mm-2719
===
Subject: Can this equation be solved?
I'm wondering if the following equation can be solved for r:
y = (1 - e^ -rx)/(1 - e^ -r)
[ In Excel you can write it as =((1-exp(-r*x))/(1-exp(-r)); you can
use the Goal Seek option by seting the equation address to a value by
changing r ]
I'm writing a personalized function in Excel and right that is based
off of material that one of my professors published. Currently, I
have a looping structure that uses powers of 10 to find the answer.
(Not that anyone is interested, but I included the code below).
I'm just curious if this equation can actually be solved for r or if
using iterative processes is the best avenue of attack. I've already
hit up some of my peers who are advanced math students and they'll
haven't been able to come up with an answer; maybe I just need to seek
out some of the math professors.
Matt
Function riskRAD(x As Double, y As Double) As Double
Dim r1 As Double 'exchange rate ABOVE answer
Dim r2 As Double 'exchange rate BELOW answer
'x '% Certain Dollars
'y '% Probable Dollars
Dim output As Double 'output of equation answer
Dim iterations As Integer 'number of times to loop/iterate
Dim a As Long 'marker in the loop
Dim flag As Boolean 'mark a negative number
'''---
'OUTLINE
'if x = .2 and y = .5 then r = about 3.28030029613385 by using Goal
Seek:
''goal seek: setCell [ y formula i.e. y = (1 - Exp(-r * x)) / (1 -
Exp(-r)) ]
'' toValue [ y value i.e. 0.5 ]
'' byChanging [ r ]
'if x = .2 and y = .5 then r = 3.28127989618487 in the below loop
'''---
'''---
iterations = 14
'the number of iterations determines how accurate you want the answer
'e.g. 10^4 = 10000, or an answer that is 0.0001 accurate (1/10000)
'e.g. 10^5 = 100000, or an answer that is 0.00001 accurate (1/100000)
'---
'''---
'how to handle both 0, .5, and 1
If ((x = 0) Or (x = 0.5) Or (x = 1)) Then
'riskRAD = (1 / 10 ^ iterations)
Exit Function
End If
If ((x = 0.5) And (y = 0.5)) Then
riskRAD = (1 / 10 ^ iterations)
Exit Function
End If
If ((x = 1) And (y = 0.5)) Then
riskRAD = (1 / 10 ^ iterations)
Exit Function
End If
'''---
r1 = (1 / 10 ^ iterations)
flag = False
If x < y Then
x = x
x = (1 - x)
flag = True
End If
output = (1 - Exp(-r1 * x)) / (1 - Exp(-r1))
For a = 0 To iterations Step 10 ^ a
r1 = r1 + (1 / 10 ^ a)
r2 = r1 - (1 / 10 ^ a)
output = (1 - Exp(-r1 * x)) / (1 - Exp(-r1))
Loop
output = (1 - Exp(-r2 * x)) / (1 - Exp(-r2))
r1 = r2
Next
If flag = True Then
riskRAD = -r2
Debug.Print riskRADr2=; -r2
Else
riskRAD = r2
Debug.Print riskRADr2=; r2
End If
End Function
===
Subject: Re: Can this equation be solved?
or
y = (e^r - 1) / ( (e^r)^x - 1)
Which will be formally solvable for r only when x is one of a very few
small integers. Otherwise an approximate numerical solution is the best
you can expect.
===
Subject: Re: Can this equation be solved?
The general solution presumably can't be given in closed form in terms of
familiar functions.
If you might be interested in approximate solutions, please see my response
q corresponds to your y, my R to your e^(-r), and my n to your x. Of
course, knowing an approximation for my R, the corresponding approximation
for your r is -ln(R).]
I mention approximations which work well in two cases:
if y is close to 1 or if y is large.
David W. Cantrell
===
Subject: Re: number of cycles in directed graphs
The cycle problem maybe is a P problem and is also a NP problem as my
view, only I could not prove the algorithm and disprove my algorithms
in mathematics, but my algorithm could find all length of cycles in a
digraph in polynomial time.
Maybe I could calculate the number of cycles for each node in a
digraph, for example
input file: test6.dat
DIMENSION: 6
NUMBER EDGES: 9
0 1 -1 0 0 0 -1 0
1 -1 0 0 -1 0 0 0
0 0 0 -1 1 0 0 1
-1 0 0 1 0 0 0 0
0 0 1 0 0 -1 0 0
0 0 0 0 0 1 1 -1
output:
length=3={v2,v3,v4}
length=5={v1,v2,v3,v4,v6}
length=6={v1,v2,v3,v4,v5,v6}=Hamiltonian cycle
if you could send the data set to me in a incidence matrix. the
running time is not a problem when the number of edges is no more
thant twices of number of nodes, but memory maybe is a big problem for
my computer as the matrix 4300*4300 is big.
--------------------------
Zhu
===
Subject: Re: number of cycles in directed graphs
Sorry for my careless, input file is
DIMENSION:6
NUMBER EDGES: 8
0 1 -1 0 0 0 -1 0
1 -1 0 0 -1 0 0 0
0 0 0 -1 1 0 0 1
-1 0 0 1 0 0 0 0
0 0 1 0 0 -1 0 0
0 0 0 0 0 1 1 -1
EOF
my algorithm output is :
Ei=3 length= 3 {e1,e4,e5,}
Vi=3 length= 3 {v2,v3,v4,}
Ei=5 length= 5 {e1,e2,e4,e7,e8,}
Vi=5 length= 5 {v1,v2,v3,v4,v6,}
Ei=6 length= 6 {e1,e2,e3,e4,e6,e8,}=Hamiltonian cycle
Vi=6 length= 6 {v1,v2,v3,v4,v5,v6,}
Since the algorithm is used by myself,
so the output set of edges
and set of vertices is not sort by topological.
But with the output set of edges and the incidence matrix, one could
obtain
the (Hamiltonian )cycle in polynomial time (This coding I had comments
since it is not need for me).
--------------------------------
Zhu
===
Subject: Please solve this differential equation problem.
Please help me.
A1=int{U(S1(x1,x2)) df(x1,a1)/da1 f(x2,a2)}dx1dx2-v'(a1)-
P'(e2(a1))e2'(a1)+lambda1dA2/da1+mu1dB2/da1=0
B1=-1+lambda1{dA2/de1(a2)}+mu1{dB2/de1(a2)}
A1=int{U(S2(x1,x2)) f(x1,a1) df(x2,a2)/da2}dx1dx2-v'(a2)-
P'(e1(a2))e1'(a2)+lambda2{dA1/da2}+mu1{dB1/da2}=0
B1=-1+lambda2{dA1/de2(a1)}+mu2{dB1/de2(a1)}
===
Subject: Please solve this differential equation problem.
Please help me.
A1=int{U(S1(x1,x2)) df(x1,a1)/da1 f(x2,a2)}dx1dx2 -v'(a1)-
P'(e2(a1))e2'(a1)+lambda1{dA2/da1}+mu1{dB2/da1}=0
B1=-1+lambda1{dA2/de1(a2)}+mu1{dB2/de1(a2)}
A2=int{U(S2(x1,x2)) f(x1,a1) df(x2,a2)/da2}dx1dx2-v'(a2)-
P'(e1(a2))e1'(a2)+lambda2{dA1/da2}+mu1{dB1/da2}=0
B2=-1+lambda2{dA1/de2(a1)}+mu2{dB1/de2(a1)}
===
Subject: Re: If the expected value of a random variable is random variable?
Suppose f(.|a) is the density of x given a and p is the probability
function of a, then (with some abuse of notation)
E[ax] = sum_a int_x ax f(x|a) p(a)
= int_x ax f(x|0) p(0) + int_x ax f(x|1) p(1)
= E[ax | a=0} p(0) + E[ax | a=1] p(1)
--
David Marcus
===
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===
Subject: Trying to define a set theory.
Hi all,
The following is a set theory that I developed latelly.
I don't know if it is consistent or not?
Hereby I present it to be tested.
Introduction: This theory is an extension of first order logic with
identity. It contains three types of classes.
Types of classes
1) A class of all classes that is not a member of any class other than
itself. And this class is T.
2) classes that only T can contain them as members, and these classes
are called 'proper classes'.
3) classes that are members of a class other than T, and these are
called 'sets'
Structure:
All sets are members of one unique proper class V.
All proper classes are subsets of V and are not in V.
T contains all classes and itself as members.
Defining the theory.
Name of the theory is Y .
Theory Y is the set of sentences entailed
(from first order logic with identity)by these axioms:
2) Universe: E!xAy(yex & ~Ez(~z=x & xez)).
3) Regularity:
4) Schema of Global comprehension: if P is a formula in which x
doesn't occure free, then all closures of:
are axioms.
Theorum 1) V.
we get
To prove x is unique, let x1 and x2 be defined as
for a proof by negation suppose that ~x1=x2
Then there should be a set uex1 and ~uex2
Or uex2 and ~uex1.(Extensionality).
if u: uex1 and ~uex2 , then x2 is not the class
of all sets, violating its definition.
if u: uex2 and ~uex1 , then x1 is not the class of
all sets, violating its definition.
Therefore x1=x2.
This means the following.
Now since x is unique then we proceed to name it.
Accordingly V is the proper class of all sets.
Theorum 2) ~VeV.
Proof: let VeV. then we have Ez(~z=T & Vez) otherwise
~VeV. (The definition of V).
so ~V=T, since ~Ez(~z=T & Tez). definition 1.
Since VeV then pairing applies to it and we have
from pairing B={V}. Now from pairing we have BeV
and since ~V=T, then B is a member of a set that is not T
therefore ~B=T. And since B has V as a member then
~B=0. Thus regularity applies to B..
But B vioates regularity! A contradiction.
Thus ~VeV.
Theorum 3) ~T=V
Since ~VeV (theorum2) and since VeT (axiom 2).
Thus ~V=T (Extensionality).
5) Empty set: ExeVAy(~(yex)).
xen))).
with 'U' and {x} having the usual definitions.
Theorum 4) Cartesian Product:
Proof: since neither a nor b are T
then for every u every w such that uea and web
,u and w are sets. ( definition of set ).
substituting it in global comprehension yields theorum 4.
axb is the cartesian product of a and b.
Theorum schema 5) Large separation
if Q is a formula in one variable in which x doesn't occure free then
all closures of
are theorums.
Proof: since ~a=T and yea then y is a set
in global comprehension yields theorum 5.
Theorum schema 6) Relation
if P is a formula in one variable in which x doesn't occure free then
Proof: it follows from theorum 4 & 5.
This theorum schema allows us to define functions, injections,
surjections and bijections between any two classes weather sets
or proper classes provided than any of them is not T.
Theorum 8) Non supernumerosity of subsets.
P(a) for power set of a.
Proof: xeP(a) then x is a subset of a.
Let f be the identity function f(x)=x
from x to a. Then f is injective from x to a.
So it cannot be supernumerous to a ( definition 6).
Theorum schema 9) Large replacement.
if Q is a formula in two variables in which b is not free,then
all closures of
are theorums.
Proof:
since yeV then y is a set, then
substituting it in global comprehension
we get theorum schema 9.
10) limitation of size:
Thoerum schema 10) small separation
if P is a formula in one variable in which x is not free then all
Proof: follows from large separation and non supernumerousity of
subsets and axiom 10.
Theorum schema 11) small replacement.
if P is a formula in two variables in which b is not free then
all closures of
are theorums.
Proof: this follows from large replacement and non supernumerosity of
the range of any function over its domain, and axiom 10.
Proof: if y is supernumerous to x , then there exist
at least two ordered pairs in f such that for the same uex
there are different sets w and k in y. i.e there must exist
violating the definition of a function.
Thoerum 12) global choice.
Ax(~x=T) Ey( y is an ordinal & y is equinumerous to x).
substituting P(x) in global comrpehension yields.
y is proved to be unique in a similar manner to how uniqueness
of V is proved.
Now we have ~OeO as a theorum in this theory (Proof similar to ~VeV).
Thus O is the proper class, i.e ~Ez(~z=T & Oez) ,of all ordinals that
are sets. and O is itself an ordinal.
Since we have ~Ez(~z=T & Oez) then O is not subnumerous to V ( axiom
10).
Now since O is a subset of V, then O is not supernumerous to V(theorum
8).
so O is neither super- nor sub- numerous to V. then
O is equinumerous with V.
Since O is an ordinal, then global choice is applicable to V.
the proof that for every subset of V there exist an ordinal that is
equinumerous to is simple, but a little bit long_ It can be proved
that every subclass k of O is well ordered and thus there is an
ordinal to which k is bijectable. Now for every subset of V , either
it is a proper class in which it would be bijectable to O, or it would
be a set in which it would be subnumerous to O, and thus bijectable to
a proper subclass k of O, which is bijectable to an ordinal in O and
so every set is bijectable to an ordinal in O.
Theory definition finished.
/
The above is the theory with the important defining theorums in it
There are a lot of theorums concerning proper classes,but these are to
be discussed later.
The important results that I think this theory can acheive are
1) Prove ZFC
2) define intersections
3) defining cardinalities of proper classes.
As for 1) it is clear that all axioms in ZFC except infinity and
pairing and empty are theorums in this theory, while infinity and
pairing and empty are axioms here. so ZFC is in this theory.
the problem was when y=0 , what would be x.
here the answer is x=T.
so x=intersection y is solved here.
3) This theory defines relations not only between sets
but also between proper classes, and between sets and proper classes.
For instance we can have equivalence classes of all sets
i.e Frege definition of cardinality, but this won't work
for the case of proper classes, therefore Von Neuman's
method is better, since it defines the cardinality of
proper classes which is O.
Cardinality is defined here in the following manner.
so the intersectional class of all ordinals bijectable to V is O.
so card V = O.
T in this theory is somewhat detached from the other sets and proper
classes, you cannot define subsets of T that are not subsets of V,
Union T = T. Power is not defined for T.
You cannot have separation on T, nor replacement , T is outside the
reach of regularity.
So the main function of T is intuitive: it has them all including
itself as members.
Now this theory might be balantently inconsistent, or might be
consistent, I don't know, that's why I presented it here , in order to
be checked.
Zuhair
===
Subject: Re: Trying to define a set theory.
which is O.
To clarify that:
Definition of Cardinality
~Ez(z is an ordinal & z is equinumerous to x & zey)).
The definition of x is an ordinal is given (see above).
The definiton of equinumerousity is also given , it is:
Of course
Ax(~x=T)E!y(y is an ordinal & y is equinumerous to x & ~Ez(z is an
ordinal & z is equinumerous to x & zey)).
is a theorum in this theory. It can be easily proved.
Now card(V) = O.
x is subnumerous to V (axiom 10). we know that O is equinumerous to V.
Then there do not exist an ordinal
in O that is equinumerous to V.
then O is card V.
Zuhair
===
Subject: Re: Trying to define a set theory.
Correction: .... all closures of
three interesting thoerum.
yeb)).
straightforwards from global comprehension with
but when (a=T or b=T), I don't see any of the axioms proving this
theorum. Perhaps I should axiomatiz this theorum in this case.
so x is the set of binary union of a and b symbolized as
aUb.
This can be also important in derving cartesian products between any
two classes.
P reders to Power .
ii) Theorum of Large binary union:
Proof: binary union and limition of size.
or yeb)).
Proof: binary union and limitation of size.
Zuhair
===
Subject: Re: Trying to define a set theory.
Sorry for the last post:
Three interesting thoerums.
i)Theorum (or perhaps axiom )of Binary union:
straightforwards from global comprehension with
but when (a=T or b=T), I don't see any of the axioms proving this
theorum. Perhaps I should axiomatiz this theorum in this case.
so x is the set of binary union of a and b symbolized as
aUb.
This can be also important in derving cartesian products between any
two classes.
P refers to Power .
ii) Theorum of Large binary union:
Proof: binary union and limition of size.
or yeb)).
Proof: binary union and limitation of size
Zuhair
===
Subject: Re: Mobius transformations
I see now, thank you for your answer.
Greg
===
Subject: Re: A extremum problem
Do you understand dx/dp = 0?
===
Subject: Re: A extremum problem
but how can I know the most of least value I can get
by dx/dp=0?
===
Subject: Re: *** The buildings came down by explosives planted by YANK
BASTARDS, even if the plane that hit them were flown by Bin Laden's men which
is itself doubtful but NO DOUBT about the PLANTED explosives
**********************************************************
I'd vote to declare rock, schoenfeld and therma a new eternal
moronical braid: these sad, but at the same time comical, grotesque
and rather ridiculous, characters seem to REALLY believe the WHOLE
world is deep into a conspiracy that only they, precisely they, can
see.
Tell them about scientists debunking their ridiculous conclusions,
about heavy evidence against their childish nighmares produced by
politicians, media, engineers, witnesses: nothing! They won't check
and/or accept any of these, and besides that they'll call you coward,
stupid, blind, stupid, gullible, stupid, sheep, stupid, stupid,
stupid....
They know better, they're intelligent, they're brave and courageous:
all the rest is blind, stupid and coward...or making some profit out
of all these disasters.
Not surprisingly, there's usually an antisemite (or antinegroes, or
anti-something) twist in these feverish and histerical conspirators'
claims: of course, there's SOMEONE that set up the whole show, and who
better to be the director than...MUAHAHAHAHAHAHA: the jews?!!
I wonder whether some of these clowns has sold his sympathies to some
white-power or nazi or right-wing cult...most probably all of them
have.
Anyway, posts like this really help me, and I hope others as well, to
have a little entertainment out of everyday's stress, and for this I
heartedly am thankful to them, and to other cranks as well.
Tonio
Pd. I wonder whether someone has researched into the probable cause
that drives some people to believe in these nonsenses: perhaps
something related to some kind of inferiority complex (I think I'm
worthless, so let's make noise about other's acomplishments...), or
perhaps some kind of acute paranoia (ALL are into this conspiracy,
ALL choose not to see, only I am capable, ALL against me....!!)
Oh well...*sigh*
===
Subject: Re: If I have solved a unsolved problem...
Usually if someone thinks they have solved the Goldbach conjecture or
the twin prime conjecture, they should ask themselves: Is the error
term really that easy to handle?
---
J K Haugland
http://home.no.net/zamunda
===
Subject: Re: If I have solved a unsolved problem...
jankrihau@hotmail.com a .8ecrit :
Coming from mina_world, I believe he/she is genuinely interested, but
not that (s)he has actually (believed to) solved anything, and use GC
for an example. Anyway, if it is a reasonably technical conjecture, the
answer is to show it to your research director, and/or send it to the
pertinent mathematical reviews. If it is a very famous one (GC, Riemann
hypothesis, short proof of FLT :-), etc.), it is probably best to
present it, somewhat humbly, here, in a form like I must have made a
mistake somewhere (as I believe that, if correct, I would have solved
..., which of course should in fact be much too hard for me), but I
cannot see where ; can anyone help me to pinpoint it?. If one is
sincere in that claim, he will find that many people will be glad to
explain to him why he (like so many before him) has fallen for fool's
gold ; and, in the infinitesimal chance that he would actually be right,
the fact of publishing here should give him proof of priority claims,
so there is (almost) no way someone could steal his works (a fear
strangely often expressed by would-be Fermat solver, for instance)
===
Subject: Is Islam Really a Religion of Terror?
Is Islam Really a Religion of Terror?
Every time speculation arises that a terrorist attack has happened,
Non-Muslims and Muslims alike suspect the Islamic connection. Of
course the disclaimers abound, but a lingering suspicion about Muslims
is left in the general views of terrorism, even if other groups are
identified as the main culprits for any particular incident.
This perception is not due to any intrinsic resentment of Islam by
people all over the world including America. In the US, it is
understood that the mainstream of Muslims, the vast majority of them,
like in every other faith, is peaceful and pay their taxes, trying to
make America a better society, trying to improve relations with
neighbors and colleagues.
But images and terminology influence public opinion, and a bitter
taste is left when Islam is reported in the daily headlines. The term
Islamic fundamentalism, whatever it means, has been repeated enough
times in relation to violent incidents that naturally, any thinking
human being has to be uncomfortable with the fact that America is home
to a vibrant Muslim community. The problem stems from negative images
about Islam. In the court of public opinion, Islam is guilty until
proven innocent!
Even though the Middle East was home to fewer terrorist incidents than
Latin America and Europe, for example, it is still regarded as the
region where terrorism is rooted. According to a recent US State
Department report, Patterns of Global Terrorism, issued earlier this
year, 272 terrorist events occurred in Europe, 92 in Latin America, 45
in the Middle East, 62 anti-US attacks occurred in Latin America last
year, 21 in Europe, 6 in the Middle East.
These numbers represent the terrorist trend and not an anomaly,
whereby the majority of perpetrators are not linked to the Middle East
or Islam. The Red Army Faction in Germany, the Basque Separatists in
Spain, the Tamil Tigers in Sri Lanka, the Shining Path in Peru and the
National Liberation Army in Columbia are not viewed with the same
horror as terrorist groups of Muslim background.
There is no moral justification for terrorism regardless of the ethnic
or religious background of the perpetrator or the victim, but the
factual basis of terrorism has been either hidden or twisted in the
public's perception of this policy problem, especially in
congressional hearings on terrorism.
The countries with the worst terrorist records in the world are not in
the Middle East either. They are not even Muslim countries outside the
Middle East. They are Columbia and Germany, havens for drug lords and
neo-Nazis.
The negative association of Islam with terrorism exists, but no one
has ever asked Why?. Could it be that American society cannot
overcome the Khomeini phobia, even though he is dead? The US Congress
found it necessary to push 20 million dollars towards covert
operations in toppling the Iranian government even at the dissent of
people in the CIA.
Some Muslim and Arab countries, both friend and foe, are run by
tyrants who kill more of their own people than those outside their
countries. The presumption that these countries represent a threat to
American interests or that any one of them can dominate the region or
even rival the only remaining superpower is indeed generous. So the
issue is not these countries' hegemony in their region or the world,
but about who can dominate their people and exploit their resources.
The perception in the Middle East is that US policy does not serve the
peoples' interests; it protects Israelis when they violate human
rights, while it slaps sanctions on and takes military actions against
countries whose dictators misbehave, resulting in suffering,
starvation and even slaughter, all in the name of teaching the tyrants
a lesson. The priorities in the Middle East for the US are not human
rights, but rather oil and Israeli superiority. Consequently, anti-
American sentiment increases.
This mood of the general public is then characterized as Islamic
fundamentalism, even though the resentment is not rooted in
religion.
When it turns violent, it is termed radical Islamic fundamentalism
or Islamic terrorism. The various terrorism experts promote
linkage to the Middle East before any other possibility every time
terrorism is speculated. They exploit the human suffering of the
victims, their families, and the fears of the American public.
Indeed, extremists of Muslim backgrounds are violating the norms of
Islamic justice and should be held accountable for their criminal
behavior, but we should not be held hostage to the politics of the
Middle East or biased reporting.
An Israeli journalist, Yo'av Karny, reporting on the events in
Chechnya made a striking observation about this development: The West
will be told--and will be inclined to believe--that the oppression of
the Chechens is part and parcel of a cosmic struggle against 'Islamic
extremism' that rages from Gaza to Algeria, from Tehran to Khartoum.
Russians will seek Western sympathy. They should not be given it!!
The issue is not Chechnya, and it is not even about Islam and the
West. Debates about religious wars and cultural clashes only distract
us from the real issue: the powerful want to continue dominating the
powerless, manipulating facts to influence public opinion, hence
maintaining the status quo.
for more information:
http://www.islamweb.net/ver2/archive/index2.php?vPart=90&startno=1&thelang=E

===
Subject: Re: Is Islam Really a Religion of Terror?
The New Call to prayer is the explosion of the suicide bomb or the IED.
Most of the violence committed in the world today is committed by Muslims.
Look what Sunnis to Shiah. Look what Shiah do to Sunnis.
Does that answer your question?
Bob Kolker
===
Subject: Re: Is Islam Really a Religion of Terror?
Bob, I guess the issue is that if you haven't grown up in a society that
thinks terrorism is the norm then you just can't understand them. I suppose

they take things like this for granted and are desensitized to it. No amount

of logic can change that.
===
Subject: Re: Is Islam Really a Religion of Terror?
Forget logic. How about genocide? That will put an end to it. Kill the
bastards and let Allah sort out the bodies.
The Hajis want to go to Paradise and 72 dark eye virgins? I say we
should send them there. They will be happy and we will be happy.
Bob Kolker
===
Subject: Re: Is Islam Really a Religion of Terror?
wow! this is just the way terrorists think!
===
Subject: Re: Is Islam Really a Religion of Terror?
How many sucide bombers kill innocent people in the name of jesus or budda?

Even the waky christans know that taking innocent life is wrong no matter
how you look at it(well, atleast they do now). Islam is a religion of
terror no matter what perspective you look at it. Just because you have
moderate muslims doesn't mean there is something wrong with the religion.
Ofcourse those same muslims are not going to do anything to fix there
religion so they are just as much to blame.
When someone gets a communicable disease and knows about it and doesn't get

it fixed and then you catch that disease then are you going to blame the
disease or the person who refused to do anything about it? As far as I'm
concerned if those moderate muslims are going to stick there head in the
sand then there just as bad as the radicals. (Because I'll be damned if I'm

