in fact I've discovered a random number generator!!!
The reality is that I show extraordinary mathematics, and posters work
their butts off to try and convince that it's ordinary and they're damn
good at that job.
The problem here is that maybe I'm not smart enough to figure out
exactly how to get surrogate factoring to work, maybe not a single
freaking Usenet poster is smart enough, but how many hackers does it
take to bring down your world, if *any* of them are smart enough?
What if terrorists figure this out?
Will all the apologies in the world make a difference then?
Will all the protestations of you didn't think a crank could come up
with anything, or you figured it was just more nuttiness make a
difference then?
Will your rationalizations bring back a single life, restore a single
person's life that has been ruined?
You people seem to not know what mathematics is.
No matter how impossible you may think it is, if the algebra says one
thing then it must be so, and your social gyrations won't change that,
and your certainty that what I say can't be right, can't change that,
and if you die as a result of ignoring what is mathematically correct
then no one will bring you back to life.
Now then, so what if I can't quite figure it out, and if all of Usenet
can't quite figure it out?
You people are betting other people's lives.  You may be giving
terrorists tools.  You may be setting up smart hackers to invade the
world's systems without anyone knowing there is even a threat until the
evidence is overwhelming.
And when the evidence is overwhelming, what will you do then?
Apologize?
James Harris
How do you measure the extraordinaryness of your work?
Maybe I'm not smart enough to get my half finished perpetuum mobile
to work but what if shell figured it out?
In what way would it help if we brought this about faster by helping
you? Wouldn't it be much better if we just shut you up by any means
available?
Look kiddie, proofs are things humans accept. They say nothing about
absolute truth, only that no one has found a mistake yet. You can yabber
about the absoluteness of maths all day long but as long as humans execute
it, the results always come with the standard disclaimer.
Then maybe it doesn't work?
Yes, definitely.
By the way, by posting all your ideas to usenet, you really provide a lot 
of
help to the terrorists, do you know that? And by inciting others to do the
same you substantially increase the risk of the end of the world. Think 
about
it.
Volker
Surely nobody has suggested that this situation is not extraordinary.
The idea that you're so proud of your discovery even though _you_
admit that it doesn't _work_ is truly extraordinary.
Really. Exactly what is it you've discovered? A method of trying
to factor numbers that doesn't work. This is really not all that
hard to do.
************************
David C. Ullrich
[JSH]
[...]
My computer experiments suggested that two of your _algorithms_ perform 
worse than picking integers k at random in 1..M-1, then seeing whether 
gcd(k, M) reveals a factor.
That's not a matter of argument, it's a much simpler matter of running the 
algorithm, counting how many gcds it does before it finds a factor, and 
comparing that to the number of gcds a random method using a very good 
random number generator requires for the same M (I used the Mersenne Twister 

here, just because that's what Python ships with).  The method requiring the 

fewer gcds then wins.  Yours lost to random-gcd, over and over and over 
again, and I trust my computer to add 1 and compare integers more than I 
trust your claims of proof.
I haven't done those experiments on your latest algorithm, by which I mean 
the one that searches over splittings of T^3 j^2, and adds 2Tj^2.  I don't 
intend to spend time on that either, unless you explain first why it doesn't 

factor some odd composites that aren't the square of a prime.  If you don't, 

then I assume your latest proof is also incorrect (beats me -- I can't make 

sense of your proofs; Rick Decker has pointed at steps he believes are 
incorrect, though), in which case you'll come up with some other algorithm. 

When you come up with an algorithm that actually works (actually factors 
integers that people give to it), then I'll pay attention again.
Here's a simple challenge for you:  you claim your latest algorithm should 
work.  At least three people (including me) have given examples of specific 

examples, and the latest method can generate an enormous number of gcds, so 

they're hard to analyze (either because the integers are huge, or the number 

of gcds is in the millions).
So here's a much easier case:  my implementation of your latest algorithm 
fails to find a factor of M=7387 at j=1.  It tries 640 gcds:  632 return 1, 

and 8 return 7387.  (BTW, if it were picking gcd candidates truly at random, 

it would be expected to find a factor of 7387 about 14 times in 640 tries.)
So you show the arithmetic:  if your algorithm is correct, show how it 
factors M=7387 at j=1.  I don't care how that turns out:  if you can show 
how it factors that, great, it will point out an error in my 
implementationm, I'll repair that, and retry the other failing examples I 
posted (and I could post many more -- literally millions of failing 
examples).
If you can't show how it factors M=7387 at j=1, you have no cause for 
complaint here.
...
You have never shown anything extraordinary that I can recall.
Your prime counting function? Oh right, you mean your prime counting 
algorithm. Hmmmm... when will you learn that point???
Well, yes, it is a slow uninteresting convoluted obfuscation of a very old 
algorithm. Why do you bring that crap up again?
No. It is not random. It is ineffective but it is still deterministic. And 
yes, it does appear to be worse (or at least no better) than random trial 
division.
I do not follow your argument here (what else is new?). But even if you are 

correct that you have some argument that would show that guarantee, that 
says nothing about the time complexity of your solution. After all, trial 
division is also able to guarantee a solution if you give it enough time. 
That is the easy part. Can you provide some Big O notation that proves that 

you complete in polynomial time as a function of bit length?
You do not seem to unsderstand what random number generatoion means. What 
else is new???
You show extraordinary stupid and invalid mathematics. Work their butts off? 

Hardly... it is more like shooting fish in a barrel.
They already have lots of ways to do harm. Cracking RSA will not be a big 
interest for them. If they do use Surrogate Factoring, the blame will be on 

your head.
Who will die?
Right. So why I are you talking to us?
When the evidence is overwhelming? Sure James... I will apologize then. But 

that will not be too likely, right? Until that happens, I will be happy to 
just laugh at you. Idiot! 
Coming from the inventor of surrogate factoring, I wouldn't be surprised if
this prime counting function of yours were indeed new, but a bunch of
random obfuscation thrown together that barely works.
All evidences point to that. And you still are unable to argue against that
fact. So how could any sane person declare that surrogate factoring is any
better than random?
It's also guaranteed that y^2 == x^2 (mod n) but x != +/- y (mod n) exist,
and that gcd(x-y,n) is a non-trivial factor of n. So what? If you can't
give an efficient and reliable method to generate the variables required to
split n. So your method is as good as useless.
And not a very good one at that, believe me.
THE TERRORISTS! RUN!!!
Let me guess, they'll steal all our credit cards to buy goat porn and ruin
Western civilization.
Oh sure. Let's all waste our time pointing out inevitable errors in methods
devised by cranks, on the off chance that they might work, instead of doing
actual, productive work.
The algebra also says that congruence of squares works. It didn't matter
until someone invented the Quadratic Sieve or the Number Field Sieve.
Don't you understand? These equations of yours are useless without a method
of reliably specifying the missing parameters so that non-trivial
factorizations come out.
Oh sure, but THE TERRORISTS CAN.
Why don't you do away with these nonsensical and grandiose rants of yours?
You sound like G. W. Bush at times. Only thing is that people here aren't
stupid sheep and won't fall for your terrorist boogeyman crap. What's next?
`Won't you think of the children?'
D.8ecio
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
Uh, yes, that is the principle behind public key cryptography.  People
relying on the fact that even the smartest mathematicians when working
for decades on the problem of factoring have not been able to get
better solutions to work than the current solutions.
And they have not started with some pretty arbitrary doodling of a
nincompoop.  Given that they have not managed to get better than the
current results in all that time, the likelihood that some hacker from
outside the field starting with the scratch pad of a mathematical
nobody with a record of falsehoods and irrelevancies under his belt
(claiming that he does not yet understand his scratch pad) will fare
better is not enough to lose any sleep over.
Not really.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
Some of you seem to think this is a game, where there are no real world
consequences, but basically I found some algebraic expressions.
That algebra is not James Harris algebra.  The equations are not marked
with JSH.
They are algebra.  The equations are simply mathematically correct.
Now if I have
x = sqrt(y^2 + n)
there are an infinite number of rational solutions for y such that the
square root is rational.
But, beyond the integer solutions for y, you just have primes that get
added in obviously, such that it's of no interest to consider other
solutions.
However, in the system of equations I've found:
Ax= Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
Az= Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
you have it that there are an infinite number of rationals, but if
there's a rhyme or reason to them, then you can consistently factor M.
If there is no rhyme or reason to them, then you have a random number
generator...a perfect random number generator.
Back to x = sqrt(y^2 + n), as imagine you have a solution y = s, where
s is an integer, then *any* other solution can be written as sf_1/f_2,
where f_1 and f_2 are integers as well, but you have to pick them.
With my equations, to find Az such that it factors M is easily done by
knowing the factors of M ahead of time, as then you can pick an integer
Az that factors M, and then you have a rational Ax.
If there's no rule to how you get a rational Ax, then the denominator
of Ax is a random number, unlike with my example with x = sqrt(y^2 + n)
where your choices tell you what the denominator must be.
If you choose f_2 then only prime factors of f_2 are in the
denominator.
But if you pick Az such that it is an integer with only one prime
factor of M, then how do you know what the denominator of Ax will be?
You have to calculate.
Now then, if there is in fact a rule for what that denominator is, then
the system is not a perfect random number generator, and *someone* can
figure out how to get it to work.
That someone would be one of the most powerful people in the world.
Right now, while some of you spend a LOT of time and effort trying to
deflect interest in my work...playing the usual social games...that
someone might be making mischief, spreading their knowledge, informing
others, until there's a gang of people capable of exploiting this
system.
It's algebra.
Some of you want to tag it Harris algebra or Harris mathematics.
Some of you want to make it where mathematical truth is some kind of
personal thing, so that you can feel safe, while ignoring this
research.
Well, you're playing Russian Roulette with the freaking world.
And God Himself won't help you if this goes bad as despite your beliefs
I can assure you that angry people will against all law if necessary
tear their rage out of your hides if it goes badly, and the story is
that some of you played stupid social games, bet the world, without its
knowledge...and lost.
You people are betting the world, without the world even knowing that
it's up for grabs, without people even having a choice as they don't
know what you're doing, as you're not informing them.
And for that, if it goes badly, I assure you they will tear some of you
limb from limb.
James Harris
Nihil ex nihilo.
Wake up Harris you have got nothing.
OK, here's the thing. I use RSA routinely to protect private information. I
suspect many of the other posters here do the same. Now why on Earth would
anyone ignore such a threat to the security of their data? I tend to follow
cryptographic developments and when the recent attack on SHA-1 was
discovered, it made me pause for thought - are any of the systems I work on
using SHA-1?
It doesn't make sense at all. The two obvious possibilities are these :
1) In blatant disregard for the security of their own data, a significant
number of people are ignoring your new factoring technique for emotional
reasons.
2) Your factoring technique is no threat to the security of encrypted data.
Which of these do you find more plausible, and why?
Duncan
You are the one who's admitted it's a game. You put half thought out 
ideas on Usenet, and you wait for others to find the errors. You enjoy 
insulting people. You enjoy the attention.
world
marked
the
get
M.
where
sf_1/f_2,
by
integer
n)
then
can
informing
I think the word you're looking for here is: Afro-engineering.
beliefs
One could take such assurance to mean you plan to incite a riot.
You better watch what you're saying.
its
you
More death threats, eh? You don't take your ISP's Terms of Service
seriously, do you?
 Discussion, linux)
What is the point of that comment?
What James S. Harris diddles with has nothing to do with his race.
-- 
Jesse F. Hughes
The sole cause of all human misery is the inability of people
to sit quietly in their rooms. -- Blaise Pascal
What is the point of James calling everyone a liar?
What is the point of James making racist comments about Germans?
What is the point of James making death threats?
Not what he diddles with but the way he diddles with it.
Is it your opinion that James' psychological problems are
totally unrelated to his race? The evidence suggests otherwise.
I agree, it is related to his race, or more accurately, it is related to his 

species. But his race and species are not what he claims them to be. In 
fact, James is a hideous mutant, hopefully the first and last member of a 
unique species. His resemblance to the black human is purely an example of 
convergent evolution. He was the result of extraterrestrial alien 
experiments involving recombinant DNA fiddling with aardvark and sea 
cucumber.
ENOUGH ALREADY! As irritating as James can be at times (as I know
from personal experience), there's no call for this behavior.
Act like adults, people, hard though it might be.
Please,
Rick
 Discussion, linux)
Ask him.
Maybe his psychological problems are related to his race (contrary to
what I said)[1], but that doesn't excuse your remark which seems to be
racist.  
I ask again, what was the point of that comment?  Was it attributing
to James's race the obvious, er, non-standard mathematical methods of
James?
Footnotes: 
[1]  I don't see any particular reason to believe so.
-- 
Customers have come to SCO asking what they can do to respect and
help protect the rights of the SCO intellectual property in Linux.
SCO has created the Intellectual Property License for Linux in
response to these customers needs. -- SCO responds to needs.
What does this mean?
 Discussion, linux)
Never post when you're running late for a train.
I meant, James's mathematical diddling has nothing to do with his
race.
-- 
So, at this time, I'd like to assure you that I am not interested in
I'll have prosecutors knocking on your doors.  I have no problem with
Let me help you to word this correctly. What you are trying to say is that:
There are an infinite number of rational y such that x is also 
rational.
I'm wondering James - are there countably infinitely many such y ? Or, are
there uncountably infinitely many such y ?
This is the most ridiculous thing I've ever read in my life.
Yes, ridiculous... And he seems to get into an orgasmic frenzy as he starts 

to talk about the physical harm that will eventually come our way... I 
imagine that he would be typing these words with the one free hand that was 

not gooped up with Vasoline. 
starts
was
I'm terribly frightened at being torn limb from limb, to wind up looking
like a disassembled stick-man. Just a collection of disconnected K2 graphs.
Oh the pain !
I think that the world would turn on you James. You were the one that got 
the ball rolling. ANd it was you who failed to provide the proper theorems, 

proofs, complexity analysis, examples, implementations, etc. You ed up. 
As you have often said, we sci.math people are not mathematicians. We are 
not capable. So nobody would consider us to be blamed. Butas you have often 

said, JSH is the only true mathematician here, so the onus is on you to 
carry Surrogate Factoring to fruition. Until you have factored the RSA 
challenge numbers, you have failed. If Osama now takes your work and 
cracks some credit card transactions, then it will be all your fault. Idiot! 

The Meyers-Serrin Theorem states that Sobolev space W^{k,p}(U) = H^{k,p}(U), 

where the latter is defined as the completion/closure of smooth function set 

{u in C^{infinity}(U), u has finite (k,p) norm} under the (k,p) norm.
My question is, what's the condition on U? Is it a general open set in R^n, 

or does it have to be bounded? maybe even regularity conditions on its 
boundary?
My book is vague about this issue. Please give me a definite answer with 
references if you could (I have no access to the original literature). 
H^{k,p}(U),
function set
norm.
in R^n,
its boundary?
with
literature).
U may be any open set.The best reference is Adams and Fournier ,Sobelev
Ziemer ,Weakly Differentiable Functions,Springer 1989
Posters have backed themselves into a corner, and I'm pointing that out
to you, as this is not a game, no matter what they think.
If surrogate factoring is random as they claim, then necessarily
solutions for Ax will give you random numbers.
It will be a perfect random number generator, where you can take an M
with a known factorization, and use it to get random numbers.
If they are *lying* as I say they are, then in fact, the denominator of
Ax for an integer Az that factors M, will have the prime factors of T,
and only the prime factors of T, as its own factors.
Now I know how this test will go, as I've said, the denominator of Ax
that follows from an Az that is an integer that factors M, will have
*only* the prime factors of T as its own prime factors.
That shows that surrogate factoring can in fact be made to work.
It's a reality test.
Now algebra is not something that you can wish away, or hope that some
crank's word doesn't mean anything about, and some of you may think
you're protecting yourselves by downplaying surrogate factorization, as
you may believe that you can block interest, and keep yourselves safe
from serious scrutiny from the world.
You are wrong.
Already there are troublesome headlines that relate to Internet
security.
I fear that soon there will be a flood, and eventually, the truth will
come out.
If someone dies, then you can be convicted of murder for blocking the
truth.
And beyond the law, if it goes badly enough, people will probably just
kill you, without regard to what the law says.
Whether you noticed or not, billions of dollars US are dependent on the
idea that factoring is a hard problem, and that's billions of dollars
daily incentive for people wanting to break that system.
They don't care about your social needs, your personal fear, or your
agony at a world that's JUST NOT FAIR.
You will die just the same, as, guess what?  You are not the center of
the world.
Some of you are betting other people's lives, other people's welfare,
in your selfish need to believe you are the center of the world.
And you will take down not only yourselves, but your families, and your
communities, or even your entire countries, in one of the worst
debacles in the history of the world.
It is foretold in most mythologies and religions.
And yes, you are the ones, the cursed ones, who destroy the world.
James Harris
 Discussion, linux)
Are you sure you won't get in trouble for this tack?  I thought
spewing on and on about random numbers was really Eray's schtick.  You
don't want to encroach on his territory.
-- 
These mathematicians are worse than communists, as how do you explain
their behavior? I *am* the American Dream, fighting for what should be
mine, having to get past weak-minded academics who are fighting to
block my success. But I shall prevail!!! -- James S. Harris
James... you desperately need to get laid!
ink
out
of
T,
some
as
will
So are you going to start refering to surrogate factoring as
The Tsunami?
just
the
of
your
Does your java program factor 666?
Started drinking early this week, eh?
No. Nobody said that that I recall. People have said that your algorithm is 

no better than random. Pseudo random may have been i bit more correct if we 

are talking about real implementations, but nobody said your algorithm would 

be a random (or even a good pseudo random) process.
Idiot.
You have not shown it to be true yet. But lets say it works. The question 
becomes how fats does it work? Can you give us some Big O notation on this?
Yes. And so far you have failed that reality test.
Apperantly, you are also not something that one just can wish away...
Can you? Can you site a legal case that set that? Where blocking a 
mathematical truth resulted in a murder conviction. James has become a crank 

lawer now? And how will cracking RSA result in a death?
That would make you so very happy. Idiot!
James the crank historian? Nope. Lots of bigger debacles in the history of 
the world.
James is now a crank cult leader, a crank lawer, a crank historian, a crank 

mathematician!
Hey... Ragnarok is here!!!! Party hardy!!! Tear James limb from limb, cause 

we all goin' ta hell anyhoooooooo!
Can a qualified skeptic prove the Truman is a hoax?
www.supernerd.com.au/~gray77/truman-challenge.html
3 days have passed and still no skeptic has picked up the $1,000 on offer.
Herc
--
The TRUeMAN CHALLENGE $1000AU
Can a qualified skeptic prove the Truman is a hoax?
www.supernerd.com.au/~gray77/truman-challenge.html
Nice! were you one of the guys in the videos? also, where was that done?
I was the guy doing the choking.  It was done in South Beach Fremantle on 
the 30th Jan.  Good fun.
Fraser 
yeah, i thought that beach looked familiar! I was imagining it further down 

the coast toward madora bay though (too many damn beaches :B). Were those 
yellow guys serious? I hope they were, nothing better than the satisfaction 

of seing people who think they are better than everyone else and 'special' 
somehow being knocked back down to size. 
Yep.  They started challenging on rec.marital-arts about 2 years ago.  They 

finally stood up for the challenge.  They honestly believed they could knock 

someone down without touching them.  Idiots.
Fraser 
Now you are scamming people.
In sci.math, W
The Mall exists, so it's not a scam.    Of course,
the Trueman probably meets a lot of people in the mall
(malls are like that), so he may not be able to prove
that he's being met for 30 seconds, especially if he's
standing in line to purchase another shrimp for the barbie...
-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
In sci.math, |-|erc
        '-693499 days have passed since this challenge was started'.
[Woo.  Competence.]
        Only registered members of official Australia Skeptic Outlets
        can apply.
[Great.  I'm nowhere near Australia, and although I'm skeptical enough,
I'm not an official anything.]
        I will pay you $1,000 AU to meet me for 30 seconds in the centre
        of Flinders Mall Townsville Qld.
[Nice place for a vacation.  Would I want to shop there?]
        This will prove I am the Truman and the Govt. spies on me and
        follows me everywhere with a satellite PA system.
[What, do they scream in your ear or something?  WE KNOW YOU'RE
IN THE MALL, TRUEMAN!  *screeeee...*  COME OUT WITH YOUR ARGUMENTS
WHERE WE CAN SEE THEM!]
        The only requirement is you are not deaf, I'm not holding my
        breath.
[Well, if I had a PA system following me around I probably would be.]
        $500 will be held by a mutually delegated umpire by paypal on
        acceptance of the challenge.
[That's nice.]
        The commitee member agrees to publish his findings.
[Now why would he need to do that?  All you'd need to do is read
his mind.]
        Challenge started Feb 16 2005
        To date no skeptic has shown up to collect the $1000 
[Well, I'm not going to, either.  So there. :-P ]
[.sigsnip]
-- 
#191, ewill3@earthlink.net -- are the PowerPuff girls nearby?  (Is that
an old joke by now?)
It's still legal to go .sigless.
you and your UNIX toys.  get a standard browser like 95% of people.
check the formula, it skips a few days here and there but overall runs 
pretty accurate.
Move, join, you'll live longer in Australia when I'm armed.
Having only bought food for 5 years I wouldn't know.
Hang on I'll ask.... THIS IS BULL WERE UP... ING HELL
one CIA agent just passed onto you..
Im wearing industrial earmuffs as we speak, once you get used to the 
pillow
trick you'd never take them off.
Honestly when I take them off, (after 2 years of using them at home and 
earplugs
when out), sounds are amazingly loud and crystal.
or a plane ticket for the skunge skeptic.
the CIA reads minds, the CIA knows everything (literally), they don't
need to convince people of anything.
mind you they will try to brainwash you into not going public by putting 
ideas in your head.
The devout skeptic, an unqualified everything, the mutually exclusive 
contradiciton in terms in the flesh.
Herc
It would take another nutcase to even think about doing this !
Why?
WHAT IS SO DIFFICULT TO GET THROUGH YOUR PINHOLE BRAIN
THAT MAYBE JUST MAYBE THE TRUMAN SHOW IS BASED ON REAL TECHNOLOGY
$1000 if any skeptic can disprove me.
Herc
Prove you have $1,000.00,  for brains.
I pay $500 up front, soon as a recognised skeptic agrees to fly up.
Herc
Okay, so prove you actually have $500.
what safety guarantee will a skeptic have from a ranting madman like
you?
You are but a disgruntled, disturbed, abusive, angry, misguided,
unstable dirty old man that is blaming everybody else for your
own problems.
And
It is not the world that has to disprove you, but it is
YOU that has to prove YOUR theories.
Creating a Web page with the only thing on it Prove me wrong,
is rather silly. But this does prove one thing, you are NUTS.
$1000 says you're wrong pansy boy
Herc
Prove me right doh boy, publish your findings in a
scientific journal and $1000 is yours.
Herc
why? why don't your prove to us that it is true troll?
But who wants to travel to that hole, Townsville, to do it??
-- 
Peter Lucas               * In a world in which we are all *
Brisbane                   * slaves to the laws of gravity,   *
Australia                   * I am proud to be called         *
                               * one of the Freedom Fighters. *
certainly not for our madman |-|erc, but sya i hear that are lots of
gorgeous aussie sheilas out there, and since it's a warm place they
habitually scrap lots of clothing. wrong?
That's why Brisbane and it's environs is so nice to visit nowadays 
-- 
Peter Lucas               * In a world in which we are all *
Brisbane                   * slaves to the laws of gravity,   *
Australia                   * I am proud to be called         *
                               * one of the Freedom Fighters. *
because you only accept words spoken by newsreaders as gospel,
no sequence of words on usenet can prove there is a real truman.
the newsreaders and practically anyone high up in media are in on it.
the only way to prove a non mainstream claim on usenet is to punch
people in the face with it.
Herc
Go to university and get a independent researcher to view your
findings and print it in a newspaper.
Making up excuses why you can't do this, is not proof you are.
tried that.
find me an email address of a Uni student who isn't a devout skeptic and I 
will.
Herc
why does it have to be Trueman or Truman, why not Madman or
knuckleheadedman or something along those lines?
A Student ?
Gotta do better than that, but then that's just another excuse so
it must be true.
Have you ever been to a University?  Do you have any idea how the research 
projects are organised?
didn't think so
Herc
prove it to us troll. you can't, can ya?
I already have.  I can lead you to water but I can't make you drink
Herc
T R U e M A N   W I T N E S S E S
Hey Trueman...
what up, my uncle told me you post here... he's did some audio on your
secret show and gave me some tapes. love the show.
You rule Truman.
I was in Townsville over the weekend, and I heard him.
Very spooky!
I'm from Townsville and YOU ARE the Truman
I'm in Townsville.  We're sick of you
I've been hearing stuff, yeah.
BTW, when I was in The Strand last weekend, I didn't hear your broadcast 
per
se. Well, at least I thought I didn't. Again, who knows? LOL.
Anyhow, you think your the Truman. So WTF, I think you're the Truman too.
Therefore 100,000 Townsville residents and 1 Melbourne tourist can't be 
wrong.
you're an idiot |-|erc, why would anyone want to join the skeptic
rubbish just to meet a madman like you for 30 seconds? nobody would do
it even with $1,000,000AU you nutcase. a better one if they see you so
that they can take you voluntarily to an asylum and place you in a
straitjacket.
Yet another thread to go into my killfile. Why do morons keep replying to 
Herc the retard? 
The TRUeMAN CHALLENGE
Can a qualified skeptic prove the Truman is a hoax?
Challenge started Feb 16 2005
To date no skeptic has shown up to collect the $1000
Herc
http://www.supernerd.com.au/~gray77/truman-challenge.html
 nobody, not even a non-skeptic will go for this 'come meet the madman
for 30 sec' challenge. raise the stakes and up the prize money to 1000
grand or so. then, maybe then only, will someone be interested.
You're probably right, but by the time I can afford that I'll be long gone
out to stud on my ranch.  I will pay the airfair up front in Aus of any
notarised skeptic.
Lots of skeptic organisations have $1000 prizes and $2,000 prizes.  I might
put it up to $10,000 by end of year depending on business.
Herc
madman
1000
Dickhead, you can't even afford $1.00 to play your ed up little
girlie game.
As you said, skeptic organizations provide rewards for people that can
prove what they are saying is TRUE. They don't offer rewards to people
to Prove them wrong.
GET A BRAIN.
I'm paying a skeptic to prove me wrong.
And all he has to do is stand next to me and say there's nothing that 
indicates
the government is spying on me.
Herc
can
people
indicates
So what is it of earth shattering uniqueness is it that would make any
government want to spy on you?
I'm better than you.
Herc
that
that
any
Madder than me would be accurate,  for brains. As for better, I
don't have spy satellites fllowing me around all day long.
Prove you are better than me,  face.
How do you know the GOV is spying on you ?, are you sure its
not aliens or have you been probed.
the spy satellites talk back.
Herc
--
The TRUeMAN CHALLENGE $1000AU
Can a qualified skeptic prove the Truman is a hoax?
www.supernerd.com.au/~gray77/truman-challenge.html
anyone could be an alien, I'm using Occams razor that the sound projected 
everywhere
around me and people talking to me constantly out of nowhere is human 
oriented.
Herc
Yip, thats pretty hard evidence alright.
its hard evidence of a Schizophrenic mind. Paranoia, hearing 'voices' when 
theres no one else around, its a textbook case. Do you know why 'in space no 

one can hear you scream'? its because sound waves can not be carried in the 

vacuume of space as theres no matter (solid/liquid/gas) for the waves to 
travel through. The reason radio waves still travel through space is because 

of the ultra long wavelengths they use. 
projected
human
when
space no
in the
to
because
Nope, the reason nobody can hear you scream in space is because sound
is a series of vibrations travelling through a solid, liquid or gas.
Spce, being a near perfect vacuum, does not have a high enough density
of gas to transmit audio freqwuency vibrations. Radio waves are, like
electromagnetic spectrum.
the point is still valid though, how can anything in space speak? wow, do i 

get my $1000 now ;P 
maybe you should try a physics text book instead.
Herc
--
Everyone gets 'out of it' 10, 20, 30 hours each and every week,
I get 5% there and 1, 2, 3 mintues later police arrive and lock
me away for 'strange behaviour ~ The Trueman
Mental Hospital.  Go. Deal. 
your kindergarten debunking charade non sequitur
Herc
IÇd much prefer you to use OccamÇs razor 
to 
slash your wrists. Please?
-- 
/Je¤us/
I'm always amazed to hear of air crash victims so badly mutilated that they
have to be identified by their dental records. What I can't understand is,
if they don't know who you are, how do they know who your dentist is?
-  Paul Merton
Has your paroll officer or Doktor heard any of these voices ?
10,000 people have heard the talking spy satellites,
though its hard to tell its a satellite in a crowded place.
when the noise follows you on the beach and in the bush
with nothing man made in sight there's only one direction left, up.
Herc
quote of the day
Hang on I'll ask.... THIS IS BULL WERE UP... ING HELL
one CIA agent just passed onto you..
The whle bloody world knows you are a bulhsitter, hercie child. Proving
you are a wit and a haox is about as useful as proving the sky is
blue and rain is wet.
You have no use for $1000?   PROVE I'm not the Truman and its yours.
Details on how to prove I'm not the truman here 
www.supernerd.com.au/~gray77/truman-challenge.html
W I T N E S S E S
Hey Trueman...
what up, my uncle told me you post here... he's did some audio on your
secret show and gave me some tapes. love the show.
You rule Truman.
I was in Townsville over the weekend, and I heard him.
Very spooky!
I'm from Townsville and YOU ARE the Truman
I'm in Townsville.  We're sick of you
I've been hearing stuff, yeah.
BTW, when I was in The Strand last weekend, I didn't hear your broadcast 
per
se. Well, at least I thought I didn't. Again, who knows? LOL.
Anyhow, you think your the Truman. So WTF, I think you're the Truman too.
Therefore 100,000 Townsville residents and 1 Melbourne tourist can't be 
wrong.
You can winge all you like, every day the counter goes up that no skeptic 
could disprove
me for $1000.
Herc
Proving
is
Only one piss weak thousand? For  sakes, little boy, I spend that
much on fuel in a week or two. One thousand, litle man, is play money
to me.
PROVE I'm not the Truman and its yours.
You're not the truman,  face, you're just a little dole bludging
fucwit. You make the claim that you are the truman, therefore it is on
you to prove your juvenile bull.
money talks, BS walks
I'm the Truman so shut up you plebian wanker.
Herc
bludging
on
And you are all bull. In fact, you're like a bull with no horns and
no balls. You can't butt, you can't , you just sit there and
bull.
So you say,  face. Now prove it.
The TRUeMAN CHALLENGE $1000AU
Can a qualified skeptic prove the Truman is a hoax?
www.supernerd.com.au/~gray77/truman-challenge.html
Verifiable witnesses referred to on the site.
Herc
Which is one explanation why you are still saving up for a scooter.
You are a raving lunatic.
-- 
Find out about Australia's most dangerous Doomsday Cult:
http://users.bigpond.net.au/wanglese/pebble.htm
You can't fool me, it's turtles all the way down.
Now I've accused some posters of lying, and I've issued warnings
mentioning hackers and terrorists, so it's only fair that I give an
easy test for people to defend themselves, as later, there may be no
defense.  My accusations alone may be enough to get some of you in
trouble, in a world where few people will be interested in listening to
reasons.
They may simply act.
The test is simple.
I say that the denominator of Ax for an integer solution of Az,
contains only prime factors of T.
Where
Ax= Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
Az= Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
and T = M^2 - j^2, and so the test would be for someone to pick an M
for which they have the factorization, and pick some integer j, to get
T.
Then use an Az which has only one prime factor of M, to solve for an
Ax, and factor its denominator.
I'm *telling* you that the denominator will only have prime factors in
common with T.
If I'm wrong, then the world has its first perfect random number
generator.
If I'm right, then for all we know by this time, hackers may be having
a field day.
It's not a hard test.
I mention for governments and their prosecutors who may come later that
some of the posters here may have acted in their own selfish interest,
possibly believing they could exact some personal gain by lying here.
Others may represent third parties.
It would make sense to check them all very carefully.
James Harris
Nihil ex nihilo.
Wake up Harris you have got nothing.
 Discussion, linux)
This is a great warning.  I'm sure government agents will later find
it very helpful.  Be sure you don't accidentally delete it from
Google, 'kay?
Third parties.  Right in our midst.  It seems so obvious now.  
I wonder which group David Ullrich represents.  PLO?  IRA?  IRS?
Could be anyone.
-- 
Jesse F. Hughes
I may have invented it, but Bill [Gates] made it famous.
       -- David Bradley, inventor of the Ctrl-Alt-Del reboot sequence
WTF.
************************
David C. Ullrich
 Discussion, linux)
World Trade Farts?  Oh.  I bet that's the Bilderbergers.
-- 
So, at this time, I'd like to assure you that I am not interested in
I'll have prosecutors knocking on your doors.  I have no problem with
Okay, let's do the test. I'm online, waiting to see if any of my
students have any problems with the WeBWorK homework. They don't seem
to be having any, so I did some surfing and found your post.
If you have an algorithm that claims to do factorization, this step
isn't allowed. A factorization algorithm is supposed to work from a
number whose factorization is unknown, and to determine whether there
is a factor.
Note that once the alleged factor is found, it's very easy to verify
that it actually is a factor: long division.
But I'm going to ignore that here, and play along anyway. I choose
M = 91 = 7 * 13.
j = 10.
T = 91^2 - 10^2 = 8181.
Once again, if you don't know what the prime factors of M are ahead of
time, this step can't be taken.
Nevertheless, I'll play along anyway. I'll choose Az = 7.
Ax = 7 * (-7 +/- sqrt((7 - 2 * 91^2)^2 - 4 * 8181 * 91^2))
        /(2 * 10^2 - 2 * 7)
Now (7 - 2 * 91^2)^2 - 4 * 8181 * 91^2 = 3080581 (I checked this on my
calculator), which is not a perfect square. So Ax is an irrational
number.
In light of what I've said about 3080581, I can't do this with either
Ax.
Another thing I could do here would be to factor
2 * 10^2 - 2 * 7 = 186 = 2 * 3 * 31.
What exactly do you mean by the denominator will only have prime
factors in common with T? Do you mean that every prime factor of T is
a prime factor of the nonexistent denominator, or every prime factor of
T is a prime factor of 186, or every prime factor of 186 is a prime
factor of T, or every prime factor of the nonexistent denominator is a
prime factor of T? None of these statements are true. T happens to
factor into 3^4 * 101.
There is no such thing as a random number generator, by definition.
Random means having no pattern or organization, and generator is a
procedure to create a set of objects in an orderly fashion. Perhaps you
mean a pseudo-random number generator, a procedure which produces a
sequence of numbers which can't be back-engineered, but there are
already plenty of those ...
A true algorithm (look up the definition) is a procedure designed to
solve a certain (fixed) problem (and to terminate). You are not allowed
to say that with certain outputs that Problem A (factorization) has
been solved and that with other outputs that Problem B (perfect random
number generation) has been solved. It has to be one or the other.
And it didn't take long, only ten minutes. And your procedure failed
the test, no matter how the vague steps are interpreted.
Lying means someone saying something is true which they know to be
false. The word you're looking for here is disagreeing, which means
that someone who has not yet been convinced will say that your work
is false, which they know (or believe) to be true.
Others may represent fourth parties, or fifth parties, or ... or
omega-th parties ...
Speaking of checking things carefully, you should go back to the
analysis you said you did (in another post) that says that the
denominator will only have prime factors in common with T, because it
clearly isn't true. (Or you haven't explained things correctly.)
Whenever there's a counterexample to a statement, any analysis
proving that statement is invalid. Better yet, why not post the
analysis? Then we can all check you very carefully.
     --- Christopher Heckman
[snippage]
You seem to have an intermittent contact on your SHIFT key. HTH :)
-- 
Larry Lard
Replies to group please
Actually, it's supposed to be capitalized that way. It's like the
typesetting languages TeX and LaTeX.
     --- Christopher Heckman
to
.... blah blah blah
WTF?  This is like daytime television... 54 channels and nothing on...
Dude, you should go invent unlimited compression and go school the
comp.compression folk.  They're so in need of some schooling.
Tom
[Nonsensical ranting garbage deleted]
Mr. Harris,
1.  If this method is so superior, factor the RSA challenge numbers!  What
is wrong, you make extraordinary claims about your work.  Just show one
example using an RSA Challenge number.
2.  If your prime counting algorithm is so superior, tell us what PrimePi 
is
for x = 10^22 through 10^25.  Come on, you are making extraordinary claims,
show us.
3.  If your diarrhea factoring method is a true RNG, test it's entropy and
give us results.  Lets make it easier, test it against DIEHARD and see how
it fairs.
Can't you realize that the entire sum of everything you have done is
worthless?
Are you blind, deaf and dumb?
Think about what you are doing to yourself.
At the end of your life, all you would have accomplished is the label of
troll and super crank.
Is that your true goal, or are you a NPD experiment from MIT?
I feel sorry for you sometimes because you have wonderful tenacity. Too bad
you have a psychological disorder that will never allow to amount to
anything.
Just for your information if you are interested in calculating pi (N) 
for many values of N in the range from 10^22 to 10^25: The Extended 
Meissel-Lehmer algorithm takes time proportional to O (N^(2/3)) to 
calculate pi (N). However, it can be modified to calculate many values 
pi (x(i)), 1 <= x <= n, with x (i) <= N for all i, in time proportional 
to n^(1/3) * N^(2/3), so if you want to calculate pi (k * 10^22) for 
_all_ 1 <= k <= 1000, that can be done in about 10 times the time it 
takes to calculate pi (10^25). 
My implementation of the prime counting function found pi (10^19) in 
about eight hours on a PowerPC G4 running at 733 MHz. The code would 
pi (10^25) could be found in 10,000 times the time, and the values pi (k 
* 10^22) for 1 <= k <= 1000 in 10 times as much time again. 
That is 800,000 hours on my CPU or about 91 years. A dual 2.5 GHz 
processor should be lets say 7 times faster, that's 13 years. About 
$50,000 worth of hardware should do it within one year. 
By pure coincidence, $50,000 is also the cost of some proposed hardware 
that could crack 512 bit RSA keys in reasonable time, according to a 
design by (I believe) two of R, S and A. A custom designed chip, built 
in such a clever way that every single transistor is continuously doing 
useful work in the cracking process. Cracking 1024 bit keys is supposed 
to be possible with about $10,000,000 worth of hardware with this 
design.
.
Is that the Hebrew Hammer?!
I think James has in mind the hammer of the Nordic god Thor. He called
himself a mythology buff around the time when he started to threaten
us with a hammer, if memory serves.
Newsgroups trimmed.
device for mathematician's around the UNIVERSE.
In one fail swoop, the not for human eyes beauty of James' work, will
flatten us all into the one-dimensional mathematics machine!
resembling beauty from James' lunatic and moronic rants!
-- 
Patrick Hamlyn   posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group (polyforms-subscribe@egroups.com)
He probably meant 'one fell swoop' but did not know how to spell it.
-- 
Gianna Stefani
Did you prove that?
You are wrong, but no, the world does not have a perfect random number 
generator. Why do you keep saying this? Do you know what a random number 
generator is? Can you give us your definition?
They might be. RSA has never been proven to be secure. But I am as sure as I 