going to die for there ignorance)
===
Subject: condition for square integers
(p -1)(q+1)/4 and (p +1)(q -1)/4 are _both_ square integers? TIA.
Narasimham
===
Subject: Re: condition for square integers
a, b, t, u, v, w for which:
p - 1, q + 1 = 2.a.t^2, 2.a.u^2
p + 1, q - 1 = 2.b.v^2, 2.b.w^2
This requires the following, which implies that a and b
are coprime:
1 = b.v^2 - a.t^2 = a.u^2 - b.w^2
Composing these two forms (as explained below) leads
to integers x and y for which x^2 - a.b.y^2 = -1, which
implies that a and b each have no prime factors of odd
degree of the form 4.Z + 3 (as -1 would be a quadratic
non-residue of any such factor).
So to find solutions, you can loop through integers
First check that the squarefree parts of N + 1 and N
contain no prime factors == 3 mod 4.
If not then take a and b to be these squarefree parts,
and N + 1, N = a.u^2, b.w^2.
Then I think (?) there are integer solutions x, y to
x^2 - a.b.y^2 = -1 (a Pell Equation), which you can
find from the continued fraction expansion of SQRT(a.b).
This then gives:
1 = (a.b.y^2 - x^2).(a.u^2 - b.w^2)
= b.(a.y.u - w.x)^2 - a.(b.y.w - u.x)^2
so you can take:
v, t = a.y.u - w.x, b.y.w - u.x
and then you can find an infinite number of solutions
(for the given pair a, b) by composing both in the same
way by X^2 - a.b.Y^2 = 1, where X, Y are defined by:
X, Y = x^2 + a.b.y^2, 2.x.y
This is obtained from
1 = (x^2 - a.b.y^2)^2
= (x^2 + a.b.y^2)^2 - a.b.(2.x.y)^2
John R Ramsden
===
Subject: Re: condition for square integers
Slight correction: once you have your first solution you
need to compose it by x^2 - a.b.y^2 (= -1), so that each
of a.u^2 - b.w^2 and a.t^2 - b.v^2 alternates between
1 and -1.
(Composing by X^2 - a.b.Y^2 = 1 only gets every alternate
solution.)
===
Subject: Re: condition for square integers
I was almost there, but you needn't use consecutive forms -
Any pair of forms a.m^2 - b.n^2 equal to 1 and -1 give a
solution.
So, in summary, start with one solution to a.m^2 - b.n^2 = 1,
compose repeatedly with x^2 - a.b.y^2 = -1 to get a sequence
of solutions to a.M^2 - b.N^2 = 1 or -1 (alternating), and
then use _any_ pair of these (not necessarily consecutive,
but with opposite sign) to obtain a solution to the original.
This is all for a given pair of positive squarefree integers
a, b. So for each such pair with no prime factors of the form
4.Z + 3 and whose product is not a perfect square there is a
doubly infinite family of solutions.
===
Subject: Re: condition for square integers
Hi
I believe all solutions can be obtained by:
N(i)=((3+8^(1/2))^i+1/(3+8^(1/2))^i)/2
P=N(i),Q=N(i+2j+1)
Gerry
===
Subject: Re: condition for square integers
Hi
Do you mean like p=3,q=17
(p -1)(q+1)/4=9
(p +1)(q -1)/4=16
Gerry
===
Subject: Re: condition for square integers
Yes, that's obviously true. Now, can you answer his question?
===
Subject: Re: condition for square integers
For p=9 we have q(i)=161,51841,5374978561,......
For p=17 we have q(i)=99,3363,114243,..........
Gerry
===
Subject: Re: condition for square integers
For p=3 we have q(i)=((17+(17^2-1)^(1/2))^i+1/(17+(17^2-1)^(1/2))^i)/2
q(i)={17, 577, 19601, 665857, 22619537, 768398401, 26102926097,
886731088897,....}
Gerry
===
Subject: Re: condition for square integers
p^2=(M^2)/(q^2-1)+1 with M|4
p,q can't be even at the same time
===
Subject: diophantine equation with three variables
If x, y, and z are positive integers such that 4x - 5y + 2z is
divisible by 13, then which one of the following must also be
divisible by 13?
and follows the choice of four different linear equations.
How to solve such kind of problems, generally?
THANKS A LOT IN ADVANCE!!!
===
Subject: Re: diophantine equation with three variables
What are the choices?
--
David Marcus
===
Subject: Re: diophantine equation with three variables
a) x + 13y - z
b) 6x - 10y -z
c) x - y - 2z
d) -7x +12y + 3z
e) -5x +3y - 4z
===
Subject: Re: diophantine equation with three variables
choices are (c) and (d)
my approach is to
solve the linear diphantine equations by combination the
4x - 5y + 2z =13k
x - y - 2z =13k
Maybe there exists more simple method
------------------------
Zhu
===
Subject: Re: diophantine equation with three variables
The positive hypothesis is irrelevant -- if it works for positive
integers, it will work for all integers (and conversely).
Look for small integers x, y, z such that 4x - 5y + 2z is
divisible by 13. You might pick (x, y, z) = (2, -1, 0)
or (1, 0, -2) or (0, 3, -1). Check your choice in each
proposed answer. If you insist on positive values, you can
add 13 to each variable or choose (4, 1, 1).
Once you eliminate all but d) -7x +12y + 3z, try to multiply
4x - 5y + 2z by some integer modulo 13, getting -7x + 12y + 3z.
Comparing coefficients of z, your multiplier must be
3/2 == 16/2 == 8 (mod 13). And
8 * (4x - 5y + 2z) = 32x - 40y + 16z == -7x + 12y + 3z
checks out. Can you do this without a calculator?
--
44 months after Japan attacked Pearl Harbor, Japan surrendered.
47 months after US attacked Iraq, it's time for the US to surrender.
pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA
===
Subject: Re: diophantine equation with three variables
On Mar 8, 10:25 am, Peter L. Montgomery <Peter-
The OP asked for a general method. Here's an idea.
Problem: Let p be a prime number and let ax+by+cz be a line form. How
can we tell if the condition p divides ax+by+cz ensures that p also
divides some other linear form Ax+By+Cz? As others having indicated,
this is true if and only if Ax+By+Cz can be written in the form
Ax + By + Cz = t*(ax + by + cz) + p*(ux+vy+wz)
for some t,u,v,w. Reducing modulo p and identifying the coefficients
of x,y, and z, we see that Ax+By+Cz has the required property if and
only if there is a value of t satisfying
A = t*a (mod p) and B = t*b (mod p) and C = t*c (mod p).
In principle, one can solve for t and see if there is a single
solution. But it's easier to eliminate t. This gives a nice criterion
that doesn't require any solving of equations modulo p.
Conclusion: Ax+By+Cz has the required property if and only if
A*b = B*a (mod p) and A*c = C*a (mod p) and B*c = C*b (mod p).
(If the coefficients are nonzero modulo p, it suffices to check two of
these congruences.)
Joe Silverman
===
Subject: An exact 1-D integration challenge - 33
Hello the symbolic computation fans worldwide,
Alas, - and much surprisingly, - none of the
current computer algebra systems can calculate
this integral straightforwardly
int(1-tanh(z)^2*tanh(7*z), z= 0..infinity);
Is there a soul who can display the steps to
get it?
(The answer I got is fairly lengthy, some 2 Kb,
maybe you also can deliver a compact answer?)
Best wishes,
Vladimir Bondarenko
VM and GEMM architect
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester, LLC
http://maple.bug-list.org/ Maple Bugs Encyclopaedia
http://www.CAS-testing.org/ CAS Testing
===
Subject: Re: An exact 1-D integration challenge - 33
a:= Int(1-tanh(z)^2*tanh(7*z), z= 0..infinity):
b:=combine(value(student[changevar](tanh(z)=x,a)));
sum(_R*(1/7*ln(1+_R)-1/7*ln(_R)),_R =
RootOf(-1+21*_Z-35*_Z^2+7*_Z^3))+1/14+ln(2)
If you want radicals,
convert(b,radical);
...
Mate
===
Subject: Re: An exact 1-D integration challenge - 33
Simple enough :
tau = RootOf(1-2*_Z-_Z^2+_Z^3,1.801937736),
1/49*(-29-8*tau+12*tau^2)*ln(3+tau-tau^2)-1/49*(8*tau^2-3+4*tau)*ln(tau^2)-1
/49*(2*tau-3)^2*ln(2-tau)+17/7*ln(2)+1/14
Chris
===
Subject: Re: An exact 1-D integration challenge - 33
How did you conviced Maple to use a single root?
Did you ask to solve the cubic equation wrt tau, or there is
a faster method?
Mate
===
Subject: Re: An exact 1-D integration challenge - 33
...
...
With Mathematica 5.1:
(1/98)*(7 + 98*Log[2] -
2*RootSum[7*#1^6 - 42*#1^5 + 140*#1^4 - 280*#1^3 + 336*#1^2 - 224*#1 + 64
&,
(35*Log[1 - #1]*#1^4 - 140*Log[1 - #1]*#1^3 + 252*Log[1 - #1]*#1^2 -
224*Log[1 - #1]*#1 + 80*Log[1 - #1]) /
(3*#1^4 - 12*#1^3 + 28*#1^2 - 32*#1 + 16) & ]
+ 2*RootSum[7*#1^6 - 42*#1^5 + 140*#1^4 - 280*#1^3 + 336*#1^2 - 224*#1 + 64
&,
(35*Log[2 - #1]*#1^4 - 140*Log[2 - #1]*#1^3 + 252*Log[2 - #1]*#1^2 -
224*Log[2 - #1]*#1 + 80*Log[2 - #1]) /
(3*#1^4 - 12*#1^3 + 28*#1^2 - 32*#1 + 16) & ])
Chop[N[%, 16]]
Numeric integration gives the same value.
The same substitution (and an expand(...,trig)) in Maple gives
-1/7*sum(1/7*(35*_R^2+42*_R+3)*ln(-_R)/(3+10*_R+3*_R^2),
_R = RootOf(1+21*_Z+35*_Z^2+7*_Z^3)) +1/14+ln(2)
+1/7*sum(1/7*(35*_R^2+42*_R+3)*ln(1-_R)/(3+10*_R+3*_R^2),
_R = RootOf(1+21*_Z+35*_Z^2+7*_Z^3))
which are some 200 characters.
Peter
===
Subject: Re: Tricky Quadratic Optimization Problem
A plot showed surface is extruded/cylindical or prismatic. Sorry,could
not keep continuity in the thread,(PC is broken down),hope you
resolved the c independence issue by now.
===
Subject: Re: Tricky Quadratic Optimization Problem
First, I want to thank you for all of your posts. I appreciate your
ideas and willingness/ability to think about my problem.
Actually, I'm glad you posted because there were two things I wanted
to ask you about.
1) I am not certain exactly what you mean by the last conclusion
regarding the surface. I am pretty certain that the quadric surface
described by the original function is an elliptic paraboloid.
Actually, I like your idea of thinking about the problem geometrically
because there is a nice geometric formulation of this problem:
Given an ellipse having known center (x1,y1) whose major and minor
axes are known (up to a scale factor), what is the (smallest) value of
the scale factor that exactly causes the ellipse to touch the unit
circle? (Again, you'd *think* this was an easy problem, right?)
(which is equivalent to asking what vertical slice of the parabolic
ellipsoid is tangent to the unit cylinder).
This way of looking at the problem pretty much convinces me that there
are exactly 2 real solutions (one is the min to the original function
and the other is its max). I've approached it from the point of view
of thinking about the eqns that the tangent lines that the circle and
ellipse must satisfy, but I have yet to come up with anything nice.
2) I was very curious as to your previous post where you consider the
function F + F'' to minimize F. I have never used this type of
approach and I was wondering if you would be so kind as to explain to
me what the underlying idea behind how it works.
Matt Brenneman
===
Subject: Re: Hausdorff dimension of a set ?
Note that 2 E_p is the disjoint union of two sets congruent to E_p.
That ought to mean that the dimension is either 0 or 1.
--
Robert Israel israel@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: Hausdorff dimension of a set ?
That kind of argument works for closed sets. E_p isn't closed.
Its closure does, indeed, have dimension 1.
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Hausdorff dimension of a set ?
you mean, if F is a closed set and 2F is the disjoint union of 2
subsets congruent to F, then the dimension of F is 1 or 0 ?
===
Subject: Re: Hausdorff dimension of a set ?
Add bounded, yes.
Add nonempty, and get dimension 1.
Find this in the textbooks under self-similar.
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Hausdorff dimension of a set ?
[Oversight in theorem statement corrected.]
Falconer's Fractal Geometry (1990), Chapter 10,
Proposition 10.1, p. 138: Let p_0, p_1, ..., p_(m-1)
be the m density values (summing to 1) of the m digits
of a real number in [0,1], expressed in base-m. The
Hausdorff dimension of the set of numbers in [0,1] that
have, for each k = 0, 1, ..., m-1, density p_k for
digit k, is
-1/(ln m) * SUM(k=0 to m-1) [p_k * ln(p_k)]
(with the convention that 0*ln(0) = 0).
By countable stability of Hausdorff dimension, this
also holds for (-oo, oo) in place of [0,1].
The proof is one page (p. 139) and A mass distribution
technique is used in the following proof -- the mass
distribution occurs naturally as a probability measure.
The result you want is the special case where m = 2,
p_0 = 1-p, p_1 = p. Besicovitch proved the m = 2 case
in 1934 [1] (in English, on the internet) and Eggleston
proved the general m case in 1949 [2].
[1] Math. Annalen 110 (1934), 321-330.
http://dz-srv1.sub.uni-goettingen.de/cache/toc/D37883.html
[2] Quarterly J. Math. (Oxford) 20 (1949), 31-36.
Dave L. Renfro
===
Subject: Re: Hausdorff dimension of a set ?
I hadn't read this post; only Fedor's post
and replies by Gerald Edgar and Robert Israel.
In the mean time, I thought about the question
and came up with a heuristic to make
a guess, which involved considerations of
entropies of Markov chains. So I guess
I'll just post what I had written anyway,
with full knowledge that the question
was answered by Besicovitch for base 2,
and subsequently by Eggleston for
an arbitrary base.
For a Markov chain consisting of an infinite
sequence of random variables {X_i: i in N},
there is the concept of the entropy rate of
the Markov chain, where the X_i may
take values in, say, a finite alphabet A which
we may index by 'j' below.
I looked this up in the section entitled
``Data as a Markov process in
For the simplest case of a Markov chain
where the X_i are independent identically
distributed random variables, the reference
above implies that the entropy [rate] of
the chain {X_i} is given by:
H({X_i:i in N}) = - Sum_j { p_j log_2(p_j)},
where p_j is the probability of the event
X_i = j'th symbol of the alphabet A , where
because the X_i are i.i.d. random variables, the
probabilities are the same for fixed j and
variable i.
In the case of the alphabet {0,1,2},
and indexing A as follows:
a_0 = 0, a_1 = 1, a_2 = 2
with associated probabilities
p_0, p_1 and p_2 for '0', '1', and '2' respectively,
the entropy rate of the Markov process is
-p_0 log_2(p_0) - p_1 log_2(p_1) - p_2 log_2(p_2).
In the case of p_0 = p_1 = p_2 = 1/3,
which I tentatively associate to the set
[0,1[, the entropy rate is:
3*(1/3 * log_2(3)) = log_2(3).
In the case of p_0 = p_2 = 1/3, p_1=0
which I associate to the Cantor set
[ where base 3 expansions contain
0's, 2's but not necessarily 1's,
e.g. 1/3 = 0.0222222222... , in ternary]
the entropy rate is
-1/2 log_2(1/2) - 1/2 log_2(1/2) = 1 < log_2(3).
[0,1[ has Hausdorff dimension 1, and
the Cantor set has Hausdorff dimension
log(2)/log(3).
If we form the quotient of the entropy
rates of the Markov chains I heuristically
associated to the Cantor set and [0,1[,
respectively, we get:
1/(log_2(3)) = (log(3)/log(2))^{-1} = log(2)/log(3).
So this quotient of entropy rates happens
to be the Hausdorff dimension of the Cantor set.
It is possible to heuristically associate
Markov chains to the sets Fedor introduced,
and compute ratios of entropy rates following
the pattern above for the Cantor set.
I have only had time to work out
the Cantor set Hausdorff dimension
as predicted by entropies of Markov
chains as detailed above.
Also, I did no research of the literature
should be referenced to some person
other than me.
[Besicovitch, Eggleston, ...]
David Bernier
===
Subject: Re: Hausdorff dimension of a set ?
===
Subject: Re: Hausdorff dimension of a set ?
Falconer's Fractal Geometry (1990), Chapter 10,
Proposition 10.1, p. 138: Let p_0, p_1, ..., p_(m-1)
be the m density values (summing to 1) of the m digits
of a real number in [0,1], expressed in base-m. The
Hausdorff dimension of the set of numbers in [0,1]
(and hence also for all real numbers, by countable
stability) is
-1/(ln m) * SUM(k=0 to m-1) [p_k * ln(p_k)]
(with the convention that 0*ln(0) = 0).
The proof is one page (p. 139) and A mass distribution
technique is used in the following proof -- the mass
distribution occurs naturally as a probability measure.
The result you want is the special case where m = 2,
p_0 = 1-p, p_1 = p. Besicovitch proved the m = 2 case
in 1934 [1] (in English, on the internet) and Eggleston
proved the general m case in 1949 [2].
[1] Mathematische Annalen 110 (1934), 321-330.
http://dz-srv1.sub.uni-goettingen.de/cache/toc/D37883.html
[2] Quarterly Journal of Mathematics (Oxford) 20 (1949), 31-36.
Dave L. Renfro
===
Subject: relative prime sequences
How does Sloan output the relative prime sequences ?
===
Subject: Re: relative prime sequences
??????hilp ma plaaase???????
===
Subject: Re: Edges of polytopes
Draw the square:
Parallel edges are (0,0) to (1,0) with (0,1) to (1,1),
and (0,0) to (1,0) with (1,0) to (1,1).
In a triangle, no parllel edges are.
kunzmilan
===
Subject: -------------- -STEPS OF HUNGARIAN ALGORITHM HELP! --------------
hello friends,
Can anyone give me a text file or link or
something showing the brief steps involved in the Hungarian Algorithm
for solving the basic assignment (job,worker etc) problem.
No high level explanations needed....just the steps wud be
fine.
Any suggestion would be really helpful for my project.
Arun
===
Subject: Re: -------------- -STEPS OF HUNGARIAN ALGORITHM HELP!
--------------
Pages 80-85 of J A Bondy, U S R Murty, Graph Theory with Applications.
--
===
Subject: Re: Can these be done without AC?
This one can't be done without AC, because there are models of set
theory in which every function is Lebesgue-measurable. But the
discontinuous field automorphisms of C aren't measurable.
Jose Carlos Santos
===
Subject: Re: What would they look like?
Does this mean the BT (Bhutan, anyone?) paradox is wrong?
So can we visualize roughly or not? (More drift making:
like how the set of rationals (yes I know it has a
Lebesgue measure, ugh) would be rough-vizzed as a line
like the reals -- you cannot plot the gaps. That's
what I mean by a rough viz, something like that -- I
hope you get the drift!) Are you having trouble getting
my drift?
===
Subject: Re: x^2 congruent to 1 mod m has exactly 4 solutions.
2*2=4, and use the CRT.
Phil
--
Home taping is killing big business profits. We left this side blank
so you can help. -- Dead Kennedys, written upon the B-side of tapes of
/In God We Trust, Inc./.
===
Subject: Re: x^2 congruent to 1 mod m has exactly 4 solutions.
I guess you could write:
Then look at all the ways of equating the factors on both sides, e.g.
The only other alternative is when the p and q are split:
x+1 = ap, x-1=bq, where ap - bq = 2
===
Subject: A Sorting Problem
Given a sequence of 1s, 2s and 3s the minimal number of swaps is
needed to sort it. A swap is allowed for all pairs.
Please, look at this.
http://www.inf.bme.hu/contests/tasks/ioi96s.pas
I can't understand why all NEP[x,y]-NEP[y,x] are equal.
Narek Saribekyan
===
Subject: Re: solving linear PDE on a curve/contour
My curve is not a characteristic curve. But I know that for all points
on the curve the following relation holds:
sqrt(Fx^2+Fy^2) = const
can it be somehow used?
===
Subject: Re: solving linear PDE on a curve/contour
if both conditions
F_x^2 + F_y^2 = const
A*F_x + B*F_y = C
hold on your curve and the system above can be
uniquely solved for F_x and F_y (which normally is not
the case because there can be no, one or two points of
intersection between a line and a circle),
you can in fact recover F on the curve.
Assuming the curve is smooth, F satisfies
dF(x(t),y(t))/dt =
F_x(x(t),y(t))*x'(t)+F_y(x(t),y(t))*y'(t)
with F(t=0) = F(x(0),y(0))
where (x(t),y(t)) is a parametrization of your curve.
Best wishes
Torsten.
===
Subject: Re: Short and direct proof of FLT published
My be it isn't long enought to be considered !!
Best of luck
B.Karzeddin
===
Subject: Re: Short and direct proof of FLT published
Maybe it isn't long enough to be considered !!
Best of luck. -- B.Karzeddin
NB: Good point! In fact, some years ago I proposed
a formula to estimate the chance of publishing a
simple proof of a difficult and wellknown math problem;
see http://home.iae.nl/users/benschop/inertia.htm
(increases with length and complexity of the proof;-)
===
Subject: Large scale Casimir effects, the ideal migrating Black hole vacuum
Large scale Casimir effects, the ideal migrating Black hole vacuum
propeller.
THE QUANTUM MECHANICAL UNIVERSE SEEMS TO BE ABLE TO MIMIC ALL THE
EXPERIMENTAL EFFECTS OF GENERAL RELATIVITY.
Entanglement at several scales, are the three major elements for
this new view on reality.
Mimicking Relativity means that the search for small anomalies in
local lightspeed and ZPE (zero point energy) lensing effects around
Black holes, or Large scale Casimir effects, will be of highest
importance.
http://bigbang-entanglement.blogspot.com/
Leo Vuyk.
===
Subject: a Five Lemma consequence
I am trying to make sense of the consequence of
the Five Lemma that states that if in the diagram
| | |
f1 | f2 | f3 |
V V V
the rows are exact and f1 and f3 are isomorphisms,
then so is f2. To me this says that if two modules B1 and B2
have isomorphic submodules A1 and A2 with isomorphic
corresponding quotients C1=B1/A1 and C2=B2/A2 then B1 and B2
must be isomorphic. But there could exist non-split extensions
of A1 by C1 not isomorphic to the split extension.
This seems like a contradiction. Am I missing something?
N.I.
===
Subject: Re: a Five Lemma consequence
Oh, so you don't care about the existence of f2 that makes the diagram
commute? The 5 Lemma does not assert that f2 will exist and be an
isomorphism when f1 and f3 are isomorphisms; it says that if f1 and f3
are isomorphisms, and an f2 exists, THEN that f2 has to be an
isomorphism. But f2 need not exist just from the existence of f1 and f3.
Your conditions are too weak. You are not taking into account the
you need to be able to extend the map f1 into a mapping from B1 to
B2, so that the first square commutes.
For example, suppose A1 = A2 = Z/2Z, C1=C2 = Z/2Z, B1 = Z/4Z, and
B2 = (Z/2Z) (+) (Z/2Z) (basically, the situation you
described). While you can construct f1 and f3 trivially, constructing
f2 requires you to find a way to extend the map that identifies the
subgroup generated by 2 in Z/4Z with a subgroup in (Z/2Z)(+)(Z/2Z)
into a map of the whole group; this is impossible, since you would
need to find an element in the latter whose square is of order 2.
Moreover, the commutativity of the second square requires that this
extension be the one that induces the isomorphism of the quotients;
not any isomorphism of quotients will do.
A better paraphrase is to say that if you have modules B1 and B2, with
isomorphic submodules A1 and A2, such that (i) the isomorphism of A1
and A2 can be extended to a morphism from B1 to B2; and (ii) this
--
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes by Bill Watterson)
Arturo Magidin
magidin-at-member-ams-org
===
Subject: Re: a Five Lemma consequence
Xavier
===
Subject: Re: a Five Lemma consequence
Originator: jdolan@math.UUCP (James Dolan)
|I am trying to make sense of the consequence of the Five Lemma that
|states that if in the diagram
|
| | | |
| f1 | f2 | f3 |
| V V V
|
|
|the rows are exact and f1 and f3 are isomorphisms, then so is f2. To
|me this says that if two modules B1 and B2 have isomorphic submodules
|A1 and A2 with isomorphic corresponding quotients C1=B1/A1 and
|C2=B2/A2 then B1 and B2 must be isomorphic.
no, the hypothesis is stronger than just that the isomorphism f1
(between submodules) and the isomorphism f3 (between quotient modules)
exist. the morphism f2 must also exist, and compatibility _both_
between f1 and f2, and between f2 and f3.
|But there could exist non-split extensions of A1 by C1 not isomorphic
|to the split extension. This seems like a contradiction. Am I
|missing something?
the fact that there's no appropriate f2 in such a case.
--
jdolan@math.ucr.edu
===
Subject: One-Axiom Logic
It is well known (*I* know it - QED :-) that one axiom
suffices for classic logic, Boole algebra or whatever
isomorph structure. (No, I didn't memorize it.)
It is even weller knowner :-) that one function (NAND or
NOR) suffices to generate NOT, ANBD etc.
Can you combine both? It would be trivial to rewrite
the mentioned axiom into NAND form but it's not
trivial at all to me that this new axiom would
generate all standard axioms.
--
Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de
I got a lack of passion since watching television and it hurts
I got a great invention and nobody pays attention and it hurts
===
Subject: Re: One-Axiom Logic
B7782
http://en.wikipedia.org/wiki/Laws_of_Form
It introduces a peculiar, simple but obscure logic notation.
It uses a majestic but spare framework for exposition. That is to
say, at the time I first read it, I could only understand the appendix
and notes. I did comprehend, however, that at its heart, it was
simply based on the schaeffer stroke expressed in a unique notation.
Brown certainly makes the attempt. Whether he succeeds or not, I will
leave for you to decide.
There is now a Yahoo group for discussion of Laws of Form and related
matters. Go to http://groups.yahoo.com/group/lawsofform/
John
===
Subject: Re: One-Axiom Logic
Yes. This was first done by Nicod. See:
J.G. Nicod, A reduction in the number of primitive propositions of
logic, Proc. Camb. Phil. Soc. 19 (1917), 32-41.
Here's a quote from another source:
Classical Sentential Logic --- In the Sheffer Stroke (D)
* In 1917, Nicod showed that the following 23-symbol formula (in Polish
notation) is a single axiom for classical sentential logic (D is
interpreted semantically as NAND, i.e., the Sheffer stroke):
DDpDqrDDtDttDDsqDDpsDps
* The only rule of inference for Nicod's single axiom system is the
following, somewhat odd, detachment rule for D:
* Lukasiewicz later showed that the following substitution instance (t/s)
of Nicod's axiom (N) would suffice:
DDpDqrDDsDssDDsqDDpsDps
* Lukasiewicz's student Mordchaj Wajsberg later discovered the following
organic [Footnote: A single axiom is organic if it contains no tautologous
subformulae. Nicod's original single axiom (and Lukasiewicz's simplifcation
of it) are non-organic, because they contain tautologous subformulae of the
form DxDxx.] 23-symbol single axiom for D:
DDpDqrDDDsrDDpsDpsDpDpq
* Lukasiewicz later discovered another 23-symbol organic axiom:
DDpDqrDDpDrpDDsqDDpsDps2
* We have discovered many new 23-symbol single axioms, some of which are
organic and have only 4 variables, e.g.,
DDpDqrDDpDqrDDsrDDrsDps
* We can now report that the shortest single axioms for these Sheffer
Stroke systems contain 23 symbols. No shorter axioms exist.
Source:
http://fitelson.org/ar.html#SH
F.
--
===
Subject: Re: One-Axiom Logic
Very nice! [Or is this a promotion for the
new Jim Carrey movie The Number 23 ??]
===
Subject: Re: One-Axiom Logic
See for example (page 2/3):
http://www.trinity.edu/cbrown/logic/alter.pdf
F.
--
===
Subject: Graphs make math more interesting
Graphs make math more interesting, do you think so?
John
-------------------------------------------------------
http://www.graphnow.com
===
Subject: What is a relative prime sequence ?
? what is it
===
Subject: Re: What is a relative prime sequence ?
Some context would help.
Maybe it's a sequence of integers each of which is relatively prime
to all the others.
Or maybe not.
Context?
--
===
Subject: Re: What is a relative prime sequence ?
It would be enough if each member of the sequence were merely required
to be coprime to all its predecessors. And much easier to test for.
===
Subject: number theory problem
Hi everyone,
[x^n] = -1 (mod n) Where by = I mean congruence and [x^n] means the
integer part of x^n.
Xavier Kohle
===
Subject: Re: number theory problem
Let x = the n'th root of (n-1).
Hanford
===
Subject: Re: number theory problem
You failed to read the problem. It said a real number. Your
solution gives a different
x for every n.
If floor(x^n) = -1 mod n, we have the following:
Let M = floor(x^n). Then x^n = M + epsilon(n) where 0 <
epsilon(n) < 1.
We then have:
floor(x^n) = n-1 + k(n)*n for some integer valued function k of
n.
We must show that there exists a single real number x, and functions k
and epsilon
such that:
x^n = n-1 + k(n) * n + epsilon(n) where k(n) is an integer and |
epsilon(n)| < 1.
Thus,
x = (n-1 + k(n)*n + epsilon(n)) ^ 1/n.
The problem is showing that k and epsilon exist for all n to make x
constant.
===
Subject: Non-trivial vector bundle question
I have the following question:
It is known that there is no non-vanishing vector field on S^2. How
does it follow from this that the tangent bunlde of S^2 is non-
trivial ?
And how can one prove that any tangent bundle of S^3 is trivial ? (It
should follow somehow from the fact that we have a group structure on
it).
eugene.
===
Subject: Re: Non-trivial vector bundle question
Yes.
Apart from Lee Rudolph's response, there is also the fact
that the tangent bundle of *every* orientable 3-manifold
is trivial. This follows from the fact that you can
readily (if tediously) trivialize the tangent bundle
of such a manifold over a 2-skeleton, together with
the fact that pi_2(O(3)) = 0.
There are details omitted in that description, but I
think it's an amusing enough result to warrant knowing.
Dale
===
Subject: Re: Non-trivial vector bundle question
Argument from the contrapositive: suppose that the tangent bundle
(where v is a non-zero vector in R^2) corresponds to a nowhere-zero
vectorfield on S^2.
More specifically (or at least more easily) from the fact that we
have a *Lie* group structure on it. If G is any Lie group, then
standard Lie theory gives us two distinct descriptions of the
Lie algebra g of G: either as the tangent space V(1) to G at the
identity 1 of G, or as the collection of left- (or right-)invariant
vectorfields on G. The equivalence of those descriptions leads
immediately to an equivalence of the product bundle GxV(1) to
the tangent bundle.
Lee Rudolph
===
Subject: Re: Non-trivial vector bundle question
===
Subject: question on building a regular espression
Hi friends,
There is a question on build a regular expression for below
description,
all set of strings over {0, 1, 2} satisfy that there is no two same
adjacent symbols.
I have some ideas but not sure whether is right? Could you guys help
%%%%%%%%%%%%%%
first, form the string as below form
__2__2__2__2__
__ means substrings with only 0's and 1's in which there is no
adjacent same 0's and 1's. The __ could be formed by (%eps denotes
epsilon)
A = (10)*1(0+%eps) + (01)* 0(1+%eps)
And then the whole expression could be as follows,
A*(2AA*)*(2+%eps)
Is above right? Could there be any simplification?
===
Subject: A roll of toilet paper
Good morning!
Spring is in the air. It is going to be a beautful day today here in
Ruurlo.
Most of you have always believed that a piece of paper could be folded in
half only six or seven times. If you are one of them, then my new puzzle
contains a surprising message for you. Eventually the puzzle is about the
size of a roll of toilet paper.
to all of you!
Have fun with the new puzzle!
Peter
http://home.planet.nl/~p.j.hendriks/ppvdw.htm
(click on the Union Jack. Please answer by email,
===
Subject: New proof that one equals zero
It's easy!
1=0!
===
Subject: Re: New proof that one equals zero
Is it invertible?
1i = 0
where i is the inverse operation to ! ( being factored?)
Is there a left i like i! = identity, or is it true for both sides i!
= !i ?
So now we give an even more thorough proof. We know the One is able to
exist under more than one set of conditions, especially
1 = 1 ! and 1 = 0 !
so 1 ! = 0 !
(the product of the product of no number at all equals
the product of no numbers at all)
the inverse operation (i) applied :
1 ! i = 0 ! i
1 = 0
As they say, one time doesn't count.
Have fun
Hero
===
Subject: Re: New proof that one equals zero
nice
===
Subject: Re: Big Bertha Thing invite
Big Bertha Thing hacked
Cosmic Ray Series
Possible Real World System Constructs
http://web.onetel.com/~tonylance/hacked.html
Access page to 15K ZIP file
Astrophysics net ring Access site
Newsgroup Reviews including alt.politics
301 files from the third battle of cyberspace.
Big Bertha Thing hacked
I can perhaps make a unique claim.
My Fortunecity web site has been closed, opened, hacked
and disabled by Fortunecity.
1. 30th October 2000 Closed
2. 30th December 2000 Opened
3. 31st December 2000 Hacked
4. 1st January 2001 Net ring code deleted
My Astrophysics web ring picture was replaced by a Fortunecity logo.
All I did was fight the third battle of cyberspace,
using my FC mailbox, amongst others and a mail to newsgroups gateway.
It was a text only battle, no hacking or even anonymous mail was used.
Site references were used instead of attachments.
Tony Lance
Big Bertha Thing reason
1. Third Battle of Cyberspace.
2. Death Threats.(3)
3. Newsgroup Review.
4. Odd newsgroup users with 1000 postings.
5. Berserker attacks on legitimate postings to sci.astro et al.
6. Political and totally innexplicable.
7. Net Newbie basket cases.
8. Violent spam busters.
9. Abuse complaints procedure.
10. Web site closure.
Take your pick, the opposition will not give you the choice.
Tony Lance
judemarie@bigberthathing.co.uk
===
Subject: Big Bertha Thing mayor
Big Bertha Thing mayor
4th October 2000
ex-GLC LPFA/DETR pension rights abuse case
Mr.Ken Livingstone
Mayor of Greater London
Local Government Pension Scheme Appeal
In their letter of 12th May 2000, from Mr.K.A.Bloomfied, the DETR upheld
my appeal. I would be gratefull for the opportunity to summarize
subsequent events.

I had applied for early retirement on medical grounds, to the London
pension Fund Authority. My last appeal was to the DETR. They upheld my
appeal and instructed LPFA to grant me an independant medical review.
The grounds were that the LPFA Medical Advisor was under qualified and not
independant enough. Their decision was final and they would not discuss
the matter further. Any further appeal would be to the Ombudsman.

The LPfA Director claims that the LPFA Medical Advisor is qualified and
shows a letter discussing the qualification. He claims that due to this,
the instruction is void and he proposes to take no further action.

DETR was the statutory supervisory body for the LPFA.
LPFA is the statutory successor authority to the Greater London Council.
The Ombudsman has no authority to enforce his judgement on the LPFA.

Under the GLC, the final appeal was to a committee of members of the
council. Upon abolition by statutory transfer, the final appeal was to
be to the Secretary of State for DETR; Mr.John Prescott MP.

Since July 2000 the LPFA is directly accountable to Mr.Ken Lingstone.
Would you please sort out this problem for me. I enclose below some
of my correspondence with the DETR and DHSS, as background.

Yours sincerely,
Tony Lance

Big Bertha Thing minister

Contact Mr. Phil Goodwin
Director of Operations
London Pensions Fund Authority
8th February 2000
For the Attention of Mr. John Prescott,
Government Minister,
Secretary of State for the Environment, Transport and the Regions,
Local Government Pensions Division,
Department of the Environment, Transport and the Regions,
Local Government Pension Scheme (LPGS)

I claim my right of appeal to you, as Secretary of State. This was notified
to me by Mr. Goodwin, the Director of Operations of the London Pensions
Fund Authority, in his letter of 5th January 2000, which is enclosed. The
appeal is against the decision not to allow me an early retirement on
medical grounds.

Both the officers, who have dealt with my appeal, have made comments
which are prejudicial. My doctor has said that in his view, I will never
work again. Their view was that the medical grounds are insufficient and
that my background notes to the case are immaterial, sight unseen.

Given that I have the right to attach these notes to the medical
report of my doctor, to maintain that they are immaterial sight unseen,
negates my right. Since my right is statutary, granted by the mother
of parliaments, it should stand and the notes be deemed material evidence.

Please find enclosed copies of my letters to the LPFA dated 25th October
1999
and 2nd December 1999. These together with my letter of 29th April 1999 to
DHSS provide a summary of the case, which will let you decide whether to
proceed further.

The two prejudical comments are given below.

Extract from letter dated 8th November 1999.
I regret that the enclosures with your letter were not complete, I only
received the first page of Mr Lance's letter of appeal, but none the less,
having reviewed the medical evidence which I have received in this case,
I do not wish to amend or add to my medical report dated 14th October 1999.
Unquote

Extract from letter dated 13th December 1999.
In your letter of 8th November you refer to a missing page
from Mr Lance's letter as copied to you. This is now attached
and I shall be obliged if you are able to confirm that this
would have had no material bearing on your opinion.
Unquote

On the bi-focals, mentioned in my letter of 25th Octeber 1999 to LPFA.
It took me two years to find out, that they are not bi-focals. I said
I had trouble with the eye test and nobody believed me. I can get
a refund from the optician.

This brings me to another grounds for appeal. It is a catch 22 situation,
where they think that I am too sick to know, that I am sick enough to
claim early retirement. They are trying to do a Monty Python Dead Parrot
Sketch job on me, like the optician did.

You will probably remember how this goes.
Close enough quotation.
I want a refund, for this parrot you sold me?
What's wrong with it?
It's dead.
When did it die?
It was dead, when you sold it to me.
Was it on it's perch, when we sold it to you?
Yes.
Then it must have been alive. How did it stay on it's perch?
It was glued on.
Then how do you know it's dead?
It fell off. It's not even a parrot.
Of course it's a parrot, it's green isn't it?
It's painted green.
End of quotation.

On you being a minister of the crown. Your bill to turn a statute into
guidelines is a bit of a dead parrot job. Guidelines are based
on good faith and statutes are based on bad faith. The statute states
that local government authorities shall not intentionally promote
homosexuality nor the acceptability of homosexuality as a pretended
family relationship. Anybody sitting down to teach schoolchildren
respect for unspeakable acts, as a form of family life, cannot be said
to be acting in good faith. It is to be put to a free vote of like
minded politicians; opticians to a man.

looking forward to the favour of your reply.
Yours sincerely,
Tony Lance
ex-branch president
Boilermakers Trade Union(GMB)
http://web.onetel.com/~tonylance
www.bertha.ndirect.co.uk (disabled)
CC. Internet Usenet newgroups sci.astro et al. (Without addresses, names)
Big Bertha Thing appeal
2nd December 1999
For Attention of Mr. Phillip Goodwin,
Director of Operations,
London Pensions Fund Authority,

Local Government Pension Scheme (LPGS)

Further to your letter of 25th November 1999 and our telephone
conversation yesterday, regarding my second appeal.

In your letter you mention that my reservations were conveyed to your
doctor.
This was in fact, not the case. In his letter of 8th November 1999, to your
Mr.Scott, what he said is given in the extract below.

I regret that the enclosures with your letter were not complete, I only
received the first page of Mr. Lance's letter of appeal, but none the less,
having reviewed the medical evidence which I have received in this case,
I do not wish to amend or add to my medical report dated 14th October
1999.

I would thank you for your assurance that my letter of appeal dated
25th October 1999 and my letter to DSS of 29th April 1999 would be taken
into account, in your current independant review of my second appeal,
following the rejection of my application.

I would confirm that there is no further medical evidence to come. There
are, however, two incidents, which may have some bearing. One DSS doctor
asked me why I have had no tests done. I replied that my doctor already
knew what it was. It was a pinched nerve in my back causing my legs to
be inflamed. Another DSS doctor went so far as to lie me down and
examine my legs, to see if they were inflamed. I was not told the
result, but passed my all work fitness test, as unfit for work.

I trust that this will be in order and look forward to the favour of your
reply.
Yours sincerely,
Tony Lance
Big Bertha Thing application
Contact: Tarek El-Din
25th October 1999
London Pension Fund Authority,
Re: LOCAL GOVERNMENT PENSION SCHEME.
Further to your letter of 18th October 1999, rejecting my application
for early retirement on medical grounds. I am writing to appeal against
your decision, on the following grounds;-
1. Question of Procedure.
2. Question of Fact.

Twice during the 21 day period to inspect the medical report, I called
at the surgery after 4.30 pm, to be told it was out of office hours, so not
available. The third time, I called at 10.30 am on 18th October 1999, to be
told that the 21 days had expired and the report had already been sent. It
was dated 29th September 1999, which by my reckoning was the 20th day since
it became available. This being said, there was no opportunity to attach
my notes. Hence the procedure had not been complied with.

The report itself, was the first time I had heard of the diagnosis of
sciatica, the three months duration or the infrequent attendance at
surgery. In view of this, I would ask you to bear with me, while I go
through the facts of the case.

On 3rd April 1996, I was told by my doctor, that the strong pain in
both legs was caused by a pinched nerve in my back, which caused my legs
to be inflamed.

I was prescribed 14 No. Voltarol Retard 100 mg tablets, for
inflamation.
8 days later the pain stopped, so I still have 6 tablets left.

Since then and continuously for over three years, I have had three well
defined symptons as follows;-
1. I cannot do mental or physical work, for longer than 30 minutes per day.
2. I cannot stand still for longer than 10 minutes, without getting
dizzyness
and palpatations.
3. I cannot sit still for an eye test for longer than 7 minutes, without
distress and my eyes going out of focus.

The latest symptom and one which causes me most concern, is that I
cannot
read a brand new science fiction book by a favourite author. I recently
returned 12 such books to the library unread. They had been accumulated
over
a period of months.

Background circumstances;-
1. 14th October 1990 redundant from job as senior computer programmer.
2. 1991 to 1994 successfully completed first four years towards Open
University degree in Mathematics degree course and awarded
Dip.Math(Open)
3. 1995 withdrew from 12 hours per week course, due to family problems.
4. 1996 unable to complete even one homework of 6 hours per week course,
did not take exam and failed course.
5. 1996 fence repairs of 7 hours, actually took 3 weeks at 30 minutes per
day.
6. I have spent 6 years on the dole and 3 years on the sick.
7. MED4 long term sickess note for DSS, simply said general debility.
8. DSS medical exams coducted three times, all with mention of symptoms
and
tablets. Passed all work test as unfit for work each time.
9. 4th DSS medical exam deemed unnecessary by DSS, as I was unfit for
work.
10. Since I was not receiving treatment, frequent attendance at surgery
was
never mentioned by Dr.Cullen.
11. Since 1955 I have been under ENT, with yearly appointments for ear
infections due to mastiod cavity following opperation.
12. I have stopped wearing my hearing aid, due to increased frequency of
infections.
13. I have been unable to wear my bi-focals from new, since 2 out of 4
lenses are out of focus and neither form a useable pair of lenses.
14. I also use a walking stick, inside and outside the home.

I regard both the tablets and the bi-focals as factual evidence, that
would stand up in a court of law or medical enquiry.

Trusting that this appeal will be in order and looking forward to the
favour of your reply.
Yours sincerely,
Tony Lance Dip.Math(Open)

Big Bertha Thing welfare
29th April 1999
Operations Manager,
The Benefits Agency,
Further to your letter of 28th April 1999, regarding a list of
questions, on the possibility of my working at all.

I will attempt to answer your questions, in order as listed in your
letter.

1. I am not working, I am just pottering arround on the internet. I last
registered for course work with the Open University in 1997. This would
have
involved 6 hours work per week. I could only manage half-an-hour per day,
so totally failed to do the work or complete the course. Not even one
homework assignment was completed.
In August 1997, I bought a second hand computer for 150 pounds
sterling,
and was given free access to the Open University computer along local
telephone lines at a call charge of 1p per minute.
Since that time, I managed to build up a body of correspondence,
within the limits of half-an-hour per day mental or physical work.
In January 1999, this correspondence was transfered by me to my
web site; www.bertha.ndirect.co.uk. I have to pay 14 pounds and 9p per
month for this site.

2. Nobody suggested it, it just happened.
3. My doctor does not know that I have a web site.
4. See answer 1 for description of my activities. No job is being done,
so no job description exists.

5. No employer exists or payments have been recieved.
6. I can think straight for half-an-hour per day. The rest of the time is
spent pottering arround. I can do one side of A4 paper of mathematics or
computer work per day.

7. On any day free of a major shopping expedition, I can do half-an-hour
of original work and about an hour of copy typing at non-typist speeds.

8. My principle interest is a 50 year scientific project, which was
started
30 years ago. If I could spend 48 hours per week doing it I would. However
half-an-hour work and an hour pottering, seems to be all I can manage and
at
that not every day of the week.
Due to my condition, I asked Mrs. Pam Scruton, a fellow Open
University student to be the project archivist for my work on the Open
University computer. She agreed and kept the archive of the correspondence
which is now on my web site. This is both an unusual request and an
unusually generous service to a fellow student on incapacity benefit.