can be that your methods are not giving them any help.
I have no gain from calling you a liar and a fool. You are a liar and a 
fool. I stand by that. If a prosecutor asked me what I thought of you and 
your nonsense, I would tell them that you are a liar and a fool. End of 
story... Idiot! 
Here's an even easier test of surrogate factoring: I gave you a list
of fifty 20-digit composite numbers a couple weeks ago and you haven't
factored a single one.  I think I know what test score to assign,
based on that result.
haven't
No, it's not a fair test.  I'll explain why, again, and I'll also
explain again, what the test is, and why my test is an easy one.
Ax= Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
Az= Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
shows that you have a set of rational Ax mapped to rational Az
solutions, where you have dependencies on the factorizations of TM^2
and Tj^2 to make the square roots rational.
Notice, you are guaranteed to have an integer Az that factors M, from
the first equation, which gives a rational Ax, and searches for
algorithms are basically about figuring out how to get all the possible
integer Az's.
However, you can do the time honored technique of working backwards to
see actual solutions, as is easily calculated if you *do* know the
prime factors of M, as then you can just pull out all the integer Az's
from the first equation.
It's a well-known technique to work backwards to figure out what's
going on.
My test is to work backwards.
If I'm wrong, then more knowledge doesn't help me, now does it?
I'd *hide* from the truth, but instead I want you to notice that
posters trying to convince you that I'm wrong are the ones who are
doing the running.
Now then, if sf is random, then the denominators of those Ax's found
that way will necessarily be random.
If it's not random, then some way exists to figure out how to get all
the integer Az's without knowing M.
Those are the two possibilities.
Now posters working hard to discount this research--apparently deciding
that it's Harris mathematics or JSH algebra--are avoiding the actual
issue from what I've seen in looking at posts in this thread.
The reason is simple, the mathematics doesn't work for them either way,
if their intent is to discount it.
That's what happens with major math results, human beings working hard
to dismiss them have to ignore facts.
Now my test is simple: calculate Ax, by taking an M with known factors,
and *look* at the prime factors of the denominator of Ax, and you will
find that they are the same prime factors as T has.
That proves that there are rules governing the value of the rationals
Ax's that results from the integer Az's.
And then, algorithms *can* be developed.
Now, if mathematicians were what they claimed then it wouldn't be
required that I actually produce a working program that can factor RSA
numbers, as that's an unreasonable request.  It's an unresaonable
request as if I were at that point I wouldn't need to convince anyone,
as I could just demonstrate.
Worse, my ability or lack of ability to write such a program does not
disprove the mathematics!
And in this case that means someone else might, or may already have.
But mathematicians are NOT what they claim, and in this case they are
setting other people up to be responsible for their failures, if things
go badly.
Now I'll work at building that program to factor an RSA number, but
while time passes, it's you who may suffer the consequences, in a world
where hackers may now have the upper hand.
And mathematicians are de facto protecting them.
James Harris
Actually, since all known implementations of your new factoring method are
slower than randomly guessing factors, any hacker following your method 
will
be LESS likely to crack net encryption.
Can you prove that? Oh... don't bother... I forgot that it does not matter. 

What matters is that if your statement above is true, does an algorithm 
result from it that is polynomial time or at least sub-exponential time. 
Where is your Big O notation on this?
Algorithms *can* be developed? I thought you already developed an algorithm 

(or is several now). And you said that it was very easy to do.
listed above.
That's true, James. And if you look and listen, I believe you'll find that
everyone here agrees that the second possibility is the case.
Heck, James, *I* can write a program to factor an RSA number. How about 
this?
int M;
M = //insert the RSA number here
for (int i=2; i<=sqrt(M);i++) {
    if (M % i = 0) {
       cout << i <<  is a factor of  << M;
    }
}
This will produce *every* factor of M less than sqrt(M) (and the only 
reason
it doesn't stop after just one factor is because I don't know enough C++).
Of course, it will take quite a while. This is precisely what others have 
been
saying: not that your math can't produce a workable algorithm, but that 
it's
(at the very best) no more efficient than the least efficient algorithms
currently existing. I think it's great that you've (apparently) arrived at
a new algorithm, but it's not important if it's not an improvement over
what's already available.
Matt
(excess newsgroup stripped)
this?
reason
First of all, you made a classis C error, confusing = with ==.
Rev 2:
M = RSA number
exit statement
{
    if (M % i == 0)    //fixed your error here
    {
        cout << i <<  is a factor of  << M;
        i = -99;        //triggered the exit
    }
}
You can also use a break or (shudder) goto to exit the loop.
That's the ugliest (and most unnecessary) hack I've seen in years. You
should have your right to program revoked.
D.8ecio
I don't think your program works, because a typical RSA number won't fit in
an integer. Try this (if you have GMP and the C++ class interface):
int main()
{
  mpz_class M(/*insert RSA number here between quotes*/);
  for(mpz_class p=2; p<=sqrt(M); mpz_nextprime(p.get_mpz_t(),p.get_mpz_t())
    if(mpz_divisible_p(M.get_mpz_t(),p.get_mpz_t())
    {
      std::cout << p <<  is a factor of  << M << std::endl;
      break;
    }
  return 0;
}
I included a few optimizations, although a key would be to try backwards
from sqrt(M), because any given factors will be closer to sqrt(M) than 2.
D.8ecio
listed above.
Oops, sorry. Apparently some have been saying that your math can't (always)
produce a workable algorithm. And you know what? After looking at some
of their examples, I have to agree with them.
Matt
Wondering whether a new and earthshattering factoring method
can be used to factor numbers is not a fair test? Huh.
Ok...
Oops, I missed the explanation of why it's not a fair test.
************************
David C. Ullrich
[...]
Well yes, factoring M is about finding all the possible integer
factors (and you can call them Az if you want).  Notice that if Az is
guaranteed (by the way you calculate it) to divide M, then Az=1 and
Az=M are possibilities, so you want to make sure that some other
alternatives actually occur.
[...]
I presume you're deliberately misreading criticisms of your
algorithms.  (The alternative is that you genuinely believe that
people claim that the algorithms are random in that sense, which seems
absurd, though given your behaviour that probably can't be ruled out.)
What people are pointing out (from experimentation) is that your
algorithms don't produce factors any faster than choosing random
integers n and looking at gcd(n,M).  Indeed, at least your latest
algorithm (with its T^3 elements) looks significantly slower than
taking gcd with random integers.
[...]
On 17 Feb 2005 17:05:50 -0800, Paul Rubin
20 digit numbers?  Where is the list? I wanna factor them!
Do the numbers have special form such as only two factors or
are they random?
Of course that is only a 2^33 problem.  How about some 30 digit
numbers?  That makes it a 2^50 problem.  That would actually be
a challenge, but still solvable by conventional means on a single
computer.
Leslie 'Mack' McBride
remove text between _ marks to respond via e-mail
OK found the message with the 20 digit numbers
They are the product of two 10 digit factors.
Should have been more specific here, it is a 2^33 problem using trial
factorization with all numbers less than the square root.  The list of
possible prime factors is much smaller (<1/20 the size).
Someone check my math on that please.
Leslie 'Mack' McBride
remove text between _ marks to respond via e-mail
... if you're using trial division or worse (say surrogate factoring) on
worst-case integers. If your integers are random the work is more like
n^0.36 not n^0.50 (see the paper by Knuth and Trabb Pardo, `Analysis of a
simple factorization algorithm'), and of course if you don't use a method
that was devised 2000 years ago, the work is even less.
Consider this in PARI/GP:
? for(i=1,1000,factorint(random(10^30)))
time = 18,490 ms.
?
for(i=1,1000,factorint(nextprime(random(10^15))*nextprime(random(10^15)),14)

)
time = 49,322 ms.
I kinda cheated on that last one, because I knew the integers were of
special form and that MPQS was the best bet in that case. Nonetheless, I
wouldn't  lose much time by letting it try a bit of Pollard rho and maybe
ECM first.
D.8ecio
Hi there:
Could anyone tell me how do i find perfect square factors of a number:
For example:
What are the perfect square factors of 2^5*3^6? (2 raised to 5 times 3
raised to 6)
Is there a trick or shortcut?
Last time I checked, there was no known way to find the square factors 
of a number -- nor indeed even a way to determine whether the number is 
square-free -- which was any faster than factoring. Is this still true?
(Note that the rings  Z  and  F[X]  (where  F  is a field)  are in
some respects similar, yet one _can_ determine easily what the 
square factors of  f  in  F[X]  are, by computing  gcd(f, f') .)
dave
Rewrite your factorization as 4^2*2*9^3, then ignore the 2.   The 
perfect square factors will be in the form 4^n*9*m where 0<=n<=2 and 
0<=m<=3.
-- 
email: wtwentyman at copper dot net
Perfect squares have prime decompositions
in which the primes are in pairs.  For 2^5*3^6,
the odd power of 2 is unusable. We can only use
(2^2)^2 * (3^2)^3 = 4^2 * 9^3.
We use this last expression.
The exponent of 4 can be 0, 1, or 2.
The exponent of 9 can be 0, 1, 2, or 3.
Including the number 1 (the exponents
are both zero) there are 3*4 = 12
perfect-square factors.
cuckoo
These equations were previously published in the 1930's by Ivan Melch, in a
German book titled Functions Theory by Fischkopf Press.  Melch has them 
as
only a side note on page 137 as part of his derivation of the famous
Gebratener Speck theory. Melch uses Z, a complex quantity instead of the
oversimplified real number A. The equations were also shown to be 
irrelevant
for either surrogate factoring, or random number generation.
Claims by James Harris that he invented them are unfounded.
Unfortunately, he made several mistakes as he copied the equations out of
the book, such as overlooking the complex to real number, which further
makes them useless.
Can You find the mistake(s)?
Zx= Zz(-Zz +/- sqrt((Zz - 2M^2)^2 - 4TM^2))/(2j^2 - 2Zz)
Zz= Zx(-Zx +/- sqrt((Zx - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Zx)
I've never heard of the Fried Bacon Theorem. What does it say? 
Rick
Melch, in a
has them as
famous
of the
irrelevant
Read the book. It is quite clearly written. I don't know the English
translation of the term, but the metaphor relates to optimally frying
bacon to a crisp without burning the fat, thus ensuring maximum lard
gain, Melch is actually exposing great insights on Number Theory all
the time. It is a fascinating read, the corollarys on factoring are
just a minor part though.
Is it related to the Ham Sandwich Theorem?
It would make more sense just in sci.math.  This has nothing to do
with relativity.
The game contains no inconsistent mathematical premises, and the
conclusion that the expected winnings are infinite is also consistent.
There is a subtle flaw in the assertion that therefore the game would
be fair if an infinite amount was paid to play.  It is neither fair
nor unfair.  It is undefined.
- Tim
The amount you should pay in order to play must be calculated as the
even bet = the Expectation value or it should be the result of a bargain
between the bank and the player.
beda pietanza
There are inconsistency basically for the abuse of the infinite and some
more, please see my other post dated 14.02.05 h 23.34
In a wager where you are asked to pay a infinite bet such a indefinite
fairness is math inconsistency in my opinion.
beda pietanza
An inconsistency is where both a statement and its negation are
asserted to be true.  Being undefined is incompleteness.  I'll agree
that it's incomplete, but not that it's inconsistent.
Does the undefined nature of 0/0 also count as an inconsistency, in
your opinion?
- Tim
Here's something off the top of my head:
The probability of winning 2^n units is (1/2)^n, so on average you
need to play 1/((1/2)^n) = 2^n times to win 2^n units. However to play
that many times, you pay x*2^n units in entrance fees, where x is the
entrance fee for one throw.. So for every amount you win from a draw,
you expect to have lost x*2^n units. So to come out on top, you'd pay
x<1 units to enter.
There's probably some silly mistake there, but I'm sure someone will
point it out soon enough.
-Sunny
play
I am not sure I got your reasoning, I only point out to you that the
SPP is meant to be played only once: the Expectation value of the game
is said to be infinite and it is expected that a bettor should be
willing to pay a infinite amount of money in order to play the game
just once !!!
This an absurd consequence of using loosely the infinite in the game
that makes the game incoherent.
I hope I helped you
Beda pietanza
Yes, I was thinking about a game which you could play more than once.
After playing with the link given before
(http://www.mathematik.com/Petersburg/Petersburg.html) I was
attempting to explain why you don't seem to come out on top, even
though the expected value of the game is infinite. Still, personally I
would pay less than 1 unit to play the game, since then I would be
garanteed to win more than the entrance fee;P
-Sunny
The amount you should pay in order to play must be calculated as the
even bet = the Expectation value or it should be the result of a bargain
between the bank and the player.
beda pietanza
bargain
Why?
I think the confusion here is using expected value too loosely as a 
decision device in a situation where only one probabilistic check is 
made. Consider the following simpler game:
You flip a coin. On the 0.0001% chance it lands on its side, I give you
$100000000000000000. On the 99.9999% chance it lands heads or tails, you 
give me $100. Now, you have no doubt that the expected value is positive 
(huge in fact, but I don't feel like doing simple arithmetic since I'm a 
lazy person), but are you really willing to play the game just once and 
pay me, say, $1000000 to do it? I don't think so, though the expected
value is much higher than that!
The St. Petersburg Paradox is not a paradox for the precise reason as
above. It has a high expectation value, infinite, actually. But you are
talking about a situation where we just play once and playing into
the psychological factors involved. There is no paradox though, for if
we keep playing, forever and ever, the player will indeed come out ahead
in the long run on average.
-Yan Z.
You don't have $10^17.  Even if you did, and if I could be guaranteed
that you would pay it, I still wouldn't play because the utility to me
of $10^17 is much less than a million times that of $100.
Even in a repeated game, I have at best $10^5 or so.  Again, the
utility to me of $10^17 is much less than a thousand times that of
$10^5.
Only in money, not in utility.
- Tim
Sorry, sir, I don't know what you are trying to say. What you say is 
correct, but I don't see how it says what I am saying is wrong. The 
purpose of my post is to show there is no paradox involved, please tell 
me what I specifically did wrong so I can offer a response.
Right. We know that. This difference between money and utility is 
game. However, there is no paradox since you can still calculate the 
expectation value for each game. I don't see any math inconsistencies 
as noted by the original author.
-Yan
Strangely enough, I'm not trying to say you are wrong.  There's enough
of that going around as it is.  I was agreeing that I would not play,
and giving the reasons why (and hence implicitly giving reasons for
the original paradox).
Indeed.  I do think it is interesting to consider whether utility can
be unbounded.  If it can, then the St Petersburg paradox still holds
(with different payoffs, not necessarily monetary).
- Tim
I don't think the SPP has really a infinite expectation value if you pay a
infinite bet in order to play it once:
the really payoff scheme for a infinite paid bet is :
1Á outcome  payoff 1- infinite net win = -infinite
2Á outcome  payoff 2-infinite net win = -infinite
3Á outcome payoff 4-infinite net win = -infinite
and so on....
if you pay a infinite amount of money you will loose it for sure
and if you play N numbers of games you surely will loose N time
the infinite bets you have  paid.
I post a intresting post in which I explain that the expectation value
1/2+1/2+1/2+.....=infinite doesn't belong properly to the SPP
but to a recursive game.
I wonder why I don't get any reply.
It ay not appear in the thread
so I add it below
beda pietanza
::::::::::::::::::::::::::
I will claim here that the calculation of the Expectation value of this 
game
doesn't belong to it.
For the Expectation value calculated as
E= 1 / 2 + 1 / 2 +1 / 2 + 1 / 2 +......= infinite
There is a iterative game that perfectly correspond to that formula:
Each possibility for which a 1 / 2 is (paid) calculated, the relative
attempt has to be effectively carried out independently of
all the others.
1Á toss, if head, 1 unit of payoff is paid; the game 
doesn't end
The game starts again to attempt for the chance that in 2 tosses
if the second one shows a head, in this case 2 unit of payoff is paid;
the game doesn't end;
The game starts again to attempt that in 3 tosses the 3Á 
one shows
a head if this happens 4 unit of payoff is paid and so on.
Of course if the head turns out before the expected toss the attempt is
aborted and the game starts again with next attempt and so on.
You can see that with all the infinite attempts there are, effectively
a infinite possibility of winning is there, also with multiple winnings,
and this is the game that really correspond to the SPP scheme of 
Expectation
value, in this case paying a infinite amount of money is justified
by the real possibility that the game warrants a real chance of infinite
gain and not as the SPP that warrants only a infinite sure lost
(remember the SPP finishes as soon as once the head shows up).
So what do you think ????
beda pietanza
Nonsense.  It's not possible to bet an infinite amount.
It is if you've got an infinite amount to begin with.  In the original
problem, at least one party must have an infinite amount of money to
begin with.  So your assertion is equivalent to saying that the
original problem cannot exist.
- Tim
You don't know what you're talking about.  You should be asking 
questions, not making statements.
Should I?
If the person offering the game has a finite amount of money, can they
cover all the possible payoffs?  If they cannot are they offering the
game as stated, or a crippled version of it?
- Tim
That's better.  Next time don't snip away the context.
My statement that you reacted to was that all bets are finite, which in 
the first place has nothing to do with the resources of the house, but 
as long as you raise the issue, no, the house doesn't need to have on 
hand at the start an infinite quantity of betting units - all that's 
required is the potential to cover any (finite) payout.  If you still 
think that means an infinite amount is needed, think again.
a
You don't have to you only calculate it.
In any case substituting infinite with a very large amount the destiny
of the bettor doesn't changes.
Please read my post dated 14.02.05
beda pietanza
The amount of any bet is necessarily finite, and for any and all bets 
the gambler's Expected gain is infinite.
Correct if IÇm wrong, but the expected premium of the game 

isnÇt
infinite, as i saw in most posts:
E(P)=1*1/2+2*1/2^2+...+n*1/2^n = sum(t/2^t), t=1,..,n
Therefore, one should bet less than 2 monetary units in order to
expect some gain with this game
You've got the payoffs wrong.  In the paradox you don't get n, you get
2^n.
- Tim
It's more like: sum(2^(t-1)/2^t) = sum(1/2) 
Technically, the expectation is undefined.  However, we can choose any
finite expectation, and find a game with that expectation for which
the payoff for every outcome is less than the original St Petersburg
game.  In that sense, the original game has infinitely positive
expectation.
There are at least two counterarguments:
1) Suppose I have an infinite number of dollar bills with serial
numbers (the writing on most of them is very small).  I give you all
the odd-numbered ones to play the game.  I still have an infinite
amount of money!  So I haven't lost anything at all.
2) No, I only lose it with probability 1.  That's not for sure when
we're talking about infinities.
For laughs, consider this variant St Petersburg game:
You flip a coin repeatedly until you get a tail, and call the number
of consecutive heads n.  If n is even (including zero), you win 3^n.
If n is odd, you must pay 3^n.  Would you play?
- Tim
Firstly, the player would be unable to afford to play such a game
for an arbitrarily large sum of money as a buy-in without risking
financial ruin. More importantly, since the expectation is infinite,
no casino (house) would be able to spread this game honestly.
The house would be unable to pay off an arbitrarily large sum,
should the bettor win it.
Under the more realistic assumption that the most the house can
lose is, say, 2^n dollars, the paradox goes away.
E = 1(1/2) + 2(1/4) + 4(1/8)+...+2^n(1/2^(n+1))
         + 2^n(1/2^(n+2) + 1/2^(n+3) +...)
= (1/2+1/2+...+1/2) + 2^n((1/2)^(n+2)/(1/2))
= (1/2+1/2+...+1/2) + 2^n((1/2)^(n+1))
= (1/2+1/2+...+1/2) + (1/2)
= (n+2)/2 
Wrong.
Why?
The paradox is that the maximum amount rational gamblers are willing to 
pay to play is much less than their Mathematical Expectation, which can 
be as much the case when the number of coin flips is limited as when 
it's unlimited.
Expectation in money, or in utility?
- Tim
In whatever - the paradox doesn't depend on the nature of the units. 
The SPP presumes the marginal value of the payouts is constant.  The 
usual resolution is to recognize that for rational gamblers, it's not.
Unless the house happens to be the government, and willing to print
more money if necessary.
(Of course, the utility of that money is still bounded)
- Tim
If this is what you want to discuss then spell out what you think are 
the inconsistencies.
Obviously, an honest casino cannot afford to offer this game to me!
(My expected winnings exceed the total assets of the casino.)
How much I am willing to pay depends on how long I think the casino
will continue until it goes bankrupt (or manages to do away with
pesky players like me)...
-- 
G. A. Edgar                               
http://www.math.ohio-state.edu/~edgar/
This is not a paradox AFAICT.
It's simply that the casino proof works only asymptotically, I think
it's in the introductory chapters of a common textbook for randomized
algorithms. So, don't sit at a roulette table thinking of this.
--
Eray
Can you explain it better: what is the casino proof ???
beda pietanza
think
randomized
Well, I suppose you want to increase the amount invested in a binary
bet, like in the red/black game on the roulette table. What happens if
you double the  bet every time you _lose_? You would definitely win the
game at some unknown distant horizon. The unfortunate thing is that
such a strategy is worthless in practice, because you can double your
bet only so few times.
Given that you don't have unbounded wealth, and that there is non-zero
probability of losing, the odds aren't good that you are going to
actually win. (Even if you have more wealth than the house!) An
interesting problem would be of course to study unfair roulette
tables on which multiple games are going on at once. In some ways, this
is reminiscient to certain underinvestigated stock market prediction
problems which depend on huge computational power to accomplish (I
think). There are interesting possibilities for a computer player when
it goes online, I believe. (But I'd rather not go into details)
The relation to randomized algorithms is not too clear to me either, a
friend had mentioned it, I don't think I read that textbook, but in
this case, it seems plain expected cost analysis, it doesn't even have
much to do with algorithms, but surely similar ideas are used
frequently in the analysis of randomized / amortized algorithms... This
is the simplest mathematical idea that Sum{k=0 to n}2^k + 1 = 2^{k+1},
no you wouldn't even win a lot, even if you won, so of course doubling
the money probably isn't a good idea if you want to win a huge amount
of money (which kind of runs contrary to the idea of a casino :) On
newsgroups, I'd heard that counting cards was illegal in some US
states, so figure...
The conclusion, if you like gambling with dice: make sure you
absolutely know how to trick the dice. :)
--
Eray
PS: I'm not a gambler, as you can tell. I can't take that many risks.
Only with probability 1, not with certainty.
- Tim
And with roulette, this seems to reinforce the gambler's
fallacy, that somehow there's a linkage between losing
this time and not-losing next time. Aint. There's
a simple inductive proof of this.
Card games are different; those are without replacement
chains. IOW, card counting works.
The thing to remember is: house always wins.
--
Les Cargill
would
Does the house always win indeed?
--
Eray
The apparent paradox arises from confusing an abstract
mathematical model with the real world.
There's no paradox in the mathematics, because you
haven't given a mathematical proof of the statement
[*]
any reasonable person would pay not more than a very 
little amoumt of money. 
Before you can do that you need a mathematical definition 
of reasonable person. _If_ by definition a reasonable
person always tries to maximize his expected gain then
[*] is simply not true, so there is no paradox. If the
definition of reasonable is something else then there
is also no paradox, because the apparent paradox depends
on the unstated assumption that a reasonable person
_does_ always try to maximize his expected gain.
About the real world:
First, in the real world it's simply not true that
a reasonable person always tries to maximize his
expected gain. For example, reasonable people buy
insurance, even though the fact that insurance
companies make money shows that your expected gain
(in the sense of probability theory) is negative
when you buy insurance.
Second:
In real life it will never happen that some casino offers
to play this game, because the casino would need to start
with infinitely much money. So the question of what an
actual real person should be willing to pay an actual
real casino to play this game simply doesn't come up.
************************
David C. Ullrich
I don't think so.  The paradox is simply that people aren't
computers.  Suppose you offer the same bet, but the maximum payout is
a zillion dollars.  Now offer it again, but the max is a jillion
dollars, which is a kadillion orders of magnitude more than a zillion.
You see the problem?  The numbers that dominate the math are
irrelevant to the person.  Not only is he not going to live long
enough to see the coin flip that pays a jillion in any case, so he
will not pay for that part of the expectation, but really, he is no
better off with a jillion dollars than a zillion.
-- Roy L
You are following the official explanation of the so called paradox,
this explanation is related to the marginal utility concept;
I don't this is the key to understand the SPP, the incongruence is
that while the bettor is asked to pay a infinite amount of money
this very infinite payment is not to increase the possibility to
win but to ensure the possibility of loosing  the infinite
bet paid.
The game warrants a maximum win if no bet is paid as the bet paid
increases the bettor looses more and more, with a infinite bet paid
there is a sure infinite lost.
The configuration of the wager is fallacious.
What is asked to be paid doesn't correspond to the real value and
mechanism of the game.
beda pietanza
It's more related to an accurate calculation of expectation.
Economics can't handle infinite quantities.
-- Roy L
Hmm,  suppose I offer you this game:
I will spin a needle on a dial evenly calibrated so that the needle will 
end up pointing to any real number in the interval [0,1).  If  it hits 0 
exactly, I will spin again, repeatedly, until I get a numbert different 
from 0.  The needle is equally likely to stop anywhere.  I will then pay 
you the reciprocal of the number indicated.  How much are you willing to 
pay to play this game?
Or are you noting that people tend to be risk averse when betting large 
amounts of money?
-- 
Stephen J. Herschkorn                        sjherschko@netscape.net
I will give you 10 bucks for said spinner which, of course, includes
the readout. 
--Lynn
at
That's assuming that people want to maximise their expected
winnings.But what people are actually trying to maximize is 'utility'
or 'happiness' - which isn't proportional to how much money you've got.
at
beda reply:
I desagree, the SPP doesn't tell you how much you are going to
win after N tosses; that depend on the unit of the money used
to pay off: you can use a unit having a infinitesimal value and the
amount you win can be of great utility.
The weirdness of the SPP lay as I said on its math inconsistency.
Keep in touch.
beda pietanza
Look at it from the casino's point of view.
In fact, give it a play and see how much the casino wins from you:
http://www.mathematik.com/Petersburg/Petersburg.html
If you are pulling money out of your wallet to play this game, you
probably don't have enough money to win.
It's analagous to the doubling sytems people keep bringing to Vegas
to break the casinos. If one doubles bets long enough eventually one
must win as a mathematical certainty. But it's no mathematical
certainly that one has enough money to double bets long enough.
With a finite, limited number of plays the average result of the St.P.
P. is a good sized loss, and in the real world that's all the most
that real people can play.
The SPP incorrectly assumes that a dollar received
at some distant point in the future is worth the
same as the dollar bet today.  This is patently
unrealistic.
If you introduce a discount rate, so that a dollar
received in the future is worth less than a dollar
received today, the paradox disappears.
Whittle shows this clearly in his his book
 Probability via Expectation 
'utility'
got.
formulation
I don't understand this.  As the St. Petersburg Paradox is usually stated,
you pay your entrance fee, you flip the coin, and you are immediately paid
your winnings.  How does receiving money in the distant future have any
relevance to this?
--Mark
If you introduce even a microscopic discount rate, then the series
converges.  Each flip takes a finite time, and so the winnings for N
flips are discounted by r^N for some 0 < r < 1.  The expected present
value of winnings is then finite (but large).
I don't think it's much of a resolution, though.  It is easily avoided
by paying (2+epsilon)^N for some small epsilon, and the conclusion is
just as absurd as the original.
- Tim
There is a bittorrent of a computer CD iso image containing the first 1.7
billion digits of pi.  The torrent file can be found here:
http://66.237.85.136/pi_1.7G.iso.torrent
The digits are contained within a file that has been compressed using
the bzip2 compression utility.  The image includes Windows command-line
tools for searching the digits along with the source code for the search
utility.  Instructions and examples for using the utilities are included.
If you have never been successful impressing your friends a parties, and
this doesn't work, then you have the wrong friends.
I am suppose to construct a set of A-orthogonal vectors with a Cholesky 
decomposition.
I use ' for transpose.
I use a space for multiplication.
ei is the ith vector of standard basis. i != j
ei' ej = 0
ei' (L inv(L)) ej = 0
... ???
ei' L L' ej = 0
ei' A ej = 0
I'm having trouble filling in ??? 
Nevermind, I figured it out...I was trying to prove the wrong thing. Rather, 