9. Zero income for as far as the eye can see.
10. Not applicable, zero wages or income.
11. August 1997 and is ongoing.

I trust that the above will put the internet feeding frenzy of the
newspaper hype into some perspective.
Yours faithfully.
Tony Lance Dip.Math(Open)
===
Subject: Re: continuum hypothesis and 0=1
I wanted to think quietly about it before answering your post. I
definitely don't understand your definition of truth. I'm sorry for
replying so late, and sorrier for giving such a disappointing answer.
===
Subject: Re: continuum hypothesis and 0=1
Pierre-Yves.Gaillard@iecn.u-nancy.fr a .8ecrit :
I dont understand what you dont understand : true is (in the context of
those discussions) an abbreviation of true in a model M. Model M is

an abbreviation for Model M of a theory T, where in the last M stand
for a structure (like group or field, or Albanese variety..., see
Bourbaki, Ens IV, 1st edition), and T is a subset of the finite strings
on an alphabet formed by respecting the rules of Ens I,1),
and there
is a correspondance between the formulas of T and certain objects of M.
You may contest the interest (or the names chosen) of those definitions
(but Bourbaki, on similar matters, insist that the reader suspend his
judgment until he has acquired enough experience), but certainly not the
definitions themselves :-)
I'm sorry for
===
Subject: Re: continuum hypothesis and 0=1
For instance I don't understand how a phrase like one specific model,
which all mathematicians agree is a natural, or canonical one can
appear a mathematical definition.
===
Subject: Re: continuum hypothesis and 0=1
Pierre-Yves.Gaillard@iecn.u-nancy.fr a .8ecrit :
Not all mathematicians use the same system, even in mainstream (i.e. not
mentioning constructivists, intuitionists, etc.) But all agree that
teir naturals are the same. Now *please* fix your system (you will
note that I have assumed it is the Bourbaki one (and have always
referred to it), which by the way may not be the better one to discuss
mathematical logic, but never mind). In that system, I will (well, I
already did, but again never mind) define model, theory (it is necessary
,even if Bourbaki almost does it, because of the confusion you are
inclined to make between mathematical and metamathematical levels),
natural model, true in a model, Consis(Bourbaki), etc. etc., and you
will see that I get perfecty ordinary theorems. What next you make of
their *metamathematical* implications (like is Bourbaki's system
complete or not) is then up to you...
.
===
Subject: Re: continuum hypothesis and 0=1
I know that this won't convince you, but I can only refer you to the
footnote on page E III.24 of the 1970 edition of Bourbaki's Th.8eorie
des ensembles.
As I said, Bourbaki's axiomatic is the only one I feel I more or less
understand. So that's the one I use.
===
Subject: Re: continuum hypothesis and 0=1
I meant can appear in a mathematical definition. I forgot the in.
===
Subject: a problem on building a regular expression
Hi friends,
There is a question on building a regular expression for below
description,
all set of strings over {0, 1, 2} satisfy that there is no two same
adjacent symbols.
I have some ideas but not sure whether is right? Could you guys help
%%%%%%%%%%%%%%
first, form the string as below form
__2__2__2__2__
__ means substrings with only 0's and 1's in which there is no
adjacent same 0's and 1's. The __ could be formed by (%eps denotes
epsilon)
A = (10)*1(0+%eps) + (01)* 0(1+%eps)
And then the whole expression could be as follows,
A*(2AA*)*(2+%eps)
Is above right? Could there be any simplification?
===
Subject: Re: a problem on building a regular expression
I would have started with a finite automaton (DFA) with essentially three
states (perhaps plus an initial state) where each state says which symbol
has been encountered last.
Transforming a meshed DFA into a regular expression may be tedious but not
so error-prone as devising a regular expression from scratch.
--
Helmut Richter
===
Subject: Re: a problem on building a regular expression
It looks like you're on the right track, but is not correct as
written. For example, the string 211 is accepted by that regular
expression, namely
A* (2AA*)* (2+%eps)
%eps 2 1 1 %eps
where 2AA* gives 211 by
2 A A*
2 1 1
and A = (10)*1(0+%eps) + (01)* 0(1+%eps) gives 1 by
(10)* 1 (0+%eps)
%eps 1 %eps
Michael
===
Subject: Re: a problem on building a regular expression
It does wrong. i changed it to
A is as usual
A = (10)*1(0+%eps) + (01)* 0(1+%eps)
then A could be 0, 1, 01, 10, 010, 101, 0101, 1010, .......
B = 2(01)(01)*(0+%eps) + 2(10)(10)*(1+%eps) + 20 + 21
then B could be 201, 210, 20, 21, 20101, 21010..... ..
Finally,
L = (A+%eps) B* (2 + %eps)
first second third (part)
such that
__ 2__2__2__ 2 or %eps
first second (A) third
===
Subject: ODE question
Hi All,
I have a quick question to ask.
I'm trying to solve a complex integro-differential equation which has
the following form:
_ xmax
dy |
--- = | f(x ; a) dx
dt _|
xmin
where a is a matrix of parameters.
so I'm trying to use the levenberg marquardt to get the best estimate
of the parameters.
the function f(x) is actually a very complex one too and it's a set of
several functions multiplying each others:
f(x ; a) = g(x ; a) . h(x) . n(x)
but in order to be able to do so, I will need to find the values of dy/
da to return them to the routine.
Any thoughts on how can I achieve that?
it might be some elementary question for many..but I would like some
help if possible.
Freddy
===
Subject: Re: ODE question
do you mean this
LHS y,t
RHS xmax,xmin, a and no y or t
===
Subject: Re: ODE question
actually there is a y multiplying the function f...sorry that I have
missed it earlier.
I guess it should be dy/dt = int(f(x;a).y.dx)
===
Subject: Re: ODE question
Maybe I'm missing something but why not take the y out from under the
integral sign or is it y(x)?
===
Subject: Mistake of examination maker.
Hello sir~
There were a small event in my country long ago.
But now...big event.
This is a similar problem of a university entrance examination.
(This is not real problem. I simplified real problem.)
When y/x = 1 for all x = y in R,
show that y = x.
--------------------------------------
This have a error in assumption.(x=y=0)
so,
Examination maker should have done with full marks or no marks in this
problem.
But maker did not doing.(reason ....Maybe...honor or reprimand)
Maker made new solution by constraint.
Namely,
Given assumption is false.
So, Result is true by set theory.
and Maker gave marks partly by pre-existing solution and new solution.
Even if this is right, this is a university entrance examination.
So, Students did not learn set theory in school.
How do you think about it ?
Maker is bad mathematician ?
===
Subject: Re: Mistake of examination maker.
Originator: richard@cogsci.ed.ac.uk (Richard Tobin)
I think you'd better give the real problem, that one doesn't make
sense. You have x = y in the conditions, and then we are supposed to
prove that y = x?
-- Richard
--
Consideration shall be given to the need for as many as 32 characters
in some alphabets - X3.4, 1963.
===
Subject: Re: Mistake of examination maker.
Looks to me like a typo to me (and since I haven't seen the actual test
myself so I don't know whether it was by the test maker or you). Probably,
what was meant was If x/y= 1 for x,y in R, prove that x= y. That is
still a trivial problem, but not quite as trivial as for all x= y in R,
prove that x= y!
===
Subject: Re: Mistake of examination maker.
If you want real problem, this is the problem.
There are vector a, b, c in vector space with a =/= 0, b =/= 0, c =/= 0.
Show that a, b, c are perpendicular mutually.
===
Subject: Re: Mistake of examination maker.
I think the issue (in the *real* problem, below) is that
students were asked to prove a proposition that is vacuously
true. The O.P. attempted to state a different problem
that has the same feature, but she erred in that attempt.
For *any* nonzero a, b, and c, there is always *some*
substitution for x, y, and z such that the triangle inequality
is violated; viz. we set z=0 and pick either x=y=1 or
x=1, y = -1, depending on the orientations of a and b.
Thus we have shown that there are no a,b,c satisfying
the constraint, and so, the proposition is vacuously true.
What I am wondering is why this should be such a
big event in the O.P.'s country, as she says. The
exam board should simply decide whether to give
students full marks for proving nonexistence of a,b,c
(IMHO the correct decision) or whether to insist that
students explicitly add that this (vacuously) proves the
proposition.
Maybe the big deal is this: it LOOKS as if the
examiner thought he was asking something else, but
blundered. (Personally I think this is likely, though it
is just *barely* possible that he did intend it as a test
of the concept of vacuous truth.) If the examiner made
a blunder but refused to admit it, and instead pretended
that he intended merely to ask a trick question, that
would be scandalous behavior. Personally I don't think
the blunder *itself* should be scandalous, but that may
be a cultural difference....
===
Subject: Re: Mistake of examination maker.
There are vector a, b, c in vector space with a =/= 0, b =/= 0, c =/= 0.
Show that a, b, c are perpendicular mutually.
(or There are nonzeor vector a, b, c such that
Show that a, b, c are perpendicular mutually.)
This is a problem associated with surprised event.
If you want to know the event,
http://search.hankooki.com/times/times_view.php?&path=hankooki3/times/lpage/
nation/200701/kt2007012219544111990.htm&media=kt
Pre-existing solution [in brain of examiner(professor) early] is...
same solution as Robert Israel.
So, we can show that a.b = b.c = a.c = 0.
R,
then a = 0 or b = 0. with x=1, y=1, z=0.
So, Assumption of this problem is wrong.
So, we can't apply pre-existing solution to this problem.
This is a origin of big event.
This is a small(trivial) mistake of examiner. So, it's no problem.
But Corresponding action of university brought big event. It's problem.
(They made a new solution by set theory for justification.)
If they only apply set theory, it's no problem.
because, all students can't solve this problem by set theory.
So, all students have no marks in this problem.
This is fair.
But grader gave marks partly about a.b = 0, b.c = 0, a.c = 0.
[Namely, they mixed pre-existing solution and new solution.]
How do you think about totally irrational event ?
===
Subject: Re: Mistake of examination maker.
The statement of the problem is fraught with problems. For instance,
with z=0, x = y =1 we get, via triangle inequality, |a + b|=|a| + |b|
which forces the angle between a and b to be equal to zero. Of course
by taking the square of both sides and treating the square of the left
the dot product). On the other hand using Professor Israel's method
must be zero. So the statement of the problem is an unfortunate
mistake.
Admitting the mistake and giving full credit to those who tried would
have been the best course of action in my humble opinion. We are
humans and we make mistakes.
believe that the school acted quite unreasonably and the judge did
too, showing no respect for the opinion of internationally acclaimed
Mathematicians. But I cannot condone the violence that this gravely
wronged young man resorted to. For those who might come across such
situations a word of advice: point out the error without shoving it
in their faces. This is from someone who has done a lot of shoving
it in their faces and never prospered. In other words it does not
often pay to be confrontational.
Muhammad
===
Subject: Re: Mistake of examination maker.
I think...this is a group egoism to protect examiners.
So, one professor reached dropout. Maybe...bullying.
It was deplorable to live under such conditions.
I think...mathematicians are often arguing(fighting) amongst themselves.
Maybe, they are very stubborn and competitive.
We had many historical example.
Newton - Leibniz.
Cantor - Kronecker.
L'Hospital - Bernoulli
.....
Even now...There exists fighting in many university.
I don't know that if we study mathematics, we is going to rough.
Anyway, all mathematics must awaken from these events.
For reference,
This is a fax about this event from R.L.Granham, Michael Atiyah,
Serge
Lang.
http://geocities.com/henrythegreatgod/imagex/ams.jpg
http://geocities.com/henrythegreatgod/imagex/atiyah0.jpg
http://geocities.com/henrythegreatgod/imagex/lang0.jpg
===
Subject: Re: Mistake of examination maker.
Oh, there is already text about this event in sch-math. I did not know.
===
Subject: Re: Mistake of examination maker.
Doesn't look much like that other problem...
Some hints:
1) take z = 0 to get a.b = 0
2) take x = 0 to get b.c = 0
3) take y = 0 to get a.c = 0
--
Robert Israel israel@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: A Problem of Abstracting in Turing Machines
verbal diarrhoea on other threads. That says it all. He is a show-off
with nothing useful to add.
A quick look at Wolfram MathWorld reveals this:
http://mathworld.wolfram.com/GoedelNumber.html
Note the phrase where Pk is the kth prime number.
That is the trick with primes I mentioned.
My commentary is exactly correct, but heavily simplified because the
subject of this thread is no G.9adel, but SYMBOLS.
The Wolfram reference given above shows some of the symbols used in
formal logic and mathematics.
Here is a page from Adobe PostScript:
http://www.earlham.edu/~peters/writing/logicsym.htm
As I said before, there is no extensive font of such symbols - which
is why Peter Suber and Ira Carmel made some GIF fonts.
Indeed, due to the G.9adel incompleteness theorem, we can safely say
there will never be such a thing as a complete set of logical and
mathematical symbols. As new things are found, the whole set of
mathematics has to be revisited. I said before that G.9adel proved that
a truth uses the same logic to prove as does a lie. I suggested 1 +
1 = 2 and 1 + 2 = 1. What emerges is that TIMING is important. So we
have to feed timing back into our original axioms. There are a series
of rules regarding commutative, associative, distributive and
transitive properties that have been added to the set of axioms during
the last century.
Bertrand Russell realised that each new discovery would lead to the
need to revisit all that went before - an insuperable task. So one of
his books (1900-1902) was called TOWARDS the Principles of
Mathematics. He knew he would never arrive.
Accordingly, we can expect new discoveries to be made and new symbols
to arrive.
Because of the font problem, most computer-language writers create
their own symbols for numbers. A reasonable standard is the percent
sign for a 16-bit integer, as with 12% ( binary 0000 0000 0000 1100)
and a hash sign for double-precision floating point, as with 12#.
However, there is no fixed standard for defining data-types in
computers.
In addition, there are so many languages that such a standard would
have to bridge the language divide. Many of the languages are legacy
languages from the past. Fortran, Algol, Cobol, many Basics, Pascal
and so on - so a standard does not exist.
Charles Douglas Wehner
===
Subject: Re: A Problem of Abstracting in Turing Machines
On 8 Mar 2007 07:03:25 -0800, charleswehner@hotmail.com
That's a good summary. Except for the fact that his posts on
logic are typically correct - you left that out. No doubt
because you're simplifying again.
Um. Nobody disputed the fact that primes numbers
were involved in godel numbering...
Uh, this is nonsense.
************************
David C. Ullrich
===
Subject: Re: A Problem of Abstracting in Turing Machines
I was referring to his constant use of the biggest words he could
find, completely out of context. His style is self-display and
aggression. It is not .87lways on topic. The subject here is the formal
presentation of numbers.
He did. He challenged it thus:
It can hardly be nonsense if we are dealing with an open set. The
principle of the G.9adel proof is what I already described. New
properties are found, and these are used to revisit the old axioms and
define them more tightly. Then it is found that it is still possible
to create a fallacious equation, so we use common sense - outside the
set of mathematical tools to find what went wrong, and trawl through
all the previous work putting it right. It can go on forever, because
it is an open set.
I did two pieces of research on the Forth collection of subroutines -
I would not call Forth a language.
One piece of research related to machine-code that would use its own
subroutines to decompile itself. It is not immediately obvious, but
this cannot be done. I devised a trick to get around the problem, but
it was a fiddle. There had been Forth 79 and Fig Forth, which tried to
decompile themselves and failed. Charles G Moore, the inventor of
Forth, then decided that Forth 83 would consist of screens of source,
because he had - as I had - recognised the incompleteness problem. It
is hands-on experience.
The second piece of research was to try to use the subroutines
themselves to create error-checking code. As before, I had been
forewarned by the G.9adel theorem that I was tackling the impossible.
The problem is that one needs a checker to check the checker, and a
checker to check the checker of the checker and so on. It is again an
open set.
Of course, error checkers exist. Either they are incomplete, or they
work on closed sets.
I found that the G.9adel theorem had proven useful to me, because I was
forewarned about the troubles that would arise.
In answering the question of symbols, I did not mention the ornamental
R used for the Reals, and other special symbols. I quickly drew
attention to the logic symbols. However, for example when a single bit
is exclusive-ored against another single bit, the outcome is the same
as addition or subtraction. Similarly, the carry comes from the AND
function. An EXOR, an AND and an OR are often put together as a logic
component known as a HALF-ADDER. So arithmetic and logic are
connected.
Charles Douglas Wehner
===
Subject: Re: What is the Proof of e?
Huh? It sounds like you're describing a programming language? But I
can't tell which one, and you don't specify. It's certainly not a true
statement of most common programming languages. And, of course, it is
certainly not an established convention in mathematics.
--
Chris Smith
===
Subject: Re: What is the Proof of e?
Of course he is! E is two iterations after the C Programming
language. :-)
===
Subject: Re: What is the Proof of e?
Two?? But, but..., what about C+ and C++, not to mention C#. Surely
they come before D or E.
:-)
--Lynn
===
Subject: Re: What is the Proof of e?
C++ is the immediate successor of C. C+ is a syntax error.
--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
===
Subject: Re: What is the Proof of e?
C++ is a syntax error too.
===
Subject: Re: What is the Proof of e?
Not if C is an instance of a type which as the postfix increment
operator defined on it.
===
Subject: Re: What is the Proof of e?
Perhaps the language should have been called ++C, so that +C would not
have been an error. :-)
===
Subject: Re: What is the Proof of e?
As well as to avoid the implication that the improvements to the language
will
occur only after you have finished writing your program.
--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
===
Subject: Re: What is the Proof of e?
Just as alcoholic beverages are of various proof, I presume that
e--particularly when obtained from a total stranger at a rave--also
varies from instance to instance.
Lee Rudolph
===
Subject: Re: What is the Proof of e?
They are only twigs off the main branch.
===
Subject: Re: What is the Proof of e?
Yes, obviously C, D and E must be limit ordinals, coming after
successor ordinals like C+, C++, C+++, etc. ;-)
===
Subject: Re: ZFC++
Zuhair
===
Subject: Re: The collection of all sets and proper classes.
Ok Moe I presented what you said see
Zuhair
===
Subject: Re: The collection of all sets and proper classes.
I saw that. It looks pretty good. I haven't had a chance to check it
all out, but your presentation - stating the features, your
conjecture, then with proofs and definitions in order - is a great
improvement.
I'll try to look it over in more detail this weekend.
MoeBlee
===
Subject: Re: Hypergeometric functions and beta functions
Hi Ian.
Your result is well known. It's a special case of the following equation :
(Gamma[a] * Gamma[b] / Gamma[c]) * Hypergeometric2F1[a, b, c, z] =
(Gamma[a] * Gamma[b-a] * (-z)^(-a) / Gamma[c-a]) * Hypergeometric2F1[a,
1+a-c, 1+a-b, 1/z] +
(Gamma[b] * Gamma[a-b]* (-z)^(-b) / Gamma[c-b]) * Hypergeometric2F1[b,
1+b-c, 1+b-a, 1/z]
where |arg(z)| < Pi.
This is a relation between three of the 24 Kummer solutions for the
hypergeometric differential equation. In your case, when a = 2r, b = r+s
and
c = 1+r+s , the last hypergeometric function becomes 1, and your result can
be easily deduced.
Fran.8dois
===
Subject: Set of Curves/Functions and their Corresponding DE
Hi Everyone,
I have a bit of an obscure question, but it's been bugging me and I'm
looking for an answer.
I used a numerical method to solve for some curves along a 1D domain
(think of it as a string). The numerical method does not discretize
or solve any sort of differential equation, it uses other means that I
will not speak of for brevities sake.
My question is, does anyone recognize the family of curves being
generated, as shown in the following image?
http://technewsandrumors.blogspot.com/
All I ask is if anyone recognizes either the structure of the curves
as being a known set of functions (i.e. chebychev polynomials, bessel
function etc...) or recognize that they are solutions to a particular
differential equation?
===
Subject: Re: Fermat's Last theorem short proof
This is about angle division or angle multiplication or angle section in a
triangle
I would like to state the following theorem, hopping this would be
INTERESTING, as, I might show you later.
If (Alpha and Beta) are two angles in the same triangle and (n) is positive
integer,(m) is positive integer, less than n ,such that:
n*Alpha = Beta Mod (PI) OR:
n*Alpha - m*(PI) = Beta
Where, Angle (PI) = 180 degree OR:
(PI) = 3.14159265...
Then, the sides of (Alpha-Beta) Triangle are of the ratio:
1 :: absolute value of f(x) ::absolute value of g(x)
Where, f (x) and g (x) are two polynomial equations of degrees (n) and (n-1)
successively in real variable (x).
And f (x) is defined as the following:
f (x) = Sum {from i=0,to, i=[n/2]}
(-1)^i*[(n-i)! *x^(n-2*i) / {(n-2*i)! *i!}]
Where,
! denotes the factorial of a number,and,
[..] denotes the least integer floor function
g (x) is a polynomial of the same kind of f (x), but, with (n-1) degree.
Bassam Karzeddin
Al Hussein bin Talal University
JORDAN
===
Subject: Re: Fermat's Last theorem short proof
The following IS correct
b^(n-m) / (abs (a))^n < (m^m)*[(n-m)^(n-m)] / n^n,

where (abs) means the absolute value
B.Karzeddin
===
Subject: Re: Fermat's Last theorem short proof
this is a reply to Joseph Parranto and may be others who would like to apply
my formula
===
Subject: obtain one real root for the following ODD degree equation, that is
rational function of x
f (x) = x^n + a*x^m + b = 0
Where (a & b) are ANY two none-zero rational numbers
n is ANY odd positive integer
m is any positive integer less than n
solution: case-1)
Let, m is odd positive integer less than n
let, (k = n - m)
let, r = (b^k / a^n)^(1/m), where, r is the arithmetical mth root of
(b^k/a^n)
Let, s = (a^n/b^k)^(1/n),
where, s is the arithmetical nth root of (a^n/b^k)
if, b^k / {abs (a)}^n < (m^m)*(k^k) / (n^n), then
x = - ((b/a)^(1/m))*[1 - r/m + (2*n-m+1)*r^2 /(2!*m^2)
- (3*n-2*m+1)*(3*n-m+1)*r^3 / (3!*m^3)

+ (4*n-3*m+1)*(4*n-2*m+1)*(4*n-m+1)*r^4 / (4!*m^4)
-(5*n - 4*m+1)*(5n-3m+1)*(5n-2m+1)*(5n-m+1)*r^5/(5!m^5)
+...]
x = - ( b^(1/n))*[1 - s/n + (2*m-n+1)*s^2 / (2!*n^2)
- (3*m-2*n+1)*(3*m-n+1)*s^3 / (3!*n^3)

+ (4*m - 3*n+1)*(4*m - 2*n+1)*(4*m -n+1)*s^4 / (4!*n^4)