I was trying to prove something that is not true.
The Lunar and Planetary abstracts are just in
on the recent Mars observations.
http://www.lpi.usra.edu/meetings/lpsc2005/pdf/program.pdf
WASHINGTON -- A pair of NASA scientists told a group of space officials
at a private meeting here Sunday that they have found strong evidence that
life may exist today on Mars, hidden away in caves and sustained by
pockets of water.
http://www.space.com/scienceastronomy/mars_life_050216.html
Lunar and Planetary Science XXXVI (2005)
EMBARGOED BY NATURE MAGAZINE UNTIL MID MARCH
EVIDENCE FROM HRSC MARS EXPRESS FOR A FROZEN SEA CLOSE TO MARS' EQUATOR.
We present evidence for a presently-existing frozen sea, with surface
pack-ice, at 5ÁN, 150ÁE, age ca. 5 million 
years. It measures ca. 800
[Times] 900 km and averages ca. 45 m deep. It has probably been protected
from complete sublimation by ash and a sublimation lag of exposed
sediment.
http://www.lpi.usra.edu/meetings/lpsc2005/pdf/1741.pdf
Here are the raw images of the Martian sea. Click on the
links at bottom of page for close ups. Incredible pics.
http://www.msss.com/moc_gallery/e19_r02/images/R02/R0200597.html
Jonathan
s
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ 
Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----
Summary: All laws/equations are Galilean invariant when expressed
  in the generalized cartesian coordinates demanded by basic
  analytic geometry, vector algebra, and measurement theory.
Originator: faqserv@penguin-lust.mit.edu
Disclaimer: approval for *.answers is based on form, not content.
    Opponents of the content should first actually find out what
    it is, then think, then request/submit-to arbitration by the
    appropriate neutral mathematics authorities. Flaming the hard-
    working, selfless, *.answers moderators evidences ignorance
    and despicable netiquette.
Archive-Name: physics-faq/criticism/galilean-invariance
Version: 0.04.03
Posting-frequency: 15 days
   Invariant Galilean Transformations (FAQ) On All Laws
     (c) Eleaticus/Oren C. Webster
        Thnktank@concentric.net
An obvious typo or two corrected.
The Brittanica section revised to less
'pussy-footing' and to more directly
anticipate the elementary measurement
theory and basic analytic geometry
that is applied to the transformation
concept.
------------------------------
The purpose of this document is to provide the student of Physics,
especially Relativity and Electromagnetism, the most basic princ-
iples and logic with which to evaluate the historic justification
of Relativity Theory as a necessary alternative to the classical
physics of Newton and Galileo.
We will prove that all laws are invariant under the Galilean
transformation, rather than some being non-invariant, after
we show you what that means.
We shall also show that another primal requirement that SR
exist is nonsense: Michelson-Morley and Kennedy-Thorndike do
indeed fit Galilean (c+v) physics.
------------------------------
 1. Foreword and Intent
 2. Table of Contents
 3. The Principle of Relativity
 4. The Encyclopedia Brittanica Incompetency.
 5. Transformations on Generalized Coordinate Laws
 6. The data scale degradation absurdity.
 7. The Crackpots' Version of the Transforms.
 8. What does sci.math have to say about x0'=x0-vt?
 9. But Doesn't x.c'=x.c?
       10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations?
       11. But Doesn't (x'-x.c+vt) Prove The Transformation Time
    Dependent?
       12. But Isn't (x'-x.c')=(x-x.c) a Tautology?
       13. But Isn't (x'-x.c')=(x-x.c) Almost the Definition of
    a Linear Transform?
       14. But The Transform Won't Work On Time Dependent Equations?
       15. But The Transform Won't Work On Wave Equations?
       16. But Maxwell's Equations Aren't Galilean Invariant?
       17. First and Second Derivative differential equations.
------------------------------
If a law is different over there than it is here,
it is not one law, but at least two, and leaves us
in doubt about any third location. This is the
Principle of Relativity:  a natural law must be the
same relative to any location at which a given event
may be perceived or measured, and whether or not the
observer is moving.
The idea of location translates to a coordinate
system, largely because any object in motion could
be considered as having a coordinate system origin
moving with it. If you perceive me moving relative
to you - who have your own coordinate system - will
your measurements of my position and velocity fit
the same laws my own, different measurements fit?
If a law has the same form in both cases it is
called covariant. If it is identical in form, var-
ables, and output values, it is called invariant.
What we're asking is that if the x-coordinate, x,
on one coordinate axis works in an equation, does
the coordinate, x', on some other, parallel axis
work?  Speaking in terms of the axis on which x is
the coordinate, x' is the 'transformed' coordinate.
The situation is complicated because we're talking
about coordinates - locations -  but in most mean-
ingful laws/equations, it is lengths/distances (and
time intervals) the equations are about, and x coord-
inates that represent good, ratio scale measures of
distances are only interval scale measures on the x'
axis. [See Table of Contents for discussion of scales.]
So, if we have an x-coordinate in one system, then
we can call the x' value that corresponds to the same
point/location the transform of x.
In particular, the Principle of Relativity is embodied
in the form of the Galilean transformation, which
relates the original x, y, z, t to x', y', z', t' by
the transform equations x'=x-vt, y'=y, z'=z, t'=t in
the simplified case where attention is focused only
on transforming the x-axis, and not y and z. In the
case of Special Relativity, the x' transform is the
same except that x' is then divided by sqrt(1-(v/c)^2),
and t'=(t-xv/cc)/sqrt(1-(v/c)^2). In either case, v
is the relative velocity of the coordinate systems;
if there is already a v in the equations being trans-
formed use u or some other variable name.
------------------------------
One example of the traditional fallacious idea
that an equation is not invariant under the galilean
transformation comes from the Encyclopedia Brittanica:
Before Einstein's special theory of relativity
was published in 1905, it was usually assumed
that the time coordinates measured in all inertial
frames were identical and equal to an 'absolute
time'.  Thus,
      t = t'.              (97)
The position coordinates x and x' were then
assumed to be related by
       x' = x - vt.         (98)
The two formulas (97) and (98) are called a
Galilean transformation. The laws of nonrelativ-
istic mechanics take the same form in all frames
related by Galilean transformations.  This is the
restricted,  or Galilean, principle of relativity.
The position of a light wave front speeding from
the origin at time zero should satisfy
       x^2 - (ct)^2 = 0          (99)
in the frame (t,x) and
      (x')^2 - (ct')^2 = 0       (100)
in the frame (t',x'). Formula (100) does not
transform into formula (99) using the transform-
ations (97) and (98),  however.
.................................................
Besides the trivially correct statement of what the
Galilean 'transform' equations are, there is exactly
one thing they got right.
I.   Eq-100 is indeed the correct basis for discussing
     the question of invariance, given that eq-99 is
     the correct 'stationary' (observer S) equation.
     [Let observer M be the 'moving'system observer.]
     In particular, eq-100 is of exactly the same
     form [the square of argument one minus the square
     of argument two equals zero (argument three).]
II.  It is nonsense to say eq-99 should be derivable from
     eq-100; for one thing, the transforms are TO x' and
     t' from x and t, not the other way around, and the
     idea that either observer's equation should contain
     within itself the terms to simplify or rearrange to
     get to the other is ridiculous. As the transform
     equations say, the relationship of t', x' to t, x
     is based on the relative velocity between the two
     systems, but neither the original (eq-99) equation
     nor the M observer equation is about a relationship
     between coordinate systems or observers. One might
     as well expect the two equations to contain banana
     export/import data; there is no relevancy. The
     'transform' equations are the relationships between
     x' and x, t' and t and have nothing to do with what
     one equation or the other ought to 'say'.  The
     equations' content is the rate at which light emitted
     along the x-axes moves.
III. Most remarkable, the True Believer SR crackpots who
     most despise the consequences of measurement theory
     (demonstrable fact) contained in this document are
     those who want to argue against our saying the Britt-
     anica got eq-100 right;
     They insist that the correct equation is derived
     directly from x'=x-vt and t'=t. Solve for x=x'+vt
     and replace t with t', then substitute the result
     in eq-99: (x'+vt')^2 - (ct')^2 = 0.
     Besides the fact that this results in an equation
     with arguments exactly equal to eq-99, they will
     insist the transform is not invariant.
IV.  A major justification they have for their idea of
     the correct M system equation on which to base the
     the discussion of invariance, is that the variables
     are M system variables, never mind the fact that
     the arguments are S system values.
     That argument of theirs is arrant nonsense. The
     velocity v that S sees for the M system relative
     to herself is the negative of what the M system
     sees for the S system relative to himself.
     In other words, x'+vt' is a mixed frame expression
     and it is x'+(-v)t' that would be strictly M frame
     notation, and that equation is far off base. [Work
     it out for yourself, but make sure you try out an
     S frame negative v so as not to mislead yourself.]
V.   In I. we said: given that eq-99 is the correct
     'stationary' equation. Let's look at it closely:
       x^2 - (ct)^2 = 0          (99)
     This whole matter is supposed to be about coordinate
     transforms. Is that what t is, just a coordinate?
     No. It isn't, in general.  Suppose you and I are both modelling
     the same light event and you are using EST and I'm using PST.
     'Just a time coordinate' is just a clock reading amd your t clock
     reading says the light has been moving three hours longer
     than my clock reading says. Well, that's what the idea that
     t is a coordinate means.
     Eq-99 works if and only if t is a time interval, and in
     particular the elapsed time since the light was emitted.
     Thus, that equation works only if we understand just
     what t is, an elapsed time, with emissioon at t=0.
     However, we don't have to 'understand' anything if we use
     a more intelligent and insightful form of the equation:
     (x)^2 - [ c(t-t.e) ]^2 = 0,
     where t.e is anyone's clock reading at the time of light
     emission, and t is any subsequent time on the same clock.
     Similarly, x is not just a coordinate, but a distance
     since emission.
     (x-x.e)^2 - [ c(t-t.e) ]^2 = 0        (99a)
VI.  In the spirit of 'there is exactly one thing
     they got right', the correct M system version
     of eq-99a is eq-100a:
     (x'-x.e')^2 - [ c(t'-t.e') ]^2 = 0   (100a)
     Every observer in the universe can derive their
     eq-100a from eq-99a and vice versa, not to mention to and
     from every other observer's eq-99a.
     Now, THAT's invariance. [You do realize that every
     eq-100a reduces to eq-99a, when you back substitute
     from the transforms, right? t.e'=t.e, x.e'=x.e-vt.]
------------------------------
The traditional Gallilean transform is correct:
     t'   = t
     x'   = x - vt.
But remember this: a transform of x doesn't effect
just some values of x, but all of them, whether they
are in the formula or not.  This is important if you
want to do things right. The crackpot position is
strongly against this sci.math verified position, and
the apparently standard coordinate pseudo-transformation
they suggest is perhaps the result. {See Table of
Contents.]
Let's use a simple equation: x^2 + y^2 = r^2, which is
the formula for a circle with radius r, centered at a
location where x=0.
But what if the circle center isn't at x=0?  Well, we'd
want to use the form analytic geometry, vector algebra,
and elementary measurement theory tells us to use, a form
where we make explicit just where the circle center is,
even if it is at x=x0=0:
   (x-x0)^2 + (y-y0)^2 = r^2.
The circle center coordinate, x0, is an x-axis coordinate,
just like all the x-values of points on the circle.
So, in proper generalized cartesian coordinate forms
of laws/equations we want to transform every occurence
of x and x0 - by whatever name we call it: x.c, x_e,
whatever.
So, what is the transformed version of (x-x0)?  Why,
(x'-x0'); both x and x0 are x-coordinates, and every
So, what is the value of (x'-x0') in terms of the original
x data?
is also true for x0'=x0-vt:
    (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0).
In other words, when we use the generalized coordinate form
specified by analytic geometry, we find that the value of
(x'-x0') does not depend on either time or velocity in any
way, shape, form, or fashion.
Similarly for (y-y0).
We can treat time the same way if necessary: (t-t0).
The above is a proof that any equation in x,y,z,t is
invariant under the galilean transforms. Just use the
generalized coordinate form, with (x-x0)/etc, in the
transformation process, not the incompetently selected
privileged form, with just x/etc.
[The form is privileged because it assumes the circle
center, point of emission, whatever, is at the origin of
the axes instead at some less convenient point. After
transform the coordinate(s) of the circle center/origin
are also changed but the privileged form doesn't make
this explicit and screws up the calculations, which
should be based on (x'-x0') but are calculated as (x'-0).]
The value of (x'-x0') is the same as (x-x0).  That makes
sense.
Draw a circle on a piece of paper, maybe to the right
side of the paper. On a transparent sheet, draw x and y
coordinate axes, plus x to the right, plus y at the top.
Place this axis sheet so the y-axis is at the left side
of the circle sheet.
Now answer two questions after noting the x-coordinate of
the circle center and then moving the axis sheet to the right:
(a) did the circle change in any way because you moved
the axis sheet (ie because you transformed the coordin-
nate axis)?
(b) did the coordinate of the circle center change?
The circle didn't change [although SR will say it did];
that means that (x'-x0') does indeed equal (x-x0).
The coordinate of the circle center did change, and it
changed at the same rate (-vt) as did every point on
circle center didn't change wrt the circle, means that
the relationship of x0' with x0 is the same as that of
any x' on the circle with the corresponding x: x'=x-vt;
x0'=x0-vt.
This is to prepare you for the True Believer crackpots that
say 'constant' coordinates can't be transformed; some even
say they aren't coordinates. These crackpots include some
that brag about how they were childhood geniuses, btw.
QED: The galilean transformation for any law on
generalized Cartesian coordinates is invariant under
the Galilean transform.
The use of the privileged form explains HOW the transformed
equation can be messed up, the next Subject explains what
the screwed up effect of the transform is, and how use
of the generalized form corrects the screwup.
------------------------------
The SR transforms and the Galilean transforms both
convert good, ratio scale data to inferior interval
scale data. The effect is corrected, allowed for,
when the transforms are conducted on the generalized
coordinate forms specified by analytic geometry and
vector algebra.
Both sets of transforms are 'translations' - lateral
movements of an axis, increasing over time in these
cases - but with the SR transform also involving a
rescaling. It is the translation term, -vt in the x
transform to x', and -xv/cc in the t transform to t',
that degrades the ratio scale data to interval scale
data.  In general, rescaling does not effect scale
quality in the size-of-units sense we have here.
SR likes to consider its transforms just rotations,
however - in spite of the fact Einstein correctly said
they were 'translations' (movements) - and in the case
of 'good' rotations, ratio scale data quality is indeed
preserved, but SR violates the conditions of good ro-
tations; they are not rigid rotations and they don't
appropriately rescale all the axes that must be rescaled
to preserve compatibility.
The proof is in the pudding, and the pudding is the
combination of simple tests of the transformations.
We can tell if the transformed data are ratio scale
or interval.
Ratio scale data are like absolute Kelvin. A measure-
ment of zero means there is zero quantity of the
stuff being measured. Ratio scale data support add-
ition, subtraction, multiplication, and division.
The test of a ratio scale is that if one measure
looks like twice as much as another, the stuff
being measured is actually twice as much. With
absolute Kelvin, 100 degrees really is twice the
heat as 50 degrees. 200 degrees really is twice
as much as 100.
Interval scale data are like relative Celsius, which
is why your science teacher wouldn't let you use it
in gas law problems.  There is only one mathematical
operation interval scales support, and that has to
be between two measures on the same scale: subtraction.
100 degrees relative (household) Celsius is not twice
as much as 50; we have to convert the data to absolute
Kelvin to tell us what the real ratio of temperatures
is.
However, whether we use absolute Kelvin or relative
Celsius, the difference in the two temperature readings
is the same: 50 degrees.
Thus, if we know the real quantities of the 'stuff'
being measured, we can tell if two measures are on
a ratio scale by seeing if the ratio of the two
measures is the same as the ratio of the known quant-
ities.
If a scale passes the ratio test, the interval scale test
is automatically a pass.
If the scale fails the ratio test, the interval scale
test becomes the next in line.
It isn't just the bare differences on an interval
scale that provides the test, however. Differences
in two interval scale measures are ratio scale, so
it is ratios of two differences that tell the tale.
Let's do some testing, and remember as we do that our
concern is for whether or not the data are messed up,
not with 'reasons', excuses, or avoidance.
------------------------------------------------------
Are we going to take a transformed length (difference)
and see whether that length fits ratio or interval scale
definitions?
Of course, not. Interval scale data are ratio after
one measure is subtracted from another. That is the
major reason the SR transforms can be used in science.
Let there be three rods, A, B, C, of length 10, 20, 40,
respectively.  These lengths are on a known ratio scale,
our original x-axis, with one end of each rod at the
origin, where x=0, and the other end at the coordinate
that tells us the correct lengths.
Note that these x-values are ratio scale only because
one end of each rod is at x=0. That may remind you of
the correct way to use a ruler or yard/meter-stick:
put the zero end at one end of the thing you are
measuring. Put the 1.00 mark there instead of the zero,
and you have interval scale measures.
Let A,B,C,   be 10, 20, 40.
Let a,b,c    be x' at v=.5, t=10.
x'=x-vt.
A   B   C         a      b      c
----------------  --------------------
10  20  40         5     15     35
----------------  --------------------
B/A = 2           b/a = 3
C/A = 4           c/a = 7
C/B = 2           c/b = 2.333
          Obviously, the transformed
          values are no longer ratio
          scale. The effect is less on
          the greater values.
C-A = 10          b-a = 10
C-A = 30          c-a = 30
C-B = 20          c-b = 20
          Obviously, the transformed
          values are now interval scale.
          This will hold true for any
          value of time or velocity.
(C-A)/(B-A) = 3   (c-a)/(b-a) = 3
(C-B)/(B-A) = 2   (c-b)/(b-a) = 2
          Obviously, the ratios of the
          differences are ratio scale,
          being identical to the ratios
          of the corresponding original
          - ratio scale - differences.
The main difference between these results and the SR
results is that the differences do not correspond so
neatly to the original, ratio scale, differences.
This is due only to the rescaling by 1/sqrt(1-(v/c)^2).
The ratios of the differences on the transformed values
do correspond neatly and exactly to the ratio scale
results.
Using the generalized coordinate form, such as (x-x0),
the transform produces an interval scale x' and an
interval scale x0'. That gives us a ratio scale (x'-x0'),
just like - and equal to - (x-x0).
------------------------------
It has become apparent - whether misleading or not -
that the crackpot responses to the obvious derive from
a common source, whether it be bandwagoning or their
SR instructors.
Below, in the sci.math subject, we see that all sci.math
respondents agree with the basic controversial position
of this faq: every coordinate is transformed, whether a
supposed constant or not.
Think about it, the generalized coordinate of a circle
center, x0, applies to infinities upon infinities of
circle locations (given y and z, too); it is a constant
only for a given circle, and even then only on a given
coordinate axis.
And even variables are often held 'constant' during
either integration or differentiation.
The utility of a variable is that you can discuss all
possible particular values without having to single out
just one.  That utility does not make particular - singled
out - values on the variable's axis not values of the
variable just because they have become named values.
In any case, all that is preamble to the incompetent idea
they have proposed for a transform of coordinates. It is
based on the idea that the circle center, point of emission,
whatever, has coordinates that cannot be transformed.
Let there be an equation, say (x)^2 - (ict)^2 = 0.
What is the transformed version of that equation?
Answer: (x')^2 - (ict')^2 = 0.   That's the one thing the
Brittanica got right. Note that the leading crackpot just
criticized this faq for presuming to correct the Britt-
anica, but it then and before poses the incompetent pseudo-
transform we analyze here in this section.
x to x' and t to t' are obviously coordinate transforms;
the x and t coordinates have been replaced by the coord-
inates in the primed system.
A tranform of an equation from one coordinate system to
another is NOT a substitution of the/a definition of x
for itself; that is not a coordinate transformation.
The most that can said for such a substitution is that
it is a change of variable.
But the crackpots are calling this a coordinate trans-
form of the original equation:
    (x'+vt)^2 - (ict')^2 = 0.
It is not a coordinate transform, of course, except
accidentally. (x'+vt) is not the primed system
coordinate, it is another form/expression of x. They
get that substitution by solving x'=x-vt for x; x=x'+vt.
So, by incompetent misnomer, they accomplish what they
have been railing against all along.
It has been the generalized coordinate form in question all
this time:
    (x-x0)^2 - (ict)^2 = 0.
Here they substitute for x instead of transforming to the
primed frame:
    (x'+vt-x0)^2 - (ict')^2.
     -----
       ^
       |
       ^
       |
It is still x ^ but see what they have accomplished
by their mis/malfeasance:
    [x'+vt-x0]=[x'+(vt-x0)]=[x'-(x0-vt)].
      =[x'-x0']
The crackpots have been bragging about how you don't
have to transform the circle center's coordinate to
transform the circle center's coordinate.  Bragging
that what they were doing was not what they said
they were doing.
This does give us insight as to some of the crackpot
variations will be discussed in later sections..
They are used to seeing the mixed coordinate form,
(x'+vt-x0) without realizing what it respresented,
so - accompanied with a lack of understanding of
the term 'dependent' - they are used to seeing just
the one vt term, and not the one hidden in the defi-
nition of x' and are used to imagining it makes the
whole expression time dependent and thus not invariant.
About which, let x=10, let, x0=20, v=10, and t
variously 10 and 23:
(x-x0)=-10.  Using their (x'+vt-x0):
For t=10, we have (x'+vt-x0) = [ (10-10*10) + (10*10) - (20) ]
        =      -90     +   100   -  20
        = -10
        = (x-x0)
For t=23, we have (x'+vt-x0) = [ (10-10*23) + (10*23) - (20) ]
        =      -220    +   230   -  20
        = -10
        = (x-x0)
The result depends in no way on the value of time;
we showed the obvious for a couple of instances of t
just so you can see that the crackpots not only do
not understand the obvious logic of the algebra
{ (x'-x0')=[ (-vt)-(x0-vt) ]=(x-x0) } - which shows
that the transform has no possible time term effect -
but they don't understand even a simple arithmetic
demonstration of the facts.
Oh. Their (x'+vt-x0) or (x'+vt'-x0) reduces the same
way since t'=t:
    (x-vt+vt-x0)=(x-x0).
Their process, which says (x'+vt') is the transform
of x, says that (x'+vt') is the moving system location
of x, but it can't be because x is moving further in
the negative direction from the moving viewpoint.
That formula will only work out with v<0 which is indeed
the velocity the primed system sees the other moving at.
However, that formula cannot be derived from x'=x-vt,
the formula for transformation of the coordinates from
the unprimed to the primed,
------------------------------
The crackpots' positions/arguments were put to sci.math
in such a way that at least two or three who posted re-
sponses thought it was your faq-er who was on the idiot's
side of the questions.
Their responses:
----------------------------------------------------------
     values on the x-axis are not subject to the transform.
  No.  x0' = x0 - vt.
  Well, if you want, you could define constant values on the x-axis, 
but
in the context of the question that is not relevant.  The relevant fact is
that if the unprimed observer holds an object at point x0, then the
primed observer assigns to that object a coordinate x0' which is
numerically related to x0 by x0'= x0 -vt.
What does this mean? The line x=x0 will give x'=x-v*t=x0-vt', so if x0'
is to give the coordinate in the (x',t',)-system, it will be given by
x0'=x0-v*t': ie., it is not given by a constant. Thus, being at rest
(constant x-coordinate) is a coordinate-dependent concept.
Sounds very false. We can say that the representation of the point X0 is
the number x0 in the unprimed system, and x0' in the primed system.
Clearly x0 and x0' are different, if vt is not zero. However one may say
that (though it sounds/is stupid) the point X0 itself is the same
throughout the transformation. However that expression sounds
meaningless, since a transform (ok, maybe we should call it a change of
basis) is only a function that takes the point's representation in one
system into the same point's representation in another system. It is
preferrable to use three notations: X0 for the point itself and x0 and
x0' for the points' representations in some coordinate systems.
------------------------------
That idea is one of the most idiotic to come up, and it does
so frequently. And in a number of guises.
we have called x0 above; the notation makes no difference.
Some crackpots have managed to maintain that position even
after graphs have illustrated that such an idea means that
after a while a circle center represented by x.c' could be
outside the circle.
The leading crackpot just make that explicit, as far as
one can tell from his befuddled post in response to a line
about active transforms, which are actually moving body
situations, not coordinate transformations:
--------------------------------------------------------------------
 Right, it is a transform of the center (in the opposite direction)
 done to effect the change of coordinates without a coordinate
 transform.  ...
E: Transform of the center?  Center of a circle?
    He really is saying a circle center moves in
    the opposite direction of the circle! Right?
--------------------------------------------------------------------
If r=10 and x.c was at x.c=0, then the points on the circle
(10,0), (-10,0), (0,10) and (0,-10) could at some time become
(-10,0), (-30,0), (-20,10), and (-20,-10), but with x.c'=x.c,
the circle center would be at (0,0) still!  The circle is here
but its center is way, way over there! Indeed, although a change
of coordinate systems is not movement of any object described in
the coordinates, the x.c'=x.c crackpottery is tantamount to the
circle staying put but the center moving away. Or vice versa.
------------------------------
One crackpot puts the (x'-x.c')=(x-vt - x.c+vt) relationship
like this:
      (x-vt+vt - x.c).
See, he says, that is transforming x (with x-vt - x.c) and then
reversing the transform (x-vt+vt - x.c).
That's just another crackpot form of the idiocy that
is that there is no transform vt term relating to x.c.
------------------------------
      Time Dependent?
That particular crackpottery is perhaps more corrupt than
moronic, since it includes deliberately hiding a vt term from
view, and pretending it isn't there.  [However, we have seen
above that it is a familiar incompetency, and not likely an
original.]
Look, the crackpots say, there is a time term in the
transformed (x' - x.c+vt). The transform isn't invariant!
It's time dependent!
Just put x' in its original axis form, also, which reveals
the other time term, the one they hide:
    (x'-x.c+vt) = (x-vt - x.c+vt) = (x-x.c).
So, at any and all times, the transform reduces to the
original expression, with no time term on which to be
dependent.
Then there is the fact that if you leave the equation
in any of the various notation forms - with or without
reducing them algebraicly - the arithmetic always comes
down to the same as (x-x.c). That means nothing to crack-
pots, but may mean something to you.
------------------------------
My dictionary relates 'tautology' to needless repetition.
The repetition involved is the vt transformation term.
Apply the -vt term to the x term, and it is needless
repetition to apply it anywhere again? The 'again' is
to the x.c term.  The x.c' = x.c crackpot idiocy.
The repetition of the vt terms is required by the presence
of two x values to be transformed.
Be sure to note the next section.
------------------------------
      a Linear Transform?
Now, how on earth can we relate a tautology to a basic
definition in math?
we get this definition:
--------------------------------------------------------------
A linear transformation, A, on the space is a method of corr-
esponding to each vector of the space another vector of the
space such that for any vectors U and V, and any scalars
a and b,
 A(aU+bV) =  aAU + bAV.
-------------------------------------------------------------
Let points on the sphere satisfy the vector X={x,y,z,1},
and the circle center satisfy C={x.c,y.c,z.c,1}. Let a=1,
and b=-1.
Let A= ( 1   0   0  -ut )
       ( 0   1   0  -vt )
       ( 0   0   1  -wt )
       ( 0   0   0   1  )
A(aX+bC) = aAX + bAC.
      aX+bC  =  (x-x.c, y-y.c, z-z.c,  0  ).
The left hand side:
     A( x - x.c ,  y - y.c,  z - z.c,  0  )
     = ( x-x.c ,  y-y.c,  z-z.c,  0  ).
The right hand side:
       aAX= ( x-ut, y-vt, z-wt, 1 ).
       bAC= (-x.c+ut, -y.c+vt, -z.c+wt, -1 ).
and
  aAX+bAC = ( x-x.c, y-y.c, z-z.c,  0  ).
Need it be said?
Sure:  QED.  On the galilean transform the
definition of a linear transform,
       A(aU+bV)=aAU + bAV,
is completely satisfied.
The generalized form transforms exactly and
non-redundantly - with ONE TRANSFORM, not a
transform and reverse transform - and non-
tautologically, just as the very definition
of a linear transform says it should.
And does so with absolute invariance, with this
galilean transformation.
------------------------------
The main crackpot that has asserted such a thing was referring
to equations such as in Subject 4, above. The Light Sphere
equation; for which we have shown repeatedly elsewhere that the
numerical calculations are identical for any primed values as
for the unprimed values.
The presence - before transformation - of a velocity term
seems to confuse the crackpots. It turns out there is ex-
treme historical reason for this, as you will see in the
subject on Maxwell's equations.
------------------------------
See Subject 17, below, for a discussion of Second Derivative
forms and the galilean transforms.
------------------------------
Oh? Just what is the magical term in them that prevents
(x'-x.c')=(x-vt - x.c+vt)=(x-x.c) from holding true?
It turns out not to be magic, but reality, that interferes
with the application of the galilean transforms to the gen-
eralized coordinate form(s) of Maxwell: there are no coordi-
nates to transform!
When True Believer crackpots are shown the simple
demonstration that the galilean transform on
generalized cartesian coordinates is invariant,
their first defense is usually an incredibly stupid
x0'=x0, because the coordinate of a circle center,
or point of emission, etc, is a constant and can't
be transformed.
The last defense is but Maxwell's equations are not
invariant under that coordinate transform.  When
asked just what magic occurs in Maxwell that would
prevent the simple algebra
    (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0)
from working, and when asked them for a demonstration,
they will never do so, however many hundreds of
times their defense is asserted.
The reason may help you understand part of Einstein's
1905 paper in which he gave us his absurd Special
Relativity derivation:
THERE ARE NO COORDINATES IN THE EQUATIONS TO BE TRANSFORMED.
Einstein gave the electric force vector as E=(X,Y,Z)
and the magnetic force vector as B=(L,M,N), where the
force components in the direction of the x axis are
X and L,  Y and M are in the y direction, Z and N in
the z direction.
Those values are not, however, coordinates, but values
very much like acceleration values.
BTW, the current fad is that E and B are 'fields', having
been 'force fields' for a while, after being 'forces'.
So, when Einstein says he is applying his coordinate
transforms to the Maxwell form he presented, he is
either delusive or lying.
(a) there are no coordinates in the transform equations
    he gives us for the Maxwell transforms, where
    B=beta=1/sqrt(1-(v/c)^2):
    X'=X.                L'=L.
    Y'=B(Y-(v/c)N).      M'=B(M+(v/c)Z).
    Z'=B(Z+(v/c)M).      N'=B(N-(v/c)Y).
    X is in the same direction as x, but is not a coordinate.
    Ditto for L. They are not locations, coordinates on the
    x-axis, but force magnitudes in that direction.
    Similarly for Y and M and y, Z and N and z.
(b) the v of the coordinate transforms is in Maxwell
    before any transform is imposed; Einstein's transform
    v is the velocity of a coordinate axis, not the velocity
    he touched it.
(c) if they were honest Einsteinian transforms, they'd be
    x, which means it is X and L that are supposed to be
    transformed, not Y and M, and Z and N. And when SR does
    transform more than one axis, each axis has its own
    velocity term;  using the v along the x-axis as the v
    for a y-axis and z-axis transform is thus trebly absurd:
    the axes perpendicular to the motion are not changed
    according to SR, the v used is not their v, and the v
    is not a transform velocity anyway.
(d) as everyone knows, the effect of E and B are on the
    direction.  Both the speed and direction are changed
    by E and B, but v - the speed - is a constant in SR.
As absurd as are the previously demonstrated Einsteinian
blunders, this one transcends error and is an incredible
example of True Believer delusion propagating over decades.
The components of E and B do differ from point to point,
and in the variations that are not coordinate free,
they are subject to the usual invariant galilean trans-
formation when put in the generalized coordinate form.
-------------------------------------------------------------
The SR crackpots don't know what coordinates are. The
various things they call coordinates include coordin-
nates, but also include a variety of other quantities.
------------------------------------------------------
1.   One may express coordinates in a one-axis-at-a-time
     manner [like x^2+y^2=r^2] but it is the use of vector
     notation that shows us what is going on. In vector
     notation the triplet x,y,z [or x1,x2,x3, whatever]
     represents the three spatial coordinates, but there
     are so-called basis vectors that underlie them. Those
     may be called i,j,k. Thus, what we normally treat as
     x,y,z is a set of three numbers TIMES a basis vector
     each.
2.   These e*i, f*j, g*k products can have a lot of meanings.
     If e, f, j are distances from the origin of i,j,k then
     e*i, f*j, g*k are coordinates: distances in the directions
     of i,j,k respectively, from their origin. That makes the
     triplet a coordinate vector that we describe as being an
     x,y,z triplet; perhaps X=(x,y,z).
     The e*i, f*j, g*k products could be directions; take any
     of the other vectors described above or below and divide the
     e,f,g numbers by the length of the vector [sqrt(e^2+f^2+g^2)].
     That gives us a vector of length=1.0, the e,f,g values of
     which show us the direction of the original vector. That
     makes the triplet a direction vector that we describe as
     being an x,y,z triplet; perhaps D=(x,y,z).
     The e*i, f*j, g*k products could be velocities; take any
     of the unit direction vectors described above and multiply
     by a given speed, perhaps v. That gives a vector of length
     v in the direction specified. That makes the triplet a
     velocity vector that we describe as being an x,y,z triplet;
     perhaps V=(x,y,z). Each of the three values, e,f,g, is the
     velocity in the direction of i,j,k respectively.
     The e*i, f*j, g*k products could be accelerations; take any
     of the unit direction vectors described above and multiply
     by a given acceleration, perhaps a. That gives a vector of
     length a in the direction specified. That makes the triplet
     an acceleration vector that we describe as being an x,y,z
     triplet; perhaps A=(x,y,z). Each of the three values, e,f,g,
     is the acceleration in the direction of i,j,k respectively.
     The e*i, f*j, g*k products could be forces (much like accel-
     erations); take any of the unit direction vectors described
     above and multiply by a given force, perhaps E or B. That
     gives a vector of length E or B in the direction specified.
     That makes the triplet a force vector that we describe as
     being an x,y,z triplet; perhaps E=(x,y,z) or B=(x,y,z). Each
     of the three values, e,f,g, is the force in the direction of
     i,j,k respectively.
Einstein's - and Maxwell's - E and B are
not coordinate vectors.
There is another variety of intellectual befuddlement that
misinforms the idea that Maxwell isn't invariant under the
galilean transform: confusions about velocities.
Velocities With Respect to Coordinate Systems.
-----------------------------------------------
Aaron Bergman supplied the background in a post to a sci.physics.*
newsgroup:
Imagine two wires next to each other with a current I in each.
Now, according to simple E&M, each current generates a magnetic
field and this causes either a repulsion or attraction between
the wires due to the interaction of the magnetic field and the
current. Let's just use the case where the currents are parallel.
Now, suppose you are running at the speed of the current between
the wires. If you simply use a galilean transform, each wire,
having an equal number of protons and electrons is neutral. So,
in this frame, there is no force between the wires. But this is a
contradiction.
First of all, the invariance of the galilean transform (x'-x.c')
=(x-x.c),  insures that it is an error to imagine there is any
difference between the data and law in one frame and in another;
the usual, convenient rest frame is the best frame and only frame
(x'-x.c')=(x-x.c).]
Second, given that you decide unnecessarily to adapt a law to
a moving frame, don't confuse coordinate systems with meaningful
physical objects, like the velocity relative to a coordinate
system instead of relative to a physical body or field.
In other words, what does current velocity with respect to a
coordinate system have to do with physics?
Nothing. Certainly not anything in the example Bergman gave.
What is relevant is not current velocity with respect to a
coordinate system, but current velocity with respect to wires
and/or a medium.  The velocity of an imaginary coordinate sys-
tem has absolutely nothing to do with meaningful physical vel-
ocity. You can - if you are insightful enough and don't violate
item (e) - identify a coordinate system and a relevant physical
object, but where some v term in the pre-transformed law is
in use, don't confuse it with the velocity of the coordinate
transform.
Velocities With Respect to ... What?
-----------------------------------------------
Albert Einstein opened his 1905 paper on Special Relativity
with this ancient incompetency:
The equations of the day had a velocity term that was taken
as meaning that moving a magnet near a conductor would create
a current in the conductor, but moving a conductor near a
wire would not.  This was belied by fact, of course.
The important velocity quantity is the velocity of the
magnet and conductor with respect to each other, not to
some absolute coordinate frame (as far as we know) and
not to an arbitrary coordinate system.
One possible cause was the idea: but the equation says the magnet
must be moving wrt the coordinate system or ... the absolute
rest frame.
There not being anything in the equation(s) to say either of
those, it is amazing that folk will still insist the velocity
term has nothing to do with velocity of the two bodies wrt
each other.
-----------------------------------------------------------
------------------------------
One of the intellectually corrupt ways of
denying the very simple demonstration of
galilean invariance of all laws expressed
in the generalized coordinate form demanded
by analytic geometry, vector analysis, and
measurement theory
    [ (x'-x.c')=[ (x-vt)-(x.c-vt) ]=(x-x.c) ]
is the assertion that those equations 'over there'
(usually Maxwell or wave) are somehow immune to
the elementary laws of algebra used to demon-
strate the invariance.  [Unfortunately, the
assertions are never accompanied by reference
to the magical math that makes elementary al-
gebra invalid.  Wonder why that is?]
Part of the time it is based on the old lore
based on the incompetent transformation of
the privileged form of an equation instead
of the correct form. [Evidence of this is
any reference to an effect due to the velocity
of the transform; it falls out algebraicly
metically - as you can see above.]
But usually it is just whistling in the dark,
waving the cross (zwastika, I'd say) at
the mean old vampire.
The most general equation that could be conjured
up is a differential with either First or Second
Derivatives.
Let's examine the plausibility of such magical
magical, non-invariance assertions.
(a) to get a Second Derivative you must have
    a First Derivative.
(b) to get a First Derivative you must have
    a function to differentiate.
(c) to get a Second Derivative you must have
    a function in the second degree.
So, let us examine the question as to whether
any such common Maxwell/wave equation will
differ for
(a) the common, privileged form, represented
    as ax^2, with a being an unknown constant
    function.
(b) the generalized cartesian form, represented
    as a(x-x.c)^2 = ax^2 -2ax(x.c) + ax.c^2,
    with a being an unknown constant function.
(c) the transformed generalized cartesian form,
    represented as a(x-vt -x.c+vt)^2, same as for
    (b), = ax^2 -2ax(x.c) + ax.c^2, of course,
    with a being an unknown constant function.
I.  for (a), remembering that x.c is a constant,
    and that this version is only correct because
    x.c=0, otherwise (b) is the correct form:
     d/dx    ax^2  = 2ax
    (d/dx)^2 ax^2  = 2a
II. for (b), remembering that x.c is a constant.
     d/dx    (ax^2 -2ax(x.c) + ax.c^2) = 2ax - 2ax.c
    (d/dx)^2 (ax^2 -2ax(x.c) + ax.c^2) = 2a
III. for (c); same as for (b).
So, what we have seen so far is
(1)  differential equations in the second degree
- the wave equations - must clearly be the same for
all forms: the privileged form in x, the generalized
cartesian form in x and the centroid, x.c, or the
transformed generalized cartesian form.
That is, anyone who imagines that correct usage
gives different results for galilean transformed
frames is at first showing his ignorance, and in
the end showing his intellectual corruption.
(2) As far as the First Derivatives are concerned, the
only cases in which there really is a difference between
use of the privileged form is obviously incompetent.
So, how do you correctly use the differential equations?
If you are using rest frame data with the centroid
at x=0, etc, you can't go wrong without trying to
go wrong.
If you are using rest frame data with the centroid
not at x=0, you must use (x-x.c) anyplace x appears
in the equation.
If you are using moving frame data, you must use the
moving frame centroid as well as the light front
(or whatever) moving frame data itself, perhaps first
calculating (x'-x.c'), which equals (x-x.c) which is
obviously correct, and which is obviously the plain old
correct x of the privileged form.
Unless, of course, there really is some magical term
or expression that invalidates the obvious and elemen-
tary algebra of the invariance demonstration.
Or maybe you just whistle when you don't want basic
algebra to hold true.
Eleaticus
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
! Eleaticus        Oren C. Webster         ThnkTank@concentric.net  ?
! Anything and everything that requires or encourages systematic   ?
!  examination of premises, logic, and conclusions                 ?
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
[snip lies]
[snip 1300 lines of trolled garbage]
eleaticus, Oren Webster, is a despised and stooopid troll,
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
 Several crimes against logic and science  Ha ha ha!
 
Originally trolled across sci.physics sci.physics.relativity
alt.physics sci.math sci.answers alt.answers news.answers
Psychotic ineducable boring troll Eleaticus,
Were there to be internal inconsistencies in SR (meaning
inconsistencies of a purely mathematical logical nature) that would
automatically lead to contradictions in number theory, itself, and
arithmetic, since the mathematics of Minkowski geometry is
equiconsistent with the theory of real numbers and with arithmetic. 
Eleaticus explicitly demonstrates that he is completely ignorant of
multivariable calculus.  He has no concept of the Chain Rule in
multivariable calculus.  Consider his Galilean Transformation goo and
dribble:
   t' = t,
   x' = x - vt,
   y' = y,
   z' = z.
His refusal to accept that t' must be introduced as a separate
variable springs from a massive emprical stupidity re space and time
are described as a four-dimensional manifold, with four coordinates
instead of a time evolution of a three-dimensional manifold,  and that
the change of coordinate system should be a change of four
coordinates, and not a time-dependent change of three coordinates. 
This is particularly vital when it comes to fields over space and time
(electric and magnetic fields for example).
The transformation law for the differential operators under the
Galilean transformation is given by:
   d/dt' = d/dt + v d/dx,
   d/dx' = d/dx,
   d/dy' = d/dy,
   d/dz' = d/dz.
This shows the necessity of introducing a new variable t', since
partial differentiation with respect to t' (constant x', y', z') is a
different operation to partial differentiation with respect to t
(constant x, y, z).  The above transformation law is determined by the
Chain Rule:
d/dt' = dt/dt' d/dt + dx/dt' d/dx + dy/dt' d/dy + dz/dt' d/dz,
d/dx' = dt/dx' d/dt + dx/dx' d/dx + dy/dx' d/dy + dz/dx' d/dz,
d/dy' = dt/dy' d/dt + dx/dy' d/dx + dy/dy' d/dy + dz/dy' d/dz,
d/dz' = dt/dz' d/dt + dx/dz' d/dx + dy/dz' d/dy + dz/dz' d/dz.
The presence of the term involving d/dx in the expression for d/dt' is
indicative of the fact that x depends on t' (x', y', z', being held
constant), as can be seen from the fact that the coefficient of d/dx
in the expression for d/dt' is dx/dt'.  Because of the now
demonstrated fact that Eleaticus has no formal education in
multivariable calculus, he has managed, somehow, to get it into his
head that the presence of the term involving d/dx in the expression
for d/dt' is indicative of t' depending on x (t, y, z, being held
constant).  Because of his stupidty Eleaticus cannot get the correct
transformation law for the differential operators under the Galilean
Transformation, and he cannot determine the invariance or otherwise of
Maxwell's Equations under the Galilean Transformation.  The first
advice to Eleaticus is to learn multivariable calculus.
Eleaticus should not pretend that he can understand how to determine
invariance or otherwise of Maxwell's Equations under the Galilean
Transformation, or under the Lorentz Transformation, until he
understands the multivariable calculus which underlies such
considerations.  Eleaticus is a loud idiot.
The homogeneous Maxwell equations are invariant under the Galilean
Transformation, with transformation laws:
  E_x' = E_x,
  E_y' = E_y - v B_z,
  E_z' = E_z + v B_y,
  B_x' = B_x,
  B_y' = B_y,
  B_z' = B_z.
The derivation of these transformation laws was determined using the
transformation laws for the differential operators given above.  These
transformation laws have the additional advantage that they determine
the correct transformation for the force law, thus providing further
evidence in favour of the transformation law for the differential
operators, as above.
The inhomogeneous Maxwell equations are also invariant under the
Galilean transformation, with transformation laws:
  E_x' = E_x,
  E_y' = E_y,
  E_z' = E_z,
  B_x' = B_x,
  B_y' = B_y + v/c^2 E_z,
  B_z' = B_z - v/c^2 E_y,
  rho' = rho,
  J_x' = J_x - v rho,
  J_y' = J_y,
  J_z' = J_z.
Note the the transformation laws for the charge density and current
density are as they should be under the Galilean transformation.
Homogeneous equations are invariant under the Galilean Transformation,
and inhomogeneous equations are invariant under the Galilean
Transformation, but Maxwell's Equations as a whole are NOT invariant
under the Galilean Transformation, since the transformation laws
required for the EM field for the two cases are inconsistent with each
other.  The transformation law for the EM field which makes the
homogeneous equations invariant will not also make the inhomogeneous
equations invariant.  The transformation law for the EM field which
makes the inhomogeneous equations invariant will not also make the
homogeneous equations invariant.
On the other hand, all of Maxwell's equations are invariant under the
Lorentz Transformation, with transformation laws:
  E_x' = E_x,
  E_y' = gamma (E_y - v B_z),
  E_z' = gamma (E_z + v B_y),
  B_x' = B_x,
  B_y' = gamma (B_y + v/c^2 E_z),
  B_z' = gamma (B_z - v/c^2 E_y),
  rho' = gamma (rho - v/c^2 J_x),
  J_x' = gamma (J_x - v rho),
  J_y' = J_y,
  J_z' = J_z,
where gamma = 1/sqrt(1 - v^2/c^2).
 Idiot Oren Webster sees himself this way,
http://www.mazepath.com/uncleal/effete6.jpg
 The entire remainder of the planet sees him this way,
http://www.mazepath.com/uncleal/effete3.png
http://www.mazepath.com/uncleal/sunshine.jpg
http://www.apa.org/journals/psp/psp7761121.html
http://insti.physics.sunysb.edu/~siegel/quack.html
Hey, stooopid troll Eleaticus - Do you want EVIDENCE?  Each of the 24
GPS satellites carries either four cesium atomic clocks or three
rubidum atomic clocks in orbit, with full relativistic corrections
being applied.
Internal inconsistencies in SR (meaning inconsistencies of a purely
mathematical logical nature) automatically lead to contradictions in
number theory, itself, and arithmetic, since the mathematics of
Minkowski geometry is equiconsistent with the theory of real numbers
and with arithmetic. 
 Mathematics of gravitation
 Equivalence Principle testing
http://arXiv.org/abs/hep-th/0111236
 Geometric structure of reality
http://arxiv.org/abs/gr-qc/0103044
http://arXiv.org/abs/hep-th/0307140
 GR structure, especially Part 4/p. 7
http://arXiv.org/abs/gr-qc/0311039
 Experimental constraints on General Relativity
http://www.eftaylor.com/pub/projecta.pdf
 Relativity in the GPS system
http://arXiv.org/abs/gr-qc/9909014
 falling light
 Hafele-Keating Experiment
http://www.hawaii.edu/suremath/SRtwinParadox.html
 Twin Paradox
http://arXiv.org/abs/astro-ph/0401086
http://arxiv.org/abs/astro-ph/0312071
 Deeply relativistic neutron star binaries
http://arxiv.org/abs/hep-th/0405160
 Black hole evaporation
http://physicstoday.org/vol-57/iss-7/p40.shtml  
 No aether
http://fsweb.berry.edu/academic/mans/clane/
 No Lorentz violation
http://arXiv.org/abs/gr-qc/0409089
 Spin-2 gravitons have problems
  (so does the proposal)
http://arXiv.org/abs/gr-qc/0301024
 Nordtvedt Effect
http://map.gsfc.nasa.gov/
http://arxiv.org/abs/astro-ph/0403292
http://arXiv.org/abs/astro-ph/0310723
 WMAP + Sloane Digital Sky Survey
http://arxiv.org/abs/hep-ph/0404175
 Dark matter candidates
 Carroll on what it all means.
Special Relativity is physics on a topologically trivial Lorentzian 
manifold with a metric whose curvature tensor is zero.  This is a 
perfectly diffeomorphism-invariant condition and does not require 
any particular coordinate choice.  It is invariant under 
the full group of diffeomorphisms.  The Poincare group is 
the group of *isometries* of the metric in special relativity. 
The Special Relativity metric is *non-dynamical* (unlike GR).  It 
defines the coupling *constants* of your theory. If you change the 
metric in any nontrivial way you are changing your theory.  An 
operation can only be called a symmetry of a special-relativistic 
(non-gravitational) theory if it preserves the metric, and therefore 
the symmetry of special-relativistic theories is the Poincare group 
only.  General Relativity (gravitation) has a dynamic metric.
NIM A 355 537 (1995)
Physics Letters B 328 103 (1994)
Physical Review Letters 64 1697 (1990)
Physical Review Letters 39 1051 (1977)
Physical Review 135 B1071 (1964)
Physics Letters 12 260 (1964)
Europhysics Letters 56(2) 170-174 (2001)
General Relativity and Gravitation 34(9) 1371 (2002)
http://fourmilab.to/etexts/einstein/specrel/specrel.pdf
 Longitudinal and transverse mass
http://arxiv.org/abs/gr-qc/0306076.pdf
http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf
http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm
http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm
http://www.trimble.com/gps/index.html
http://sirius.chinalake.navy.mil/satpred/
http://www.phys.lsu.edu/mog/mog9/node9.html
http://egtphysics.net/GPS/RelGPS.htm
http://www.schriever.af.mil/gps/Current/current.oa1
http://edu-observatory.org/gps/gps_books.html
-- 
Uncle Al
http://www.mazepath.com/uncleal/
 (Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
Summary: The Lorentz transforms themselves are proof t' and x'
  cannot possibly be just coordinates.  Examination
  of their derivation verifies their identity as 
  intervals.
Originator: faqserv@penguin-lust.mit.edu
Disclaimer: approval for *.answers is based on form, not content.
    Opponents should first actually find out what the content is, 
    then think, then request/submit-to arbitration by the 
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    working, selfless, *.answers moderators evidences ignorance 
    and atrocious netiquette.
Version: 0.02.1
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Posting-frequency: 15 days
       