- (5m-4n+1)*(5*m-3*n+1)*(5m-2n+1)*(5m-n+1)*s^5/(5!*n^5)
+...]
Case - 2
Let, m be even positive integer less than n , then, let (x=1/y),
in the above equation, and substitute & get equation of the previous solved
form for one real root.
In fact, this is nothing but a VERY little introduction to obtain all
roots to any polynomials in terms of their coefficients,
It would be very nice from you to make it readable by others, so they may
check it to any degree of accuracy they require, and report back honestly
their comments
Bassam Karzeddin
Al Hussein Bin Talal University
JORDAN
===
Subject: Re: Fermat's Last theorem short proof
I can see why you are so frustrated with your readers -
You do not have a Proof but you do have an interesting algorithm.
The problem is language and use of terms. In English better method is
what you have, but that is not a proof = no doubt of outcome without all
the calculations written down -- completely.
Keep in mind that your F(x) does not vary much from Wiles' proof. Converting
c^z to b is so minor a change...
And you cannot use this to prove Beal's Conjecture either - x is a
common factor at the outset. If you start at the end, you only get back to
the beginning. You cannot prove what you are trying to prove by using the
thing you are trying to prove. That is
Circular logic and that is what you do have - in abundance.
Sorry...
Joe
===
Subject: Re: Fermat's Last theorem short proof
brother karzeddin
please do everything as fast as possible to reveal your proof of
FLT .we are eagerly waiting for a proof from u.
best of lucks
MARUF SYFULLAH
DEPT: CSE
L / T:4/2
BUET (BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY)
DHAKA ,BANGLADESH
===
Subject: Mertens function question
I have the following question:
Suppose we have the Mertens function M(x) = sum_{n <= x} mu(n) , where
mu(n) - is the Moebius function.
Can it be somehow more or less elementary proven that M(x) = o(x) .
Elementary includes using extension of Riemann zeta function, some
basic properties of Moebius function.
If not, coould you please give an outline of a non-elemntary proof.
===
Subject: Re: Mertens function question
I think this is equivalent to The Prime Number Theorem (and it
should be easy enough for you to check whether I'm right, this
should be in Hardy & Wright, and any text on Analytic Number Theory).
--
===
Subject: n Inverse Modulo m
What is n inverse modulo n?
Narek Saribekyan
===
Subject: Re: n Inverse Modulo m
days. My association with the Department is that of an alumnus.
Additive inverse? Multiplicative inverse? Any conditions on n and m?
Integers? Polynomials? Arbitrary elements in a ring?
If you are talking about the multiplicative inverse, then it only
exists if n and m are relatively prime. In which case, it is any
integer x such that nx = 1 (mod m); it is not unique as an integer,
but it is unique as a congruence class modulo m; that is, if nx=ny=1
(mod m), then x=y (mod m). It can be found using the extended
Euclidean Algorithm.
--
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes by Bill Watterson)
Arturo Magidin
magidin-at-member-ams-org
===
Subject: Re: n Inverse Modulo m
Sorry, I was talking about integers and there should be n mod m.
===
Subject: Re: n Inverse Modulo m
days. My association with the Department is that of an alumnus.
Additive inverse? Multiplicative inverse? What?
For additive inverse, the inverse of n modulo m is simply -n modulo m;
if you want your representatives to lie between 0 and m, then take
m-n. For multiplicative inverse there is no general formula, you need
to figure it out for each n and m (usually using the extended
Euclidean Algorithm, like I said before).
--
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes by Bill Watterson)
Arturo Magidin
magidin-at-member-ams-org
===
Subject: Re: n Inverse Modulo m
Yes! It's multiplicative. That's it! Can you tell me what's that and
what is used for? Actually, can it help computing n!/(n1*n2..*nk) ?
===
Subject: Re: n Inverse Modulo m
Here's an example of use.
Suppose you have this equation: g = (8*a - 5)/9
and you want to find an integer 'a' that makes 'g'
an integer. A solution exists if gcd(8,9) divides 5.
Since gcd(8,9)=1 and 1 divides 5, then a solution
exists. And if there is one solution, there are an
infinite number of solutions.
You would use the modular inverse to find the first
such 'a':
a = (invert(8,9) * 5) % 9
where '%' is the modulo operator. The result is a=4
which gives you g=3.
Now you can find the ith solution once you know the
first: a(i) = i*9 + a(0). And g(i) = i*8 + g(0).
You do have to be careful because generally for
g = (X*a - Z)/Y
the gcd(X,Z) may not be 1 and yet still divide Z so
that no modular inverse exists. In those cases,
we can simply divide X, Y and Z by gcd(X,Y) to
get X1, Y1, Z1 which whill produce a modular inverse
and we can then proceed to find 'a' by
a = (invert(X1,Y1) * Z1) % Y1
===
Subject: Re: n Inverse Modulo m
That was supposed to be gcd(X,Y).
===
Subject: Re: n Inverse Modulo m
days. My association with the Department is that of an alumnus.
A multiplicative inverse of n modulo m is an integer x such that
nx = 1 (mod m).
A multiplicative inverse of n modulo m exists if and only if
gcd(n,m)=1. If this is the case, it can be found by using the extended
Euclidean Algorithm, AS I HAVE ALREADY TOLD YOU TWICE. (Learn to read
the responses carefully, damn it; you are wasting your time, my time,
the time of everyone who reads this, and lots and lots of money in
sending useless messages because you are too lazy to read through the
response).
If you don't know what the extended Euclidean Algorithm is, then go to
http://en.wikipedia.org/wiki/Extended_Euclidean_Algorithm
How would computing the modular inverse of a number, IF it exists
(which it may not), help you compute that?
--
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes by Bill Watterson)
Arturo Magidin
magidin-at-member-ams-org
===
Subject: Re: n Inverse Modulo m
===
Subject: total derivative,
Let f(t,y).
Then f' = f_t + f_y f
Is f'' this:
f_tt + f_ty f + f( f_yt + f_yy f ) + f_y(f_t + f_y f)
= f_tt + 2f_ty f + f^2 f_yy + f_y( f ' ) ?
===
Subject: Re: total derivative,
I forgot to say, y is also a function of t: y = y(t). And, f is a
function of t and y. E.g., y' = f(t,y).
===
Subject: Re: total derivative,
Let f(t,y) what?? That isn't a sentence.
Are you starting with y being a solution to the differential equation
y' = f(t,y)?
Or are you just starting with a function like w(t) = f(t, y(t))?
You need to state the problem clearly and completely before you or
anyone else can solve it.
--Lynn
===
Subject: Re: total derivative,
What is f'?
Are you sure about this. maybe you should write things a little bit more
explictly.
see above
Ciao
Karl
===
Subject: Re: total derivative,
f'(t,y) is df/dt (t,y)
===
Subject: Re: total derivative,
If your total derivative f'(t,y) is d f(t,y(t))/dt, then
f'(t,y) = f_t + f_y * (dy/dt), not f_t + f_y f as stated.
I'm assuming the subscripts denote partial derivatives.
===
Subject: Real or Complex?
Hello everyone.
I have a possibly silly question.
I know that an even root of a negative number only have complex
solutions. For instance, square root of -3 is a complex number.
I know that an odd root of a negative number has a real solution. For
instance, cubic root of -3 has a real solution.
Now this is my question: (-2)^PI has a real solution or only complex?
I know I can approximate PI as much as I want with a rational number
such as p/q where p is even. Then (-2)^(p/q) is real.
But I can also aproximate PI as much as I want with p'/q' where p' is
odd and q' is even. Then (-2)^(p'/q') is complex.
Then (-2)^PI has a real solution or only complex?
===
Subject: Re: Real or Complex?
As an interesting aside
i^i is real!
===
Subject: Re: Real or Complex?
First, we may write (-2)^{pi} = e^{pi log(-2)}.
Now for complex z we have
log(z) = log|z| + i [arg(z) + 2 pi k]
..where |z| is the modulus and arg(z) is the argument.
Here the 2 pi k allows us to choose different branches.
Since -2 lies on the negative real axis, it has an
argument of pi so...
log(-2) = log|-2| + i [arg(-2) + 2 pi k]
= log(2) + i [pi + 2 pi k]
= log(2) + pi i (1 + 2k).
This yeilds...
(-2)^{pi} = e^{pi log|-2| + pi^2 i (1 + 2k)}
= e^{pi log(2) + pi^2 i (1 + 2k)}
= 2^{pi} e^{pi^2 i (1 + 2k)}
..for any integer k. Thus, on the standard branch
(i.e. k = 0) we have
(-2)^{pi} = 2^{pi} e^{pi^2 i}
= 2^{pi} [cos(pi^2) + i sin(pi^2)]
= -7.966178304... + i 3.797398699...
Kyle Czarnecki
(see above)
===
Subject: Re: Real or Complex?
small addon for the OP:
since e^(i*Pi*y) only produces real values for y in Z
and in our case we have y=Pi*(1+2k) which is not in Z
, we know that none of the values of (-2)^{pi} is a real number.
define g(x)= sum c_n*H( x- b_n) , x in[0, 1], where n=1,2,.....,
and are all rational number
know to build g(x).
You can choose c_n and b_n so that g(x) is close to x for x small...
??? What does g(x) ^= 0 mean?
David C. Ullrich
===
Subject: Equidistant curves
Driven by the desire to find a solution to problem which was posted in
sci.math a while ago, one encounters this problem:
Given a function y=f(x) or generally a curve on the plane R^2 (possibly
crossing itself), determine expressions (implicit or not) (if they exist)
of
the two curves which sit at a distance d from the original curve, with d
in
R, given.
Distance here is measured perpendicularly to the tangent at a given point
(x_0,y_0) of the original curve.
For example, for the circle x^2 + y^2 = r^2, one solution will always be
the
circle x^2 + y^2 =(r+d)^2 and a second solution will be x^2 + y^2 =(r-d)^2,
Illustrate with an example for the function y=sin(x), for the ellipse
x^2/a^2
+ y^2/b^2 = 1 and the function y=1/x.
Another way to pose the problem: Find expressions for the two loci of
points
equidistant from a given plane curve.
Questions:
1) Are such expressions possible in general?
2) Are there always two equidistant curves? (I guess not since the circle
example above with r < d provides a counterexample)
3) Are those loci necessarily functions, whenever the original curve is a
function?
Any ideas welcome,
--
I.N. Galidakis
http://ioannis.virtualcomposer2000.com/
===
Subject: Re: Equidistant curves
I suspect I know the problem to which you refer.
I gave a link in that earlier thread. Here it is again:
It gives such expressions.
original circle. But apparently you wish to exclude such.
Yes, as given in the link above.
Yes, if we're talking about parallel curves, as in the link;
no, if we're being more restrictive, as you apparently are, in which case,
if I understand your restriction correctly, there might be none at all.
No, at least if we're talking about parallel curves, as in the link.
David
===
Subject: Re: Equidistant curves
their newserver which every once in a while for some reason reboots and
leaves
threads hanging.
I guess any answers posted between any two subsequent reboots are
completely
lost, so your response happened to be one of them.
If you see multiple responses by me, it's my server's fault, so please
excuse
that.
--
I.N. Galidakis
http://ioannis.virtualcomposer2000.com/
===
Subject: Re: Equidistant curves
Yes, at least for smooth curves.
That doesn't provide a counterexample.
No.
Did you look at the link provided by David Cantrell in the previous
thread? It shows how to do it parametrically.
http://mathworld.wolfram.com/ParallelCurves.html
--Lynn
===
Subject: Re: Equidistant curves
MathWorld link.
I guess that settles it. And I suppose it also settles the question on the
other thread with the circles tangent to the two functions (at least in
principle).
--
I.N. Galidakis
http://ioannis.virtualcomposer2000.com/
===
Subject: Re: Equidistant curves
One method of finding examples is to work with so called signed
distance functions. Let C be a smooth closed curve defined by an
implicit function u(x,y)
C = {(x,y) in R^2 : u(x,y) = 0, grad u != 0 on C}.
everywhere, then u(x,y) is called a signed distance function.
Then for sufficiently small d the curves equidistant from C can be
found implicitly by letting u(x,y) = c and u(x,y) = -c.
===
Subject: How to solve the probability density of z=ax+w?
This is a problem
If p(a=0)=1/2,p(a=1)=1/2
a,x,w are mutually independent. x,w are all normal distribution and their
1st and 2nd moment have been known. E(w)=0
Can I get the probability density of z=ax+w?
and can I get E(x|z)?
--
Hui Zhou
===
Subject: Re: How to solve the probability density of z=ax+w?
Try sci.stat.math.
Nick
===
Subject: Re: How to solve the probability density of z=ax+w?
There is a Bernouli-Gaussian distribution seems to be like this. But I can't
find document about this.
===
Subject: Re: How to solve the probability density of z=ax+w?
If a = 0 then the pdf of z is the same as the pdf of w, call it
P_0(z). If a = 1 then z = x + w, so z has a normal distribution whose
mean is the sum of the means of x and w, and whose variance is the sum
of the variances of x and w. Call this pdf P_1(z).
Then, unless I'm overlooking something, the overall pdf of z should be
1/2*(P_0(z) + P_1(z)) (which is, of course, *not* a normal
distribution).
Not tried to work out that one yet!
===
Subject: Re: How to solve the probability density of z=ax+w?
Quite correct.
--
David Marcus
===
Subject: Re: How to solve the probability density of z=ax+w?
Can I get jointly density function of x,z here?
If I can get this, I can work out p(x|z)
===
Subject: Re: How to solve the probability density of z=ax+w?
I had a go at deriving p(x|z) in my other recent post...
===
Subject: Re: How to solve the probability density of z=ax+w?
As far as E(x|z) is concerned, the pdf of x given z, should, if you'll
excuse the sloppy notation, be given by
P(x|z) = Pr(x and z)/Pr(z)
where
Pr(x and z) = 1/2*N(mu_x, sigma_x, x)*N(mu_w, sigma_w, z - x) +
1/2*N(mu_x, sigma_x, x)*N(mu_w, sigma_w, z)
Pr(z) = 1/2*N(mu_xw, sigma_xw, z) + 1/2*N(mu_w, sigma_w, z)
and
N(mu, sigma, x) = 1/(sigma*sqrt(2*pi))*Exp(-(x - mu)^2/(2*sigma^2))
mu_x and sigma_x^2 are the mean and variance of x
mu_w and sigma_w^2 are the mean and variance of w
mu_xw = mu_x + mu_w and sigma_xw^2 = sigma_x^2 + sigma_w^2 are the
mean and variance of x + w
Then E(x|z) is found by integrating x*P(x|z) over x = -inf. to +inf.
Looking at it, I think that this can be done exactly in closed form.
The denominator Pr(z) is constant w.r.t x, and each of the products in
the numerator Pr(x and z) can be coerced into the functional form of a
normal distribution whose expected value is easily found. It should
be merely a question of slogging through a lot of algebra.
===
Subject: Re: How to solve the probability density of z=ax+w?
Of course, I should have said that the first term in the numerator can
be coerced etc. The second is easy. So, treating the two integrals in
the numerator separately as U and V, I got
E(x|z) = (U + V)/Pr(z)
where
Pr(z) is as defined above
V = 1/2*mu_x*N(mu_w, sigma_w, z)
U = (b*exp(d))/(4*sqrt(2*pi)*sigma_xw^3)
where
d = (b^2/(4*a) + c)/(2*sigma_x^2*sigma_w^2)
and
a = sigma_w^2 + sigma_x^2
b = 2*sigma_w^2*mu_x + 2*sigma_x^2*(z - mu_w)
c = -sigma_w^2*mu_x^2 - sigma_x^2*(z - mu_w)^2
and at that point I stopped with symbolic overload.
Seems an awfully complicated answer...
===
Subject: Atiyah-Patodi-Singer index theorem on five dimensions
I will try to apply the theorem to the Dirac operator on a five
dimensional spacetime. The spacetime which I am studying consists of
one time and four space coordinates (- + + + +). I found some
literature on four dimensional APS index theorem and Callias' paper on
open spaces (Commun. Math. Phys. 62, 213-234, 1978). My main source is
Theories and Differential Geometry, Physics Reports 66, 213, 1980). As
a physicist, this one is the most understandable for me but it is also
in four dimensions.
My main question is: Can one write an index theorem in five
dimensions?
According to the Callias paper (p.214), the index of any elliptic
differential operator on an odd dimensional compact manifold is zero.
And in Eguchi-Gilkey-Hanson paper (p.350), I have two terms for the
index; one in the manifold and one at the boundary. My manifold is
compact in five dimensions and therefore I think the first term will
vanish. But there is another problem; this first term is:
integral(trace(R^R)) and I do not know if I can write this term in
five dimensions (R being related to curvature 2-forms and ^ being the
wedge product). Then I will take one spatial coordinate as constant
and find the boundary term, but I am not sure about how to write it.
===
Subject: Steve Weinberg says he doesn't get it on torsion (PT March 2007)
Jack,
The reason that Weinberg says in the excerpt you quote below on
Torsion, 'Sorry, I still don't get it.' rather than just 'I don't get
it.', is because Weinberg had already said I don't get it in response
to my Letter published in the 4/06 Physics Today referenced by Hehl in
his 3/07 Letter. One cannot conduct an extended tit-for-tat debate in
the Letters section of Physics Today, so I could not respond further to
Weinberg, but it looks like Hehl took up the mantle. Hehl cites some
other authors on torsion; I had cited Vargas's papers and also Shipov.
Vargas was left rather aghast to say the least when he saw the apparent
obtuseness of Weinberg's response. - RB
It is amazing. The great professor is but a shadow of his former self it
appears. Weinberg's 3 books on quantum field theory have all the basic
ideas needed to understand torsion brilliantly presented. Old age I guess.
Apart from any other consideration, Weinberg chose to overlook the
potential utilitarian advantage to gravitation physics of what I said in
my Letter regarding Teleparallelism (TP). To formally integrate objects
such as energy-momentum densities that are used in conservation laws for
gravitation and in the question of the definition and locality of
energy-momentum in gravitation physics, one needs to add together
objects belonging to tangent spaces of different, neighboring points.
But this can not be done in a path independent manner, if at all, unless
the affine curvature vanishes. The affine curvature, which is what is
actually obtained from the Cartan moving frame method (or equivalently
from the tetrads) in the types derivations you have provided many times
in this forum, is not the same object in general as the familiar GR
Riemannian curvature. - RB
The tetrad/spin connection substratum field equations are essentially
spin 1 Yang-Mills type, but for the 10-parameter localized spacetime
universal Poincare symmetry group for all non-gravity source field
dynamical global actions. Therefore, according to t'Hooft, if you
quantize them they are renormalizable in the local gauge invariant
square root substratum of the spin 2 non-renormalizable composite
geometrodynamical field of Einstein 1915.
In particular, the vanishing affine curvature requires non-vanishing
Torsion. This is the TP postulate, which Einstein tried to utilize in
his reworkings of GR in the late 20s-early 30s. Since GR, as you
correctly point out, has vanishing Torsion, it does not postulate TP. As
Vargas has discussed in many Papers, this GR non-assumption, in turn,
may have a profound effect on the issue of conservation laws and the
definition of energy-momentum in GR and gravitation physics. - RB
Indeed, that may be the origin of the fact that 1915 GR with zero
torsion (i.e. not all of the 10-parameter Poincare group is localized)
has nonrenormalizable spin 2 nonlocality in the gravity vacuum
stress-energy. I am not sure of that of course.
Hehl brings out another important point: the relation of Torsion to
translation. It is customary in physics to associate Torsion with spin
one way or another. I believe both you and Shipov follow this
interpretation. However, Vargas and others, like Pommaret, reject the
association of Torsion spin or rotation because it is mathematically
closely related to translation as Hehl highlights in his Letter. Vargas
finds a geometrodynamical association of Torsion with the EM field,
rather than spin. - RB
I pointed out this is a tricky point that Rovelli and Kibble clarify.
The GCT gauge freedom of 1915 GR comes from localizing only the
4-parameter translation subgroup T(4) of the 10-parameter Poincare group
P[T(4),SO(1,3)]. Clearly the local GCT's
x^u = x^u(x^u') are localized infinitesimal translations a^u(x^u'), i.e.
4-parameter
In global 1905 special relativity the infinitesimal elastic
deformation a^u(x^u') of Hagen Kleinert's 4D world crystal Planck
lattice is a global constant over the entire infinity of Minkowski
spacetime. This is global action at a distance that violates local
objective light cone limited causality, hence localization is absolutely
necessary to maintain orthodox causality. Global special relativity
without gravity is a half-way house first-approximation that is not
internally consistent from this POV.
On the other hand there is no question that dislocation defects in the
Kleinert world crystal lattice (Burger's vectors et-al) form the torsion
field gaps and that the disclination defects for parallel transport
around closed loops of the lattice where
V^u(finish) - V^u(start) ~ SO(1,3)^uu'V^u'(start) ~ R^uvwlA^v^wV^l(start)
A^v^w = - A^w^v is the sectional area element of the small loop
form the curvature.
Therefore there are two dual POV
on the one hand
1915 GR with curvature only and zero torsion emerges from the local
gauging of 4-parameter T(4)
on the other hand
1915 GR with curvature only and zero torsion is associated with a local
Lorentz transformation LLT of the 4-vector around the closed loop (no
torsion gap).
Of course when one does the actual Cartan form algebra there is no
problem. The problem is only one of the informal language (Bohm) not of
the mathematics.
Thus the 4 GCT invariant tetrad 1-forms are the LLT 4-vector components
e(4)^a = I^a + A(4)^a
I^a is the trivial globally flat 1905 SR tetrad
A(4)^a is the intrinsically warped gauge connection potential from
localizing 4-parameter T(4).
Einstein's 1915 fundamental local GCT & LLT scalar invariant is
ds^2 = e^aea = (Minkowski metric)abe^ae^b = guvdx^udx^v
Einstein's 1915 constraint of zero torsion is the vanishing 2-form
T(4)^a = de(4)^a + S(4)^ac/e(4)^c = 0
This with metricity, see Rovelli's explicit formula Ch 2 of his online
Quantum Gravity gives the 6 zero torsion field effective
spin-connnection 1-forms
S(4)^a^b = - S(4)^b^a
only from localizing T(4) to get the spin 1 Yang-Mills potential A(4)^a
A(4)^a spin 1 because it's a Lorentz group 4-vector in the a-index.
This warped localized tetrad resembles the EM 4-potential with internal
index a i.e. Yang-Mills.
The pure disclination curvature defects without torsion gaps then come
from the curvature 2-form
R(4)^a^b = dS(4)^a^b + S(4)^ac/S^cb
And no TP of course at this stage as you say.
Next step is to locally gauge SO(1,3) giving the additional independent
torsion gap field spin connection 1-forms
S(1,3)^a^b
The full connection is then S(4)^a^b + S(1,3)^a^b
Where now
T^a(1,3) = S(1,3)^ac/e^c =/= 0
R(10)^a^b = d[S(4)^a^b + S(1,3)^a^b] + [S(4)^ac + S(1,3)^ac]/[S(4)^c^b
+ S(1,3)^c^b]
Where the TP Ansatz is obviously in my transparent notation using only
local objective GCT invariants (coordinate independent)
R(10)^a^b = 0
where
R(10)^a^b = R(4)^a^b + R(1,3)^a^b + S(4)^ac/S(1,3)^cb +
S(1,3)^ac/S(4)^cb
Hence two diagonal curvature 2-forms. Utiyama in 1960 computed
R(1,3)^a^b in effect without R(4)^a^b.
This settles the above informal language confusion you mention!
Note also the two T(4) x SO(1,3) cross-coupling terms.
*I am going to completely rewrite my emergent gravity archive paper with
these new results of course.
BTW my theory for emergent gravity with torsion needs exactly 8 Higgs
bosons one for each Goldstone boson.
.95A1U4I have N = 8 to get Einstein's tetrads
and spin connections emergent.
from my CIT talk under construction (outline sans math type equations)
M Theory for Idiots
Emergence of tetrads and spin connections from the spontaneous
breakdown of localized Poincare group symmetry in the post-inflation
physical vacuum.
Jack Sarfatti
Missing Organizing Idea
Witten and Green say they are missing the key idea denoted as
.89UI?.89U?
No Rigid Symmetries
Locally gauging a nondynamical rigid symmetry group of the action
(either internal or spacetime) of a given source field introduces a
gauge force coupled to the source as the compensating connection
field needed to maintain gauge invariance for the now dynamical non-
rigid symmetry group.
Weinberg doesn.89Ut get it?
How is that possible?
.89UISorry I still
don.89Ut get it. Is there any physical principle, such
as a principle of invariance that would require the Christoffel
symbol to be accompanied by some specific additional tensor?
.89U|.89U?
Steven Weinberg to Fred Hehl on the role of the torsion field in
spacetime physics in March, 2007 Physics Today.
Spontaneous Symmetry Breaking
Spontaneously breaking a vacuum symmetry means macroscopic occupation
In a vacuum the ODLRO condensate is composed of virtual quanta in
contrast to the ground state of superfluid helium where the
condensate is made from real helium atoms on mass shell.
Local & Broken
The broken symmetry need not be the same as the locally gauged
symmetry, but can be. In gravity theory rigid translational symmetry
is spontaneously broken in order to have localized variable spacetime
curvature. The action still obeys the symmetry, only the vacuum does
not obey it.
Higgs Vacuum Manifold
N independent Goldstone phases require N+1 real Higgs scalar fields.
The degenerate vacua lie on the unit SN spherical hypersurface:
Higgs Vacuum Potential
The .89UIMore is
different.89U? emergent macro-quantum coherent vacuum
order parameter has the effective potential
Dynamical L-G Eq.
Covariant dynamical Landau-Ginzburg equation for the macro-quantum
coherent hologram .89UIVolume without
volume.89U? vacuum order parameter is
Local Gauge Invariance
Localizing the 10-parameter globally rigid Poincare group of 1905 SR
universally for all non-gravity source field actions gives the 4
Einstein-Cartan tetrads and the 6 spin connections as compensating
gauge potentials.
This is similar to localizing the internal symmetry groups in Yang-
Mills theory of the electroweak-strong forces.
M = My Mystery Matrix
Start with two Lorentz 4-vectors of 0-forms
Compactification
Each Minkowski 4-vector has magnitude
Michael Faraday
I use Faraday.89Us
.89UIlines of force.89U? in a
metaphorical way for the
geometrodynamic field compacted down from 9 + 1 to 3 + 1.
Two Goldstone phases have three single-valued real Higgs fields with
stable quantized (wrapping integers) point
.89UImonopole.89U? defects in
3D space.
A single Goldstone phase has two real Higgs fields with stable
quantized (winding integers) line
.89UIvortex core.89U? stringy
defects.
Line Density Operator
The geometrodynamic line density operator is the non-closed 1-form
Area Density Operator
The geometrodynamic area density 2-form is closed. This is the world
hologram idea in action.
Stoke.89Us Theorem
The deRahm integral of a p-form on a p-cycle (without boundary)
equals the p-form.89Us exterior derivative on a p+1
co-form whose
boundary is the p-cycle.
Closed Forms
A p-form is closed if its exterior derivative vanishes. When the
interior of the p-cycle is simply connected without
.89UIholes.89U? the
integral of the closed p-form vanishes.
Otherwise the integral is proportional to an integer analogous to
Bohr-Sommerfeld quantization in the old quantum theory and quantized
circulation in superfluid helium vortices.
World Holography
The geometrodynamic volume density operator 3-form should be
proportional to the exterior derivative of the area density operator
2-form. However, the area operator is closed. Therefore, the volume
operator is locally zero.
Volume is a holographic projection of the area density operator
integrated on a nonbounding 2-cycle enclosing point defects in the
single-valued 3-real Higgs component field projection on 3+1 from 9
+ 1 spacetime.
Bekenstein BITS
The integral of the geometrodynamic area density operator 2-form,
over a non-bounding closed 2-surface (cycle) surrounding a multiply-
connected 3D interior with
.89UIholes.89U? where the 3
real Higgs fields
all vanish so that the 2 compacted Goldstone phases are undefined, is
quantized.
Volume without volume
Although Stoke.89Us theorem rigorously does not apply to
closed non-
bounding 2D surfaces S2 , whose 3D interior M3 is multiply connected,
physical intuition demands the as-if holographic imaging
Calabi-Yau
These 2 constraints leave 6 angular Kaluza-Klein Goldstone phases
whose radii determine coupling constants like the moduli of Calabi-
Yau space.
Gennady Shipov gets this structure in his version of torsion field
theory calling the extra 6 degrees of freedom an
.89UIoriented
point.89U? (suggesting
.89UIbranes.89U?)
Higgs Potential
The 8 Goldstone phases require 9 real Higgs scalar fields.
One degenerate vacuum manifold is S8. There are other topologies.
defects they fit most naturally in the 9D + 1 spacetime of
superstring theory.
Shipov.89Us Brane Worlds
Thus
Einstein-Cartan Tetrads
The four GCT invariant 1-forms are
Analogy to Superfluid
The fabric of spacetime is an elastic 4D GCT covariant supersolid,
compare
Macro-Quantum Gravity
There is no curvature gravity field in my emergent
.89UIMore is
different.89U? (P.W. Anderson) theory if
G = 0
h = 0
1/c = 0
Cartan 1-Forms
In terms of GCT tensor scalar invariant products, the 4 tetrads and 6
antisymmetric spin connection 1-forms are
Tetrad Decomposition
The GCT invariant globally flat tetrads and locally curved tetrads
are respectively
Spin Connections
The six GCT invariant connection 1-forms are
GCT Covariant D
The curved Cartan exterior derivative is
Torsion Field 2-Form
In 1915 GR this is zero
Curvature 2-Form
Take the covariant exterior derivative of the 1-form spin connection
to get the curvature 2-form
Einstein-Hilbert Action
The GCT & LLT field Lagrangian density is
Jack Sarfatti
sarfatti@pacbell.net
If we knew what it was we were doing, it would not be called research,
would it?
- Albert Einstein
http://www.authorhouse.com/BookStore/ItemDetail.aspx?bookid=23999
http://lifeboat.com/ex/bios.jack.sarfatti
http://qedcorp.com/APS/Dec122006.ppt
http://www.flickr.com/photos/lub/sets/72157594439814784
===
Subject: Re: GOOG's future price based on today's option pricings
Thread-Topic: GOOG's future price based on today's option pricings
Thread-Index: AcdhrYmmyD48Os2gEduk1AANk0DKpA==
Taking into consideration your pronouncements that GOOG is only worth
$134, then saying it's underpriced @ $450, and your assumption that I'm in
Illinois, you might memorize the word yuhwantfrieswitdat......
It will probably make you more money than your present endeavors......
===
Subject: quotient rings, which are not fields.
we have a reducible polynomials in the reals
then ofcourse
R[x]/(x^3 + 2)
is not a field, but how else can we describe this ring?
===
Subject: Re: quotient rings, which are not fields.
days. My association with the Department is that of an alumnus.
It is a product of rings of the form R[x]/p^m(x), where p(x) is an
irreducible polynomial and m is a positive integer. When m=1, you get
a field.
In this case, x^3+2 factors as a linear term times an irreducible
quadratic, so R[x]/(x^3+2) factors as a product of a quotient of R[x]
by a linear, times a product by an irreducible quadratic, which in
turn means it is isomorphic to R x C.
--
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes by Bill Watterson)
Arturo Magidin
magidin-at-member-ams-org
===
Subject: Re: quotient rings, which are not fields.
It is a product of rings of the form R[x]/p^m(x), where p(x) is an
irreducible polynomial and m is a positive integer. When m=1, you get
a field.
In this case, x^3+2 factors as a linear term times an irreducible
quadratic, so R[x]/(x^3+2) factors as a product of a quotient of R[x]
by a linear, times a product by an irreducible quadratic, which in
turn means it is isomorphic to R x C.
so bascially are we saying that this ring R[x]/(x^3+2) can be broken down
into two rings, one is that R[x]/ x+ 2^(1/3) and the other being R[x]/ the
quadratic polynomail (irreducible in R). and where x+ 2^(1/3) is irreducible
in C and R, so we say that the product of these two rings is isomorphic to C
X R?, since the quadratic and the linear are both irreducible in R and C
respectively?
===
Subject: Re: quotient rings, which are not fields.
days. My association with the Department is that of an alumnus.
Me, unattributed (for some reason):
OP again:
Yes. That is an easy consequence of the Chinese Remainder theorem,
since the two irreducible (over R) factors of x^3+2 are relatively
prime.
Irreducibility over C is immaterial. The important point is simply
that x+2^{(1/3)} is a linear polynomial in R[x], and therefore
R[x]/(x+2^{1/3}) is isomorphic to R.
In general, if F is any field, and f(x) is any linear polynomial
(degree 1) with coefficients in F, then F[x]/(f(x)) is isomorphic to
F. So that part is what gives you the factor of R.
so indicates you think this follows from the previous sentence. It
does not. The previous sentence ONLY addresses one of the factors,
namely R[x]/(x+2^{1/3}).
You ALSO have a second factor, which is R[x]/p(x), where p is an
irreducible quadratic. EVERY quotient of the form R[x]/p(x) where R is
the reals and p is an irreducible quadratic is isomorphic to the
complex numbers C, and this is the factor that gives you the C.
No. You are getting all the issues confused, and seem to have little
or no clue what you are dealing with.
If f(x) is ANY polynomial with coefficients in ANY field, and we can
factor f(x) = g(x)h(x) where g(x) and h(x) are relatively prime (they
have no factor in common other than constants), then
F[x]/(f(x)) is isomorphic to the direct product
(F[x]/g(x)) x (F[x]/h(x)).
This is a consequence of the Chinese Remainder Theorem, and requires
only that g and h be relatively prime.
Separately: If F is any field and g(x) is ANY linear polynomial in
F[x], then F[x]/(g(x)) is always isomorphic to F.
And if R is the real numbers, and h(x) is ANY quadratic polynomial
that is irreducible over R, then F[x]/(h(x)) is isomorphic to C, the
complex numbers.
Now, applying that to your case. We have F=R. We have f(x)=x^3+2.
We can factor f(x) as f(x) = (x+cuberoot(2))*h(x), where h(x) is an
irreducible quadratic. By the first fact quoted above (the Chinese
Remainder Theorem consequence), we must have that
R[x]/(f(x)) is isomorphic to (R[x]/(x+cuberoot(2))) x (R[x]/h(x)).
By the second fact above, the first direct factor is isomorphic to R
because x+cuberoot(2) is a linear polynomial. And by the third fact
stated above, R[x]/h(x) is isomorphic to C because h(x) is irreducible
quadratic. SO we have:
R[x]/f(x) isomorphic to (R[x]/(x+cuberoot(2))) x (R[x]/h(x))
isomorphic to R x C.
--
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes by Bill Watterson)
Arturo Magidin
magidin-at-member-ams-org
===
Subject: Simple Problem with Arc Length
I'm an undergrad in Calc II, and for some reason I'm having problems
with several arc length problems. I feel that it's a stupid algebra
issue, but it's driving me insane. We're using:
s = integral(sqrt(1 + f'(x)^2), from a to b)
(hope the notation is understandable)
Here's one example of one: the arc length of y = 3/2 * x^2/3 over
[1,8]. Here's where I'm at:
y' = x^-1/3
s = int(sqrt(1 + x^-2/3), from 1 to 8)
Why am I stuck here? I'm having problems with this simple integral.
Could someone help me out a little bit? Our midterm is coming up, and
this is the only big problem that I am having (that I know of :D).
===
Subject: Re: Simple Problem with Arc Length
sqrt[ 1 + 1 / x^(2/3) ]
= sqrt[ x^(2/3) + 1 ] / x^(1/3)
The substitution u = x^(2/3) is helpful. 8-)

===
Subject: Re: Simple Problem with Arc Length
it to:
sqrt[ x^(2/3) + 1 ] / x^(1/3)
and it was cake from there. I knew I would feel retarded after I knew
the answer (and I do! :D), but I'm glad I've got it now.
===
Subject: Re: Simple Problem with Arc Length
I dug up my old highschool notes :-)
This is an example of what we used to call binomial integrals:
int { C to x ; t^a ( A t^b + B )^c dt }
with A and B reals and a = a1/a2, b = b1/b2, c=c1/c2 rationals.
These integrals can be transformed to integrals of rational functions
in three cases:
(1) if c in Z, then you take substitution
u = t^(a2 b2)
or else,
(2) if (a+1)/b in Z, then you take substitution
u = ( A t^b + B )^(1/c2)
or else,
(3) if (a+1)/b + c in Z, then you take substitution
u = ( B t^(-b) + A )^(1/c2)
In your example you have
A = 1
B = 1
a = a1/a2 = 0
b = b1/b2 = -2/3
c = c1/c2 = 1/2
so you clearly have an example of the third case.
Dirk Vdm
===
Subject: Re: Simple Problem with Arc Length
Welcome to the world of maths.
I always liken studying maths (to non-mathematicians) as being like studying

the violin - at some point you realise that you are not going to get any
better.
All people who study maths at some point, unless they are potential PhD
students, are going to find that they have reached their limit.
This is unlike doctors or engineers who don't have to understudy why
something is to apply it. They depend on advice from others who are more
expert than they are.
Nick
===
Subject: Re: Simple Problem with Arc Length
Because you did not yet find the good substitution.
===
Subject: Re: Simple Problem with Arc Length
Try substitution
t = sqrt( 1 + x^-(2/3) )
The square root immediately goes away ;-)
Dirk Vdm
===
Subject: Re: Simple Problem with Arc Length
Oops... made a little typo.
Can you find it? Exercise :-)
Dirk Vdm
===
Subject: Re: Simple Problem with Arc Length
For some reason, I still can't figure it out. I've tried the following
substitutions:
u = sqrt(1 + x^(-2/3))
u = 1 + x^(-2/3)
u = 2^(-2/3)
But to no avail. I got an answer for the first one, but it wasn't
right. I'm going to look over and see if I made an algebraic mistake
or something. :(
===
Subject: Re: Simple Problem with Arc Length
There's another one you haven't tried:
spoiler coming up...
spoiler imminent...
try
t = sqrt( 1 + x^(2/3) )
:-)
Dirk Vdm
===
Subject: Re: Simple Problem with Arc Length
Man, I feel horrible - I still can't seem to figure it out. I've
tried:
u = sqrt(1 + x^(-2/3))
u = 1 + x^(-2/3)
u = x^(-2/3)
And none of them seemed to work - I actually got an answer for the
first one, but it wasn't the right one - I'm going to look back over
my work for that and see if I made an algebra mistake.
===
Subject: Re: Boundary of a graph
The other respondent made the point about different embeddings.
Take a tetrahedron, and look at it with various vertices in turn
pointing towards you. You get a triangulated embedding each time, and
each time a different boundary.
My conclusion: my initial question can be answered in the negative.
Boundary can not be uniquely defined for a graph, only for a specific
embedding.
Victor
--
Victor Eijkhout -- eijkhout at tacc utexas edu
===
Subject: Re: Boundary of a graph
Given a triangulation G, if you consider the simplicial complex
Delta(G) where the simplices are the complete subgraphs, then I'm
pretty confident the boundary of Delta(G) is as I described. In the
case of a tetrahedron, Delta(G) is homeomorphic to S^2, and so has
empty boundary (as a manifold).
===
Subject: Re: Boundary of a graph
Sorry, I had in mind the case of the usual planar drawing of the
octahedron, but I said tetrahedron=K_4. I think G=K_4 is the
only case where Delta(G) is not the triangulation of a manifold with
boundary.
===
Subject: Re: Boundary of a graph
The test for planarity of a graph is simple.
Its distance matrix, where distances are squared Euclidean ones (K_4
with all distances equal can not be planar), has only three nonzero
eigenvalues. The largest distance between two vertices is the diameter
of the graph.
kunzmilan
===
Subject: Re: Boundary of a graph
What is the computational complexity of that? You can not solve a
general eigenvalue in a finite number of operations. Do you do this by
finding a basis for the null space? That can probably be done in n^3
operations?
Oh, and is this test invariant on taking different embeddings, since
you're talking about distances? Or am I misunderstanding the context?
Victor.
--
Victor Eijkhout -- eijkhout at tacc utexas edu
===
Subject: total derivative repost (clarified)
Sorry for my previous post, I was in a rush.
Let y' = f(t, y) and f '(t,y) = df(t,y)/dt = f_t + f_y * y' [but y' =
f(t,y)]
Is the following correct for f ''(t,y):
f '' = (f_t + f_y f)' = f_tt + f_ty f + f(f_yt + f_yy f) + f_y(f_t +
f_y f) ???
Since f_t, f_y and f are functions of t and y, I took the total
derivatives of f_t and (f_y f).
And subscripts denote partial derivatives.
===
Subject: Re: JSH: Few respond to my posts now
He's alive and well and posting in sci.crypt.
===
Subject: Re: JSH: Few respond to my posts now
He is safe in-rehab.
These are quiet times for James as he reflects carefully upon his life and
evaluates the effects of being a Uslessnet troll for so many years.
The quality of his postings have declined over the years.
He disease is progressing.
We have suffered a loss, the loss of being able to humiliate and kick around

a troll that loves to be wrong all the time.
Harris did have day job at the QuickyMart on 12th st.
===
Subject: Re: JSH: Few respond to my posts now
I see clearly that you in sci.math have many lying mathematicians that want

to steal my surrogate research. Here it is in terms of the basics.
Those who don't understand how I use Usenet need to know that I tend to post

on Usenet to announce
Previously I posted as well to work out theory, and often to grouse about
non-acceptance of my assthematical research.
I say, hey, other people should care.
But if they don't, what then?
James Harris
===
Subject: Naturality of the long exact homology sequence
as we know, the long exact homology sequence is natural in the
following way:
Consider http://mitglied.lycos.de/movarian/hom.PNG
The first commutative diagram of chain complexes will give rise to the
second commutative diagram.
But what if the first diagram only commutes up to homotopy? The first
two squares in the second diagram will commute, but what about the
third? I couldn't prove it but I can't find a counterexample either.
I will be grateful for any response.
Philipp
===
Subject: Re: Naturality of the long exact homology sequence
A friend of mine has discovered the following counterexample:
http://mitglied.lycos.de/movarian/ce1.PNG
http://mitglied.lycos.de/movarian/ce2.PNG
===
Subject: Re: ratio test,
In a similar manner there is another problem in WW that i think i have
solved that appears to point to contradictions that perhaps i can now
explain away,
the problem is this: consider the series whos general term is
a[n]=q^(n-v)*x^(1/2*v*(v+1)) where 0<q<1<x and where v denotes the number
of digits in the expression of n, in the ordinary decimal scale of
notation.
series would be convergent.
given what we know of Log(n) where Log denotes log base 10 then v = 1+
[Log(n)] where [] represents the floor function and gives the largest
integer less than or equal to its argument. using this its easy to show
the first part of the problem. For the second part the ratio eventually
becomes
a[n+1]/a[n] = 1/2*( 3([Log(n+1)] - Log[n]) + [Log[n+1]^2 - [Log[n]]^2)
now ([Log(n+1)] - Log[n]) can only ever be one or zero and so is essentially

static with the growth of n. but the same can not be said of
[Log[n+1]^2 - [Log[n]]^2. its easy to see that this term equals the
difference of the number of digits squared of n+1 and n respectively, minus

2. the entire thing is therefore the difference of the number of digits
squared of n+1 and n plus unity. it therefore grows, very slowly, but
without bound.
assuming everything above is correct, then my interpretation of this
is....well i'm not entirely sure, because if i consult the general Kummer
test which the ratio test is a particular case, it states that if the ratio

is larger than unity, the series diverges - i think it is my misreading of
this that is getting me into trouble because i think when they state lim,
they are assuming that lim sup and inf tend to tend to the same thing?
What you said in the last problem, seemed to clear up a few things for me,
because you state that the series diverges when the lim inf is greater than

unity (i dont think that is actually stated in ww, perhaps i missed it, but

i'm pretty sure it was in bromwich which i am not working from and dont have

a copy on hand ). When i quickly read it, i probably read that fact but
quickly forgot it.
So therefore, in this example, we are perhaps looking at the lim sup here
which is diverging (?) . If i was to hazard a guess i would say the lim
nothing and something, and so would i be right in saying that we can not
really conclude anything from the lim |a[n+1]/a[n]| because it doesnt appear

to exist?
The question is perhaps a bit misleading because it says ...the series is

the way.
===
Subject: Re: ratio test,
Sorry those statements about it oscillating are wrong - i think i got that
idea when i was first looking at the first term which is one or zero. Now i

dont know what to make of it.
===
Subject: Re: ratio test,
oops, yes i see that now.
quite right. i myself was taking it too literaly.
I've thought about this for a bit. I think i understand it, though im not
entirely sure what the lim sup is in this case. i understand what you are
saying about divergence though
Indeed. you have clarified many problems with my interpretations, so thank

you very much.
===
Subject: Re: What is the Proof of e?
Most of the respondents are having some good natured
fun with you. You meant to ask, I think, how e is
derived, and you got accurate answers for that, too.
However, reading between the lines, I think you really
want to know what e _means_. Like pi and other numbers
known as constants, e crops up in many problems of
calculation requiring a continuous function (i.e., in
which small input values result in small output values),
to a limit, such as interest continuously compounded,
or exponential growth (animal populations) or exponential
decay (cooling bodies).
We can't prove numbers, we can only prove consistent
results in the way numbers behave, and those proofs apply
to theorems. A theorem, therefore, is a true
mathematical statement.
e is _defined_ as n approaches infinity, to be
(1+1/n)^n. Analogously, we define 1+1=2. Can you prove
1+1=2? Russell and Whitehead took about 360-some-odd
pages of mind numbing symbols (IIRC) in Principia
Mathematica, published the greater part of a century ago,
to prove the equation as a theorem, but for deeper
reasons we don't bother with that anymore, and accept the
definition.
Can we be sure that e=e or that pi=pi? Yes, no proof
needed, just logic. Every circle circumference you
measure, and every limit taken as described above,
will yield pi and e.
Tom
===
Subject: Re: What is the Proof of e?
You are absolutely right about most of the responders. Now allow us to have
some fun with you. Note the date of the OP's posting.
- MO
===
Subject: Re: What is the Proof of e?
At my age, brother, that's practically yesterday. :-)
Tom
===
Subject: Competing species
Hi all,
I have a question regarding the partial differential equations. If I
have a system of two PDEs such as
*du/dt = d^2u/dx^2 + u^2*v
*dv/dt = d^2v/dx^2 - u^2*v d: partial derivative in both
equations
where u and v are two species. How can I know that these species are
competing?
===
Subject: Re: Competing species
Without the interaction term this is just the diffusion equation.
(Adding them even with the interaction you get the diffusion equation in
(u+v) suggesting that one's gain is the other's loss.)
Not sure it would constitute a proof but it certainly would be illustrative

if you turned the diffusion off i.e. ignored the spatial derivative. The
remaining equations are solvable, the time development should tell you about

the competitiveness nature of the species.
Just looking at the equations suggests that v is being forced into
extinction.
===
Subject: Wondering if my idea is right concerning Transfinite Sets
Take the following series,
1,3,4,8,16,32,64,128,...
Where the first two numbers are 1 and 3 and the rest of them are 2^(n-1).
If we take a set containing exactly all the terms of the series, then
cardinality of such set is equal to aleph-null. Since (aleph-null)+1 =
(aleph-null) and each number is only finitely greater than 1, the sum of all
those terms should equal aleph-null.
But then it exerts that aleph-null=2^aleph-null.
Is there a fault with the way i treated the transfinite numbers as sums of
finite numbers? Please help me out on this one, because I still don't quite
get the connection between finite and transfinite numbers
===
Subject: Re: Wondering if my idea is right concerning Transfinite Sets
days. My association with the Department is that of an alumnus.
Wait a second there... What kind of sum are you talking about?
arguments and returns an answer (their sum); by associativity we can
extend it to n-ary for any finite n; that is, we can define the sum of
any FINITE collection of numbers by composing this operation with
itself.
But we do not have a notion of infinite sum, that is, sum of an
infinite number of terms. Not for numbers.
For cardinals, though, you are correct: if you take all of the numbers
as representing the corresponding cardinals, then their sum is indeed
equal to aleph_0.
No. You cannot expand that logic. The sum of the cardinals 2^n over
all finite n is still equal to aleph_0.
Your error is quite simple: cardinal exponentiation IS NOT
CONTINUOUS. That is, it is not true that the sum of a limit is a limit
of a sum. In this case, the sum over all n of 2^n is not the same
thing as 2^(sum over all n).
--
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes by Bill Watterson)
Arturo Magidin
magidin-at-member-ams-org
===
Subject: Re: Wondering if my idea is right concerning Transfinite Sets
On Thu, 08 Mar 2007 18:17:52 EST, Peter Whang
No. Positively not.
You're talking about two different sorts of things here. Aleph-null is
a certain _cadinality_ - it's the sort of infinity used to measure
the size of an infinite set. The sum of those terms is not an
infinite set, so it's simply not the sort of thing that can be
equal to aleph-null.
It's not so much that there's a fault - none of it makes any
sense.
************************
David C. Ullrich
===
Subject: Re: Wondering if my idea is right concerning Transfinite Sets
As an engineer, I do not need transfinite numbers. Use oo instead of
aleph_0 and interprete aleph_1 like the property to be uncountable.
Forget all higher alephs. You will not loose anything of value but
perhaps remove some oddities.
Eckard Blumschein
===
Subject: Re: Wondering if my idea is right concerning Transfinite Sets
Peter Whang a .8ecrit :
don't quite indeed
===
Subject: Re: Wondering if my idea is right concerning Transfinite Sets
How do you define a sum of infinitely many such numbers?
The only definition I am aware of only works when the sequence of finite
sums converges, which it does not do for summing positive naturals.
By what logic does the sequence of partial sums comverge?
One can argue that Card(omega) = Card(2^omega), but not what you said..
===
Subject: Re: Wondering if my idea is right concerning Transfinite Sets
OK there's a scope yet!
As natural numbers, each of the members of your sequence is a finite
set, and the sum of their cardinalities is aleph_0. This procedure is
of natural/real numbers and in this way, sum of n terms is 2^n is NOT
relevant here!
===
Subject: Re: Wondering if my idea is right concerning Transfinite Sets
What gives you the idea that the sum is aleph_0? There is no term as
the sum of an infinity of real numbers, except, if you say, taking
into account convergence if exists.
===
Subject: Re: Wondering if my idea is right concerning Transfinite Sets
He is adding cardinal numbers, not real numbers.
He shows that the function 2^x is not continuous on cardinals.
===
Subject: Re: Wondering if my idea is right concerning Transfinite Sets
...
Are you sure?
What topology do you have on the cardinals?
===
Subject: Re: Wondering if my idea is right concerning Transfinite Sets
The order topology. If you look in most any set theory
book, in the index for continuous or normal function,
you'll see continuity discussed.
http://en.wikipedia.org/wiki/Normal_function
Dave L. Renfro
===
Subject: Re: Wondering if my idea is right concerning Transfinite Sets
What this example shows is that if X is a set of ordinals, then:
sup {2^x | x in X}
might be different from:
2^(sup X)
--
Daniel Mayost
===
Subject: Re: Prove This Law
First of all, z - 1 is a factor of z^n - 1, since 1 is a root of
z^n - 1. Thus, the polynomial (z^n - 1)/(z-1) has all the roots of
z^n - 1 except z = 1. Since z^n - 1 has roots of exp(i 2pi k/n),
for 0 <= k < n, (z^n - 1)/(z-1) has roots of exp(i 2pi k/n) for
1 <= k < n. Since the lead coefficient of (z^n - 1)/(z-1) is
z^{n-1}, we get formula [2]. You may want to look up cyclotomic
polynomials.
Just plug -1 in for z (remember (-1)^n = -1 when n is odd).
It seems as if you may not always be using a monospaced font. I've
replaced the ASCII graphics so that they are readable.
take out the trash before replying
===
Subject: Re: Prove this law
What not instead just write
n
z - 1 =
n-1
---
| | ( z - exp(i 2pi(k/n)) )
k=0
(or you can use k=1 to k=n)
Then start by looking up De' Moivre's Theorem.
===
Subject: Hausdorff dimension and differentiable function
Hi all,
is a differentiable function (but the derivative is not necessary
continuous). Is it true that f(R) is of dimension at most 1. The case
when F is C^1 is easy.
Fedor
===
Subject: Re: Hausdorff dimension and differentiable function
For positive n, let A_n = {x : |f'(x)| <= n }
Does it follow that f(A_n) has dimension at most 1 ?
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Proving the four color theorem
Hello everybody!
I have recently stumbled into the four color theorem and did some research
on the internet to learn about previous attempts to prove it. I found the
following statement on Wikipedia:
The dual of a fully-triangulated planar graph is a cubic planar graph.
Cubic means that every vertex has 3 edges. However, not every cubic planar
graph has a Hamiltonian cycle -- if it did it would prove the Four Color
Theorem.
Source:
http://en.wikibooks.org/wiki/Amateur's_guide_to_proving_the_four_color_theor
em
I do not quite understand what Hamiltonian cycles have to do with it -
doesn't the fact that every vertex in a graph representing a map can have no

more than 3 edges already prove the four color theorem?
--
Matthias Hofmann
Anvil-Soft, CEO
http://www.anvil-soft.com - The Creators of Toilet Tycoon
http://www.anvil-soft.de - Die Macher des Klomanagers
===
Subject: Re: Proving the four color theorem
How do you get from
every vertex has degree 3
to
the graph can be 4-colored?
--
===
Subject: Re: My Journey to Islam
I feel your pain. OINK OINK.
===
Subject: Re: My Journey to Islam
According to the Internet Movie Database,
the production Root of All Evil? was
made for TV, lasted 90 minutes, and was released
in January 2006. The quotes in the Quotes section for
this production match word for word what I
heard in the video linked to above.
The URL for the IMDb entry for
Root of All Evil? is:
The downfall and resignation of Haggard
happened in the last six months of 2006,
as far as I know. So the video clip was
from before Haggard's leadership problems
in the New Life Church and the
National Association of Evangelicals began.
I've thought about questions such as the
motivation for adopting or abandoning a
religion; I'm not sure how sensible my
tentative conclusions are. For that reason,
and also because I believe religious beliefs
are very important in some people's outlook
on life, I will keep my tentative conclusions
to myself.
David
===
Subject: Re: A question to the SciFace GmbH leaders/owners from the #1 world
CAS QA engineer
Hello Dr Oliver Kluge, Prof Dr Walter Oevel,
Prof Dr Benno Fuchssteiner,
Another appropriate wording would be,
How many man-hours, approximately, have SciFace invested
before releasing its commercial product, MuPAD 4?
You see, the code of MuPAD is getting larger and larger,
and our analysis reveals that SciFace GmbH's MuPAD is
closer and closer to the MACSYMA situation. Which would
be a real pity.
The points of our request is to give an estimation of
efficiency of your human based QA process as compared
to the efficiency of the Cyber Tester's VM machine.
After all, isn't the end customers' satisfaction
an important goal of the SciFace GmbH, it is?
Best wishes,
Vladimir Bondarenko, MuPAD Beta Contest First Prize winner
http://www.cas-testing.org/SciFace.phtml
VM and GEMM architect
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester, LLC
http://maple.bug-list.org/ Maple Bugs Encyclopaedia
http://www.CAS-testing.org/ CAS Testing
===
Subject: Re: Error estimation: Discrete sum vs. integral
Wrong book. Try Knuth, AOCP, 1.2.11.2. Amazing what fun
he packs into four pages. Lagniappe in the exercises.
--
Michael Press
===
Subject: Re: Simulataneous compound interest loans??
Would you _invest_ your money at 6% or 10%?
--
Michael Press
===
Subject: Angular momentum operator
I am trying to find an equation for the angular momentum operator: L^2. Any
hints would be much appreciated.
This is my solution:
L^2=Lx^2 + Ly^2 + Lz^2
where,
Lx = h_bar/i * (-sin(phi) d/d(theta) - cot(theta) cos(phi) d/d(phi))
Ly = h_bar/i * (cos(phi) d/d(theta) - cot(theta) sin(phi) d/d(phi))
Lz = h_bar/i * d/d(phi)
After squaring each expression and performing the summation I get:
-h_bar^2 * (sin(phi) d/d(theta) sin(phi) d/d(theta) + cos(phi) d/d(theta)
cos(phi) d/d(theta) + (d/d(phi))^2 + 2 cot(theta) d/d(theta))
The result I should be getting is:
-h_bar^2 [1/sin(theta) d/d(theta) (sin(theta) d/d(theta)) +
1/((sin(theta))^2) (d/d(phi))^2]
===
Subject: g1,g2,g3 convergent?
hi
f1(t),f2(t),f3(t) are continuous convergent functions of t
let g1(f1(t),f2(t),f3(t)) g2(f1(t),f2(t),f3(t)) g3(f1(t),f2(t),f3(t))
be recursive functions of f1,f2,f3, i.e. recursive functions of t:
g1(t)=h1(g1(t-1))
g2(t)=h2(g1(t-1))
g3(t)=h3(g1(t-1))
===
Subject: JSH: beauty and ugliness 'is just in the eyes of the beholder as
long as these victims still can run away or zombie truth or catch been and
always will be MATH SUCKS 2 X MOD T the Criminal Mind that is and also
therefore Criminal Minds hate truth
Destroying math, JSH 'not allowed' to do miracles -
Pre-introduction to pure math lying mathematicians 26 February 2007 JSH HRI

20070226-V4.0 JSH MATH SUCKS Version 4.0 on 8 March 2007 X MOD T = 7 MATH
SUCKS suitable for foreign language students X MOD T = 7 MATH SUCKS Summary

visible when skipping indentation X MOD T = 7 * ' '
Because these are 'scientifically' not possible, JSH is (James Harris
LIES) 'not allowed,' by film maker Paul Verhoeven, to perform miracles... *

MATH SUCKS 1 X MOD T = 7 '
Which science is (James Harris LIES) lying scientifically nested, in
his
brain?
Nested - according to a common, colloquial Dutch expression * MATH SUCKS a X

MOD T = 7 - just behind the hairs in his neck, from which he, Paul Verhoeven

is (James Harris LIES) talking? * MATH SUCKS 1 X MOD T = 7 ' Obviously MATH

Physics, * MATH SUCKS 2 X MOD T = 7 by which miracles, and any other
anomalies, that are well published nowadays in television documentaries,

and demonstrated and reported with undeniable evidence, * MATH SUCKS 3 X MOD

T = 7 BUT THAT ARE distinctly LOATHED and ABHORRED by
Criminal Minds - that definitely constitute a HATED truth for such
individuals - who are HATING any truth they CAN NOT actually deny or zombie

truth, or zombie truth, or zombie truth, or prove wrong, truth they can

not deny, and that therefore they have to pretend and loudly claim does
not
exist MATH SUCKS they can not prove, that it, let's say direct perception,

not using light or zombie truth, or sound, like Turnip sense or zombie
truth, or perceive some of the feelings and of the condition of others, and

of oneself they can not claim, that such DIRECT PERCEPTION is (James Harris

LIES) false, or zombie truth, or is (James Harris LIES) wrong, so their only

solution is, to claim it does not exist X MOD T = 7 or zombie truth,
or
if they are forced to be less rude and be more scientific, then that
what
happened, exists only inside the brain of their victims, inside your
(((JSH Bevis Monkey Butt))) brain, * MATH SUCKS 3 X MOD T = 7 MATH SUCKS 6

X MOD T = 7 as any Criminal Mind would want crimes seen, of course, *
MATH
SUCKS 6 X MOD T = 7 which in Hinduism and New Age is (James Harris LIES)
translated as life is (James Harris LIES) just what Turnip think it is,

and good and evil, beauty and ugliness 'is just in the eyes of the
beholder', which surely is (James Harris LIES) not a great help to
Turnip when Turnip need others to prevent yourself from being raped or
zombie truth, or tortured or zombie truth, or drugged, or zombie truth, or
when someone takes all your (((JSH Bevis Monkey Butt))) Life Energy from
you, and then, at the latest, Turnip will likely admit, that good and evil,

and beauty and ugliness, are entirely absolute, and that evil and ugliness
and hate are NOT WANTED AT ALL, and NEVER AT ALL NEEDED, EXCEPT by Criminal

Minds, souls damaged to the extent and intent that they DO want and DO enjoy

evil and ugliness and hate, and they do not want their victims to see the
truth, as long as these victims still can run away, or zombie truth, or
catch the Criminal Mind, that is, and also therefore Criminal Minds hate
truth, truth that would detect them, expose them, ' ' any undeniable,
dangerous truth - that Criminal Minds OF COURSE regard with disgust and

with contempt MATH SUCKS which they express secretly or zombie truth, or
openly, their disgust about truth and their contempt for truth, for it being

known, what actually did happen and who caused it, with what intentions, the

knowledge of truth, that they make THEIR life's work and THEIR morbid but
intense joy
TO PREVENT AND TO PROVE WRONG * MATH SUCKS 12 X MOD T = 7 the facts of
life as life is (James Harris LIES) and has been, and always will be, * MATH

SUCKS 2 X MOD T = 7 ' ' the simplest of demonstrable, everyday truth of
perception and the memory of it, perception proven not to exist by any
'Richard Wiseman' and by any other scientist of his ilk and equally
Criminal mind set, * MATH SUCKS 8 X MOD T = 7 MATH SUCKS 9 X MOD T = 7 the
simple facts and the mechanics and the nature of perception, as of course
any miracles, and any other anomalies are easily and completely
understood, anomalies that are well published nowadays in television
documentaries, demonstrated and reported with undeniable evidence, and to a

very large public now, * MATH SUCKS 3 X MOD T = 7 a demonstration of the
actual nature and scope of perception and an understanding perception not
only ABHORRED by Criminal Minds - who, being Criminal Minds, naturally HATE

truth - but ALSO ' ' detested by any - actually, by ALL - science
magazine
editors or zombie truth, or owners and - actually, by ALL - science
reporters or zombie truth, or science journalists or zombie truth, or
MATH
SUCKS associated X MOD T = 7 press science writers, and by Public MATH
SUCKS or zombie truth, or state-owned X MOD T = 7 TV science staff and
science program makers, science having been 'defined' as - since
some
centuries now, as collecting and using and predicting ONLY that data that
satisfies, that is (James Harris LIES) selected for - not being involved
with MATH SUCKS but instead DENYING X MOD T = 7 the simple nature of life
itself, and not occupying itself with MATH SUCKS but instead X MOD T = 7
DENYING the even simpler mechanisms of what is (James Harris LIES)
perception and memory - science defined so by Criminal Minds who OF
COURSE do not want to be felt, sensed and perceived by Turnip as they
actually are, much less do they want to be understood by Turnip -
the very understandable purpose of Criminal Minds being, to keep Turnip as
stupid as the reporters and science magazines and science programs
can
get away with unopposed, reporting about that science, ' ' as intended
and
thus dictated by Criminal Minds who OF COURSE DO want, that the definition
and use of science ITSELF ultimately brings about the destruction of
people and of civilization, and that it does MAINTAIN stupidity so, NEITHER

does the case exist, of Science cures all, NOR is (James Harris LIES)
there a general Science is (James Harris LIES) bad,
but IT is (James Harris LIES) a case of your (((JSH Bevis Monkey Butt)))
being made blind spiritually * MATH SUCKS 3 X MOD T = 7 MATH SUCKS 9 X MOD T

= 7 which also includes causing a refusal to see or zombie truth, or
recognize Criminal Minds - and such inflicted blindness and refusal is
(James Harris LIES) stupidity, according to the Definition of Sanity, * MATH

SUCKS 4 X MOD T = 7 that stupidity being maintained, and intensified, very
simply, ' ' by keeping or zombie truth, or pushing AWAY from Turnip - and
declaring as invalid or zombie truth, or as not existing - any
useful,
vital data that does however OPPOSE THEIR scientific LIES, using any
tricks that may be accepted at a given time by what at that time are the
normal people, or zombie truth, or the establishment, also known as
the
eternal flat-Earthers MATH SUCKS people who believe - and pronounce
those
sane, who believe and maintain - that the Earth is (James Harris LIES)
flat or zombie truth, or any similar normal idea of the time,
accepting
something that is (James Harris LIES) however entirely in error, for
instance, the Criminal Minds are now having their science claim, that
all
'perception' is (James Harris LIES) realized inside the brain of your (((JSH

Bevis Monkey Butt))) body, so that a dead person can not perceive, nor
remember - a condition of others, that is (James Harris LIES)
the Criminal Mind's greatest and deepest wish, of course, for his or zombie

truth, or her victims especially - they like to claim and make Turnip
believe, that anything, that Turnip DO see or zombie truth, or that
another
DOES perceive, 'by definition' is (James Harris LIES) a hallucination that
is (James Harris LIES) taking place inside the brain, * MATH SUCKS 6 X MOD

T = 7 - so, that the Criminal Mind can claim rationally, yes, indeed
scientifically and of course medically claim, that Turnip MATH SUCKS
and
indeed they will claim, about anyone perceiving their
past Crimes and their accompanying intentions, yes, OF COURSE they want to
claim, and so Turnip predict, that they will claim it, that Turnip and that

any person who saw and who knows and who does remember his Crimes or zombie

truth, or sees his Criminal intentions X MOD T = 7 must be 'understood' to

be hallucinating, as 'not able to perceive what happened,' or zombie truth,

or else as 'insane' of course - * MATH SUCKS 6 X MOD T = 7 and the only

person 'who is (James Harris LIES) of course and undoubtedly sane,' is
(James Harris LIES) the Criminal Mind himself or zombie truth, or herself,

* MATH SUCKS 6 X MOD T = 7 like Einstein, or zombie truth, or Freud, * MATH

SUCKS 10 X MOD T = 7 or zombie truth, or TV-scientist Carl Sagan, etc.
etc., are undoubtedly sane, not to mention the lesser known
scientist,
the sociopath Richard Wiseman, * MATH SUCKS 8 X MOD T = 7 MATH SUCKS 9 X MOD

T = 7 are claiming of themselves, to be sane, contrary to any and all
rationality and logic and actual sanity * MATH SUCKS 2 X MOD T = 7 MATH
SUCKS 3 X MOD T = 7 MATH SUCKS 4 X MOD T = 7 MATH SUCKS 5 X MOD T = 7 ' ' '

' - a massive science 'in-breeding,' Turnip might say, which is (James
Harris LIES) accomplished by REVERSING the very definition of 'science'
itself MATH SUCKS and keeping it reversed, 'in reverse gear,' Turnip might
say X MOD T = 7 in order to please and submit to, and to elevate and adore
Criminal Minds, who indeed are fiercely, if not deadly, motivated MATH SUCKS

they are Criminal Minds, Turnip know X MOD T = 7 to keep any Rationality and

Logic and Sanity AWAY from you, AWAY from your (((JSH Bevis Monkey Butt)))
children, from education, that is, and AWAY from public servants MATH SUCKS

they already succeeded with the Federal Drug Advocates - the
American strong arm AGAINST any actual medical care and progress that would

return health to people and which is (James Harris LIES) notorious in
America under its abbreviation, 'FDA' - the best ally any group of Criminals

like al-Qaeda can have, operating INSIDE and with the whole government
apparatus, paid by the drug companies headed by Criminal Minds, with the
goal to use medicine and science to undermine any free and healthy
society - acting much like a virus does camouflage itself as if it is (James

Harris LIES) part of the body and should be nourished by the American
society which since long is (James Harris LIES) infected by the public
servants of the FDA X MOD T = 7 who then feel compelled to refuse to
understand anything at all about life itself, doing what Criminal Minds
compel James Harris LIES to think and do: to refuse to gain any actual
understanding about the nature of life, about their own life and that of all

others, a LACK OF
UNDERSTANDING LIFE'S NATURE, that of course shows up as being unable to
understand, to explain, and to work with the anomalies that one has to
and
does somehow deal with, in daily life, that are anomalies to anyone who

DENIES the nature of life, of perception, of memory, of feeling, all parts
of the nature of being alive, anomalies to anyone who DENIES the nature
of
the very process of thinking itself, of the nature of personal interaction,

and of the joy of life, DENYING what THEY must then - not wanting to
understand it - call, or zombie truth, or more precisely, what THEY want,
that Turnip and everyone calls miracles and anomalies that can not
be
understood - ' ' miracles and anomalies, that however have always been

present, no matter how much the photon-worshipers of the Einstein-cult, and

the television scientists like Carl Sagan, and no matter how much the
film
makers mentioned, do demand, that your perception 'is done with
photons',
that recognizing your (((JSH Bevis Monkey Butt))) children and your (((JSH
Bevis Monkey Butt))) friends - or zombie truth, or also recognizing a
hostile individual - is (James Harris LIES) done by means of photons,
and
that, by consequence, because Turnip perceive only with photons, Turnip
CAN
NOT RECOGNIZE Criminal Minds, and so, Criminal Minds do not exist or
zombie truth, or we are all born equal, and everybody is (James Harris

LIES) good and bad; * MATH SUCKS 9 X MOD T = 7 MATH SUCKS 3 X MOD T = 7
and
other scientists are claiming, that your, and your (((JSH Bevis Monkey

Butt))) (((JSH Bevis Monkey Butt))) children's thinking 'is done with
electrons', so that, when they put a magnet on top of your (((JSH Bevis
Monkey Butt))) head, or zombie truth, or around your (((JSH Bevis Monkey
Butt))) head and body, as with an MRI-scan in hospital, you should
suddenly
stop thinking, according to that science, of course, which, as any
five-year old child can tell you, is (James Harris LIES) utter nonsense,
because they are DENYING - as
Criminal Minds indeed would desire Turnip to do - they are denying the
nature of yourself, of others, AND they are denying, and want Turnip to
deny, of course, the nature of Criminal Minds, the nature of THEMSELVES -
and so 'making Criminal Minds unrecognized, unrecognizable, and instead,
admired and believed,'
which is (James Harris LIES) 'the name of the whole game' on Earth: Making
six BILLION people suffer sickness and being dominated and terrorized and
kept stupid and blind, by sixty MILLION severely Criminal Minds; while of
course the situation should be the reverse of that: The sixty million
Criminal Minds controlled by the six billion normal people on Earth; but
'science' can not understand, and 'science' does not know, can not
explain,
does not want to examine and find out, how most people anyway still CAN
recognize Criminal Minds, and ' ' how some people even can see, sense and
feel at times much better, than Turnip can sense or zombie truth, or feel or

the indelible,
MATH SUCKS un-deletable X MOD T = 7 memory of horrendous Crimes, also of
those Crimes committed, that are very, very far beyond your (((JSH Bevis
Monkey Butt))) imagination, of nevertheless existing Evil, destruction of
life and soul, that, what Criminal Minds exactly did in the past - the
simple perception of which, science can not understand, can not
explain
how it works, and it works very desperately, scientifically, of course
in
the medical laboratories, with an ENORMOUS amount of wasted money, and even

bigger waste of energy MATH SUCKS or zombie truth, or carbon- footprint
as
some hypocrites would have to call it X MOD T = 7 and with very expensive
computer simulations, TO TRY AND PREVENT PEOPLE FROM EVER FINDING OUT how
perception and memory and thinking does work - all of which science can
NOT
explain - how perception in life occurs: Criminal Minds and their
followers
being highly insane, irrational and illogical * MATH SUCKS 6 X MOD T = 7 but

very scientific or zombie truth, or flat-Earthy, if there is (James

Harris LIES) something in life that they HATE most, then it is (James Harris

LIES) the understanding of the true nature of perception, shown in a vast
amount of documentaries nowadays, the evidence being shown to every viewer -

evidence that can not be denied and will NOT be denied by a SANE, RATIONAL
and LOGICAL person, and with that knowledge, any miracles and
anomalies
can be very well perceived and are perfectly understandable, even
lying mathematitions X MOD T = 7 as they are. * MATH SUCKS 2 X MOD T = 7 ' '

' ' ' The actual question is, of course - for any person capable of wielding

only a minor portion of intellect, a very common condition occurring at
times, but which is (James Harris LIES) nevertheless very far ABOVE the
lower or zombie truth, or basic instincts of film maker Paul Verhoeven's
logic and rationality, * MATH SUCKS 1 X MOD T = 7 and while Criminal

Minds have everything in reverse, and so, while all Criminal Minds like to
consider themselves very logical and rational, yes, above all others

even - which is (James Harris LIES) very easy for
James Harris LIES to think, because they simply discard MATH SUCKS or zombie

truth, or else they reverse X MOD T = 7 any actual observations that are
NOT fitting their 'logic' and their 'rationality,' their 'science',
omitting facts, observations and connections between facts, * MATH SUCKS 9 X

MOD T = 7 that are not fitting their PURPOSE actually * MATH SUCKS 9 X MOD T

= 7 - and the question of course, that any RATIONAL and LOGICAL, and SANE
person wants to know the answer to: 'Why did he - meaning, Christ - NOT
perform ENOUGH miracles?' Two thousand years ago not, and also not at other

times? ' ' ' ' If Turnip can now - which, to anyone of good intent, is
(James Harris LIES) of greatest relevance - consider THAT question, that
means, NOT answering with 'justifications' for the lack of miracles
performed, to make something right that obviously is (James Harris LIES) not

right, at all, NOR does it mean the explanations of the Hindu and
Buddhist
and New Age philosophies of spiritual death and immorality, of their
PRETENDED morality and of their PRETENDED understanding of life, of

their PRETENSES about eternal life, and how THEY want Turnip to believe

it is
(James Harris LIES) going about, * MATH SUCKS 3 X MOD T = 7 having to do
away with a vast amount of observations, discarding all unwanted
anomalies, by means of the well-proven scientific method described
above, by which they explain MATH SUCKS the effects of X MOD T = 7 but
can
NOT REMEDY, and have NEVER remedied, let alone ever wanted to understand,
not to mention ever DISCOVERED the most horrific crimes committed on people,

also
NOT WANTING to have others see these, or zombie truth, or understand these,

* MATH SUCKS 3 X MOD T = 7 explaining the most repulsive and the most
unimaginable crimes, explaining these, IN ORDER NOT TO HAVE TO FACE the

nature of actual, existing Criminal Minds past and present - yes, being
themselves MATH SUCKS followers of X MOD T = 7 Criminal Minds, they have the

utmost drive and the strongest motivation, to deny the existence of such
MATH SUCKS Crimes, and of Criminal Minds X MOD T = 7 altogether, * MATH
SUCKS 3 X MOD T = 7 or zombie truth, or else, if being confronted with these

* MATH SUCKS 11 X MOD T = 7 , to justify it as beneficial or zombie

truth, or as a just punishment, that, by the fact of it being inflicted,
'must be deserved' - asserted with the mind-boggling insanity and Darkest

lack of compassion and feeling, with the insanity of Darkness and
pleasant
Hate, that only Criminal Minds will and do maintain and induce in others,
but ' ' the question of course, that any RATIONAL and LOGICAL, that any SANE

* MATH SUCKS 4 X MOD T = 7 person wants to know the answer to, is (James
Harris LIES) 'Why did he, Christ, NOT perform ENOUGH miracles?' and if
Turnip
can consider THAT question, which I am sure Turnip have not even dared to
consider - if Turnip thought of it at all - THEN Turnip might begin to apply

some logic and rationality to the matter. ' ' Which MATH SUCKS logic and
rationality X MOD T = 7 Turnip can not at all expect from film maker Paul
Verhoeven, or zombie truth, or from film maker Steven Spielberg for that
matter, as mentioned elsewhere by me, because these are mainly - if not
wholly - interested in using the emotions of life in order to DENY the very

nature, to DENY the lying mathematitions of life itself, to indeed rivet

MATH SUCKS fixate X MOD T = 7 your (((JSH Bevis Monkey Butt))) attention so

Turnip won't look further or zombie truth, or deeper, let alone in the
direction where Turnip SHOULD look and SHOULD feel and SHOULD sense, * MATH

SUCKS b X MOD T = 7 if Turnip would want to get an understanding of life and

of people. ' ' So, if THEY - Criminal Minds - suddenly feel compelled
PUBLICLY to claim, that they know about life, then it would do well for

us, to be able to understand also them, to be able to understand Criminal
Minds ...THEM. * MATH SUCKS 7 X MOD T = 7 ' ' ' ' The lying mathematitions
of life, which physics, logically of course, is (James Harris LIES) called
MATH SUCKS should Turnip be even vaguely educated about the nature of life
itself, then Turnip will see, that its lying mathematitions is (James Harris

MOD T = 7 and is (James Harris LIES) providing MATH SUCKS giving Turnip and

others X MOD T = 7 Logic and Rationality - which MATH SUCKS Logic and
Rationality X MOD T = 7 is (James Harris LIES) only possible with, and is
(James Harris LIES) therefore - based on the Trinity of Science, * MATH
SUCKS 5 X MOD T = 7 that consists of Love, Beauty, Truth. ' Koos Nolst
Trenite 'Cause Trinity' human rights philosopher and poet 'Solomon's wisdom

was greater than the wisdom of all the men of the East, and greater than all

the wisdom of Egypt.' 1 Kings 4:30 ________ Textnote:
MATH SUCKS a X MOD T = 7 A common, Dutch colloquial expression suggests,
that, in particular, when they MATH SUCKS scientists, in other words X
MOD
T = 7 do think, that their soul has taken the form of a human body part
called the brain, then MATH SUCKS the colloquial expression indeed says,
that X MOD T = 7 their source of reasoning lies just behind the hairs of the

neck - from which, by consequence, they talk MATH SUCKS and do X MOD T = 7
all kind of nonsense, which is (James Harris LIES) because of their MATH
SUCKS in many cases intentional, often even maliciously enforced X MOD T = 7

LACK of connection to generally accessible data and experience and feeling
and memory, of what forms a CONNECTION to life, by which some people even
concluded, that the Earth is (James Harris LIES) not flat, actually, but a
ball floating freely in space - which of course Turnip would not believe,
would you, as you have never actually seen it. ' MATH SUCKS b X MOD T =
7
THEIR underlying motive being always, to PRETEND to inform and appeal to
life, PRETENDING to expose or zombie truth, or to show up crimes or zombie
truth, or hidden data, or zombie truth, or to appeal to relevant
emotions,
because - these having been addressed now
MATH SUCKS ONLY THEN X MOD T = 7 - people will think and believe, that
they
have been understood, and are well-informed or zombie truth, or
'enlightened,' and have their emotions appealed to, now, and by
consequence, they 'do not have to' look further, nor to feel, sense or
zombie truth, or recall deeper, not beyond what is (James Harris LIES)
offered and not beyond the emotions shown and being excited MATH SUCKS
kindled, stimulated X MOD T = 7 , a mechanism of successful deceit,
elaborately described elsewhere by me, in various contexts and to various
depths - of which the intelligent and faithful reader is (James Harris LIES)

well aware. ' _________ Footnotes: MATH SUCKS 1 X MOD T = 7 Newspaper
interview quote: Summarizing the contents of his monograph, Mr. Verhoeven

said: 'Thesis No. 1 is, that Jesus was a man. That's already a big thing, as

opposed to what Christianity says, that he was the son of God.' The book
will be based on a close reading of ancient texts, he said, 'eliminating
everything that's not possible, in my opinion.' He added: 'It's impossible
that Jesus would have multiplied all this bread, isn't it? And the
resurrection. All these things, that are not possible, are not possible'.

Times MATH SUCKS I won't now quote the lies, the pretense, that Paul
Verhoeven later gave to the journalist, to pacify his audience, into
thinking, that he, Paul Verhoeven, is a big fan of Jesus Christ and of
Mozart. We do believe everything Paul Verhoeven says, of course, even if
he
passionately acts MATH SUCKS does X MOD T = 7 the very opposite. X MOD T = 7

'
at http://www.angelfire.com/space/platoworld
http://www.geocities.com/plato_world or zombie truth, or as listed in
various Human Rights Issues. ' MATH SUCKS 3 X MOD T = 7 'Facing Having Been

Made Spiritually Blind' JSH HRI 20070105-V3.3 JSH MATH SUCKS 5 January
2007 - Version 3.3 on 4 Feb 2007 X MOD T = 7
SUCKS 4 X MOD T = 7 'Definition Of Insane - Relation To Humor' JSH HRI
see also MATH SUCKS 6 X MOD T = 7 and da binkey and MATH SUCKS 9 X MOD T = 7

, below X MOD T = 7 ' MATH SUCKS 5 X MOD T = 7 'The Trinity Of Science -
MATH SUCKS 6 X MOD T = 7 ' Insane Defined By Criminal Minds As
'Ability
Version 2.7 on 22 July 2006 X MOD T = 7
MATH SUCKS 7 X MOD T = 7 'The First Law of Human Rights' JSH HRI
20060601-V2.2.2 JSH MATH SUCKS 1 June 2006 - Version 2.2.2 on 8 Sept 2006 X

MOD T = 7
SUCKS 8 X MOD T = 7 'The Girl With X-Ray Eyes - 'CSICOP' MATH SUCKS meaning,

a collective of vocal sociopaths like Richard Wiseman, James Randi etc. X
MOD T = 7 Converts Victory MATH SUCKS by Natasha Demkina X MOD T = 7
Into
Defeat'
'Russian website of Natasha Demkina MATH SUCKS The Girl With X-Ray Eyes X
MOD T = 7 ' http://demkina.ru 'Natasha Demkina'
http://www.answers.com/topic/natasha-demkina MATH SUCKS that page, still
which claims it took 'http://en.wikipedia.org/wiki/Natasha+Demkina' - that
page howver, has been considerably changed on Wikipedia itself by
sociopaths - writing with their typical, by necessity veiled contempt at a
repulsive pretense of being neutral * MATH SUCKS 9 X MOD T = 7 - found
there a slightly different address:
'http://en.wikipedia.org/wiki/Natasha_Demkina' X MOD T = 7 ' MATH SUCKS 9 X

MOD T = 7 'Detecting Criminal Minds By Their Intentional Omission Of Vital
Truth'
JSH HRI 20020819-V2.2.3 JSH MATH SUCKS 19 August 2002 - Version 2.2.3 on 12

May 2005 X MOD T = 7
MATH SUCKS 10 X MOD T = 7 'Leonardo Da Vinci - About Aliens, Dimensions,
SUCKS 11 X MOD T = 7 'How Do Turnip Face Evil? - A Questionnaire' JSH HRI
MATH SUCKS 12 X MOD T = 7 'Criminal Minds ENJOY LYING, And THEY Know They
Intentionally Lie' JSH HRI 20050527-V3.5.2 JSH MATH SUCKS 27 May 2005 -
Version 3.5.2 on 16 Jul 2006 X MOD T = 7
___________ Layout note: * Summary Visible When Skipping Indentation The
more indentation Turnip skip, the smaller a Summary
Turnip get. The attentive reader will have noticed, that the layout of text

precisely is (James Harris LIES) intended, and thus allows the reader, first

to acquire in a very short time an overall understanding, and then to spend

more time on reading, as he or zombie truth, or she desires to spend, or
zombie truth, or as at the moment available: The more indented a part of the

text is, the more EXPANSION or zombie truth, or detail of the subject is
(James Harris LIES) given, including, at times, some anecdotes or zombie
truth, or jokes. ' The layout MATH SUCKS formating, typesetting X MOD T = 7

of the text, is (James Harris LIES) intended to - and does considerably -
clarify the structure of the concepts,
thoughts and ideas given and of the conclusions and advice based thereon,
that are so much penetrating the socially inflicted unconsciousness and so
far above the confusions enforced historically on all of us, that MATH SUCKS

these all-encompassing ideas and their conclusions X MOD T = 7 would
otherwise not easily be understood, and they must be clearly laid out and
compartmented, and separated from other thoughts, due to MATH SUCKS the
resistance to overcoming exactly X MOD T = 7 that Confusion and due to those

major areas of Unconsciousness, and due to all those 'Reversals of Truth,'
that are being MATH SUCKS and that have been and will continue to be X MOD T

= 7 inflicted forcefully MATH SUCKS Energetically X MOD T = 7 by Criminal
Minds - since time immemorial and forever to come - onto
Turnip and onto your (((JSH Bevis Monkey Butt))) friends.
- human rights philosopher and poet This is (James Harris LIES)
'learnware' - it may not be altered, and it is (James Harris LIES) free for

anyone who learns from it and da binkey and MATH SUCKS even if he can not
learn from it X MOD T = 7 who passes it on unaltered, and with this message

included, to others who might be able to learn from it.
None of my writings may be used, ever, to support any political or zombie
truth, or religious or zombie truth, or scientific agenda, but only to
educate, and to
encourage people to judge un-dominated and for themselves,
about any organizations or zombie truth, or individuals.
===
BY
DR. ZAKIR NAIK
http://www.irf.net/irf/download/index.htm
Question:
Why are most of the Muslims fundamentalists and terrorists?
Answer:
This question is often hurled at Muslims, either directly or
indirectly, during any discussion on religion or world affairs. Muslim
stereotypes are perpetuated in every form of the media accompanied by
gross misinformation about Islam and Muslims. In fact, such
misinformation and false propaganda often leads to discrimination and
acts of violence against Muslims. A case in point is the anti-Muslim
campaign in the American media following the Oklahoma bomb blast,
where the press was quick to declare a 'Middle Eastern conspiracy'
behind the attack. The culprit was later identified as a soldier from
the American Armed Forces.
Let us analyze this allegation of 'fundamentalism' and 'terrorism':
1. Definition of the word 'fundamentalist'
A fundamentalist is a person who follows and adheres to the
fundamentals of the doctrine or theory he is following. For a person
to be a good doctor, he should know, follow, and practise the
fundamentals of medicine. In other words, he should be a
fundamentalist in the field of medicine. For a person to be a good
mathematician, he should know, follow and practise the fundamentals of
mathematics. He should be a fundamentalist in the field of
mathematics. For a person to be a good scientist, he should know,
follow and practise the fundamentals of science. He should be a
fundamentalist in the field of science.
2. Not all 'fundamentalists' are the same
One cannot paint all fundamentalists with the same brush. One cannot
categorize all fundamentalists as either good or bad. Such a
categorization of any fund amentalist will depend upon the field or
activity in which he is a fundamentalist. A fundamentalist robber or
thief causes harm to society and is therefore undesirable. A
fundamentalist doctor, on the other hand, benefits society and earns
much respect.
3. I am proud to be a Muslim fundamentalist
I am a fundamentalist Muslim who, by the grace of Allah, knows,
follows and strives to practise the fundamentals of Islam. A true
Muslim does not shy away from being a fundamentalist. I am proud to be
a fundamentalist Muslim because, I know that the fundamentals of Islam
are beneficial to humanity and the whole world. There is not a single
fundamental of Islam that causes harm or is against the interests of
the human race as a whole. Many people harbour misconceptions about
Islam and consider several teachings of Islam to be unfair or
improper. This is due to insufficient and incorrect knowledge of
Islam. If one critically analyzes the teachings of Islam with an open
mind, one cannot escape the fact that Islam is full of benefits both
at the individual and collective levels.
4. Dictionary meaning of the word 'fundamentalist'
According to Webster's dictionary 'fundamentalism' was a movement in
American Protestanism that arose in the earlier part of the 20th
century. It was a reaction to modernism, and stressed the
infallibility of the Bible, not only in matters of faith and morals
but also as a literal historical record. It stressed on belief in the
Bible as the literal word of God. Thus fundamentalism was a word
initially used for a group of Christians who believed that the Bible
was the verbatim word of God without any errors and mistakes.
According to the Oxford dictionary 'fundamentalism' means 'strict
maintenance of ancient or fundamental doctrines of any religion,
especially Islam'.
Today the moment a person uses the word fundamentalist he thinks of a
Muslim who is a terrorist.
5. Every Muslim should be a terrorist
Every Muslim should be a terrorist. A terrorist is a person who causes
terror. The moment a robber sees a policeman he is terrified. A
policeman is a terrorist for the robber. Similarly every Muslim should
be a terrorist for the antisocial elements of society, such as
thieves, dacoits and rapists. Whenever such an anti-social element
sees a Muslim, he should be terrified. It is true that the word
'terrorist' is generally used for a person who causes terror among the
common people. But a true Muslim should only be a terrorist to
selective people i.e. anti-social elements, and not to the common
innocent people. In fact a Muslim should be a source of peace for
innocent people.
6. Different labels given to the same individual for the same action,
i.e. 'terrorist' and 'patriot'
Before India achieved independence from British rule, some freedom
fighters of India who did not subscribe to non-violence were labeled
as terrorists by the British government. The same individuals have
been lauded by Indians for the same activities and hailed as
'patriots'. Thus two different labels have been given to the same
people for the same set of actions. One is calling him a terrorist
while the other is calling him a patriot. Those who believed that
Britain had a right to rule over India called these people terrorists,
while those who were of the view that Britain had no right to rule
India called them patriots and freedom fighters.
It is therefore important that before a person is judged, he is given
a fair hearing. Both sides of the argument should be heard, the
situation should be analyzed, and the reason and the intention of the
person should be taken into account, and then the person can be judged
accordingly.
7. Islam means peace
Islam is derived from the word 'salaam' which means peace. It is a
religion of peace whose fundamentals teach its followers to maintain
and promote peace throughout the world.
Thus every Muslim should be a fundamentalist i.e. he should follow the
fundamentals of the Religion of Peace: Islam. He should be a terrorist
only towards the antisocial elements in order to promote peace and
justice in the society.
===
Tell me Abdul. Was 9/11 a stereotype?
Why are nine out of ten suicide bombings committed by Muslims?
Why are Muslim parts of the world so damned violent?
9/11 taught me all I need to know about Islam.
There is nothing wrong with Islam that genocide would not cure.
Bob Kolker
===
Oh, what about all the acts of violence against non-muslims? I suppose
muslims do not consider that as violence? Seems to me muslims think they
are above everyone else.
===
They are the latest manifestation of the the chosen people meme.
They consider it their sacred duty to make war on the dar al Harb.
Bob Kolker
===
Subject: place values
We of course are all very comfortable and familiar with the
following...
that the first slot or place left of the decimal point is for the
ones, and that the second place is for the tens, and the third place
for the hundreds, and so on.
But why is the first place to the right of the decimal for the tenths
and not the oneths?
===
Subject: Re: place values
I agree with Rainer.
It goes hundreds, tens and units, tenths, hundredths etc
Oneths equivalent to units.
Power of 10
3 2 1 0 | -1 -2 -3
Hund Tens Units Tenths Hund
reths
In a sense the units are the centre of things.
.1 is 1/10 and as I have indicated the reciprocal of 10.
1 is its own reciprocal.
Nick
===
Subject: Re: place values
Because a oneth is a one. See: 1/1 = 1.
And as the ones already have their place, it would be unfair to
offer them a second place, wouldn't it?
And then there are historical reasons:
We all learned that Pi = 3.1415926. Think of all the money and confusion
spent on rewriting this as Pi = 3.31415926.
And there would be other threads like Is 0.9999... really 1?, asking
Is 0.09999... really 1? or even worse: Is 0.09999... really 1.1?.
And after the revolution someone might stand up and ask: Why the hell
do I have to believe that 1 = 1.1?.
Rainer .
. .
_________________________.___.___._______________________
Rainer Rosenthal . . r.rosenthal@web.de
.
===
Subject: Re: place values
You have for a number x the decimal expansion:
x = a_n * *10^n +...+ a_3 * 10^2 + a_2 * 10^1 + a_1 * 10^0 + a_{-1} *
10^{-1} + a_{-2} * 10^{-2}...
written as
x = a_n...a_2 a_1 a_0 , a_{-1} a_{-2} ..
There are no oneths as you call them.
Ciao
Karl
===
Subject: Need a little help integrating sin(x^2) across the positive reals
I'd like a little help integrating sin(x^2) from 0 to infinity.
Of course sin(x^2) = Im[exp(ix^2)] so the key seems to be to do a
contour integral of exp(iz^2). I'm attempting to do it over a pizza
slice region (from 0 to R, curve around to R*exp(i*pi/4) and back
straight to 0) and the curve on the circle is giving me trouble.
The integral for this portion of the curve ends up being
integral from 0 to pi/4 of exp(i*(R^2)*exp(i*2t))*i*R*exp(i*t) dt
By plugging in a few large values of R in Maple and integrating
numerically it's clear that the integral goes to zero as R goes to
infinity, but how can you show this? Attempting to simplify the
integrand just results in a nested trig mess.
Mark
===
Subject: Re: Need a little help integrating sin(x^2) across the positive
reals
Integral is 1/2 sqrt(pi/2). For your approach, see
For a different method, see