    (SR) Lorentz t', x' = Intervals                    
     (c) Eleaticus/Oren C. Webster
        Thnktank@concentric.net
------------------------------
     of the use of 'just coordinates' in any
     scientific formula.
Defenders of the Special Relativity faith are especially
fond of telling opponents of their space-time fairy tales
that they do not know the difference between coordinates
and magnitudes.  
That may often be so, but the fault lies ultimately with 
SR dogma. The Lorentz-Einstein transformations cannot 
possibly be 'just coordinates', which is the interpre-
tation required to support the many sideshow carnival acts
with which the SR faithful bedazzle the public, and establish
their moral and intellectual superiority.
If I get in my car and drive steadily for a few hours at 50 
kilometers per hour, is 50t the distance I travel? 
Of course not. The last time my hours-counting 'just coord-
inates' clock was set to zero was when Zeno first reported 
one of his paradoxes to Parmenides. 
That was a long time ago,  so my t is not useful for such 
purposes unless you also use my clock to established the starting 
time, perhaps t0,  and use the formula 50(t-t0) to calculate the 
distance. 
In any case, my t is even then not 'just a coordinate' because
it always represents particular elapsed times that can be
used in the (t-t0) form to calculate perfectly good time
intervals (elapsed times).
Alternatively, I could (re)set my clock to zero at the start
of some meaningful time interval, in which case my t shows a 
scientifically perfect current and/or end time. 
In which case, the Lorentz-Einstein t'=(t-vx/cc)/g is a function 
of an elapsed time interval (not 'just a coordinate') and a time 
interval (-vx/cc; the interval amount the t' clock is being 
screwed up at time t) and thus cannot be 'just a coordinate' 
since neither of the independent variables is such a 'just' thing. 
{Their meaning is shown below, step-by-step.]
If it takes me 50 minutes to cross the Interstate highway,
was x/50 my velocity crossing it?
Of course not.  The origin of all my axes is at the very
spot where Zeno first presented his first paradox to 
Parmenides. That makes my x equal a couple of thousands of
miles, plus, and is not useful for such purposes unless 
you establish the starting x value, perhaps x0, and use the
formula (x-x0)/50 to calculate my velocity.  
In any case,  even then my x is not 'just a coordinate' 
because it always repesents particular distance intervals
that can always be used in the (x-x0) form for any and every
scientific purose.
Alternatively, I could move my x-axis origin to the starting
(zero) point of some meaningful distance, in which case my x 
shows a scientifically perfect current and/or end distance.
In which case, the Lorentz-Einstein x'=(x-vt)/g is a function 
of a current/ending distance interval (not 'just a coordinate') 
and a distance interval (-vt; the interval amount the x' axis
is being screwed up at time t) and thus cannot be 'just a coordinate' 
since neither of the independent variables is such a 'just' thing. 
{Their meaning is shown below, step-by-step.]
------------------------------
 
 1. Introduction with the obvious debunking
    of the use of 'just coordinates' in any
    scientific formula.
 2. Table of Contents.
 3. The Lorentz-Einstein transforms.   
 4. The 'just coordinates' argument.
 5. Single-system, little-purpose ambiguity.
 6. Relating two coordinate measures/systems. 
 7. Distances and moving coordinate axes.
 8. Time intervals.                          
 9. Einstein's (1905) derivations.
       10. A word about intervals.
       11. Intervals versus the Twins Paradox.
       12. Summary
------------------------------
 
Special Relativity's space-time circus is based on
the 'transformation' equations by which it is believed
one can relate a nominally 'stationary' system's space
and time coordinates to those of an inertially (not
accelerating) moving other observer. 
That moving observer's own physical body and coordinate 
system might have been identical in size to those of the 
stationary observer before the traveller began moving, 
but are 'seen' as very different by the stationary observer 
when the relative velocity of the two is great enough, a 
high percentage of the velocity of light.
Concerning ourselves - as is customary - with just
the spatial coordinate axis that lies parallel to
the direction of motion, and with time, Einstein
arrived at these formulas that relate the moving
system measures or coordinates (x' and t') to the
stationary system coordinates (x and t):
      x' = (x - vt)/sqrt(1-vv/cc)      (Eq 1x)
      t' = (t - vx/cc)/sqrt(1-vv/cc)   (Eq 1t)
The v is for the two systems' relative velocity as seen 
by the stationary observer, and is positive if the dir-
ection is toward higher values of x.  By concensus,
the moving system x'-axis higher values also lie in
that direction, and all axes parallel the other system's
corresponding axis.
We used vv to mean the square of v but might use v^2
for that purpose below. Similarly for c.
Because it is believed that no physical object can
reach or exceed c, the square-root term in both
denominators is presumed always less than one, which 
means that the formulas say both x' and t' will tend to
be greater than x and t, respectively.  However,
SRians call the x' result 'contraction' - which means
shortening - and the t' result 'dilation' - which
means increasing. 
------------------------------
 
The 'just coordinates' argument is so patently ridiculous
that even opponents have a hard time accepting just how
simple and obvious its debunking can be, as shown in this
section.  However, further sections take a more arithmet-
ical approach that you'll maybe find more professorial.
The 'just coordinates' argument is that t is mot a
duration, not a time interval; it's just an arbitrary
clock reading.  But what if the moving system observer
comes speeding by while you make your annual 'spring
forward' or 'fall back' change?  The formula says that
the moving system clock's 'just coordinate' reading 
can be calculated from yours:
      t' = (t - vx/cc)/sqrt(1-vv/cc)   (Eq 1t)
Imagine the moving system oberver's confusion if his 
clock changes its reading while he's looking at it!  
If his clock doesn't change when yours does, the formula 
is wrong;  if it is truly a 'just coordinates' formula. 
And then what happens if you realize you were a day 
early and put your clock back to what it had said 
previously?
And suppose you are in NYC and your twin in LA and
both are watching the moving observer. You'll both be
using the same v because you are at rest wrt (with
respect to) each other. You're on Eastern Standard
Time and your twin is on Pacific Standard Time
maybe. You have three hours more on your clock than 
does your twin. 
On which 'just coordinate' clock will the Lorentz 
transforms base the 'just coordinate' time the moving 
system clock says?  The formula applies to both of
your t-times:
      t' = (t - vx/cc)/sqrt(1-vv/cc)   (Eq 1t)
Sure, the idea that you can change someone else's
clock with no connection of any kind is really 
ridiculous, but Eqs 1x and 1t aren't MY equations. 
Are they yours?  And we aren't the ones to say x, t, 
x', and t' are just coordinates.
If the t' formula is actually either an elapsed
time formula, or the basis of a t'/t ratio, then
there is no implication that one clock's reading
has anything to do with the other's. 
It can only be rates of clock ticking, or how one 
time INTERVAL compares to the other that the formula 
is about.
------------------------------
 
Since we're going to be comparing measurements on two
coordinate systems in the next section, let's go to
our supply cabinet and get our yard-stick (which we
use to measure things in inches) and our meter-stick
(which we use to measure things in centimeters).
Here, I'm getting mine. Oh! Oh!
There's an ant on mine, and he ... she ... sure is
hanging on, right at the 3.5 inch mark of the yard-
stick.
Let's see if I can wave the stick around enough that
she'll let go. Nope.
However, before I gave up I waved the stick and the
ant 'all over the place.
Always, however, the ant was at the 3.5 mark on the 
yard-stick, and always 3.5 away from the end of the
stick, however far and wide I have transported her.
Neither of those 3.5 facts means very much. Of the
two, the distance aspect meant almost nothing. So
the distance was 3.5 from the end. So what? That
length, distance, was not in use. And only maybe
the ant might have been concerned with just what
location, 'just coordinate', on the stick she was
at.
Just so with x and t.
So, is the 3.5 reading just a coordinate? Or a
distance/length?  It's ambiguous in and of itself,
and really makes no difference what you say until
you try to make use of the number. 
Hey, my address is 5047 Newton Street. If you
are looking for me and you're at 4120 Newton, it
is helpful information, because it tells you which
direction to go.  Is that 'just coordinate'? 
Where it really becomes useful, perhaps, is in
telling you how far away I am. That's not just
a coordinate value, that's a distance, length,
interval.
However, it is subtracting 4120 from 5047 that 
tells you which direction and how far. It is only 
because both 5047 and 4120 are distances from the 
same point - ANY same point - that the result means 
anything.
My x - my yardstick reading - is always a distance
or length; it is impossible to be otherwise with
an honest, competently designed yardstick.
Whether or not its reading is of good use in some 
particular scientific formula depends on whether 
I put the zero end of the yardstick at some useful
place. As in the introduction, we should either
put it at the starting location/end, or use two
readings from it: (x-x0).
------------------------------
Taking care to not damage our brave little ant, I place
my yard-stick onto the table, zero end to the left, 36
end to the right.
Now I place the 'just coordinate' meter-stick on the table
in the same orientation,  in a random location, and find
that the ant's coordinate on the meter-stick is 51.
The formula relating centimeters to inches is cm=i*2.54
but we want a formula similar to x'=(x-vt)/sqrt(1-vv/cc).
That would be c=i/.03937 approximately, but let's use x'
for the meter-stick reading, and x for the inch reading:
    x'=x/.3937.
    3.5/.3937 = 8.89  
    
Wait a minute.  It's not just science but definition 
that says c=i/.3937=8.89, so something is wrong.  8.89
is not 51.
We already knew that 51 cm was just an arbitrary coordinate. 
Arbitrary not because that point isn't 51 cm from the zero 
end of the meter-stick, but because the zero point was in an 
arbitrary position.
Let's put the meter-stick in a position where it's 
zero point is at the yard-stick zero point.
What is the centimeter coordinate now?  Hey. 8.89,
just like the formula says.
The only way for a 'transform' like x'=x/g to work, 
whatever g might be, is for both coordinate systems
to have their zero points aligned, in which case
saying the two measures are not intervals is pure
idiocy.
Noe that with both zero points at the same position
both x' and x are great measures for scientific
purposes, in any and every case where we were smart
enough to put those zero points at a useful location.
There is one extension of x'=x/g that will let us
use the meter-stick in arbitrary position. 
When the cm reading was 51, the zero point of the
yard-stick read (51-8.89=) 42.11 cm. If we call that
point x.z' we get 
     x' = x.z' + x/.3937.
 = 42.11 + 3.5/.3937
 = 42.11 + 8.89
 = 51.
Obviously, in this formula x/.3937 is the distance
from the x' coordinate of the location where x=0. 
An interval.
Just as obviously, the fact that we now have the
correct formula for relating an x interval to an
arbitrary x' coordinate, does not mean that x'
is anything more than nonsense for use in any
scientific formula.
Unless we were smart enough to put the x zero
point in a useful location, and use (x'-x.z') in
the scientific formula. (x'-x.z') equals the useful,
Ratio Scale value x/.3937.
So, we have discovered a basic fact: a transformation
formula like x'=x/g works only if the two zero points
of the coordinate systems coincide. That makes it non-
sense to say the two coodinates are only coordinates
and not intervals.  Both must be values that represent
distances from their respective zero points unless you
take the proper steps to adjust for the discrepancy.
Make sure you understand that although the inclusion
of x.z' made it possible to correctly calculate x',
the result is nonsense when it comes to use of x'
for general length/distance purposes; it is x'-x.z' 
that is a useful number in such cases. It could be
that we're measuring a sheet of paper with one end
at x=0 and the other at x=3.5; x'=51 is nonsense as
a centimeter measure of the paper.
But, you say, the Lorentz transform contain a -vt term.
------------------------------
 
We discovered x'=x.z' + x/g as the correct formula
for relating one coordinate to another system's.
But the Lorentz transform contains another term, 
-vt/sqrt(1-vv/cc). What is it?
Let's start with our x'=51 cm, x=3.5, x.z'=42.11 example.
Every minute, let's move the meter-stick one inch to our
right.
At minute 0, the cm reading  was   51 cm.
At minute 1, the cm reading is now 50 cm.
At minute 2, the cm reading is now 49 cm.
In this instance, v=1 inch/minute. And t was 0, 1, 2.
What has happened is that we have made our x.z' a lie,
and increasingly so.  -vt/.3937 is the change in x.z'.
   x' = (x.z - vt/.3937) + x/.3937.
Obviously, vt/.3937 is not a coordinate; even most SRians
wouldn't imagine it was. It is an interval, the distance
over which the moving system has moved since t=0.
And, of course, x/.3937 is the distance of our brave
little ant from the point where x=0 and the centimeter
reading is x.z'-vt/.3937. Yes, every minute the meter-
stick moves to the right and the meter-stick coordinate
of the spot where x=0 gets less and less - and eventually
negative.
Make sure you understand that every minute the x' 
coordinate, because of -vt/g, becomes a better measure 
of, say, the  3.5  paper we might be measuring with 
the yard-stick, given that 51 was too big a number and
-vt is negative.  That is, until the two origins coincide 
at x'=x=0,  and then it gets worse and worse.
With -vt positive (because v<0) the situation is different.
With 51 and -vt positive, x' just gets worse and worse
over time.
Quite obviously, the fact that we now have the
correct formula for relating an x interval to an
arbitrary x' coordinate even when the x' axis is
moving, does not mean that x'is anything more than 
nonsense for use in any scientific formula.
Unless we were smart enough to put the x zero point 
in a useful location, and use (x'-x.z'+vt/.3937) in
the scientific formula. (x'-x.z'+vt/.3937) equals the 
useful, Ratio Scale value x/.3937.
------------------------------
 
Instead of using our sticks, let's get out two clocks.
Mind you, we're not going to deal with different clock
rates here, just establish the same basics as for distance.
Your clock says 9:00 Eastern Standard Time (EST) and we 
note that t=540 minutes when we put down the clock.
Blindly, let's turn the setting knob of your twin's Pacific 
Standard Time clock and put it down before us.
According to what we see, EST's 540 minutes (9:00) corre-
sponds to PST's 14:30; t'=870.
We know the formula relating PST to EST is t' (pacific)
= t (eastern) - 180 (minutes). Thus, it is not correct 
that the second clock can have an arbitrary setting, 
We know that the two clocks are related by t' = t/1 since 
both are using the same second, hour, etc units. But 870 
(14:30 in minutes) is not 540/1-180, so once again we know 
something is wrong. 
However, t'=t.z' + t/1 works. EST midnight equals PST 0.0 
(midnite)  - 180,  so t.z' = -180, and
     t' = -180 + 540/1  = 360.
Since EST-180=PST, 9:00 EST is 6:00 PST = 360 minutes.
We see thus that like distance measures/coordinates, time
axis origins (zero points) must either be 'lined up' or 
adjusted for. 
So, the Lorentz/Einstein t'=t/sqrt(1-vv/cc) must be the moving 
system elapsed time interval since the time axes were both at 
a common zero. There is no t.z' adjustment:
      t' = (t - vx/cc)/sqrt(1-vv/cc)   (Eq 1t)
Make sure you understand that in the clock case, if the
EST is showing a good number for elapsed time since the
travelling observer passed NYC, then the PST clock is
silliness. t.z' must be zero or must be taken out of
time lapse calculations for the PST clock to be used
intelligently, just as was true for x.z'.
What is lacking as yet for Lorentz t' is the -vx/cc term that
corresponds to the x' formula -vt term.
Break it up into two parts: v/c and x/c.  
v/c is a scaling factor that changes velocity from whatever 
kind of unit you are using over to fractions of c.
x/c is distance divided by velocity, which is time. x/c
is thus the time interval since the two time axes
had a common zero point - which they have to have in the
Lorentz transforms which do not have the t.z' term we
learned to use above.
Thus, (-vx/cc)/sqrt(1-vv/cc) is the interval amount the 
moving system clock has been changed - since the common/
adjusted time - over and beyond the elapsed time interval
represented by x/sqrt(1-vv/cc).
We have discovered that the only way for t' to be t/g
is for t' and t to have a common zero point, just as
for x' and x. It would be otherwise if the t' formula
contained an adjustment t.z' under some name or other,
but the necessity to include such a term correlates
100% with t' numbers that aren't directly usable.
As for x and x', our knowledge of how to setup a proper
formula relating t and t' is of no use unless we use
the knowledge in scientific formulas; (t'-t.z'+xv/gcc)
gives us the only directly useful value: t/g.
------------------------------
 