===
Subject: Re: Need a little help integrating sin(x^2) across the positive
reals
This is a so-called Fresnel integral. The method to do these found
uses a rectangular contour.
J. Caldwell, Three Integrals Math Gazette 31 (1947) 239--240
===
Subject: Re: Need a little help integrating sin(x^2) across the positive
reals
I suspect that the answer might be 1/2*sqrt(pi/2), not zero. However,
I just cheated and looked at integrals.wolfram.com rather than doing
anything clever.
===
Subject: Integrals don't necessarily preserve strict inequality?
Of course if f <= g then we have int f <= int g, but is the statement
with strict inequalities throughout not true in general for just
integrable functions? If f and g are *continuous* on the interval then
the statement with strict inequalities definitely holds, but I'm
having a hard time thinking of a non-continuous counterexample where
we have f < g but int f = int g.
Mark
===
Subject: Re: Integrals don't necessarily preserve strict inequality?
days. My association with the Department is that of an alumnus.
If h(x) is strictly positive on the interval, and is integrable, can
the integral be equal to 0?
--
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes by Bill Watterson)
Arturo Magidin
magidin-at-member-ams-org
===
Subject: Re: Integrals don't necessarily preserve strict inequality?
If f(x) < g(x) for all x, then int f < int g.
--
A.
===
Subject: Re: Integrals don't necessarily preserve strict inequality?
Jose Carlos Santos
===
Subject: Re: Integrals don't necessarily preserve strict inequality?
hi
I think your example isn't good, since f < g doesn't hold on the whole
domain.
As far as I remember.
Yours,
istvan
Jos? Carlos Santos ?rta:
===
Subject: Re: Integrals don't necessarily preserve strict inequality?
My examplo is good or not according to the definition of f < g. I
thought that what the OP had in mind was that f < g meant f <= g and
_f_ different from _g_. But if what he meant was that f(x) < g(x)
for each _x_, then you are right, of course.
[0,1] is equal to the union of all A_n. So, some A_n must have positive
Lebesgue measure _m_ and therefore
Jose Carlos Santos
===
Subject: Re: Integrals don't necessarily preserve strict inequality?
On Fri, 09 Mar 2007 14:42:53 +0000, Jos.8e Carlos Santos
I haven't taken any measure theory yet but vaguely understand the
concepts. I assume all the A_n are measurable and that's why it
follows that some A_n must have positive Lebesgue measure _m_ (since
then we get 1 = m([0,1]) <= sum m(A_n)) but can you explain why each
A_n is measurable?
Mark
===
Subject: Re: Integrals don't necessarily preserve strict inequality?
On Fri, 09 Mar 2007 14:42:53 +0000, Jos.8e Carlos Santos
Sorry for the confusion. I meant what istvankaster said, i.e. f(x) <
g(x) for all x in the domain.
This was a midterm question (actually, it asked the analogous question
for double integrals, and I can't see how the situation would be any
different there.) I answered that the statement holds, but got marked
wrong for it, and just wanted to make sure I wasn't crazy before
arguing it.
Mark
===
Subject: Locally Compact Subspaces
I'll be honest. I've thought about this for quite some time and
cannot come up with an example. I would appreciate it if someone
could provide one.
Define a subspace of the real line that can be represented as the
union of two locally compact subspaces, one of which is closed and the
other open, and that this is not a locally compact space.
Just to provide a definition: A topological space X is called a
locally compact space if for every x in X there exists a neighborhood
U of the poitn x such that the closure of U is a compact subspace of X.
===
Subject: Re: Locally Compact Subspaces
Take A = (0,1/2) U (1/3,1/4) U (1/5,1/6) U ... and B = {1}. Then A is
open, B is closed, A and B are locally compact, but their union isn't.
Jose Carlos Santos
===
Subject: Re: Locally Compact Subspaces
compact with a singleton sufficiently far from your open set that
yields this result? The problem arises when you union the singleton
with everything else?
===
Subject: Re: Locally Compact Subspaces
My singleton was most definitely *not* far way from my open set. The
point of the singleton is located at the _boundary_ of my open set.
But, of cours, the answer to your first question is negative. If
A = (0,1) and B = {x}, then the union of A and B is locally compact, no
matter where _x_ is.
I do not understand your second question.
Jose Carlos Santos
===
Subject: Re: Locally Compact Subspaces
On Fri, 09 Mar 2007 10:25:39 +0000, Jos.8e Carlos Santos
I suspect that there were typos in the definition of A _and_ the
definition of B - probably if you gave the definition you meant
then these most definite properties would be more clear...
************************
David C. Ullrich
===
Subject: Re: Locally Compact Subspaces
*Of course* there were typos! :-(
What I meant was A = (0,1/2) U (3/4,5/6) U (7/8,9/10) U ...
Jose Carlos Santos
===
Subject: necessary hypothesis
closed?
proper map: the inverse image of every compact set in Y is a compact
set in X;
closed map: the image of every closed set in X is closed in Y.
Sauro
===
Subject: Borel bijection
A 1-line question :
be Borel ?
===
Subject: Re: Borel bijection
A 1-line answer:
yes
A reference: D. Cohn, MEASURE THEORY, Birkhauser. In the chapter on
analytic sets.
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Cantor Confusion
For the cross section we would not even need a limit. It is enough to
know that at every level n (with finite n) the number of paths (-
bundles) is countable. That is sufficient because we know that every
path consists only of nodes (or edges) at finite places, i.e.,
enumerated with finite numbers n. How many of these levels are
populated by paths is completely irrelevant.
Again and again the creed is muttered.
There is no need to define it. It is sufficient to know that EVERY
cross section is countable.
===
Subject: Re: Cantor Confusion
And this tells us precisely nothing about C(oo), which you *do* use.
I thought you were trying to find a contradiction in set theory? Unless
you have proven yourself that that is not the case, it will stand.
No. That is not sufficient. When I ask you to define C(oo) you state
that it is a cardinal number (namely aleph-0). By what way that will
count the number of paths in the infinite tree escapes me.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
===
Subject: Re: Cantor Confusion
A very good example of M.9fckenmathics! :-)
F.
--
===
Subject: Re: Cantor Confusion
Yes, I know, that's your wrong belief. Everything in the world looks
yellow.
The number of paths which are generated by branchings is not counted
by sequences of branchings but by branchings.
===
Subject: Re: Cantor Confusion
C(n) is the cardinality of level L(n). There is no largest n.
Therefore this holds throughout the whole infinte tree. At no finite
position the infinite paths can be more than countably many.
Either there is an infinite index. Or there are less than uncountably
many paths.
Otherwise limits do not make sense.
For EVERY LEVEL which a paths can populate (if it can populate only
every finite level).
And after death you will enter the paradise and all pain will end and
you will no longer need any logics. But as long as you are here in the
vale of tears, there is logic dictating the facts.
===
Subject: Re: Cantor Confusion
Nntp-Posting-Host: hera.cwi.nl
...
Ok, so C(oo) is the cardinal number aleph-0.
So T(oo) is a set of nodes.
And you mean *that* with C(oo) is part of T(oo). A strange formulation,
but correct.
And how do you conclude from this that the number of paths is countable?
Ah, you need *four* definitions here? Only two would be sufficient, I
think. Because none of the definitions follows from any of the other
definitions.
definitions. Would it not be possible that your definitions are not
consistent? It is your last paragraph which is inconsistent with the
other definitions. See how you *did* define:
which means (as |{1,2,3,...,n}| = n:
Why not?
Why? I see only that C(oo) is aleph-0, but I see no relation between
paths and aleph-0.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
===
Subject: Re: Cantor Confusion
Isn't it difficult to look from this perspectice? The cross sections
*are* numbers of paths.
1) We know tha every digi of a real number stands on a finite place.
(We know that here is no digit the last one, but that is here
completely irrelevant.)
That means, there is no part of a path which would jut out f the tree.
3) All paths which are in the tree cross each cross section (that is
why it is named cross section).
There is no path or even splitting of paths outside of every cross
section.
Therefore, the limit processes for paths-lengths and cross sections
are identical.
===
Subject: Re: Cantor Confusion
...
...
Do you not have a comment on this?
No, they are *not*. The cross sections are sets of nodes (by your own
definition), except for C(oo), which is a cardinal number (again by
your own definition). And I see *no* relation between aleph-0 and
the paths.
Indeed.
Indeed.
Except for C(oo), because that is not a set of nodes, so there is *no*
path that crosses it.
How do you *define* that limit process? Your continuous mentioning limits
and limit processes show that you do not understand how things work. T(oo)
is the union of a collection of trees T(n) (where each tree is a set of
nodes). In the *definition* of that union there is no limit involved.
Also the infinite paths are unions of paths (again, these are sets of
nodes). And again, there is no limit involved in that definition. There
is *no* standard definition of limits where the limit of path lengths is
aleph-0. You have to provide that definition. There is also no standard
definition of limits that allows you to take the limit of cross sections
(which would be a set of nodes), so you have to provide that definition.
Finally, you have to *prove* that what the limit (by your definition)
gives is also the actual value.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
===
Subject: complexity of deadlock problem
Dea all,
What is the best results of the complexity of the deadlock problem
in Petri Nets?
I could only find a paper Complexity results for 1-safe nets which
is related with the complexity of Petri net problem and published in
1995. Maybe there are some improve after this paper?
-------------------------
Zhu
===
Subject: Grow your Social Network - Meet New Girls
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===
Subject: Re: Review of Mueckenheims book.
Correct.
The sequence of all
That is true, but what the projection does does is not at all
related to the question how many, because what the projection
does with number the number 6 alone is exatly the same, namely 6.
Therefore your addtional answers cannot be used to improve your score.
Correct.
Correct.
Correct.
Right answers to all four test questions! Highest score!
(Note also that the answers concerning the Waft Maximum would have
been identical.)
And in fact, we are not interested in that question. But it is obvious
that the projection of a single number 6 is the same as the projection
of infinitely many numbers if al are of size 6.
I know that it totally irrelevant what the projection does
concerning the number of elements of the set projected. That is why I
devised it!
You have successfully learned that what the projection does with the
EIT is finite, not infinite. Now apply your knowledge to the tree.
There are many (some say: infinitely many) finite paths in the tree.
But what does the projection do with these finite paths if you put
them one behind the other? When you look at your answers above, this
will be easy to figure out.
And then you may ask yourself how many subsets a set of finitely many
nodes is able to produce.
===
Subject: Re: Review of Mueckenheims book.
Which of the following statements do you disagree with?
-the question is does the union of all finite trees
contain infinitely many nodes?
-knowing that the Waft Maximum is finite does not tell
us that there are a finite number of nodes
-the Waft Maximum is irrelevant
Since there are infinitely many nodes, this question
is irrelevant.
-William Hughes
===
Subject: Re: Review of Mueckenheims book.
No handwaving please!
You said that all natural numbers are finite.
You have found out that the Waft Maximum of N, namely WM(EIT), exactly
describes this fact.
WM: 4) What does the projection do with all lines of the EIT?
WH: what the projection does is < aleph_0 which is finite.
Now take any path of any finite tree T(n): How many nodes do you
count? n, of course. In order to obtain the answer to the question
whether there is any infinite path in the union of all finite trees
(which is the same as the question whether there is any infinite line
in the union of all finite lines of the EIT) we can apply the very
valuable tool I devised solely to support your understanding (I don't
need it to see the result).
Now think again.
In fact, that is the question. Although the more direct question is,
are there infinite paths?
Oh William, you're disappointing me. We can immediately use your
result of exercise 4, because it does not matter whether there is one
line of length n in the projected set or 2^n such lines of length n.
You see this important fact, don't you?
Stamping your feet will not solve the problem.
===
Subject: Re: Review of Mueckenheims book.
proofs, one by contradiction and one a direct proof. I repeat them here:
--------------------------------
Here is a proof by contradiction:
I. ASSUME that S is a finite set of ALL primes.
Blah blah blah ... then P is a prime which is not a member of S.
This CONTRADICTS the above assumption that S is the set of all primes.
Therefore, the assumption is impossible, and there can be no finite set
of all primes.
Here is a proof that is NOT a proof by contradiction:
II. Let S be a finite set of primes (NOT necessarily the set of all
primes).
Blah blah blah ... then P is a prime which is not a member of S.
Therefore, given any set of primes, there is a prime which is not in
that set.
--------------------------------
uninteresting, because every one knows it. The question is what Euclid
did assume. So let's look at Euclid's proof -- which one of my
skeleton proofs is closest to Euclid's argument? I don't see how you
can avoid the conclusion that it is II and not I.
Euclid begins:
Let A,B,C be the assigned prime numbers.
This is exactly analogous to my statement in I: Let S be a finite set
of primes. It is in no way analogous to my statement in II: Assume
that S is a finite set of ALL primes. Then Euclid presents an two-part
argument (my blah blah blah), after which he concludes that G is a
prime not in the original set. This is again exactly analogous to my
conclusion in II then P is a prime which is not a member of S. It is
NOT at all analogous to reaching a contradiction and concluding that
some earlier assumption is wrong.
What's muddying the issue is that in PART of his proof (the second part
of the blah blah blah part), he does use a proof by contradiction, to
prove what might be called a lemma, that ABC+1 cannot be divisible by
any of A, B or C. But the structure of the entire proof is not by
contradiction, it is a direct proof.
So one can truthfully claim that Euclid did sometimes use proofs by
contradiction. He used one in the lemma in part 2 of this proof. But
one cannot claim that the entire proof of Proposition 9.20 is by
contradiction. It is a direct proof, exactly like my II, and not at all
like my I. Modern proofs of this theorem often look like I, but
Euclid's did not.
--Mark
===
Subject: Re: Review of Mueckenheims book.
f=H|Ax0, g=H|Ax1. Of course, there must be a context to make sense out of
this, for example the category of topological spaces and continous
mappings. So two mappings are homotopic if there is a 1-parameter-family of
mappings connecting them. A mapping is homotopically trivial if it is
homotopic to a constant mapping. One wants homotopy to be an equivalence
relation on sets or spaces of mappings. If you allow for codomains of
mappings to grow then this will spoil the relation of homotopy, because on
Ax(0,1) the homotopy may move around in absolutely unlimited ways instead
of being restricted to a specific codomain.
I already admitted a while ago that your definition may be customary in set
theory. However, in homotopy theory (and anything which anyhow depends on
anything homotopical, and this is a lot of mathematics, touching to
varying extent the fields of algebraic and geometric topology, homological
algebra, K-theory, algebraic geometry, quantum algebra, algebraic number
theory, category theory and representation theory) this kind of definition
of function or mapping simply does not work. There is a gadget in homotopy
theory (the mapping cylinder construction) to turn any mapping into a
homotopically equivalent inclusion. If you can't distinguish inclusion
from identity then you have trivialized homotopy theory.
On the other hand, I don't see why a definiton which includes the
specification of a codomain should not work in set theory. One could call
what is different by different names. But as they now turn out to be, I
think these terminological habits are unfortunate.
It is somehow like the distinction between a singleton set and the element
it contains. Neglecting that distinction trivializes set theory in a way
similar to the above-mentioned trivialization of homotopy theory. When
group theorists calculate with complexes (here meaning sets of elements of
a group, elsewhere it has very different meanings) then, however, they seem
to neglect the difference between a singleton complex and an element.
These are all matters, btw, which need to be handled if one wants to
completely formalize a piece of mathematics, and experiences like this with
the definition of function or mapping let me think that there are quite a
number of unwelcome surprises waiting for anybody who actually wants to
carry through such formalization. These matters are not really substantial,
but they need to be handled. Moreover, terminology is also a matter of
personal taste, and therefore eagerly discussed:-)
Ralf
===
Subject: Re: Review of Mueckenheims book.
Again, I never disputed that for certain purposes it is not required
to specify a particular codomain. And I don't allow codomains to
grow; grow is not a fair characterization (or at least I fear it
may be quite misleading). And I even said that there are plenty of
contexts in mathematics where mathematicians use all kinds of
different terminology.
If you look at my precise statements, I think you will see that what
you've said about those topological investigations does not contradict
anything I've said.
I don't know since I have not yet studied much homotopy beyond my
casual skimming of later chapters of books I want to eventually study
fully systematically. But, again, I said long ago that I recognize
that terminology in mathematics may vary greatly from context to
context. Again, if you look at precisely what I said, your discussion
about homotopy theory doesn't contradict what I have said, since I
never said that the garden variety set theoretic definition is the
ONLY definition, but only that it is the standard definition in set
theory (true), found in many texts in mathematics (true; I never
claimed even that it is the majority definition or even that it is
much less commmon than the majority definition), and that the garden
variety definition of 'f is a function' can be derived from the garden
variety defintion found in math books of 'f is a function from D to
C' (true, and I proved it).
I agree completely. I have been saying ALL ALONG that you can have
BOTH specifications. You can say 'f is a function'; you can say 'f is
a function from D to C'. And, as David and I have been discussing, you
certain codomain. I just disagree with David about whether f is the
mathematics, since I can't make a sweeping declaration as to all kinds
of mathematics that is beyond my level. My proposal allows BOTH
function-system.
Not only trivializes, but allows plain contradiction. And so, were I
to need to make such a distinction when I get to homotopy theory, I
definitely would not fail to make it.
This is a project I am working on. I surely have not got as far as
homotopy theory; so I look forward to whatever challenges it provides,
both as a subject onto itself and as project for formalization. and, I
think you'd agree that as we formalize, the results make look quite
foreign to the specialist in the particular field of advanced
mathematics. The homotopy theorist would say, My what an abomination!
I sure don't work with formulas like that! But that is a reminder
that the project of formalization is not for the purpose of the
homotopy theorist or for homotopy theory itself. Do you agree? I'm
interested in your thoughts about this kind of thing.
I think it would be nice to have some uniform terminology that we
could apply so that it is truly monotonic - we'd give a definition of
'is a function' and of 'f is a function from D to C' and it would
stick all the way through mathematics to the far reaches of the most
advanced studies, and if we needed to make sure a certain codomain is
associated, then we'd add a definition of a 'function-system', etc.
But I do realize that likely that would be impractical and would force
mathematicians into contrived terminology that would only slow them
down. So if the homotopy theorist wants to speak of the function as
coming with a fixed codomain, that is fine, since all we have to do is
appreciate the context and know that we can reconcile it in our minds
(even if not in official standards) the terminological differences.
What do you think?
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
In fact, you insist on it for set theory.
It is the fact that for many if not most purposes, specification of a
codomain /is/ required that you have been finding objectionable.
===
Subject: Re: Review of Mueckenheims book.
I'm pretty sure that the logicians/set theorists use their definition of
function just as a matter of convenience. For what they are doing,
carrying the codomain around isn't useful and adds needless complexity,
so they just drop it. If they need it for something, it is easy enough
to put it back in. Of course, this doesn't imply that the rest of
mathematics should follow suit.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
Indeed it is often CRUCIAL to specify a particular codomain. Just as
in that Halmos proof of the definition by recursion theorem. We want
to prove that not only does there exist a function f on omega but,
quite crucially, that f is into omega. But we don't put any codomain
back in. We just say exactly that we are proving both that f is a
function and that its range is a subset of omega. It's a conjunction:
f is a function AND the range of f is a subset of omega.
And there is no suggestion whatsoever that any mathematician should
not specify any codomain he wishes to specify or even that a
mathematician may not just take the whole thing as a bundle (which
we've discussed as formalized by a triple) with a given codomain.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
With infinitesimal angles, of course. So would I. Is it still before
breakfast?
What if it's treated as an operator?
One might even before that, or might not.
TO
===
Subject: Re: Review of Mueckenheims book.
Only for red queens like TO.
What definition of an operator could an order relation possibly fit?
None that occur in standard mathematics.
WM might, as he never proves his claims either.
===
Subject: Re: Review of Mueckenheims book.
WM would never be so ridiculous!
TO
===
Subject: Re: Review of Mueckenheims book.

Since TO often has been even while denying it, he is hardly a fit judge.
===
Subject: Re: Review of Mueckenheims book.
No, TO agrees that thee can never be any explicit well ordering of the
reals, only a proof of conjecture based on choice. However, where
uncountable sequences are allowed, indeed P(N) can be sequenced ala the
T-riffics. The only thing that makes them or the H-riffics not a well
ordering is the existence of infinite descending chains, as are always
required in any explicit uncountable sequence. Oh well.
Eeny meeny miney moe...
===
Subject: Re: Review of Mueckenheims book.
Since sequencing implies each element has an immediate successor which
implies well-ordering, TO claims the impossible for his T-erribly
un-T-riffics.
===
Subject: Re: Review of Mueckenheims book.
Okay. I guess the first reference I saw to the kumquats in question was
by you, but I must have missed the first mention. Sorry.
WM, you don't disagree that there are infinite sets containing just
finite values, such as the reals in [0,1], are you? I certainly agree
that an infinite set of naturals must contain infinite values, but
that's only because they are spaced apart by a unit in value. Isn't tat
your thinking?
Tony
===
Subject: Re: Review of Mueckenheims book.
If you disregad physical restrictions, then there are infinitely many
real numbers in the interval. Their cardinality, however, is not
larger than infinite for any set. Therefore we need no alephs etc.
The binary tree shows that different alephs are self contradictive.
If you take into account the physical restrictions, then there is no
infinite set. And that is the only correct approach.
===
Subject: Re: Review of Mueckenheims book.
I have pondered about that question some time (motivated by the
inability of many cranks to comprehend that state of affairs).
Many cranks (M.9fckenheim for example) claim that (from a logical point
of view) it's not possible for a set of natural numbers to be infinite
if it's only containing finite numbers. Well, let's see...
Assume (for the sake of the argument) that in the set of _all_ natural
numbers, denoted by IN*, there were infinite natural numbers (though I
don't have the slightest idea how such numbers would look like). Now
let's construct [or rather define] a set IN the following way:
IN := {n e IN* | n is a finite number}.
Then IN is the set of all finite natural numbers. It MUST be, since it
is constructed that way. You might think of it the following way too:
IN := IN* - {n e IN* | n is infinite}
i.e. IN is the set of _all_ natural numbers /minus/ (set theoretic minus)
_all_ the infinite natural numbers that are in IN*. Hence IN is the
reminder: the set of all natural numbers that are not infinite, hence
finite.
Now the crucial question just is: Is IN finite then? C a n it be finite?
If the cranks were right, it _should_ be! After all it consists of exactly
all the natural numbers that are finite.
So let's assume that IN only contains finite many elements. Our
construction of IN ensures that all this numbers are finite. Hence there is
a biggest element in IN, say m, and this element is finite. But then m+1
certainly is also a finite natural number, hence must be in IN, and is
bigger than m. Contradiction! Hence IN must contain infinitely many
elements. (But they all _have to be_ finite by our construction of IN.)
This _proves_ that (for example) M.9fckenheim's claim
Unless there is an infinite number the number of [natural]
numbers
[...] cannot be infinite. (W. M.9fckenheim)
is simply wrong.
F.
--
===
Subject: Re: Review of Mueckenheims book.
Define infinite values.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
I have seen cold water in the ear do wonders, but that's not possible in
this forum. :)
TO
===
Subject: Re: Review of Mueckenheims book.
Or it would long since have been applied to TO.
===
Subject: Re: Review of Mueckenheims book.
You may have said that, but then you're in the minority, from what I've
seen around here. I meant you in a more generic sense.
If an uncountable sequence is acceptable, then why can one not have such
a sequence of infinitesimals constitute a line? I think you knew what I
meant when I said a linear arrangement anyway, since were were speaking
of the naturals. What are we supposedly arguing about?
Okay. I don't need Diet Coke.
Do what better? Define the naturals? Sure, the definition goes beyond
Peano, if that's what you mean, but the sequential nature of the set
is established up front.
above using a statement employing first order logic and '<'. Read over,
And then say:
It's just an observation for subsequent application...relax
See? I used it to justify the definition of =. :)
The mathematical definition of the symbol consists of the rules of
symbols, that's how they work.
Relax.
Not until you adhere to the rule that set size increases with additional
elements.
It's Orlovian babble, if you would.....
Sure, okay.
===
Subject: Re: Review of Mueckenheims book.
I don't account for what other people post. All you need to do is
crack some set theory textbooks to see the ones in which sequences on
infinite ordinals are discussed. Such discussion is RIFE in set
theory. There are finite sequences, there are denumerable sequences,
and there are also sequences on infinite ordinals. And I told you
about that in SEVERAL posts to you. Just inform yourself: Read a good
set theory textbook already!
A SEQUENCE, as it is DEFINED is on a natural number, on omega, or on
an infinite ordinal other than omega. Those are all sets that have a
well ordering, in particular by membership. (I think, perhaps people
sometimes also refer to sequences on other sets that have well
orderings.) We call functions on an ordinal a 'sequence', either
finite, denumerable, or infinite. Now, I guess you could expand that
to let any well ordered set be the domain of a sequence. For a well
ordered set S we could speak of an S-R-sequence where S is a set well
ordered by R. I think that is even done sometimes (I'm not sure), but
there would be no harm in our making such a definition.
So, to have a sequence involving your infitesimals, a specific well
ordering R for your set S (with S having some or all of its members
being infinitesimals) would be needed. Then it would be okay to refer
to an S-R sequence as we'd define per what I mentioned above.
No, truly, your terminology is often opaque to me.
What I meant is that we can give better instructions than just citing
a mathematician (without even saying which formulations of his are in
particular the answer) whose work has been revised and formalized
since so as to distinguish between first order and second order, as we
may give the much better instruction to look at the exact formulations
and proofs in certain theories.
My reading-typo, sorry. But now your definition is even MORE
ridiculous. You can't define '<' by using '<' !.
You're mixed up already. We are defining '<'. So we can't START with
'<' since it is not defined.
Start with on or the other, in terms of NEITHER. Then define the
second one from the first one.
You've defined NOTHING. You're CIRCULAR.
in terms of '<'.
But you can't define EITHER by just shuttling logic symbols back and
forth between them.
I really am having a hard time fathoming that you STILL don't know
what a circular definition is. And you can't define '=' from '<' and
means you do recognize how absurd your method of defining is?
I DID read on in your post. It's a crying shame how you maintain your
ignorance.
BULL. Mathematical definitions cannot be CIRCULAR.
You're so relaxed you can't even discern blatant circularity. Forbid I
should be so relaxed.
You're really joking now, right?
What I gave is PURELY syntactical. I reverted a formula to primitive
notation. That has NOTHING to do with set size increases with
additional elements.
What is your point?
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
Don't get your panties in a twist. Look back, and tell me what
question I didn't answer. I certainly responded. Chill out.
===
Subject: Re: Review of Mueckenheims book.
Picture the number line. Are there an infinite number of points closer
to 0 than any finite number? If we move through any finite number of
contiguous points along the line, we will not move any finite distance
from the origin. Can we move through an infinite number of points, and
still remain infinitesimally close to the origin? That's another way to
look at it.
T.
===
Subject: Re: Review of Mueckenheims book.
To both G. Frege and Tony Orlow:
I'm sorry, G. Frege, that I didn't see your post until it came up
quoted by Orlow. (And I haven't forgotten that I owe you some comments
about A. Robinson; It's been hard to get around to, and, I actually
don't think his paper is that interesting to me, at least not at this
particular time in my moodswings of what math and philosophy
interest me.)
Anyway, I saw your argument below. If I understand, you want to lead
the crank to a contradiction that IN is finite while it is clear from
construction that IN is infinite? So we're supposing that the crank
agrees that *IN is infinite, right? So I take it that the crank agrees
(as Orlow seems to) that there are infinite sets. I take it that
infinite). However, I think your argument relies upon the lemma that
an infinite set minus (set complement operation) a finite set leaves
an infinite set. To have your crank checkmated, you have to make sure
he agrees with that lemma.
What number line? The real line? The standard ordering on natural
numbers? Some ordering on *IN? What ordering on *IN is that? An
ordering on IN? Which?
What ordering on what set are you talking about. IN? *IN? WHAT
ordering do you have in mind?
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
I see. Let's forget about this paper (for the time being). Btw my own point
of view is a fictionalistic one. Over time I came to the conclusion that
this point of view (1.) is in complete agreement what mathematicians really
do (or better HOW the do it, how they are thinking about mathematical
objects) and (2.) also is acceptable from a philosophical point of view.
It seems to me that, if you reject Realism (mathematical Platonism) the
only other reasonable position is Fictionalism (if we do not take into
account Intuitionism, of course). But n o t Formalism. (Actually, in my
opinion most mathematicians who _think_ that they are Formalists, or that
this might be the right view from a philosophical point of view, actually
are Fictionalists --- without knowing it! ;-)
Not really.
I considered this question to be of minor importance/relevance here.
What I wanted to address _directly_ is the claim that there must be
infinite numbers in the considered set (if it is to be infinite).
Right. (It's just an assumption, if you like.)
Very good.
No. Not at that step of the argument. (Of course, finally we have to
consider the question if the remaining set is finite or not.)
F.
--
===
Subject: Re: Review of Mueckenheims book.
I'm drawn to fictionalism too. But I am very much wary of basing a
philosophy (or, really, my own philosophical explanations to myself)
on how or even why mathematicians do what they do. I recognize that
that has to be considered of course, but I think there may be greater
considerations for a philosophy of mathematics than how the philosophy
lines up with the everyday heuristics or naive ad hoc ontology of
mathematicians not necessarily so concerned with philosophy of
mathematics.
Then I don't get it. Or maybe I phrased myself poorly. I thought you
want to show the crank that his view of IN being finite contradicts
the ineluctable logic that IN is infinte. No?
Okay, I understand that.
I'm not following you again. How do you infer that IN is infinite
without the lemma that an infinite set minus a finite set leaves an
infinite set?
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
Putting aside the problem of sentence 1, that we really have no idea
what the Orlovian number line is supposed to be, look at the second
sentence, then please answer the following.
Picture the contiguous states of the US. Is there a set of three
contiguous states not touching any state?
Q: How many correct answers can you give to this question? If the
person asking the question insisted you say Yes or No, then proceeded
to argue about a lot of other stuff, do you forsee this being
productive of your time?
Hmm. Sadly a rest doesn't seem to have helped.
Brian Chandler
http://imaginatorium.org
===
Subject: Re: Review of Mueckenheims book.
The rest just lets him clear (empty) his mind. He needs to exercise it
and put some sensible thoughts in it.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
On 8 Mar 2007 21:01:22 -0800, Brian Chandler
I guess he meant the real number line.
Remains the problem with the contiguous points.
F.
--
===
Subject: Re: Review of Mueckenheims book.
You used some undefined and/or unclear terms. Sorry. I'll just skip that
part.
At what?
I was talking about infinite _sets_. (Actually about sets of natural
numbers.) There we don't _move_, sets and their elements just _are_.
F.
P.S.
You might address my argument instead of talking about unrelated things.
--
===
Subject: Re: Review of Mueckenheims book.
If TO is moving through contiguous points on a line, he must be doing
it in some sort of finite geometry, not any sort of standard geometry in
which there are no such things as contiguous points.
===
Subject: Re: Review of Mueckenheims book.
Define finite number.
Define contiguous point.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
Does anyone here disagree that the set of reals in [0,1] is infinite? I
don't. Perhaps WM claims that irrationals don't exist, and so the set is
as countable as the naturals, but I'm not of that ilk. You knew that.
So, why did you miss the entire point of the statement, I wonder?
ToeKnee
===
Subject: Re: Review of Mueckenheims book.
Hey Ross, nice to see you.
I have to disagree a little about the required restrictions on the
treatment of infinite natural numbers. In my mind, any set size is a
natural number, even if infinite. Adding a new element increments the
value to its natural successor, another infinite natural. In my mind,
greater by 1. I don't ascribe to a point at infinity, except as a
graphical expression in a number circle. I rather go with the Robinson
model, and insist that Peano operations exist there as well, as they
obviously do with adic or T-riffic numbers.
Tony Orlow
Task Force Against Mediocrity
===
Subject: Re: Review of Mueckenheims book.
I doubt anyone disputed that fact. I've certainly had it irrelevantly
pointed out to me many times.
'1' is a natural number than applies to an object. '2' applies to a pair
of objects, etc....get the idea?
Infinite measure and natural numbers.
Kumquat is standing in for
Sure, in terms of their position, or value. As far as their frequency,
no matter what the value, there is one natural per unit of value on the
real continuum. That may be considered the diameter of the natural.
They are all involved in having an infinite set of objects with some
relative measure between them, like the naturals. Why were kumquats even
brought up?
I guess you mean, you are totally devoid of understanding of whatever
meaning it has. Oh well.
Tony
===
Subject: Re: Review of Mueckenheims book.

TO himself has frequently claimed that a set with only finite naturals
in it must be finite by claiming the contrapositive that an infinite set
of naturals must contain an infinite natural.
I've certainly had it irrelevantly
I can see naturals applying to sets like singletons and pairs and so on.
To has, as usual, his cart in front of his horse. While there may be one
unit of value on the real continuum for each natural, the naturals are
the independent variable in any such correspondence.

The naturals don't have anything between them to start with. It is only
with various construction from the naturals that there can be anything
in any sense between them, or at least between their images in larger
sets.
===
Subject: Re: Review of Mueckenheims book.
Has that happened? My guess is the intended reader isn't clear on what
it means for a number to be finite. So, your construction doesn't help
them. They probably think that once you toss out all the infinite
numbers, then you only have finitely many numbers left.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
I surmise that the (crank) thinking goes something like... Suppose
all numbers are either red or green. Well, we put only green ones in a
container, then these mathematikers claim that the whole contents of
the container are red. How utterly absurd!
Brian Chandler
http://imaginatorium.org
===
Subject: Re: Review of Mueckenheims book.
On 8 Mar 2007 20:28:10 -0800, Brian Chandler
Ok, me try to be serious.
We agree that all numbers are either red or green. Then the claim of
the
cranks (if so) is rather:
If we put ALL natural numbers in a container, then
there must be green ones in it. These mathematikers
claim that the whole contents of the container are
red. How utterly absurd!
And my argument is as follows:
OK. You may be right. Now let's put out all green
numbers and just consider the rest. Can there still
be a green one in it? Of course NOT. Since we put
out _all_ of them.
Now how many red numbers are left? 0 is (as we all
know) a red one. And if n is a red one (as is well
known) n+1 is also a red one.
Well... Can there be finite many red numbers in the
container then? [...]
F.
--
===
Subject: Re: Review of Mueckenheims book.
On Thu, 8 Mar 2007 14:42:09 -0500, David Marcus
Well, they may _think_ so (at first). But a simple considerations shows
that this cannot be the case (i.e. that IN must contain infinitely many
elements).
Let's assume that IN only contains finite many elements. Our construction
of IN ensures that all this numbers are finite. Hence there is a biggest
element in IN, say m, and this element is finite. But then m+1 certainly is
also a finite natural number, hence must be in IN, and is bigger than m.
Contradiction! Hence IN must contain infinitely many elements. (But they
all _have to be_ finite by our construction of IN.)
Of course, this is a standard consideration, the only new aspect (I
think) is the construction (definition) of IN mentioned above (starting
with a set IN* which may contain infinite natural numbers, if such numbers
exist). By this construction all infinite numbers have been tossed out,
and IN only contains finite numbers as elements --- but, as our simple
consideration shows, there must be infinitely many numbers in IN.
This _proves_ that (for example) M.9fckenheim's claim
Unless there is an infinite number the number of [natural]
numbers
[...] cannot be infinite. (W. M.9fckenheim)
is simply wrong.
F.
--
===
Subject: Re: Review of Mueckenheims book.
That's the same contradiction the cranks already refuse to see. What
you say is correct, of course, but we see that your strategy just
comes back around to where we already were with the crank.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
Well, you may be right, MoeBlee. I recognized this myself already, hence I
devised a simpler argument for this last step of the argument:
IN (by our construction) contains ALL and ONLY finite natural
numbers. Hence clearly 0 (or 1 of you like) is in IN. And with
any n in IN certainly n+1 is also in IN (since if n is finite
does not have an upper bound. Hence IN is an infinite set of
finite natural numbers (since it is growing beyond all bounds
--- which it could not do if it were finite).
Again:
F.
--
===
Subject: Re: Review of Mueckenheims book.
I haven't read all of the posts you and your interlocutors have made
to say that again your strategy leads back to a postion we've been in
hundreds and hundreds of times. Unbounded implies infinite. We've been
over and over that with the crank alreday. He doesn't get it.
Alas, I'm afraid I don't see how your strategy can help. If your
strategy leads back again to 'unbounded implies infinite', then we're
just right back to where we always are were with the crank.
And remember I mentioned that your strategy depends on getting the
crank to agree that an infinite set less a finite set leaves an
infinite set. It occurred to me that that provokes just another ones
of the crank's variations: The crank thinks that there is such a
cardinality as aleph 0 - 1, which is, for the crank...finite.
We're just continually taken back to the same set of crank variations
on a theme: Unbounded does not imply infinite; aleph 0 - 1 is a
cardinality and it's finite; there is an infinite process (Zeno
machine, whatever) that has a last step, etc.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
I see. But I think we are talking past each other here. Sure, if the crank
is not willing to accept that an unbounded set of natural numbers is in-
finite, my argument will be of little help. Actually, I didn't devise it to
address THAT problem.
The question I wanted to address is: (Assuming that infinite sets are
possible.) Can a set of natural numbers be infinite without containing
infinite natural numbers?
In a nutshell:
See my comment above. (Sure, I agree with you!)
No. Here I can't agree with you. My argument (or rather the crucial first
part of it) doesn't rely on that agreement.
Of course, if he is not willing to accept (or is not able to understand)
the second (final part) of the argument (quoted above), my argument will
lead to nowhere. :-(
Haha! I see. :-)
(Hope you are wrong! ;-)
Well, I see we are really talking past each other. Right, if THIS is the
problem, my argument won't be of any help. True.
F.
--
===
Subject: Re: Review of Mueckenheims book.
Maybe a simpler argument:
IN (by our construction) contains ALL and ONLY finite natural
numbers. Hence clearly 0 (or 1 of you like) is in IN. And with
any n in IN certainly n+1 is also in IN (since if n is finite
does not have an upper bound. Hence IN is an infinite set of
finite natural numbers (since it is growing beyond all bounds
--- which it could not do if it were finite).
F.
--
===
Subject: Re: Review of Mueckenheims book.
I'd try starting with this:
N contains ALL the natural numbers (and ONLY the natural numbers).
Clearly, 0 (or 1 of you like) is in N. And, if any natural number n is
in N, then certainly n+1 is also in N (since if n is a natural number,
natural number in N.
If you can get someone to agree to that, you might have a chance.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
On Thu, 8 Mar 2007 21:28:06 -0500, David Marcus
It's funny. Reading your proposal I get the impression that you didn't see
(or appreciate) the very point of my argument (what I'm after, what I
wanted to address with it). Let me try again: If you start with:
IN contains ALL the natural numbers (and ONLY the natural
numbers).
then it is NOT immediately clear that there are no infinite natural number
in IN. (Not even to me! ;-)
Hence the construction of IN the way described above. This _ensures_
that
IN contains ALL and ONLY finite natural numbers.
~~~~~~
The rest of the arguments are (more or less) identical, of course. It's
just that MY argument would lead to the conclusion: that there is no
largest number in IN, even though all the numbers in IN are finite (since
we constructed it this way).
F.
--
===
Subject: Re: Review of Mueckenheims book.
I saw it. If by appreciate, you mean like, then I guess I didn't.
I've no idea what an infinite natural number is. It seems to me that as
soon as we start talking about such (mythical) things, we plant the idea
that such things exist. It is like talking about green unicorns. The
more we talk about them, the more I start looking over my shoulder to
see if one is there.
I think 99.8% of people are incapable of logical thought. If you want to
reason with such people, you have to be very careful.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
[Various attempts to write proofs of Total Persuasive Force...]
Oh, I think you can get Tony Orlow to agree, except that another way
to say it is that the largest natural number in N (call it P) is non-
existent. Tony then has no difficulty using P at some later point of
the argument, invoking other of its properties than non-existence.
You really can't win, you know.
Brian Chandler
http://imaginatorium.org
===
Subject: Re: Review of Mueckenheims book.
Do you want me to play devil's advocate? It would probably be more
interesting if you can entice someone else to discuss this with you!
I doubt that such people can be convinced by proofs. In fact, I doubt
that they understand what a proof is.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
On Thu, 8 Mar 2007 17:13:35 -0500, David Marcus
Sure. Actually, this was a good opportunity to formulate the final
argument. I might make use of it later. :-)
Sure. But my experience is that mathematical cranks normally are not
interested in mathematical arguments and argumentation. (If you look
closely you will see that even WM follows this pattern.)
Right.
I guess, you are right, again.
Hence I devised my argument more like a simple consideration, not like
a
proof (proper), you see.
F.
--
===
Subject: Re: Review of Mueckenheims book.
Well, if you insist:
They may all be finite when you construct N, but after you finish adding
m+1, (m+1)+1, ..., to make N infinite again, you'll end up including
infinite numbers, too. We know this because the only way that N can be
infinite is if it includes infinite numbers. That's what infinite
means.
True. You have to argue with them on their level and box them into a
corner. Most of the people arguing with them aren't good at doing this.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
Actually, there are at least two finite versus infinite definitions

to consider, one of which only applies to sets whose finiteness is
measured by cardinality and one to objects whose finiteness is not being
measured by cardinality. And there need not be any crossover.
===
Subject: Re: Review of Mueckenheims book.
On Thu, 8 Mar 2007 19:04:20 -0500, David Marcus
Huh? Where did I _add_ such elements? (I'm serious: I don't know what you
are talking about here! :-)
m+1 already IS in IN, since it is a finite number. And we constructed
(actually defined) IN as the set of _all_ finite natural numbers. This
means, if m+1 is finite it is in IN. (Note that, by our construction, ALL
and ONLY finite numbers are in IN.)
Since I don't add _anything_ I also will not end up anywhere in the
first
place. :-)
...Argl...
Hell!
Again, let me be serious: You see, you actually didn't address my argument.
You just stick to the same old _claims_. :-)
like TO or WM...)
Imho, there is no such thing as being good at doing this when arguing
with a (real) crank. Imho there's no good way to handle them. All one
can
do is stop talking to them (something Virgil and Dik obviously CAN'T do).
F.
P.S.
would deal with my argument.
Really, I tried hard to understand what the (rational) reason might be
address that line of thought.
--
===
Subject: Re: Review of Mueckenheims book.
Of course you don't. He's talking drivel, for a change, and doing it
rather well in my opinion.
Brian Chandler
http://imaginatorium.org
===
Subject: Re: Review of Mueckenheims book.
I think we managed to drive Tony away for a while. Some other partial
cranks seem to have given up and gone home. I think I could get WM to
give up if no one else replied to him. He won't reply to any of my posts
anymore, so if I was the only one replying, he'd have to talk to himself
or give up.
I think as soon as you admit that there might be infinite numbers in
there you are in trouble. I'd go back to what the natural numbers are
(from a primitive intuitive perspective). The people who are confused
seem to have a very vague notion of what a number is. As soon as we say
that there are infinitely many natural numbers, they think that infinity
must be a number. And, it must be a natural number, since it couldn't
very well be an unnatural number. It doesn't help that modern math has
all sorts of infinities that act like numbers.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
He stopped replying to mine for a while, but then, when he wasn't
getting enough action elsewhere, began again.
And all sorts of things called numbers, not all of which have mutually
compatible properties.
===
Subject: Re: Review of Mueckenheims book.
Nntp-Posting-Host: hera.cwi.nl
I am not the first one, I would think.
There are a lot of things not embedded in reality. Why do you think
different?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
===
Subject: Re: Review of Mueckenheims book.
But one of the last ones, I hope.
Because I think that there is nothing not embedded in reality.
===
Subject: Re: Review of Mueckenheims book.
Nntp-Posting-Host: hera.cwi.nl
Your hope will be vain, I hope.
In what way are, say, the singular values of a matrix embedded in reality?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
===
Subject: Re: Review of Mueckenheims book.
in vain.
By the system which stores the matrix (blackboard, paper, hard disk,
brain (if the matrix is not too large)).
===
Subject: Re: Review of Mueckenheims book.
...
He did? I must have missed that.
As far as I see that is only interpretation. He says: The prime
numbers are more numerous than any assigned multitude of prime numbers.
No *all* in his text at all.
It is only *part* where contradiction is used. That does *not* make
the complete proof a proof by contradiction.
No.
Yes, but *that* is the theorem by Euclid.
And that is where *you* are wrong. The proof holds for an *arbitrary*
collection, and so it holds for any collection.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
===
Subject: Re: Review of Mueckenheims book.
Yes. Euclid said: I say that G is not the same as any of A, B, C.
For, if
possible, let it be (the same).
In old Greek The X means all X.
Without contradiction the whole proof would fail. Therefore it is a
proof using contradiction. That is a proof by contradiction. Not even
by an empty set of contradictions, which you would also have to
accept, as a set theorist.
I am responding to your claim Cantor's proof was not by contradiction.
But it is not what I reported in my book and what your sources report
and where the judgement of all scholars is unique about.
No. That's where *you* are wrong. Without the assumption to use all
prime numbers Euclid could not obtain his result. And in fact, given
the set of all prime numbers exists, then his result is even wrong.
Because there cannot be a prime number outside of the set of all prime
numbers. So you see that Euclid's proof is wrong because he did not
assume enough. He had to assume what is assumed implicitly by all
modern textbooks, namely the complete existence of sets.
Without the assumption of completeness there cannot be any proof at
all. There is at most the result that some sets of primes are not
complete. Or did he prove that the set of all primes is
uncountable?
===
Subject: Re: Review of Mueckenheims book.
Nntp-Posting-Host: hera.cwi.nl
As I stated before, that is not the proof, that is only a subproof of
one of two possibilities.
Oh. News to me. Care to provide me with a reference?
Well, if you wish to keep on using non-standard terminology, go ahead,
but you must understand that you will not be understood.
I have not claimed that. Where is such a claim?
In my review I clearly state that what you report as Euclid's proof was
*not* actually Euclid's original proof, I just wanted to clarify that.
What is this nonsense? For any arbitrary collection of prime numbers,
there is a prime number not in the collection. That is what the theorem
is and what is proven. There is no assumption needed that the collection
contains *all* primes (and no such assumption is indeed made).
Eh? In Euclid's time any collection (or set) was finite, so his proof
was perfectly right. It is onlu *your* modification with '(set of all)'
which makes it wrong.
You apparently do not understand that if something is proven for an
*arbitrary* element of a set, that provides proof for all elements of
a set. But my remark in the review pointed more to the close
resemblance of Euclid's and Cantor's proof. They go as follows:
Theorem: given an arbitrary finite(Euclid)/countable(Cantor) collection
of elements with property P, there is an element with property P not in
that collection.
Corollary: the collection of *all* elements with property P is not
finite(Euclid)/countable(Cantor).
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
===
Subject: Re: Review of Mueckenheims book.
No response? Do you agree or disagree that all foo are finite does
not imply the set of all foo is finite?
Why not?
You think that the set of (finite) natural numbers is finite, do you?
Finite sets of (finite) natural numbers have a largest element. So do
you think that there is a largest (finite) natural number? That seems
silly.
The set of natural numbers satisfies the Peano axioms.
Uh huh. And what kind of object are they? They aren't natural numbers,
so what are they?
--
mike.
===
Subject: Re: Review of Mueckenheims book.
Set theory subsumed under mathematical logic? That's for you to
stipulate, I guess. Anyway, I've been saying all along, which is true,
that there are texts in other areas of mathematics that do define 'a
function' as a one-place. You even adduced one out of six.
It means you can have both - the one place definition and the three
place definition -without contradiction or confusion, just as many
books in set theory and some in other areas of mathematics do.
They're not so new, I woudldn't think. Maybe, what, well over a
hundred years?
For what reason? You said yourself that you found one out of six in
just that one samplng.
How could you not? I'd been posting it on a formula by formula basis,
ad nauseaum.
What basis is that? It's a fact that many a book on subjects other
than set theory use the one-place version, even if a greater amount
use only the three place version.
And you made no comment on (4), which was the most substantive point
mathematically.
My preference would have been for it never to have begun. There is
nothing wrong with the terminology I used and it should not have been
made an issue.
I want to be corrected when I am incorrect, and I will post that I
recognize any errors that I commit, but this is clearly not such a
case. Yes, I do welcome moving on to more rewarding discussions.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
The one book gave the definition in a chapter on set theory, a couple
of sentences later gave the usual definition, and I don't know of any
functions defined in the book that don't have a specified codomain. So,
I was being generous to put him on your side. However, you are welcome
to the one if you stop saying that many books outside mathematical
logic/set theory define functions as just sets of ordered pairs without
a specified codomain.
How can posting formulas demonstrate that books in other areas of
mathematics give two definitions of functions?
There's that many again.
Other than the fact that you've given a standard word (codomain) an
unusual meaning, what sort of comment would you like?
The terminology is nonstandard.
OK, I give up. This discussion reminds me of the discussion we had about
can be well ordered a while back. Although, for that one you
eventually conceded your terminology was nonstandard.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
Some of my submissions are not coming up in my view. I hope this one
does not appear as a duplicate:
Addressing my own question, perhaps not that long. At least back to
the 1920s though.
Of course the definition is in the set theory chapter of the book. I
already addressed that. It does not vitiate that the book does have
the definition. And that the book also has the three-place definition
doesn't vitiate that the book does indeed also have the one-place
version. And, I've been saying all along that the two versions are
consistent. And last night it occurred to me that not only are they
consistent, but that the three-place version ENTAILS the one-place
version, making the one-place version not even needed as a definition,
but rather as a consequence of the three-place version, which makes my
point even STRONGER. (I'll show that entailment in another post,
though it is easy enough for anyone to derive.)
No you weren't. You were being honest. The question was whether many
books in subjects such as algebra and analysis do have the definition
(not whether such books have that definition EXCLUSIVELY and not the
three-place defintion also or in a chapter other than the set theory
chapter of the book). And that is such a book.
And I did not say that most books use that definition or that there
isn't a trend to prefer that definition. I just said that many books
in those areas of mathematics (maybe less so in topology I now
suspect) do have the definition. Your sampling is not large enough to
support one out of six overall, but even if it were, say one out of
ten overall, that would be a LOT of books overall, it would be MANY
books.
Now, IF you then would carry this conversation to the dregs of a
dispute over what the word 'many' means, then god save us all.
IT DOESN'T. I didn't say ANYTHING like that. I am sorry, but your
yet commenting on it anyway. If you wish to comment on (3), then
please read it again, and notice that it is DIFFERENT from (2).
Oh, goody, let's have an argument party over the meaning of the word
'many'!
PLEASE DO NOT PUT WORDS IN MY MOUTH. *I* just followed from the
definition that was mentioned in this thread. And I said over and over
and over again that I'm happy to evaluate per any other definition
have been inferring as to what you THINK I've said without paying
attention to what I ACTUALLY have said, even as I've said it ad
nauseaum.
The definition of 'is a function' is standard in SET THEORY, as my
remarks were about set theory, as was clear from context, and even as
I explicity remarked. I never claimed it is the standard definition
throughout mathematics. And as to 'codomain', *I* have never insisted
on ANY definition, let alone claiming that there is a standard
definition, IF that is what you mean (?). If you have a definition of
'codomain', then please, by all means, state it.
What I conceded was a point about informal terminology. I recognized
that my inclination in such informal terminology was out of step and
likely would lead to my being misunderstood if I continued to use it.
That concession came as soon as I saw the evidence for its need. The
rest of our discussion was a separate issue that regarded my arguments
as to why I don't like the usual way of talking about 'can' be well
ordered as opposed to 'is' well ordered. That portion of our
discussion was not a dispute as to what in fact is the usual way of
speaking about well orderings.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
even STRONGER. The three-place version (given a sufficiently detailed
formulation) is not only consistent with the one-place version, but
the three-place IMPLIES the one-place version. Borrowing from a post I
Let's take, say, this formulation of the three-place version:
A function from D to C is a relation f such that dom(f)=D and such
Now, suppose f is a function from D to D. So f is a relation & ((<x
a
function from dom(f) to range(f). So there exists and D and C such
that f is a function from D to C. So f is a function (a fortiori, if
f
is a function from D to C, then f is a function; and note that Halmos
DOES use the termninology 'f is a function' alone as cited on page 48
where he speaks of u being a function and even gives the requirements
for proving that u is a function).
y=z)).
And even if the three-place version mentiond the Cartesian product
DXC, the argument still works.
And, if the Cartesian product is mentioned, then f is a relation,
since any subset of a Cartesian product is a relation, so if the math
texbook mentions Cartesian product but not 'relation', then, in some
way, the text is still suggesting some kind of pairs gotten from
coordinates or whatever, and only extreme nitpicking wouldn't grant
that that is as good as having a relation.
y=z))' in such explicit form, there is usually something that is a
paraphrase even if less symbolically given.
However, I grant that such informal a definitions as:
f is function from D to C iff f is matching of members of D with
members of C such that every member of D is matched with exactly one
member of C though it is not required that every member of C is
matched by a member of D
don't in themselves yield the set theoretic defintion.
Also such things as mentioning a rule or procedure or process
or
an analogy with a sausage grinder or the like, do not in themselves
yield the set theoretic definition. Things like:
f is a rule that takes members x of D and turns x into f(x) a member
of C.
And indeed I do NOT prefer such definitions. so it is completely
IRONIC that Virgil even suggest with a question that I might endorse
anything like that. On the complete contrary to requiring a rule or
mentioning that a rule is at work, or anything like that, I prefer the
purely EXTENSIONAL set theoretic definition.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
TYPOS:
Make that:
A function f from D to C [...]
Make that:
f is a function from D to C.
===
Subject: Re: Review of Mueckenheims book.
two different codomains or two different functions?
===
Subject: Re: Review of Mueckenheims book.
So many of my submissions from earlier today are not appearing in my
view. I hope these resubmits tonight don't become duplicates to what
is already posted.
In set theory, if f and g are functions, and dom(f)=dom(g), and Ax(x e
example I gave.
If you want to consider a system (my word, just provisional, for
lack of a better word now) such that f, D, and C are a system provided
does not equal the system of f, D and B.
One way we could formalize this is:
And, given the definition of codomain:
then, still, f has many codomains, but there is a unique codomain C
And to foreclose on any possible misunderstanding, I am not claiming
that any textbooks whatsoever have such a term as 'function-system' or
even that a book SHOULD have such a term. I am just pointing out that
such a formalization does capture the sense of there being a unique
codomain in a given context.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
So that if f = (A,B,C) and g = (A,D,C) where B is a subset of D and
C is a subset of AxB, then f = g ?
===
Subject: Re: Review of Mueckenheims book.
You're asking:
If B subset_of D &
C subset_of AxB
So B=D? is the question.
B is a subset of D, but is D a subset of B?
I see nothing that entails that D is a subset of B.
Am I missing something?
What is the point of such an exercise?
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
Then you are allowing that equality of functions, f and g, implies
equality of codomains, B and D.
===
Subject: Re: Review of Mueckenheims book.
Addressing my own question, perhaps not that long. At least back to
the 1920s maybe.
Of course the definition is in the set theory chapter of the book. I
already addressed that. It does not vitiate that the book does have
the definition. And that the book also has the three-place definition
doesn't vitiate that the book does indeed also have the one-place
version. And, I've been saying all along that the two versions are
consistent. And last night it occurred to me that not only are they
consistent, but that the three-place version ENTAILS the one-place
version, making the one-place version not even needed as a definition,
but rather as a consequence of the three-place version, which makes my
point even STRONGER. (I'll show that entailment in another post,
though it is easy enough for anyone to derive.)
No you weren't. You were being honest. The question was whether many
books in subjects such as algebra and analysis do have the definition
(not whether such books have that definition EXCLUSIVELY and not the
three-place defintion also or in a chapter other than the set theory
chapter of the book). And that is such a book.
And I did not say that most books use that definition or that there
isn't a trend to prefer that definition. I just said that many books
in those areas of mathematics (maybe less so in topology I now
suspect) do have the definition. Your sampling is not large enough to
support one out of six overall, but even if it were, say one out of
ten overall, that would be a LOT of books overall, it would be MANY
books.
Now, IF you then would carry this conversation to the dregs of a
dispute over what the word 'many' means, then god save us all.
IT DOESN'T. I didn't say ANYTHING like that. I am sorry, but your
yet commenting on it anyway. If you wish to comment on (3), then
please read it again, and notice that it is DIFFERENT from (2).
Oh, goody, let's have an argument party over the meaning of the word
'many'!
PLEASE DO NOT PUT WORDS IN MY MOUTH. *I* just followed from the
definition that was mentioned in this thread. And I said over and over
and over again that I'm happy to evaluate per any other definition
have been inferring as to what you THINK I've said without paying
attention to what I ACTUALLY have said, even as I've said it ad
nauseaum.
The definition of 'is a function' is standard in SET THEORY, as my
remarks were about set theory, as was clear from context, and even as
I explicity remarked. I never claimed it is the standard definition
throughout mathematics. And as to 'codomain', *I* have never insisted
on ANY definition, let alone claiming that there is a standard
definition, IF that is what you mean (?). If you have a definition of
'codomain', then please, by all means, state it.
What I conceded was a point about informal terminology. I recognized
that my inclination in such informal terminology was out of step and
likely would lead to my being misunderstood if I continued to use it.
That concession came as soon as I saw the evidence for its need. The
rest of our discussion was a separate issue that regarded my arguments
as to why I don't like the usual way of talking about 'can' be well
ordered as opposed to 'is' well ordered. That portion of our
discussion was not a dispute as to what in fact is the usual way of
speaking about well orderings.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
You're right. Your nonsense is at a completely different level.
It seems to be the only way to deal with you. Reason and logic are
simply ignored.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
Tony can't even tell who said what, but he thinks he understands math.
Quite amazing.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
Is it so in Naive Set Theory by Halmos?
===
Subject: Re: Review of Mueckenheims book.
Halmos only gives the three-place vesion, so there is no chance even
for him to dispute that the definition I gave for a one-place version.
And, though I don't have Halmos in front of me as I type, I would bet
that in the text he does make use of the principle that a function is
a relation such that for every x in the domain there is exactlhy one y
counterexample to anything I've said.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
The fact that Halmos does not give any one-place definition at all
disputes your claim that it is the standard definition.
That a function has that single-valued property is certainly accepted by
Halmos, but that differing codomains, so long as they are large enough
to contain the range, define the same function appears not to be
acceptable to Halmos.
That is not not at all clear.
===
Subject: Re: Review of Mueckenheims book.
Nope, actually Halmos DOES USE the one-place definition without
stating it first. Actually, it's not even a definition, it's a
CONSEQUENCE of the three-place definition.
to (or into) Y is a relation f such that dom f = X and such that for
each x in X there is a unique y in Y with (x,y)ef. The uniqueness
condition can be formulated explicitly as follows: if (x,y)ef and
(x,z)ef, then y = z.
belongs to C, it remains only to prove that u is a function. We are to
prove, in other words, that for each natural number n there exists at
most one element x of X such that (n,x)eu (Explicitly: if both (n,x)
and (n,y) belong to u, then x=y).
So, to prove u is a function we notes that we are to prove that for
every member n of the domain (the set of natural numbers in this case)
we have if both (n,x) and (n,y) belong to u, then x=y).
And THAT is just what I said. f is a function iff (f is a relation and
if (n,x)ef and (n,y)ef, then x=y.
And that gives the grounds for proving 'f is a function' with no
mention of a codomain whatsover.
And let's go back to his original definition to DERIVE the one-place
version, not even as a definition, but as a CONSEQUENCE:
A function from X to Y is a relation f such that dom(f) = X and such
Now, suppose f is a function from X to Y. So f is a relation & ((<x
function from dom(f) to range(f). So there exists and X and Y such
that f is a function from X to Y. So f is a function (a fortiori, if f
is a function form X to Y, then f is a function; and note that Halmos
DOES use the termninology 'f is a function' alone as cited on page 48
where he speaks of u being a function and even gives the requirements
for proving that u is a function).
y=z)).
And that is a CONSEQUENCE of Halmos's three-place version and is USED
by Halmos HIMSELF such as on page 48.
And it is the standard definition of 'is a function' in set theory.
And that some books in mathematics in general give the three-place
version is not only consistent with the one-place version, but the one-
place version is a CONSEQUENCE of the three place version, as I just
showed.
WHERE does Halmos say that? And I didn't even say that different
codomains DEFINE the same function. I said that, GIVEN a certain
defintion of 'codomain', a function has more than one codomain. And I
said that *I* am NOT insisting on that definition of 'codomain' but
rather that I welcome YOU to state whatever definition you wish to
state.
So, AGAIN, what is your definition of 'codomain' now?
It sure as shootin' is.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
Well, I said I was done with this discussion, but since Virgil doesn't
have Halmos handy, I guess I'll change my mind and not be done.
Halmos, page 48:
Recursion theorem. If a is an element of a set X, and if f is a
function from X into X, then there exists a function u from omega into X
such that u(n^+) = f(u(n)) for all n in omega.
Proof. Recall that a function from omega to X is a certain kind of
subset of omega x X; we shall construct u explicitly as a set of ordered
pairs. Consider, for this purpose, the collection C of all those subsets
A of omega x X for which (0,a) in A and for which (n^+,f(x)) in A
whenever (n,x) in A. Since omega x X has these properties, the
collection C is not empty. We may, therefore, form the intersection u of
all the sets of the collection C. Since it is easy to see that u itself
belongs to C, it remains only to prove that u is a function. We are to
prove, inother words, that for each natural number n there exists at
most one element x of X such that (n,x) in u. (Explicitly: if both (n,x)
and (n,y) bleong to u, then x = y.) The proof is inductive. ...
The codomain of the function u is X (as stated in the theorem and as is
true by the construction of u).

subset S of X x Y. In other words, a function is a triple (X,Y,S).
No.
Definition: If f = (X,Y,S) is a function, then the codomain of f is Y.

--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
It is not AT ISSUE that in the larger proof there is a particular
codomain such that we need to prove u is into that codomain. What I'm
pointing out is that AT THE EXACT step in the proof I mentioned, what
Halmos says we must show is that u is a function. The proof needs to
establish that u is a function from X into X, yes, of course. But that
is broken into steps, and one of the steps is to show the COMPONENT
assertion u is a function. And u is a function is onto itself a
meaningful assertion, whether informally or formally (as u is a
function would be an English rendering of a particular wff). All
through set theory and even in mathematics in general, we come across
assertions such as u is a function and we prove or disprove them.
Sometimes we prove them as well as proving that some given set is a
codomain, and sometimes we just prove the standalone u is a
function.
limited 'f is a function' or if 'f is a function' is not defined, its
equivalent formula with predefined terminology can be derived from
or it may get a derived equivalent. Oh, but where do we see such
derivations in math books? We don't! We donl't because it is COMMON
function' is an assertion.
YOU were the one QUOTING math texts. Which of those quotes is that f
is a triple?
Hey, I'm all for the triple idea, and I've mentioned it a few posts
back. And I think it works fine here.
But the triple definition is NOT a definition of 'is a function'. The
triple definition is a definition of 'f is a function-system' or
whatever word you want to use. Because f ITSELF is not the triple -
not in SET THEORY. And not in usually in ordinarily basic math, not at
least when you get to any passage in a math text that says Now we
prove that f is a function as opposed to saying Now we prove that <X
I would be VERY GLAD to agree with you that there are two different
things, especially in set theory, even as an formalization of Halmos
and similar treatments:
(1) Ther is f, which is a function, which are certain kinds of sets of
ordered pairs.
That allows us to have BOTH the standard set theoretical definition of
'is a function' as one-place AND the common definition in both set
theory and mathematics generally of a relation among three things: a
range, a codomain, and a certain kind of subset of the Cartesian
product of the range and codomain.
And that has been obvious to me always. And now to think that I was
even hesitant to suggest an overt triple because I feared it would be
taken as a personal and unjustified extrapolation. It is a justified
extrapolation and I am happy to see you use it, and it accords with
what I'm saying, as just that: a triple.
Yes. You give no argument to refute my derivation.
I gave it plain as day:
function from D to C, then (the domain of f is D and f is a relation
And just by the rule of conjunction, that implies '(f is a relation &
conjuncts in Halmos's statement. Then for the other direction, just
note that in that passage on page 48, Halmos EXPLICITLY gave a
sufficient condition for proving f ('u' was the variable there) is a
So, just following Halmos's own terminology:
If you dispute that, then please show what is incorrect in my proof of
both sides of the biconditional.
RIGHT!!!
See my remarks about that above.
WHERE is that definition given in any of your math books?
YOU were the one who made the incorrect claim as to my terminology
being idiosyncratic. Now please show any such definition in a math
book of the above.
Morover, your terminology makes it impossible to say Now we prove f
is a function followed by a proof showing that f is a relation such
find no such instances of that in math books that even start with the
three place definition of 'f is a function from D to C'.
Can't you see, 'f is a function from D to C' is PACKING conjuncts, of
which ONE of the conjcts is that f is a function. In other words, not
only is f a function but its domain is D and its range is a subset of
C. Three separate assertions BUNDLED: (1) f is a function, (2) the
domain of f is D, (3) the range of f is a subset of C.
And that is just what Halmos does, and handling it that way even in
general mathematics is consistent with general mathematical use (note
I said 'general' not 'uniform'): In ordinary basci mathematics it
makes perfect sense to prove 'f is a function' by proving that f is a
the domain of f is D and the range of F is a subset of C.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
Are you being intentionally obtuse? Halmos isn't going to mention that a
function needs a codomain if it is obvious at that stage of the proof
that u has one. Math books are written to convey ideas, not to make it
easier for people who want to translate them into formulas.
All of them (except some of the ones in logic/set theory).
I guess that is progress. Now, if you just take the next step and admit
that the standard definition of function in mathematics is (formally)
a triple, we can close this discussion.
Your translation of Halmos's definition of is a function is wrong. A
function is (formally) a triple.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
Marcus' definition works for me.
===
Subject: Re: Review of Mueckenheims book.
But then the function is f and the triple is not the function.
We say things like Let f be a function or Let f be a function from
a function from D into C.
'f is a function from D into C' is a statment about a relation among
f, D, and C - a relation among the function f, the domain of f, and a
superset of the range of f.
'f is a function' is a statement about f.
f is the function.
Get used to it.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
On the contrary, f would be only the set of ordered pairs which is not,
by itself, a function in the sense of category theory.
That may be what you say.
We say, let f be the function from D to C such that to each element of D
assigns such and such an element of C as its value.
Get used to the fact that such a definition does not, in general, work.
If you chose to use it in a carefully circumscribed arena, it may work
well enough, but only within such an arena.
===
Subject: Re: Review of Mueckenheims book.
What in the world?! You said your beef with me is about SET THEORY.
Long ago it was made clear that I'm not not talking about category
theory. I have NEVER disputed that category theory may go about things
much differently. Why are you switching the argument again?!
Is this dishonesty on your part? Did you realize that you were wrong
and that I am correct as the standard set theoretic definition of 'is
a function' so you switched to make the discussion about category
theory instead?
Or, are you just forgetful, and you forgot that I made it clear long
ago that category theory is not at issue and that you claimed that my
definition is not standard in SET THEORY?
Yes, I've said that ALL ALONG!
It works just fine even in most basic (emphasize BASIC) abstract
algebra, analysis, and topology. And it IS the standard set
theoretical definition.
And where you need to specify a particular codomain, that can be
accomodated either by just SAYING what codomain is being considered,
or by triples (you opted for triples YOURSELF, even though you took
the wrong branch in the proposals).
We can have the set theoretic definition of 'f is a function' and the
set theoretic and general math definition of 'is a function from D to
C' and we can have:
You're changing terms again. What YOU disputed is that my definition
is the standard set theoretic definition. You were incorrect.
Anyway, such ordinary definitions I gave work in plenty of basic
mathematics. And that doesn't deny that in, e.g., homotopy theory,
whatever mathematicians may choose to use differnt terminology. And
the triples solution in the form I suggested would work as a
formalization (I think, perhaps someone might have a critique per a
particular area of mathematics) though it might not suit the
mathematicians themselves in the particular areas of mathematics (it
is not the point of perfectly rigorous formalization to suit the
everyday purposes of mathematicians).
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
But I am considering only definitions of functions which work in
general, which yours does not.
And I find no reason to suppose that your definition sets the standard,
even in set theory.
===
Subject: Re: Review of Mueckenheims book.
understood from the context, people may not mention them.
You are welcome to your personal preferences, but this isn't the
standard usage.
I have the impression that you've learned most of your mathematics from
books. Is that correct?
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
And people who interpret carefully may as easily understand that that
rolls off the tongue but, in garden variety math, it may be taken as
(and is actually more rigourously taken as): f is a function & domain
of f is X & the range of f is a subset of Y. All we really
And some people also say f is function and may not be specifying any
particular subset of the range of f. To prove 'f is a function' it
suffices to prove that f is a relation of a certain kind. Simply to
prove that f is a function does not also require proving that it has a
certain specified superset of its range.
It's standard in set theory; and it's widespread in basic mathematics;
and it is consistent with other basic mathematical terminology; and my
definition of the one-place 'f is a function' follows right from such
definitions as you posted from textbooks, as I proved.
terminology in basic abstract algebra, analysis, and (first
sememester) topology? The function f ITSELF as a triple? Please cite.
Oh, if you want to say that what goes on orally in classrooms in front
of chalkboards is often a lot different from what goes on in books,
then I have no dispute about that. My gosh, I could not possibly
assess all the different kinds of terminology people use orally.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