When we return to Einstein's derivations of the transform
formulas with a well-focused eye, we find he was a wee bit
confused - or at least self-contradictory.
When he set up his (at first unknown) tau=moving system
time formulas, he created three particular instances of tau.
Tau.0 is the time at which light is emitted at the moving
origin toward a mirror to the right that is moving at rest 
wrt that moving origin and at a constant distance from that 
origin. He lets the stationary time slot have the value t,
a constant, the stationary system starting time.
Tau.1 is the time at which the light is reflected. He
lets the stationary time be t+x'/(c-v); t is still a
constant and x'/(c-v) is the time interval since t.
Tau.2 is the time at which the light gets (back) to the
moving origin. The stationary time value is put as t +
x'/(c-v) + x'/(c+v);  t is still a constant and x'/(c-v)
+ x'/(c+v) is the time interval since t.
On the thesis that the moving observer sees the time to
the mirror as the same as the time back to the origin,
he sets
   .5[ tau.0 + tau.2 ] = tau.1.
Tau.0 completely drops out of the analysis and leaves
no trace, and has no effect.
Further, the t you see in tau.0, tau.1, and tau.2 also 
completely drops out with no trace and no effect, leaving 
us with exactly what you'd get if you had explicilty said 
t' is an interval and so is t.
What doesn't drop out in the stationary time values is
x'/(c-v) and x'/(c+v), the time interval it takes for
light to get to the fleeing mirror, and the time interval
it takes for light to get back to the approaching origin.
Thus, his resultant t' formula is strictly based on time 
intervals in the stationary system. Time intervals since 
some starting time, yes, but time intervals.
There is absolutely nothing in the derived formulas that 
depends on arbitrary coordinates like the constant t in 
the stationary time arguments.
Let's look at the x dimension; it is x'=x-vt [as x increases
by vt, the effect over time is x'=(x+vt)-vt)], which Einstein 
explicitly sets up as a constant stationary distance.
He uses that x' not just in the time interval parts of the 
stationary time arguments,  but also in the x (distance) 
stationary system argument for the tau at the time light 
is reflected. 
x' can't be the stationary system coordinate of the mirror 
at that time.  That value is x'+vt.
x' is explicitly an interval, distance. 
Thus, the whole tau derivation of the t' formula is fully and
explicitly based on x'  - a spatial length/distance/interval -
and the two time interals x'/(c-v) and  x'/(c+v).
While we're at it, if the starting t is not zero, his 
x'=x-vt formula is complete nonsense also. Given that
there was some L that was the mirror x-location and length
when the light is emitted, if t was already, say, 500, then
x'=L-vt could have been a very negative length.
------------------------------
 
There are intervals, and there are intervals.
If we put our yard stick zero point at one end
of a piece of paper and read off the coordinate
at the other end of the paper, we have a good
measure of the paper's length, a Ratio Scale
measure. [Absolute temperature scales are ratio
scale.]
If instead we put the one end of the paper at the
one inch mark (or the zero end of the stick one
inch 'into' the length of the paper) we get measures
that are one inch off the true, ratio scale length.
The two messed up measures are still intervals,
but they are Interval Scale measures. [Household
temperature scales are interval scale, which is
why your physics and chemistry professors won't
let you use them without first converting to the
ratio scale absolute temperatures.)
t'=t/g and x'=x/g represent ratio scale measures,
given that t and x were ratio scalae to start with.
t'=t.z'+t/g and t'=t/g-vx/gcc are both interval 
scale measures, even given a good ratio scale t
and a good ratio scale x.
x'=x.z'+x/g and x'=x/g-vt/g are both interval 
scale measures, even given a good ratio scale x
and a good ratio scale t.
Look for the (SR) Lorentz t', x' = degraded measures
document soon at a newsgroup near you.
------------------------------
 
t'=(t-vx/cc)/g shows t' being greater than t.
The reason Special Relativity will not allow the
use of its basic time equation in determining what
SR has to say about the twins' ages, is that t' and
x' are supposedly just coordinates, and they say you 
have to take the coordinate pairs (t',x') and (x,t)
into consideration in both the time and place the 
twins' separation started and the time and place the 
twins reunited.
Since t' and x' are actually both intervals, not
just coordinates, the 'excuse' is spurious, and is 
so even without use of the obvious (x_b-x_a) and
(t_b-t_a) usages.
However, SR is right to be embarrassed by their
transformation formulas.
Look for the (SR) Lorentz t', x' = degraded measures
document at a newsgroup near you.
 
------------------------------
 
A.  t'=t/g and x'=x/g can be almost 'just coordinates'
    in the sense that the values obtained may not be
    of much use except in the most primal and useless
    way: how long and how far since/from the time/
    place they were zero. Even here, however, the zero
    points within each of the two scale pairs (t',t) 
    and (x'.x) must have been lined up.  If the zero
    points have been intelligently selected (such as
    at the starting point and time of a trip) they 
    can be rationally used 'as is' in any valid sci-
    entific equation.
B.  Even the interval scale t'=t.z' - xv/gcc + t/g and 
    x'=x.z' - vt/g + x/g are not 'just coordinates'. They 
    can be used to good effect by establishing the relevant 
    starting times/points and using (t'-t.z'+xv/gcc) and 
    (x'-x.z'+vt/g), as the situation may require.
C.  When you see vx/gcc or vt/g in use in any guise with non-zero
    values, you know the resultant t' or x' is a degraded, interval
    scale value.
E-X: Anytime you do not see what amounts to t.z' and xv/gcc in
     the time case, or x.z' and vt/g in the distance case, you
     know that the t' and/or x' in use are intervals. Period.
Y:   Either set your clock to zero at the start of the relevant
     time interval, or use (t-t0), with both being readings on
     the same clock. Either move your x-axis origin to the starting
     end or point, or use (x-x0), with both being readings on
     the same axis.
Z:   In _(SR) Lorentz t', x' = Degraded (Interval) Scales_ we see 
     that t' and x' satisfy the mathematical tests for/of interval
     scales when -vt and -vx/cc are not zero; thus, they must
     be intervals. When -vt and -vx/cc are zero, t' and x'
     satisfy the much better mathematical definition of 
     ratio scales, and are thus not just mere intervals,
     but (rescaled) good ones.
Eleaticus
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
! Eleaticus        Oren C. Webster         ThnkTank@concentric.net  ?
! Anything and everything that requires or encourages systematic   ?
!  examination of premises, logic, and conclusions                 ?
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
[snip lies]
[snip 700 lines of trolled garbage]
eleaticus, Oren Webster, is a despised and stooopid troll,
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
 Several crimes against logic and science  Ha ha ha!
 
Originally trolled across sci.physics sci.physics.relativity
alt.physics sci.math sci.answers alt.answers news.answers
Psychotic ineducable boring troll Eleaticus,
Were there to be internal inconsistencies in SR (meaning
inconsistencies of a purely mathematical logical nature) that would
automatically lead to contradictions in number theory, itself, and
arithmetic, since the mathematics of Minkowski geometry is
equiconsistent with the theory of real numbers and with arithmetic. 
Eleaticus explicitly demonstrates that he is completely ignorant of
multivariable calculus.  He has no concept of the Chain Rule in
multivariable calculus.  Consider his Galilean Transformation goo and
dribble:
   t' = t,
   x' = x - vt,
   y' = y,
   z' = z.
His refusal to accept that t' must be introduced as a separate
variable springs from a massive emprical stupidity re space and time
are described as a four-dimensional manifold, with four coordinates
instead of a time evolution of a three-dimensional manifold,  and that
the change of coordinate system should be a change of four
coordinates, and not a time-dependent change of three coordinates. 
This is particularly vital when it comes to fields over space and time
(electric and magnetic fields for example).
The transformation law for the differential operators under the
Galilean transformation is given by:
   d/dt' = d/dt + v d/dx,
   d/dx' = d/dx,
   d/dy' = d/dy,
   d/dz' = d/dz.
This shows the necessity of introducing a new variable t', since
partial differentiation with respect to t' (constant x', y', z') is a
different operation to partial differentiation with respect to t
(constant x, y, z).  The above transformation law is determined by the
Chain Rule:
d/dt' = dt/dt' d/dt + dx/dt' d/dx + dy/dt' d/dy + dz/dt' d/dz,
d/dx' = dt/dx' d/dt + dx/dx' d/dx + dy/dx' d/dy + dz/dx' d/dz,
d/dy' = dt/dy' d/dt + dx/dy' d/dx + dy/dy' d/dy + dz/dy' d/dz,
d/dz' = dt/dz' d/dt + dx/dz' d/dx + dy/dz' d/dy + dz/dz' d/dz.
The presence of the term involving d/dx in the expression for d/dt' is
indicative of the fact that x depends on t' (x', y', z', being held
constant), as can be seen from the fact that the coefficient of d/dx
in the expression for d/dt' is dx/dt'.  Because of the now
demonstrated fact that Eleaticus has no formal education in
multivariable calculus, he has managed, somehow, to get it into his
head that the presence of the term involving d/dx in the expression
for d/dt' is indicative of t' depending on x (t, y, z, being held
constant).  Because of his stupidty Eleaticus cannot get the correct
transformation law for the differential operators under the Galilean
Transformation, and he cannot determine the invariance or otherwise of
Maxwell's Equations under the Galilean Transformation.  The first
advice to Eleaticus is to learn multivariable calculus.
Eleaticus should not pretend that he can understand how to determine
invariance or otherwise of Maxwell's Equations under the Galilean
Transformation, or under the Lorentz Transformation, until he
understands the multivariable calculus which underlies such
considerations.  Eleaticus is a loud idiot.
The homogeneous Maxwell equations are invariant under the Galilean
Transformation, with transformation laws:
  E_x' = E_x,
  E_y' = E_y - v B_z,
  E_z' = E_z + v B_y,
  B_x' = B_x,
  B_y' = B_y,
  B_z' = B_z.
The derivation of these transformation laws was determined using the
transformation laws for the differential operators given above.  These
transformation laws have the additional advantage that they determine
the correct transformation for the force law, thus providing further
evidence in favour of the transformation law for the differential
operators, as above.
The inhomogeneous Maxwell equations are also invariant under the
Galilean transformation, with transformation laws:
  E_x' = E_x,
  E_y' = E_y,
  E_z' = E_z,
  B_x' = B_x,
  B_y' = B_y + v/c^2 E_z,
  B_z' = B_z - v/c^2 E_y,
  rho' = rho,
  J_x' = J_x - v rho,
  J_y' = J_y,
  J_z' = J_z.
Note the the transformation laws for the charge density and current
density are as they should be under the Galilean transformation.
Homogeneous equations are invariant under the Galilean Transformation,
and inhomogeneous equations are invariant under the Galilean
Transformation, but Maxwell's Equations as a whole are NOT invariant
under the Galilean Transformation, since the transformation laws
required for the EM field for the two cases are inconsistent with each
other.  The transformation law for the EM field which makes the
homogeneous equations invariant will not also make the inhomogeneous
equations invariant.  The transformation law for the EM field which
makes the inhomogeneous equations invariant will not also make the
homogeneous equations invariant.
On the other hand, all of Maxwell's equations are invariant under the
Lorentz Transformation, with transformation laws:
  E_x' = E_x,
  E_y' = gamma (E_y - v B_z),
  E_z' = gamma (E_z + v B_y),
  B_x' = B_x,
  B_y' = gamma (B_y + v/c^2 E_z),
  B_z' = gamma (B_z - v/c^2 E_y),
  rho' = gamma (rho - v/c^2 J_x),
  J_x' = gamma (J_x - v rho),
  J_y' = J_y,
  J_z' = J_z,
where gamma = 1/sqrt(1 - v^2/c^2).
 Idiot Oren Webster sees himself this way,
http://www.mazepath.com/uncleal/effete6.jpg
 The entire remainder of the planet sees him this way,
http://www.mazepath.com/uncleal/effete3.png
http://www.mazepath.com/uncleal/sunshine.jpg
http://www.apa.org/journals/psp/psp7761121.html
http://insti.physics.sunysb.edu/~siegel/quack.html
Hey, stooopid troll Eleaticus - Do you want EVIDENCE?  Each of the 24
GPS satellites carries either four cesium atomic clocks or three
rubidum atomic clocks in orbit, with full relativistic corrections
being applied.
Internal inconsistencies in SR (meaning inconsistencies of a purely
mathematical logical nature) automatically lead to contradictions in
number theory, itself, and arithmetic, since the mathematics of
Minkowski geometry is equiconsistent with the theory of real numbers
and with arithmetic. 
 Mathematics of gravitation
 Equivalence Principle testing
http://arXiv.org/abs/hep-th/0111236
 Geometric structure of reality
http://arxiv.org/abs/gr-qc/0103044
http://arXiv.org/abs/hep-th/0307140
 GR structure, especially Part 4/p. 7
http://arXiv.org/abs/gr-qc/0311039
 Experimental constraints on General Relativity
http://www.eftaylor.com/pub/projecta.pdf
 Relativity in the GPS system
http://arXiv.org/abs/gr-qc/9909014
 falling light
 Hafele-Keating Experiment
http://www.hawaii.edu/suremath/SRtwinParadox.html
 Twin Paradox
http://arXiv.org/abs/astro-ph/0401086
http://arxiv.org/abs/astro-ph/0312071
 Deeply relativistic neutron star binaries
http://arxiv.org/abs/hep-th/0405160
 Black hole evaporation
http://physicstoday.org/vol-57/iss-7/p40.shtml  
 No aether
http://fsweb.berry.edu/academic/mans/clane/
 No Lorentz violation
http://arXiv.org/abs/gr-qc/0409089
 Spin-2 gravitons have problems
  (so does the proposal)
http://arXiv.org/abs/gr-qc/0301024
 Nordtvedt Effect
http://map.gsfc.nasa.gov/
http://arxiv.org/abs/astro-ph/0403292
http://arXiv.org/abs/astro-ph/0310723
 WMAP + Sloane Digital Sky Survey
http://arxiv.org/abs/hep-ph/0404175
 Dark matter candidates
 Carroll on what it all means.
Special Relativity is physics on a topologically trivial Lorentzian 
manifold with a metric whose curvature tensor is zero.  This is a 
perfectly diffeomorphism-invariant condition and does not require 
any particular coordinate choice.  It is invariant under 
the full group of diffeomorphisms.  The Poincare group is 
the group of *isometries* of the metric in special relativity. 
The Special Relativity metric is *non-dynamical* (unlike GR).  It 
defines the coupling *constants* of your theory. If you change the 
metric in any nontrivial way you are changing your theory.  An 
operation can only be called a symmetry of a special-relativistic 
(non-gravitational) theory if it preserves the metric, and therefore 
the symmetry of special-relativistic theories is the Poincare group 
only.  General Relativity (gravitation) has a dynamic metric.
NIM A 355 537 (1995)
Physics Letters B 328 103 (1994)
Physical Review Letters 64 1697 (1990)
Physical Review Letters 39 1051 (1977)
Physical Review 135 B1071 (1964)
Physics Letters 12 260 (1964)
Europhysics Letters 56(2) 170-174 (2001)
General Relativity and Gravitation 34(9) 1371 (2002)
http://fourmilab.to/etexts/einstein/specrel/specrel.pdf
 Longitudinal and transverse mass
http://arxiv.org/abs/gr-qc/0306076.pdf
http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf
http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm
http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm
http://www.trimble.com/gps/index.html
http://sirius.chinalake.navy.mil/satpred/
http://www.phys.lsu.edu/mog/mog9/node9.html
http://egtphysics.net/GPS/RelGPS.htm
http://www.schriever.af.mil/gps/Current/current.oa1
http://edu-observatory.org/gps/gps_books.html
-- 
Uncle Al
http://www.mazepath.com/uncleal/
 (Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
can someone help me how to construct the generating function for
this and similar problems ?  See A070030 for a similar problem
with a smaller matrix and a nice quotient of 2 polynomials as
generating
function.
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?An

um=A070030
How large may the matrices grow so
it's still feasable ? Suppose M=M1*..*M4 where each Mi is a sparse
1e6*1e6 matrix with only 10-20 nonzero entries per row.
Can it be done ? Where is the limit with today's computers ?
Guenter, http://magictour.free.fr/enum
c 12 comment-lines including this one, then 2 + 1(empty) + 145 data
lines
c this file contains the data to compute the number of open knight's
tours
c on a 3*b board, denoted by kto(3,b), for even b. A tour and its
c reversed tour are counted twice.
c The first line gives the (1*145) startvector S, the second line
gives
c the (1*145) endvector E, the 3rd line is empty , 
c the next 145 lines give the rows of the
c (145*145) transfer-matrix M. The formula is:
c      kto(3,2*b+6) = E * (M^b)(S) 
c where * is the scalar dot-product of vectors
c and M^b is the b-th power of M by matrix-multiplication.
c -------------------------------------------------------------------------
0,0,0,0,22,4,0,0,2,4,4,2,4,2,4,0,2,2,0,0,0,0,0,0,0,0,0,0,0,0,2,4,0,2,2,0,0,0

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,
0,0,0,2,0,0,2,2,2,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0
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0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,4,0,4,4,0,0,0,0,0,0,0,0,

0,0,0,0,0,4,0,0,0,0,0,0,8,0,0,8,0,4,0,8,0,0,0,4,0,0,2,2,0,2,4,2,4,10,0,0,0,0
,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,4,0,0,4,0,4,0,0,0,0,4,0,4,0,4,0,4,0,0,0,0,8,
0
,4,4,8,4,8,4,20,0,0,4,0,0,4,0,4,0,0,8,0,8,0,8,4,8,0,12,16,4,12,40
0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,2,0,0,0,0,0,4,0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,

0,0,0,0,0,0,0,6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,4,0,0,0,0,0,4,0,0,0,0,0,0,0,0,0,
0
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,
0,0,0,0,0,0,0,0,6,0,0,0,0,4,0,0,0,0,9,0,0,0,0,0,4,0,0,0,0,0
0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,

0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,
0
,0,0,0,1,0,2,0,0,3,0,1,0,2,0,0,0,0,0,0,3,0,0,0,0,3,0,0,0,0,0,3,0,0,0,0,4,0,0
,
0,0,0,0,0,0,0,0,2,0,0,0,0,4,0,0,0,0,3,0,0,0,0,0,4,0,0,0,0,0
0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,

0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,3,0,0,
0
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,
0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0
... (145 lines) , full matrix at http://magictour.free.fr/KTO3
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0
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,
0,0,0,1,0,0,0,0,0,0,0,0,0,0,2,0,1,0,0,1,2,0,0,0,0,2,0,0,0,0
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0
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,
0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,2,1,0,0,1,2,0,0,0,0,2,0,1,1,0