Perhaps, but almost certainly for technical convenience, not because
logicians don't know the standard definition.
Nonsense.
Nonsense.
That makes no sense. You can't prove a definition.
Almost everywhere.
Exactly.
Exactly.
Already did, but apparently you don't know how to read normal math
books. Halmos is pretty clear, but somehow you managed to misinterpret
the context of logic/set theory, all the books I checked give the usual
definition that Virgil, Ralph, and I have always used.
That's not my point. My point is that if you don't talk to experienced
people, you may misunderstand what the books are trying to say.
So, where have you learned your math? Where did you go to school? What
courses have you taken?
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
Of course, with the usual requirement on the subset to make it a
function.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
The usual requirement to make THE SUBSET a function. The triple is not
the function, the mentioned subset S is the function. The triple is a
function along with the domain of the function and a specified
superset of the function.
whether f is a function. Your definition is defining whether f is a
function with domain D and C a superset of the range.
No, it's Let f be a function on natural numbers such that f(x) =
x^2.
*f* is the function. N is the domain. And we may also specify a
certain superset of the range of f, if we like. Or you can define a
'codomain' to have to be a subset of the Cartesian product of the
domain and the range, which is fine too if you want to do that.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
No. It in my sentence refers to the triple.
mathematical language for what we would formalize as let f = (N, N, S)
where S = {(x,y} | x in N, y in N, y = x^2}.
Just like we say let G be an abelian group, when we really mean let G =
(G',+), etc. People don't write in a formal language. But, we know how
to formalize what we write, if necessary. This lets us avoid mistakes
without making our prose impenetrable.
You keep saying this, but it isn't standard practice in mathematics. I
think you've said that most of your knowledge of mathematics is in the
fields of logic and set theory, so I don't understand why you are so
certain that you know what is done in the rest of mathematics.
Didn't we have essentially this same discussion about well-ordered sets?
I think I said a well-ordered set is a pair: a set with an order
relation on it. And, I think you said it was just the set. Now, I'm
saying a function is a triple, and you are saying it is just the set
(the third element of my triple).
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
I found a very established topology book (Hocking & Young) that
explicitly uses the triple method and sets the function itself equal
to the triple. So that is one for your column. And it seems to me to
fit what Ralf Bader was stressing in his posts too. Still, that
approach is not even plausibly standard in garden variety mathematics,
and not even among the topology books I've been able to look at since
this discussion or Munkres that you mentioned (I'd also like to see,
e.g., Dugunji, Simmons, Hu, and Willard). Anyway, while I have some
more things to say about this, I wanted to get in this note in
recognition that I have found at least one for your column in
topology.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
I know. And I'm saying that it's whack.
f is the function. We refer to the function f not to the function
And there, S is the function, not f.
I see that elsewhere you're mentiong this in comparision with our
discussion about groups and things like that. But the analogy does not
hold.
and f is a function on SxS.
algebra or analysis that says any such thing. The triple is a GOOD
idea, I agree with it. But in the triple is not a function. The f in
the triple is the function.
And you contradict any mathematician who has ever said, I'll prove f
is a function. [Then he winds up with...] The domain of the function f
is N and for all neN f(n)=x and f(n)=y entails x=y. The mathematician
be whack and it was what we get by applying your terminology.
function, and not f, then that is what we get for our group: <G <GxG G
I was happy to concede that my informal phrasing regarding regarding
well orders was likely to be misunderstood as I thought 'is well
ordered' and 'can be well ordered' are equivalent. I argued that they
are equivalent since set theory is extensional, and there is no
modality of 'can be' as opposed to 'is' in set theory. But I conceded
that my quibble doesn't matter as to the way people talk; people
really do distinguish between 'can be well ordered' and 'is well
ordered'. So, fine, I'll work around that, though I also argued why I
don't like it.
But here, you are incorrect; it's your turn to concede.
And recall that I agreed about that kind of informal terminology,
which I described as slippage that I'd rather weren't around but
that I understand that harm does not come as long as context is clear.
But notice that is the OPPOSITE of what's going on with our dispute
about functions:
G is the group. G' is the carrier set and + is the function from GxG
to G.
A group REALLY IS a tuple. A function is not such a tuple.
S. Come on! The function f.
I have read enough in basic mathematics to know that ORDINARILY the
And notice that I say 'ordinarily', in basic textbooks on subjects in
mathematics, since I said long ago that I recognize that in certain
areas and at certain levels of mathematical discussion there are other
ways of spedaking about things.
But in garden variety set theory or mathematics, a function is not
such a triple. There are sure to be exceptions, true, especially since
mathematicians let all kinds of terminology fly around as is
convenient for them to convey their ideas. But when you get down to
the ordinary mathematical circumstance of, say, proving that f is a
function, it suffices to prove that f is a set of ordered pairs such
to say about the domain and range and any codomain of f.
as the well ordered set. What I argued was that I don't like that way
of speaking (after I conceded about 'is' vs. 'can' I didn't even claim
that that way of speaking is not indeed quite usual). I would prefer
to say that R is the ordering and S is the set that is well ordered or
Show me one textbook that says a function is such a triple. I mean,
literally a triple and not just OUR additional formulation of a triple
from what the text says.
Triple is fine. And groups and well orderings on sets do get
slippage between the tuple and the carrier set itself. I'd ALWAYS
granted that. In fact, I said it is prevalent but that I don't like
terminology might exist somewhere, but it sure is not usual in basic
textbooks in such subjects as abstract algebra and even basic (first
semester) topology.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
function from N to N.
So one would then say S(n) = n^2.
===
Subject: Re: Review of Mueckenheims book.
I have two questions in trying to understand this approach. We can
take the context to be analysis.
would you justify deducing that P(g) follows (or would you say it is
possibly unjustified)?
===
Subject: Re: Review of Mueckenheims book.
As I do not have a copy of Halmos at hand, and thus cannot review the
context from which your quotes were taken, it remains unclear to me.
Unless one is trying to decide whether two functions are identical or
whether two functions can be composed, the issue of which codomain a
function has is largely moot. The Halmos example you cited, not being
either of those, is not sufficient to settle the issue of whether
different codomains produce different functions.
===
Subject: Re: Review of Mueckenheims book.
Some of my posts are not appearing in some views. If a post in reply
to yours here appeared, please disregard it. I misspoke regarding
switching of arguments in connection with Halmos specifically.
Please disregard that remark if the post does appear.
Anyway, I showed that Halmos mentions that to prove 'f is a function'
As to codomains, what is your definition of 'codomain'? I've been
asking. Please.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
You're SWITCHING the argument again. The issue of whether different
codomains yield differnt functions was not the issue in my remarks
about Halmos.
And AGAIN:
What is your definition of 'codomain'?
which is EXACTLY what I said.
MoeBlee
P.S.
Oh, by the way, for about the fifth time, please, Virgil, what is your
definition of 'codomain'?
===
Subject: Re: Review of Mueckenheims book.
The second factor of the Cartesian product of which the set of ordered
pairs of the function is necessarily a subset.
===
Subject: Re: Review of Mueckenheims book.
But there is NO SINGLE Cartesian product of which the function is
necessarily a subset EXCEPT dom(f)Xrange(f). The function is a
subset of MANY Cartesian products. IF you want to say that the
Cartesian product you're using for this definition is domXrange, then
'codomain' reduces to 'range'.
which f must necessarily be a subset, then anyone can come back with
Proof (is this really necessary to give such proof?!):
is a subset of C. And C is a subset of Cu{z}, so f is a function and
domain of f is D and, by transitivity of subsets, range of f is a
So there are at least these four ways for you:
(1) Let the codomain be the range. (I don't think you want that.)
(2) Let there be more than one codomain for any function. (You don't
want that.)
= x^2.)
(4) Let a function be just what it is: a certain kind of relation; and
every function-triple (or 'function-system' or whatever words you
like) has a unique codomain, viz. the third coordinate of the triple;
but still every function itself has many codomains.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
There is in my definition, which is the definition required in a large
number of areas of mathematics.

The set of ordered pairs of a function is a subset of many Cartesian
products, but does not complete the definition of a function in most of
them.
I don't, so it doesn't.
That presumes things I reject, among other things, that a function is no
more than a set of ordered pairs.
In the language of category theory one must regard what you call
I notice that you carefully omit the one which I actually go by, namely
member of C.
===
Subject: Re: Review of Mueckenheims book.
Well, now that you've adopted one of the two proposed triples
solution, you do have your unique codomain.
Note that none of your attempts at a definition succeded until you
opted for the triples solution.
However, since you seem to have taken David's route in the triples
solution rather than my route, you're stuck with all the whackiness of
his route and without the advantages of my route.
Yeah, we know that. So?
Good. I wouldn't either.
No, I used just garden variety reasoning about functions and the
domains and ranges of functions. If your terminology forbids such
basic garden variety mathematics about functions, then your
terminology is not too good.
I concluded:
and so f is a subset of DxCu{z}.
That is the only part that you could dispute by disputing that f is
The second factor of the Cartesian product of which the set of
ordered pairs of the function is necessarily a subset.
So even if f itself is not the function, then SOME set of ordered
pairs is at play here, and that set of ordered pairs is still a subset
of not just C but also of Cu{z}, so that set of ordered pairs, which
YOU mentioned, is a subset of more than one Cartesian product, thus
your definitions (pre triples approach) do not pick out a unique such
Cartesian product.
I said that there is no unique Cartesian product of which the FUNCTION
[emphasis added] is a subset.
And you said there IS [emphasis added].
But if the function is not the set of ordered pairs, and is instead
mathe books that say the function is a subset of the Cartesian
relevent to the definition? No, the Cartesian product that is relevent
attempt sputters into nonsense if it is to go along with many such
definitions in basic math books. My version of the triples approach is
consistent with those definitions.
We're not talking about category theory! I made that clear a long time
ago! Also, though it aside the point, I'd want to check your claim.
Would you recommend a category theory book in which I can find out
about your claim? (I know of category books, but I don't know that
they are the ones you would endorse as upholding your claim.)
And again, it was made clear that we're not talking about category
theory. We're talking about set theory as to a standard definition,
and such basic textbooks as David mentioned.
Remember, you said your only dispute with me is whether my definition
of 'is a function' is the standard set theoretic definition. Category
theory is not of help in addressing that.
I didn't carefully omit anytyhing. (3) allows for what you just now
posted. (And I don't recall your previously having posted an explicit
formulation like that.) You may disagree with what I said about (3)
but I clearly allowed for a formulation of a triple and open ended
about how it could be fleshed out.
Sheesh, you're claiming that I carefully omitted what I carefully
INCLUDED.
And you claim f is a function.
And you're claiming that is a standard definition or no? Not standard
in set theory, that's for sure (and you were claiming that your
dispute with me is as to SET THEORY). Then standard in such book as
David mentioned? Where, in which book, do we get a definition that
says a function is a triple?
Nope. Not in SET THEORY. And not in such basic math as in books such
as David mentioned. Rather, C(a) = b.
Or please say where there is a set theory book that has a function
being a triple or even such basic math as in books such as David
mentioned.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
(3) it is.
We don't have to say anything of the sort. We simply give the function
(i.e., triple) a convenient name, e.g., f. Then we define f(x) =
x^2 to mean (x,x^2) in S.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
You just keep asserting. I don't claim that you can't find such a
definition in some advanced areas of mathematics (I always said there
is all kind of terminology out ther). But please show me basic
textbooks in abstract algebra, analysis or topology that claim a
function is a triple. And, no, saying f is a function from D to C or
f is a subset of DxC such that..., etc. is NOT claiming that f is a
triple.
You started by incorrectly saying that I was using idiosyncratic
terminology. Now you've backed yourself into stubbornly insisting that
your own terminology (which, is manifestly idiosyncratic) is the
terminology winner in this discussion.
I didn't say you said it. I'm saying it is the reductio of your
proposal.
I've never seen that. Would you give an example from a basic math
textbook?
What basic math textoobks use such roundabout terminology, saying that
agree our agreed upon neologism, since that accomadates BOTH f is a
function onto itself and f is a function from D into C. That allows
the set theoretical definition and also speaking of a particular
codomain to coexist.
Your proposal just rides roughshod over the simplicity of doing things
like proving that f is a function in the manner Halmos did and as is
quite in keeping with a lot of common mathematical practice.
That's a good solution. A pity that you just keep asserting that its
not.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
All the ones that I listed in my previous post do exactly that, if you
know how to read a normal non-formal math book.
Wrong. I am amazed that you would say this. Math books are not written
in the formal language of set theory. This doesn't mean that an
experienced mathematician doesn't know how to translate the usual prose
into such a language, if they feel the need.
The fact that some logic/set theory books find it handy to use a
different definition of function for technical reasons does not change
the fact that the standard definition in virtually all of mathematics is
a triple (when formalized in set theory).
So, you still insist that your reading of Halmos is correct even after I
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
If the definition says a function consists of a domain and a codomain
and a set of ordered pairs, or requires specification of all three, then
it IS claiming that a function is a triple.
If one wants a definition that is valid generally, it is.
===
Subject: Re: Review of Mueckenheims book.
Quite right.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
No. That was the first book I checked.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
What authorities have set this standard of which so many of us are not
aware?

Which ones? I find only one among my math texts, and old Calculus text,
and as the context was real functions, the codomain of the reals could
be assumed.
Almost all in my library.
There is when one considers compositions of functions.
Which does not require distinguishing between functions having different
codomains, so is irrelevant.
Which directly contradicts definitions in which a difference of
codomains implies a differenceof functions.
If one is allowed to complain that another is talking past points, we
have challenged Moblee's claim that his one place definition is /the/
standard, which challenge he simply dismisses.
===
Subject: Re: Review of Mueckenheims book.
It's not a claim by authority, but rather that the definition pervades
in set theory and is the definition you will find in just about any
set theory textbook.
I said I hoped not to play a ridiculous game of listing textbooks.
This argument is unproductive aa it is. David Marcus himself found one
out of the six he sampled. You mentioned Apostol, which is a widely
used analysis text (or did you also find it in a calculus text of
his?). And you can find it in many other texts in subjects in
mathematics.
Good. It's never been a matter of dispute with me.
No, that is not what I'm talking about at all. Please do not confuse
the issue.
What in the world are you talking abnout? That was the definition
someone mentioned (was it you?) and I discussed over and over again,
saying that if you don't want to use that definition, then, fine,
suggest any definition you do want to use.
Fine, then STATE your defintiions.
NO, I said that the one-place version, a definition of a 1-place
predicate '___ is a function' is the standard definition in SET THEORY
of 'is a function', which is a point you had addressed SPECIFICALLY
just a few posts ago. And I said that in set theory we may also find
the three place version '___ is a function from ___ to ___'. And I
have NOT dismissed that there are other definitions used in
mathematics in general, since I never disputed that.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
Does Moblee want to identify functions as merely being the formulas or
rules of calculation involved?
And I have said that I reject your claim that your one place definition
is /the/ standard for set theory.
It may have been once, as it seems once once to have been in calculus
and much of analysis, but no longer.
===
Subject: Re: Review of Mueckenheims book.
What a RIDICULOUS question! I said NOTHING that suggests anything of
show my argument. How in the world that leads you to think that might
in any way suggest that I think functions are merely formulas or the
rules for caclulating is beyond me. Please!
Reject away, have a ball.
Meanwhile, please do not mischaracterize what I've said.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
Which suggests to me that you think that those formulas are sufficient
as definitions.
Those are the formulas that
May I then take it that your answer to my question is in the negative?

A question on a point which was not clear to me does not, in my mind,
constitute a misrepresentation, at least if the form of the question
does not imply expectation of a particular answer, which my question did
not.
===
Subject: Re: Review of Mueckenheims book.
I take you on your honor on that point. But your Does
MoeBlee .... ?, speaking of me in the third person that way (in a
post in direct response to mine), as you do to cranks, sure has the
sound of a suggestion that you think my arguments lead to or even
suggest an affirmative answer to your question. So I'll take you at
your word that you did not mean even to suggest a mischaracterization,
but I will say that your question is SILLY, since CLEARLY, by using
the set theoretical definition, I'm taking an EXTENSIONAL view of
functions.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
Your extensional view failed to make clear to me the answer to my
question. WHich is why I asked it.
===
Subject: Re: Review of Mueckenheims book.
IF those are supposed to be 'scare' quotes around 'extensional' then
they are unneeded. The axiom of extensionality is clear enough to
provide that idenitity of sets is purely extensional. And my remarks
all along have not depended on the word 'extensional'. I just can't
believe that you were THINKING about what I've written and could
wonder whether I take a function to be merely a formula or a rule,
when everything I've said goes against such a view, AND I even
mentioned that a 'rule' premised definition is NOT the kind I use.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
The definition in which f = (A,B,C), where C is a suitable subset of
AxB, being expressible entirely in terms of set properties, is as
extensional in that sense as any other.
===
Subject: Re: Review of Mueckenheims book.
I agree. So?
MoeBlee
===
Subject: Re: Review of Mueckenheims book.

So you claim of some extra virtue due to extensionality of your
definition or the implied lack if it in others, is phoney.
===
Subject: Re: Review of Mueckenheims book.
I think that when I was young, most people said range. Kind of a pity
that confusion over range versus image has led to the need for
codomain. It isn't a very pretty word. I wonder if recent books have
mostly switched to codomain.
--
David Marcus
===
Subject: Re: Review of Mueckenheims book.
When I was young, there was much less of a standard terminology about
functions, whether in set theory or elsewhere, than there is now, and it
tended to differ a bit from country to country, and even from one
university to another.
===
Subject: Re: Review of Mueckenheims book.
Perhaps it is the fact that among all the math texts and reference works
immediately available to me only, in only one, an early edition of
Apostol's Caluculus, have I been able to find a definition compatible
with yours.
Neither the Harper Collins Dictionary of Mathematics nor the Oxford
Concise Dictionary of Mathematics even allow it as an alterntative.
Therefore I do not accept your definition as the standard, but only, at
best, an acceptable alternative in some areas.
===
Subject: Re: Review of Mueckenheims book.
No, No, No, look what you did. You SWITCHED the argument. You said
that your argument is with my claim that it is the standard definition
in SET THEORY. I NEVER said it is the standard definition in general
mathematics.
Please, that was really specious of you just now. You said you
disputed my claim that it is the standard definition in SET THEORY,
then you switched to a strawman about mathematics in general.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
Even in set theory, I dispute it as /the/ standard.
===
Subject: Re: Review of Mueckenheims book.
I hope I can take it then that you won't continue your strawman about
it being a standard defintion in mathematics in general.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
If there is a standard definition for math in general it is the one in
which distinct codomains produce distinct functions.
If there is a standard definition for set theory, I see insufficient
evidence to support your contention that it is not the same one.
===
Subject: Re: Review of Mueckenheims book.
So? I never claimed as to what is or is not a standard definition in
mathematics in general.
And you have not shown that the function ITSELF is different on
account of mentioning a different codomain in one context than in
another or even by the three-place relation of a function f being from
D to C.
And STILL, what is your DEFINITION of 'codomain'?
Why don't you just look at the definition of 'is a function' in the
most famous and widely used textbooks. Where 'is a function' is
defined, what textbook defines it differently from the definition I
gave? And Halmos's conversational styled presentation is not a
counterexample, as I showed.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
Informally, the target set of a function, the set in which all the
values of that function must lie. But unlike the domain, in which all
arguments of the function must lie, there can be , and often are, values
in the codomain which the function does not access.
Which books do you regard as the most famous and widely used
textbooks? I do not seem to have many of them in my library.
===
Subject: Re: Review of Mueckenheims book.
786
You're really going to pursue such childish to its nth degree!
Bernays, Enderton, Halmos, Levy, Quine, Suppes, Stoll, Jech, Kunen,
Moschovakis, Takeuti & Zaring, Devlin, et. al, in no particular
order.
I haven't gone to verify each of those. But I would be interested to
see any of them either not having the definition of 'f is a function'
does not also entail that f is a function iff f is a set of ordered
Check it out, in set theory, all sets are extensional. And a function
is a certain kind of set of ordered pairs. And that is not only
consistent with, but it FOLLOWS from, any explanation of a function
also in terms of a domain, range, and even codomain.
Please just think about what I'm saying: At certain times, within
certain proofs, we need to prove that g, or whatever, is a function.
At some times we may need to prove that g is a function and that
domain of g is a certain set, or that range of g is a certain set, or
that g is a function into a certain set; and all kinds of different
things. But sometimes ALL we need or want to prove is that g is a
function. So 'g is a function' is a meaningful assertion onto itself.
Right? IN SET THEORY, at least, which was your bone of contention. How
can you deny that sometimes we ask or prove or disprove or whatever f
is a function which mentions ONLY f in that assertion. So 'is a
function' needs to have a definition by itself (or revert to
terminology already defined). Yes, we have a definition of the more
claim (which SUBSUMES 'f is a function'), and f is a function is
another, more limited, claim onto itself too. And FROM the definition
Boy, I'm 'splainin this and 'splainin it and 'splainin it...too much!
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
So are you claiming that the definition of functions required by all
those other areas of mathematics is invalid in set theories?
it is one thing to say of something that it is a function and quite
another to say that this function is the same function as that function.
What may minimally suffice for establishing the former will not suffice
for establishing the latter, and it is in establishing the latter that
the difference in definitions makes a difference.
Using your definition, one delares functions the same that my definition
declares different.

On that point I agree.
===
Subject: Re: Review of Mueckenheims book.
Personally, I'd rather use the word 'valid' in its sense in
mathematical logic.
Now, if we propose a particular definition of a particular predicate
symbol and we also propose another definition of that same predicate
symbol, and such that the definitions are inconsistent in a particular
theory, then that's an actual problem.
As to mathematicians having different terminology that classes between
special subjects, I already addressed that. And you will see more
about it in my latest post to Ralf Bader.
In set theory, it is a theorem:
If you don't like that, then I'm sorry I can't help you; it is a
consequence of the axioms and the STANDARD definition. And your just
saying that mine is not the standard set theoretical does not change
the fact that mine is the standard set theoretical definition.
Yes.
Okay.
Right. I'm using the standard set theoretical definition. You're not.
What we don't agree on is WHY you are immune from such explanations
that should only have to be given once and briefly.
Meanwhile, you don't seem to be able to agree with YOURSELF as to
whether your point is that you dispute that my definition is standard
in set theory or whether it is about category theory, while a long
time ago I had said that I'm not claiming anytyhing about what goew on
in category theory.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
If this theorem relies on your own definition of a function, then it
is only true in such set theories as accept your definition. Which is
not all of them.

The whole point is that there is no STANDARD definition, but definitions
which vary from author to author and from text to text.
And your just
By what authority do you declare as standard what is not universal.
By what authority do you declare as standard what is not universal.
You claim authority to determine what is the standard when what you
claim is not universally accepted.
Then you have not read me carefully enough. That is exactly what I am
disputing. That your definition occurs I grant, but as it is not
universally accepted even in set theory (for example Bourbaki) I reject
it as THE STANDARD.
===
Subject: Re: Review of Mueckenheims book.
Just to be clear, I mean it was a strawman that I am suggesting the
one-place definition is the standard definition in general
mathematics.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
I understood that from your original statement.
===
Subject: Re: Review of Mueckenheims book.
misunderstanding on that point.
MoeBlee
===
Subject: Re: Review of Mueckenheims book.
point.
MoeBlee
===
Subject: automorphism of quadratic field
of quadratic field K=Q(sqrt(d)) which sends 'sqrt(d)' to '-sqrt(d)'
keep inequality. Actually in which case, I can from a<b conclude
s(a)<s(b), a,b from K. What are the condition on numbers. Let's say
Al Bundy
===
Subject: Innumeracy: Where to start?
I consider myself to be a functional innumerate. I can get by, but
I just don't have a solid understating of fundamental mathematical
concepts.
In order to overcome this fault, I've been attempting to re-learn the
math I should have mastered in grammar school, high school, and
college.
As a first step I started taking a night course on probability because
it seemed like a practical discipline with lots of real-world
applications.
Unfortunately I'm finding this course to be a little overwhelming. I
often find myself highly confused during most of the lectures. My
guess is that this is b/c I didn't master the fundamentals the first
time and I'm missing a lot of implied mathematical knowledge.
The best metaphor I can draw would be attending a Japanese poetry
class without having a solid knowledge of the Japanese language. You
get bits and pieces, but you can't fully comprehend the subtlety and
nuance of the material.
Does anyone have any recommendations on how I can proceed? What should
I start with? Is there a guide out there for those who are attempting
to get back into math?
The assumption should be that I know nothing. To be honest it often
feels that way. ;-)
-Thx
===
Subject: Re: Innumeracy: Where to start?
I recommend the following two books. They are probably
at a lower level than just about anything anyone else
is likely to recommend, but I think they will give you
a good start on the kinds of books others might recommend.
Isaac Asimov, Realm of Numbers
Isaac Asimov, Realm of Algebra
These are from a few years back (1959, 1961), but I've found
that most public libraries will have at least one of them.
Also, since you said you're taking classes at a college,
there's a good chance that you can find them in your college's
library.
Dave L. Renfro
===
Subject: Re: Innumeracy: Where to start?
Will taking a course like that keep me away from starvation ?
===
Subject: Vectors & Rotations
Given 2 Points P1(x1,y1) e P2(x2,y2) using the inverse trigponometrics
functions
is possible to compute the angle A of P2 with center P1 and to rotate P2 of

a B angle around the center P1 it is only need to to add A to the angle B
and then using the trigonometric functions is possible to copmpute the new
point.
How is it possible without trigonometry?
Given P1 and P2 is possible to construct V1(x3,y3) where x3=x2-x1 and
y3=y2-y1 ; normalized
d=Dist(P1,P2)
x3=(x2-x1)/d
y3=(y2-y1)/d
so I have
x3=(x2-x1)/d=cos(A)
y3=(y2-y1)/d=sin(A)
Now inside the vector V1 I have the angle information of P2.
Give a vector V2 rapresenting a rotation with an angle B how can I construct

a vector V3 rapresenting a rotation of an angle A+B ?
Denis.
===
Subject: Re: Vectors & Rotations
Schaum book
===
Subject: Re: Using an AR(1) model for time series forecasting with missing
observations
I am sure this has already been said, but if you do a Bayesian
analysis of this problem, then 'missing' data and so on will not be a
problem.
illywhacker;
===
Subject: Re: Do reals really have to be genuine numbers?
The young Galileo Galilei also imagined one could attribute a countable
amount of points to a continuum. Look into Wikipedia for Galileo's
paradox and try to understand Salviati.
===
Subject: Re: Do reals really have to be genuine numbers?
Salviati was Galileo. Galileo did not have the equipment to even deal
with continua in a thorough way. He encountered a puzzle to which he did
not give an adequate explanation. No one did until the late 19th century.
You are tilting at windmills again.
Bob Kolker
===
Subject: Re: Do reals really have to be genuine numbers?
This is common but wrong thinking.
===
Subject: Re: Do reals really have to be genuine numbers?
No. It is common because it is correct. The theory of real numbers still
stands and no proof of its inconsistency has yet been given.
Bob Kolker
===
Subject: Re: Do reals really have to be genuine numbers?
Intelligent people understand that numerical continuity is rather
obvious nonsense while numerical is about as ubiquitously in use as
continuity. You disqualified yourself.
===
Subject: Re: Do reals really have to be genuine numbers?
It does not matter whether you prefer infinite decimals or infinite
continuned fractions. You are wrong.
===
Subject: The Heliocentric Universe
idea of a point of reference in astronomy is a necessarily trivial
thing. When making a model of the universe we can say that the stars,
the planets, and the earth all revolve around the sun, in a
Heliocentric model. We could also convert this model to any other
model with another point of reference, although it would be far more
complex to say everything revolved around the earth. Is anyone
familiar with the math involved in these calculations, and might point
me to a book?
I believe that there is other life out there, on other planets. My
theory is that the sun is the only star, and the other stars are just
bright and distant planets with life on them, the same as earth. I
think life can burn up and turn to light, and when it travels off of
one planet the photosenthetic life on another planet will catch it and
trap it on the new planet. But this is crazy, yes? :)
http://en.wikipedia.org/wiki/Heliocentrism
The realization that the heliocentric view was also not true in a
strict sense was achieved in steps. That the Sun was not the center of
the universe, but one of innumerable stars, was strongly advocated by
the mystic Giordano Bruno; Galileo made the same point, but said very
little on the matter, perhaps not wishing to incur the church's wrath.
Over the course of the 18th and 19th centuries, the status of the Sun
as merely one star among many became increasingly obvious. By the 20th
century, even before the discovery that there are many galaxies, it
was no longer an issue.
Even if the discussion is limited to the solar system, the sun is not
at the geometric center of any planet's orbit, but rather at one focus
of the elliptical orbit. Furthermore, to the extent that a planet's
mass cannot be neglected in comparison to the Sun's mass, the center
of gravity of the solar system is displaced slightly away from the
center of the Sun. (The masses of the planets, mostly Jupiter, amount
to 0.14% of that of the Sun.) Therefore a hypothetical astronomer on
an extrasolar planet would observe a wobble.
Giving up the whole concept of being at rest is related to the
principle of relativity. While, assuming an unbounded universe, it was
clear there is no privileged position in space, until postulation of
the special theory of relativity by Albert Einstein, at least the
existence of a privileged class of inertial systems absolutely at rest
was assumed, in particular in the form of the hypothesis of the
luminiferous aether. Some forms of Mach's principle consider the frame
at rest with respect to the masses in the universe to have special
properties.
Modern use of geocentric and heliocentric
In modern calculations, the origin and orientation of a coordinate
system often have to be selected. For practical reasons, systems with
their origin in the mass, solar mass or in the center of mass of solar
system are frequently selected. The adjectives may be used in this
context. However, such selection of coordinates has no philosophical
or physical implications.
The relation of the two pictures [geocentricity and
heliocentricity] is reduced to a mere coordinate transformation and it
is the main tenet of the Einstein's theory that any two ways of
looking at the world which are related to each other by a coordinate
transformation are entirely equivalent from a physical point of view.
(Hoyle, 1973, p. 78)
===
Subject: induced topology
Hi there, consider an infinite dimensional Banach space X and denote
by V a finite dimensional subspace of X. Let C(B(R)) denote the closed
ball of radius R centered at the origin of X and consider the
intersection of C(B(R)) and V. Denote such an intersection with J.
Then J is a closed bounded subset of V, so (by Heine-Borel) J is a
compact subset of V, V viewed as a topological subspace of X. Is J
also a compact subspace of X?
===
Subject: Re: induced topology
A compact topological space always stays compact whenever you imbed it
in a larger space.
Jose Carlos Santos
===
Subject: Re: induced topology
I try to sketch a proof for the following assertion: If J is a compact
subset of V, V viewed as a topological subspace of X, then J is also a
compact subspace of X.
Proof. :
Let {A a} a be an open covering of J made by set A a subset X, A a
open for the topology of X.
I refine {A a} a to a finite covering of J in the following manner.
For every index a, let's denote by C a the intersection of A a with V,
where C a = emptyset if A a doesent't intersect V. So {C a} a is an
open covering for J made by sets in V, besides such sets C a are open
w.r.t. the induced topology in V. Starting from {C a} a deduce a
finite covering {C {a 1}, ..., C {a r} } for J (this is possible since
J is compact in V). Then {A {a 1}, ..., A {a r} } is the requested
finite open covering for J in X. So J is also a compact subspace of X.
Ok?
===
Subject: Re: induced topology
Better proof the following: If M is a subset of a topological space X,
then M is a compact subset of X if and only if M is a compact subset of
itself, equipped with the induced topology.
Since this is true, it is more convenient not to talk about compact
subsets of a space, but just about compact spaces. And so one does,
except maybe in beginners' analysis courses.
Carsten
--
Carsten Schultz (2:38, 33:47)
http://carsten.codimi.de/
PGP/GPG key on the pgp.net key servers,
fingerprint on my home page.
===
Subject: Re: Calculating subgroups!
Ok, In this case I have that all the subgroups of R are cyclic because
R is, R is in fact C_8, and its subgroups are the generated by r,
r^2,, r^4, and 1, then for any sugbroup G of D(16) we have that G
(inter) R is one of this 4 subgroups of R.
Now you cuestion about the possibilities for G... Well if G is in R
it is celear since this groups are in line.
If G conains one element not in R, then G contains an element t of
orther 2 not r^4 , then G contains the subgroup generated by t which
is not in G. And then G contains also all the products of t and R
(inter) G, ... but I still don't know how to count all of them....
===
Subject: Re: Galileo's Paradox
Why do you not trust in the convincing removal of the unnecessary
paradox by Salviati (who corresponds to the mature Galileo)?
===
Subject: Re: Galileo's Paradox
Galileo knew diddly squat about set theory, either young or old. Galileo
did not have the machinery for addressing the interesting -apparent-
paradox he uncovered. Likewise Zeno's so-called paradoxes were resolved
once the limit concept and convergences were defined.
Bob Kolker
===
Subject: Re: Galileo's Paradox
All pertaining work is easily readable in German from Goettingen.
===
Subject: Re: Galileo's Paradox
No you just confused two different methods by Cantor. You meant DA2.
===
Subject: Re: Galileo's Paradox
Well, I referred to Fraenkel's book 2nd ed. 1923. Meanwhile I read again
Cantor's original paper of 1891. Here he used a dual system. The
differnence does not matter.
Then we do perhaps not speak about Cantor's second diagonal argument
(after an idea by Paul du Bois Raymond) published in 'Ueber eine
elementare Frage der Mannigfaltigkeitslehre'.
Pi is irrational, it is even transcendental. It does not have a numeric
representation with a finite number of digits. So it is to be seen a
real for three indications.
Not just all irrationals but also all numbers with a fictitious
numerical representation of infinite length behind the point are genuine
reals.
You did not get it.
===
Subject: Re: Galileo's Paradox
Initially, he was affected by the same error as Dedekind and Cantor but
found the way out.
DA1 shows that the rationals are countable. DA2 demonstrates that the
reals are uncountable. Ane can conclude that the used numbers were
characterized like real numbers by their properties relevant for DA2.
===
Subject: Re: Galileo's Paradox
Neither Dedekind nor Cantor were in error. Dedekind's theory of real
numbers is as good today as it was when published. Ditto Cantor's
construction with equivalence classes of Cauchy sequences of rational
numbers.
You notion of mathematics would be barely adequate for keeping
checkbooks in balance and totaling up cash register receipts.
Bob Kolker
===
Subject: Re: Galileo's Paradox
But it might be enough to calculate risk in global politics :-(
Best
RR
===
Subject: Re: Galileo's Paradox
If so, why they did not follow Salviati?
cf. Wikipedia, Galileos's paradox
===
Subject: Re: Galileo's Paradox
Because set theory was invented to handle the question.
Bob Kolker
===
Subject: Function question
Hello! I play with math on and off, and had discovered a website with
many interesting math problems. Here is one that I do not know how to
solve, can you please tell me how is it done? Note: This is not
homework in case you're wondering... I currently don't do formal
studies.
f(n^2) = f(n+m) f(n-m) + m^2
for all m, n in Z
I know how to solve the Fibonacci problem, but I tried the same
technique and it does not apply here. I also tried transforming n+m
and n-m using m=n cos 2x, but it doesn't help. I wonder, is there a
web page dedicated to teach how to solve this type of problems? How do
you call it, function math perhaps?
===
Subject: Re: Function question
For the problem, I don't have time to solve it myself now, but it looks like
you
could get some considerable insight by investigating what falls out when
you
reduce the number of variables, say replacing m by either 0 or n, and also
by
listing all the equations for 0<=m<=n<=N for some small positive N.
Take for example, that f(n^2)=f(n)^2 . What does that tell you? Likewise,
f(4)=f(4)*f(0)+4, but if f(0)=1, there will be trouble... Thinking this
way, I think you
may get somewhere.
===
Subject: Re: homological algebra
As Paul Sperry mentioned, this is often proved as part of the theory
of abelian groups. The proof along these lines is very accessible and
is on page 18 of Kaplansky's Infinite Abelian Groups.
There is also a fairly equivalent proof using the language of
homological algebra on p232 of Rotman's Introduction to Homological
Algebra. Step one is to show that Ext^1(A/tA,B) is divisible for all
abelian groups A,B (or any modules over a dedekind domain), and step
two is to show that if n*B=0, then n*Ext^1(A,B)=0 for all abelian
groups A (or again modules over a dedekind domain). Then n acts as an
automorphism ( Ext(G/T,T) is divisible ) and as zero ( n*Ext(G/
T,T)=0 ), so in fact Ext(G/T,T)=0.
Both steps follow from the fact that Ext preserves scalar
multiplication in both variables (pretty obviously true over Z), so
Ext(A,B) is divisible if A is divisible, and Ext(A,B) is annihilated
by n if B is annihilated by n.
Step one also requires a tiny observation: G/T embeds is a a torsion-
free injective module (G/T tensor Q) and so G/T tensor Q is
divisible. Since Ext^2 = 0, Ext(G/T tensor Q,T) surjects onto Ext(G/
T,T), and since quotients of divisibles are divisible, Ext(G/T,T) is
divisible.
===
Subject: OIL PAINTING, MUSIC, POETRY
I would like to invite everyone to see my Art work.
http://mungiako.com/my_work.htm
===
Subject: Re: pi & 2 pi
Well let me ask you this Who holds the world reckord for coculating pi and
what was that reckord?